Leif Otto Nielsen. Concrete Plasticity Notes BYG DTU T E K N I S K E UNIVERSITET. Undervisningsnotat BYG DTU U ISSN

Transcription

1 BYG DTU Lei Otto Nielsen Conrete Plastiit Notes DANMARKS T E K N I S K E UNIVERSITET Undervisningsnotat BYG DTU U ISSN

2 CONCRETE PLASTICITY NOTES Lei Otto Nielsen 6 de 008 CONTENTS. COULOMB PARAMETER RELATIONS. COMMENTS TO PROBLEM 8. NUMERICAL DETERMINATION OF YIELD CONDITION FOR A REINFORCED DISK MATERIAL 4. DISSIPATION IN YIELD LINE OF MODIFIED COULOMB MATERIAL 5. RIGID, PERFECTLY PLASTIC MATERIAL 6. ADDITIONS TO CP 7. EQUILIBRIUM STRESSES IN DISK BY MEANS OF CONSTANT STRESS TRIANGLES 8. NUMERICAL DETERMINATION OF YIELD LOAD FOR A REINFORCED DISK 9. ALTERNATIVE RECIPE FOR DISK REINFORCEMENT DESIGN 0. PROBLEMS a,a,6a,0a,6a,a 0,0,0,04 E06lonpr,,. ANSWERS TO SOME PROBLEMS. EQUIVALENT ORTHOTROPIC REINFORCEMENT IN A SOLID. NUMERICAL DETERMINATION OF YIELD CONDITION FOR A REINFORCED SOLID MATERIAL

3 . COULOMB PARAMETER RELATIONS Figure. Mohr s irle at sliding ailure. To obtain the prinipal stress orm o Coulomb s ield ondition, COB is projeted on CD giving ( ) ( + ) + osϕ (.) (.) is rearranged to ( + ) ( ) osϕ (.) This is multiplied b + ( + ) os ϕ osϕ( + sin ϕ ) (.) and is net divided b os ϕ giving ( + ) + (.4) os ϕ osϕ Introduing the rition parameter k ( + ) k (.5) os ϕ simpliies (.4) to (.6) k k (.6) Dividing sin ϕ + os ϕ b os ϕ gives tan ϕ +. With this and μ -epression os ϕ in CP,ig.. tan ϕ μ is obtained rom (.5) ( + ) os ϕ osϕ Epanding the μ -epression in (.7) gives k ( + tanϕ) (tanϕ + + tan ϕ ) ( μ + + μ ) (.7)

4 k μ k (.7a) Dividing (.) b ( sin ) gives + osϕ (.8) Comparing this with (.6) gives two epressions or k + osϕ k ( ) (.9) From (.9 ) is obtained + + ( ) k i.e. k (.0) In the same wa is obtained rom (.9 ) k + (.) and k + (.) Multipling (.9) with (.) gives k + k + (.) Rearranging (.) gives k k + (.4) (.9 ) and (.) gives osϕ k k ( ) k + (.5) (.4-5) gives k tanϕ k (.6) From the -epression CP(..9) is obtained using (.5) respetivel (.9 ) osϕ k ( k + )osϕ (.7) With the trigonometriall ormulas sin ± sin tan ( ± ) os + os sin ± sin ot ( ) os os (.8) where, are arbitrar angles a set o hal rition angle ormulas are easil obtained. (.8) with π, ϕ in the irst ormula and ϕ in the seond ormula gives ± 0 ϕ 0 ϕ tan(45 ± ) ot(45 ± ) 0 + osϕ 0 osϕ ϕ

5 and then the ratios + 0 ϕ tan (45 + ) Inserting (.9) in (.9 ) gives ϕ tan + 0 ϕ (45 ϕ 0 0 k tan (45 + ) tan (45 ) k i.e. 0 ϕ 0 ϕ k os (45 + ) os (45 ) k + k + 0 ϕ k 0 ϕ sin (45 + ) sin (45 ) k + k + ) (.9) (.0) (.) (.) 4

6 . COMMENTS TO PROBLEM 8 Without loal strengthening o a lateral loaded plate near a olumn support, a loalized ield line pattern at the olumn is oten essential. In the atual ase two possibilities are shown on the igure. m + p For ase b is obtained the upper bound or the ield load py.7, i.e. a smaller value than the a best value rom CPans, problem 8. Figure : Yield line patterns with loalization (CP, p505, ig 6.5.9p) at the olumn support. However, an epansion o the loal ield line pattern to a non-loal ield line pattern is more m + p dangerous here. A smaller upper bound py. is obtained or the ield line pattern a indiated on igure (positive ield line solid, negative ield line dashed). A lower bound based on a moderate number o equilibrium elements with moment degree or higher has been determined to m p py 0.7. a Figure : Yield line pattern or hal-plate non-loal at the olumn support in node. 5

7 . NUMERICAL DETERMINATION OF YIELD CONDITION FOR A REINFORCED DISK MATERIAL The determination o the ield ondition is as in CP, se... based on the lower bound theorem. The notation rom CP is used as ar as possible. Figure : Reinored disk The stresses on the disk boundar, see Figure, are written as λ (,, τ ), where λ is the load ator. The stress in a rebar in the -diretion is alled s and or a rebar in the -diretion is used s. The stress equilibrium on the disk boundar gives A A 0 s 0 s 0 λ s + λ s + λτ τ () t t The ield onditions or both onrete and rebars must not be violated 0 0 () Y s Y Y s Y () The relation between the stress omponents and the prinipal stresses in the onrete is ( + ) ± ( ( )) + τ (4) Combining (4) and () means that the greatest prinipal stress must be 0 and that the smallest must be, i.e. + ( ) + ( ( )) + τ 0 (5) ( + ) ( ( )) + τ (6) Given (,, τ ), a solution ( s, s,,, τ, λ) to (,,5,6) is obviousl a lower bound solution. Maimizing the load parameter must give the ield load λ Y ma λ λ Y (7) The onditions (,,5,6,7) deine a problem in non-linear programming. It is non-linear beause not all onditions are linear ((5,6) are non-linear). I (5,6) are linearized a problem in linear programming is obtained. The linearization o (5) is now desribed. Introduing the auiliar variables b m and d deined 6

8 m ( + ) d ( ) (8) simpliies (5) to d + τ (9) m A linearization o (9), whih is a one or m 0, see Figure, is easil made. The one is approimated b n planes as auratel as wanted. Plane j ( j,... n) o these planes ontains the points A : (0,0,0), j : (os(( j ) Δθ ),sin(( j ) Δθ ),) and j : (os jδθ,sin jδθ,), where Δ θ π / n. An outward direted plane normal n is determined b n Aj j j and then is obtained the plane equation (, τ, ) n 0.. d m Figure : Linearization o one. Introduing the auiliar variable a deined b a + m (0) simpliies (6) to d + τ () a i.e. the orm (9). () is then linearized analogousl to (9). 7

9 4. DISSIPATION IN YIELD LINE OF MODIFIED COULOMB MATERIAL The notation rom CP,p60-6,65-66 is used as ar as possible and will not be redeined. Figure : Yield line. Figure : Modiied Coulomb material. A ield line is shown on Figure and we have the relations onerning the displaements in the ield line un u n u sinα u t u osα tan α () ut The urved part o the modiied Coulomb ield ondition is onsidered Figure, i.e. π v < ϕ. For a stress point (, τ ) on this part, the low rule (normalit ondition) determines the ratio between the displaements in the ield line u, u ) u ( n t t tan v () un Comparing () and () gives π v α () i.e. the urved part o the modiied Coulomb ield ondition is ative i ϕ < α < π ϕ (4) From Figure is seen + (, τ ) (,0) + (os v,sin v) (5) 8

10 The internal work per length unit o the ield line (and one length unit perpendiular to the Figure plane) is obviousl W u n + τu t (6) Inserting (5) and () in this gives + W ( + os v) u sinα + (sin v) u osα whih is urther redued b () + W u sinα + u (7) With the, -irle through the points D m and (, τ ) ( t,0), is obtained t and then k t giving kt. With these epressions or, we get + t ( t + kt ) ( ( k + )) m (8) where last equal sign deines m. Moreover we get t ( t kt + ) ( ( k )) l (9) where last equal sign deines l. Now (7) with (8-9) inserted gives W u( msin l α + ) (0) (0,8,9) are idential with CP(.4.86,85,84) used with thikness b. In at (0) also determines the dissipation outside the urved part o the modiied Coulomb ield ondition. Obviousl the dissipation in point D m an be used. Using (0) or α ϕ + or α π ϕ gives the limit value t t W u( ( ( k + ))sin + ( k ϕ )) and with CPnotes,(.0-) t W u( + ( )) u( ) () Inserting CPnotes,(.7) in () gives W u osϕ () idential with CP(.4.77) used with thikness b. 9

11 5. RIGID, PERFECTLY PLASTIC MATERIAL 0

12 6. ADDITIONS TO CP p, line4b: I < 0, reinorement -> Reinorement p4, line6t: see below -> see below or see (..49)

13 7. EQUILIBRIUM STRESSES IN DISK BY MEANS OF CONSTANT STRESS TRIANGLES In order to obtain a lower bound solution or a rigid-plasti reinored disk, an equilibrium stress ield in the disk is needed. The disk onsidered here is onl loaded b boundar load. How to obtain equilibrium stresses is eplained b eample. The disk shown in Figure is onsidered. The notation deined in CP is used as ar as possible. Figure : Disk with triangle mesh. Utilizing smmetr the hal disk is partitioned in three triangles,, as shown on Figure. In eah triangle is used a et unknown onstant stress ield,, τ, i.e. the equilibrium onditions are satisied inside eah triangle. With these onstant triangle stresses, the stresses on the triangle boundaries must be onstant too, i.e. the give a resultant ore vetor F, F in the midpoint o eah triangle edge. In Figure a is shown the ores rom the boundar and smmetr onditions. Net the equilibrium onditions are utilized or eah triangle (projetion, projetion and moment) and between neighboring triangles (projetion, projetion ) to determine as ar as possible the still unknown ores on the triangle edges. In the atual eample these are statiall determined. For triangle is introdued the notation t and the projetion equations determine the unknown ores on the edge - as shown on Figure b. The moment equation is ulilled. For triangle is introdued the notation and the projetion equations determine the unknown ores on the edge - as shown on Figure b. The moment equation is ulilled. Triangle is now loaded b ores rom triangle and as shown on Figure b. For triangle the -projetion equation is ulilled, while the -projetion equation and the moment equation (here about the midpoint o edge -) give

14 ( h t t () t 0 ) 0 pl L h t t ( h 0 ) t () 4 Figure : Fores on triangle edges. Most o the stresses in the three triangles are determined b uniorm distribution o the ores on the ais parallel triangle edges. However, in triangle annot be determined this wa so a ut is L L L 0 made in triangle as shown in Figure b and -projetion then gives t p t + p t in h triangle. In total we have :,, τ ) (,0,0) () ( t p0 pl : (,, τ ) (0, p +, ) (4) h h : (,, τ ) (, p,0) (5)

15 8. NUMERICAL DETERMINATION OF YIELD LOAD FOR A REINFORCED DISK Optimization ormulation based on onstant stress triangles. Notation as in CP with the etension that the rebar ield stress Y ma be given dierent values in the,-diretion and in tension respetivel ompression. With onstant stresses in eah triangle Fig., the interior equilibrium is satisied in eah triangle or zero bod loads. Fig.. Equilibrium or eah intertriangle edge Fig. with load shape nij nji IJ IJ t and load ator λ t t + λt 0 (8.) nji (triangle boundar edge inluded: triangle J vanish and then t reations (0 i none) ). Stress vetor (trations) stress tensor relation in inde notation and with summation onvention: ( n) i t jinj (8.) (8.) and (8.) ombined with ji ij > I J I J IJ ( ) n + ( τ τ) n λt (8.) I J I J IJ ( τ τ) n + ( ) n λt In eah triangle: A A s s s + s t + t τ τ (8.4) (8.5) Y s Y Y s Y ( ) τ ( + ) + ( ) + 0 ( ) ( ) ( ) + + τ Lower bound theorem > ma λ λy (8.7) Utilizing (8.) or all element edges in the struture to be analzed, (8.4)-(8.6) or all elements and (8.7) gives a nonlinear optimization (programming) problem, whih is onve. It an be handled as a seond order one optimization problem, whih an be solved b Mosek (see I (8.6) is linearized a linear optimization problem is obtained and the simple method an be used or its solution. (8.6) 4

16 To get a load ator λ lose to the ield load λ Y, the meshing with onstant stress triangles o the struture to be analzed must allow a stress distribution in the meshed struture lose to one o the stress distributions, whih an eist in the original struture at the ield load. 9. ALTERNATIVE RECIPE FOR DISK REINFORCEMENT DESIGN Compute t + τ t + τ τ () Ok i t 0 and t 0. I not ompute d τ () I d 0 no reinorement is needed and is determined b the smallest prinipal stress, i.e. t 0 t 0 ) ( + ) 4 ( + τ () Otherwise ( d < 0) the negative value o t or t is hanged to zero and the relevant ormula set below is used τ τ t 0 t + + (4) τ τ t + t 0 + (5) 5

17 0. PROBLEMS Problem 6a ompared with problem 6 is modiied suh that the answers to problem 6 are relevant also or problem 6a. This priniple has been ollowed or all problems with a number ollowed b a. Problems with numbers above 00 are not ound in CPpr and CPans. Problem a Modiiations ompared with CPpr, problem : Plane stress or plane strain an be onsidered (depends on the bar thikness). The ield planes are perpendiular to the igure plane. Moreover the question has been detailed. Case a. Question : Determine the internal work in the ield line pattern using the, τ orm o the Coulomb ield ondition. + Question : Determine an upper bound p Y or the ield load p Y o p. Question : Optimise the upper bound solution in regard to the ield line pattern. Show that the obtained value equals the ompressive strength o the Coulomb material and spei the value o angle β or the rition parameter value k 4 ( tpial value or onrete). Case b. Question 4: Determine the ratio between u and u using the low rule. Question 5: Determine an upper bound or the ield load, optimise it and ompare it with the value obtained in ase a. 6

18 Problem a π Modiiations ompared with CPpr, problem : The disk thikness is t and α. Moreover the question has been modiied. Question : Determine, b means o the work equation, an upper bound or v. Question : Minimize v and ompare the result with (..0) in CP. Has the ield load been determined? 7

19 Problem 6a Modiiations ompared with CPpr, problem 6: The irst question has been detailed. Question 0: Determine the stresses in a oordinate sstem with aes in the reinorement diretions. 8

20 Problem 0a Modiiations ompared with CPpr, problem 0: Question 0 has been inserted. Question 0: a) Based on a onstant stress triangle mesh is wanted a set o equilibrium ores on the triangle edges (in agreement to CP, p58 uniorm normal stress distribution should be assumed on the olumn ross setion in a depth below the line load equal to the relevant olumn side length (here 40 m)). b) Determine the neessar reinorement area in the smmetr plane to arr the tension (result as in CPans, pr0, q). ) Determine the ompression zone depth in the smmetr plane (result as in CPans, pr0, q). 9

21 Problem 6a Modiiations ompared with CPpr, problem 6: The questions have been detailed and the igure orreted. Question : Determine Question : Determine p pi utilizing CPans, pr. p piii. Question : Determine l I and l III onsidering a strip in the -diretion. Question 4: Determine the neessar ield moment in the -diretion in area II. 0

22 Problem a Modiiations ompared with CPpr, problem : The igure and the question have been modiied. Question Determine an upper bound solution or the load arring apait o the beam based on the ield 0 line pattern shown on the igure with β 60 and displaements orresponding to the translation mode with α 0. Compare the result with CPans and omment the dierene. Question Determine an upper bound solution or the load arring apait o the beam based on the ield line pattern shown on the igure and displaements orresponding to the rotation mode speiied b the rotation η about the upper side load point o the let beam part.

23 Problem 0 A vertial retangular h * b unreinored onrete disk ( in-plane loaded plate) ABCD - see the igure - o thikness t is loaded b a rather onentrated vertial load (load resultant P) uniorml distributed over the retangular area a * t. The onrete is modelled as a rigid-plasti material based on the modiied Coulomb ield ondition with 0. t Question + Determine an upper bound P Y or the ield load PY o P based on the indiated ield line pattern with vertial displaement u o the wedge part o the disk and horizontal displaement u u o the other movable disk parts.

24 Problem 0 An orthogonal reinored onrete disk is onsidered. The onrete is modelled as a rigid-plasti material based on the modiied Coulomb ield ondition with tension strength t 0. A rebar is modelled as a uniaial rigid-plasti material with ield stress Y in both tension and ompression. The MatLab untion YsReDisk determines a point on the ield surae used in the disk, while the MatLab sript eeysredisk applies YsReDisk to determine a setion in the ield surae. Question Appl YsReDisk to determine the point (,, τ ) (0,0, τ ) on the ield surae used in a disk with Φ Φ 0. and test the result omparing with CP. Question Appl eeysredisk to determine the setion 0 o the ield surae. Consider the ases a) Φ Φ 0. b) Φ Φ 0. 6 ) Φ 0.6 Φ 0. and test the results as ar as possible omparing with CP. Problem 0 A parallelogram shaped reinored onrete disk ABCD o thikness t, see the igure, is reinored parallel with the edges. The reinorement in eah diretion has the area A per unit length. The ield stress o the reinorement in one diretion is Y and in the other diretion. Question Determine the diretions and the areas per unit length o the equivalent orthogonal reinorement with ield stress Y. Y

25 Problem 04 On the igure is shown the isotropi reinored onrete disks d), g), h) o thikness t. As ield ondition or the disk material is used CP (..7). The load is a pressure load p > 0 per area unit in equilibrium with the reations (the reation pressure is indiated b q on d) and h)). Sel-weight is negleted. For eah disk is wanted: Question Determine an equilibrium stress ield based on a onstant stress triangle mesh. Question For L a h, 0.h, Φ 0. determine a lower bound solution or the load arring apait and determine the deviation rom the eat solution. Question Determine an upper bound solution or the load arring apait assuming a bending tpe mehanism in the smmetr setions and determine the deviation rom the eat solution. 4

26 E06, LONpr (40% o 4 hours eam) A onrete disk ABC, see the igure, o thikness t is reinored homogeneous and orthotropi in the, oordinate diretions. The onrete and the reinorement are modelled as rigid, peretl plasti materials. The onrete is desribed as a modiied Coulomb material with tension strength t 0. The reinorement strength in ompression is negleted. The onrete has the eetive ompressive design strength ν d 5MPa and the reinorement has the tensile design strength d 400MPa. The minimum reinorement ratio is set to r min The load is a pressure load p > 0 per area unit. Sel-weight is negleted. Question Based on a onstant stress triangle mesh with triangles ABD and BCD, determine a set o equilibrium stresses in the disk. Question For h a set o equilibrium stresses in the disk is determined b (,, ) ( 4 τ p,,0) in ABD and (,, τ ) 4 p(,, ) in BCD. For p MPa, h 800mm, t 50mm determine the neessar tensile strengths t, t and investigate i the ompression stress in the onrete is sae in eah onstant stress triangle. Question With a straight ield line AE, see the igure, and a displaement u > 0 with an angle α 0 with the ield line o disk part ABE (no displaement o disk part AEC) is wanted an upper bound solution 0 0 or the load arring apait o the disk or β 45, α 0, 4MPa, 0. t t 5

27 E06, LONpr (40% o 4 hours eam) A onrete disk ABCD, see the igure, o thikness t is reinored homogeneous and orthotropi in the, oordinate diretions. Rebars in the -diretion ross the supporting setion AD and are anhored in the support. The onrete and the reinorement are modelled as rigid, peretl plasti materials. The onrete is desribed as a modiied Coulomb material with tension strength t 0. The reinorement strength in ompression is negleted. The onrete has the eetive ompressive design strength ν d 5MPa and the reinorement has the tensile design strength d 00MPa. The minimum reinorement ratio is set to r min The load is a pressure load p > 0 per area unit. Sel-weight is negleted. Question Based on a onstant stress triangle mesh with triangles ABC, ACE and CDE, determine a set o equilibrium stresses in the disk. Question A set o equilibrium stresses in the disk is determined b (,, τ ) p(0,,0) in ABC, (,, ) 4 τ p,, τ ) p(4, 5, ) in ACE and (,, ) in CDE. For ( 6 p 5 MPa, a 900mm, t 00mm, determine the neessar tensile strengths t, t and investigate i the ompression stress in the onrete is sae in eah onstant stress triangle. Question For t 4MPa determine an upper bound solution or the load arring apait o the disk based on a rotation η > 0 o the disk about point E see the igure. 6

28 E06, LONpr (40% o 4 hours eam) A onrete disk ABCDEF, see the igure, o thikness t is reinored homogeneous and orthotropi in the, oordinate diretions. Rebars in the -diretion ross the supporting setion CDEF and are anhored in the support. The onrete and the reinorement are modelled as rigid, peretl plasti materials. The onrete is desribed as a modiied Coulomb material with tension strength t 0. The reinorement strength in ompression is negleted. The onrete has the eetive ompressive design strength ν d 0MPa and the reinorement has the tensile design strength d 400MPa. The minimum reinorement ratio is set to r min The load onsists o a onstant shear stress p > 0 per area unit along AB. Sel-weight is negleted. Question Based on a onstant stress triangle mesh with 4 triangles AEF, ABE, BDE and BCD with zero stresses in triangle BCD, determine a set o equilibrium stresses in the disk. Question Smmetrization o the question solution determines a more eiient set o equilibrium stresses in the disk. In AEF is obtained (,, ) ( τ p 4,, ) and in BCD (,, ) ( τ p 4,, ). For eah o these two triangles and or p 6 MPa, a 000mm, t 00mm, determine the neessar tensile strengths, and investigate i the ompression stress in the onrete is sae. t t Question With a straight ield line AG, see the igure, and a translation u > 0 with an angle α 0 with the ield line o disk part ABG is wanted an upper bound solution or the load arring apait o the 0 disk or α β 0 and MPa in ABCDE. t t 6 7

29 .ANSWERS TO SOME PROBLEMS + 5 Prob 0. q: PY t( a + b) + Prob 0. q: θ 5 0, A ξ A A. 8, A η A A p ph E06,LONpr. q: ABD: (,, τ ) (, p,0), BCD: (,, τ ) (,, ) ; q: / h (beore onsidering min reinorement) ABD: 4MPa, 0, MPa( sae), 0 t t + BCD: 0, 0, 4MPa( sae) ; q: p Y 4. 5MPa t t E06,LONpr. q: ABC: (,, τ ) (0, p,0), ACE:,, τ ) ( p, 5 p, p ), 4 (,, ) ( 4, 4 p p, p τ t 0, t 0, 5MPa( sae 0 ( 6 CDE: ) ; q: (beore onsidering min reinorement) ABC: ), ACE: 4Mpa, 0, 4.8MPa( sae), t t + CDE: 0, 0,.MPa( sae) ; q: p Y 7. 75MPa t t (,, ) ( p, p, p τ E06,LONpr. q: AEF: ), ABE:,, τ ) ( p,0, p), ( BDE: (,, τ ) (0, p,0) ; q: (beore onsidering min reinorement) AEF: 4.5MPa, 9MPa, 6MPa( sae), BCD: 0, 0, 7.5MPa( sae) ; q: t t + p Y 4. 6MPa t t 8

30 . EQUIVALENT ORTHOTROPIC REINFORCEMENT IN A SOLID In CP se... is onsidered the determination o the orthotropi reinorement in a disk, whih - onerning the tension ield load - is equivalent to a given arbitrar reinorement in the disk. Here is onsidered the same problem or a three dimensional solid. The notation rom CP is used as ar as possible. Figure : Reinorement in diretion i. The solid is reinored in m arbitrar diretions. In diretion ii,,... mwith unit diretion () i vetor n see Fig., the reinorement has the ross setion area A si per unit area o the solid and the tensile ield stress. Y We onsider now the ase where all reinorement ields in tension. First step is to determine the equivalent stress on a setion with outward direted unit normal vetor n. From reinorement i is ni () () i () i obtained the stress vetor ontribution t A ( ni n ) n, i.e. or the total reinorement we get m m n n() i () i () i YAsi( ) i i Y si t t n i n n () Net we want to determine those setions i an, where there are onl normal stresses and no n shear stresses on the setion. The ondition is obviousl t λn, where λ is a et unknown onstant. Combining this with () gives m () i () i YAsi( nn i ) n λn () i whih as we shall see below is an eigenvalue problem or determination o n,λ. () () () () Using the omponent orms n ( n, n, nz) and n i ( n i, i, i n nz ) the summation in () is turned into m m () i () i () i () i () i () i YAsi( nn i ) n YAsi( nn + nn + nznz ) n i i m m m () i () i () i () i () i () i n YAn si n + n YAn si n + nz YAn si z n i i i 9

31 and then () an be written as m m m () i () i () i () i () i () i YAsin n YAsin n YAsinz n i i i n m m m n () i () i () i () i () i () i YAsin n YAsin n YAsinz n n λ n () i i i m m m () i () i () i () i () i () i n z n z YAsin nz YAsin nz YAsinz nz i i i The oeiient matri in () is obviousl a smmetri matri. Then the eigenvalue problem () has real solutions λ, λ, λ with the orthogonal eigenvetors n,n,n. Then in onlusion: the equivalent orthotropi reinorement with ield stress must go in the diretions n Y,n,n and have strengths deined b YAe λ YAe λ YAe λ (4) where Ae, Ae, A e are the ross setional areas o the equivalent orthotropi reinorement in diretion,,. Eample For non-orthogonal reinorement as shown on Fig. with ield stress Y, determine the equivalent orthotropi reinorement. Answer: The unit diretion vetors o the rebars are Figure : Non-orthogonal reinorement. 0

32 n () [ 0 0] () n 0 () n 4 4 Then () gives n n Y A 0 n λ n nz nz sm 4 The solution o this eigenvalue problem is here obtained b the Matlab omputation S[+/4+*/6 0+sqrt()/4+*sqrt()/6 0+0+*/8 0 0+/4+/6 0+0+*sqrt()/ */4]; % smmetrize S(,)S(,); S(,)S(,); S(,)S(,); % solve [V,D]eig(S) V e e e e e e e e e-00 D e e e+000 Then the equivalent orthotropi reinorement is determined b () e YAe D(,) YA in diretion n V (:,) ( e) YAe D(,) YA in diretion n V (:, ) ( e) YAe D(,) YA in diretion n V (:,) Test: For test is utilized the smmetr o the reinorement as indiated on Fig.. Then one o the equivalent rebar diretions must be normal to the smmetr plane i.e. 0 0 ( sin 0,os0,0) ( 0.500,0.866,0) orresponding to V (:,). The area o the equivalent reinorement in this diretion has onl ontributions rom reinorement and and beause their angle with this diretion is , the ontribute to the strength with YA(os60 ) YA orresponding to D (,). Another test is Ae + Ae + Ae As + As + As. As both sides gives 4A, this test is satisied.

33 . NUMERICAL DETERMINATION OF YIELD CONDITION FOR A REINFORCED SOLID MATERIAL The determination o the ield ondition is as in CP, se... based on the lower bound theorem. The notation rom CP is used as ar as possible. The usual oordinate sstem, z, on Fig. has aes in the (equivalent) orthotropi reinorement diretions. Figure : Reinored solid The stresses on the boundar o the solid are written as λ(,, z, τ z, τz, τ ), where λ is the load ator. The stress in a rebar in the -diretion is alled s, or a rebar in the -diretion is used s and or a rebar in the z-diretion sz. The stress equilibrium on the boundar o the solid gives λ A + λ A + λ A + () s s s s z τz λτz τz λτ τ z sz sz z λτ () where A s is the ross setion area o the rebars in the -diretion per unit area o the solid. As, A sz are deined analogousl reerring to the, z -diretions. The ield onditions or both onrete and rebars must not be violated. For the rebars we have Y s Y Y s Y Yz sz Yz () where Y, Y, Yz are the ield stresses o the rebar material in the, z-diretions., Modiied Coulomb ormulation For the onrete is here used the modiied Coulomb ield ondition rom CP, i.e. when the prinipal onrete stresses are ordered we have k A (4) The relation between the prinipal onrete stresses and the onrete stress omponents are obtained via the stress invariants deined b I + + z z z z z I + + τ τ τ (5)

34 I z + τzτzτ τz τz zτ Appling (5) in the prinipal oordinate sstem or onrete stresses gives I + + I + + I Inserting these epressions in (5) gives the wanted relations z z z z z z + z z z z z τ τ τ (6) τ τ τ τ τ τ With (6) the prinipal onrete stresses,, are not ordered as (4) presupposes. Then (6) must be ombined with the ield ondition on the orm k k k k k k (7) A A A O ourse (7) an be replaed b the more simple ormulation k A (7a) Given (,,, τ, τ, τ ), a solution (,,,,,, τ, τ, τ, λ ) to (-,6-7) is z z z s s sz z z z obviousl a lower bound solution. Maimizing the load parameter must give the ield load λ Y ma λ λ Y (8) The onditions (-,6-8) deine a problem in non-linear onve optimization. It is non-linear beause not all onditions are linear ((6 - ) are non-linear). It is onve as a peret plastiit problem. As a non-linear onve optimization problem it ma be preerable rom a omputational point o view to use a smooth ield ondition. This possibilit is onsidered below. Smoothed Mohr-Coulomb ormulation Here is onsidered the same problem as above, but the modiied Coulomb ield ondition is ehanged with a smoothed Mohr-Coulomb ield ondition. This ield ondition ( I, I, I ), whih an be losed, is epressed b a set o stress invariants γ A ki + ( I ) I + ( ) ( k 9) 0 (9) s The stress invariants I, I, I are deined rom the shited non-dimensional stresses τ z z τ τ z,,, τ, τ, τ, (0) s s z s z i.e. I + + z I + z + z τ z τ z τ () I z + τ zτ zτ τ z τ z zτ The ield ondition (9) ontains our material onstants: k > 9 is the rition parameter, γ the losing parameter (not losed or γ ), s > 0 the shit stress and A the separation resistane. A urther disussion o this ield ondition is ound in the notes LON: Computational plastiit. s z s s

35 Inserting (0) in () and the obtained result in (9) epress the ield ondition b the stress omponents, i.e. (,, z, τz, τz, τ ) 0 () thus avoiding invariant epressions in the optimization ormulation Given (,, z, τ z, τz, τ ), a solution ( s, s, sz,,, z, τz, τz, τ, λ ) to (-,) is obviousl a lower bound solution. Maimizing the load parameter must give the ield load λ Y. The onditions (-,8,) deines a problem in non-linear onve optimization. Numerial solution On Figure is shown a setion in the τ -plane o the ield ondition or the reinored solid. Both the modiied Coulomb (mc) and the smoothed Mohr-Coulomb (smc) has been used or the s Y solid. The reinorement is isotropi with reinorement degrees Φ Φ Φ z 0. ( Φ et.). The mc-parameters are k 4, A 0.. For smc with γ is used the equivalent parameters k.5, s, A 0.. With the applied solution untion MATLAB s minon, it is diiult to obtain a solution or mc and the urve irregularities indiate onvergene problems. Contraril it is eas to get a solution with the smooth smc using a zero solution vetor as starting point. Compared with a disk solution minor three-dimensional eets are seen orresponding to utilization o the reinorement in the,z-diretions in tension improving the ompression state o the solid. A Figure : Setion in ield suraes or reinored solid. 4

36 Modiied Coulomb positive semideinite ormulation The Coulomb ield ondition ( 4 ) an be written on positive semideinite orm Krabbenhøt et al, Int. J. Solids Strut. 44 (007) This is interesting rom a omputational point o view, beause eiient numerial solution methods eist or positive semideinite optimization Sturm, Optimization Methods and Sotware - (999) with SeDuMi and Löberg, Proeedings o the CACSD Conerene, 004 with YALMIP. Below is made a positive semideinite ormulation o the modiied Coulomb ield ondition ( 4 ). This is inluded in the ormulation o the ield ondition or a reinored onrete solid. Finall a setion in the ield surae is determined numeriall b positive semideinite optimization using the lower bound theorem. We are working in the real number domain. A positive semideinite smmetri matri A satisies T A 0 or arbitrar vetors () This is written as A 0 (4) With this notation the Coulomb ondition is written as + kαi 0 (5) + ( α) I 0 k where is the stress tensor or the onrete τ τ z τ τz (6) τz τz z with prinipal stresses, α an auiliar variable and I a unit matri 0 0 I 0 0 (7) 0 0 In order to prove (5) this is obviousl equivalent to + kα 0 + ( α) 0 k Net α is isolated (Dines method) giving α + k k i.e. + k k or k whih is the Coulomb ondition ( 4). With ( 4 ) written as + AI 0, the modiied Coulomb ield ondition ( 4 ) on positive semideinite orm is 5

37 + AI 0 + kαi 0 (8) + ( α) I 0 k Given (,, z, τ z, τz, τ ), a solution ( s, s, sz,,, z, τz, τz, τ, λ, α ) to (-,8) is obviousl a lower bound solution. Maimizing the load parameter must give the ield load λ Y ma λ λ Y (9) The onditions (-,8-9) deine a problem in positive semideinite optimization. The numeriall determined solution on Figure shows a setion in the τ -plane o the ield ondition or the reinored solid or the same data as used or Figure and now without the omputational problems mentioned in onnetion with Figure or mc. Figure : Setion in ield surae or reinored solid obtained b positive semideinite optimization. The onditions (8) an be simpliied somewhat as proposed b Löberg. First (8, ) with α named α are written AI and ( α) I k and then olleted to min A, α I αi (0) k where last equal sign deines the salar variable α. Now (0) is rewritten 6

38 αi α A () α α k Using () the alternative ormulation o (8) is obviousl + αi 0 + kαi 0 α () A α α k i.e. ompared with (8) the number o semideinite onditions has been redued b, while etra salar inequalities have been inluded. O ourse replaing (8) with () still gives the ield surae setion shown on ig.. 7