FEM-Design. Verification Examples. version Motto: ,,There is singularity between linear and nonlinear world. (Dr.

Transcription

1 FEM-Design version.3 8 Motto:,,There is singularity between linear and nonlinear world. (Dr. Ire Bojtár)

2 StruSoft AB Visit the StruSoft website for copany and FEM-Design inforation at Copyright 8 by StruSoft, all rights reserved. Tradearks FEM-Design is a registered tradeark of StruSoft.

3 Contents. Linear static calculations...6. Bea with two point loading at one-third of its span...6. Calculation of a circular plate with concentrated force at its center A siply supported square plate with unifor load.... Second order analysis...5. A colun with vertical and horizontal loads...5. A plate with in-plane and out-of-plane loads Stability analysis Flexural buckling analysis of a bea odell with different boundary conditions Buckling analysis of a plate with shell odell Lateral torsional buckling of an I section with shell odell Lateral torsional buckling of a cantilever with elongated rectangle section Calculation of eigenfrequencies with linear dynaic theory Continuous ass distribution on a cantilever colun Free vibration shapes of a claped circular plate due to its self-weight Seisic calculation Lateral force ethod with linear shape distribution on a cantilever Lateral force ethod with fundaental ode shape distribution on a cantilever Modal analysis of a concrete frae building Calculation considering diaphrags A siple calculation with diaphrags The calculation of the shear center Calculations considering nonlinear effects Uplift calculation A trusses with liited copression ebers A continuous bea with three supports Cracked section analysis by reinforced concrete eleents Cracked deflection of a sily supported bea Cracked deflection of a statically indeterinate bea Cracked deflection of a cantilever bea Cracked deflection of a cantilever bea with copressed reinforceent bars Cracked deflection of a cantilever with bending oent and noral forces Cracked deflection of a siply supported square slab Nonlinear soil calculation Elasto-plastic calculations Elasto-plastic point support in a bea Elasto-plastic line support in a plate Elasto-plastic surface support with detach in an ebedded plate Elasto-plastic trusses in a ultispan continuous bea Elasto-plastic point-point connection between cantilevers Elasto-plastic point-point connection with uplift in a ultispan continuous bea Elasto-plastic edge connections in a building braced by shear walls Elasto-plastic edge connections with detach in a shear wall Elasto-plastic line-line connections in a square plate Calculation with construction stages A steel frae building with construction stages calculation

4 8. Cross section editor Calculation of a copound cross section Design calculations Foundation design Design of an isolated foundation Design of a wall foundation Design of a foundation slab Reinforced concrete design Moent capacity calculation for beas under pure bending Under-reinforced cross section Noral-reinforced cross section Over-reinforced cross section Required reinforceent calculation for a slab Elliptic bending Hyperbolic bending Shear capacity calculation Shear capacity of a bea Shear capacity of a slab Crack width calculation of a bea Crack width calculation of a slab Elliptic bending Hyperbolic bending Punching calculation of a slab Bended bars Circular stirrups Open stirrups Interaction of noral force and biaxial bending in a colun Noinal stiffness ethod Noinal curvature ethod Calculation of a statically indeterinate bea with post tensioned cables Steel design Interaction of noral force, bending oent and shear force Buckling of a doubly syetric I section Buckling of a doubly syetric + section Buckling of a ono-syetric channel section Lateral torsional buckling of a doubly syetric I section Interaction of biaxial bending and axial copression in an RHS section Interaction calculation with a Class 4 section Tiber design...3 References...3 Notes

5 In this verification handbook we highlighted the analytical results with green and the finite eleent results with blue background for better coparison. The analytical closed forulas are highlighted with a black frae. The coparisons between the hand calculations and FEM-Design calculations are highlighted with yellow. If the finite eleent esh is not entioned during the exaple it eans that the autoatically generated esh was used. WARNING: We are continuously developing this verification book therefore soe discrepancy in the nubering of the chapters or soe issing exaples can occur. 5

6 . Linear static calculations. Bea with two point loading at one-third of its span Fig... left side shows the siple supported proble. The loads, the geoetry and aterial properties are as follows: Force F 5 kn Length L6 Cross section Steel I bea HEA 3 The second oent of inertia in the relevant direction I.864ˑ-4 4 The shear correction factor in the relevant direction ρ.597 The area of the cross section A.53 c Young's odulus E GPa Shear odulus G GPa F L/3 F L/3 δf L/ L/3 -F L/ -.5 δv V +.5 +F δm M M+FL/3 +L/4 Figure.. The bea theory and the application of a virtual force The deflection of the id-span based on the hand calculation (based on virtual force theore [], see Fig... right side also): e M L L L L L L F L 3 FL3 FL EI ρ GA EI 3 ρ GA e ˑ [ ] [ ] 6

7 The first part of this equation coes fro the bending deforation and the second part coes fro the consideration of the shear deforation as well, because FEM-Design is using Tioshenko bea theory (see the Scientific Manual). The deflection and the bending oent at the id-span based on the linear static calculation with three -noded bea eleents (Fig... and Fig...3): e FEM 3.5 and the bending oent M FEM 3 kn The theoretical solution in this case (three -noded bea eleents) ust be equal to the finite eleent solution because with three bea eleents the shape functions order coincides with the order of the theoretical function of the deflection (the solution of the differential equations). Figure.. The finite eleent odel Figure..3 The id-span deflection [] Therefore the difference between the results of the two calculations is zero. Download link to the exaple file: Bea with two point loading at one-third of its span.str 7

8 . Calculation of a circular plate with concentrated force at its center In this chapter a circular steel plate with a concentrated force at its center will be analyzed. First of all the axiu deflection (translation) of the plate will be calculated at its center and then the bending oents in the plate will be presented. Two different boundary conditions will be applied at the edge of the plate. In the first case the edge is claped (Case I.) and in the second case is siply supported (Case II.), see Fig.... P P Figure.. Claped (Case I.) and siply supported (Case II.) circular plate with concentrated force The input paraeters are as follows: The concentrated force P kn The thickness of the plate h.5 The radius of the circular plate R5 The elastic odulus E GPa The Poisson's ratio ν.3 The ratio between the diaeter and the thickness is R/h. It eans that based on the geoetry the shear deforation only have negligible effects on the axiu deflections. It is iportant because FEM-Design uses the Mindlin plate theory (considering the shear deforation, see Scientific Manual for ore details), but in this case the solution of Kirchhoff's plate theory and the finite eleent result ust be close to each other based on the entioned ratio. The analytical solution of Kirchhoff's plate theory is given in a closed for [][3]. 8

9 Case I.: For the claped case the axiu deflection at the center is: w cl P R E h3 6 π ( ν ) ( ) The reaction force at the edge: P π R Qr And the bending oent in the tangential direction at the edge: M cl P 4π With the given input paraeters the results based on the analytical and the finite eleent results (with the default finite eleent esh size, see Fig...) are: w cl Qr π (.3) ( ) kn.38 π 5 M cl w clfem.4 QrFEM.38 P kn.796 4π 4π kn M clfem.796 kn Figure.. The claped (Case I.) and the siply supported (Case II.) plate with the default esh 9

10 Case II.: For the siply supported case the axiu deflection in the center is: w ss P R E h3 6 π ( ν ) ( ) ν ( 3+ +ν ) The reaction force at the edge: P π R Qr With the given input paraeters the results based on the analytical and the finite eleent results (with the default finite eleent esh size, see Fig...) are: w ss Qr π (.3 ) ( ) ( ) kn.38 π 5 w ssfem 5. Q rfem.38 kn Figure..3 The deflected shape of Case I. (claped) and Case II. (siply supported) with the default esh [] Fig...3 shows the two deflected shape in side view. The different boundary conditions are obvious based on the two different displaceent shape. The differences between the analytical solutions and finite eleent solutions are less than 5% but the results could be ore accurate if the applied esh is ore dense than the default size. Based on the analytical solution the bending oents in plates under the concentrated loads are infinite. It eans that if ore and ore dense esh will be applied the bending oent under the concentrated load will be greater and greater. Thus the following diagra and table (Fig...4 and Table..) shows the convergence analysis of Case I. respect to the deflection and bending oent. The deflection converges to the analytical solution (w cl.7 ) and the bending oent converges to infinite.

11 3, 6 4,5,,5 8 6, Deflection Bending oent,5, Bending oent [kn/] Deflection [] 35 Nuber of eleents Figure..4 Convergence analysis regarding to deflection and bending oent Nuber of eleents [pcs.] Deflection Bending oent Average eleent size [] [] [kn/],34,73,5,57 3,9,4,6 3,6,3,68 4,33,,7 5,49,,73 6,4,5,75,7 Local refineent,76 4,3 Local refineent Table.. The convergence analysis Download links to the exaple files: Claped: Calculation of a circular plate with concentrated force at its center claped.str Siply supported: Calculation of a circular plate with concentrated force at its center siplysup.str

12 .3 A siply supported square plate with unifor load In this exaple a siply supported concrete square plate will be analyzed. The external load is a unifor distributed load (see Fig..3.). We copare the axiu displaceents and axiu bending oents of the analytical solution of Kirchhoff's plate theory and finite eleent results. The input paraeters are in this table: The intensity of the unifor load p 4 kn/ The thickness of the plate h.5 The edge of the square plate a5 The elastic odulus E 3 GPa Poisson's ratio ν. The ratio between the span and the thickness is a/h. It eans that based on the geoetry the shear deforation ay have effect on the axiu deflection. It is iportant because FEM-Design uses the Mindlin plate theory (considering the shear deforation, see Scientific Manual for ore details), therefore in this case the results of Kirchhoff's theory and the finite eleent result could be different fro each other due to the effect of shear deforations. Figure.3. The square plate with siply supported edges, unifor load and the default esh Based on Kirchhoff's plate theory [][3] the axiu deflection is in the center of the siply supported square plate and its intensity can be given with the following closed for: 4 w ax.46 ( pa 3 Eh ( ν ) )

13 The axiu bending oent in the plate if the Poisson's ratio ν.: M ax.469 p a According to the input paraeters and the analytical solutions the results of this proble are the following: The deflection at the center of the plate: w ax.46 ( (. ) ) w axfem.63 The bending oent at the center of the plate: M ax kn M axfem kn Next to the analytical solutions the results of the FE calculations are also indicated (see Fig..3. and.3.3). The difference is less than 3% and it also coes fro the fact that FEM-Design considers the shear deforation (Mindlin plate theory). Download link to the exaple file: A siply supported square plate with unifor load.str Figure.3. The deflected shape [] and the reaction forces [kn/] with the default esh 3

14 Figure.3.3 The internal forces; x y xy [kn/] 4

15 . Second order analysis. A colun with vertical and horizontal loads We would like to analyze the following colun (see Fig...) with second order theory. First of all we ake a hand calculation with third order theory according to Ref. [6] and [8] with stability functions. After this step we copare the results with FEM-Design. In this oent we need to consider that in FEM-Design second order analysis is ipleented and the hand calculation will be based on third order theory therefore the final results won't be exactly the sae. By FEM-Design calulation we splitted the colun into three bar eleents thus the finite eleent nuber of the bars was three for ore precise results. The input paraeters: Elastic odulus E 3 GPa Noral force P 468 kn Horizontal load q kn/ Cross section. x.4 (rectangle) Second oent of inertia in the relevant direction I Colun length L4 P L q Figure.. The colun with vertical and horizontal loads According to Ref. [6] and [8] first of all we need to calculate the following assistant quantities: ρ P P PE π EI π L 4 ( )( ) 5

16 The constants based on this value for the appropriate stability functions: s3.94 ; c.666 ; f.4 With these values the bending oents and the shear forces based on third order theory and FEM-Design calculation: M claped f (+c) ql 4.4(+.666) 4.5 kn M roller. kn [ V claped + M ndfemclaped 5.55 kn M ndfemroller. kn f (+c ) 6 ]( ) [.4(+.666) ql + 6 ]( ) kn V ndfemclaped 6.38 kn [ V roller f (+c ) 6 ]( ) [.4 (+.666) ql 6 ]( ) kn V ndfemroller 3.6 kn The differences are less than 5% between the hand and FE calculations. Download link to the exaple file: A colun with vertical and horizontal loads.str Figure.. The shear [kn] and bending oent [kn] diagra with st and nd order theory 6

17 Figure..3 The lateral translations [] with st and nd order theory 7

18 . A plate with in-plane and out-of-plane loads In this chapter we will analyze a rectangular plate with single supported four edges. The load is a specific noral force at the shorter edge and a lateral distributed total load perpendicular to the plate (see Fig...). The displaceent and the bending oent are the question based on a nd order analysis. First of all we calculate the results with analytical solution and then we copare the results with FE calculations. In this case the aterial and the geoetric properties are the following: The thickness of the plate h.5 The diensions of the plate a 8 ; b 6 The elastic odulus E GPa Poisson's ratio ν.3 The specific noral force nx kn/ The lateral distributed load qz kn/ y nx qz nx a b x Figure.. The single supported edges, the lateral distributed load and the specific noral force The axiu displaceent and oents based on the st order linear calculation: w ax 35.38, x, ax8.5 kn kn kn, y, ax 5.6, xy, ax 3.68 Based on Chapter 3. the critical specific noral force for this exaple is: n cr 86 kn If the applied specific noral force is not so close to the critical value (now it is lower than the 8

19 half of the critical value) we can assue the second order displaceents and internal forces based on the linear solutions with the following forulas (with blue highlight we indicated the results of the FE calculation): w ax,nd w ax , nx 86 ncr ( ) x, ax,nd x, ax ( ) y, ax, nd y, ax ( nx ncr 8.5 ( kn kn 7.76, x, ax,nd, FEM ) kn kn , y, ax, nd, FEM 4.3 n x 86 ncr ( ) xy, ax,nd xy, ax ) w ax,nd, FEM ( ) nx n cr ( 3.68 ( ) kn kn.4, xy, ax,nd, FEM ) The differences are less than 3% between the hand and FEM-Design calculations. Figure.. shows the proble in FEM-Design with the default esh. Figure.. The single supported slab with in-plane and out-of-plane loads The following figures show the oent distribution in the plate and the displaceents with FEM-Design according to st and nd order theory. Download link to the exaple file: A plate with in-plane and outof-plane loads.str 9

20 Figure..3 The x [kn/] oent with st and nd order analysis Figure..4 The y [kn/] oent with st and nd order analysis Figure..5 The xy [kn/] oent with st and nd order analysis

21 Figure..6 The vertical translations [] with st and nd order analysis

22 3. Stability analysis 3. Flexural buckling analysis of a bea odell with different boundary conditions The cross section is a rectangular section see Fig. 3.. The aterial C /5 concrete The elastic odulus E 3 GPa The second oent of inertia about the weak axis I.667ˑ-4 4 The length of the colun L4 The boundary conditions see Fig. 3.. b.4 L4 a. Figure 3.. The buckling proble with the different boundary conditions and the cross section The critical load paraeters according to the Euler's theory are as follows and next to the analytical solutions [] the relevant results of the FEM-Design calculation can be seen. By the calculation we splitted the beas to five finite eleents to get ore accurate buckling ode shapes (see Fig. 3..). Pinned-pinned boundary condition: π E I F cr kn L F crfem49.9 kn

23 Fixed-pinned boundary condition: F cr πe I 94. kn (.699 L) F crfem kn Fixed-fixed boundary condition: π E I F cr kn (.5 L) F crfem kn Fixed-free boundary condition: π E I F cr kn ( L) F crfem4 33. kn The differences between the two calculations are less than 3% but keep in ind that FEMDesign considers the shear deforation therefore we can be sure that the Euler's results give a bit higher critical values in these cases. Fig. 3.. shows the first ode shapes of the probles with the different boundary conditions. Figure 3.. The buckling ode shapes for different boundary conditions pinned-pinned; fixed-pinned; fixed-fixed; fixed-free 3

24 Download links to the exaple files: Flexural buckling analysis of a bea odell with different boundary conditions fixed-free.str Flexural buckling analysis of a bea odell with different boundary conditions pinned-pinned.str 4

25 3. Buckling analysis of a plate with shell odell In this chapter we will analyze a rectangular plate with siply supported four edges. The load is a specific noral force at the shorter edge (see Fig. 3..). The critical force paraeters are the questions due to this edge load, therefore it is a stability proble of a plate. In this case the aterial and the geoetric properties are the following: The thickness of the plate h.5 The diensions of the plate a 8 ; b 6 The elastic odulus E GPa Poisson's ratio ν.3 The solutions of the differential equation of the plate buckling proble are as follows [6]: ( n cr b n a + a b π ) ( E h3 ( ν ) ),,, 3..., n,, 3... b y nx nx a b x Figure 3.. The siply supported edges and the specific noral force Figure 3.. shows the proble in FEM-Design with the default esh. 5

26 Figure 3.. The stability proble of a plate with siply supported edges According to the analytical solution the first five critical load paraeters are:, n ( n cr π ) (.53 (.3 ) ) kn n crfem86.58 kn kn n crfem kn kn n crfem kn kn n crfem kn, n ( 6 8 n cr π ) (.53 (.3 ) ) 6 3, n ( n cr π ) (.53 (.3 ) 6 ) 4, n ( n cr ) π (.53 (.3 ) 6 ) , n ( n cr ) π (.53 (.3 ) 6 ) 69.4 kn n crfem kn Next to these values we indicated the critical load paraeters what were calculated with FEMDesign. 6

27 The difference between the calculations less than 5%. Figure 3..3 shows the first five stability ode shapes of the rectangular siply supported plate. Figure 3..3 The first five stability ode shapes of the described proble Download link to the exaple file: Buckling analysis of a plate with shell odell.str 7

28 3.3 Lateral torsional buckling of an I section with shell odell The purpose of this exaple is calculate the lateral torsional critical oent of the following siply supported bea (see Fig. 3.3.). M M L Figure 3.3. The static frae of a siply supported bea loaded with bending oents at both ends The length of the bea L The cross section see Fig The warping constant of the section Iw 584 c6 The St. Venant torsional inertia It 5.34 c4 The inor axis second oent of area Iz 6.7 c4 The elastic odulus E GPa The shear odulus G 8.77 GPa Figure 3.3. The diensions of the double syetric cross section In this case the critical oent can be calculated with the following forula based on the analytical solution [6]: M cr π E Iz L M cr π knc 6.7 π 6.7 I w L G It + Iz πe Iz 8

29 In FEM-Design a shell odell was built to analyze this proble. The oent loads in the shell odel were considered with line loads at the end of the flanges (see Fig ). The supports provides the siple supported bea effects with a fork support for the shell odel (see Fig ). Figure The FEM odel with the supports and the loads (oents) at the ends Fro the FEM-Design stability calculation the critical oent value for this lateral torsional buckling proble is: M crfem4363 knc The critical shape is in Fig The finite eleent esh size was provided based on the autoatic esh generator of FEM-Design. Figure The critical ode shape of the proble The difference between the two calculated critical oents is less than %. 9

30 Download link to the exaple file: FEM-Design file: Lateral torsional buckling of an I section with shell odell.str Section Editor file for cross-sectional properties: Lateral torsional buckling of an I section with shell odell.sec 3

31 3.4 Lateral torsional buckling of a cantilever with elongated rectangle section The purpose of this exaple is calculate the critical force at the end of a cantilever bea (see Fig. 3.4.). If the load is increasing the state of the cantilever will be unstable due to lateral torsional buckling. h Pcr t L Figure 3.4. The cantilever bea with concentrated load The input paraeters: The length of the bea is L The cross section t 4 ; h 438 ; see Fig The St. Venant torsional inertia It The inor axis second oent of area Iz The elastic odulus E GPa The shear odulus G 8.77 GPa In this case (elongated rectangle cross section with cantilever boundary condition) the critical concentrated force at the end can be calculated with the following forula based on analytical solution [ask for the reference fro Support tea]: P cr 4. E I z G I t E Iz L P cr N3.69 kn 336 3

32 In FEM-Design a shell odell was built to analyze this proble. The concentrated load at the end of the cantilever was considered at the top of the bea (see Fig. 3.4.). Figure 3.4. The FE odel of the cantilever bea with the default esh With the FEM-Design stability calculation the critical concentrated force value for this lateral torsional buckling proble is: P crfem4. kn The critical shape is in Fig The finite eleent esh size was provided based on the autoatic esh generator of FEM-Design. Figure The critical ode shape of the proble The difference between the two calculated critical load paraeters is less than %. Download link to the exaple file: Lateral torsional buckling of a cantilever with elongated rectangle section.str 3

33 4. Calculation of eigenfrequencies with linear dynaic theory 4. Continuous ass distribution on a cantilever colun H4 The cross section square with.4 edge The second oent of inertia I.33 4 The area of the cross section A.6 The shear correction factor ρ 5/ The elastic odulus E 3 GPa The shear odulus G.5 GPa The specific self-weight of the colun γ 5 kn/3 The ass of the colun.63 t H4 continuous ass distribution Colun height Figure 4.. The cantilever with continuous ass distribution Based on the analytical solution [4] the angular frequencies for this case is: ω B μ Bi EI H3 ; μ B3.5; μ B.3; μ B36.7 if only the bending deforations are considered. The angular frequencies are [4]: 33

34 ω S μ Si ρ GA H ; μ S.5 π ; μ S.5 π ; μ S3.5 π if only the shear deforations are considered. Based on these two equations (considering bending and shear deforation) using the Föppl theore the angular frequency for a continuous ass distribution colun is: + ω n ω B ω S Based on the given equations the first three angular frequencies separately for bending and shear deforations are: ω B s.63 4 ω B s.63 4 ω B s.63 4 ω S.5 π ω S.5 π ω S3.5 π s s s According to the Föppl theore the resultant first three angular frequencies of the proble are: ω n86.639, ω n53.64, ω n345.8 s s s And based on these results the first three eigenfrequencies are (f ω/(π)): f n3.789, f n84.63, f n36.93 s s s 34

35 In FEM-Design to consider the continuous ass distribution bea eleents were used for the cantilever colun. The first three planar ode shapes are as follows according to the FE calculation: f FEM 3.78, f FEM , f FEM s s s The first three ode shapes can be seen in Fig Figure 4.. The first three ode shapes for the cantilever with continuous ass distribution The differences between the analytical and FE solutions are less than %. Download link to the exaple file: Continuous ass distribution on a cantilever colun.str 35

36 4. Free vibration shapes of a claped circular plate due to its self-weight In the next exaple we will analyze a circular claped plate. The eigenfrequencies are the question due to the self-weight of the slab. In this case the aterial and the geoetric properties are the following: The thickness of the plate h.5 The radius of the circular plate R5 The elastic odulus E GPa Poisson's ratio ν.3 The density ρ 7.85 t/3 The solution of the dynaic differential equation for the first two angular frequencies of a claped circular plate are [5]: ω n π β n R ( E h3 ( ν ) ρh ), β.5, β.468 Figure 4.. shows the proble in FEM-Design with the claped edges and with the default esh. Figure 4.. The claped circular plate and the default finite eleent esh 36

37 According to the analytical solution the first two angular frequencies are: ( ω π.5 5 ω π ( ).5 3 (.3 ) 3.83, f 5.66, f FEM 5.9 s s s ).53 (.3 ) 66.58, f.6, f FEM.73 s s s Based on the angular frequencies we can calculate the eigenfrequencies in a very easy way. Next to these values we indicated the eigenfrequencies which were calculated with FEM-Design. The differences between the calculations are less than %. Figure 4.. shows the first two vibration ode shapes of the circular claped plate. Figure 4.. The first two vibration shape ode of a claped circular plate Download link to the exaple file: Free vibration shapes of a claped circular plate due to its self-weight.str 37

38 5. Seisic calculation 5. Lateral force ethod with linear shape distribution on a cantilever Inputs: Colun height H The cross section square with.4 edge The second oent of inertia I.33 4 The elastic odulus E 3 GPa The concentrated ass points pieces of. t (see Fig. 5..) The total ass. t Figure 5.. The cantilever colun with the concentrated ass points, the first vibration shape [T.765 s], the equivalent forces [kn], the shear force diagra [kn] and the bending oent diagra [kn] with FEM-Design First of all based on a hand calculation we deterine the first fundaental period: The first fundaental period of a cantilever colun (length H) with a concentrated ass at the end ( ass) and EI bending stiffness [4]: T i π 3 EI i H i3 38

39 The fundaental period separately for the ass points fro botto to top: T T 3 T 5 T 7 T 9 π π.4s ; T.399 s ; π π.733 s ; T 4.9 s ; π π.577 s ; T 6.73s ; π s ; T 8 π s ; π π.389 s ; T.446 s The approxiated period based on these values according to the Dunkerley suary and the result of FE calculation: T HC i T i.7758 s T FEM.765s The difference between the hand calculation and FEM-Design calculation is less than %, for further inforation on the fundaental period calculation see Chapter 4. The base shear force according to the fundaental period of vibration (see Fig. 5..) and the response spectru (see Fig. 5..): F bs d (T ) λ kn We considered the response acceleration based on the period fro FE calculation to get a ore coparable results at the end. Thus the equivalent forces on the different point asses are: F i F b z i i z j j z i i z i i

40 The equivalent forces fro the botto to the top on each point ass: F. kn ; F.4 kn ; F kn ; F kn ; F kn ; F 6.79 kn ; F kn ; F kn ; F 9.78 kn ; F.98 kn These forces are identical with the FEM-Design calculation and the shear force and bending oent diagras are also identical with the hand calculation. Figure 5.. The response spectru [T.765 s; Sd.6588 /s] Download link to the exaple file: Lateral force ethod with linear shape distribution on a cantilever.str 4

41 5. Lateral force ethod with fundaental ode shape distribution on a cantilever Inputs: Colun height H The cross section square with.4 edge The second oent of inertia I.33 4 The elastic odulus E 3 GPa The concentrated ass points pieces of. t (see Fig. 5..) The total ass. t Figure 5.. The cantilever colun with the concentrated ass points, the first vibration shape with the value of the eigenvector [T.765 s], the equivalent forces [kn], the shear force diagra [kn] and the bending oent diagra [kn] with FEM-Design The base shear force according to the fundaental period of vibration (see Fig. 5..) and the response spectru (see Fig. 5..): F bs d (T ) λ kn We considered the response acceleration based on the period fro FE calculation to get a ore coparable results at the end. Thus the equivalent forces on the different point asses are: 4

42 F i F b s i i sjj si i s i i The equivalent forces fro the botto to the top on each point ass: F.46 kn ; F.939 kn ; F 3. kn ; F kn ; F 5.5 kn ; F kn ; F kn ; F 8.8 kn ; F 9.89 kn ; F.498 kn These forces are identical with the FEM-Design calculation and the shear force and bending oent diagras are also identical with the hand calculation. Figure 5.. The response spectru [T.765 s; Sd.6588 /s] Download link to the exaple file: Lateral force ethod with fundaental ode shape distribution on a cantilever.str 4

43 5.3 Modal analysis of a concrete frae building In this chapter we show a worked exaple for odal analysis on a concrete frae building according to EN 998-:8 with hand calculation and copare the results with FEM-Design. This exaple is partly based on [4]. The geoetry, the diensions, the aterial and the bracing syste are in Fig and in the following table. As a bracing syste we used trusses with very large noral stiffness (EA) to reach pure eigenvectors by the fundaental period calculation (see Fig and Fig ). Inputs: Colun height/total height h 3. ; H The cross sections Coluns: 3/3 c; Beas: 3/5 c The second oent of inertia Ic.675 4; Ib.35 4 The elastic odulus E 8.8 GPa The concentrated ass points pieces of t on st storey and pieces of.68 t on nd storey (see Fig. 5.3.) The total ass st storey: 6.3 t nd storey: 35. t total ass: M 95.5 t Reduction factor for elastic odulus considering the cracking according to EN 998:8 α.5 Behaviour factors q.5, qd.5 Accidental torsional effect was not considered ξ.5 (daping factor) Figure 5.3. The concrete frae building with the coluns and beas 43

44 The first exercise is the deterination of the fundaental periods and ode shapes. There are several hand calculation odes to get these values but in this chapter the details of the odal analysis are iportant therefore we considered the first two fundaental periods based on FEMDesign calculation (see Fig ). See the details and exaple on the eigenfrequency calculation in Chapter 4. The dead loads and the live loads are considered in the ass points (see Fig and the input table). Figure 5.3. The frae building with the asses and bracings Figure The side view of the building 44

45 Figure The considered design response specra according to EN 998-:8 Figure The first two fundaental ode shapes [-], T.74 s; T.5 s For this figure we splitted the beas/coluns to 5-5 finite eleents for better visualization of the eigenvectors but the default esh was used during the whole FEM calculation According to the fundaental periods in Fig the response accelerations fro Fig are: T.74 s S d.5 T.5 s S d.57 ; s. s The second step is to calculate the effective odal asses based on this forula: ( Φi T ι ) * i Φ it Φ i 45

46 During the hand calculation we assue that the structure is a two degrees of freedo syste in the x direction with the two storeys, because the first two odal shapes are in the sae plane see Fig Thus we only consider the seisic loads in one direction because in this way the hand calculation is ore coprehensible. * * ( ][ ]) [ [ ] [ 7.3 t ; ][ ] ][ ]) ][ ] 4.9 [ ] ( [ [ ] [ * % M [ ] t ; * % M 95.5 According to the assuption of a two degrees of freedo syste the su of the effective odal asses is equal to the total ass: * * % M M Calculation of the base shear forces: * F bs d kn ; F bs d * kn The equivalent forces coe fro this forula: p i Φi Φ it ι S di Φ it Φ i The equivalent forces at the storeys respect to the ode shapes considering the entioned two degrees of freedo odel: p [ ][ ] [ ][ ] [ ] [ ].5.6 kn [ ] [ ][ ] 46

47 [ p ][ [ ][ ] [ ] ] [67.6 [ ] kn ] [ ][ ] The equivalent forces on one frae fro the six (see Fig. 5.3.): [ ][ ] p [ 8.98/6 ][ 3.66 ] kn 44.5 /6 7.4 pf.6 /6. kn 83. /6 3.5 f The shear forces between the storeys respect to the two different ode shapes: [ ][ ] [ V kn ][ ] V kn The shear forces in the coluns respect to the two different ode shapes: [ ][ ] V c 5.6 / 5.3 kn 3.5/ 5.5 [ ][ ] V c 6.4 / 3.3 kn 7.4 / 3.7 The bending oents in the coluns respect to the two different ode shapes fro the relevant shear forces (by the hand calculation we assued zero bending oent points in the iddle of the coluns between the storeys): [ ][ ] M c / 4.48 kn / 4.4 [ ][ ] M c / 5.8 kn / The bending oents in the beas respect to the two different ode shapes: [ ][ ] M b kn [ ][ ] M b kn The SRSS suation on the internal forces: 47

48 [ [ ][ ] ][ ] kn 5.5 +( 3.7) 5.69 V c M b [ M c ][ ] kn (.98) kn ( 5.936) The CQC suation on the internal forces: α T T.74 r [ r 8 ξ (+α ) α 3 / 8.5 (+.358).3583/ ( α ) +4 ξ α (+α ) (.358 ) (+.358) ] And based on these values the results of the CQC suation: [ [ [ ][ ] V c 5.5 kn ( 3.7) ( 3.7).7588 M c kn M b (.98) (.98) kn ( 5.936)+ 4.4 ( ) ][ ] ][ ] The following displaceents coe fro the FEM-Design calculation on the coplete frae structure to ensure the coprehensible final results on the P-Δ effect. The displaceents at the storeys respect to the two different ode shapes considering the displaceent behaviour factor: [ ] [ ][ ] uq d [ ] [ ][ ] u q d Based on these values the storey drifting respect to the two different ode shapes: 48

49 4.3 [ ][4.3.4 ] Δ.7.7 [.8.7 ][.37 ] Δ SRSS suation on the story drifting: [ Δ ][ ] (.7) P-Δ effect checking on the total building: [ ][ ][ ] P tot (+ )g ( ) kn g [ V tot θ ][ ] kn ( 7.4) P tot Δ P Δ ; θ tot.55 V tot h V tot h After the hand calculation let's see the results fro the FEM-Design calculation and copare the to each other. Fig shows the effective odal asses fro the FE calculation. Practically these values coincide with the hand calculation. Figure The first two fundaental periods and the effective odal asses fro FEM-Design Fig and the following table shows the equivalent resultant shear forces and the base shear forces respect to the first two ode shapes. The differences between the two calculations are less than %. Storey equivalent resultant [kn] Storey equivalent resultant [kn] Hand FEM Hand FEM Hand FEM Mode shape Mode shape Base shear force [kn]

50 Figure The equivalent forces respect to the storeys and the base shear forces for the first two ode shapes [kn] Fig and the following table shows the internal forces after the different suation ethods (SRSS and CQC). The differences between the two calculation ethods are less than %. Here by the oents the difference between the hand and FEM calculations (%) coes fro the siplified oent hand calculation. Colun shear force [kn] Colun bending oent [kn] Bea bending oent [kn] Storey Storey Storey Storey Storey Storey SRSS Hand SRSS FEM ( )/ CQC Hand CQC FEM ( )/ Figure The shear force [kn] and bending oent diagra [kn] after the SRSS suation rule 5

51 Figure The shear force [kn] and bending oent diagra [kn] after the CQC suation rule Fig shows the Θ values fro FEM-Design. The differences between the hand calculation and FEM-Design calculations are less than 3%. Figure 5.3. The θ values at the different storeys fro FEM-Design Download link to the exaple file: Modal analysis of a concrete frae building.str 5

52 6. Calculation considering diaphrags 6.. A siple calculation with diaphrags If we apply two diaphrags on the two storeys of the building fro Chapter 5.3 then the eigenfrequencies and the periods will be the sae what we indicated in Chapter 5.3 with the bracing syste. 6.. The calculation of the shear center In this exaple we show that how can we calculate the shear center of a storey based on the FEM-Design calculation. We analyzed a botto fixed cantilever structure ade of three concrete shear walls which are connected to each other at the edges (see Fig. 6..). The diaphrag is applied at the top plane of the structure (see also Fig. 6.. right side). If the height of the structure is high enough then the shear center will be on the sae geoetry point where it should be when we consider the coplete cross section of the shear walls as a thin-walled C cross section (see Fig. 6.. left side). Therefore we calculate by hand the shear center of the C profile assued to be a thin-walled cross section then copare the results what we can get fro FEM-Design calculation with diaphrags. Secondly we calculate the idealized bending stiffnesses in the principal rigidity directions by hand and copare the results what we can calculate with FEM-Design results. Inputs: Height of the walls H 63 The thickness of the walls t c The width of wall nuber and 3 w w3 4. The width of wall nuber w 6. The applied Young's odulus of concrete E GPa Figure 6.. The geoetry of the bracing core and the height of the botto fixed structure (the diaphrag is lying on the top plane, see the red line and hatch) 5

53 First of all let's see Fig The applied cross section is a syetric cross section. In the web the shear stress distribution coes fro the shear forula regarding bending with shear (see Fig. 6..) therefore it is a second order polyno. In the flanges the shear stress distribution is linear according to the thin walled theory. With the resultant of these shear stress distribution (see Fig. 6.., V, V and V3) the position of the shear center can be calculated based on the statical (equilibriu) equations. τ w τ V S xs xg τ G w τax V t V V3 w τ Figure 6.. The shear stress distribution in a thin-walled cross section if the shear force acting on the shear center The shear stress values (see Fig. 6..): τ (. 4 3) VS kn It τ ax ( ) V S ax ( ) kn It ( ) Based on these stresses the resultant in the flanges and in the web: V V 3 τ t w kn V (τ ax τ ) w t+τ w t ( ) kn 3 3 Respect to the equilibriu (su of the forces): V kn V.9997 kn And also respect to the equilibriu (if the external load is acting on the shear center, see Fig. 53

54 6..) the su of the oents: V w V x S xs V w V.9997 Thus the shear center is lying on the syetry axis and it is x S.6 fro the web (see Fig. 6..). In FEM-Design the global coordinate syste does not coincide with the syetry axis of the structure (see Fig. 6..). Therefore we need no transfor the results. Lets be a selected key node at the diaphrag in the global coordinate syste (see Fig. 6..): x ; y Based on the unit forces ( kn) and oent ( kn) on the key node the displaceents of the key node are as follows based on the FEM-Design calculation (see the Scientific Manual Calculation considering diaphrag chapter also): According to unit force on key node in X direction: u xx.585 u yx.766 φ zx rad According to unit force on key node in Y direction: u xy.766 u yy φ zy.38 rad According to unit oent on key node around Z direction: φ zz rad Based on these finite eleent results the global coordinates of the shear center of the diaphrag are: φ.38 x S x φ zyzz φ zx y S y +φ zz In FEM-Design the coordinates of the iddle point of the web are (see Fig. 6..): x id 4.99 ; y id With the distance between these two points we get a coparable solution with the hand calculation. 54

55 x SFEM ( x x S ) +( y S y id ) ( ( 4.99)) +( ).59 id The difference between FEM and hand calculation is less than %. The gravity center of the cross section (Fig. 6..) can be calculated based on the statical oents. And of course the gravity center lying on the syetry axis. The distance of the gravity center fro the web is: x ' G Sy ' (. 4 ).43 A.(4+6+4) With the input Young's odulus and with the second oents of inertia the idealized bending stiffnesses in the principal directions can be calculated by hand. ( ( E I ) kn ) E I (. 4(.43) )+ +. 6(.43) kn 3 With the finite eleent results we can calculate the translations of the shear center according to the unit forces and oent on the key node (see the forer calculation ethod). The distances between the shear center and the selected key node are: Δ xx S x Δ y y S y The translations of the shear center are as follows: 4 u Sxx u xx φ zx Δ y u Syx u yx+φ zx Δ x ( 6343).533 u Sxy u xy φ zy Δ y u Syy u yy+φ zy Δ x ( 6343).783 Based on these values the translations of the shear center in the principal directions: u u Sxx +u Syy + ( ) u Sxx u Syy u Sxy + 55 ( )

56 u +u u Sxx Syy ( ) u Sxx u Syy usxy ( ) According to these values the angles of the principal rigidity directions: α FEM arctan u u Sxx arctan 6.57o u Sxy.533 α FEM arctan u u Sxx arctan 63.43o u Sxy.533 The directions coincide with the axes of syetries (see Fig. 6..-) which is one of the principal rigidity direction in this case. Then with FEM-Design results we can calculate the idealized bending stiffnesses of the structure: EI FEM H u 3 (.538/) EI FEM H 63 3 u 3 (.8463/) E I FEM.6 kn E I FEM kn The difference between FEM and hand calculation is less than 4%. Download link to the exaple file: FEM-Design file: A siple calculation with diaphrags.str Section Editor file: The calculation of the shear center.sec 56

57 7. Calculations considering nonlinear effects 7. Uplift calculation 7.. A trusses with liited copression ebers In this exaple a truss will be analyzed. First of all we calculate the noral forces in the truss ebers and the axiu deflection for the given concentrated loads. After this step we calculate the load ultiplier when the vertical truss ebers reaches its liit copression bearing capacity what we set. See the inputs in the following table. After the hand calulation we copare the results with the FEM-Design nonlinear calculation results. Inputs: Colun height/span length H. ; L 8. The cross sections KKR 8x8x6 The area of the cross sections A 65 The elastic odulus E GPa, structural steel The concentrated loads F 4 kn Liited copression of the vertical truss ebers Pcr 7 kn F 4 3 F F F 5 7 H F 4 L8 δf δf δf Figure 7... The truss with the concentrated loads and with the virtual loads for the translation calculation 57

58 The noral forces in the truss ebers based on the hand calculation (without further details) are: N N 7 kn ; N N 4 kn ; N 3 N kn ; N 4 N 5 N 3 N 6 6. kn ; N 6 N +6. kn ; N 7 N +8.8 kn ; N 8 N 8. kn ; N 9 4. kn. The noral forces in the truss ebers according to the vertical virtual force (see Fig. 7...): N, N,7.5 kn ; N, N,4N,9 kn ; N,3 N,5N,7 N,+.77 kn ; N,4 N,5 N,3N,6.5 kn ; N,6 N, +.5 kn ; N,8 N,. kn. The noral forces in the truss ebers according to the horizontal virtual force (see Fig. 7...): N, N,6 N, N,4+. kn ; N, N,3 N,4 N,5 N,7 N,8 N,9 N, N,N,3N,5 N,6 N,7 kn. The hand calculation of the vertical translation at the id-span with the virtual force ethod: 7 e z N δ N,i li EA i i The hand calculation of the horizontal translation at right roller with the virtual force ethod: 7 e x N δ N,i li EA i i Figure 7... The truss with the concentrated loads in FEM-Design 58

59 Figure The reaction forces Figure The noral forces in the truss ebers Figure The vertical translation at the id-span and the horizontal translation at the right roller [] 59

60 The translations and the noral forces in the truss ebers based on the hand calculation are identical with the FEM-Design calculation, see Fig After this step we would like to know the axiu load ultiplier when the vertical truss ebers reaches its liit copression bearing capacity what we set, Pcr 7 kn. The axiu copression force arises in the side coluns, see the hand calculation, N ( ) kn. Therefore the load ultiplier based on the hand calculation is λ 7.. Let's see the FEM-Design uplift calculation considering the liit copression in the vertical ebers. Figure Large nodal displaceents when the side truss ebers reached the liit copression value [] Figure The two different analyzed load ultiplier in FEM-Design With λfem 7. ultiplier the FEM-Design analysis gives the accurate result but with λfem 7. (see Fig ) large nodal displaceents occurred, see Fig Thus by this structure if we neglect the effect of the side ebers the coplete truss becae a statically overdeterined structure. FEM-Design solve this proble with iterative solver due to the fact that these kind of probles are nonlinear. Based on the FEM-Design calculation the load ultiplier is identical with the hand calculation. Download link to the exaple file: A trusses with liited copression ebers.str 6

61 7.. A continuous bea with three supports In this exaple we analyse non-linear supports of a bea. Let's consider a continuous bea with three supports with the following paraeters: Inputs: Span length L, total length x 4 The cross sections Rectangle: x5 The elastic odulus E 3 GPa, concrete C/5 Intensity of distributed load (total, partial) p kn/ In Case I. the distribution of the external load and the nonlinearity of the supports differ fro Case II. See the further details below (Fig and Fig ). a) Case I. In this case the distributed load is a total load (Fig. 7...). In the first part all of the three supports behave the sae way for copression and tension. In the second part of this case the iddle support only bears tension. We calculated in both cases the deflections, shear forces and bending oents by hand and copared the results with FEM-Design uplift (nonlinear) calculations. Figure 7... The bea with three supports and unifor distributed load In first part of this case the axiu deflection coes fro the following forula considering only the bending deforations in the bea: 4 e ax 4. p L EI / The relevant results with FEM-Design: Figure 7... The deflection [] of the bea with three supports (total load) 6

62 The extreus of the shear force without signs: V p L 7.5 kn ; V p L.5 kn The relevant results with FEM-Design: Figure The shear diagra [kn] of the bea with three supports (total load) The extreus of the bending oent without signs: M idspan 9 9 pl.8 kn ; M iddle p L 5. kn The relevant results with FEM-Design: Figure The bending oent diagra [kn] of the bea with three supports (total load) When the iddle support only bear tension (second part of this case) basically under the total vertical load (Fig. 7...) the iddle support is not active (support nonlinearity). Therefore it works as a siply supported bea with two supports. The deflection, the shear forces and the bending oents are the following: The axiu deflection coes fro the following forula considering only the bending deforations in the bea: 5 p ( L+ L)4 5 (+)4 e ax EI / 6

63 The relevant results with FEM-Design: Figure The deflection [] of the bea when the iddle support only bear tension (total load) The axiu of the shear force without sign: V p ( L+L) (+) kn The relevant results with FEM-Design: Figure The shear diagra [kn] of the bea when the iddle support only bear tension (total load) The extreu of the bending oent without sign: M ax p (L+ L) (+) kn 8 8 The relevant results with FEM-Design: Figure The bending oent diagra [kn] of the bea when the iddle support only bear tension (total load) The differences between the calculated results by hand and by FEM-Design are less than %. Download link to the exaple file: A continuous bea with three supports case a.str 63

64 b) Case II. In this case the distributed load is a partial load (Fig ). In the first part all of the three supports behave the sae way for copression and tension. In the second part of this case the right side support only bears copression. We calculate in both cases the deflections, shear forces and bending oents by hand and copared the results with FEM-Design calculations. Figure The bea with three supports and partial load The extreus of the deflection coe fro the following forulas considering only the bending deforations in the bea (without signs): e ax. ( p /)L 4 5 ( p/ ) L 4. / 4 5 / EI 384 EI 384 EI 384 EI e in 5 ( p / ) L4 ( p /) L 4 5 / 4 / EI 384 EI 384 EI 384 EI The relevant results with FEM-Design: Figure The deflection [] of the bea with three supports (partial load) The extreus of the shear force without signs: V p L 8.75 kn ; V p L.5 kn ; V 3 p L.5 kn

65 The relevant results with FEM-Design: Figure 7... The shear diagra [kn] of the bea with three supports (partial load) The extreus of the bending oent without signs: ( ) ( ) 7 7 pl 6 6 M idspan 3.88 kn ; M iddle p L.5 kn 6 6 p The relevant results with FEM-Design: Figure 7... The bending oent diagra [kn] of the bea with three supports (partial load) When the right side support only bear copression (second part of this case) basically under the partial vertical load (Fig ) the right side support is not active (support nonlinearity). Therefore it works as a siply supported bea with two supports. The deflection, the shear forces and the bending oents are the following: The axiu deflection coes fro the following forula considering only the bending deforations in the bea: 5 p L4 5 4 e idspan EI 384 EI p L4 4 e right EI 4 EI 65

66 The relevant results with FEM-Design: Figure 7... The deflection [] of the bea when the right support only bear copression (partial load) The extreu of the shear force without sign: V p L kn The relevant results with FEM-Design: Figure The shear diagra [kn] of the bea when the right support only bear copression (partial load) The extreu of the bending oent without sign: M ax p L 5. kn 8 8 The relevant results with FEM-Design: Figure The bending oent diagra [kn] of the bea when the right support only bear copression (partial load) The differences between the calculated results by hand and by FEM-Design are less than %. Download link to the exaple file: A continuous bea with three supports case b.str 66

67 7. Cracked section analysis by reinforced concrete eleents 7.. Cracked deflection of a sily supported bea Inputs: Span length Leff 7. The cross section Rectangle: b 3 ; h 45 The elastic odulus of concrete Ec GPa, C5/3 The creep factor φ8.35 Effective elastic odulus of concrete Eceff Ec/(+φ8) GPa Mean tensile strength fct.565 MPa Elastic odulus of steel bars Es GPa Characteristic value of dead load gk 8.5 kn/ Characteristic value of live load qk. kn/ Live load cobination factor ψ.6 Diaeter of the longitudinal reinforceent ϕl 8 Diaeter of the stirrup reinforceent ϕs 8 Area of longitudinal reinforceent Al 5x8π/4 7.3 Noinal concrete cover cno Effective height d h cno ϕs ϕl/ 43 Shrinkage strain εcs.5 The cross sectional properties without calculation details (considering creep effect): I. stress stadiu second oent of inertia II 3.75x9 4 II. stress stadiu second oent of inertia III.8x9 4 I. stress stadiu position of neutral axis xi 56.4 II. stress stadiu position of neutral axis xii 97.3 ψqk gk Leff 5ϕ8 M Max Figure 7... The siply supported RC bea 67

68 The calculation of deflection according to EN 99--: The load value for the quasi-peranent load cobination: p qpg k +ψ qk kn The axiu deflection with cross sectional properties in Stadiu I. (uncracked): 4 5 pqp Leff w k.i E ceff I I The axiu deflection with cross sectional properties in Stadiu II. (cracked): 4 5 p qp L eff w k.ii E ceff I II The axiu bending oent under the quasi-peranent load (see Fig. 7...): M ax p qp L eff kn 8 8 The cracking oent with the ean tensile strength: M cr f ct II kn h x I The interpolation factor considering the ixture of cracked and uncracked behaviour at the ost unfavourable cross-section: [ ( )] [ M cr 4.74 ζ ax.5, ax.5 M ax.74 ( ), ].998 This value is alost., it eans that the final deflection will be closer to the cracked deflection than to the uncracked one. The final deflection with the ai of interpolation factor: w k ( ζ )w k.i +ζ w k.ii (.998)

69 This deflection is even greater if we are considering the effect of shrinkage. The curvatures in uncracked and cracked states due to shrinkage: κ I,cs ε cs Es S s, I ( ) E c, eff I I κ II,cs ε cs E s S s, II ( ) E c, eff I II The additional deflection in the two different states due to shrinkage: 4 w k.i.cs Leff κ I, cs w k.ii.cs Leff κ II,cs The total deflection considering cracking and the effect of shrinkage: w k.cs( ζ )( wk.i +w k.i.cs )+ζ (w k.ii +w k.ii.cs ) (.998)( )+.998( )

70 First we odelled the bea with bea eleents. In FEM-Design we increased the division nuber of the bea finite eleents to ten to get the ore accurate results. Figure 7... The cross section of the RC bea in FEM-Design Fig shows the applied cross section and reinforceent with the defined input paraeters. Figure The deflection [] of the RC bar in FEM-Design with cracked section analysis (without shrinkage [above], with shrinkage [below]) Fig shows the deflection after the cracked section analysis without and with considering shrinkage. The deflection of the bea odel in FEM-Design: 7

71 Cracked section analysis without shrinkage: w kfem 7.39 Cracked section analysis with shrinkage: w k.csfem 35.9 The difference between the hand and FEM-Design calculations is less than 3%, but keep in ind that FEM-Design considers the interpolation factors individually in every finite eleents one by one to get a ore accurate result. Secondly we odelled the bea with shell finite eleents. Fig shows the applied specific reinforceent with the defined input paraeters with slab odel. Figure The specific reinforceent with the shell odel in FEM-Design Figure The deflection [] of the RC shell odel in FEM-Design with cracked section analysis (without shrinkage [above], with shrinkage [below]) Fig shows the deflection and the finite eleent esh after the cracked section analysis 7

72 without and with considering shrinkage. The deflection of the shell odel in FEM-Design: Cracked section analysis without shrinkage: w kfem 7.6 Cracked section analysis with shrinkage: w kfem 35.5 The difference between the hand and FEM-Design calculations is less than 4% but keep in ind that FEM-Design considers the interpolation factors individually in every finite eleents one by one to get a ore accurate result. Fig shows the effect of tension stiffening (without shrinkage effects) in FEM-Design at the relevant SLS load interval. We indicated the load level where the first crack occured. Effect of tension stiffening in FEM-Design 8 6 Load [kn/] w 7.3 p 5.7 kn/ 4 First crack at id-span 8 6 Shell odell Bea odell Uncracked Fully cracked Deflection [] Figure The effect of tension stiffening by a siply supported bea Download link to the exaple files: Bea odel: Cracked deflection of a siply supported bea.bea.str Shell odel: Cracked deflection of a siply supported bea.shell.str 7

73 7.. Cracked deflection of a statically indeterinate bea Inputs: Span length Leff 7. The cross section Rectangle: b 3 ; h 45 The elastic odulus of concrete Ec GPa, C5/3 The creep factor φ8.35 Effective elastic odulus of concrete Eceff Ec/(+φ8) GPa Mean tensile strength fct.565 MPa Elastic odulus of steel bars Es GPa Characteristic value of dead load gk 8.5 kn/ Characteristic value of live load qk. kn/ Live load cobination factor ψ.6 Diaeter of the longitudinal reinforceent ϕl 8 Diaeter of the stirrup reinforceent ϕs 8 Area of longitudinal reinforceent Al 5x8π/4 7.3 Noinal concrete cover cno Effective height d h cno ϕs ϕl/ 43 Shrinkage strain εcs.5 The cross sectional properties without calculation details (considering creep effect): I. stress stadiu second oent of inertia II 3.75x9 4 II. stress stadiu second oent of inertia III.8x9 4 I. stress stadiu position of neutral axis xi 56.4 II. stress stadiu position of neutral axis xii 97.3 In this chapter we will calculate the cracked deflection of a statically indeterinate structure (see Fig. 7...). 73

74 ψqk gk Leff Max M 5ϕ8 5ϕ8 Figure 7... The fixed end and a roller boundary conditions RC bea (statically indeterinate structure) The axiu deflection with cross sectional properties in Stadiu I. (uncracked): w k.i 4. pqp Leff E ceff I I The axiu deflection with cross sectional properties in Stadiu II. (cracked): w k.ii 4. p qp L eff E ceff I II The axiu bending oent under the quasi-peranent load at the fixed end (see Fig. 7...): M ax p qp L eff kn 8 8 The cracking oent with the ean tensile strength: M cr f ct II kn h x I The interpolation factor considering the ixture of cracked and uncracked behaviour at the ost unfavourable cross-section: 74

75 [ ( )] [ M cr 4.74 ζ ax.5, ax.5 M ax.74 ( ), ].998 This value is alost., it eans that the final deflection will be closer to the cracked deflection than to the uncracked one with the hand calculation. The final deflection with the ai of interpolation factor: w k ( ζ )w k.i +ζ w k.ii (.998) This deflection is even greater if we are considering the effect of shrinkage. The curvatures in uncracked and cracked states due to shrinkage: κ I,cs ε cs Es S s, I ( ) E c, eff I I κ II,cs ε cs E s S s, II ( ) E c, eff I II The additional deflection due to shrinkage: w k.i.cs Leff κ I,cs w k.ii.cs L κ eff II,cs The total deflection considering cracking and the effect of shrinkage: w k.cs( ζ )( wk.i +w k.i.cs )+ζ (w k.ii +w k.ii.cs ) (.998)( )+.998(.+6.)7.74 First we odelled the bea with bea eleents. In FEM-Design we increased the division nuber of the bea finite eleents to ten to get the ore accurate results. 75

76 Figure 7... The cross section of the RC bea in FEM-Design Fig shows the applied cross section and reinforceent with the defined input paraeters. Figure The deflection [] of the RC bar in FEM-Design with cracked section analysis (without shrinkage [above], with shrinkage [below]) Fig shows the deflection after the cracked section analysis without and with considering shrinkage. The deflection of the bea odel in FEM-Design: Cracked section analysis without shrinkage: w kfem.6 Cracked section analysis with shrinkage: w k.csfem 5.55 The difference between the hand and FEM-Design calculations is around %, but keep in ind that FEM-Design considers the interpolation factors individually in every finite eleents one by one to get a ore accurate result and by a statically indeterinate structure it causes greater difference. 76

77 Secondly we odelled the bea with shell finite eleents. Fig shows the applied specific reinforceent with the defined input paraeters with slab odel. Figure The specific reinforceent with the shell odel in FEM-Design Figure The deflection [] of the RC shell odel in FEM-Design with cracked section analysis (without shrinkage [above], with shrinkage [below]) Fig shows the deflection and the finite eleent esh after the cracked section analysis without and with considering shrinkage. The deflection of the shell odel in FEM-Design: Cracked section analysis without shrinkage: w kfem.63 Cracked section analysis with shrinkage: w k.csfem

78 The difference between the hand and FEM-Design calculations is around %, but keep in ind that FEM-Design considers the interpolation factors individually in every finite eleents one by one to get a ore accurate result and by a statically indeterinate structure it causes greater difference. Fig shows the effect of tension stiffening (without shrinkage effects) in FEM-Design at the relevant SLS load interval. We indicated the load level where the first crack occurred at the fixed end and at the id-span. Effect of tension stiffening in FEM-Design 8 6 p 5.7 kn/ First crack at id-span w.6 Load [kn/] 4 First crack at the fixed end 8 6 Shell odell Bea odell Uncracked Fully cracked Deflection [] Figure The effect of tension stiffening by a statically indeterinate structure Download link to the exaple files: Bea odel: Cracked deflection of a statically indeterinate bea.bea.str Shell odel: Cracked deflection of a statically indeterinate bea.shell.str 78

79 7..3 Cracked deflection of a cantilever bea Inputs: Span length Leff 4 The cross section Rectangle: b 3 ; h 45 The elastic odulus of concrete Ec GPa, C5/3 The creep factor φ8.35 Effective elastic odulus of concrete Eceff Ec/(+φ8) GPa Mean tensile strength fct.565 MPa Elastic odulus of steel bars Es GPa Characteristic value of dead load gk 8.5 kn/ Characteristic value of live load qk. kn/ Live load cobination factor ψ.6 Diaeter of the longitudinal reinforceent ϕl 8 Diaeter of the stirrup reinforceent ϕs 8 Area of longitudinal reinforceent Al 5x8π/4 7.3 Noinal concrete cover cno Effective height d h cno ϕs ϕl/ 43 Shrinkage εcs.4 The cross sectional properties without calculation details: I. stress stadiu second oent of inertia II 3.75x9 4 II. stress stadiu second oent of inertia III.8x9 4 I. stress stadiu position of neutral axis xi 56.4 II. stress stadiu position of neutral axis xii 97.3 ψqk gk Leff Max M Figure The cantilever RC bea 79 5ϕ8

80 The calculation of deflection according to EN 99--: The load value for the quasi-peranent load cobination: p qpg k +ψ qk kn The axiu deflection with cross sectional properties in Stadiu I. (uncracked): 4 w k.i 4 p qp Leff E ceff I I The axiu deflection with cross sectional properties in Stadiu II. (cracked): 4 w k.ii 4 pqp Leff E ceff I II The axiu bending oent under the quasi-peranent load: M ax pqp Leff kn The cracking oent with the ean tensile strength: M cr f ct II kn h x I The interpolation factor considering the ixture of cracked and uncracked behaviour: [ ( )] [ ) ] M cr 4.74 ζ ax.5, ax.5,.9474 M ax 5.6 ( This value is alost., it eans that the final deflection will be closer to the cracked deflection than to the uncracked one. The final deflection with the ai of interpolation factor: w k ( ζ )w k.i +ζ w k.ii (.9474) This deflection is even greater if we are considering the effect of shrinkage. 8

81 The curvatures in uncracked and cracked states due to shrinkage: κ I,cs ε cs Es S s, I ( ) E c, eff I I κ II,cs ε cs E s S s, II ( ) E c, eff I II The additional deflection due to shrinkage: 4 w k.i.cs Leff κ I,cs w k.ii.cs Leff κ II, cs The total deflection considering cracking and the effect of shrinkage: w k.cs( ζ )( wk.i +w k.i.cs )+ζ (w k.ii +w k.ii.cs ) (.9474)( )+.9474( )

82 We odelled the bea with bea finite eleents. In FEM-Design we increased the division nuber of the bea finite eleents to five to get the ore accurate results. Figure The cross section of the RC cantilever in FEM-Design Fig shows the applied cross section and reinforceent with the defined input paraeters. Figure The deflection [] of the RC bar odel in FEM-Design with cracked section analysis (without shrinkage [above], with shrinkage [below]) 8

83 Fig shows the deflection after the cracked section analysis. The deflection of the bea odel in FEM-Design: Cracked section analysis without shrinkage: w kfem 4.3 Cracked section analysis with shrinkage: w k.csfem 3.5 The difference between the hand and FEM-Design calculations is around 9%, but keep in ind that FEM-Design considers the interpolation factors individually in every finite eleents one by one to get a ore accurate result. Download link to the exaple file: Cracked deflection of a cantilever bea.str 83

84 7..4 Cracked deflection of a cantilever bea with copressed reinforceent bars Inputs: Span length Leff 3 The cross section b ; h 4 The elastic odulus of concrete Ec 3.5 GPa, C5/3 The creep factor φ8. Effective elastic odulus of concrete Eceff Ec/(+φ8).5 GPa Mean tensile strength fct.6 MPa Elastic odulus of steel bars Es GPa Characteristic value of the point oent at the end Mk 4 kn Diaeter of the longitudinal reinforceent ϕl Diaeter of the stirrup reinforceent ϕs Area of longitudinal reinforceent (tension) Al 4xπ/ Area of longitudinal reinforceent (copression) Al' xπ/4 6. Noinal concrete cover cno Effective height (tension) d h cno ϕs ϕl/ 364 Effective height (copression) d' cno + ϕs + ϕl/ 36 Shrinkage εcs.4 The cross sectional properties without calculation details: I. stress stadiu second oent of inertia II.49x9 4 II. stress stadiu second oent of inertia III 6.563x8 4 I. stress stadiu position of neutral axis xi 7.6 II. stress stadiu position of neutral axis xii 8. 4ϕ Mk Leff ϕ Max M Figure The cantilever RC bea with copressed RC bars 84

85 The calculation of deflection according to EN 99--: The load value for the quasi-peranent load cobination (see Fig ): M k 4 kn The axiu deflection with cross sectional properties in Stadiu I. (uncracked): M k Leff 4 3 w k.i.7.7 E ceff I I 5.49 The axiu deflection with cross sectional properties in Stadiu II. (cracked): M k Leff 4 3 w k.ii.6 6. E ceff I II The axiu bending oent under the quasi-peranent load (see Fig ): M ax M k 4 kn The cracking oent with the ean tensile strength: M cr f ct II kn h x I.4.76 The interpolation factor considering the ixture of cracked and uncracked behaviour: [ ( )] [ ( ) ] M cr 9.4 ζ ax.5, ax.5,.8867 M ax 4 The final deflection with the ai of interpolation factor: w k ( ζ )w k.i +ζ w k.ii (.8867) This deflection is even greater if we are considering the effect of shrinkage. The curvatures in uncracked and cracked states due to shrinkage: κ I,cs ε cs Es S s, I ( ) 6. (7.6 36) E c, eff I I.5.49 κ II,cs ε cs E s S s, II (364 8.) 6. (8. 36) E c, eff I II The additional deflection due to shrinkage: 85

86 w k.i.cs Leff κ I,cs w k.ii.cs Leff κ II, cs The total deflection considering cracking and the effect of shrinkage: w k.cs( ζ )(wk.i +w k.i.cs )+ζ (w k.ii +w k.ii.cs ) (.8867)( )+.8867( )8.6 We odelled the bea with bea finite eleents. In FEM-Design we increased the division nuber of the bea finite eleents to five to get the ore accurate results. Figure The cross section of the RC cantilever in FEM-Design Fig shows the applied cross section and reinforceent with the defined input paraeters. 86

87 Figure The deflection [] of the RC bar odel in FEM-Design with cracked section analysis (without shrinkage [above], with shrinkage [below]) Fig shows the deflection after the cracked section analysis. The deflection of the bea odel in FEM-Design: Cracked section analysis without shrinkage: w kfem 3.3 Cracked section analysis with shrinkage: w k.csfem 7.3 The difference between the hand and FEM-Design calculations is around 7%. Download link to the exaple file: Cracked deflection of a cantilever bea with copressed reinforceent bars 87

88 7..5 Cracked deflection of a cantilever with bending oent and noral forces Inputs: Span length Leff The cross section Rectangle: b ; h 4 The elastic odulus of concrete Ec 3.5 GPa The creep factor φ8. Effective elastic odulus of concrete Eceff Ec/(+φ8).5 GPa Ratio between the oduli α Es/Eceff 9.5 Mean tensile strength fct. MPa Elastic odulus of steel bars Es GPa Diaeter of the longitudinal reinforceent ϕl Diaeter of the stirrup reinforceent ϕs Area of longitudinal reinforceent (top) Al 4xπ/ Area of longitudinal reinforceent (botto) Al' 4xπ/ Noinal concrete cover cno Effective height (botto) d h cno ϕs ϕl/ 364 Effective height (top) d' cno + ϕs + ϕl/ 36 4ϕ M5 kn N -5 5 kn Leff 4ϕ Figure A RC cantilever with the applied loads In this chapter we will calculate the deflection (vertical translation) of the end of a cantilever (see Fig ). According to the behaviour of the reinforced concrete aterial this deflections will depend on the aount of the applied noral force. At the free end of the cantilever we applied a constant concentrated bending oent (M5kN) and we changed the intensity of the applied concentrated noral force at the end fro -5 kn copression to +5 kn tension. The force is acting on the centroid of the uncracked RC section. Now it is in the iddle. During the hand calculation (and by the FEM-Design calculation as well) we considered eccentricity caused by cracking in cracked section analysis. We are going to calculate the copressed concrete zone by hand at five notable copressed zone condition and based on these values the inhoogeneous inertias and areas are also can be calculated. The deflections depend on these inertias. After these we calculate the interpolation factors to get the accurate deflection considering the tension stiffening effect. At the end of this chapter we copare the results with FEM-Design solutions. 88

89 Case a) M 5 kn; N - 5 kn (copression): In this case the coplete section is uncracked, the total concrete zone is active. Therefore we only need to calculate an inhoogeneous cross-section. Aa b h+α ( Al +Al ') ( ).974 I a b h3 h h +α Al d +α Al ' d ' ( ) ( ) ( ) ( ) The stresses at the extree fibres are as follows: σ top N M h MPa < f ct. MPa Aa I a σ botto (tension) N M h MPa (copression) Aa I a Based on these equation it is trivial that the first crack occur at the following noral force: ( N crack f ct ) ( ) M h 5.4 Aa kn (copression) Ia.533 The deflection (vertical translation) of the free end of the cantilever: M Leff E ceff I a w a Case b) M 5 kn; N - kn (copression): The copressed zone (easured fro the copressed side) in this case coes fro the solution of a third order polynoial: M N α Al (d x b )(d h /)+α Al '(x b d ')(h/ d ')+ b x b (h/ x b ) 3 b x b α Al (d x b) α Al '( x b d ') The relevant solution of this equation: x b

90 Based on this, the position of the centroid easured fro the copressed side: x b b +α Al d +α Al ' d ' x Sb 3. b x b+α (Al + Al ') According to this the cracked section area and inertia: Abb x b+α ( Al + Al ') ( ).578 I b b x 3b +b xb ( x Sb xb /)+α Al ( d x Sb) +α Al ' ( x Sb d ' ).77 4 The deflection (vertical translation) of the free end of the cantilever with this cracked condition considering the decreased oent at the cracked centroid fro the additional eccentricities (because the original loads are acting on the centroid of the uncracked RC section): w bii (M +N (h/ x Sb )) Leff (5 (..3)) 9.56 E ceff I b 5.77 But this is not the final deflection, because we need to consider the tension stiffening to get a ore coparable result. The stress in the tension reinforceent calculated on the basis of a cracked section and the additional oent due to cracked eccentricity: σ sb α [ N M +N (h/ x Sb ) + (d x Sb) Ab Ib [ 5 (..3) + (.364.3) 57. MPa ] ] The load conditions causing first crack: ηb [ ] [ ] N M 5 + (h/) η b + (.4 /) f ct. MPa ; thus η b.493 Aa I a Thus the noral force: η b N.493 ( ) 98.6 kn (copression) And the bending oent: η b M kn 9

91 The stress in the tension reinforceent calculated on the basis of a cracked section under the loading conditions causing first cracking considering the additional oent due to cracked eccentricity: [ η b N η b M +η b N (h/ x Sb ) + (d x Sb ) Ab Ib [ (..3) + (.364.3) 77.4 MPa σ srb α 9.5 ] ] The interpolation factor considering the ixture of cracked and uncracked behaviour: [ ] [ ( ) ] σ srb 77.4 ζ bax.5 σ, ax.5,.879 sb 57. ( ) The final deflection with the ai of interpolation factor: w b( ζ b ) w a+ζ b wbii (.879) Case c) M 5 kn; N kn (pure bending): In this case the cross section is under pure bending. In this situation the following equation gives the position of the copressed zone (cracking occur on the tension side). x b+α (x c d ') Al '+α ( x c d ) Al c The relevant solution of this equation: x c 8.5 Based on this value the cracked cross-sectional area and inertia: Ac b x c +α ( Al + Al ') ( ).494 b x 3c I c +α Al ( d x c ) +α Al ' ( x c d ' ) ( ) ( ) The deflection (vertical translation) of the free end of the cantilever with this cracked condition: w cii M Leff E ceff I c

92 But this is not the final deflection, we need to consider the tension stiffening to get a ore coparable solutions. The interpolation factor based on the pure bending condition: The cracking oent with the ean tensile strength: M cr f ct Ia kn h/.4/ The interpolation factor considering the ixture of cracked and uncracked behaviour: [ ( ) ] [ ( ) ] M cr ζ c ax.5 M 6.83, ax.5, The final deflection with the ai of interpolation factor: w c ( ζ c )w a +ζ c wcii (.9434) Case d) M 5 kn; N kn (tension): The copressed zone (easured fro the copressed side) in this case coes fro the solution of a third order polynoial: b x d α Al (d x d )(d h /)+α Al '( x d d ')(h/ d ')+ (h / x d ) M 3 N b xd α Al (d x d ) α Al '( x d d ') The relevant solution of this equation: x d Based on this the position of the centroid easured fro the copressed side: x d b +α Al d +α Al ' d ' x Sd 3. b x d +α ( Al + Al ') According to this the cracked section area and inertia: Ad b x d +α ( Al +Al ') ( ).89 3 b x d I d +b x d ( x Sd x d / ) +α Al ( d x Sd ) +α Al ' ( x Sd d ' )

93 The deflection (vertical translation) of the free end of the cantilever with this cracked condition considering the increased oent at the cracked centroid fro the additional eccentricities (because the original loads are acting on the centroid of the uncracked RC section): (M +N (h/ x Sd )) L eff ( 5+(..3)) w dii 8.4 E ceff I d 5.67 But this is not the final deflection, we need to consider the tension stiffening to get a ore coparable solution. The stress in the tension reinforceent calculated on the basis of a cracked section and the additional oent due to cracked eccentricity: σ sd α [ N M +N (h/ x Sd ) + (d x Sd ) Ad Id [ 5+(..3) + (.364.3) MPa ] ] The load conditions causing first crack: ηd [ ] [ ] N M 5 + ( h/ ) η d + (.4/) f ct. MPa ; thus η d.56 Aa I a Thus the noral force: η d N kn kn And the bending oent: η d M kn The stress in the tension reinforceent calculated on the basis of a cracked section under the loading conditions causing first cracking considering the additional oent due to cracked eccentricity: [ η d N η d M +η d N (h / x Sd ) + ( d x Sd ) Ad Id [.8+5. (..3) 5. + (.364.3) 4.9 MPa σ srd α 9.5 ] ] The interpolation factor considering the ixture of cracked and uncracked behaviour: [ ] [ ( ) ] σ srd 4.9 ζ d ax.5 σ, ax.5,.9673 sd ( ) 93

94 The final deflection with the ai of interpolation factor: w d ( ζ d ) w a +ζ d w dii (.9673) Case e) M 5 kn; N 5 kn (tension): In this case the cross section is fully cracked (therefore this is only a theoretical solution). The whole concrete zone is cracked. Practically in eans that only the reinforceent bars are active in the section, thus the cross-sectional properties: Ae α ( Al + Al ')9.5 ( ).74 I eα Al ( d h/ ) +α Al ' ( h / d ') ( 364 ) ( 36 ) The deflection (vertical translation) of the free end of the cantilever with this fully cracked condition (only the RC bars are working): M Leff 5 w eii.54 E ceff I e But this is not the final deflection, we need to consider the tension stiffening to get a ore coparable solution. The stress in the tension reinforceent calculated on the basis of a cracked section: σ se α [ ] [ ] N M (d h/) (.364.) MPa Ae I e The load conditions causing first crack: ηe [ ] [ ] N M (h/) η e + (.4/ ) f ct. MPa ; thus η e.479 Aa I a Thus the noral force: η e N kn And the bending oent: η e M kn The stress in the tension reinforceent calculated on the basis of a cracked section under the loading conditions causing first cracking: 94

95 σ sre α [ ] [ ] ηn ηm ( d h/) (.364.) 38.7 MPa Ae Ie The interpolation factor considering the ixture of cracked and uncracked behaviour: [ ] [ ( ) ] σ 38.7 ζ e ax.5 σ sre, ax.5,.984 se ( ) The final deflection with the ai of interpolation factor: w e ( ζ e )w a +ζ e weii (.984) Figure A RC cantilever with the applied loads in FEM-Design Fig shows the FEM-Design odel with bars. Fig shows the deflections (vertical translations) of the free end of cantilever under the constant bending oent regarding different noral forces (copression and tension as well). In Fig in addition to the hand calculation we indicated the FEM-Design results with bea odel and with shell odel also. The differences between the hand and FEM-Design calculations are less than 5%, thus we can say that the results are identical to each other. 95

96 Cracked deflection under bending and noral force 8 Fully cracked concrete zone case e) 6 case d) Noral force [kn] 4 Pure bending case c) First crack -6 Fully active concrete zone case a) -8 case b) Shell odell Bea odell Hand calculation Deflection (vertical translation) [] Figure A results with FEM-Design (bea and shell also) copared to the hand calculation NOTE: The cracked section analysis in FEM-Design is based on a non-linear calculation but only accurate by SLS cobinations because the used aterial odel for the reinforceent is linear (both copression and tension) and for the concrete is a non-tension aterial (only copression, assued to be linear) if the extree fibres reached the ean tensile strength in the cross-section. Download link to the exaple files: Bea odel: Cracked deflection of a cantilever with bending oent and different noral forces.bea.str Shell odel: Cracked deflection of a cantilever with bending oent and different noral forces.shell.str 96

97 7..6 Cracked deflection of a siply supported square slab Inputs: Span length Leff 6 (Fig ) The thickness h The elastic odulus of concrete Ec 3 GPa, C/5 The Poisson's ratio of concrete ν. The creep factor φ8.35 Effective elastic odulus of concrete Eceff Ec/(+φ8) 8.96 GPa Mean tensile strength fct. MPa Elastic odulus of steel bars Es GPa Characteristic value of dead load gk 6 kn/ Characteristic value of live load qk kn/ Live load cobination factor ψ.6 Diaeter of the longitudinal reinforceent ϕl / The specific reinforceent As.393 / Noinal concrete cover cx 3 ; cy 4 Average effective height d 6 The ratio of the elastic odulus αs Es/Eceff.3 y' pqpgk+ψqk 6 6 x' Figure Siply supported RC slab with constant total distributed load The load value for the quasi-peranent load cobination: p qpg k +ψ qk 6+.6 kn 97

98 y' Φ/393 / c3 x' Φ/393 / c4 Figure The reinforceent in the slab The deflection of an isotropic sily supported square slab under unifor load (see Chapter.3): p a4 w ax.46 E h3 ( ν ) ( ) The uncracked unreinforced specific inertia (only with the concrete): I c h The deflection based on this inertia: w ax,c.46 ( ) The uncracked reinforced specific inertia (Stadiu I.): h +α s a s d x I.5 h+α s a s x I (h x I ).5 (.5) I I + +α s a s (d x I ) (6.5) I I The deflection based on this inertia: 4 w ax, I.46 ( ) 98

99 The cracked reinforced specific inertia (Stadiu II.): x II x II +α s a s d x II x II +α s a s x II x II x II I II +α s a s ( d x II ) ( ) The deflection based on this inertia: 64 w ax, II ( ) The cracking oent: cr f ct II N kn h x I.5 ax.469 p a kn The interpolation factor considering the ixture of cracked and uncracked behaviour: [ ( )] [ ) ] cr 5.73 ζ ax.5, ax.5,.6986 ax.6 ( The deflection based on this interpolation factor: w ax,cr ( ζ ) w ax, I +ζ w ax, II (.6986) Figure The FE odel 99

100 Figure The finite eleent esh [average eleent size:. ] Figure The reactions [kn/] without reinforceent and with reinforced cracked section analysis

101 Figure The x [kn/] without reinforceent and with reinforced cracked section analysis

102 Figure The y [kn/] without reinforceent and with reinforced cracked section analysis Figure The xy [kn/] without reinforceent and with reinforced cracked section analysis

103 Figure The deflection [] without reinforceent and with reinforced cracked section analysis This hand calculation ethod is very very conservative, thus do not considering the realistic crack pattern and the torsional effects. In addition the shear deforations in the slab is also neglected. Fig shows the deflection based on FEM-Design. Here the difference between the deflections is quite large but this coes fro the entioned very very conservative hand calculation ethod. Download link to the exaple file: Cracked deflection of a siply supported square slab.str 3

104 7.3 Nonlinear soil calculation This chapter goes beyond the scope of this docuent, therefore additional inforations are located in: FEM-Design Geotechnical odul in 3D, Theoretical background and verification and validation handbook 4

105 7.4 Elasto-plastic calculations 7.4. Elasto-plastic point support in a bea Inputs: Span length L6 Distributed load q 9 kn/ Structural steel S35, fy 35 MPa Young's odulus of steel E GPa Shear odulus G 8.77 GPa Cross section IPE 4 Cross-sectional area A 8446 The first principal inertia I The shear correction factor in the relevant direction ρ.4 The elastic cross sectional odulus Wel The first oent of the half of the cross sectional area about the centroid S Plastic load-bearing oent capacity of the section Mpl Sfy 37. kn q9 kn/ L The bea with a fixed support and a roller The proble is a bea with fixed end on the left side and a roller on the right side (see Fig ). The input paraeters are in the table above. First of all we calculate the deflection and the internal forces according to the external total distributed load based on linear elastic theory (Case ). For the second calculation we assue that the fixed support can only bear axiu Mpl bending oent (Case ). Thus we assue a plastic hinge after a certain aount of load level when the bea reaches this plastic liit oent in the fixed support. Based on the plastic hinge the distribution of the internal forces and the deflection will be different fro the linear elastic calculation. At the hand calculation we neglect the shear deforation (Euler-Bernoulli bea theory) to get a siple solution for this proble. In FEM-Design the applied bea theory considers the shear deforation (Tioshenko bea theory). It eans that when we copare the solution of the hand calculation to the finite eleent solution the deflection will be a bit larger in case of the 5

106 FEM calculation. To avoid this difference in the FEM analysis of the bea then the crosssectional area ust be rewritten with a large nuber (e.g.: A 6 ). Practically it eans that the shear stiffness is alost infinite. In this case the solutions will be the sae than the results of the hand calculations. At the end of this exaple we will show two finite eleent results. One of the will neglect and the other will consider the shear deforation. Case : The characteristic shear forces, bending oents, rotations and translations are indicated in Fig left side according to linear elastic calculation. Here are soe hand calculation results without further details based on the classical theory of elasticity: 5 5 V ax q L kn ; 8 8 ϕ roller M ax q L kn 8 8 q L rad 48 E I ql 9 6 e idspan E I Case : To easily consider the bending oent liit (plastic analysis) at the fixed support the following siplification need to be declared at the hand calculation. Until the certain aount of load level (q) which causes Mpl bending oent at the fixed support the behaviour is linear elastic: M M pl 8M q L 37. kn > q pl 68.7 kn/ 8 L q68.7 kn/ Mpl37. kn q9 kn/ + q.73 kn/ The bea behaviour with the plastic bending oent liit at the fixed end After this load level the fixed end will be a plastic hinge therefore the statical syste differs fro the original one (see Fig ). The rest of the load level which will act on a siply supported bea (according to the plastic hinge): q q q kn / 6

107 With this assuption we can superpose the two different results based on the two different statical systes. The specific shear forces, bending oents, rotations and translations are as follows without furher details (Fig right side): 5 5 V ax q L+ q L kn 8 8 M ax M pl q L kn 8 8 q L3 q L ϕ roller E I 4 E I 48 4 ( )[ ] q L.73 6 (.47 rad 4 E I 4 )[ ] q L 5 q L E I 384 E I ( )[ ] 3 3 ϕ pl hinge 6 4 e idspan rad Figure left side shows the linear elastic and right side shows the plastic results V [kn] V [kn] M [kn] M [kn] φ [rad] φ [rad] e [] e [] The internal forces, the rotations and translations in the bea according to linear elastic (left) and plastic calculation (right) by hand Figure shows the FEM-Design odel with the input data. Figure shows the FEM-Design results without shear deforation (A 6 ). The left side shows the linear elastic calculation and the right side shows the plastic analysis. 7

108 We can say that the results are identical with the hand calculations The odel with the input data in FEM-Design The internal forces, the rotations and translations in the bea according to linear elastic (left) and plastic liit calculation (right) with FEM-Design without shear deforation (A 6 ) 8

109 Figure shows the FEM-Design results with shear deforation (A 8446 ). The left side shows the linear elastic calculation and the right side shows the plastic analysis. In this case the result differs a bit fro the hand calculations according to the shear deforation (Tioshenko bea theory) The internal forces, the rotations and translations in the bea according to linear elastic (left) and plastic liit calculation (right) with FEM-Design with shear deforation (A8446 ) Download links to the exaple files: Without shear deforation: Elasto-plastic point supports in a bea_without_shear def.str With shear deforation: Elasto-plastic point supports in a bea_with_shear def.str 9

110 7.4. Elasto-plastic line support in a plate Inputs: Diensions of the slab Lx 4 ; Ly 8 Distributed load qz kn/ Concrete slab C 3/37 Young's odulus of concrete E 33 GPa Poisson's ratio ν. Thickness t Line support pl x' 3 kn/ (around line) The proble is a concrete slab with a fixed support on its edge and with a total distributed load (see Fig ). y' e fix ine dl p su p s thi on t r o qz e edg x' A concrete slab with a fixed support on its edge Case : In this case we assue that the line support has a plastic liit oent capacity (pl, around the edge) along the whole support (blue and red line equally, see Fig ). The slab behaves siilarly as a cantilever bea. If the plastic liit oent capacity is valid along the whole line support the load bearing capacity can be assued with the help of the following equation: q L x pl This is the approxiated specific oent value at the fixed end. It eans that the axiu of the total distributed load which the structure can bear is: q pl Lx kn / 4

111 Thus if we apply qz kn/ load on the slab there is no equilibriu on this final load level. The last converged load level ust be around: η q % q z We odelled this case in FEM-Design. Let's see the results based on the finite eleent analysis. Fig shows the odel with the adjusted paraeters The concrete slab with a fixed support on its edge in FEM-Design (Case ) After the elasto-plastic analysis the last converged load level based on the progra is 37% (see Fig ). Fig also shows the reactions. We can see that the reaction oent around the line support is a constant 3. kn/ which is the expected value according to the adjusted liit oent value (pl x' 3 kn/). Thus we can say that the progra gives the sae solution as the hand calculation.

112 Last converged load level 37%, the deflection [] and the reactions by this load level Case : In this case we assue that the line support has a plastic liit oent capacity (pl, around the edge) along the red part (see Fig ). The blue parts behave in a linear elastic way. In this case the slab behaves also as a cantilever bea. The slab behaves linearly until load level q3.75 kn/. Above this load level there will be a plastic hinge line along the red part of the fixed support (see Fig ). This reaining load is: q q z q kn/ Since the su of the length of the blue parts is equal to the length of the red one (see Fig ) we can estiate that (as a conservative assution) the reaining load level is redistributing on the blue linear elastic fixed supports. To assue the axiu deflection of the slab first of all we need to calculate the deflection according to q3.75 kn/ (linear elastic behaviour): The approxiated bending stiffness of the slab: D E t kn 97 ( ν ) (. ) Based on this value the deflection (as a cantilever): q L x e D 8 97

113 Since the reaining part of the load is redistributing to the blue (linear elastic) parts of the support we can assue that the reaining iddle part of the distributed total load acts on the outer part. Therefore the reaining distributed load which has affect on the linear elastic parts of the line support: q q kn / Because the length of the red part equals to blue ones. According to this load the deflection of the slab after the linear elastic behaviour can be assued as: e q L x D 8 97 Thus the approxiation of the total deflection with elasto-plastic calculation is: e pl,tot e +e This is a very conservative result. We odelled this case in FEM-Design. Let's see the results based on the finite eleent analysis. Fig shows the odel with the adjusted paraeters The concrete slab with a fixed support on its edge in FEM-Design (Case ) After the elasto-plastic analysis the last converged load level based on the progra is % (see Fig ). Fig also shows the reactions. We can see that the reaction oent around the line support of the iddle part is a constant 3. kn/ which is the expected value according to the adjusted liit oent value (pl x' 3 kn/). The reaining part of the line support behaves linearly as expected. At the end of the linear behaviour (load level 37%) 3

114 the deflection is 5.74 (see Fig ). The difference between the FEM-Design and hand calculation is less than % (5.74 vs ). This difference coes fro the very conservative hand calculation forula and fro the fact that FEM-Design considers the shear deforations (Mindlin plate theory). At load level % the deflection is 9.36 (see Fig ). The results based on the hand calculation was.69. The difference is around 5%. This difference coes fro the very conservative hand calculation forula and fro the fact that FEM-Design considers the shear deforations (Mindlin plate theory). Thus we can say that the progra gives an accurate elasto-plastic solution Last converged load level %, the deflection [] and the reactions [kn/, kn/] by this load level Download link to the exaple files: Case : Elasto-plastic line supports in a plate case.str Case : Elasto-plastic line supports in a plate case.str 4

115 7.4.3 Elasto-plastic surface support with detach in an ebedded plate In this exaple the structure is an ebedded rectangle plate (a6, b4). The plate is assued to be infinitely rigid. There are three different load positions of copression point load (N3 kn, see Fig ). The behaviour of the bedding of the plate will be considered in three different ways (see Fig ). The table below shows the nine analyzed cases: No. Eccentricity of the copression point load (N 3 kn) Behaviour of the bedding in y direction, ey linear elastic in x direction, ex linear elastic 3 in y direction, ey and in x direction, ex linear elastic 4 in y direction, ey linear elastic, non-tension 5 in x direction, ex linear elastic, non-tension 6 in y direction, ey and in x direction, ex linear elastic, non-tension 7 in y direction, ey elasto-plastic, non-tension 8 in x direction, ex elasto-plastic, non-tension 9 in y direction, ey and in x direction, ex elasto-plastic, non-tension b4 x y N b4 N ey b4 y x ex a6 N ey y x ex a6 a Diensions of the ebedded plate and the positions of the copression point load on it f f f K copression tension tension d copression tension d K copression K d fli,cop The aterial behaviour of the bedding of the plate (K 5 kn//, fli,cop kpa) 5

116 Case ): In this case the behaviour of the bedding is linear elastic therefore the reaction forces can be calculated according to the superposition of pure copression and uniaxial bending: σ ax() N Mx y ax kpa A Ix 4 3 σ in() N Mx 3 3 y in 3.5 kpa A Ix 4 3 Case ): In this case the behaviour of the bedding is linear elastic therefore the reaction forces can be calculated according to the superposition of pure copression and uniaxial bending: σ ax() N My x kpa A I y ax 4 7 σ in() N My 3 6 x in kpa A Iy 4 7 Case 3): In this case the behaviour of the bedding is linear elastic therefore the reaction forces can be calculated according to the superposition of pure copression and biaxial bending: σ ax(3) M N Mx y ax + y x ax kpa A Ix Iy σ in(3) M N Mx y in + y x in kpa A Ix Iy The specific results of the hand calculation can be seen in Fig y y - N3 kn y N3 kn N3 kn x x + x The distribution of the reaction forces [kpa] of the linear elastic cases 6

117 Case 4): In this case the behaviour of the bedding is linear elastic, non-tension therefore the reaction forces can be calculated according to the theory of the stress volue. The aount of the resultant and the centroid of the linear stress volue ust be equal to the acting point load: N 6 3 σ () in σ in() N kpa Case 5): In this case the behaviour of the bedding is linear elastic, non-tension therefore the reaction forces can be calculated according to the theory of the stress volue. The aount of the resultant and the centroid of the linear stress volue ust be equal to the acting point load: N 4 3 σ () in σ in() N 3 5 kpa Case 6): In this case the behaviour of the bedding is linear elastic, non-tension therefore the reaction forces can be calculated according to the theory of the stress volue. The aount of the resultant and the centroid of the linear stress volue ust be equal to the acting point load: [ ] N σ in(3) 3 σ in(3) 6N kpa 6 4 The specific results of the hand calculation can be seen in Fig y N3 kn - N3 kn y N3 kn x x y x The distribution of the reaction forces [kpa] of the linear elastic, non-tension cases 7

118 In Case 7-9 the bahaviours are elasto-plastic, non-tension thus it eans that according to the given fli,cop kpa liit force (see Fig ) there are different values of the load-bearing capacity due to the positions of the point load. Case 7): In this case the behaviour of the bedding is elasto-plastic, non-tension therefore the reaction forces can be calculated according to the theory of the stress volue. The aount of the resultant and the centroid of the constant stress volue (with fli,cop kpa) ust be equal to the acting point load: N liit()6 f li,cop 6 4 kn η() N liit() 4 8 % N 3 Case 8): In this case the behaviour of the bedding is elasto-plastic, non-tension therefore the reaction forces can be calculated according to the theory of the stress volue. The aount of the resultant and the centroid of the constant stress volue (with fli,cop kpa) ust be equal to the acting point load: () N liit 4 f li,cop 4 6 kn N liit( ) 6 η 53.3 % N 3 ( ) Case 9): In this case the behaviour of the bedding is elasto-plastic, non-tension therefore the reaction forces can be calculated according to the theory of the stress volue. The aount of the resultant and the centroid of the constant stress volue (with fli,cop kpa) ust be equal to the acting point load: [ ( + N liit(3) )] 3 f li,cop 9 kn η(3) N liit(3) 9 3 % N 3 The specific results of the hand calculation can be seen in Fig y y N4 kn - N9 kn - N3 kn x x / y x The distribution of the reaction forces [kpa] of the elasto-plastic, non-tension cases 8

119 In the second part of this chapter we ake a FEM-Design odel for the proble and copare the results with the hand calculation. Fig shows the ain input properties of the odel (concrete slab with thickness). In FEM-Design the reaction result of a surface support eleent is the average value. For ore precise results the average eleent (esh) size was.. By the calculation of the surface support reactions we exrapolated the eleent average FEMDesign nueric values to the extree fibre (edge of the plate) to get ore coparable results. You can find below one by one the 9 different cases and their results based on FEM-Design The FEM-Design odell (plate with surface support) with the input data and the different position of the point loads and the finite eleent esh Case ) FEM (see Fig ): σ ax FEM()+64.7 kpa σ in FEM() 37.5 kpa The reaction results in the surface support with the specific nuerical values (eleent average) for Case ) [kpa] 9

120 Case ) FEM (see Fig ): σ ax FEM() +3.4 kpa σ in FEM( ) 379 kpa The reaction results in the surface support with the specific nuerical values (eleent average) for Case ) [kpa] Case 3) FEM (see Fig ): σ ax FEM(3)+95.5 kpa σ in FEM(3) 56 kpa The reaction results in the surface support with the specific nuerical values (eleent average) for Case 3) [kpa]

121 Case 4) FEM (see Fig ): σ in FEM() 338 kpa The reaction results in the surface support with the specific nuerical values (eleent average) for Case 4) [kpa] Case 5) FEM (see Fig ): σ in FEM( ) 5.5 kpa The reaction results in the surface support with the specific nuerical values (eleent average) for Case 5) [kpa]

122 Case 6) FEM (see Fig ): σ in FEM(3) 95 kpa The reaction results in the surface support with the specific nuerical values (eleent average) for Case 6) [kpa] Case 7) FEM (see Fig ): Last converged load level and the bearing capacity. η FEM () 8 % N liit FEM().8 34 kn The reaction results in the surface support with the specific nuerical values (eleent average) for Case 7) [kpa]

123 Case 8) FEM (see Fig ): Last converged load level and the bearing capacity. η FEM ( )5 % N liit FEM() kn The reaction results in the surface support with the specific nuerical values (eleent average) for Case 8) [kpa] Case 9) FEM (see Fig ): Last converged load level and the bearing capacity. η FEM (3) 3 % N liit FEM(3).3 39 kn The reaction results in the surface support with the specific nuerical values (eleent average) for Case 9) [kpa] The differences between the hand calculations and the FEM-Design results are less than 5%. 3

124 Download link to the exaple file: Elasto-plastic surface supports with detach in an ebedded plate.str 4

125 7.4.4 Elasto-plastic trusses in a ultispan continuous bea Inputs: Structural steel (trusses with plastic liit force) S35, fy 35 MPa Young's odulus of steel Es GPa Cross-sectional area (CHS -4.) A Cross-sectional area (CHS -.) A 3 Plastic liit force (CHS -4.) Fpl,li fya 47. kn Plastic liit force (CHS -.) Fpl,li fya 6.6 kn Concrete (bea, linear elastic aterial odel) C 5/3 Young's odulus of concrete Ec 3 GPa Cross-sectional area (rectangle) b ; h 3 Inertia Ib Lb 4 x Lc q kn/ CHS -4. Fpl,li 47. kn Total distributed load CHS -4. Fpl,li 47. kn Lc CHS -. Fpl,li 6.6 kn Truss length CHS -4. Fpl,li 47. kn Lb 8 CHS -4. Fpl,li 47. kn Bea length q The trusses with the plastic liit forces support a concrete bea The proble is a concrete bea with five trusses as supports (see Fig ). The bea is linear elastic. The trusses are linear elastic, perfectly plastic (with liit force, see the input table above). Truss nuber,, 4 and 5 have Fpl,li 47. kn plastic liit force. Truss nuber 3 has Fpl,li 6.6 kn plastic liit force. The external load is a total distributed load q kn/ (see also Fig for the geoetry). At this proble (loads and supports) the degree of static indeterinacy is three. We will solve the proble with force ethod. The optial solution with the force ethod is when the priary structure is essentially a series of siply supported beas (see Fig ). Apply the unit redundant forces Xi (i,, 3) in the lines of the reoved constraints, pairs of opposite oents at the end cross-sections of beas connected to hinges created above the interediate supports (see the priary structure Fig ). 5

126 We apply vertical springs to represent the trusses as supports. The flexibility of the trusses (in the elastic region): k Lc for truss nuber,, 4 and E s A kn k Lc for truss nuber E s A kn L L k L k k L k EcIb k q q S() +ql/ S() +ql q L 8 S3() +ql q L 8 X() ± S() +/ S4() +ql S5() +ql/ q L 8 X() ± S() - M() [kn] q L 8 X3() ± S3() +/ M() [-] + S() +/ S3() - S4() +/ M() [-] + S33() +/ S43() - S53() +/ M3() [-] The priary structure and the redundant forces with the reactions and internal forces for the force ethod solution (at stage ) The bending stiffness of the bea: E c I b kn 6

127 Fig shows the statical syste of the analyzed proble with the priary structure. By force ethod first of all we need to calculate the flexibilty atrix. Flexible coefficients (see Fig for the values which provide these coefficients according to the virtual force ethod): a a 33 a L rad +k +k +k Ec I b kn L rad + k +k Ec I b kn a 3 a 3k rad.7 5 kn a a 3a a 3 L rad k k Ec I b kn These coefficients are independent fro the external loads. The second step is the calculation of the load constants based on the total load intensity q. Load constants: 3 q L q L q L q L a a3 +k k +k. 3 rad 4 E c I b 3 q L q L q L a + k k 5. 4 rad 4 E c I b The equation syste of the force ethod: A X +a where: [ [ ][ ] ] a a a rad A a a a and kn a 3 a 3 a a 5. 4 rad. 3 7

128 The inverse atrix for the solution: [ ] kn A rad The solution of the equation syste: [ ] 6.94 X.3467 kn 6.94 The calculation of the truss forces based on the force ethod (see also Fig ): 3 S k S k + S ki X i i ' ' S S 5 S +S X +S X +S 3 X 3 q L + X + X + X 36.9 kn S ' S '4S +S X +S X +S 3 X 3 ' S 3S 3+S 3 X +S 3 X +S 33 X 3 q L X + X + X 346. kn q L + X X + X kn Since the plastic liit force for truss nuber 3 is Fpl,li 6.6 kn the elastic behaviour with the described statical syste ended at load level q: q F pl, li q S3 ' 33.8 q.787 q5.74 kn/ At this load level the noral force in truss nuber 3: S 3 q S '6.6 kn q 3 After this load level the statical syste is changing (see Fig ). For this syste the degree of static indeterinacy is. We will solve the proble also with force ethod. The optial solution with the force ethod is when the priary structure is essentially a series of siply 8

129 supported beas (see Fig ). Apply the unit redundant forces Xi (i, ) in the lines of the reoved constraints, pairs of opposite oents at the end cross-sections of beas connected to hinges created above the interediate supports (see the priary structure Fig ). L L 4 L q q S() +ql/ S4() +q(l+l)/ S() +q(l+l)/ S5() +ql/ M() [kn] q L 8 q L 8 q L 8 X() ± S() +/ X() ± S() -(/+/4) S4() +/4 M() [-] + S() +/4 S5() +/ S4() -(/+/4) M() [-] The priary structure and the redundant forces with the reactions and internal forces (at stage ) Flexible coefficients (see Fig for the values which provide these coefficients according to the virtual force ethod): L L rad a a + +k +k +k Ec I b 3 Ec I b 44 4 kn ( ) a a L rad k Ec I b 4 4 kn ( ) The calculation of the load constants based on the total load intensity q. 9

130 Load constants: q L 3 q L3 q L q L q L a a + +k k + 4 E c I b 4 E c I b ( )( ) ( ) q L q L + +k rad The equation syste is now: A X +a where: A [ ][ ] a a rad a a kn [ 5 ] a rad The inverse atrix for the solution: [ ] kn A rad The solution of the equation syste: [ ] X 5.6 kn 5.6 The calculation of the truss forces: '' q L + X + X. kn '' S S 5 S +S X +S X ( S '' S '4' S +S X +S X )( ) q L q L + + X + X 67.8 kn 4 4 Since the plastic liit force for truss nuber and 4 is Fpl,li 47. kn this elastic behaviour for this second load step with the described statical syste ended at load level q (see also Fig ): q q q 5.74 S ' q.599 S'' 67.8 F pl,li q.599 q3.98 kn/ 3

131 On this load level the noral force in truss nuber and 4: S S 4 q q S '+ S ' '47. kn q q The reaining load is: q 3q q q.6 kn/ This load is applied now on a new statical syste too (see Fig ). ''' S q Lb 8 kn Thus the final noral force in truss nuber : S q q q S '+ S ' '+ 3 S '' '9.5 kn q q q k k L k k L EcIb L k L k EcIb L 4 k k L k k L EcIb q q k q The statical systes of the structure at different load steps (stage, stage and stage 3) Figure shows the truss noral forces and the internal forces for the different load levels with the different statical syste. Fig also shows the final results of the proble (truss noral forces and internal forces). After the hand calculation we provide the FEM-Design calculation applying these plastic liit forces for the trusses. Fig shows the FEM-Design odel with the geoetry and with the ain odeling issues. 3

132 k k k k EcIb k q5.74 kn/ L L S' +3.3 kn S' kn L L S4' S3' +6.6 kn S5' +3.3 kn M' [kn] k k L S'' +.95 kn k EcIb L 4 S'' +.84 kn k L q3.98 kn/ S5'' +.95 kn S4'' +6.6kN M'' [kn] k k EcIb q3.6 kn/ L 8 S5''' kn S''' kn M''' [kn] k k k k EcIb k q kn/ L S +9.5 kn L S +47. kn L S kn L S S kn M [kn] The superposition of the different load level results and the final truss forces and internal forces for the proble 3

133 The FEM-Design odel After the nonlinear plastic calculation Fig shows the noral forces in the trusses. Fig shows the final oent diagra. The last converged equilibriu load level is at % of the total load, thus the equilibriu exists for the total load level The noral forces in the trusses according to the plastic calculation [kn] The bending oent distribution in the bea according to the plastic calculation [kn] We can say that the hand calculation and the FEM calculation results are identical! 33

134 Download link to the exaple file: Elasto-plastic trusses in a ultispan continuous bea.str 34

135 7.4.5 Elasto-plastic point-point connection between cantilevers Inputs Colun height H3 Point load F 5 kn or F 6 kn Structural steel S 35 Young's and shear oduli of steel Es GPa, G 8.77 GPa Cross-section (both vertical coluns) HEA 6 Cross-sectional area Ac 646 Relevant inertia Ic,.4 4 Relevant shear correction factor ρc,.334 Point connection with plastic copression liit Npl kn We have two vertical coluns (HEA 6, S35) with fixed botto and hinges at the top ends (see Fig ). We applied a horizontal point load (with different intensity, see Fig ) at the top of the left colun. The tops of the coluns are connected to each other with a pointpoint connection with a plastic copression liit Npl kn. The bending oent distributions in the three different cases (see Fig ) according to the sae bending stiffnesses and the plastic copression liit in the point-point connection are trivial. Point connection: Plastic liit copression force: kn Point connection: No plastic liit F 5 kn F 6 kn M [kn] 9 6 M [kn] H3 H3 H3 F 5 kn Point connection: Plastic liit copression force: kn 6 M [kn] The three different probles and the different bending oent distributions in the coluns 35

136 If there is no liit copression force in the point-point connection the distribution of the point load is equal on both top ends of the coluns: () M F / H 5/ 375 kn M () F / H 5 / 375 kn M ()(F N pl ) H (5 )39 kn () M N pl H 36 kn M (3)(F N pl ) H (6 )3 kn M (3 )N pl H 36 kn The displaceents of the top points of the coluns: 3 e () ( F /) H F / H (5/) 33 5 / Es I c ρ c, G Ac e () (F /) H F / H (5 /) 33 5/ EsIc ρ c, G Ac e () e N pl H 3 N pl H E s I c ρ c, G Ac () e (3) e (F N pl ) H 3 (F N pl ) H (5 ) 33 (5 ) Es I c ρ c, G Ac ( F N pl ) H 3 (F N pl ) H (6 ) 33 (6 ) Es I c ρ c, G Ac N pl H 3 N pl H E s I c ρ c, G Ac (3 ) Fig shows the FEM-Design odel and their input paraeters. The results of the FE calculations according to the three different cases are in Fig and (see also Fig ). The bending oent diagras and the displaceent values of the top of the coluns are identical with the hand calculation. 36

137 The FEM-design odel of the proble with the inputs The bending oent diagras of the FEM solution for the three different cases bending oents [kn], connection force [kn] 37

138 The displaceent diagras of the FEM solution for the three different cases [] Download link to the exaple file: Elasto-plastic point-point connection between cantilevers.str 38

139 7.4.6 Elasto-plastic point-point connection with uplift in a ultispan continuous bea Inputs: Bea length L 4 ; L L3 Partial distributed load q kn/ Point load F 5 kn Concrete (bea, linear elastic aterial odel) C 5/3 Young's odulus of concrete Ec 3 GPa Cross-sectional area (rectangle) b ; h 5 Inertia Ib Plastic hinge above support B (point-point connection) Mpl,Rd 5 kn C support only bears copression uplift A ultispan continuous bea, a plastic hinge with liit oent capacity (as a point-point connection) above support B and only copression resistance at support C with different types of loads are given in Fig q kn/ A B L4 F 5 kn C plastic hinge Mpl,Rd 5 kn L only copression support L Multispan continuous bea with plastic hinge above support B and uplift at support C First of all we ake a hand calculation and then ake a FEM-Design plastic calculation. At the end of this chapter these two results will be copared. According to the uplift and plastic behaviour we need to analyze and calculate the results in load steps. The first load step q, F are at a certain load level where the plastic hinge reaches its liit oent value. On this load level there will be zero reaction force in support C, because this support only can bear copression. The 5% of the total load level: q 5 kn/, F.5 kn. At this load level the specific bending oent values (see also Fig ): M B F L.5 5 kn this is the plastic liit oent capacity above support B. M ABid BCid M F L q L kn 8 8 kn 39

140 At this load level the specific reaction values (see also Fig ): L 4 q F L 5.5 A 8.75 kn L 4 L q +F ( L +L ) B 3.75 kn L C kn At this load level the specific deflection values according to virtual force ethod (see also Fig ): e ABid [ )] ( L L M B q L L 5 L +.5 Ec I b e BCid [ ] L L B M B L q L L M + L L 6.37 Ec I b [ ( ] L+ L3 (L +L3 ) B q L L +L 3 B L L +L 3 e M + + M L Ec Ib C ) After this load step the reaining loads (q, F) are acting on a different statical syste (see Fig ). The second 5% part of the total load level: q 5 kn/, F.5 kn are acting on separate siply supported beas. q 5 kn/ A F.5 kn B 6.37 C e [].5 5 M [kn] 8.75 kn kn The deflection and the bending oent diagra at the first load step (5%) which causes Mpl,Rd oent above support B, and uplift at support C At this load step the specific bending oent values (see also Fig ): 4

141 M ABid BCid M q L 5 4. kn 8 8 F L +L kn 4 4 At this load step the specific reaction values (see also Fig ): A q L 4 5. kn B q L F kn C F.5.5 kn At this load step the specific deflection values according to virtual force ethod (see also Fig ): e ABid 5 q L 4 F ( L+ L3)3 BCid 5.93 ; e E c I b 48 E c I b q 5 kn/ A F.5 kn B C e [] M [kn].5 kn. kn.5.5 kn Second load step (5%) which is acting on a different statical syste with the plastic hinge above support B and copression in support C The final results coe fro the superposition of the two forer calculated cases. The final specific bending oent values (see also Fig ): B B M M 5 kn M ABid M ABid +M ABid kn 4

142 M BCid BCid M BCid +M kn The final specific reaction values (see also Fig ): A A+ A kn B B+B kn CC +C kn The final specific deflection values (see also Fig ): e ABid eabid +e ABid At the final deflection calculation of the iddle point of BC span we need to consider the fact that above support B a plastic hinge occurred and hence in support C copression is arising in the second load step. e BCid e BCid e C ( 9.558) +e BCid q kn/ A B F 5 kn C plastic hinge Mpl,Rd kn L4 L only copression support L3 e [] M [kn] 5. kn 8.75 kn.5.5 kn The final deflection and the final bending oent diagra with the reactions 4

143 After the hand calculation we odelled this proble in FEM-Design. We adjusted a point-point connection with plastic liit oent value above support B (see Fig ) and uplift at support C (see Fig ) The FEM-Design odell for the proble Fig shows the reactions, bending oents and deflection values at load level 5%. The differences between hand and FEM calculations are less than % The FEM-Design results at load level 5% 43

144 Fig shows the reactions, bending oents and deflection values at load level %. The differences between hand and FEM calculations are less than % The FEM-Design results at load level % Download link to th exaple file: Elasto-plastic point-point connection with uplift in a ultispan continuous bea.str 44

145 7.4.7 Elasto-plastic edge connections in a building braced by shear walls Inputs: Geoetry Fig Horizontal loads (on one storey, on one wall plane) V kn Concrete (walls and slabs) C 5/3 Young's odulus of concrete Ec 3 GPa Thickness of the walls/slabs t c Plastic edge connection shear force liit (above Wall4) vpl.5 kn/ Edge connection behaviour Without detach The shear wall proble is given in Fig The given loads are only two concentrated horizontal loads on the two storeys (see Fig ). The top of Wall4 only can bear vpl.5 kn/ plastic liit force. We defined the proble and the odel in FEM-Design (7.4.7.). Edge connection: Plastic liit force in x direction:.5 kn/ kn Wall Slab Wall4 Wall3 z H3 kn Wall H3 Slab x L L4 L One planar wall syste with geoetry and loads If all of the eleents behaviour are linear elastic then the solution of this proble based on the finite eleent ethod (according to the echanical odel of the eleents) is exact. Fig shows the finite eleent surface esh of the proble. 45

146 The odel with the equivalent loads, supports and the plastic edge connections Fig shows the resultants of the reactions and the edge connection forces/oents according to the linear elastic calculation. We can see that the horizontal shear force in the edge connections (above Wall4, see Fig and ) is: V el 5.6 kn The ultiate plastic resultant shear force bearing capacity of the edge connection above Wall4 is: V ult v pl L.5 4. kn Fig shows the resultants of the reactions and the edge connection forces/oents according to the elasto-plastic calculation. The edge connection resultant shear force based on the FEM-Design calculation (Fig ): V ult FEM. kn 46

147 The resultants of the reactions and the edge connections according to linear elastic calculation (forces [kn], oents [kn]) The resultants of the reactions and the edge connections according to elasto-plastic calculation (forces [kn], oents [kn]) It is obvious that according to the elasto-plastic calculation due to the plastic deforations on the edge connections the final displaceents need to be greater than the displaceents of the linear elastic calculation. 47

148 Fig shows the displaceents of the linear elastic (above) and the elasto-plastic (below) calculations The displaceents according to linear elastic (above) and elastoplastic (below) calculation [] Download link to the exaple file: Elasto-plastic edge connections in a building braced by shear walls.str 48

149 7.4.8 Elasto-plastic edge connections with detach in a shear wall Inputs: Geoetry Fig Vertical distributed load qv 5 kn/ (N kn) Horizontal distributed load interediate levels qh 5 kn/ (V 6 kn) Horizontal distributed load top level qh 7.5 kn/ (V/ 3 kn) Concrete (wall) C 5/3 Young's odulus of concrete Ec 3 GPa Thickness of the wall t c Plastic edge connection shear force liit vpl, kn/ Plastic support shear force liit vpl, 5 kn/ Edge connection / Fixed support behaviour Detach in vertical direction Fig shows the shear wall proble with the geoetry, external loads and the behaviour of the edge connections and the support. N kn V/ 3 kn H3 Edge connection : Detach for tension in z direction Plastic liit force in x direction: kn/ N kn V 6 kn H3 Edge connection : Detach for tension in z direction Plastic liit force in x direction: kn/ N kn V 6 kn H3 Fixed Line support: Detach for tension in z direction Plastic liit force in x direction: 5 kn/ z x L4 Figure The properties of the supports and edge connections of a shear wall According to the loads and the geoetry the internal forces and the eccentricities can be calculated (see Fig ). e M 9 M M ; e.9 ; e N N 4 N

150 V/ 3 kn 9 kn 4 kn V 6 kn e 36/4.9 6 kn 5 kn V 6 kn - + V [kn] - + N [kn] - + M [kn] 8 kn 36 kn N kn e 9/.45 9 kn N kn 3 kn kn N kn e3 8/6.35 Figure The internal force distribution and the eccentricities of the noral forces The eccentric noral force at the top level (see Fig ) is inside the Culann's kernel therefore the specific noral force distribution can be calculated with the following equations and approxiations: n+z n z N M 9 t+ x ax t kn / A I / N M 9 t x t kn / A I ax / The eccentric noral forces at the interediate levels (see Fig ) are outside the Culann's kernel therefore the specific noral force distribution can be calculated with the following equations and approxiations (based on the resultant of the stress volue): ( 3 ) L e n zn ( L e ) n N 3 3 z3 3 N n z kn/ ( L e ) ( 3 N3 n z kn/ ( L e ) ( 3 ) 5 )

151 The average shear forces at the different levels: n xz,average V 3 t.7.5 kn/ A. 4 n xz,average V t Aactive kn / n xz3,average V3 t Aactive kn/ ( ( ) ) You can see these calculated values in Fig also. e 9/.45 e 36/4.9 e3 8/6.35 Average shear force: 7.5 kn/ -6.5 kn/ kn/ kn/.95 Average shear force in the active copressed zone: 7.7 kn/ Average shear force in the active copressed zone: 76.9 kn/ kn/ Figure The eccentricities and the specific noral forces and average shear forces The plastic liit shear forces differ fro each other at the edge connections and the fixed supports hence one-by-one the load-bearing capacities are (see Fig and ): η η η 3 v pl, n xz, average v pl, n xz,average v pl, n xz3, average 66.7 % % % 76.9 Thus the significant value which gives us the load-bearing capacity of the structure is the last value: 65% of the external load. 5

152 After the hand calculation we have ade a finite eleent calculation with FEM-Design. The odel can be seen with the adjusted paraeters in Fig (see also the input table for the data). Figure FEM-Design odel with fixed support and the vertical and horizontal forces The first FEM-Design results can be seen in Fig In this case the results are based on the uplift (detach) calculation (without plastic calculation). The results (Fig ) are in good agreeent with the hand calculation. Do not forget that by the hand calculation we assued prisatic bea behaviour by the specific noral force calculation but by the FEM calculation the behaviour is ore copound. 5

153 Figure Distribution of the reactions and connection forces with detach calculation specific noral forces [red and blue, kn/] and shear forces [green lines, kn/] In the second calculation we considered plastic analysis with detach (uplift) in FEM-Design. According to the plastic behaviour the last converged (equilibriu) load level was: η FEM6 % Fig shows the edge connection forces and the reactions for the last converged solution. The difference between the hand and FEM calculation is 4.6 %. FEM η HAND η 65 % 6 % 3 Δ 4.6 % HAND 65 % η3 53

154 Figure Distribution of the reactions and connection forces with plastic detach calculation specific noral forces [red and blue, kn/] and shear forces [green lines, kn/] Download link to the exaple file: Elasto-plastic edge connections with detach in a shear wall.str 54

155 7.4.9 Elasto-plastic line-line connections in a square plate Inputs: Square slab L Lx Ly 8 Slab thickness t c Applied distributed load p 5 kn/ Concrete C 5/3 Young's odulus of concrete Ec 3 GPa Isotropic positive plastic oent capacity + x+ y+ 3 kn/ Line-line connection, with plastic liit specific oent (see the adjusted local coordinate syste in Fig ) y'+pl.liit 3 kn/ In this exaple line-line connections with plastic liit force/oent will be used to calculate the plastic load-bearing capacity of a sily supported square plate with isotropic positive plastic oent capacity. If the slab has constant isotropic positive plastic oent capacity (x+ y+ 3 kn/) the yield-line layout is known in this case [] (see Fig and ). y p Ly L x 8 8 x Siply supported square slab with distributed load If the correct (real) yield-line layout is known the plastic load-bearing capacity can be calculated according to a kineatically adissible virtual displaceent field which belong to this layout. The axiu total distributed load which causes the plastic failure of this square slab coes fro the equality of the external and the internal virtual work. W i n t W ext 55

156 The external work done by loads: ax W ext pax p ax HAND δ V p HAND L x L y HAND L L 3 3 The internal work done by resisting oents: W i n t +x pl δ Θ y L y ++y pl δ Θ x L x +x pl 4 4 L y + +y pl L x 8 + Lx Ly Hence the plastic load-bearing capacity: kn.5 L 8 δθx 4/Ly p ax HAND y δe Ly 8 x+pl y+pl 3 kn/ x Lx 8 δe δθy 4/Lx Siply supported slab with the virtual displaceent syste In FEM-Design along the yield lines line-line connections will be applied with adjusted plastic liit capacity y'+pl.liit 3 kn/. Fig shows the local co-ordinate syste of the lineline connections. We applied these systes because the resisting positive oents were assued to be isotropic, hence the adjusted plastic liit capacity of the line-line connection was specified along the line (see Fig ). Fig also shows the adjusted paraeters, geoetry and the applied p 5 kn/ unifor distributed load. 56

157 FEM-Design odell with plastic line-line connections along the yield-lines With FEM-Design the last converged equilibriu load level was at 7%. It eans that the loadbearing capacity due to the plastic calculation is: p ax FEM.7 p kn/ The difference between the hand calculation and FEM analysis is 4%. ax ax p p.5.8 Δ HANDax FEM 4 %.5 p HAND Fig shows the plastic connection forces in the adjusted line-line connections (which are the expected values according to the specific liit oent capacity) The relevant plastic connection oents [kn/] 57

158 Download link to the exaple file: Elasto-plastic line-line connections in a square plate.str 58

159 7.5 Calculation with construction stages 7.5. A steel frae building with construction stages calculation Inputs: Strength of the steel S35; E GPa Coluns HEA 4 Beas IPE 6 Line load on each floors ped kn/ In this chapter we will show a short exaple about the construction stage calculation through a steel frae building with the Increental Tracking Method. Fig shows the geoetry of the frae building. In this figure we also indicated the loads and the three different stages which were considered in this analysis. In this exaple there were three different stages by the three different storeys. As a siplification by the three different stages there were only a kn/ line load on the beas one-by-one on each stages (see also Fig ) for the better coparison. By the increetal tracking ethod of the construction stage calculation FEM-Design handles the different stages as different structures. We apply the given loads on these different structures and apply a superposition by the internal forces and displaceents respectively. Thus basically the construction stage calculation is a nonlinear calculation with nonlinear boundary conditions and statical systes through the whole analysis. First of all we will show the different stage calculations one-by-one with the different boundary conditions and statical systes and superpose the results to verify the FEM-Design construction stage calculation ethod. In the end we will copare the results and show the differences between the regular and the constuction stage calculation ethod as well. Stage 3 Lc 3 qed kn/ Stage Lc 3 qed kn/ Stage Lc 3 qed kn/ Lb 6 Lb The geoetry and the considered loads on a steel frae building 59

160 7.5.. The translational displaceents [] separately according to the three different stages and the different load situations Fig shows the different stages with the different statical systes and loads. Next to the structural view you can see the translational displaceents for each stages one-by-one. With the construction stage calculation we accuulate these results therefore for the verification we need to superpose these values to get the final stage results. We calculated the displaceents at specific points of the beas by the syetry axis of the structure (see Fig ): e Be STB+e STB+e ST3B e B e STB+e STB+e ST3B e B3e STB3+e STB3+e ST3B

161 The bending oent diagras [kn] separately according to the three different stages and the different load situations Fig shows the different stages with the different statical systes and loads. Next to the structural view you can see the bending oent diagras for each stages one-by-one. With the construction stage calculation we accuulate these results therefore for the verification we need to superpose these values to get the final stage results. We calculated the bending oents at specific points of the beas by the syetry axis of the structure (see Fig ): M BM STB+M STB+M ST3B kn M B M STB+M STB+M ST3B kn M B3M STB3+M STB3+M ST3B3++(.8).8 kn 6

162 Fig shows the final accuulated displaceent field at the end of the stage calculation in FEM-Design. Here you can see the results at specific points of the beas by the syetry axis of the structure: e BFEM CS.8 e BFEM CS 4.9 e B3FEM CS The final accuulated translational displaceents [] according to the construction stage increental tracking ethod Fig shows the final accuulated bending oents at the end of the stage calculation in FEM-Design. Here you can see the results at specific points of the beas by the syetry axis of the structure: M BFEM CS 9.8 kn M BFEM CS kn M B3FEM CS.8 kn The final accuulated bending oent diagra [kn] according to the construction stage increental tracking ethod 6

163 The verification results are identical with the FEM-Design calculation. At the end of this chapter let's copare the final construction stage results with the results of a calculation without construction stage calculation ethod (regular calculation on the frae with one statical syste and load distribution). Fig show the displaceents and the bending oents after a regular calculation. The results based on a regular calculation: e BFEM withoutcs 4.9 ; M BFEM withoutcs 8.78 kn e BFEM withoutcs 4.93 ; M BFEM withoutcs 83.9 kn e B3FEM withoutcs 4.94 ; M B3FEM withoutcs 65.5 kn In this verification exaple the axiu displaceent fro the construction stage calculation is.7 ties greater at the first floor (copare Fig with Fig ). The bending oent value is.35 ties greater at the first floor by this verification exaple (copare Fig with Fig ) The translational displaceents [] according to the regular calculation ethod The bending oent diagra [kn] according to the regular calculation ethod 63

164 By a real frae structure the difference in the final results are not that uch if we consider and add the live loads to the final stage results. Here in this exaple to show and ephasize the calculation ethod only a self-weight-like load was considered. Download link to the exaple file: A steel frae building with construction stages calculation.str 64

165 8. Cross section editor 8. Calculation of a copound cross section An exaple for copound cross section is taken fro [7] where the authors calculated the cross sectional properties with the assuption of thin-walled siplifications. The welded cross section is consisting of U3 and L6x8x (DIN) profiles. In the Section Editor the exact cold rolled geoetry was analyzed as it is seen in Figure 8... Figure 8.. The analyzed cross section The following table contains the results of the two independent calculations with several cross sectional properties. Notation Ref. [] Section Editor A [c] yg [c]..44 zg [c] y's [c] z's [c].6.36 Iy [c4] Iz [c4] It [c ] Iw [c6] Iyz [c ] 4 Table 8.. The results of the exaple 65

166 Download link to the exaple file: Calculation of a copound cross section.sec 66

167 9. Design calculations This chapter in unfinished. According to Eurocode standard! 9. Foundation design This chapter in unfinished. 9.. Design of an isolated foundation This chapter in unfinished. 9.. Design of a wall foundation This chapter in unfinished Design of a foundation slab This chapter in unfinished. 67

168 9. Reinforced concrete design In this chapter we will show soe detailed verification calculations regarding to reinforced concrete design according to EN Moent capacity calculation for beas under pure bending In this chapter we calculate the oent capacity of reinforced concrete cross sections under pure bending (uniaxial bending). The first exaple will be an under-reinforced cross section. The second one will be a noralreinforced and the third one an over-reinforced section. After independent hand calculations we copare the results with FEM-Design values. The following input paraeters are coon for the different cross sections regarding to this subchapter. Inputs: Concrete characteristic copressive strength fck 3 N/ or N/ The end of the parabolic part (aterial odel, see EC-) c. % Ultiate liit strain of concrete (see EC-) cu.35 % Partial factor of concrete γc.5 Reinforcing steel characteristic yield strength fyk 4 N/ Elastic odulus of reinforcing steel Es GPa Ultiate liit strain of reinforcing steel uk.5 % Partial factor of reinforcing steel γs.5 Behaviour of plastic part (see Fig ) k.5 Concrete cover (on stirrups) c Stirrup diaeter ϕs 8 The external diensions are the sae for every cross section (b 3, h 5 ) Under-reinforced cross section In this exaple we put two longitudinal rebars with 6 diaeter at the botto left and right corner of the strirrups (see Fig. 9..., the concrete is C3/37). We neglect the effect of the hangers. Two different hand calculation ethods provided here. First of all with ai of the ost siple aterial odels for concrete and reinforcing steel and secondly with iproved aterial odels, which FEM-Design uses also. First we calculate the oent capacity with the following aterial odels (see Fig. 9...). 68

169 h 5 ϕ6 b 3 Figure 9... The cross section of the under-reinforced case σc σs fcd fyd.7 εsyfyd/es εcu3.5 εc εud.9εuk εs Figure 9... The aterial odels for the first calculation left: concrete (only copression), right: rebars (both tension and copression) Due to the under-reinforced section behaviour we assue that the rebars strains are at the design ultiate liit strain value. Based on the su of the forces we can get the height of the active copression concrete zone (see Fig ). b x c f cd As f yd The copressed zone: 6 π 4 As f yd 4.5 xc 4.48 b f cd

170 εc ε fcd σ xc d x εud fyd Figure The assued strains and stresses We need to check that the assuption of the rebar strains were proper or not. Based on the assuption that the concrete reaches its ultiate copression strain liit then the strain in the rebars (see Fig ): ε s ( d.5 x c ) ε cu ((5 8 8) ) %>ε ud.5 %.5 x c Thus the axiu concrete strain cannot be εcu.35%, due to this the rebars reach the ultiate strain, thus the assuption was correct, the failure ode is the rupture of the rebars (under-reinforced section). But it has no effect on the forer equilibriu equation. The oent capacity of the section with the siple aterial odels: ( ) M Rd,b x c f cd d xc kn.5 ( ) Secondly we calculate with iproved aterial odels, see the following equations. Concrete (see Fig left side): σ c (ε c ) f cd σ c (ε c ) f cd [ ε ε cc ( )] if ε c ε c if ε c <ε c ε cu 7

171 Rebars (see Fig right side): σ s ( ε s )ε s E s if ε s f yd Es σ s ( ε s ) f yd +(k ) f yd f ε ud yd Es ε s if f yd Es f yd <ε s ε ud Es σc σs fcd kfyd fyd εc εcu εc εsy fyd/es εud εs Figure The aterial odels for the second calculation left: concrete (only copression), right: rebar (both tension and copression) The assued stress and strain distributions (Fig ). εc x ε σ σc d464 xn εud kfyd Figure The assued strains and the stresses The concrete strains depends on the curvature and based on the iproved aterial odels this led to a nonlinear equations. ε (x )κ ( x x n ) The su of the forces: N c +N s Resultant force in concrete: h N c b σ c ( ε c )dx 7

172 Resultant force in rebars: N s A si σ si ( ε si ) i We solved the equation syste with independent nuerical ethod as a hand calculation. The position of the neutral axis: x n38.8 The stress and strain values which belong to the equilibriu state are shown in Fig εc -.7 ε σc - MPa σ xn d464 x εud.5 kfyd MPa Figure The characteristic values of strains and stresses The resultant of the concrete stress volue according to the independent hand calculation: h N c b σ c ( ε c )dx 53.4 kn The resultant force in the rebars: N s A si σ si ( ε si ) i 6 π kn 4.5 The difference between the copression and tension forces is less than % therefore this is the correct position of the neutral axis and curvature according to the iproved aterial odels. The centroid of the concrete stress volue easured fro the top of the section: h b x σ c (ε c )dx xc Nc

173 The oent around neutral axis provided by the concrete: M c N c ( x n x c )53.4 ( )3.657 kn Rebars oent around neutral axis: M s A si σ si ( ε si )( d i x n) ( )65.64 kn i The oent capacity with iproved aterial odels: M Rd,M c +M s kn Ratio between the two hand calculations with the different aterial odels: M Rd, M Rd, The oent capacity of the sae section with FEM-Design (Fig ): M Rd, FEM kn The difference between the two hand calculations is 4%. The FEM-Design results of the stresses and strains are shown in Fig The strain values and neutral axis position value between the hand and FE calculations are under 3%, the oent capacity difference is less than %. Figure The FEM-Design results (strains [ ], neutral axis []) 73

174 Download link to the exaple file: Moent capacity calculation for beas under pure bending.str 74

175 9... Noral-reinforced cross section Here we put 6 longitudinal rebars with 6 diaeter at the botto of the cross section (see Fig , the concrete is C3/37). We neglect the effect of the hangers. The following verification calculations will be perfored with the iproved aterial odels (see Chapter 9...). The assued stress and strain distributions in the section are shown in Fig εcu x ε fcd σ d xn 6 ϕ6 εs> εsy σs > fyd Figure Cross section and assued strains and stresses The concrete strain depends on the curvature and based on the iproved aterial odels this led to a nonlinear equation. ε (x )κ ( x x n ) The su of the forces: N c +N s Resultant force in concrete: h N c b σ c ( ε c )dx Resultant force in rebars: 6 N s A si σ si ( ε si ) i We solved the equation syste with independent nuerical ethod as a hand calculation. The neutral axis position is: x n93.6 The stress and strain values which belong to the equilibriu state are shown in Fig

176 x ε fcd - MPa σ xn d 464 εcu 3.5 εud σs MPa Figure The calculated strains and the stresses The resultant of the concrete stress volue according to the independent hand calculation: h N c b σ c ( ε c )dx 45.3 kn The resultant force in the rebars: 6 N s A si σ si ( ε si )6 i 6 π kn 4 The difference between the copression and tension forces is less than % therefore this is the correct position of the neutral axis and curvature according to the iproved aterial odels. The centroid of the concrete stress volue easured fro the top of the section: h b x σ c (ε c )dx xc Nc 38.7 The oent around neutral axis provided by the concrete: M c N c ( x n x c )45.3( )4.57 kn Rebars oent around neutral axis: 6 M s A si σ si ( ε si )( d i x n) ( )66.8 kn i The oent capacity with iproved aterial odels: M Rd,M c +M s kn 76

177 The oent capacity with FEM-Design (Fig. 9...): M Rd, FEM 87.4 kn The FEM-Design results of the stresses and strains are shown in Fig The strain values and neutral axis position value difference between the hand and FE calculation are less than 4%, the oent capacity difference is %. Figure 9... The FEM-Design results (strains [ ], neutral axis []) Download link to the exaple file: Moent capacity calculation for beas under pure bending.str 77

178 9...3 Over-reinforced cross section We put longitudinal rebars with diaeter at the botto (see Fig. 9..., the concrete is C/5). We neglect the effect of the hangers. The following verification calculations will be perfored with the iproved aterial odels (see Chapter 9...). The assued stress and strain distributions in the section are shown in Fig εcu x ε fcd σ d d xn 6ϕ ϕ6 εs< εsy σs < fyd Figure 9... Cross section and assued strains and the stresses The concrete strain depends on the curvature and based on the iproved aterial odels this led to a nonlinear equation. ε (x )κ ( x x n ) The su of the forces: N c +N s Resultant force in concrete: h N c b σ c ( ε c )dx Resultant force in rebars: N s A si σ si ( ε si ) i We solved the equation syste with independent nuerical ethod as a hand calculation. The neutral axis position is: x n37.95 The stress and strain values which belong to the equilibriu state are shown in Fig

179 x εcu 3.5 ε fcd -3.3 MPa xn d 4 d 46 σ εs.45 εs.585 σs 9.8 MPa σs 37.3 MPa Figure 9... The calculated strains and the stresses The resultant of the concrete stress volue according to the independent hand calculation: h N c b σ c ( ε c )dx 9.6 kn The resultant force in the rebars: N s A si σ si ( ε si )6 i π π kn 4 4 The difference between the copression and tension forces less than % therefore this is the correct position of the neutral axis and curvature according to the iproved aterial odels. Centroid of the concrete stress volue easured fro the top of the section: h b x σ c (ε c )dx xc Nc 3.3 The oent around neutral axis provided by the concrete: M c N c ( x n x c )9.6 ( )9. kn Rebars oent around neutral axis: M s A si σ si ( ε si )( d i x n)6 34. [ 37.( )+9.( ) ]3. kn i The oent capacity with iproved aterial odels: M Rd,M c +M s kn 79

180 The oent capacity with FEM-Design (Fig ): M Rd, FEM 3. kn The FEM-Design results of the stresses and strains are shown in Fig The strain values and neutral axis position value difference between the hand and FE calculation are less than %, the oent capacity difference is less than 4%. Figure The FEM-Design results (strains [ ], neutral axis []) Download link to the exaple file: Moent capacity calculation for beas under pure bending.str 8

181 9.. Required reinforceent calculation for a slab In this exaple we calculate the required reinforceents of a slab due to elliptic and hyperbolic bending conditions. First of all the applied reinforceent is orthogonal and then the applied reinforceent is non-orthogonal. We calculate the required reinforceent with hand calculation and then copare the results with FEM-Design values. Inputs: The thickness h The elastic odulus of concrete Ec 33 GPa, C3/37 The Poisson's ratio of concrete ν. The design value of copressive strength fcd MPa Elastic odulus of steel bars Es GPa The design value of yield stress of steel bars fyd MPa Diaeter of the longitudinal reinforceent ϕl Noinal concrete cover cx ; cy 3 Effective heights dx 75 ; dy Elliptic bending In the first case the bending condition is an elliptic bending. In FEM-Design the odel is a slab with statically deterinant support syste and specific oent loads at its edges for the pure internal force state (see Fig. 9...). Figure 9... The slab with the edge loads for pure stress state 8

182 Fig shows the constant internal forces in the slab due to the loads. Fig shows the principal oents and their directions based on the FEM-Design calculation. According to the pure stress state the principal oents and the directions are the sae in each eleents. Figure 9... The x, y, and xy internal forces in the slab [kn] Figure The and principal oents and their directions in the slab [kn] First of all the reinforceent is orthogonal and the hand calculation and the coparison are the following:. Orthogonal reinforceent (φ9o) The reinforceent is orthogonal and their directions concide with the local syste (xξ, yϑη). The oents in the slab (tensor of the applied oents): x ξ +6 kn / y ϑ η +8 kn / xy ξ ϑ ξ η +6 kn / 8

183 The first invariant of the tensor: x + y +4 kn / The calculation of the principal oents and their directions: + y x + + y x α arctan ( ( ) ) x y 6+8 +xy + x y xy ( ( ) kn / ) kn / x 9. 6 arctan 8.5o xy 6 Copare these results with Fig The difference is %. The design oents (according to [9][]) if the reinforceent (ξ,η) is orthogonal and their directions concide with the local co-ordinate syste (x,y): Case a) ud ξ ξ ϑ ud η ϑ cos φ cos φ cos 9o cos 9 o +ξ ϑ kn / +cos φ sin φ +cos 9 o sin 9 o + ξ ϑ kn / o +cos φ sin φ +cos 9 sin 9o This is a valid solution! Because ud ξ +ud η +36 kn /> x + y +4 kn / ud ξ + kn / ud η +4 kn / Case b) ud ξ ξ +ϑ ud η ϑ cos φ + cos φ cos 9o + cos 9 o ξ ϑ kn / cos φ sin φ cos 9 o sin 9 o ξ ϑ kn / o cos φ sin φ cos 9 sin 9o Invalid solution! Because ud ξ +ud η + kn /< x + y +4 kn / Case ξ) ξ ϑ 6 ud ξ ξ kn/ ϑ 8 ud η 83

184 Invalid solution! Because ud ξ +ud η +.5 kn /<x + y +4 kn / Case η) ud ξ ξ ϑ ξ ϑ kn ud η o o o ξ sin φ + ϑ cos φ ξ ϑ sin φ 6 sin 9 +8 cos 9 6 sin ( 9 ) Invalid solution! Because ud ξ +ud η kn /< x + y +4 kn / The results of the design oents based on FEM-Design are in Fig and The difference between the hand and FE calculation is %. Figure The ud ξ design oent for elliptic bending with orthogonal reinforceent [kn] 84

185 Figure The ud η design oent for elliptic bending with orthogonal reinforceent [kn] Calculation of the required reinforceent based on the valid design oents: In x (ξ) direction: Su of the oents: ( ud ξ f cd x c d x xc ) ( ; x c 75 xc ) ; x c 6.43 Su of the forces: x c f cd a sξ f yd ; 6.43 a sξ ; a s ξ.945 /94.5 / Fig shows the required reinforceent in the relevant direction based on FEM-Design calculation. The difference is less than %. Figure The asξ required reinforceent at the botto for elliptic bending with orthogonal reinforceent [/] 85

186 In y (η) direction: Su of the oents: ( ud η f cd x c d y xc ) ( ; 4 x c 65 xc ) ; x c 4.98 Su of the forces: x c f cd a sη f yd ; 4.98 a sη ; a s η.977 /97.7 / Fig shows the required reinforceent in the relevant direction based on FEM-Design calculation. The difference is less than %. Figure The asη required reinforceent at the botto for elliptic bending with orthogonal reinforceent [/] Secondly the reinforceent is non-orthogonal and the hand calculation and the coparison are the following:. Non-orthogonal reinforceent (φ75o between ξ and η) The reinforceent is non-orthogonal and the ξ direction concides with the local x direction. Thus yϑ. The angle between the ξ directional reinforceent and η directional reinforceent is φ75o. The oents in the slab (tensor of the applied oents): x ξ +6 kn / y ϑ +8 kn / xy ξ ϑ +6 kn / The first invariant of the tensor: x + y +4 kn / 86

187 The design oents (according to [9][]) if the reinforceent (ξ,η) is non-orthogonal: Case a) o o cos φ cos φ cos 75 cos 75 ud ξ ξ ϑ +ξ ϑ kn / o o +cos φ sin φ +cos 75 sin 75 ud η ϑ + ξ ϑ kn / o +cos φ sin φ +cos 75 sin 75o This is a valid solution! Because ud ξ +ud η +9.9 kn / >x + y +4 kn / ud ξ kn / ud η +.57 kn / Case b) cos φ + cos φ cos 75o + cos 75o ud ξ ξ +ϑ ξ ϑ kn / cos φ sin φ cos 75o sin 75o ud η ϑ ξ ϑ kn / o cos φ sin φ cos 75 sin 75o Invalid solution! Because ud ξ +ud η kn /<x + y +4 kn / Case ξ) ξ ϑ 6 ud ξ ξ kn/ ϑ 8 ud η Invalid solution! Because ud ξ +ud η +.5 kn /<x + y +4 kn / Case η) ud ξ ξ ϑ ξ ϑ kn ud η o o o ξ sin φ + ϑ cos φ ξ ϑ sin φ 6 sin cos 75 6 sin( 75 ) Invalid solution! Because ud ξ +ud η kn / < x + y +4 kn / The results of the design oents based on FEM-Design are in Fig and The difference between the hand and FE calculation is %. 87

188 Figure The ud ξ design oent for elliptic bending with non-orthogonal reinforceent [kn] Figure The ud η design oent for elliptic bending with non-orthogonal reinforceent [kn] Calculation of the required reinforceent based on the valid design oents: In x (ξ) direction: Su of the oents: ( ud ξ f cd x c d x xc ) ( ; 735 x c 75 xc ) ; x c 5.9 Su of the forces: x c f cd a sξ f yd ; 5.9 a sξ ; a s ξ.33 /3.3 / Fig shows the required reinforceent in the relevant direction based on FEM-Design calculation. The difference is less than %. 88

189 Figure 9... The asξ required reinforceent at the botto for elliptic bending with non-orthogonal reinforceent [/] In η direction: Su of the oents: ( ud η f cd x c d y xc ) ( ; 57 x c 65 xc ) ; x c Su of the forces: x c f cd a sη f yd ; a sη ; a s η.773 /77.3 / Fig shows the required reinforceent in the relevant direction based on FEM-Design calculation. The difference is less than %. Figure 9... The asη required reinforceent at the botto for elliptic bending with non-orthogonal reinforceent [/] 89

190 Download links to the exaple files: Elliptic bending, orthogonal reinforceent: Required reinforceent calculation in a slab with elliptic bending and orthogonal reinforceent.str Elliptic bending, non-orthogonal reinforceent: Required reinforceent calculation in a slab with elliptic bending and skew reinforceent.str 9

191 9... Hyperbolic bending In the second case the bending condition is a hyperbolic bending. In FEM-Design the odel is a slab with statically deterinant support syste and specific oent loads at its edges for the pure internal force state (see Fig. 9...). Figure 9... The slab with the edge loads for pure stress state Fig shows the constant internal forces in the slab due to the loads. Fig shows the principal oents and their directions based on the FEM-Design calculation. According to the pure stress state the principal oents and the directions are the sae in each eleents. Figure The x, y, and xy internal forces in the slab [kn] 9

192 Figure The and principal oents and their directions in the slab [kn] Firstly the reinforceent is orthogonal and the hand calculation and the coparison are the following:. Orthogonal reinforceent The reinforceent is orthogonal and their directions concide with the local syste (xξ, yϑη). The oents in the slab (tensor of the applied oents): x ξ +6 kn / y ϑ η 8 kn / xy ξ ϑ ξ η +6 kn / The first invariant of the tensor: x + y +8 kn / The calculation of the principal oents and their directions: x + y + + y x α arctan ( ( ) ) x y 6+( 8) +xy + x y 6+( 8) + xy ( ( ) ) 6 ( 8) kn / x arctan 3.3o xy 6 Copare these results with Fig The difference is %. 9 6 ( 8) kn /

193 The design oents (according to [9][]) if the reinforceent (ξ,η) is orthogonal: Case a) o o cos φ cos φ cos 9 cos 9 ud ξ ξ ϑ +ξ ϑ kn / o o +cos φ sin φ +cos 9 sin 9 ud η ϑ + ξ ϑ 8 +6 kn / o +cos φ sin φ +cos 9 sin 9 o Invalid solution! Because their have different signs. Case b) ud ξ ξ +ϑ ud η ϑ cos φ + cos φ cos 9o + cos 9 o ξ ϑ kn / cos φ sin φ cos 9 o sin 9 o ξ ϑ kn / o o cos φ sin φ cos 9 sin 9 Invalid solution! Because their have different signs. Case ξ) ud ξ ξ ξ ϑ kn / ϑ 8 ud η This is a valid solution at the botto! ud ξ +.5 kn / ud η kn / Case η) ud ξ ξ ϑ ξ ϑ 6 ( 8) 6 kn ud η.5 o o o ξ sin φ + ϑ cos φ ξ ϑ sin φ 6 sin 9 +( 8) cos 9 6 sin( 9 ) This is a valid solution at the top! ud ξ kn / ud η.5 kn / 93

194 The results of the design oents based on FEM-Design are in Fig and The difference between the hand and FE calculation is %. Figure The ud ξ design oent for hyperbolic bending with orthogonal reinforceent [kn] Figure The ud η design oent for hyperbolic bending with orthogonal reinforceent [kn] Calculation of the required reinforceent based on the valid design oents: In x (ξ) direction at the botto: Su of the oents: ( ud ξ f cd x c d x xc ) ( ; 5 x c 75 xc 94 ) ; x c 5.959

195 Su of the forces: x c f cd a sξ f yd ; a sξ ; a s ξ.74 /74. / Fig shows the required reinforceent in the relevant direction based on FEM-Design calculation. The difference is less than %. Figure The asξ required reinforceent at the botto for hyperbolic bending with orthogonal reinforceent [/] In y (η) direction at the top: Su of the oents: ( ud η f cd x c d y xc ) ( ; 5 x c 65 xc ) ; x c 3.36 Su of the forces: x c f cd a sη f yd ; 3.36 a sη ; a s η.443 / 44.3 / Fig shows the required reinforceent in the relevant direction based on FEM-Design calculation. The difference is less than %. 95

196 Figure The asη required reinforceent at the top for hyperbolic bending with orthogonal reinforceent [/] Secondly the reinforceent is non-orthogonal and the hand calculation and the coparison are the following:. Non-orthogonal reinforceent (φ75o between ξ and η) The reinforceent is non-orthogonal and the ξ direction concides with the local x direction. Thus yϑ. The oents in the slab (tensor of the applied oents): x ξ +6 kn / y ϑ 8 kn / xy ξ ϑ +6 kn / The first invariant of the tensor: x + y +8 kn / The design oents (according to [9][]) if the reinforceent (ξ,η) is non-orthogonal: Case a) cos φ cos φ cos 75o cos75o ud ξ ξ ϑ + ξ ϑ kn / +cos φ sin φ +cos 75o sin 75o ud η ϑ +ξ ϑ kn / o o +cos φ sin φ +cos 75 sin 75 Invalid solution! Because their have different signs. 96

197 Case b) cos φ + cos φ cos 75o + cos 75o ud ξ ξ +ϑ ξ ϑ kn / cos φ sin φ cos 75o sin 75o ud η ϑ ξ ϑ kn / o cos φ sin φ cos 75 sin 75o Invalid solution! Because their have different signs. Case ξ) ud ξ ξ ξ ϑ kn / ϑ 8 ud η This is a valid solution at the botto! ud ξ +.5 kn / ud η kn / Case η) ud ξ ud η ξ ϑ ξ ϑ ξ sin φ + ϑ cos φ ξ ϑ sin φ 6 ( 8) 6 kn 4.4 o o o 6 sin 75 +( 8) cos 75 6 sin( 75 ) This is a valid solution at the top! ud ξ kn / ud η 4.4 kn / The results of the design oents based on FEM-Design are in Fig and 9... The difference between the hand and FE calculation is %. 97

198 Figure The ud ξ design oent for hyperbolic bending with non-orthogonal reinforceent [kn] Figure 9... The ud η design oent for hyperbolic bending with non-orthogonal reinforceent [kn] Calculation of the required reinforceent based on the valid design oents: In x (ξ) direction at the botto: Su of the oents: ( ud ξ f cd x c d x xc ) ( ; 5 x c 75 xc ) ; x c Su of the forces: x c f cd a sξ f yd ; a sξ ; a s ξ.74 /74. / Fig shows the required reinforceent in the relevant direction based on FEM-Design calculation. The difference is less than %. 98

199 Figure 9... The asξ required reinforceent at the botto for hyperbolic bending with non-orthogonal reinforceent [/] In η direction at the top: Su of the oents: ( ud η f cd x c d y xc ) ( ; 44 x c 65 xc ) ; x c 4.43 Su of the forces: x c f cd a sη f yd ; 4.43 a sη ; a s η.34 /3.4 / Fig shows the required reinforceent in the relevant direction based on FEM-Design calculation. The difference is less than %. Figure 9... The asη required reinforceent at the top for hyperbolic bending with non-orthogonal reinforceent [/] 99

200 Download links to the exaple files Hyperbolic bending, orthogonal reinforceent: Required reinforceent calculation in a slab with hyperbolic bending and orthogonal reinforceent.str Hyperbolic bending, non-orthogonal reinforceent: Required reinforceent calculation in a slab with hyperbolic bending and skew reinforceent.str

201 9..3 Shear capacity calculation In this chapter we will show detailed calculations of beas and slabs regarding to shear force Shear capacity of a bea In this exaple we check the shear capacity of a cantilever bea (see Fig ). The input paraeters and details are in the following table. Inputs: Concrete characteristic copressive strength fck 5 N/ Bea height h 35 Bea width bw 5 Partial factor of concrete γc.5 Reinforcing steel characteristic yield strength fyk 5 N/ Partial factor of reinforcing steel γs.5 Stirrup distance s Longitudinal rebar diaeter ϕ Stirrup diaeter ϕs Concrete cover c FEd VEd kn h 4ϕ ϕ/ bw Figure The cantilever, the cross section and the design value of the shear force First of all we need to check that we need any designed shear reinforceent or not: V Rd, c { } { } ( )3 k ρ f ( ) ax γ c l ck b w d ax kn ν in.44

202 where: ϕ d h c ϕ s 35 3 { } { } + + k in d in 3.83 ;.. { } 4 π Asl 4 ρ l in b w d in.6 ; { } 3 3 ν in.35 k f ck It is necessary to use designed shear reinforceent because: V Rd, c 57.3 kn < V Ed kn Before the calculation of the designed shear reinforceent we need to check that the diensions of the cross section is enough to bear the designed value of the shear force or not. Thus the upper liit of the shear force bearing capacity: V Rd, ax α cw bw z ν f cd cot(θ )+cot( α ) kn +cot (θ ) +.3 where: α cw. the noral force is zero in the cantilever; z.9 d ( ν.6 ) ( f ck ) α 9 cot(θ ).3 this value is adjustable in the progra. The copressed concrete strut can bear the acting shear force thus we can calculate the designed shear reinforceent. V Ed kn < V Rd, ax 33.4 kn

203 The next step is to calculate the shear capacity of the bea according to the defined shear reinforceent (see Fig ). z V Rd, s Asw f s yd (cot (α )+cot (θ )) sin α 79 π 435(+.3) 3.85 kn 4 V Ed kn 97 % V Rd, s 3.8 kn The nuerical results based on FEM-Design are shown in Fig The difference between the two calculations is %. Figure The FEM-Design results Download link to the exaple file: Shear capacity of a bea.str 3

204 9..3. Shear capacity of a slab In this chapter we will calculate the shear capacity of a slab (see Fig ). In FEM-Design based on the internal forces we calculate the required shear capacity and based on the paraeters of the slab and the applied longitudinal reinforceent in it the actual shear capacity is also coputable. Inputs: Concrete characteristic copressive strength fck N/ Partial factor of concrete γc.5 Reinforcing steel characteristic yield strength fyk 5 N/ Partial factor of reinforcing steel γs.5 Longitudinal bar diaeter in x direction ϕlx 4 Distance in x direction between the bars sx 5 Longitudinal bar diaeter in y direction ϕly Distance in y direction between the bars sy 5 Concrete cover (on longitudinal bars in x direction) c Slab thickness t The effective depth of the x reinforceents dx 73 The effective depth of the y reinforceents dy 6 Noral force in x direction nx kn/ Noral force in y direction ny kn/ Vertical surface total load on the slab qz,ed kn/ Fig shows the slab. Two opposite edges of the slab are siply supported and the other two are free. The longitudinal reinforceents are also indicated in the figure. n x,ed Z / kn dx dy c qz,ed kn/ ϕ4/5 ϕ/5 t c Ly ny, Ed kn / X Lx 6 Y X Figure The slab, the reinforceents and the design values of external forces 4

205 First of all we need to calculate the required shear capacity of the slab which depends on the internal shear forces in the two perpendicular directions of the slab local coordinate syste which were also indicated in Fig The internal forces of the slab at a corner point (based on finite eleent analysis with.5 average eleent size (see Fig ): Shear forces: v xz 69.5 kn kn ; v yz 7.3 Noral forces: n x kn kn kn ; n y ; n xy The axiu shear force in the slab according to the two shear forces in the two directions: v ax v xz +v yz ( 69.5) +(7.3) 7.38 kn The direction of the axiu shear force (right-handed coordinate syste): α arctan ( ) v yz 7.3 arctan +4.3 v xz 69.5 ( ) The axiu shear force value could be different in every nodes and therefore the angle of it also. It eans that the shear force capacity ust be coputed in every nodes in different directions (see the relevant forulas and equations below). Here is the calculation ethod for the entioned corner point: The shear capacity with the applied paraeters in the calculated ain direction: { }.8 3 v Rd, c ax γ c k ( ρ α f ck ) +k σ cp d eff ν in +k σ cp { α α }.8 ( )3 kn ax , where: The longitudinal reinforceent in the ain directions: ϕ x π 4 π a sx 6 4 sx 4 5 ϕ π π a sy y sy 4 5 5

206 The effective reinforceent area in the forer calculated ain direction: a α a sx cos (α o)+a sy cos ( α 9 o)6 cos (4.3 )+53.6 cos (4.3 9 ) Effective height: d eff d x +d y Reinforceent ratio in the ain direction: { } aα ρ α in d eff The effective noral force in the direction of the axiu shear force: nα n x cos α +n y sin α + n xy cos α sin α ( ) cos 4.3o +( )sin 4.3o + cos 4.3o sin 4.3o 9.4 kn (copression) Noral stress in the axiu shear force direction (the transfored noral force is copression): σ α nα 9.4 N.975 t σ cp in α N.975 {. f }in { } σα cd Modifying factors: k in { } { } + d eff in ν in.35 k f ck

207 Figure The slab, loads and the finite eleent esh for the proble Figure The shear forces at the corner point vxz69.5 [kn/] (left side); vyz7.3 [kn/] (right side); Figure The required shear force at the corner point: 7.38 [kn/] (left side) The applied shear force at the corner point: 94.3 [kn/] (right side) The results of the hand calculation are equal to the FEM-Design results (see Fig ). Download link to the exaple file: Shear capacity of a slab.str 7

208 9..4 Crack width calculation of a bea In this exaple we calculate the crack width of a siple supported noral-reinforced concrete bea under the given external load. Inputs: Concrete characteristic copressive strength fck N/ Concrete effective tension strength fct,eff. N/ Concrete's Young odulus Ec 3 GPa Bea height h 4 Bea width b Partial factor of concrete γc.5 Reinforcing steel characteristic yield strength fyk 5 N/ Young's odulus of rebars Es GPa Partial factor of reinforcing steel γs.5 Longitudinal rebar diaeter ϕ Stirrup diaeter ϕs Longitudinal bars effective height d 36 Cover (on stirrups) c Partial factor of rebars γs.5 M 4ϕ h 4 L 4. d 36 F kn b Mqp kn Figure The statical syste, cross section and external loads First we calculate the crack width with linear-elastic non-tension concrete aterial odel. The crack width is calculated with the ai of the distance of the cracks and the difference between concrete and rebar strains. w k sr, ax Δ ε

209 In order to calculate strains we need to calculate the cracked reinforced cross-sectional data (Stadiu II.: concrete and steel are linear-elastic and the cross-section is cracked). The position of the neutral axis according to the cracked section (Stadiu II.): α e E s x ; bx II II α e A s (d x II ) thus: x II E c 3 The oent of inertia: 3 bx π I II II + As α e ( d x II ) ( ) I II The concrete and rebar strain difference: }{ {. f ( ) σ s k t ρ ct, eff (+α e ρ p, eff ).76 p, eff Δ ε ε s ε c ax ax Es σs Es Δ ε.48 Rebar stress: σ s M qp (d x II ) N 6 ( ) α e ; σ s I II kt depends the durability of the load, the given load is long-ter loading: k t.4 Effective tensile rebar ratio: 4ϕ π As 4 ρ p, eff Ac, eff b hc, eff 4ϕ π 4 4 π 4.76 { } { }.5(h d ) h x II b in 3 h.5 (4 36) in }

210 First we need to check if the longitudinal bars are close to each other or not: ( t liit 5 (c+ϕ s )+ ) ( ) ϕ 5 (+)+ ; t actual 4 tliit > tactual therefore the rebars are close to each other, so the distance between cracks calculated as follows: ϕ s r, ax 3.4(c+ϕ s )+.45 k k ρ 3.4 (+) eff.76 k depends the cohesion between the rebars and concrete, the reinforceents are ribbed: k.8 k depends the strains in the cross-section, in this case we have pure bending: k.5 Nuerical results: w k, FEM. The difference between the nuerical and hand calculation is less than 3%. Figure The FEM-Design detailed results for crack width (strains [ ], neutral axis [])

211 Now we calculate the crack width with non-linear non-tension concrete aterial odel: The following figure shows (Fig ) the concrete aterial odel (dashed curve) but FEMDesign odifies it a bit due to nuerical stability (continuous curve). σc fck.95 fck εc εcu εc Figure The nonlinear concrete aterial odel First we need to calculate the neutral axis position and the bea curvature where the axiu oent occurs (in this case the iddle cross-section). The x is easured fro the top of the cross-section and its positive direction eant to the botto of the cross-section. The equilibriu equation respect to the forces: N c N s The force in the rebars according to that the rebars are in the linear elastic region: N sa s E s ε sa s E s κ (d x n) The force in the concrete according to the nonlinear copression aterial odel: h h N c b σ c ( ε )dx b f ck ( ( h κ (x x ) ε ( ε c ) dxb f ck εnc ( ) ) ) dx With nuerical calculation the neutral axis and the curvature are: 6 x n8.6 ; κ The copression and the tension forces: h N c b f cd ( ( κ (x x) εn c )) 4 ( ( (8.6 x) dx. N c 48.4 kn N s (36 8.6)48.4 kn )) dx

212 The crack width with non-linear concrete aterial odel: w k sr, ax Δ ε The concrete and rebar strain difference: }{ {. f ( ) σ s k t ρ ct, eff (+α e ρ p, eff ).859 p, eff Δ ε ε s ε c ax ax Es σs Es Δ ε.544 Rebar stress: σ se s ε se s κ ( d x n) (36 8.6)34.96 N kt depends the durability of the load, this load is long-ter loading: k t.4 Effective tensile rebar ratio: 4 ϕ π A 4 ρ eff s Ac,eff b h c,eff 4 ϕ π 4 4 π 4 { } { }.5(h d ) h x II b in 3 h.5 (4 36) in First we need to check if the longitudinal bars are close to each other: ( t liit 5 (c+ϕ s )+ ) ( ) ϕ 5 (+)+ ; t actual 4 tliit > tactual therefore the rebars are close to each other, so the distance between cracks calculated as follows: ϕ s r, ax 3.4(c+ϕ s )+.45 k k ρ 3.4 (+) eff.859 }

213 k depends the cohesion between the rebars and concrete, the reinforceents are ribbed: k.8 k depends the strains in the cross-section, in this case we have pure bending: k.5 The difference between the hand calculations are under 3% and the results of the second hand calculation coincide with the FEM-Design results. Download link to the exaple file: Crack width calculation of a bea.str 3

214 9..5 Crack width calculation of a slab In this exaple we calculate the crack width of a slab due to elliptic and hyperbolic bending conditions. First of all the applied reinforceent is orthogonal and then the applied reinforceent is non-orthogonal. We calculate the crack width with hand calculation and then copare the results with FEM-Design values. The crack width calculation is relevant in SLS cobination, thus the internal forces (oents) in the exaples coe fro a quasi-peranent serviceability cobination. The calculation of the crack directions is based on [] and [] according to the tensor of the reserve forces due to an arbitrary internal force and reinforceent distribution in the slab. Fig shows the notation syste of the applied angles in this chapter. The figure is valid for botto and top reinforceent separately. The x-y syste denotes the direction of the local co-ordinate syste of the slab. The ξ and η directions are the directions of the reinforceents. The ξ and η angles are the angles between the reinforceent directions and axis x. The φ angle is the angle between the two directional reinforceents. The α angle is the angle between axis x and the direction of the crack. The α angle is the angle perpendicular to the direction of the crack. y η φ α c cra v k α η ξ ξ x Figure 9..5 The notation of the angles The concrete covers in the different directions are valid on top and botto equally even if there is no need for reinforceent in one direction. 4

215 Inputs: The thickness of the slab h The elastic odulus of concrete Ec 3 GPa, C/5 Mean tensile strength of concrete fct. N/ The Poisson's ratio of concrete ν. Elastic odulus of steel bars Es GPa The characteristic value of yield stress of steel bars fyk 5 MPa Diaeter of the longitudinal reinforceent (top and bot.) ϕl Noinal concrete cover (top and botto as well) cξ ; cη 3 Average concrete cover c 5 Effective heights (top and botto as well) dξ 75 ; dη 65 Effective heights (top and botto as well) d'ξ 5 ; d'η 35 Average effective height (botto and top) d 7 ; d' 3 Lever ar of internal forces z d d' Elliptic bending The SLS oents in the slab in shell local syste due to elliptic bending: x +6 kn / the resultant of the x directional noral stresses. y +8 kn / the resultant of the y directional noral stresses. xy +6 kn / the resultant of the x-y directional shear stresses. Orthogonal reinforceent (φ9o between ξ and η) The reinforceent is orthogonal and their directions concide with the local syste (xξ, yη). Fig shows the applied reinforceents and the concrete covers. Thus ξ o ; η 9o. sη 8 y,η sξ y,η Φ/654.5 / cξ x,ξ Φ/ / x,ξ cη3 Figure The applied orthogonal reinforceent for elliptic bending 5

216 The applied reinforceent in the slab: Only botto reinforceent is necessary (see Chapter 9.. for further inforation). ϕ ξ π π as ξ sξ 4 ϕη π π a s η sη 4 8 Calculation of the direction of the crack based on the tensor of the reserve forces. The tensor of the applied forces (effect) based on the internal forces in the quasi-peranent cobination: [ ][ ][ E E Ex E E xy E x x 6 kn, 4.3 z.4 E y y 8 kn, 57.4 z.4 E xy ] E xy kn , where Ey xy 6 kn z.4 The tensor of the resisting (yield) forces based on the resistance of the reinforceent: [ R ] [ ][ ] R R x R xy kn 37.3, where 8. R R xy R y A sξ a sξ f yk kn, A sη a s η f yk kn, o o R x A sξ cos ( ξ )+As η cos (η )37.3 cos ( )+8. cos (9 )37.3 R y As ξ sin ( ξ )+As η sin (η )37.3 sin ( o )+8. sin (9o )8. o o kn, kn, o o R xy As ξ cos( ξ )sin( ξ )+A sη cos (η ) sin(η )37.3 cos ( )sin ( )+8. cos( 9 )sin(9 ) kn. 6

217 The tensor of the reserve forces: [ ] [ r ][ r r x r xy Rx r r xy r y R xy ] [ ][ R xy E E xy R r E x r x x Ry E xy E y R xy r E xy R xy r E xy R y r E y ], where r is a scalar ultiplier of the internal forces. Yielding occurs (described by Gvozdiev []), when the saller principal value of the reserve force tensor is equal to zero: r It gives the following equation based on the well known calculation ethod of the saller principal values of a tensor: ( ( r ) ( r x +r y ) ) ( r x r y +r xy R r E x +R y r E y r x ) Rx r E x R y +r E y +( R xy r E xy ) This equation gives two solutions for the r scalar internal force ultiplier. The sallest positive r value has physical eaning. Without further detailed calculation the relevant r scalar load ultiplier is: r. Based on this scalar value the reserve force tensor: [ ][ r ] [ ][ ] r 8. r x r xy kn r r xy r y The principal direction of the first principal reserve force gives the direction of the crack: α arctan r r x arctan 47.o r xy 9.86 Now we can calculate in this direction the crack width based on the standard forulas. The internal forces and the reinforceent need to considered in the perpendicular direction of the crack: α α +9o43o 7

218 The effective reinforceent area in this direction: a α a s ξ cos ( α ξ )+a s η cos (α η )654.5 cos (43 ) cos ( 43 9 )553. o o o o The bending oent in the direction perpendicular to the crack direction: α x cos α + y sin α + xy cos α sin α o o o o (6) cos 43 +( 8) sin 43 + (6) cos 43 sin kn The position of the neutral axis according to the uncracked section (Stadiu I.): h +α e aα d E s α e ; x I.3 h+α e aα E c 3 The oent of inertia (Stadiu I.): 4 x 3I (h x I )3 8 I I + +α e aα (d x I ) Concrete tensile stress (Stadiu I.) to check if the crack exist or not: σ c,α α (h x I ) (.3) N.63 > f ct. MPa crack occurred. 5 II The position of the neutral axis according to the cracked section (Stadiu II.): x II x II α e a α (d x II ) thus: x II 3.9 The oent of inertia (Stadiu II.): x 3.9 I II II +α e aα (d x II ) (7 3.9) Rebar stress (Stadiu II.): σ s, α α e α (d x II ) I II (7 3.9) N

219 Effective tensile rebar ratio: ρ p, eff aα aα a c,eff h c,eff { }{ }.5(h d ) h x II in 3 h.5 ( 7) 3.9 in in 56.3 { } The concrete and rebar strain difference: { } f σ s, α k t ρ ct,eff (+α e ρ p,eff ) p,eff Δ ε ε s ε c ax Es σ s,α.6 Es { } ( ).987 ax ax { } The criteriu of the spacing of the bonded bars that they are close to each other or not: sp cr5 (c+ϕ l /)5( 5+/)5 The effective spacing of the bonded bars in the direction of the crack: sp ϕ l π / 4 π / 4 4. aα.553 Now the axiu crack spacing: sp spcr thus: s r, ax 3.4 c+.45 k k ρ ϕl p,eff Thus the crack width: w k sr, ax Δ ε Nuerical results: w k, FEM.559 9

220 Figure The crack width [] and the direction of the cracks at the botto Fig shows the FEM-Design results. The difference between the hand and FEM-Design calculations is less than 3%. Download link to the exaple file: Crack width calculation in a slab with elliptic bending and orthogonal reinforceent.str

221 Non-orthogonal reinforceent (φ75o between ξ and η) The reinforceent is non-orthogonal and the ξ direction concides with the local x direction. The angle between the ξ directional reinforceent and η directional reinforceent is φ75o. Fig shows the applied reinforceents and the concrete covers. Thus ξ o ; η 75o. y sη y sξ 6 η Φ/649.9 / cξ x,ξ φ Φ/39.7 / x,ξ cη3 Figure The applied skew reinforceent for elliptic bending The applied reinforceent in the slab: Only botto reinforceent necessary (see Chapter 9.. for further inforation). ϕ ξ π π as ξ sξ 4 6 ϕη π π sη 4 a s η Calculation of the direction of the crack based on the tensor of the reserve forces. The tensor of the applied forces (effect) based on the internal forces in the quasi-peranent cobination: [ ][ ][ E E Ex E E xy E x x 6 kn, 4.3 z.4 E y y 8 kn, 57.4 z.4 E xy ] E xy kn , where Ey xy 6 kn z.4 The tensor of the resisting (yield) forces based on the resistance on the reinforceent: [ R ] [ ][ ] R R x R xy kn , where R R xy R y

222 A sξ a sξ f yk kn, A sη a s ξ f yk kn, R x A sξ cos ( ξ )+As η cos (η )45.5 cos (o )+96.4 cos (75o )58.7 R y As ξ sin ( ξ )+As η sin (η )45.5sin (o )+96.4 sin (75o )83. kn, kn, R xy As ξ cos( ξ )sin( ξ )+A sη cos (η ) sin(η )45.5cos ( o) sin( o)+96.4 cos( 75o)sin(75 o) 49. kn. The tensor of the reserve forces: [ ] [ r ][ r r x r xy Rx r r xy r y R xy ] [ ][ R xy E E xy R r E x r x x Ry E xy E y R xy r E xy R xy r E xy R y r E y ], where r is a scalar ultiplier of the internal forces. Yielding occurs (described by Gvozdiev []), when the saller principal value of the reserve force tensor is equal to zero: r It gives the following equation based on the well known calculation ethod of the saller principal values of a tensor: ) ( ( ( r x +r y r ) ) ( r x r y +r xy R r E x +R y r E y r x ) Rx r E x R y +r E y +( R xy r E xy ) This equation gives two solutions for the r scalar internal force ultiplier. The sallest positive r value has physical eaning. Without further detailed calculation the relevant r scalar load ultiplier is: r.58 Based on this scalar value the reserve force tensor: [ ][ r ] [ ][ r 89.8 r x r xy kn r r xy r y ]

223 The principal direction of the first principal reserve force gives the direction of the crack: r r x α arctan arctan 59.o r xy 39. Now we can calculate in this direction the crack width based on the standard forulas. The internal forces and the reinforceent need to considered in the perpendicular direction of the crack: α α +9o3.8o The effective reinforceent area in this direction: a α a s ξ cos ( α ξ )+a s η cos (α η )49.9 cos (3.8o o )+39.7 cos (3.8o 75o) 564. The bending oent in the direction perpendicular to the crack direction: α x cos α + y sin α + xy cos α sin α (6) cos 3.8o +(8)sin 3.8o+ (6)cos 3.8o sin 3.8o9.8 kn The position of the neutral axis according to the uncracked section (Stadiu I.): h +α e aα d E s α e ; x I.3 E c 3 h+α e aα The oent of inertia (Stadiu I.): 4 x 3I (h x I )3 8 I I + +α e aα (d x I ) Concrete tensile stress (Stadiu I.) to check if the crack exist or not: σ c,α α (h x I ) (.3) N.76 > f ct. MPa crack occurred. 5 II

224 The position of the neutral axis according to the cracked section (Stadiu II.): x II x II α e a α (d x II ) thus: x II 3.9 The oent of inertia (Stadiu II.): 3 4 x 3II I II +α e aα (d x II ) (7 3.9) Rebar stress: σ s, α α e 3 α (d x II ) 9.8 (7 3.9) N I II 8.53 Effective tensile rebar ratio: ρ p, eff aα aα a c,eff h c,eff { }{ }.5(h d ) h x II in 3 h.5 ( 7) 3.9 in in { } The concrete and rebar strain difference: { } f σ s, α k t ρ ct,eff (+α e ρ p,eff ) p,eff Δ ε ε s ε c ax Es σ s,α.6 Es { } ( ).8 ax ax { }

225 The criteriu of the spacing of the bonded bars that they are close to each other or not: sp cr5 (c+ϕ l /)5( 5+/)5 The effective spacing of the bonded bars in the direction of the crack: sp ϕ l π / 4 π / aα.564 Now the axiu crack spacing: sp spcr thus: s r, ax 3.4 c+.45 k k ρ ϕl p,eff Thus the crack width: w k sr, ax Δ ε Nuerical results: w k, FEM.579 Figure The crack width [] and the direction of the cracks at the botto Fig shows the FEM-Design results. 5

226 The difference between the hand and FEM-Design calculations is less than 3%. Download link to the exaple file: Crack width calculation in a slab with elliptic bending and skew reinforceent.str 6

227 9..5. Hyperbolic bending The SLS oents in the slab in shell local syste due to hyperbolic bending: x +3 kn / the resultant of the x directional noral stresses. y 6 kn / the resultant of the y directional noral stresses. xy + kn / the resultant of the x-y directional shear stresses. Orthogonal reinforceent (φ9o between ξ and η) The reinforceent is orthogonal and their directions concide with the local syste (xξ, yη). Fig shows the applied reinforceents and the concrete covers. Thus ξ o ; η 9o. sη y,η sξ 6 y,η Φ/639 / x,ξ cξ cη3 Φ/654.5 / x,ξ Figure The applied orthogonal reinforceent for hyperbolic bending The applied reinforceent in the slab: Top and botto reinforceent are also necessary but in this case only in one direction (see Chapter 9.. for further inforation). Botto: ϕ ξ π π a 39 4 sξ 4 6 bot sξ a bot s η Top: a top s ξ ϕ η π π a sη 4 top sη Crack width on botto: Calculation of the direction of the crack based on the tensor of the reserve forces. 7

228 The tensor of the applied forces (effect) based on the internal forces in the quasi-peranent cobination: [ ][ ][ E E Ex E E xy E x x 3 kn, 8.6 z.4 E y y 6 kn, 4.3 z.4 E xy ] E xy kn , where Ey xy kn z.4 The tensor of the resisting (yield) forces based on the resistance on the reinforceent: [ R ] [ ][ ] R R x R xy kn 654.5, where 6.5 R R xy R y A sξ a bot sξ f yk kn. If there is no reinforceent on botto in the other direction we need to consider soehow the tensile resistance of the concrete, because this ay effect the relevant direction of the crack. A sη c f ct5.6.5 kn, R x A sξ cos ( ξ )+As η cos (η )654.5cos ( o)+6.5 cos (9o)654.5 R y As ξ sin ( ξ )+As η sin (η )654.5 sin ( o )+6.5 sin (9 o)6.5 kn, kn, R xy As ξ cos( ξ )sin( ξ )+A sη cos (η ) sin(η )654.5 cos (o )sin (o )+6.5 cos (9o )sin (9 o) kn. The tensor of the reserve forces: [ ] [ r ][ r r x r xy Rx r r xy r y R xy ] [ ][ R xy E E xy R r E x r x x Ry E xy E y R xy r E xy R xy r E xy R y r E y ], where r is a scalar ultiplier of the internal forces. Yielding occurs (described by Gvozdiev []), when the saller principal value of the reserve force tensor is equal to zero: r 8

229 It gives the following equation based on the well known calculation ethod of the saller principal values of a tensor: ( ( r ) ( r x +r y ) ) ( r x r y +r xy R r E x +R y r E y r x ) Rx r E x R y +r E y +( R xy r E xy ) This equation gives two solutions for the r scalar internal force ultiplier. The sallest positive r value has physical eaning. Without further detailed calculation the relevant r scalar load ultiplier is: r.337 Based on this scalar value the reserve force tensor: [ ][ r ] [ ][ r 45. r x r xy.5.3 kn r r xy r y ] The principal direction of the first principal reserve force gives the direction of the crack: α arctan r r x arctan 58.7o r xy.3 Now we can calculate in this direction at the botto the crack width based on the standard forulas. The internal forces and the reinforceent need to considered in the perpendicular direction of the crack: α α +9o3.9o The effective reinforceent area in this direction: bot bot a bot α a sx cos (α ξ )+a sy cos ( α η ) 39 cos (3.9o o)+ cos (3.9o 9 o)955.9 top top top a α a sx cos (α ξ )+a sy cos ( α η ) o o o o cos (3.9 ) cos (3.9 9 )86.8 9

230 The bending oent in the direction perpendicular to the crack direction: α x cos α + y sin α + xy cos α sin α (3)cos 3.9 o+( 6)sin 3.9o+ cos 3.9o sin 3.9o 9.7 kn The position of the neutral axis according to the uncracked section (Stadiu I.): h ' +α e aαbot d ξ +α e atop α d η E s α e ; x I.9 top E c 3 h+α e a bot α +α e aα The oent of inertia (Stadiu I.): 4 x 3I (h x I )3 ' 8 + +α e aαbot (d ξ x I )+α e atop (x d ) 7.7 α η I 3 3 I I Concrete tensile stress (Stadiu I.) to check the crack exist or not: σ c,α α (h x I ) II (.9) N 4. > f ct. MPa crack occurred The position of the neutral axis according to the cracked section (Stadiu II.): In case of this hyperbolic bending the applied reinforceents are only in one direction on one side, thus we considered the real effective height of the tensile and copressed bars instead of the average values which ones were introduced at the beginning of this chapter. x II x II ' bot +α e a top α (x II d η )α e a α (d ξ x II ) thus: x II 4.3 The oent of inertia (Stadiu II.): I II x 3II top ' +α e abot α ( d ξ x II ) +α e a α ( x II d η ) (75 4.3) ( ) Rebar stress: σ s, α α e α (d ξ x II ) I II (75 4.3) N

231 Effective tensile rebar ratio: a bot a bot α α ρ p, eff a c,eff h c,eff { }{ }.5(h d ξ ) h x II in 3 h.5 ( 75) 4.3 in in 5.96 { } The concrete and rebar strain difference: { } f σ s, α k t ρ ct,eff (+α e ρ p,eff ) p,eff Δ ε ε s ε c ax Es σ s,α.6 Es { } ( ).85 ax ax { } The criteriu of the spacing of the bonded bars that they are close to each other or not: sp cr5 (cξ +ϕ l /)5(+ /)5 The effective spacing of the bonded bars in the direction of the crack: ϕ l π / 4 π / 4 sp bot aα Now the axiu crack spacing: sp spcr thus: s r, ax 3.4 c ξ +.45 k k ρ ϕl p, eff Thus the crack width: w bot k s r,ax Δ ε Nuerical results: 3

232 bot w k, FEM.9 Figure The crack width [] and the direction of the cracks at the botto Fig shows the FEM-Design results. The difference between the hand and FEM-Design calculations is less than %. 3

233 Crack width on top: Calculation of the direction of the crack based on the tensor of the reserve forces. The tensor of the applied forces (effect) based on the internal forces in the quasi-peranent cobination: [ ][ ][ ] E E E E xy kn x , where E E xy E y E x x 3 kn, 8.6 z.4 E y y +6 kn, +4.3 z.4 E xy xy kn z.4 The tensor of the resisting (yield) forces based on the resistance on the reinforceent: [ R ] [ ][ ] R R x R xy kn 6.5, where 37.5 R R xy R y A sη a top s η f yk kn. If there is no reinforceent on top in the other direction we need to consider soehow the tensile resistance of the concrete, because this ay effect the relevant direction of the crack. A sξ c f ct kn, R x A sξ cos ( ξ )+As η cos (η )6.5cos ( o)+37.5 cos (9o)6.5 kn, R y As ξ sin ( ξ )+As η sin (η )6.5 sin (o )+37.5 sin (9 o)37.5 kn, R xy As ξ cos( ξ )sin( ξ )+A sη cos (η ) sin(η )6.5 cos (o )sin (o )+37.5 cos (9o )sin (9 o) kn. The tensor of the reserve forces: [ ] [ r ][ r r x r xy Rx r r xy r y R xy ] [ ][ R xy E E xy R r E x r x x Ry E xy E y R xy r E xy R xy r E xy R y r E y ], where r is a scalar ultiplier of the internal forces. Yielding occurs (described by Gvozdiev []), when the saller principal value of the reserve 33

234 force tensor is equal to zero: r It gives the following equation based on the well known calculation ethod of the saller principal values of a tensor: ( ( ) ( ) ) ( r r x +r y r x r y +r xy r Rx r E x +R y r E y ) Rx r E x R y +r E y +( R xy r E xy ) This equation gives two solutions for the r scalar internal force ultiplier. The sallest positive r value has physical eaning. Without further detailed calculation the relevant r scalar load ultiplier is: r.9 Based on this scalar value the reserve force tensor: [ ][ ] [ ][ r r x r xy kn r r r xy r y ] The principal direction of the first principal reserve force gives the direction of the crack: r r x α arctan arctan 8.4 o r xy 96.4 Now we can calculate in this direction at the botto the crack width based on the standard forulas. The internal forces and the reinforceent need to considered in the perpendicular direction of the crack: α α +9o8.4 o The effective reinforceent area in this direction: bot bot bot a α a sx cos (α ξ )+a sy cos ( α η ) 39 cos (8.4 o o)+ cos (8.4o 9 o)3.4 top top a top α a sx cos (α ξ )+a sy cos ( α η ) o o o o cos (8.4 ) cos (8.4 9 )

235 The bending oent in the direction perpendicular to the crack direction: α x cos α + y sin α + xy cos α sin α (3) cos 8.4 o+( 6)sin 8.4 o+ cos 8.4o sin 8.4o 8.4 kn The position of the neutral axis according to the uncracked section (Stadiu I.): h +α e aαbot d 'ξ +α e atop α dη E s α e ; x I.9 top E c 3 h+α e a bot α +α e aα The oent of inertia (Stadiu I.): x 3I (h x I ) α e aαtop ( d η x I )+α e aαbot ( x I d 'ξ ) I I Concrete tensile stress (Stadiu I.) to check the crack exist or not: σ c,α α ( h x I ) (.9) N.65 > f ct. MPa crack occurred. 5 II 6.88 The position of the neutral axis according to the cracked section (Stadiu II.): In case of this hyperbolic bending the applied reinforceents are only in one direction on one side, thus we considered the real effective height of the tensile and copressed bars instead of the average values which ones were introduced at the beginning of this chapter. x II x II ' top +α e a bot α ( x II d ξ )α e aα (d η x II ) thus: x II 3. The oent of inertia (Stadiu II.): I II x 3II bot ' +α e atop α (d η x II ) +α e a α ( x II d ξ ) (65 3.) (3. 5) Rebar stress: σ s, α α e α (d η x II ) I II (65 3.) N

236 Effective tensile rebar ratio: atop atop α α ρ p, eff a c,eff h c,eff { }{ }.5(h d η ) h x II in 3 h.5 ( 65) 3. in in { } The concrete and rebar strain difference: { } f σ s, α k t ρ ct,eff (+α e ρ p,eff ) p,eff Δ ε ε s ε c ax Es σ s,α.6 Es { }..7.4 ( ).53 ax ax { } The criteriu of the spacing of the bonded bars that they are close to each other or not: sp cr5 (cη +ϕ l /)5(3+/)75 The effective spacing of the bonded bars in the direction of the crack: ϕ l π / 4 π / 4 sp bot aα Now the axiu crack spacing: sp spcr thus: s r, ax 3.4 cη +.45 k k ρ ϕl p,eff Thus the crack width: w top k sr, ax Δ ε Nuerical results: 36

237 top w k, FEM.6 Figure The crack width [] and the direction of the cracks at the top Fig shows the FEM-Design results. The difference between the hand and FEM-Design calculations is %. Download link to the exaple file: Crack width calculation in a slab with hyperbolic bending and orthogonal reinforceent.str 37

238 Non-orthogonal reinforceent (φ75o between ξ and η) The reinforceent is non-orthogonal and the ξ direction concides with the local x direction. The angle between the ξ directional reinforceent and η directional reinforceent is φ75o. Fig shows the applied reinforceents and the concrete covers. Thus ξ o ; η 75o. y sη 9 y sξ 6 η Φ/639 / x,ξ φ cξ cη3 Φ/987.7 / x,ξ Figure The applied skew reinforceent for elliptic bending The applied reinforceent in the slab: Top and botto reinforceent are also necessary but in this case only in one direction (see Chapter 9.. for further inforation). Botto: ϕ ξ π π a 39 4 sξ 4 6 bot sξ a bot s η Top: a top s ξ ϕ η π π a sη 4 9 top sη Crack width on botto: Calculation of the direction of the crack based on the tensor of the reserve forces. The tensor of the applied forces (effect) based on the internal forces in the quasi-peranent cobination: [ ][ ][ ] E E E E xy kn x , where E E xy E y E x x 3 kn, 8.6 z.4 38

239 y 6 kn, 4.3 z.4 E y E xy xy kn z.4 The tensor of the resisting (yield) forces based on the resistance on the reinforceent: [ R ] [ ][ ] R R x R xy kn 654.5, where 6.5 R R xy R y A sξ a bot sξ f yk kn. If there is no reinforceent on top in the other direction we need to consider soehow the tensile resistance of the concrete perpendicular to the rebars, because this ay effect the relevant direction of the crack. A s( ξ +9 )c f ct5.6.5 o kn, R x A sξ cos ( ξ )+As (ξ +9 ) cos (ξ +9o )654.5 cos (o )+6.5 cos (9o )654.5 o R y As ξ sin ( ξ )+As (ξ +9 ) sin ( ξ +9 o)654.5sin (o )+6.5 sin (9o )6.5 o kn, kn, R xy As ξ cos( ξ )sin( ξ )+A s(ξ +9 ) cos ( ξ +9o )sin (ξ +9o) o cos( o) sin(o )+6.5 cos(9o )sin (9o ) kn. The tensor of the reserve forces: [ ] [ r ][ r r x r xy Rx r r xy r y R xy ] [ ][ R xy E E xy R r E x r x x Ry E xy E y R xy r E xy R xy r E xy R y r E y ], where r is a scalar ultiplier of the internal forces. Yielding occurs (described by Gvozdiev []), when the saller principal value of the reserve force tensor is equal to zero: r It gives the following equation based on the well known calculation ethod of the saller principal values of a tensor: ( r ) ( r x +r y ) r x r y +r xy 39

240 ) ( ( ) R r E x +R y r E y r x Rx r E x R y +r E y +( R xy r E xy ) This equation gives two solutions for the r scalar internal force ultiplier. The sallest positive r value has physical eaning. Without further detailed calculation the relevant r scalar load ultiplier is: r.337 Based on this scalar value the reserve force tensor: [ ][ r ] [ ][ ] r 45. r x r xy.5.3 kn r r xy r y The principal direction of the first principal reserve force gives the direction of the crack: r r x α arctan arctan 58.7o r xy.3 Now we can calculate in this direction at the botto the crack width based on the standard forulas. The internal forces and the reinforceent need to considered in the perpendicular direction of the crack: α α +9o3.9o The effective reinforceent area in this direction: bot bot bot a α a s ξ cos (α ξ )+a sη cos ( α η ) o o o o 39 cos (3.9 )+ cos ( )955.9 top top a top α a s ξ cos (α ξ )+a sη cos ( α η ) cos (3.9o o )+87.7 cos (3.9o 75 o)456. The effective internal forces in the direction perpendicular to the crack direction: α x cos α + y sin α + xy cos α sin α (3)cos 3.9 o+( 6)sin 3.9o+ cos 3.9o sin 3.9o kn

241 The position of the neutral axis according to the uncracked section (Stadiu I.): h ' +α e aαbot d ξ +α e atop α d η E s α e ; x I.3 top E c 3 h+α e a bot α +α e aα The oent of inertia (Stadiu I.): x I (h x I ) bot top ' 8 I I + +α e aα (d η x I ) +α e a α ( x I d η ) Concrete tensile stress (Stadiu I.) to check the crack exist or not: σ c,α α (h x I ) II (.3) N 4. > f ct. MPa crack occurred The position of the neutral axis according to the cracked section (Stadiu II.): In case this of hyperbolic bending the applied reinforceents are only in one direction on one side, thus we considered the real effective height of the tensile and copressed bars instead of the average values which ones were introduced at the beginning of this chapter. x II x II ' bot +α e a top α (x II d η )α e a α (d ξ x II ) thus: x II 4.9 The oent of inertia (Stadiu II.): I II x 3II bot top ' +α e aα ( d ξ x II ) +α e a α ( x II d η ) (75 4.9) (4.9 35) Rebar stress: σ s, α α e α (d ξ x II ) I II (75 4.9) N

242 Effective tensile rebar ratio: a bot a bot α α ρ p, eff a c,eff h c,eff { }{ }.5(h d ξ ) h x II in 3 h.5 ( 75) 4.9 in in 53.3 { } The concrete and rebar strain difference: { } f σ s, α k t ρ ct,eff (+α e ρ p,eff ) p,eff Δ ε ε s ε c ax Es σ s,α.6 Es { } ( ).83 ax ax { } The criteriu of the spacing of the bonded bars that they are close to each other or not: sp cr5 (cξ +ϕ l /)5(+ /)5 The effective spacing of the bonded bars in the direction of the crack: ϕ l π / 4 π / 4 sp bot aα Now the axiu crack spacing: sp spcr thus: s r, ax 3.4 c ξ +.45 k k ρ ϕl p, eff Thus the crack width: w bot k s r,ax Δ ε Nuerical results: 4

243 bot w k, FEM. Figure The crack width [] and the direction of the cracks at the botto Fig shows the FEM-Design results. The difference between the hand and FEM-Design calculations is less than %. 43

244 Crack width on top: Calculation of the direction of the crack based on the tensor of the reserve forces. The tensor of the applied forces (effect) based on the internal forces in the quasi-peranent cobination: [ E ][ ][ ] E E E xy kn x , where E E xy E y E x x 3 kn, 8.6 z.4 E y y +6 kn, +4.3 z.4 E xy xy kn z.4 The tensor of the resisting (yield) forces based on the resistance on the reinforceent: [ R ][ ][ ] R R R xy kn x , where R R xy R y A sη a top s η f yk kn. If there is no reinforceent in the other direction on top we need to consider soehow the tensile resistance of the concrete perpendicular to the rebars, because this ay effect the relevant direction of the crack. A s(η +9 )c f ct5.6.5 o kn, R x A s(η +9 ) cos (η +9 o)+ Asη cos (η )6.5cos (65o ) cos ( 75o)87.55 o o o o R y As (η+9 ) sin (η +9 )+ Asη sin (η )6.5 sin (65 ) sin ( 75 )4.4 o R xy As(η +9 ) cos(η +9 o)sin(η +9o )+ Asη cos(η )sin(η ) o 6.5 cos(65o )sin (65o ) cos(75 o) sin(75o ) kn. The tensor of the reserve forces: [ ][ r ][ ] [ r r x r xy R x R xy Ex r r r xy r y R xy R y E xy ][ E xy R r E x R xy r E xy x Ey R xy r E xy R y r E y 44 ], kn, kn,

245 where r is a scalar ultiplier of the internal forces. Yielding occurs (described by Gvozdiev []), when the saller principal value of the reserve force tensor is equal to zero: r It gives the following equation based on the well known calculation ethod of the saller principal values of a tensor: ) ( ( ( r x +r y r ) ) ( r x r y +r xy R r E x +R y r E y r x ) Rx r E x R y +r E y +( R xy r E xy ) This equation gives two solutions for the r scalar internal force ultiplier. The sallest positive r value has physical eaning. Without further detailed calculation the relevant r scalar load ultiplier is: r.375 Based on this scalar value the reserve force tensor: [ ][ r ][ ][ ] r 77. r x r xy kn r r xy r y The principal direction of the first principal reserve force gives the direction of the crack: r r α arctan x arctan 5.4o r xy 97 Now we can calculate in this direction at the botto the crack width based on the standard forulas. The internal forces and the reinforceent need to considered in the perpendicular direction of the crack: α α +9o5. o The effective reinforceent area in this direction: bot bot a bot α a s ξ cos (α ξ )+a sη cos ( α η ) 39 cos (5. o o)+ cos (5. o 75o )37.3 top top a top α a s ξ cos (α ξ )+a sη cos ( α η ) 45

246 cos (5. )+87.7 cos (5. 75 )59. o o o o The effective internal forces in the direction perpendicular to the crack direction: α x cos α + y sin α + xy cos α sin α (3) cos 5.o +( 6)sin 5. o+ cos 5.o sin 5.o 6.54 kn The position of the neutral axis according to the uncracked section (Stadiu I.): h +α e aαbot d 'ξ +α e atop α dη E s α e ; x I.5 top E c 3 h+α e a bot α +α e aα The oent of inertia (Stadiu I.): 4 x 3I (h x I )3 top bot ' 8 + +α e aα ( d η x I ) +α e aα ( x I d ξ ) I I Concrete tensile stress (Stadiu I.) to check the crack exist or not: σ c,α α ( h x I ) (.5) N.39 > f ct. MPa crack occurred. 5 II The position of the neutral axis according to the cracked section (Stadiu II.): In case of this hyperbolic bending the applied reinforceents are only in one direction on top side, thus we considered the real effective height of the tensile and copressed bars instead of the average values which ones were introduced at the beginning of this chapter. x II x II bot ' top +α e a α ( x II d ξ )α e aα (d η x II ) thus: x II 3. The oent of inertia (Stadiu II.): I II x 3II bot ' +α e atop α (d η x II ) +α e a α ( x II d ξ ) (65 3.) ( 3. 5)

247 Rebar stress: σ s,α α (d η x II ) (65 3.) N ; α e I II 7.9 Effective tensile rebar ratio: atop atop α α ρ p, eff a c,eff h c,eff { }{ }.5(h d η ) h x II in 3 h.5 ( 65) 3. in in { } The concrete and rebar strain difference: { } } f σ s, α k t ρ ct,eff (+α e ρ p,eff ) p,eff Δ ε ε s ε c ax Es σ s,α.6 Es { ( ).8985 ax ax { } The criteriu of the spacing of the bonded bars that they are close to each other or not: sp cr5 (cη +ϕ l /)5(3+/)75 The effective spacing of the bonded bars in the direction of the crack: sp ϕ l π / 4 π / aαbot Now the axiu crack spacing: sp spcr thus: s r, ax 3.4 cη +.45 k k ρ ϕl p,eff

248 Thus the crack width: w top k sr, ax Δ ε Nuerical results: w top k, FEM.87 Figure The crack width [] and the direction of the cracks at the top Fig shows the FEM-Design results. The difference between the hand and FEM-Design calculations is less than %. Download link to the exaple file: Crack width calculation in a slab with hyperbolic bending and skew reinforceent.str 48

249 9..6 Punching calculation of a slab In this section we will check three different types of punching reinforceent: bended bar circular stirrups open stirrups Inputs: Concrete Concrete characteristic copressive strength fck 5 N/ Plate height h Cover c Partial factor of concrete γc.5 Reinforceent x' direction Reinforcing steel characteristic yield strength fyk 5 N/ Rebars Young odulus Es GPa Partial factor of reinforcing steel γs.5 Longitudinal rebar diaeter ϕx Distance between longitudinal reinforceents sx 5 Longitudinal bars effective distance dl,x 7 y' direction Reinforcing steel characteristic yield strength fyk 5 N/ Rebars Young odulus Es GPa Longitudinal rebar diaeter ϕy Distance between longitudinal reinforceents sy 5 Longitudinal bars effective distance dl,y 5 Geoetry Plate width in x direction Lx 4. Plate width in y direction Ly 4. Circle colun diaeter dcolun 5 c Specific noral force in x direction nx,ed 4 kn/ Specific noral froce in y direction ny,ed 4 kn/ Vertical surface total load qz,ed 5 kn/ 49

250 ϕ/5 c qz,ed 5 kn/ nx,ed 4 kn/ Lx 4 Z dcolun 5 c Y X Figure The geoetry, loads and the reinforceents Fig shows the analyzed punching proble Bended bars Inputs (see Fig ): Reinforceent Reinforcing steel characteristic yield strength fywk 5 N/ Bended bar diaeter ϕbb 4 Partial factor of reinforcing steel γs.5 Bended bar distance sr VEd.4d 6 8 dcolun 5 c 85 α α α Figure The applied punching reinforceent in x direction (the sae valid in y direction) 5 dl,y dl,x 4 ϕ/5 ny,ed 4 kn/ t c Ly

251 The effective longitudinal reinforceent ratios: ϕ x π π ϕ y π π 4 sx 4 sy ρ l, x.3 ; ρ l, y.396 d l, x 7 d l, y 5 { } { } ρ ρ ρ l in l, x l, y in in { } The average effective height of the longitudinal reinforceents: d l, x +d l, y 7+5 d 6 Concrete copression resistance The colun reaction: V Ed L x L y q z, Ed kn The perieter of the colun: u d colun π 5 π Because this colun is an inner colun the β is: β.5 The specific shear force at u: V 4 N v Ed β Ed d u The concrete copression resistance: f 5 N 5 v Rd, ax.5ν f cd where: ν.6 ck ( ) ( ) Concrete copression utilization: v Ed % v Rd,ax 4.5 Utilization is less than % thus the slab can bear the acting shear force. Fig shows the relevant results with FEM-Design. The results are identical with the hand calculation. Shear reinforceent resistance Punching perieter at d distance fro the edge of the colun: u ( ) ( d colun 5 + d π + 6 π 796 ) 5

252 The shear force at u perieter: V 4 N v Ed β Ed.5.8 d u Modifying factors for concrete shear resistance:.8.8 C Rd,c γ. ; k. c.5 Average noral stress in the concrete: } { } 4 4 n x, Ed n y, Ed + + N σ cp in σ cx +σ cy h h in. 5.. f cd.5 { { } { } k in +. + in d 6 in { }.5.5 ν in.35 k.5 f.5 ck N The shear resistance of concrete: { { } } v Rd, c ax C Rd,c k ( ρ l f ck )3 +k σ cpax. (.3 5) ν in.495 N ax { } We need to check if we need any punching reinforceent or not: v Ed.8.35>. v Rd,c.788 Thus we need punching reinforceent. Bended bars area in the u perieter: ϕ bb π 4 π A sw Effective tensile strength of the bended bars: 5 f ywk f ywd γs N.5 f y, sw,eff in in in d in d { } { } { } 5 { }

253 The punching reinforceent resistance: v Rd, sw.5 d A f s r sw y, sw,eff 6 N sin α sin u d Punching shear resistance: v Rd, cs.75 v Rd, c +v Rd, sw N Punching shear resistance utilization: v Ed % v Rd,cs.55 Fig and the table below shows the relevant FEM-Design results. Concrete copression Shear reinforceent Concrete shear Hand calculation 4.5 N/.55 N/.788 N/ FEM-Design calculation 4.5 N/.9 N/.79 N/ Difference.% 3. %. % The differences between the nuerical and hand calculations are less than 3%. 53

254 Figure The punching detailed results in FEM-Design 54

255 9..6. Circular stirrups Inputs (see Fig ): Reinforceent Reinforcing steel characteristic yield strength fywk 5 N/ Circular stirrup diaeter ϕstirrup Stirrup height hs Stirrup width ws sr Partial factor of reinforcing steel γs.5 ϕ hs ws ws ws VEd.4d 6 Figure The applied punching reinforceent Concrete copression resistance The calculation and the results are identical with the relevant part of the forer chapter (see Chapter 9..6.). Shear reinforceent resistance The results of the shear resistance of the conrete without punching reinforceent are identical with the relevant part of the forer chapter (see Chapter 9..6.). Thus it eans that we need punching reinforceent. 55

256 Punching perieter at d distance fro the edge of the colun: u ( ) ( d colun 5 + d π + 6 π 796 ) The area of the stirrups at the u perieter (see Fig ): ϕ stirrup π π A sw Effective tensile strength of the stirrups: { } { } 5 N.5 f y, sw,eff in in in d in d f ywk γs { } f ywd { } The punching reinforceent resistance: v Rd, sw.5 d A f s r sw y, sw,eff 6 N sin α sin 9. u d Punching shear resistance: v Rd, cs.75 v Rd, c +v Rd, sw N Shear reinforceent utilization: v Ed % v Rd,cs.83 Fig shows the relevant FEM-Design results. Concrete copression Shear reinforceent Concrete shear Hand calculation 4.5 N/.83 N/.788 N/ FEM-Design calculation 4.5 N/.8 N/.79 N/ Difference.%. %. % The differences between the nuerical and hand calculations are under.5%. 56

257 Figure The punching detailed results in FEM-Design 57

258 Open stirrups Inputs: Reinforceent Reinforcing steel characteristic yield strength fywk 5 N/ Open stirrup diaeter ϕos 4 Stirrup height sx 6 Stirrup width sy 8 Partial factor of reinforcing steel γs.5 sy. sx 8 7. VEd Figure The applied punching reinforceent Concrete copression resistance The calculation and the results are identical with the relevant part of the forer chapter (see Chapter 9..6.). Shear reinforceent resistance The results of the shear resistance of the conrete without punching reinforceent are identical with the relevant part of the forer chapter (see Chapter 9..6.). Thus it eans that we need punching reinforceent. Effective stirrup area: ϕ os π 4 π 4 4 a sw.963 sx s y

259 Effective tensile strength of the open stirrups: { } { } 5 N.5 f y, sw,eff in in in d in d { } f ywd f ywk γs { } The punching reinforceent resistance: v Rd, sw.5 d a sw f y,sw, eff N sin α sin d 6 Punching shear resistance: v Rd, cs.75 v Rd, c +v Rd, sw N Open stirrups utilization: v Ed % v Rd,cs.56 Fig shows the relevant FEM-Design results. Concrete copression Shear reinforceent Concrete shear Hand calculation 4.5 N/.444 N/.788 N/ FEM-Design calculation 4.5 N/.44 N/.79 N/ Difference.%.%. % The differences between the nuerical and hand calculations are under.5%. Download link to the exaple file: Punching calculation of a slab.str 59

260 Figure The punching detailed results in FEM-Design 6

261 9..7 Interaction of noral force and biaxial bending in a colun In this section we will calculate the utilization and the load-bearing capacity of a cantilever colun under biaxial bending and noral force (see Fig ) according to non-linear concrete and reinforcing steel aterial odel. Inputs: Concrete characteristic copressive strength fck N/ Partial factor of concrete γc.5 Effective creep factor φef Partial factor of concrete Young's odulus γce. Reinforcing steel characteristic yield strength fyk 5 N/ Partial factor of reinforcing steel γs.5 Ultiate liit strain of reinforcing steel uk.5 %, ud.5 % Slope of plastic part (aterial odel) k.5 Longitudinal bar diaeter ϕ Stirrup diaeter ϕs 8 Concrete cover (on stirrups) c Colun height L 3. Colun width (square section) b 3 ; h 3 The effective depth d 3 8 / 6 Noral force NEd 5 kn Bending oent in y' direction MEd,yI kn Bending oent in z' direction MEd,zI kn NEd 5 kn MEd,z kn ϕ8 L3 MEd,y NEd 3 MEd,z MEd,y kn ϕ 3 Figure Cantilever colun with the external forces and oents and the RC section the x direction is the axis of the colun, y and z are respect to the oent directions 6

262 In Eurocode there are different ethods to calculate the load-bearing capacity of the colun. One of the is the Noinal stiffness ethod and an other one is the Noinal curvature ethod. We will calculate the utilizations and the load-bearing capacities with independent hand calculations respect to both entioned ethods then we will copare the results with FEMDesign calculations Noinal stiffness ethod Here we will calculate the noinal bending stiffness of the colun and then increase the acting oents to consider second order effects. The Eurocode suggests to increase the oent only in the unfavorable principal direction. In FEM-Design to ensure the ost unfavourable result the first order oents will be increased with the effect of the iperfections and second order effects in both principal directions to be on the safe side and get the ost unfavourable condition. The noinal bending stiffness of the colun: 8 7 EI K c E cd I c +K s E s I s N, where: K c k k k is a factor for effects of cracking, creep etc. +φ ef + f ck is a factor which depends on concrete strength. { } { } 69.8 n λ in is a factor which depends on axial force and.. slenderness. k in n N Ed N 5 Ed.467 is the relative axial force. Ac f cd f ck 3 b γ c.5 l L 3 λ 69.8 is the slenderness ratio. i i b4 3 4 I i c is the radius of gyration of the uncracked concrete Ac b 3 6

263 section. 4 4 b 3 I c is the inertia of the uncraked concrete section. E cd E c / γ ce 3/.5 MPa is the design value of the odulus of elasticity of concrete. K s is a factor for contribution of reinforceent. b ϕ π b π 3 I s As d 4 d ( ) [ ] [ ] is the second oent of area of reinforceent, about the centre of area of the concrete. After this we need to increase the first order oents in both direction due to iperfection and second order effects. Considering the effect of the iperfection (see the forer underlined coent also): M, Ed, y M IEd, y +ei N Ed kn M, Ed, z M IEd, z+ei N Ed kn The eccentricity according to the iperfection in both direction: e i l L l value ay be reduced in the function of the colun height, 4 but FEM-Design do not consider this effect thus we are on the safe side. NOTE: In Eurocode the The increased design oent values according to the second order effects (see the forer underlined coent also): M Ed, y M Ed, z M, Ed, y kn and N Ed 5 6. NB M, Ed, z kn, N Ed 5 6. NB 63

264 where NB is the buckling load based on noinal stiffness of the colun: π EI π 4.7 N B 6 N6 kn l ( 3) Now we need to calculate the resistance (load-bearing capacity). For this the M-N interaction curve (bending oent noral force interaction curve)in the principal directions or the M-N interaction surface is necessary. In Eurocode there is a siplified ethod to check the utilization with the aid of the M-N interaction curves in the principal directions but there is a ore accurate solution for the proble with the help of the M-N interaction surface. Let's see what is the different between these calculation ethods and then we show you what is the FEMDesign solution for this proble. First of all with independent nuerical calculation we provided the M-N interaction curve in the principal directions. According to the square and double-syetric cross section these two curves will be identical each other (see Fig ). At the given design noral force value (NEd 5 kn) the oent resistance with nuerical calculation in both y and z direction is: M Rd, y M Rd, z 4. kn (see Fig ). 8 6 Noral force resistance [kn] 4 8 [3.47; 5] 6 [4.; 5] M oent resistance [kn] Figure The M-N interaction curve in the x-y or in the x-z plain The utilization based the approxiation forula of Eurocode is the following: a a ( )( )( ) ( ) M Ed, y M Ed, z M Rd, y M Rd, z %

265 The noral force utilization is: N Ed N Ed N Rd Ac f cd + As f yd N Ed b f cd +4 ϕ π f yd π The a value can be linearly interpolated based on the following table: a.55 N Ed N Rd..7. a..5. The M-N curves were calculated based on the aterial odels what you can see in Fig The strain values for concrete coincide with the values what were entioned in Chapter NOTE: Due to nuerical stability FEM-Design uses a bit odified concrete aterial odel according to Fig but by ultiate liit state with the design stress values. σs σc fcd kfyd fyd εc εcu εsy fyd/es εc εud Figure The aterial odels for the M-N curve calculations left: concrete (only copression), right: rebars (both tension and copression) Concrete: σ c (ε c ) f cd [ ε ε cc ( )] σ c (ε c ) f cd if ε c ε c if ε c <ε c ε cu Rebars: σ s ( ε s )ε s E s if ε s f yd Es σ s ( ε s ) f yd +(k ) f yd f ε ud yd Es ε s if f yd Es f yd <ε s ε ud Es 65 εs

266 Fig shows a slice of the M-N surface in the direction of 45 o of angle between y and z axes. Based on this figure we can calculate the utilization by hand in a ore accurate way. At the NEd 5 kn level the axiu oent capacity is 8.96 kn for the 45o direction (see Fig ). Based on this oent capacity we can calculate the axiu oent capacity in the principal directions according to the equal design oents in y and z directions: precise precise M Rd, y ( N Ed )M Rd, z ( N Ed ) kn Due to this value we can calculate the a value ore precisely based on tha fact that in this exaple the increased oent values are the sae in the two directions. The following expression ust be true: a ( ) a precise. and based on this: a precise.83. Thus the ore precise utilization is: precise a precise ( ) ( ) ( ) ( ) M Ed, y M Rd, y M Ed, z + M Rd, z % This eans that the Eurocode forula is on the safe side based on this utilization type. 8 6 Noral force resistance [kn] 4 [7.63; 86.] 8 [8.96; 5] [44.5; 5] Moent resistance [kn] Figure A slice of the M-N surface between y and z axes with 45o of angle 66 9

267 Fig shows a slice of the M-N surface in the 45 o of angle direction between y and z axes (which is the relevant direction in this exaple, see the increased design oent values in the two directions). In FEM-Design the utilization is the following: N Ed η N Rd ; M Ed, y η M Rd, y ; M Ed, zη M Rd, z where η is the utilization, it eans that if we increase the noral force and the two bending oents linearly at the sae tie we will reach the failure surface of the cross-section (see the red line in Fig ). With other words the eccentricity is constant. According to this scenario the utilization with the independent hand calculation is the following: The red line breaks through the M-N curve at (see Fig ): N Rd 86. kn, M Rd 7.63 kn The utilization of the colun based on the red line geoetrical point of view according to Fig : η M Ed, y +M Ed, z +N Ed M Rd +N Rd % Utilization based on the FEM-Design detailed results is (see Fig ): η 64 % The difference between the hand and nuerical calculations is less than 5%. This difference coes fro the fact that FEM-Design uses a bit odified concrete aterial odel (see. Fig , but here the stress values were replaced by the design values) to ensure the nuerical stability for every case. Download link to the exaple file: Interaction of noral force and biaxial bending in a colun.str 67

268 Figure The detailed results of the RC colun in FEM-Design with noinal stiffness ethod 68

269 9..7. Noinal curvature ethod Here we will calculate the noinal curvature of the colun and then increase the acting oents based on this curvature to consider second order effects. The Eurocode suggests to increase the oent only in the unfavorable principal direction. In FEM-Design to ensure the ost unfavourable result the first order oents will be increased with the effect of the iperfections and second order effects in both principal directions to be on the safe side and get the ost unfavourable condition. Considering the effect of the iperfection: M, Ed, y M IEd, y +ei N Ed kn I M, Ed, z M Ed, z+ei N Ed kn The eccentricity according to the iperfection in both direction: e i l L 3 5 (see also the relevant NOTE in Chapter 9..6.) The second order eccentricity: l r e 65.38, c where the curvature: K r K φ , r r where: f yd 435 ε E s yd r.45 d.45 d.45 6 { } nu n K rin n u nbal.984 is a correction factor depending on axial load ϕ π f yk π 5 4 A f 4 γs n u+ω ; ω s yd Ac f cd f ck 3 b γ.5 c 4 69

270 The relative axial noral force: n N Ed N 5 Ed.467, n bal.4 Ac f cd f ck 3 b γ c.5 { } K φ ax + β φ ef ax.976 is a factor for taking account of creep. { } Where: β.35+ f ck 69.8 λ l L 3 λ 69.8 the slenderness ration (see Chapter also). i i The increased design oent values according to the second order effects: M Ed, y M, Ed, y +N Ed (e i+e )+5( )5.9 kn and M Ed, zm,ed, z+ N Ed (e i +e )+5( )5.9 kn. 8 6 Noral force resistance [kn] 4 8 [7.98; 5] 6 [8.35; 566.] 4 [8.96; 5] Moent resistance [kn] Figure A slice of the M-N surface between y and z axes with 45o of angle 7 9

271 Fig shows a slice of the M-N surface in the 45 o of angle direction between y and z axes (which is the relevant direction in this exaple, see the increased design oent values in the two directions). In FEM-Design the utilization is the following: N Ed η N Rd ; M Ed, y η M Rd, y ; M Ed, zη M Rd, z where η is the utilization, it eans that if we increase the noral force and the two bending oents linearly at the sae tie we will reach the failure surface of the cross-section (see the red line in Fig ). With other words the eccentricity is constant. According to this scenario the utilization with the independent hand calculation is the following: The red line breaks through the M-N curve at (see Fig ): N Rd 566. kn, M Rd 8.35 kn The utilization of the colun based on the red line geoetrical point of view according to Fig : M η Ed, y +M Ed, z +N Ed M Rd +N Rd % Utilization based on the FEM-Design detailed results (see Fig ): η 9.7 % The difference between the hand and nuerical calculations is around 5%. This difference coes fro the fact that FEM-Design uses a bit odified concrete aterial odel (see. Fig , but here the stress values were replaced by the design values) to ensure the nuerical stability for every case. Download link to the exaple file: Interaction of noral force and biaxial bending in a colun.str 7

272 Figure The detailed results of the RC colun in FEM-Design with noinal stiffness ethod 7

273 9..8 Calculation of a statically indeterinate bea with post tensioned cables Inputs: Characteristic tensile strength of steel fpk 86 MPa Cross-section of one strand Ap 5 Nuber of strands n Curvature coefficient μ.5 Wobble coefficient κ.7 / Anchorage set slip g 4 Young's odulus of the strand Ep 95 MPa Jacking side point A Young's odulus of concrete when post tensioning applied Ec Ec(t) 3 GPa The final value of creep coefficient φ(t,t). Final value of shrinkage εcs.4 Relaxation class of the strands Class Value of relaxation loss ρ.5 % The cross-section of the bea is rectangle b 3, h 6 In this exaple we would like to calculate the equivalent forces of a post tensioned cable syste before and after the long ter stress losses and copare these results with FEM-Design results. For these calculations we need the data which were indicated above in the table and we need to know the shape of the cables. The geoetry of the statically indeterinate bea and the shape of the cables are shown in Fig e B A f C Parabola ; L f 33 D 6 Par. ; L.5 Parabola ; L 4.4 L 5. L 5. Figure The statically indeterinate bea with the cable profile The angular deviation of the cables is a function starting at point A and it depends on the shape of the cables (e.g. base points and inflections). Now we have parabolic shapes on each parts (see Fig ). 73

274 The angular deviation function: α ( x ) The values of the angular deviation function at soe typical points: α ( A) rad ; α (B)α ( A)+ atan α (C )α (B)+ atan 4f.639 rad ; L 4f 4f.5 rad ; α ( D)α (C )+ atan.89 rad L L Based on these values the function of the angular deviation, see Fig Figure The function of the angular deviation [α(x)] Let's be the stress at the active jacking end (point A) during post tensioning: σ p.8 f pk 488 MPa NOTE: By a real design task according to the Eurocode we need to consider an upper bound for this value, e.g.: σ p,ax in(.8 f pk ;.9 f p,k ), but by this exaple we do not consider it because it does not affect the ethod of the following calculation. In FEM-Design the user should give the σ p value and all of the calculations will consider this input stress as jacking stress. Before the calculation of the equivalent forces which coe fro the post tensioning of the bea, first of all we need to calculate the different stress losses. Fisrtly we consider the stress losses which coe fro the technology of post tensioning. The stresses in the strands are also a function considering the losses due to friction. 74

275 The function of the stresses considering losses due to friction: σ p (x ) The values of the function of the stresses considering losses due to friction at soe typical points: σ p ( A)σ p 488 MPa σ p ( B)σ p e μ (α σ p (C )σ p e μ (α σ p ( D)σ p e μ (α B +κ L ) C +κ (L +L )).5 ( ) 488 e D 44 MPa 488e.5 (.5+.7 (4.4+.5))397 MPa +κ ( L+ L +L )).5( ( )) 488 e 354 MPa Based on these values the function of the stresses considering the losses due to friction, see Fig Figure The function of the stresses considering losses due to friction [σp(x)] Now we need to calculate the stresses considering the losses due to the anchorage set slip. The jacking is applied at the start point A. The length of the effect of the anchorage set slip (Ls) coes fro the following equation: L s L s g E P ( σ p( x) σ pl (L s ) ) dx (σ p (x ) σ p e μ (α (L s ) dx )+κ L s ) The solution of this equation after soe iterations is: L s The stress loss at the active jacking side due to anchorage set slip: Δ σ s [ σ [ e μ ( α (L )+κ L ) ] ] [ 488 [ e.5( ) ] ] MPa s p s The stresses in the strands are also a function considering the anchorage set slip. The function of the stresses considering losses due to anchorage set slip and friction: σ p ( x) 75

276 The values of this function at soe typical points: σ p ( A) σ p ( A)+ σ p Δ σ s MPa σ p (B) σ p ( B)+ σ p Δ σ s MPa σ p (C ) σ p( C)+ σ p Δσ s MPa σ p ( Ls ) σ p ( L s)+ σ p Δ σ s MPa σ p (D)σ p( D)354 MPa Based on these values the function of the stresses considering the losses due to anchorage set slip and friction, see Fig Figure The function of the stresses considering losses due to anchorage slip and friction [σp(x)] The average stress value in the strands after the losses due to anchorage set slip and friction: L σ p( x)dx σ L MPa NOTE: By a real design task according to the Eurocode we need to consider an upper bound for this value, e.g.: σ,ax in(.75 f pk ;.85 f p, k ), but by this exaple we do not consider it because it does not affect the reaining calculation. In FEM-Design the progra is also calculating an average stress after considering the losses due to anchorage set slip and friction and using this value to estiate the short and long ter stress losses. We need to consider the short and long ter stress losses in the strands after the jacking. 76

277 Next to the frictional and anchorage set slip losses another short ter loss is the losses due to elastic shortening. The average noral stress in the concrete cross-section at the level of the anchorages: σ c nσ Ap nσ Ap e e MPa 3 Ac Ic / The average elastic stress losses according to Eurocode : Δ σ el Ep n 95 σc MPa n E c (t) 3 The average stress in the strands after the elastic shortening: σ ti σ Δσ el MPa The average stress in concrete at the level of the anchorages considering the elastic shortening: σ c,ti n σ ti A p n σ ti A p e e +.98 MPa 3 Ac Ic / After the short ter stress losses we can calculate the long ter stress losses is the strands. The stress losses due to creep: Δ σ cr Ep 95 φ (t, t ) σ c,ti MPa E c 3 The stress losses due to shrinkage: Δ σ se p ε cs MPa The stress losses due to relaxation: σ 34.5 μ p ti.7 f pk 86 Δ σ pr.66 ρ e 9. μ p.75( μ p ) ( ) t 5 σ ti.66.5 e 9..7 (Relaxation class:.) Δ σ pr 57.9 MPa 77 ( 5.75(.7) )

278 These three effects, naely the creep, shrinkage and the relaxation are in interaction. The stress losses due to the interaction: Δσ p, c+ s+r Δσ p, c+ s+r Δσ s+.8 Δ σ pr+ Δσ cr E n Ap A + p (+ c z cp )[+.8 φ (t, t )] E c Ac Ic MPa (+ )[+.8.] / The average stress in the strands before the long ter losses (T): σ ti σ Δσ el MPa The average stress in the strands after the long ter losses (T ): σ t σ ti Δ σ p,c+s+r MPa Based on these values we can calculate the equivalent forces which will represent the effect of the post tensioning on the statically indeterinate bea. The equivalent forces at T tie before the long ter losses: The concentrated forces at the ends (at the centroid of the concrete cross-section): P n A p σ ti kn The angle of the tangent of the cable at the ends: α atan 4f 4 33 atan 7.9 o L 44 The horizontal and vertical coponents of the concentrated forces at the ends: P H P cos α 4.5 cos 7.9 o384.3 kn P V P sin α 4.5 sin 7.9o9.5 kn The concentrated oents at the ends according to the eccentricities at the ends: M PH e kn 78

279 The intensity of the distributed load according to the different parabola shapes: u u 8 P f L 8 P f L kn kn The equivalent forces at T tie after the long ter losses: The concentrated forces at the ends (at the centroid of the concrete cross-section): P n A p σ t kn The horizontal and vertical coponents of the concentrated forces at the ends: P H P cos α cos 7.9 o34.3 kn P V P sin α sin 7.9o 5.9 kn The concentrated oents at the ends according to the eccentricities at the ends: M P H e kn The intensity of the distributed load according to the different parabola shapes: u u 8 P f L 8 P f L kn kn Fig shows the equivalent forces on the statically indeterinate bea before and after the long ter stress losses in the strands. 79

280 u 67.3 kn/ u 59. kn/ u P4.5 kn P B A C D M M38.4 kn u 48. kn/ u 5.3 kn/ u P kn P B A C M D M 34. kn Figure The equivalent forces before [above] and after [below] the long ter stress losses Considering the equivalent forces after the long ter stress losses (T, Fig below) the caber (at the id-span) of the statically indeterinate bea (without detailed calculation according to the theory of elasticity) is: e p,caber.86 By this caber calculation we considered only the effect of the equivalent forces (Fig below) and the effective odulus of elasticity of concrete: Ec,eff Ec/(+φ(t,t)) GPa. Figure The shape of the post tensioned cables in FEM-Design Fig shows the shape of the post tensioned cable in FEM-Design. 8

281 67 kn/ 43 kn/ 47 kn/ 45 kn 385 kn 36.8 kn 49.3 kn/ 57.5 kn/ 5.7 kn/ 5. kn/ 6.3 kn/ 38.6 kn 5.8 kn/ 47 kn/ 6 kn/ 3 kn/ 338 kn 3.3 kn 358 kn 43.3 kn/ 5.7 kn/ 45. kn/ 46.7 kn/ 53.5 kn/ 34. kn 45.8 kn/ Figure The equivalent forces before [above, T] and after [below,t ] the long ter stress losses in FEM-Design Figure The vertical translations [] in FEM-Design fro the equivalent post tensioning loads at T tie considering the effective odulus of eleasticity of concrete We can say that the results of the hand calculation and the autoatic post tensioned cable calculation of FEM-Design are identical. See the FEM-Design results about the equivalent forces in Fig and the caber in Fig Keep in ind that FEM-Design post tensioned cable odul calculates the equivalent forces in ore precise way than this hand calculation and considers the curvatures of the shape of the cable in ore accurate way. Theoretically the calculated equivalent forces are in equilibriu but the presented hand calculation ethod does not consider the friction force (and its eccentricities), but FEM-Design is checking the equilibriu of the equivalent forces and autoatically applying a distributed axial force and a distributed bending oent on the bea if it is necessary. These forces and oents are also indicated in the progra by the post tensioned equivalent load cases. Download link to the exaple file: Calculation of a statically indeterinate bea with post tensioned cables.str 8

282 9.3 Steel design 9.3. Interaction of noral force, bending oent and shear force In this sub-chapter an IPE bea will be investigated under the interaction of noral force, shear force and bending oent around its strong axis (see Fig ). Stability analysis will not be considered here only strength resistance calculations. Inputs: Yield strength of structural steel fy 35 N/ Cross-sectional width b Cross-sectional height h Flange thickness tf 8.5 Web thickness tw 5.6 Web height hw 59 Radius of root fillet r Cross-sectional area A 848 Plastic cross-sectional odulus Wpl,y Noral force NEd 7 kn (copression) Bending oent around strong (y') axis MEd 37.5 kn Shear force VEd 5 kn FEd3 kn Z NEd7kN.5 X L Z Y -7 N [kn] -5 V [kn] M [kn] Figure The statical syste and internal forces 8

283 First we need to ake the classification of the cross section individually for every internal forces: The coefficient depending on fy: ε fy 35 Classification due to noral force Flanges: c b t w r 5.6 c c 35. c 4.4 ; <9 ε 35. ; t t 8.5 t f Because 4.4<9 thus the flanges are in Class. Web: c c h w ; <33 ε t t t w 5.6 Because 8.39<33 thus the web is in Class. Therefore the cross section is in Class under noral force. Classification due to bending oent around strong axis Flange: c b t w r c c 35. c 4.4 ; <9 ε t t f 8,5 t because 4.4<9 thus the flange is in Class. Web: c h w t t w 5.6 c <7 ε t because 8.39<7 thus the web is in Class. Therefore the cross section is in Class under bending oent around strong axis. 83

284 Interaction of bending, shear and axial force Noral force resistance under copression: f 35 N c, Rd A γ y kn M. Bending oent resistance around strong axis: fy 35 M c, Rd W pl, y γ kn M. Shear resistance: Av A b t f +(t w +r )t f (5.6+ ) 8.54 f / 3 35 / 3 V pl, Rd Av γy kn M. The oent resistance should be reduced if V Ed >.5 V pl, Rd. Now 5 kn>94.95 kn therefore the reduction factor is: ( )( V Ed 5 ρ.336 V pl, Rd 89.9 ) The reduced bending resistance: ( ) ( ρ Aw f y M y, V, Rd W pl, y kn 4 t w γ M ) The utilization according to the interaction: N Ed M y, Ed <. thus the resistance is adequate. N c, Rd M y,v, Rd Fig shows the statical syste and the internal forces in FEM-Design. 84

285 Figure The statical syste and internal forces in FEM-Design The detailed results with the interaction utilization based on FEM-Design are shown in Fig

286 Figure The detailed results fro FEM-Design The difference between the two calculations is 3%. Download link to the exaple file: Interaction of noral force, bending oent and shear.str 86

287 9.3. Buckling of a doubly syetric I section The buckling stability analysis will be investigate in an IPE 4 siply supported bea (see Fig ). Inputs: Yield strength of structural steel fy 355 N/ Cross-sectional width b Cross-sectional height h 4 Flange thickness tf 9.8 Web thickness tw 6. Web height hw 9.4 Radius of root fillet r 5 Cross-sectional area A 39 Inertia around strong axis Iy Inertia around weak axis Iz St. Venant torsional constant It Warping constant Iω Buckling length in both directions Lcr 6. NEd4 kn X Z Lcr 6. Y Z Y NEd4 kn X Figure The statical syste and the cross-section First of all we need to ake the classification of the cross section for noral force: The coefficient depending on fy: 87

288 ε fy 355 Classification due to noral force Flanges: c b t w r 6. 5 c c 4.9 c 4.76 ; <9 ε 4.9 ; t t f 9.8 t Because 4.76<7.3 thus the flanges are in Class. Web: c c h w ; <38 ε t t tw 6. Because 3.7<3.9 thus the web is in Class. Therefore the cross section is in Class under noral force. Flexural buckling around strong axis The radius of gyration (y-y axis): i y The non-diensional slenderness: λ y λ π Es π fy Iy A 39 Lcr , where i y λ The iperfection α factor value based on EN Table 6.: h 4 >. rolled section (y-y axis) and t f 9.8 <4 therefore a buckling b curve is relevant thus the iperfection factor is α y.. Φ y.5 [ +α y ( λ y. )+ λ y ].5 [ +. ( ) ].876 Reduction factor: χ y.89 Φ y + Φ y λ y

289 χ y A f y Flexural buckling resistance: N b, y, Rd γ 5 kn M Flexural buckling around weak axis The radius of gyration (z-z axis): i z Iz A 39 The non-diensional slenderness: λ z λ π Es π fy Lcr 6.96, where i z λ The iperfection α factor value based on EN Table 6.: h 4 >. rolled section (z-z axis) and t f 9.8 <4 therefore b buckling b curve is relevant thus the iperfection factor is α z.34. Φ z.5 ( +α z ( λ z. )+λ z ).5 ( +.34 (.96. )+.96 )5.3 Reduction factor: χ z.49 Φ z + Φ z λ z χ A f y Flexural buckling resistance: N b, z, Rd γz 45.7 kn M Torsional buckling The elastic torsional buckling critical force: ( ) ( ) π E Iω π N cr, T G I t kn i L cr where i is the polar radius of gyration: i i y +i z + y + z , where y and z are the distances between the center of gravity and the shear center of the crosssection respect to the principal directions. The torsional-flexural buckling is not relevant because the cross-section is doubly syetric. 89

290 The non-diensional slenderness for torsional buckling: λ T Afy N cr, T 6 The iperfection α factor value based on EN Table 6.: h 4 >. rolled section (z-z axis) and t f 9.8 <4 therefore b buckling b curve is relevant thus the iperfection factor is α T.34. Φ T.5 ( +α T ( λ T. )+λ T ).5 (+.34 (.93. )+.93 ).49 Reduction factor: χ T.5395 Φ T + Φ T λ T χ A f y Torsional buckling resistance: N b,t, Rd Tγ 749. kn M The statical syste and the internal forces shown in Fig Figure The statical syste and the internal force in FEM-Design Fig shows the detailed results about flexural and torsional buckling in FEM-Design. The nuerical result are alost identical with the hand calculations. The difference is less than.5%. Download link to the exaple file: Buckling of a doubly syetric I section.str 9

291 Figure Detailed results based on FEM-Design 9

292 9.3.3 Buckling of a doubly syetric + section Yield strength of structural steel fy 355 N/ Cross sectional width b Thickness of the parts t 4 Cross-sectional area A 584 Inertia around strong axis Iy Inertia around weak axis Iz St. Venant torsional constant It Warping constant Iω Buckling length Lcr NEd3 kn Z Z X Lcr. Y Y NEd3 kn X Figure The statical syste and the cross-section First of all we need to ake the classification of the cross section for noral force: The coefficient depending on fy: ε fy 355 Classification due to noral force Outstand flanges: c b t 4 c 98 c 98 ; 4.45 ; >4 ε t 4 t Because 4.45>.39 thus the flanges (and the whole section) are in Class 4. 9

293 Calculation of the effective cross-section k σ.43 c 98 t 4 λ p ε k σ λ because λ p.67>.748 thus: ρ p.5465 λ p.67 beff ρ c Aeff t +4 b eff t Flexural buckling The radius of gyration (based on the gross section): i y The non-diensional slenderness: λ λ π Es π fy Lcr i y λ Iy A 584 Aeff , where A In EN Table 6. this type of section is not included thus c buckling curve was chosen. The iperfection factor is: α.49. Φ.5 [+α ( λ. )+λ ].5 [ +.49 ( ) ].679 Reduction factor: χ Φ + Φ λ χ Af Flexural buckling resistance: N b, Rd γ y 65.8 kn M Torsional buckling The elastic torsional buckling critical force: ( ) ( ) π E Iω π N cr, T G I t kn i L cr 58.4 where i is the polar radius of gyration: i i y +i z + y + z , 93

294 where y and z are the distances between the center of gravity and the shear center of the crosssection respect to the principal directions. The torsional-flexural buckling is not relevant because the cross-section is doubly syetric. The non-diensional slenderness for torsional buckling: λ T Aeff f y N cr, T 56 In EN Table 6. this type of section is not included thus c buckling curve was chosen. The iperfection factor is: α T.49. Φ T.5 ( +α T ( λ T. )+λ T ).5 (+.49 (.8. )+.8 ).56 Reduction factor: χ T Φ T + Φ λ T T Torsional buckling resistance: N b,t, Rd χ T Aeff f y kn γ M The statical syste and the noral forces shown in Fig Figure The statical syste and the noral force diagra [kn] in FEM-Design Fig shows the detailed results about flexural and torsional buckling in FEM-Design. The nuerical result are alost identical with the hand calculations. The difference is less than.%. 94

295 Figure Detailed results based on FEM-Design 95

296 Download link to the exaple file: Buckling of a doubly syetric + section.str 96

297 9.3.4 Buckling of a ono-syetric channel section Yield strength of structural steel fy 75 N/ Thickness of the parts t Cross-sectional area A 9 Inertia around strong axis Iy Inertia around weak axis Iz St. Venant torsional constant It Warping constant Iω Buckling length Lcr 6 Figure The channel section with the diensions [] and the positions of the gravity and shear centers NEd5 kn X Z Lcr 6. Y NEd5 kn X Figure The statical syste and design forces First of all we need to ake the classification of the cross section for noral force. The coefficient depending on fy: 97