Structural Steelwork Eurocodes Development of A Trans-national Approach

Size: px
Start display at page:

Download "Structural Steelwork Eurocodes Development of A Trans-national Approach"

Transcription

1 Structural Steelwork Eurocodes Development of A Trans-national Approach Course: Eurocode 3 Module 7 : Worked Examples Lecture 20 : Simple braced frame Contents: 1. Simple Braced Frame 1.1 Characteristic Loads 1.2 Design Loads F d = γ F F k 1.3 Partial Safety Factors for Strength 2. Floor Beam - Fully Restrained 2.1 Classification of Cross-section 2.1.1Flange buckling Web buckling 2.2 Shear on Web 2.3 Deflection Check 2.4 Additional Checks if Section is on Seating Cleats 2.5 Crushing Resistance 2.6 Crippling Resistance 2.7 Buckling Resistance 2.8 Summary 3. Roof Beam Restrained at Load Points 3.1 Initial Section Selection 3.2 Classification of Cross Section Flange buckling Web buckling 3.3 Design Buckling Resistance Moment 3.4 Shear on Web 3.5 Deflection Check 3.6 Crushing, Crippling and Buckling 3.7 Summary 4. Internal Column 4.1 Loadings 4.2 Section properties SSEDTA 2001

2 4.3 Classification of Cross-Section Flange (subject to compression) Web (subject to compression) 4.4 Resistance of Cross-Section 4.5 Buckling Resistance of Member 4.6 Determination of Reduction Factor χy 4.7 Determination of Reduction Factor χz 5. External Column 5.1 Loadings 5.2 Section properties 5.3 Classifcation of Cross-Section Flange (subject to compression) Web (subject to compression) 5.4 Resistance of Cross-Section 5.5 Buckling Resistance of Member 5.6 Determination of Reduction factor χy 5.7 Determination of Reduction factor χz 6. Design of Cross-Bracing 6.1 Section Properties 6.2 Classification of Cross-Section 6.3 Design of Compression Member Resistance of Cross-section Design Buckling Resistance Determination of Reduction Factor χ? 6.4 Design of Tension Member 6.4.1Resistance of Cross-Section 7. Concluding Summary 23/02/07 2

3 1. Simple Braced Frame The frame consists of two storeys and two bays. The frames are at 5 m spacing. The beam span is 7,2 m. The height from column foot to the beam at floor level is 4,5 m and the height from floor to roof is 4,2 m. It is assumed that the column foot is pinned at the foundation. Roof Beam External Column Internal Column 4,2 m Floor Beam 4,5 m 7,2 m 7,2 m Figure 1 Typical Cross Section of Frame It is assumed that resistance to lateral wind loads is provided by a system of localised cross-bracing, and that the main steel frame is designed to support gravity loads only. The connections are designed to transmit vertical shear, and it is also (2) assumed that the connections offer little, if any, resistance to free rotation of the beam ends With these assumptions, the frame is classified as simple, and the internal forces and moments are determined using a global analysis which assumes the members to be effectively pin-connected. 1.1 Characteristic Loads Floor: Variable load, Q k = 3,5 kn/m 2 Permanent load, G k = 8,11 kn/m 2 Roof: Variable load, Q k = 0,75 kn/m 2 Permanent load, G k = 7,17 kn/m Design Loads F d = γ F F k (1) Floor: G d = γ G G k. At ultimate limit state γ G = 1,35 (unfavourable) G d = 1,35 x 8,11 = 10,95 kn/m 2 Q d = γ Q Q k. At ultimate limit state γ Q = 1,5 (unfavourable) Q d = 1,5 x 3,5 = 5,25 kn/m (2) (2) /02/07 3

4 Roof: G d = γ G G k. At ultimate limit state γ G = 1,35 (unfavourable) G d = 1,35 x 7,17 = 9,68 kn/m 2 Q d = γ Q Q k. At ultimate limit state γ Q = 1,5 (unfavourable) Q d = 1,5 x 0,75 = 1,125 kn/m 2 The steel grade selected for beams, columns and joints is Fe360. (f y = 235 N/mm 2 ) 1.3 Partial Safety Factors for Strength The following partial safety factors for strength have been adopted during the design: Resistance of Class 1,2 or 3 cross-section, γ M0 = 1,1 Resistance of member to buckling, γ M1 = 1,1 Resistance of bolted connections, γ Mb = 1,25 The following load case, corresponding to permanent and variable actions (no horizontal loads) is found to be critical. 2. Floor Beam - Fully Restrained The beam shown in Figure 2 is simply supported at both ends and is fully restrained along its length. For the loading shown, design the beam in grade Fe360, assuming that it is carrying plaster, or a similar brittle finish. F d = γ G G k + γ Q Q k Design load, F d = (5 x 1,35 x 8,11) + (5 x 1,5 x 3,5) = 81 kn/m 81 kn/m (2) (2) (1) 5.1.1(2) 5.1.1(2) 6.1.1(2) 2.1 7,2 m Figure 2 Loading on Fully Restrained Floor Beam 2 FL d Design moment, MSd = 8 Where M Sd is the design moment in beam span, F d is the design load = 81 kn/m, and L is the beam span = 7,2m. 2 81x7,2 MSd = = 525 knm 8 FL d 81x7,2 Design shear force, VSd = = = 292 kn 2 2 To determine the section size it is assumed that the flange thickness is less than 40 mm so that the design strength is 235 N/mm 2, and that the section is class 1 or /02/07 4

5 The design bending moment, M Sd, must be less than or equal to the design moment resistance of the cross section, M c.rd : M Sd M c.rd M c.rd = M pl.y.rd = Wf pl y γ M0 Where W pl is the plastic section modulus (to be determined), f y is the yield strength = 235 N/mm 2, and γ M0 is the partial material safety factor = 1,1. Therefore, rearranging: 3 Msdγ M0 525x10 x1,1 3 Wpl.required = = = 2457 cm f 235 Try IPE 550 Section properties: Depth, h = 550 mm, Web thickness, t w = 11,1 mm y Width, b = 210 mm Flange thickness, t f = 17,2 mm Plastic modulus, W pl = 2787 cm 3 This notation conforms with Figure 1.1 in Eurocode 3: Part Classification of Cross-section As a simply supported beam is not required to have any plastic rotation capacity (only one hinge required), it is sufficient to ensure that the section is at least class 2 to develop the plastic moment resistance. Figure 3 shows a typical cross-section for an IPE. IPE sections have been used in this example to reflect the European nature of the training pack (1) (2) and t t w f c d Figure 3 A Typical Cross-Section Flange buckling Class 1 limiting value of c/t f for an outstand of a rolled section is 10ε. ε = 235 / f y and f y = 235 N/mm 2, therefore ε =1. Calculate the ratio c, where c is half t f the width of the flange = 105 mm, and t f is the flange thickness = 17,2 mm (if the flange is tapered, t f should be taken as the average thickness). c 105 = = 6,10 t 17,2 f (Sheet 3) 23/02/07 5

6 2.1.2 Web buckling Class 1 limiting value of d/t w for a web subject to bending is 72ε. ε = 235 / f y and f y = 235 N/mm 2, therefore ε =1. Calculate the ratio d, where d is the depth between root radii = 467,6 t w mm and t w is the web thickness = 11,1 mm. d 467,6 = = 42,1 t w 11,1 c < 10ε and d < 72ε t t f w Section is Class 1 and is capable of developing plastic moment. 2.2 Shear on Web The shear resistance of the web must be checked. The design shear force, V Sd, must be less than or equal to the design plastic shear resistance, V pl.rd : (Sheet 1) (Sheets 1 and 3) V Sd V pl.rd Where V pl.rd is given by A f / 3 y v γ M0 For rolled I and H sections loaded parallel to the web, Shear area, A v = 1,04 h t w, f y is the yield strength = 235 N/mm 2, and γ M0 is the partial material safety factor = 1, (4) (2) 1,04ht f Vpl.Rd = 3xγ w y M0 1,04 x 550 x 11,1 x 235 = = 783 kn 3 3 x 1,1x10 This is greater than the shear on the section (292 kn). The shear on the beam web is OK. If the beam has partial depth end-plates, a local shear check is required on the web of the beam where it is welded to the end-plate. V A f / 3 y pl.rd = v γ M0 where A v = t w d, and d is the depth of end-plate = (for example) 300 mm. 11,1 x 300 x 235 Vpl.Rd = = 411 kn 3 3 x 1,1 x 10 This is greater than the shear on the section (292 kn). The local shear on the beam web is OK. 23/02/07 6

7 Other simple joints may be used instead, e.g. web cleat joints or fin plate joints. A further check is sometimes required, especially when there are significant point loads, cantilevers or continuity, to ensure that the shear will not have a significant effect on the moment resistance. This check is carried out for the moment and shear at the same point. The moment resistance of the web is reduced if the shear is more than 50% of the shear resistance of the section. With a uniform load, the maximum moment and shear are not coincident and this check is not required for beams without web openings. 2.3 Deflection Check Eurocode 3 requires that the deflections of the beam be checked under the following serviceability loading conditions: Variable actions, and Permanent and variable actions. Figure 4 shows the vertical deflections to be considered (3) 4.2 δ δ 1 2 δ δ 0 max L Figure 4 Vertical Deflections δ 0 is the pre-camber (if present), δ 1 is the deflection due to permanent actions, δ 2 is the deflection caused by variable loading, and δ max is the sagging in the final state relative to the straight line joining the supports. For a plaster or similar brittle finish, the deflection limits are L/250 for δ max and L/350 for δ 2. Deflection checks are based on the serviceability loading. For a uniform load δ = x FL 3 k EIy where F k is the total load = Q k or (G k + Q k ) as appropriate, L is the span = 7,2 m, E is the modulus of elasticity ( N/mm 2 ), and I y is the second moment of area about the major axis = x 10 4 mm Figure /02/07 7

8 For permanent actions, F k = 5 x 8,11 x 7,2 = 292 kn. Therefore, deflection due to permanent actions, x 292x10 x 7200 δ 1 = = 10,1 mm x x 67120x10 For variable actions, F k = 5 x 3,5 x 7,2 = 126 kn. Therefore, deflection due to variable actions, x 126x10 x 7200 δ 2 = = 4,3 mm x x 67120x10 The maximum deflection, δ max = δ 1 + δ 2 = 10,1+ 4,3 = 14,4 mm L 7200 Deflection limit for δ 2 = = = 20,6 mm The actual deflection is less than the allowable deflection: 4,3 mm < 20,6 mm L 7200 Deflection limit for δ max = = = 28,8 mm The actual deflection is less than the allowable deflection: 14,4 mm < 28,8 mm OK. The calculated deflections are less than the limits, so no pre-camber is required. It should be noted that if the structure is open to the public, there is a limit of 28 mm for the total deflection of δ 1 + δ 2 (neglecting any pre-camber) under the frequent combination, to control vibration. This is based on a single degree of freedom, lumped mass approach. For the frequent combination the variable action is multiplied by ψ, which has a value of 0,6 for offices. 2.4 Additional Checks if Section is on Seating Cleats There are cases where the beams may be supported on seating cleats, or on other materials such as concrete pads. A similar situation arises when a beam supports a concentrated load applied through the flanges. In these cases the following checks are required: Crushing of the web Crippling of the web Buckling of the web Generally, these checks need only be carried out for short beams or beams with concentrated loads. For the sake of completeness, these checks are included in this worked example. The following detailed checks are for an 85 mm stiff bearing. (The actual length of stiff bearing will depend on the detail of the connection - see Figure 5.7.2) 4.3.2(2) Lecture 3, section (2) /02/07 8

9 2.5 Crushing Resistance The crushing resistance is given by (ss + s y)twfyw Ry.Rd = γ M1 where s s is the length of stiff bearing = 85 mm, t w is the web thickness = 11,1 mm, f yw is the yield strength of the web = 235 N/mm 2, γ M1 is the partial material safety factor for buckling resistance = 1,1, and s y is the length over which the effect takes place, based on the section geometry and the longitudinal stresses in the flange. s y = 2t f (b f /t w ) 0,5 (f yf /f yw ) 0,5 [1 - (σ f.ed /f yf ) 2 ] 0,5 At the support, the stress in the beam flange, σ f.ed, is zero, f yf = f yw but the value of s y is halved at the end of the member (2) 5.7.3(1) 5.7.3(3) 0,5 2 x 17,2 x (210 / 11,1) sy = = 75 mm 2 ( ) x 11,1 x 235 Crushing resistance, R y.rd = = 379,4 kn 3 1,1x10 This is greater than the reaction (292 kn). The crushing resistance is OK 2.6 Crippling Resistance The crippling resistance is given by 5.7.4(1) R a.rd = 2 w 0,5 0,5 0,5t (Ef ) [(t / t ) + 3(t / t )(s / d)] yw f γ w M1 w f s where t w is the thickness of the web = 11,1 mm, E is the modulus of elasticity = N/mm 2, f yw is the yield strength of the web = 235 N/mm 2, t f is the flange thickness = 17,2 mm, s s is the length of stiff bearing = 85 mm, but s s is limited to a maximum of 0,2d (467,6 mm x 0,2 = 93,5 mm), d is the depth between root radii = 467,6 mm, and γ M1 is the partial material safety factor buckling resistance = 1, (1) 5.1.1(2) 2 0,5 0,5 0,5 x 11,1 ( x 235) [(17,2 / 11,1) + 3(11,1/ 17,2)(85 / 467,6)] Ra,Rd = 3 1,1x10 = 626 kn This is greater than the reaction (292 kn). The crippling resistance is OK. 23/02/07 9

10 2.7 Buckling Resistance The buckling resistance is determined by taking a length of web as a strut. The length of web is taken from Eurocode 3, which in this case, gives a length: 2 2 0,5 ss beff = 0,5(h + s s ) + a+ but [h 2 + s 2 s ] 0,5 2 where h is the height of the section = 550 mm, s s is the length of stiff bearing = 85 mm, and a is the distance to the end of the beam = 0 mm ,5 85 beff = 0,5( ) + = 320,5 mm 2 Provided that the construction is such that the top flange is held by a slab and the bottom by seating cleats, against rotation and displacement, the effective height of the web for buckling should be taken as 0,7 x distance between fillets. l = 467,6 mm x 0,7 = 327 mm t w 11,1 Radius of gyration for web, i = = = 32, l 327 Slenderness of the web, λ = = = 102 i 3,2 λ Non-dimensional slenderness of the web, λ = λ β 0,5 A 1 Where λ 1 = 93,9 ε = 93,9 x 1 = 93,9, and β A = Non - dimensional slenderness of the web, λ = = 109, 93, 9 Using buckling curve c, the value of the reduction factor, χ may be determined from Reduction factor, χ = 0,49 AAfy Buckling resistance of a compression member, Nb.Rd = χβ γ M1 A is the cross-sectional area = b eff t w, f y is the yield strength = 235 N/mm 2, and γ M1 is the partial material safety factor for buckling resistance = 1,1. 0,49 x 1 x 320,5 x 11,1 x 235 N b.rd = = 372,4 kn 3 1,1x10 This is greater than the reaction (292 kn). The buckling resistance is OK. 2.8 Summary The trial section IPE 550 is satisfactory if the section is on a stiff bearing 85 mm long. If it is supported by web cleats or welded end plates, the web checks, except for shear, are not required and the section is again satisfactory. The beam is satisfactory Figure (4) (3) (3) 5.7.5(3) and (1) (2) 23/02/07 10

11 3. Roof Beam - Restrained at Load Points The roof beam shown in Figure 5 is laterally restrained at the ends and at the points of application of load. The load is applied through purlins at 1,8 m spacings. Internal point load = 1,8 [(5 x 1,35 x 7,17) + (5 x 1,5 x 0,75)] = 97,2 kn External point load = 0,9 [(5 x 1,35 x 7,17) + (5 x 1,5 x 0,75)] = 48,6 kn It is assumed that the external point loads will be applied at the end of the beams, and will contribute to the maximum shear force applied to the end of the beam, and the moment induced in the column due to the eccentricity of connection. For the loading shown, design the beam in grade Fe360 steel. 48,6 kn 97,2 kn 97,2 kn 97,2 kn 48,6 kn A B C D E 1,8 m 1,8 m 1,8 m 1,8 m 7,2 m Figure 5 Beam Restrained at Load Points Reactions V Sd (at supports) = [(2 x 48,6) + (3 x 97,2)] / 2 = 194,4 kn Design Moment Figure 6 shows the distribution of bending moments. 1,8 m 1,8 m 1,8 m 1,8 m ,4 knm 262,4 knm 349,9 knm Figure 6 Bending Moment Diagram Moment at mid-span (maximum) M Sd = [(194,4-48,6) x 3,6] - (97,2 x 1,8) = 349,9 knm 23/02/07 11

12 3.1 Initial Section Selection Assume that a rolled I beam will be used and that the flanges will be less than 40 mm thick. For grade Fe360 steel, f y = 235 N/mm 2. Because the beam is unrestrained between load points, the design resistance, M c.rd, of the section will be reduced by lateral torsional buckling. The final section, allowing for the buckling resistance moment being less than the full resistance moment of the section, would have to be determined from experience. Try IPE O 450 Section properties: Depth, h = 456 mm, Width, b = 192 mm Web thickness, t w = 11 mm Flange thickness, t f = 17,6 mm Plastic modulus, W pl = 2046 cm 3 This notation conforms with Figure 1.1 in Eurocode 3: Part Classification of Cross-Section As a simply supported beam is not required to have any plastic rotation capacity (only one hinge required), it is sufficient to ensure that the section is at least class 2 to develop the plastic moment resistance Flange buckling Class 1 limiting value of c/t f for an outstand of a rolled section is 10ε. ε = 235 / f y and f y = 235 N/mm 2, therefore ε =1. Calculate the ratio c, where c is half the width of the flange = 96 mm, t f and t f is the flange thickness = 17,6 mm (if the flange is tapered, t f should be taken as the average thickness). c 96 = = 5,5 t 17,6 f Web buckling Class 1 limiting value of d/t w for a web subject to bending is 72ε. ε = 235 / f y and f y = 235 N/mm 2, therefore ε =1. Calculate the ratio d, where d is the depth between root radii = 378,8 t w mm and t w is the web thickness = 11,0 mm. d 378,8 = = 34,4 t w 11,0 c < 10ε and d < 72ε t t f Section is Class 1. w and (Sheet 3) (Sheet 1) (Sheets 1 and 3) 23/02/07 12

13 3.3 Design Buckling Resistance Moment The design buckling resistance moment of a laterally unrestrained beam is given by: χ M = b.rd β W γ f LT W pl.y y M1 in which χ LT is the reduction factor for lateral torsional buckling, from 5.5.2, for the appropriate value of λ LT, using curve a for rolled sections (4) In this case full lateral restraint is provided at the supports and at the load points B, C and D. In general, all segments need to be checked, but in this case they are all of equal length. The segments B - C and C - D are subject to the most severe condition, but with symmetrical loading only one segment needs to be checked. Segment B - C The value of λ LT can be determined using Annex F. For segment B - C it is assumed that the purlins at B and C provide the following conditions: restraint against lateral movement, restraint against rotation about the longitudinal axis (i.e. torsional/twisting restraint), and freedom to rotate in plan. i.e. k = k w = 1,0 For this example, the general formula for λ LT has been used, as the section is doubly symmetric and end-moment loading is present. The following formula for λ LT may be used: λ LT = C 05, L/i LT ( L/a ) LT 25, , where L is the length between B and C = 1800 mm, I z is the second moment of area about the z - z axis = 2085 x 10 4 mm 4, I w is the warping constant = 998 x 10 9 mm 6, W pl.y is the plastic modulus about the y - y axis = 2046 x 10 3 mm 3, and I t is the torsion constant = 89,3 x 10 4 mm 4. i a LT LT = II W z w 2 pl.y 0,25 0, x10 x 998x10 = 47,2 mm 3 2 = (2046x10 ) 9 Iw 998x10 = 957 mm 4 = I = 109x10 t 0,5 0,5 Annex F F.1.2(2) Equation F.15 F.2.2(3) F.2.2(1) 23/02/07 13

14 Note: These properties will probably be tabulated in handbooks. See also appendix A at the end of this example. C 1 is the correction factor for the effects of any change of moment along the length, L. ψ = 262,4/349,9 = 0,75, k = 1, therefore C 1 = 1,141 Substituting into the above equation: 1800 / 47, 2 λ LT = = 34, , ( ) 05, 1800 / , , 66 λ Non-dimensional slenderness, λ LT LT = λ β w 1 Where λ 1 = 93,9 ε = 93,9 x 1 = 93,9, β w = 1 for class 1 sections. 34, 6 0,5 Therefore, λ LT = (,) 10 = 037, 93, 9 For rolled I sections, buckling curve a should be used. From 5.5.2, the reduction factor, χ LT = 0,96. (This represents a 4% strength reduction due to moment) W pl.y is the plastic modulus about the y - y axis = 2046 x 10 3 mm 3, f y is the yield strength of the steel = 235 N/mm 2, and γ M1 is the partial material safety factor for buckling resistance = 1,1. The design buckling resistance moment for segment B - C is: χ M = b.rd 3 β W f 096, x 1 x 2046x10 x 235 = = 419, 6 knm 6 γ 1,1 x 10 LT W pl.y y M1 0,5 F (4) 5.5.2(5) (2) M b.rd = 419,6 knm > M Sd = 349,9 knm, therefore the section is satisfactory. 23/02/07 14

15 3.4 Shear on Web The maximum shear occurs at the supports, V Sd = 194,4 kn. The design shear resistance for a rolled I section is: V pl.rd 1,04htw = γ ( fy / 3) M0 where h is the height of the section = 456 mm, t w is the web thickness = 11 mm, f y is the yield strength of the steel = 235 N/mm 2, and γ M0 is the partial material safety factor for the resistance of the crosssection = 1, (1) 5.4.6(4) (2) V 1,04 x 456 x 11 x 235 = = 643 kn 3 x 1,1 x 10 pl.rd 3 V Sd = 194,4 kn < V pl.rd = 643 kn, therefore the section is satisfactory. Inspection shows that V Sd < (V pl.rd / 2), so there is no reduction in moment resistance due to the shear in the web. 3.5 Deflection Check Eurocode 3 requires that the deflections of the beam be checked under the following serviceability loading conditions: 5.4.7(2) 4.2 Variable actions, and Permanent and variable actions. For a general roof, the deflection limits are L/200 for δ max and L/250 for δ 2. Deflection checks are based on the serviceability loading. Consider the deflection from the permanent loading. For a point load, distance a from the end of the beam: 4.1 Figure 4.1 Central deflection, δ = 2 2 Fa k L a EIy (1) where F k is the value of one point load = (7,17 x 5 x 1,8) = 64,5 kn, E is the modulus of elasticity = N/mm 2, I y is the second moment of area about the major axis = x 10 4 mm 4, L is the span of the beam = 7,2 m, and a is the distance from the support to the adjacent load = 1,8 m. Central deflection, δ = 3 64,5x10 x x 40920x = , mm 23/02/07 15

16 For a central point load: Central deflection, δ = FL 3 k 48EI y where F k is the value of one point load = (7,17 x 5 x 1,8) = 64,5 kn, L is the span of the beam = 7,2 m, E is the modulus of elasticity = N/mm 2, and I y is the second moment of area about the major axis = x 10 4 mm (1) ,5x10 x 7200 Central deflection, δ = = 58, mm 4 48 x x 40920x10 Total deflection due to permanent loading, δ 1 = 5,8 + (2 x 4,0) = 13,8 mm Consider the deflection from the variable loading. For a point load, distance a from the end of the beam: 2 2 Fa k L a Central deflection, δ = EIy where F k is the value of one point load = (0,75 x 5 x 1,8) = 6,75 kn, E is the modulus of elasticity = N/mm 2, I y is the second moment of area about the major axis = x 10 4 mm 4, L is the span of the beam = 7,2 m, and a is the distance from the support to the adjacent load = 1,8 m ,75x10 x Central deflection, δ = x 40920x10 = 04, mm (1) For a central point load: Central deflection, δ = FL 3 k 48EI y where F k is the value of one point load = (0,75 x 5 x 1,8) = 6,75 kn, L is the span of the beam = 7,2 m, E is the modulus of elasticity = N/mm 2, and I y is the second moment of area about the major axis = x 10 4 mm (1) 3 3 6,75x10 x 7200 Central deflection, δ = = 06, mm 4 48 x x 40920x10 23/02/07 16

17 Total deflection due to variable loading, δ 2 = 0,6 + (2 x 0,4) = 1,4 mm Therefore, the total central deflection, δ max = δ 1 + δ 2 = 13,8 + 1,4 = 15,2 mm. The limit for δ 2 is L/250 = 7200/250 = 28,8 mm. The limit for δ max = L/200 = 7200/200 = 36 mm. 13,8 mm < 28,8 mm and 15,5 mm < 36 mm, therefore the deflections are within limits and no pre-camber of the beam is required. 3.6 Crushing, Crippling and Buckling If the beam is supported on seating cleats, the checks for web crushing, crippling and buckling must be made. To satisfy the assumptions made in the design, both flanges must be held in place laterally, relative to each other. If seating cleats are used then the top flange must be held laterally. There is no requirement to prevent the flanges from rotating in plan, as k has been taken as 1, Summary All Eurocode recommendations are satisfied, therefore this beam is satisfactory. The beam is satisfactory. 4.1 and Figure Internal Column The internal column shown in Figure 7 is subject to loads from the roof and one floor. Design the column for the given loading, in grade Fe360 steel, as a member in simple framing. 23/02/07 17

18 4.1 Loadings (54 x 7,2) At roof level, the applied axial load = 2 x = 389 kn 2 (81 x 7,2) At first floor level, the applied axial load = 2 x = 583 kn 2 Maximum load, from the first floor to the base, = kn = 972 kn Roof Internal Column Floor 4,2 m 4,5 m Figure 7 Internal Column Consider the column from ground floor to first floor. The size of the column must be determined from experience and then checked for compliance with the Eurocode rules. 4.2 Section Properties Try an HE 240 A Grade Fe360 (terminology in accordance with EC3) h = 230 mm b = 240 mm t w = 7,5 mm t f = 12 mm d/t w = 21,9 c/t f = 10 A = 7680 mm 2 I y = 77,63 x 10 6 mm 4 I w = 328 x 10 9 mm 6 I z = 27,7 x 10 6 mm 4 I t = 41,6 x 10 4 mm 4 W pl.y = 745 x 10 3 mm 3 W el.y = 675 x 10 3 mm 3 i y = 101 mm i z = 60 mm i Lt = II W z w 2 pl.y 0,5 0,25 0, ,7x10 x 328x10 = 63,6 mm 3 2 = (745x10 ) 0,5 9 Iw 328x10 alt = 888 mm 4 = It = 41,6x10 All the above properties can be obtained from section property tables. F.2.2(3) F.2.2(1) 23/02/07 18

19 4.3 Classification of Cross Section This section is designed to withstand axial force only. No moment is applied as the connecting beams are equally loaded Flange (subject to compression) Class 1 limiting value of c/t f for an outstand of a rolled section is 10ε. ε = 235 / f y where f y = 235 N/mm 2, therefore ε = 1. From section properties, c/t f = Web (subject to compression) Class 1 limiting value of d/t w for a web subject to compression only is 33ε. From section properties, d/t w = 21,9 c/t f 10ε and d/t w 33ε Class 1 section (Sheet 3) (Sheet 1) (Sheets 1 and 3) Class 1 section 4.4 Resistance of Cross-Section It is highly unlikely that the resistance of the cross-section will be the critical case - it is generally the buckling resistance that governs the suitability of a cross-section. For the sake of completeness, the check is included in this worked example. The resistance of the cross-section will only be critical if a short, stocky column is used. For members in axial compression, the design value of the compressive force, N Sd, at each cross-section shall satisfy N Sd N c.rd For a class 1 cross-section, the design compression resistance of the cross-section, N c.rd, may be determined as: Afy Nc.Rd = γ M0 where A is cross-sectional area = 7680 mm 2, f y is the yield strength = 235 N/mm 2, and γ M0 is the partial material safety factor = 1,1. N 7680 x 235 = = 1641 kn 1,1x10 c.rd 3 N Sd = 972 kn, therefore N sd N c.rd. The section can resist the applied axial load (1) 5.4.4(2) (2) 23/02/07 19

20 4.5 Buckling Resistance of Member A class 1 member should be checked for failure due to flexural and lateral torsional buckling. Here, since M y = M z = 0, only failure due to flexural buckling needs to be checked. The design buckling resistance of a compression member shall be taken as: N b.rd AAf = χβ γ M1 y where χ is the reduction factor for the relevant buckling mode, β A = 1 for class 1 cross-section, A is the cross-sectional area = 7680 mm 2, f y is the yield strength of the steel = 235 N/mm 2, and γ M1 is the partial material safety factor for buckling resistance = 1,1. The magnitude of the reduction factor, χ depends on the reduced slenderness of the columns. χ is the lesser of χ y and χ z, where χ y and χ z are the reduction factors from clause for the y-y and z-z axes respectively. Values of χ for the appropriate value of non-dimensional slenderness, λ may be obtained from λ Non-dimensional slenderness, λ = λ β 0,5 A Where the slenderness, λ = l i l is the column buckling length, and i is the radius of gyration about the relevant axis. The braced frame is designed as a simple pinned structure. Therefore, the buckling length ratio l/l is equal to 1 - the buckling length is equal to the system length. 0,5 E λ1 = π 93,9ε f = y 4.6 Determination of Reduction Factor, χ y Slenderness, λ y = l/i y = 4500/101 = 44,6 λ 1 = 93,9ε = 93,9 x 1 = 93,9 λ Non-dimensional slenderness, λ y = y λ β 0,5 44,6 0,5 A = 1 93,9 x 1 = 0,47 From 5.5.2, using buckling curve b, the reduction factor, χ y = 0, (1) (1) (2) (1) (3) (2) (1) (3) /02/07 20

21 4.7 Determination of Reduction Factor, χ z Slenderness, λ z = l/i z = 4500/60 = 75 λ 1 = 93,9ε = 93,9 x 1 = 93,9 λ Non-dimensional slenderness, λ = z λ β 0,5 75 0,5 y A = 1 93,9 x 1 = 0,8 From 5.5.2, using buckling curve c,the reduction factor, χ z = 0,6622 Therefore χ = χ z = 0,6622. Design buckling resistance of member: χβ AAfy 0,6622 x 1 x 7680 x 235 Nb.Rd = = = 1086 kn 3 γ M1 1,1x10 The design buckling resistance of the member is greater than the applied load (972 kn), therefore the column is satisfactory. The column is OK (3) (1) 5. External Column The external column shown in Figure 8 is subject to loads from the roof and one floor. Design the column for the loading given below, in grade Fe360 steel, as a member in simple framing. 5.1 Loadings (54 x 7,2) At roof level, the applied axial load = = 194 kn 2 (81 x 7,2) At first floor level, the applied axial load = = 292 kn 2 Maximum load, from first floor to base, = kn = 486 kn The beams in the frame are designed to span from column centre to column centre, therefore all axial load is applied at the mid-point of the column. No moment due to eccentricity of applied load is therefore applied to the column. See Annex H 23/02/07 21

22 Roof 4,2 m First Floor 4,5 m Figure 8 External Column Consider the column from ground floor to first floor. The size of the column must be determined from experience and then checked for compliance with the Eurocode rules. 5.2 Section Properties Try an HE 200 A Grade Fe360 h = 190 mm b = 200 mm t w = 6,5 mm t f = 10 mm d/t w = 20,6 c/t f = 10 A = 5380 mm 2 I y = 36,92 x 10 6 mm 4 I w = 108 x 10 9 mm 6 I z = 13,36 x 10 6 mm 4 I t = 21,0 x 10 4 mm 4 W pl.y = 429 x 10 3 mm 3 W el.y = 389 x 10 3 mm 3 i y = 82,8 mm i z = 49,8 mm i Lt = II W z w 2 pl.y 0,5 0,25 0, ,36x10 x 108x10 = 52,9 mm 3 2 = (429x10 ) 0,5 9 Iw 108x10 alt = 717 mm 4 = It = 21,0x10 All the above properties can be obtained from section property tables. F.2.2(3) F.2.2(1) 23/02/07 22

23 5.3 Classification of Cross Section This section is designed to withstand axial force only Flange (subject to compression) Class 1 limiting value of c/t f for an outstand of a rolled section is 10ε. ε = 235 / f y where f y = 235 N/mm 2, ε = 1. 10ε = 10 x 1 = 10 From section properties, c/t f = Web (subject to compression) Class 1 limiting value of d/t w for a web subject to compression only is 33ε. ε = 235 / f y where f y = 235 N/mm 2, ε = 1. 33ε = 33 x 1 = 33 From section properties, d/t w = 20,6 c/t f 10ε and d/t w 33ε Class 1 section (Sheet 3) (Sheet 1) (Sheets 1 and 3) 5.4 Resistance of Cross-Section It is highly unlikely that the resistance of the cross-section will be the critical case - it is generally the buckling resistance that governs the suitability of a cross-section. For the sake of completeness, the check is included in this worked example. The resistance of the cross-section will only be critical if a short stocky column is used. For members in axial compression, the design value of the compressive force, N Sd, at each cross-section shall satisfy N Sd N c.rd For a class 1 cross-section, the design compression resistance of the cross-section, N c.rd, may be determined as: Afy Nc.Rd = γ M0 where A is cross-sectional area = 5380 mm 2, f y is the yield strength = 235 N/mm 2, and γ M0 is the partial material safety factor = 1, x 235 Nc.Rd = = 1149 kn 3 1,1x10 N Sd = 486 kn, therefore N sd N c.rd. The section can resist the applied axial load (1) 5.4.4(2) (2) 23/02/07 23

24 5.5 Buckling Resistance of Member A class 1 member should be checked for failure due to flexural and lateral torsional buckling. Here, since M y = M z = 0, only failure due to flexural buckling needs to be checked. The design buckling resistance of a compression member shall be taken as: AAfy Nb.Rd = χβ γ M1 where χ is the reduction factor for the relevant buckling mode, β A = 1 for class 1 cross-section, A is the cross-sectional area = 5380 mm 2, f y is the yield strength of the steel = 235 N/mm 2, and γ M1 is the partial material safety factor for buckling resistance = 1,1. The magnitude of the reduction factor, χ depends on the reduced slenderness of the columns. χ is the lesser of χ y and χ z, where χ y and χ z are the reduction factors from clause for the y-y and z-z axes respectively. Values of χ for the appropriate value of non-dimensional slenderness, λ may be obtained from λ Non-dimensional slenderness, λ = λ β A Where the slenderness, λ = l i l is the column buckling length, and i is the radius of gyration about the relevant axis. The braced frame is designed as a simple pinned structure. Therefore, the buckling length ratio l/l is equal to 1 - the buckling length is equal to the system length. 0,5 E λ1 = π 93,9ε f = y 5.6 Determination of Reduction Factor, χ y Slenderness, λ y = l/i y = 4500/82,8 = 54,3 λ 1 = 93,9ε = 93,9 x 1 = 93,9 λ Non-dimensional slenderness, λ y = y λ β 0,5 54,3 0,5 A = 1 93,9 x 1 = 0,58 From 5.5.2, using buckling curve b, the reduction factor, χ y = 0,84 1 0, (1) (1) (2) (1) (3) (2) Annex E Figure E (1) (3) /02/07 24

25 5.7 Determination of Reduction Factor, χ z Slenderness, λ z = l/i z = 4500/49,8 = 90,4 λ 1 = 93,9ε = 93,9 x 1 = 93,9 λ Non-dimensional slenderness, λ = z λ β 0,5 90,4 0,5 y A = 1 93,9 x 1 = 0,96 From 5.5.2, using buckling curve c,the reduction factor, χ z = 0,55 Therefore χ = χ z = 0,55. Design buckling resistance of member: χβ AAfy 0,55 x 1 x 5380 x 235 N b.rd = = = 632,2 kn 3 γ M1 1,1x10 The design buckling resistance of the member is greater than the applied load (486 kn), therefore the column is satisfactory. The column is OK. 6. Design of Cross-Bracing All horizontal loading will be resisted by bracing. For the purpose of illustration assume this will be present on every other frame (i.e. at 10 m spacing). It is more likely that bracing will be located at each end of the building or perhaps in a stair/lift well. The forces may therefore be greater than here but the principles would remain the same. For the loading shown, design the bracing members in grade Fe360 steel (3) (1) (2) 2.2 Characteristic wind load, Q k = 1,6 kn/m 2. Design wind load, Q d = γ Q Q k At ultimate limit state, γ Q = 1,5 (unfavourable), Q d = 1,5 x 1,6 = 2,4 kn/m 2. Therefore, the total load (per m height of frame) = 10 x 2,4 = 24 kn/m. 23/02/07 25

26 24 kn/m 4,2 m 4,5 m 7,2 m Figure 9 Wind Load on Frame It is assumed that the uniformly distributed load acts as two point loads on the frame. Top load = (24 x 2,1) = 50,4 kn. Middle load = (24 x 2,1) + (24 x 2,25) = 104,4 kn. Assume that all horizontal load is resisted by the bracing only. Therefore, the load in the top brace = 50,4 / cos 30,3º = 58,4 kn, and the load in the bottom brace = (104,4 +50,4)/ cos 32º = 182,5 kn. 23/02/07 26

27 50,4 kn 58,4 kn 4,2 m 104,4 kn 182,5 kn 4,5 m 7,2 m Figure 10 Equivalent Point Wind Loads and Loads Within Bracing Design the bottom brace, as this carries the greater load. Try a CHS 175 x 5,0 6.1 Section Properties Depth of section, d = 175 mm, Thickness, t = 5 mm Area of section, A = 2670 mm 2, Ratio for local buckling, d/t = 35, Radius of gyration, i = 60,1 mm. 6.2 Classification of Cross-Section As the bracing is axially loaded, check the section classification is at least class 1, 2 or 3. Figure 11 shows a typical CHS cross-section (1)a t d Figure 11 Typical CHS Cross-Section 23/02/07 27

28 Class 1 limiting value of d/t for a tubular section is 50ε 2. ε = 235 / f y where f y = 235 N/mm 2, ε = 1. 50ε 2 = 50 x 1 = 50 From section properties, d/t =35, therefore the section is Class Design of Compression Member The bracing members need to be checked as axially loaded Resistance of Cross-Section It is highly unlikely that the resistance of the cross-section will be the critical case - it is generally the buckling resistance that governs the suitability of a cross-section. For the sake of completeness, the check is included in this worked example. The resistance of the cross-section will only be critical if a short, stocky column is used. The applied axial load, N Sd, must be less than the design compressive resistance of the cross-section, N c.rd (Sheet 4) 5.4.4(1) Applied axial load, N Sd = 182,5 kn. Design compressive resistance of cross-section, N c.rd = N Where A is the cross-sectional area = 2670 mm 2, f y is the yield strength of the steel = 235 N/ mm 2, and γ M0 is the partial material safety factor = 1,1. N 2670 x 235 = = 570, 4 kn 11, x 10 pl.rd 3 pl.rd Af = γ y M (2) The design compressive resistance of the cross-section, N c.rd = 570,4 kn, is greater than the applied axial load, N Sd = 182,5 kn. Therefore the section is satisfactory Design Buckling Resistance A class 1 member subject to axial compression should be checked for failure due to buckling. The design buckling resistance of a compression member shall be taken as: AAfy Nb.Rd = χβ γ M1 where χ is the reduction factor for the relevant buckling mode, β A = 1 for class 1 cross-section, A is the cross-sectional area = 2670 mm 2, f y is the yield strength of the steel = 235 N/mm 2, and γ M1 is the partial material safety factor for buckling resistance = 1,1. Values of χ for the appropriate value of non-dimensional slenderness, λ may be obtained from (1) (1) (2) 23/02/07 28

29 λ Non-dimensional slenderness, λ = λ β A Where the slenderness, λ = l i l is the member buckling length, and i is the radius of gyration. The bracing is designed as a simple pinned member. Therefore, the buckling length ratio l/l is equal to 1 - the buckling length is equal to the system length , 5 + 7, 2 = 8500 mm Length of member = ( ) 0,5 E λ1 = π 93,9ε f = y Determination Of Reduction Factor, χ Slenderness, λ = l/i = 8500/60,1 = 141 λ 1 = 93,9ε = 93,9 x 1 = 93,9 λ Non-dimensional slenderness, λ = λ β 0,5 = 141 0,5 A = 1 93,9 x 1 1,50 From 5.5.2, using buckling curve b, the reduction factor, χ = 0,3422. Design buckling resistance of member: χβ AAfy 0,3422 x 1 x 2670 x 235 N b.rd = = = 195,2 kn 3 γ M1 1,1x10 The design buckling resistance of the member is greater than the applied load (182,5 kn), therefore the bracing is satisfactory. 1 0, (1) (3) (2) Annex E Figure E (1) (3) (1) 23/02/07 29

30 6.4 Design of Tension Member When the wind load is applied in the opposite direction, the bracing will be loaded in tension. The section therefore needs to be checked, for the same magnitude of loading, to ensure it is also satisfactory in tension. 50,4 kn 4,2 m 58,4 kn 104,4 kn 4,5 m 182,5 kn 7,2 m Figure 12 Equivalent Point Wind Loads and Loads Within Bracing Resistance of Cross-Section The applied axial load, N Sd, must be less than the design tension resistance of the cross-section, N t.rd. Applied axial load, N Sd = 182,5 kn. Afy Design tension resistance of the cross-section, N t.rd = N pl.rd = γ M0 where A is the cross-sectional area = 2670 mm 2, f y is the yield strength of the steel = 235 N/ mm 2, and γ M0 is the partial material safety factor = 1, x 235 N pl.rd = = 570, 4 kn 3 11, x 10 The design tension resistance of the cross-section, N t.rd = 570,4 kn, is greater than the applied axial load, N Sd = 182,5 kn. Therefore the section is satisfactory (1) (2) The bracing fulfils all the Eurocode requirements for members in tension and in compression, and is therefore satisfactory. The frame is satisfactory for all EC3 checks 23/02/07 30

31 7. Concluding Summary 23/02/07 31

32 Structural Steelwork Eurocodes Development of A Trans-national Approach Course: Eurocode 3 Module 7 : Worked Examples Lecture 22 : Design of an unbraced sway frame with rigid joints Summary: NOTE This example is a draft version Pre-requisites: Notes for Tutors: This material comprises one 60 minute lecture. Objectives: To explain the main principles of EC3 by practical worked example. References: Eurocode 3: Design of steel structures Part 1.1 General rules and rules for buildings Contents: 23/02/07 32

33 WORKED EXAMPLE 3 Design of a Sway Frame The frame consists of three storeys and three bays. The frames are at 10 m spacing. The beam span is 6,5 m and the total height is 10,5 m, each storey being 3,5 m high. It is assumed that the column foot is pinned at the foundation. Roof beams Floor beams 3,5 m Internal columns Internal columns 3,5 m External columns External columns 3,5 m 6,5 m 6,5 m 6,5 m Figure 1 Typical Cross Section of Frame The structure is assumed to be braced out of its plane and to be unbraced in its plane. In the longitudinal direction of the building, i.e. in the direction perpendicular to the frame plane, a bracing does exist so that the tops of 23/02/07 33

34 the columns are held in place. The lateral support for the floor beams is provided by the floor slabs. All the beam-to-column joints are assumed to be perfectly rigid. The connections must be capable of transmitting the forces and moments calculated in design. With these assumptions, the frame is classified as continuous, and the internal forces and moments are determined using a global elastic analysis which assumes the members to be effectively held in position. The steel grade selected for beams, columns and joints is Fe360 (f y = 235 N/mm 2 ). Characteristic Loads Floor: Variable actions, Q k = 1,8 kn/m 2, Permanent actions, G k = 3,0 kn/m 2 Roof: Variable actions, Q k = 0,6 kn/m 2, Permanent actions, G k = 2,0 kn/m 2 The wind loads are applied as point loads of 10,5 kn at roof level and 21 kn at the first and second storey levels. The basic loading cases, shown in Figure 2, have been considered in appropriate combinations. 20 kn/m 30 kn/m 10,5 kn 21 kn (3) kn/m 21 kn Permanent Loading (G) Wind Loading (W) 6 kn/m 6 kn/m 6 kn/m 6 kn/m 18 kn/m 18 kn/m 18 kn/m 18 kn/m 18 kn/m 18 kn/m 18 kn/m 18 kn/m loading case 1 (N1) loading case 2 (N2) loading case 3 (N3) Imposed Loading Cases Figure 2 Loading Cases 23/02/07 34

35 Frame Imperfections Frame imperfections are considered by means of equivalent horizontal loads. The initial sway imperfection is given as: φ = k c k s φ (1) where k = k s c 0,5+ 1 n c but k c 1,0, 1 = 02, + but k s 1,0, and n s 1 φ 0 =. 200 In this case, the number of full height columns per floor, n c, is 4 and the number of storeys in the frame, n s, is 3. Therefore k c = 05, + 1 = 0, 866, and 4 1 k s = 02, + = 073,. 3 Substituting into the above equation: 1 φ = 0,866 x 0,73 x 200 = The equivalent horizontal load, H, at each storey of the frame is derived from the initial sway, φ, and the total design vertical load, N, in any storey for a given load case. Therefore H = φn. The relevant values are listed in 1 for all the basic loading cases. Basic loading case (see Figure 2) Storey N (kn) φn (kn) G Roof 390 1,24 2nd Floor 585 1,86 1st Floor 585 1,86 N1 Roof 117 0,37 2nd Floor 351 1,11 1st Floor 351 1,11 N2 Roof 39 0,12 2nd Floor 234 0,74 1st Floor 117 0,37 N3 Roof 78 0,25 2nd Floor 117 0,37 1st Floor 234 0,74 1 Equivalent Horizontal Forces The equivalent horizontal forces must also be multiplied be the appropriate partial safety factors for actions. 23/02/07 35

36 Load Combinations It was decided to use the simplified combinations for the ultimate limit state and the serviceability limit state. The basic load cases are combined at the ULS as summarised in 2. Load Case 1 1,35G + 1,5W Load Case 2 1,35G + 1,5N1 Load Case 3 1,35G + 1,5N2 Load Case 4 1,35G + 1,5N3 Load Case 5 1,35G + 1,35W + 1,35N1 Load Case 6 1,35G + 1,35W + 1,35N2 Load Case 7 1,35G + 1,35W + 1,35N3 2 Load Combination Cases at the Ultimate Limit State The basic load cases are combined at the SLS as summarised in 3. Load Case 1 1,0G + 1,0W Load Case 2 1,0G + 1,0N1 Load Case 3 1,0G + 1,0N2 Load Case 4 1,0G + 1,0N3 Load Case 5 1,0G + 0,9W + 0,9N1 Load Case 6 1,0G + 0,9W + 0,9N2 Load Case 7 1,0G + 0,9W + 0,9N3 3 Load Combination Cases at the Serviceability Limit State Partial Safety Factors for Strength The following partial safety factors for strength have been adopted during the design: Resistance of Class 1,2 or 3 cross-section, γ M0 = 1,1 Resistance of member to buckling, γ M1 = 1,1 Resistance of bolted connections, γ Mb = 1,25 Trial Sections In order for a global elastic analysis of the structure to be carried out, initial section sizes must be assumed and allocated to the structural members. The analysis must then be carried out and the members checked for the relevant failure modes. The sections will then need to be modified and the structure re-analysed. This can be a long winded iterative process. The engineer may have his own method of selecting initial section sizes. As a guideline, for this example, columns were selected by assuming an average stress of approximately 100 N/mm 2 under axial forces. Axial forces can be estimated by approximating the floor area supported by that column. Generally, the bending moments withing the beams are critical. Simple bending moment diagrams can be constructed, assuming fixed end moments, and the maximum bending moment can then be estimated. An initial section size can then be determined. The trial member sizes for this example are: Inner columns: HEB 260 Outer columns: HEB (5) and 2.3.4(5) (1) 5.1.1(2) 5.1.1(2) 6.1.1(2) 23/02/07 36

37 Floor beams: IPE 450 Roof beams: IPE 360 Roof beams IPE 360 Floor beams IPE 450 3,5 m Internal columns HEB 260 Internal columns HEB 260 3,5 m External columns HEB 220 External columns HEB 220 3,5 m 6,5 m 6,5 m 6,5 m Figure 3 Trial Member Sizes Determination of Design Moments and Forces This worked example uses the amplified sway moments method of analysis. An alternative method is to calculate the sway-mode buckling lengths of members, and then carry out a first order linear elastic analysis. However, it may be possible to directly use a second order elastic analysis. A global linear elastic analysis is carried out on the sway frame in order to determine the moments, axial forces and shear forces in each member. Props are then applied to the structure in order to prevent any horizontal movement (i.e. as if the structure was now braced), and the analysis carried out again. The moments induced in the members from the propped case are then subtracted from the moments determined from the sway case. The resultant moments are those induced by pure sway. The pure sway moments are then amplified to take account of second order effects ignored by the linear elastic analysis. The amplified pure sway moments are then added to the moments obtained from analysis of the propped structure. These are the design moments which each member must be able to resist. The members must also be checked for the axial and shear forces determined from the initial analysis of the sway frame, and in certain cases the interaction of the design moments with the axial forces and/or shear forces must also be checked. This needs to be carried out for every load case so that the critical conditions for each individual member can be identified (1) (1) (1) 23/02/07 37

38 Original Sway Case Moments Minus Propped Case Moments Equals "Pure Sway" Moments (To Be Amplified) 23/02/07 38

39 Figure 4 Determination of Pure Sway Moments Calculation of Amplification Factors The sway moments should be increased by multiplying them by the ratio: 1 1 VSd / Vcr where V Sd is the design value of the total vertical load, and V cr is its elastic critical value for failure in a sway mode. Instead of determining V Sd / V cr directly, the following approximation may be used: VSd V = V δ cr h H where δ is the horizontal displacement at the top of the storey, relative to the bottom of the storey, h is the storey height, V is the total vertical reaction at the bottom of the storey, and H is the total horizontal reaction at the bottom of the storey. This approximation may not be used if V Sd / V cr is greater than 0,25. If V Sd / V cr is greater than 0,25, then the frame may be more susceptible to buckling. It is therefore necessary to carry out analysis using a direct second order analysis. It may also be necessary to stiffen the frame - for example increasing the column sizes. Alternatively, if the value of V Sd / V cr is less than 0,1 then the structure is classified as non-sway. The amplification factor will be different for each storey of the structure. The maximum factor should be used to multiply the moments at all levels of the structure. This is essentially a conservative method as it corresponds to the critical elastic load of the whole structure. The amplification factor for load case 5 of this example was determined as follows: (3) (6) (4) (4) (3) 23/02/07 39

40 δ 2 δ = δ2 - δ = 15,6 mm 1 h = 3500 mm V = 3296 kn H = 81 kn h H V δ1 Figure 5 Determination of Amplification Factor Therefore V Sd V 15,6 x 3296 = V δ cr h = = 018, H 3500 x The amplification factor = = = 122, 1 VSd / Vcr 1 018, All the pure sway moments for load case 5 were amplified by a factor of 1,22. The global linear elastic analysis and amplification of sway moments was carried out for all seven load cases. s 4a and 4b shows the maximum forces in each member. N 19 P 20 Q 21 R K 13 L 14 J M F 6 G 7 E H 9 = Element G = Node A B C D Figure 6 Labelling of Members within the Structure 23/02/07 40

Structural Steelwork Eurocodes Development of A Trans-national Approach

Structural Steelwork Eurocodes Development of A Trans-national Approach Structural Steelwork Eurocodes Development of A Trans-national Approach Course: Eurocode Module 7 : Worked Examples Lecture 0 : Simple braced frame Contents: 1. Simple Braced Frame 1.1 Characteristic Loads

More information

Structural Steelwork Eurocodes Development of A Trans-national Approach

Structural Steelwork Eurocodes Development of A Trans-national Approach Structural Steelwork Eurocodes Development of A Trans-national Approach Course: Eurocode Module 7 : Worked Examples Lecture 22 : Design of an unbraced sway frame with rigid joints Summary: NOTE This example

More information

Design of Beams (Unit - 8)

Design of Beams (Unit - 8) Design of Beams (Unit - 8) Contents Introduction Beam types Lateral stability of beams Factors affecting lateral stability Behaviour of simple and built - up beams in bending (Without vertical stiffeners)

More information

STEEL BUILDINGS IN EUROPE. Multi-Storey Steel Buildings Part 10: Technical Software Specification for Composite Beams

STEEL BUILDINGS IN EUROPE. Multi-Storey Steel Buildings Part 10: Technical Software Specification for Composite Beams STEEL BUILDINGS IN EUROPE Multi-Storey Steel Buildings Part 10: Technical Software Specification for Composite Beams Multi-Storey Steel Buildings Part 10: Technical Software Specification for Composite

More information

Eurocode 3 for Dummies The Opportunities and Traps

Eurocode 3 for Dummies The Opportunities and Traps Eurocode 3 for Dummies The Opportunities and Traps a brief guide on element design to EC3 Tim McCarthy Email tim.mccarthy@umist.ac.uk Slides available on the web http://www2.umist.ac.uk/construction/staff/

More information

Made by SMH Date Aug Checked by NRB Date Dec Revised by MEB Date April 2006

Made by SMH Date Aug Checked by NRB Date Dec Revised by MEB Date April 2006 Job No. OSM 4 Sheet 1 of 8 Rev B Telephone: (0144) 45 Fax: (0144) 944 Made b SMH Date Aug 001 Checked b NRB Date Dec 001 Revised b MEB Date April 00 DESIGN EXAMPLE 9 - BEAM WITH UNRESTRAINED COMPRESSION

More information

University of Sheffield. Department of Civil Structural Engineering. Member checks - Rafter 44.6

University of Sheffield. Department of Civil Structural Engineering. Member checks - Rafter 44.6 Member checks - Rafter 34 6.4Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic rafter sections, as it is the normal practice to use a UB cutting of

More information

Job No. Sheet No. Rev. CONSULTING Engineering Calculation Sheet. Member Design - Steel Composite Beam XX 22/09/2016

Job No. Sheet No. Rev. CONSULTING Engineering Calculation Sheet. Member Design - Steel Composite Beam XX 22/09/2016 CONSULTING Engineering Calculation Sheet jxxx 1 Member Design - Steel Composite Beam XX Introduction Chd. 1 Grade 50 more common than Grade 43 because composite beam stiffness often 3 to 4 times non composite

More information

3. Stability of built-up members in compression

3. Stability of built-up members in compression 3. Stability of built-up members in compression 3.1 Definitions Build-up members, made out by coupling two or more simple profiles for obtaining stronger and stiffer section are very common in steel structures,

More information

COMPARISON BETWEEN BS 5950: PART 1: 2000 & EUROCODE 3 FOR THE DESIGN OF MULTI-STOREY BRACED STEEL FRAME CHAN CHEE HAN

COMPARISON BETWEEN BS 5950: PART 1: 2000 & EUROCODE 3 FOR THE DESIGN OF MULTI-STOREY BRACED STEEL FRAME CHAN CHEE HAN i COMPARISON BETWEEN BS 5950: PART 1: 2000 & EUROCODE 3 FOR THE DESIGN OF MULTI-STOREY BRACED STEEL FRAME CHAN CHEE HAN A project report submitted as partial fulfillment of the requirements for the award

More information

Advanced Analysis of Steel Structures

Advanced Analysis of Steel Structures Advanced Analysis of Steel Structures Master Thesis Written by: Maria Gulbrandsen & Rasmus Petersen Appendix Report Group B-204d M.Sc.Structural and Civil Engineering Aalborg University 4 th Semester Spring

More information

THE EC3 CLASSIFICATION OF JOINTS AND ALTERNATIVE PROPOSALS

THE EC3 CLASSIFICATION OF JOINTS AND ALTERNATIVE PROPOSALS EUROSTEEL 2002, Coimbra, 19-20 September 2002, p.987-996 THE EC3 CLASSIFICATION OF JOINTS AND ALTERNATIVE PROPOSALS Fernando C. T. Gomes 1 ABSTRACT The Eurocode 3 proposes a classification of beam-to-column

More information

PLATE GIRDERS II. Load. Web plate Welds A Longitudinal elevation. Fig. 1 A typical Plate Girder

PLATE GIRDERS II. Load. Web plate Welds A Longitudinal elevation. Fig. 1 A typical Plate Girder 16 PLATE GIRDERS II 1.0 INTRODUCTION This chapter describes the current practice for the design of plate girders adopting meaningful simplifications of the equations derived in the chapter on Plate Girders

More information

Fundamentals of Structural Design Part of Steel Structures

Fundamentals of Structural Design Part of Steel Structures Fundamentals of Structural Design Part of Steel Structures Civil Engineering for Bachelors 133FSTD Teacher: Zdeněk Sokol Office number: B619 1 Syllabus of lectures 1. Introduction, history of steel structures,

More information

C6 Advanced design of steel structures

C6 Advanced design of steel structures C6 Advanced design of steel structures prepared b Josef achacek List of lessons 1) Lateral-torsional instabilit of beams. ) Buckling of plates. 3) Thin-walled steel members. 4) Torsion of members. 5) Fatigue

More information

Job No. Sheet 1 of 7 Rev A. Made by ER/EM Date Feb Checked by HB Date March 2006

Job No. Sheet 1 of 7 Rev A. Made by ER/EM Date Feb Checked by HB Date March 2006 Job No. Sheet of 7 Rev A Design Example Design of a lipped channel in a Made by ER/EM Date Feb 006 Checked by HB Date March 006 DESIGN EXAMPLE DESIGN OF A LIPPED CHANNEL IN AN EXPOSED FLOOR Design a simply

More information

MODULE C: COMPRESSION MEMBERS

MODULE C: COMPRESSION MEMBERS MODULE C: COMPRESSION MEMBERS This module of CIE 428 covers the following subjects Column theory Column design per AISC Effective length Torsional and flexural-torsional buckling Built-up members READING:

More information

Basis of Design, a case study building

Basis of Design, a case study building Basis of Design, a case study building Luís Simões da Silva Department of Civil Engineering University of Coimbra Contents Definitions and basis of design Global analysis Structural modeling Structural

More information

1C8 Advanced design of steel structures. prepared by Josef Machacek

1C8 Advanced design of steel structures. prepared by Josef Machacek 1C8 Advanced design of steel structures prepared b Josef achacek List of lessons 1) Lateral-torsional instabilit of beams. ) Buckling of plates. 3) Thin-walled steel members. 4) Torsion of members. 5)

More information

Made by PTY/AAT Date Jan 2006

Made by PTY/AAT Date Jan 2006 Job No. VALCOSS Sheet of 9 Rev A P.O. Box 000, FI-0044 VTT Tel. +358 0 7 Fax +358 0 7 700 Design Example 3 Stainless steel lattice girder made Made by PTY/AAT Date Jan 006 RFCS Checked by MAP Date Feb

More information

BRACING MEMBERS SUMMARY. OBJECTIVES. REFERENCES.

BRACING MEMBERS SUMMARY. OBJECTIVES. REFERENCES. BRACING MEMBERS SUMMARY. Introduce the bracing member design concepts. Identify column bracing members requirements in terms of strength and stiffness. The assumptions and limitations of lateral bracing

More information

Accordingly, the nominal section strength [resistance] for initiation of yielding is calculated by using Equation C-C3.1.

Accordingly, the nominal section strength [resistance] for initiation of yielding is calculated by using Equation C-C3.1. C3 Flexural Members C3.1 Bending The nominal flexural strength [moment resistance], Mn, shall be the smallest of the values calculated for the limit states of yielding, lateral-torsional buckling and distortional

More information

Karbala University College of Engineering Department of Civil Eng. Lecturer: Dr. Jawad T. Abodi

Karbala University College of Engineering Department of Civil Eng. Lecturer: Dr. Jawad T. Abodi Chapter 05 Structural Steel Design According to the AISC Manual 13 th Edition Analysis and Design of Beams By Dr. Jawad Talib Al-Nasrawi University of Karbala Department of Civil Engineering 71 Introduction

More information

Design of Compression Members

Design of Compression Members Design of Compression Members 2.1 Classification of cross sections Classifying cross-sections may mainly depend on four critical factors: 1- Width to thickness (c/t) ratio. 2- Support condition. 3- Yield

More information

Plastic Design of Portal frame to Eurocode 3

Plastic Design of Portal frame to Eurocode 3 Department of Civil and Structural Engineering Plastic Design of Portal frame to Eurocode 3 Worked Example University of Sheffield Contents 1 GEOMETRY... 3 2 DESIGN BRIEF... 4 3 DETERMINING LOADING ON

More information

Influence of residual stresses in the structural behavior of. tubular columns and arches. Nuno Rocha Cima Gomes

Influence of residual stresses in the structural behavior of. tubular columns and arches. Nuno Rocha Cima Gomes October 2014 Influence of residual stresses in the structural behavior of Abstract tubular columns and arches Nuno Rocha Cima Gomes Instituto Superior Técnico, Universidade de Lisboa, Portugal Contact:

More information

Karbala University College of Engineering Department of Civil Eng. Lecturer: Dr. Jawad T. Abodi

Karbala University College of Engineering Department of Civil Eng. Lecturer: Dr. Jawad T. Abodi Chapter 04 Structural Steel Design According to the AISC Manual 13 th Edition Analysis and Design of Compression Members By Dr. Jawad Talib Al-Nasrawi University of Karbala Department of Civil Engineering

More information

Design of Steel Structures Prof. S.R.Satish Kumar and Prof. A.R.Santha Kumar

Design of Steel Structures Prof. S.R.Satish Kumar and Prof. A.R.Santha Kumar 5.4 Beams As stated previousl, the effect of local buckling should invariabl be taken into account in thin walled members, using methods described alread. Laterall stable beams are beams, which do not

More information

Equivalent Uniform Moment Factor for Lateral Torsional Buckling of Steel Beams

Equivalent Uniform Moment Factor for Lateral Torsional Buckling of Steel Beams University of Alberta Department of Civil & Environmental Engineering Master of Engineering Report in Structural Engineering Equivalent Uniform Moment Factor for Lateral Torsional Buckling of Steel Beams

More information

ENCE 455 Design of Steel Structures. III. Compression Members

ENCE 455 Design of Steel Structures. III. Compression Members ENCE 455 Design of Steel Structures III. Compression Members C. C. Fu, Ph.D., P.E. Civil and Environmental Engineering Department University of Maryland Compression Members Following subjects are covered:

More information

Design of Steel Structures Prof. S.R.Satish Kumar and Prof. A.R.Santha Kumar

Design of Steel Structures Prof. S.R.Satish Kumar and Prof. A.R.Santha Kumar 6. BEAMS 6.1 Introduction One of the frequently used structural members is a beam whose main function is to transfer load principally by means of flexural or bending action. In a structural framework,

More information

Failure in Flexure. Introduction to Steel Design, Tensile Steel Members Modes of Failure & Effective Areas

Failure in Flexure. Introduction to Steel Design, Tensile Steel Members Modes of Failure & Effective Areas Introduction to Steel Design, Tensile Steel Members Modes of Failure & Effective Areas MORGAN STATE UNIVERSITY SCHOOL OF ARCHITECTURE AND PLANNING LECTURE VIII Dr. Jason E. Charalambides Failure in Flexure!

More information

STEEL BUILDINGS IN EUROPE. Multi-Storey Steel Buildings Part 8: Description of member resistance calculator

STEEL BUILDINGS IN EUROPE. Multi-Storey Steel Buildings Part 8: Description of member resistance calculator STEEL BUILDINGS IN EUROPE Multi-Store Steel Buildings Part 8: Description of member resistance calculator Multi-Store Steel Buildings Part : Description of member resistance calculator 8 - ii FOREWORD

More information

Design of Reinforced Concrete Structures (II)

Design of Reinforced Concrete Structures (II) Design of Reinforced Concrete Structures (II) Discussion Eng. Mohammed R. Kuheil Review The thickness of one-way ribbed slabs After finding the value of total load (Dead and live loads), the elements are

More information

STEEL MEMBER DESIGN (EN :2005)

STEEL MEMBER DESIGN (EN :2005) GEODOMISI Ltd. - Dr. Costas Sachpazis Consulting Company for App'd by STEEL MEMBER DESIGN (EN1993-1-1:2005) In accordance with EN1993-1-1:2005 incorporating Corrigenda February 2006 and April details type;

More information

SECTION 7 DESIGN OF COMPRESSION MEMBERS

SECTION 7 DESIGN OF COMPRESSION MEMBERS SECTION 7 DESIGN OF COMPRESSION MEMBERS 1 INTRODUCTION TO COLUMN BUCKLING Introduction Elastic buckling of an ideal column Strength curve for an ideal column Strength of practical column Concepts of effective

More information

9-3. Structural response

9-3. Structural response 9-3. Structural response in fire František Wald Czech Technical University in Prague Objectives of the lecture The mechanical load in the fire design Response of the structure exposed to fire Levels of

More information

FLOW CHART FOR DESIGN OF BEAMS

FLOW CHART FOR DESIGN OF BEAMS FLOW CHART FOR DESIGN OF BEAMS Write Known Data Estimate self-weight of the member. a. The self-weight may be taken as 10 percent of the applied dead UDL or dead point load distributed over all the length.

More information

Steel Frame Design Manual

Steel Frame Design Manual Steel Frame Design Manual Eurocode 3-1:2005 with 8:2004 Eurocode 3-1:2005 with Eurocode 8:2004 Steel Frame Design Manual for ETABS 2016 ISO ETA122815M13 Rev 0 Proudly developed in the United States of

More information

Application nr. 3 (Ultimate Limit State) Resistance of member cross-section

Application nr. 3 (Ultimate Limit State) Resistance of member cross-section Application nr. 3 (Ultimate Limit State) Resistance of member cross-section 1)Resistance of member crosssection in tension Examples of members in tension: - Diagonal of a truss-girder - Bottom chord of

More information

APPENDIX 1 MODEL CALCULATION OF VARIOUS CODES

APPENDIX 1 MODEL CALCULATION OF VARIOUS CODES 163 APPENDIX 1 MODEL CALCULATION OF VARIOUS CODES A1.1 DESIGN AS PER NORTH AMERICAN SPECIFICATION OF COLD FORMED STEEL (AISI S100: 2007) 1. Based on Initiation of Yielding: Effective yield moment, M n

More information

Structural Steelwork Eurocodes Development of a Trans-National Approach

Structural Steelwork Eurocodes Development of a Trans-National Approach Course: Eurocode 4 Structural Steelwork Eurocodes Development of a Trans-National Approach Lecture 9 : Composite joints Annex B References: COST C1: Composite steel-concrete joints in frames for buildings:

More information

to introduce the principles of stability and elastic buckling in relation to overall buckling, local buckling

to introduce the principles of stability and elastic buckling in relation to overall buckling, local buckling to introduce the principles of stability and elastic buckling in relation to overall buckling, local buckling In the case of elements subjected to compressive forces, secondary bending effects caused by,

More information

Tekla Structural Designer 2016i

Tekla Structural Designer 2016i Tekla Structural Designer 2016i Reference Guides (Australian Codes) September 2016 2016 Trimble Solutions Corporation part of Trimble Navigation Ltd. Table of Contents Analysis Verification Examples...

More information

Steel Structures Design and Drawing Lecture Notes

Steel Structures Design and Drawing Lecture Notes Steel Structures Design and Drawing Lecture Notes INTRODUCTION When the need for a new structure arises, an individual or agency has to arrange the funds required for its construction. The individual or

More information

Finite Element Modelling with Plastic Hinges

Finite Element Modelling with Plastic Hinges 01/02/2016 Marco Donà Finite Element Modelling with Plastic Hinges 1 Plastic hinge approach A plastic hinge represents a concentrated post-yield behaviour in one or more degrees of freedom. Hinges only

More information

Compression Members. ENCE 455 Design of Steel Structures. III. Compression Members. Introduction. Compression Members (cont.)

Compression Members. ENCE 455 Design of Steel Structures. III. Compression Members. Introduction. Compression Members (cont.) ENCE 455 Design of Steel Structures III. Compression Members C. C. Fu, Ph.D., P.E. Civil and Environmental Engineering Department University of Maryland Compression Members Following subjects are covered:

More information

Mechanics of Materials Primer

Mechanics of Materials Primer Mechanics of Materials rimer Notation: A = area (net = with holes, bearing = in contact, etc...) b = total width of material at a horizontal section d = diameter of a hole D = symbol for diameter E = modulus

More information

7.3 Design of members subjected to combined forces

7.3 Design of members subjected to combined forces 7.3 Design of members subjected to combined forces 7.3.1 General In the previous chapters of Draft IS: 800 LSM version, we have stipulated the codal provisions for determining the stress distribution in

More information

Design of Steel Structures Prof. S.R.Satish Kumar and Prof. A.R.Santha Kumar

Design of Steel Structures Prof. S.R.Satish Kumar and Prof. A.R.Santha Kumar 5.10 Examples 5.10.1 Analysis of effective section under compression To illustrate the evaluation of reduced section properties of a section under axial compression. Section: 00 x 80 x 5 x 4.0 mm Using

More information

Example 4: Design of a Rigid Column Bracket (Bolted)

Example 4: Design of a Rigid Column Bracket (Bolted) Worked Example 4: Design of a Rigid Column Bracket (Bolted) Example 4: Design of a Rigid Column Bracket (Bolted) Page : 1 Example 4: Design of a Rigid Column Bracket (Bolted) Determine the size of the

More information

Bridge deck modelling and design process for bridges

Bridge deck modelling and design process for bridges EU-Russia Regulatory Dialogue Construction Sector Subgroup 1 Bridge deck modelling and design process for bridges Application to a composite twin-girder bridge according to Eurocode 4 Laurence Davaine

More information

Sabah Shawkat Cabinet of Structural Engineering Walls carrying vertical loads should be designed as columns. Basically walls are designed in

Sabah Shawkat Cabinet of Structural Engineering Walls carrying vertical loads should be designed as columns. Basically walls are designed in Sabah Shawkat Cabinet of Structural Engineering 17 3.6 Shear walls Walls carrying vertical loads should be designed as columns. Basically walls are designed in the same manner as columns, but there are

More information

Design of AAC wall panel according to EN 12602

Design of AAC wall panel according to EN 12602 Design of wall panel according to EN 160 Example 3: Wall panel with wind load 1.1 Issue Design of a wall panel at an industrial building Materials with a compressive strength 3,5, density class 500, welded

More information

Downloaded from Downloaded from / 1

Downloaded from   Downloaded from   / 1 PURWANCHAL UNIVERSITY III SEMESTER FINAL EXAMINATION-2002 LEVEL : B. E. (Civil) SUBJECT: BEG256CI, Strength of Material Full Marks: 80 TIME: 03:00 hrs Pass marks: 32 Candidates are required to give their

More information

VALLOUREC & MANNESMANN TUBES. Design-support for MSH sections

VALLOUREC & MANNESMANN TUBES. Design-support for MSH sections VALLOUREC & MANNESMANN TUBES Design-support for MSH sections according to Eurocode 3, DIN EN 1993-1-1: 2005 and DIN EN 1993-1-8: 2005 Design-Support for MSH sections according to Eurocode 3, DIN EN 1993-1-1:

More information

BASE PLATE CONNECTIONS

BASE PLATE CONNECTIONS SKILLS Project BASE PLATE CONNECTIONS LEARNING OUTCOMES Design process for pinned and fixed column base joints Base-plate resistance Anchor bolt resistance Concrete resistance Weld resistance Application

More information

Job No. Sheet No. Rev. CONSULTING Engineering Calculation Sheet

Job No. Sheet No. Rev. CONSULTING Engineering Calculation Sheet CONSULTING Engineering Calculation Sheet E N G I N E E R S Consulting Engineers jxxx 1 Structural Description The two pinned (at the bases) portal frame is stable in its plane due to the moment connection

More information

: APPLIED MECHANICS & STRENGTH OF MATERIALS COURSE CODE : 4021 COURSE CATEGORY : A PERIODS/ WEEK : 5 PERIODS/ SEMESTER : 75 CREDIT : 5 TIME SCHEDULE

: APPLIED MECHANICS & STRENGTH OF MATERIALS COURSE CODE : 4021 COURSE CATEGORY : A PERIODS/ WEEK : 5 PERIODS/ SEMESTER : 75 CREDIT : 5 TIME SCHEDULE COURSE TITLE : APPLIED MECHANICS & STRENGTH OF MATERIALS COURSE CODE : 4021 COURSE CATEGORY : A PERIODS/ WEEK : 5 PERIODS/ SEMESTER : 75 CREDIT : 5 TIME SCHEDULE MODULE TOPIC PERIODS 1 Simple stresses

More information

Compression Members Columns II

Compression Members Columns II Compression Members Columns II 1. Introduction. Main aspects related to the derivation of typical columns buckling lengths for. Analysis of imperfections, leading to the derivation of the Ayrton-Perry

More information

Structural Mechanics Column Behaviour

Structural Mechanics Column Behaviour Structural Mechanics Column Behaviour 008/9 Dr. Colin Caprani, 1 Contents 1. Introduction... 3 1.1 Background... 3 1. Stability of Equilibrium... 4. Buckling Solutions... 6.1 Introduction... 6. Pinned-Pinned

More information

CHAPTER 4. ANALYSIS AND DESIGN OF COLUMNS

CHAPTER 4. ANALYSIS AND DESIGN OF COLUMNS 4.1. INTRODUCTION CHAPTER 4. ANALYSIS AND DESIGN OF COLUMNS A column is a vertical structural member transmitting axial compression loads with or without moments. The cross sectional dimensions of a column

More information

DEPARTMENT OF CIVIL ENGINEERING

DEPARTMENT OF CIVIL ENGINEERING KINGS COLLEGE OF ENGINEERING DEPARTMENT OF CIVIL ENGINEERING SUBJECT: CE 2252 STRENGTH OF MATERIALS UNIT: I ENERGY METHODS 1. Define: Strain Energy When an elastic body is under the action of external

More information

Design of reinforced concrete sections according to EN and EN

Design of reinforced concrete sections according to EN and EN Design of reinforced concrete sections according to EN 1992-1-1 and EN 1992-2 Validation Examples Brno, 21.10.2010 IDEA RS s.r.o. South Moravian Innovation Centre, U Vodarny 2a, 616 00 BRNO tel.: +420-511

More information

CHAPTER 4. Design of R C Beams

CHAPTER 4. Design of R C Beams CHAPTER 4 Design of R C Beams Learning Objectives Identify the data, formulae and procedures for design of R C beams Design simply-supported and continuous R C beams by integrating the following processes

More information

QUESTION BANK SEMESTER: III SUBJECT NAME: MECHANICS OF SOLIDS

QUESTION BANK SEMESTER: III SUBJECT NAME: MECHANICS OF SOLIDS QUESTION BANK SEMESTER: III SUBJECT NAME: MECHANICS OF SOLIDS UNIT 1- STRESS AND STRAIN PART A (2 Marks) 1. Define longitudinal strain and lateral strain. 2. State Hooke s law. 3. Define modular ratio,

More information

APOLLO SCAFFOLDING SERVICES LTD SPIGOT CONNECTION TO EUROCODES DESIGN CHECK CALCULATIONS

APOLLO SCAFFOLDING SERVICES LTD SPIGOT CONNECTION TO EUROCODES DESIGN CHECK CALCULATIONS Alan White Design APOLLO SCAFFOLDING SERVICES LTD SPIGOT CONNECTION TO EUROCODES DESIGN CHECK CALCULATIONS Alan N White B.Sc., M.Eng., C.Eng., M.I.C.E., M.I.H.T. JUL 2013 Somerset House 11 Somerset Place

More information

Slenderness Effects for Concrete Columns in Sway Frame - Moment Magnification Method (CSA A )

Slenderness Effects for Concrete Columns in Sway Frame - Moment Magnification Method (CSA A ) Slenderness Effects for Concrete Columns in Sway Frame - Moment Magnification Method (CSA A23.3-94) Slender Concrete Column Design in Sway Frame Buildings Evaluate slenderness effect for columns in a

More information

COURSE TITLE : APPLIED MECHANICS & STRENGTH OF MATERIALS COURSE CODE : 4017 COURSE CATEGORY : A PERIODS/WEEK : 6 PERIODS/ SEMESTER : 108 CREDITS : 5

COURSE TITLE : APPLIED MECHANICS & STRENGTH OF MATERIALS COURSE CODE : 4017 COURSE CATEGORY : A PERIODS/WEEK : 6 PERIODS/ SEMESTER : 108 CREDITS : 5 COURSE TITLE : APPLIED MECHANICS & STRENGTH OF MATERIALS COURSE CODE : 4017 COURSE CATEGORY : A PERIODS/WEEK : 6 PERIODS/ SEMESTER : 108 CREDITS : 5 TIME SCHEDULE MODULE TOPICS PERIODS 1 Simple stresses

More information

Where and are the factored end moments of the column and >.

Where and are the factored end moments of the column and >. 11 LIMITATION OF THE SLENDERNESS RATIO----( ) 1-Nonsway (braced) frames: The ACI Code, Section 6.2.5 recommends the following limitations between short and long columns in braced (nonsway) frames: 1. The

More information

Civil Engineering Design (1) Design of Reinforced Concrete Columns 2006/7

Civil Engineering Design (1) Design of Reinforced Concrete Columns 2006/7 Civil Engineering Design (1) Design of Reinforced Concrete Columns 2006/7 Dr. Colin Caprani, Chartered Engineer 1 Contents 1. Introduction... 3 1.1 Background... 3 1.2 Failure Modes... 5 1.3 Design Aspects...

More information

Annex - R C Design Formulae and Data

Annex - R C Design Formulae and Data The design formulae and data provided in this Annex are for education, training and assessment purposes only. They are based on the Hong Kong Code of Practice for Structural Use of Concrete 2013 (HKCP-2013).

More information

ε t increases from the compressioncontrolled Figure 9.15: Adjusted interaction diagram

ε t increases from the compressioncontrolled Figure 9.15: Adjusted interaction diagram CHAPTER NINE COLUMNS 4 b. The modified axial strength in compression is reduced to account for accidental eccentricity. The magnitude of axial force evaluated in step (a) is multiplied by 0.80 in case

More information

Metal Structures Lecture XIII Steel trusses

Metal Structures Lecture XIII Steel trusses Metal Structures Lecture XIII Steel trusses Contents Definition #t / 3 Geometry and cross-sections #t / 7 Types of truss structures #t / 15 Calculations #t / 29 Example #t / 57 Results of calculations

More information

PES Institute of Technology

PES Institute of Technology PES Institute of Technology Bangalore south campus, Bangalore-5460100 Department of Mechanical Engineering Faculty name : Madhu M Date: 29/06/2012 SEM : 3 rd A SEC Subject : MECHANICS OF MATERIALS Subject

More information

9.5 Compression Members

9.5 Compression Members 9.5 Compression Members This section covers the following topics. Introduction Analysis Development of Interaction Diagram Effect of Prestressing Force 9.5.1 Introduction Prestressing is meaningful when

More information

Unit III Theory of columns. Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE, Sriperumbudir

Unit III Theory of columns. Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE, Sriperumbudir Unit III Theory of columns 1 Unit III Theory of Columns References: Punmia B.C.,"Theory of Structures" (SMTS) Vol II, Laxmi Publishing Pvt Ltd, New Delhi 2004. Rattan.S.S., "Strength of Materials", Tata

More information

ENG1001 Engineering Design 1

ENG1001 Engineering Design 1 ENG1001 Engineering Design 1 Structure & Loads Determine forces that act on structures causing it to deform, bend, and stretch Forces push/pull on objects Structures are loaded by: > Dead loads permanent

More information

SERVICEABILITY LIMIT STATE DESIGN

SERVICEABILITY LIMIT STATE DESIGN CHAPTER 11 SERVICEABILITY LIMIT STATE DESIGN Article 49. Cracking Limit State 49.1 General considerations In the case of verifications relating to Cracking Limit State, the effects of actions comprise

More information

FRAME ANALYSIS. Dr. Izni Syahrizal bin Ibrahim. Faculty of Civil Engineering Universiti Teknologi Malaysia

FRAME ANALYSIS. Dr. Izni Syahrizal bin Ibrahim. Faculty of Civil Engineering Universiti Teknologi Malaysia FRAME ANALYSIS Dr. Izni Syahrizal bin Ibrahim Faculty of Civil Engineering Universiti Teknologi Malaysia Email: iznisyahrizal@utm.my Introduction 3D Frame: Beam, Column & Slab 2D Frame Analysis Building

More information

Design of Steel Structures Prof. Damodar Maity Department of Civil Engineering Indian Institute of Technology, Guwahati

Design of Steel Structures Prof. Damodar Maity Department of Civil Engineering Indian Institute of Technology, Guwahati Design of Steel Structures Prof. Damodar Maity Department of Civil Engineering Indian Institute of Technology, Guwahati Module 7 Gantry Girders and Plate Girders Lecture - 3 Introduction to Plate girders

More information

= = = 1,000 1,000 1,250. g M0 g M1 g M2 = = = 1,100 1,100 1,250 [ ] 1 0,000 8,000 HE 140 B 0,0. [m] Permanent Permanent Variable Variable Variable

= = = 1,000 1,000 1,250. g M0 g M1 g M2 = = = 1,100 1,100 1,250 [ ] 1 0,000 8,000 HE 140 B 0,0. [m] Permanent Permanent Variable Variable Variable Project Job name Part Author Date Steel Products and Solutions Standard 29.01.2018 Standard EN 199311, EN 199314/Czech Rep.. Factors for steel structures Section capacity Section resistance when checking

More information

Chapter 8: Bending and Shear Stresses in Beams

Chapter 8: Bending and Shear Stresses in Beams Chapter 8: Bending and Shear Stresses in Beams Introduction One of the earliest studies concerned with the strength and deflection of beams was conducted by Galileo Galilei. Galileo was the first to discuss

More information

QUESTION BANK DEPARTMENT: CIVIL SEMESTER: III SUBJECT CODE: CE2201 SUBJECT NAME: MECHANICS OF SOLIDS UNIT 1- STRESS AND STRAIN PART A

QUESTION BANK DEPARTMENT: CIVIL SEMESTER: III SUBJECT CODE: CE2201 SUBJECT NAME: MECHANICS OF SOLIDS UNIT 1- STRESS AND STRAIN PART A DEPARTMENT: CIVIL SUBJECT CODE: CE2201 QUESTION BANK SEMESTER: III SUBJECT NAME: MECHANICS OF SOLIDS UNIT 1- STRESS AND STRAIN PART A (2 Marks) 1. Define longitudinal strain and lateral strain. 2. State

More information

STRUCTURAL ANALYSIS CHAPTER 2. Introduction

STRUCTURAL ANALYSIS CHAPTER 2. Introduction CHAPTER 2 STRUCTURAL ANALYSIS Introduction The primary purpose of structural analysis is to establish the distribution of internal forces and moments over the whole part of a structure and to identify

More information

General Comparison between AISC LRFD and ASD

General Comparison between AISC LRFD and ASD General Comparison between AISC LRFD and ASD 1 General Comparison between AISC LRFD and ASD 2 AISC ASD and LRFD AISC ASD = American Institute of Steel Construction = Allowable Stress Design AISC Ninth

More information

2. (a) Explain different types of wing structures. (b) Explain the advantages and disadvantages of different materials used for aircraft

2. (a) Explain different types of wing structures. (b) Explain the advantages and disadvantages of different materials used for aircraft Code No: 07A62102 R07 Set No. 2 III B.Tech II Semester Regular/Supplementary Examinations,May 2010 Aerospace Vehicle Structures -II Aeronautical Engineering Time: 3 hours Max Marks: 80 Answer any FIVE

More information

7 STATICALLY DETERMINATE PLANE TRUSSES

7 STATICALLY DETERMINATE PLANE TRUSSES 7 STATICALLY DETERMINATE PLANE TRUSSES OBJECTIVES: This chapter starts with the definition of a truss and briefly explains various types of plane truss. The determinancy and stability of a truss also will

More information

Balcony balustrades using the SG12 laminated glass system: PAGE 1 (SG12FF010717) Structural Calculations for SG12 System balustrades using 21.5mm laminated toughened glass without the need for a handrail

More information

CIVL473 Fundamentals of Steel Design

CIVL473 Fundamentals of Steel Design CIVL473 Fundamentals of Steel Design CHAPTER 4 Design of Columns- embers with Aial Loads and oments Prepared B Asst.Prof.Dr. urude Celikag 4.1 Braced ultistore Buildings - Combined tension and oments Interaction

More information

R13. II B. Tech I Semester Regular Examinations, Jan MECHANICS OF SOLIDS (Com. to ME, AME, AE, MTE) PART-A

R13. II B. Tech I Semester Regular Examinations, Jan MECHANICS OF SOLIDS (Com. to ME, AME, AE, MTE) PART-A SET - 1 II B. Tech I Semester Regular Examinations, Jan - 2015 MECHANICS OF SOLIDS (Com. to ME, AME, AE, MTE) Time: 3 hours Max. Marks: 70 Note: 1. Question Paper consists of two parts (Part-A and Part-B)

More information

Level 7 Postgraduate Diploma in Engineering Computational mechanics using finite element method

Level 7 Postgraduate Diploma in Engineering Computational mechanics using finite element method 9210-203 Level 7 Postgraduate Diploma in Engineering Computational mechanics using finite element method You should have the following for this examination one answer book No additional data is attached

More information

Critical Load columns buckling critical load

Critical Load columns buckling critical load Buckling of Columns Buckling of Columns Critical Load Some member may be subjected to compressive loadings, and if these members are long enough to cause the member to deflect laterally or sideway. To

More information

MECHANICS OF STRUCTURES SCI 1105 COURSE MATERIAL UNIT - I

MECHANICS OF STRUCTURES SCI 1105 COURSE MATERIAL UNIT - I MECHANICS OF STRUCTURES SCI 1105 COURSE MATERIAL UNIT - I Engineering Mechanics Branch of science which deals with the behavior of a body with the state of rest or motion, subjected to the action of forces.

More information

Design of Steel Structures Prof. S.R.Satish Kumar and Prof. A.R.Santha Kumar. Local buckling is an extremely important facet of cold formed steel

Design of Steel Structures Prof. S.R.Satish Kumar and Prof. A.R.Santha Kumar. Local buckling is an extremely important facet of cold formed steel 5.3 Local buckling Local buckling is an extremely important facet of cold formed steel sections on account of the fact that the very thin elements used will invariably buckle before yielding. Thinner the

More information

SIMPLE MODEL FOR PRYING FORCES IN T-HANGER CONNECTIONS WITH SNUG TIGHTENED BOLTS

SIMPLE MODEL FOR PRYING FORCES IN T-HANGER CONNECTIONS WITH SNUG TIGHTENED BOLTS SIMPLE MODEL FOR PRYING FORCES IN T-HANGER CONNECTIONS WITH SNUG TIGHTENED BOLTS By Fathy Abdelmoniem Abdelfattah Faculty of Engineering at Shoubra, Zagazig University, Banha Branch Mohamed Salah A. Soliman

More information

Structural Steelwork Eurocodes Development of a Trans-National Approach

Structural Steelwork Eurocodes Development of a Trans-National Approach Structural Steelwork Eurocodes Development of a Trans-National Approach Course: Eurocode 4 Lecture 9 : Composite joints Annex A Summary: Traditionally structural joints are considered as rigid or pinned

More information

Structural Calculations for Juliet balconies using BALCONY 2 System (Aerofoil) handrail. Our ref: JULB2NB Date of issue: March 2017

Structural Calculations for Juliet balconies using BALCONY 2 System (Aerofoil) handrail. Our ref: JULB2NB Date of issue: March 2017 Juliet balconies using BALCONY 2 System (Aerofoil) handrail PAGE 1 (ref: JULB2NB280317) Structural Calculations for Juliet balconies using BALCONY 2 System (Aerofoil) handrail Our ref: JULB2NB280317 Date

More information

D : SOLID MECHANICS. Q. 1 Q. 9 carry one mark each. Q.1 Find the force (in kn) in the member BH of the truss shown.

D : SOLID MECHANICS. Q. 1 Q. 9 carry one mark each. Q.1 Find the force (in kn) in the member BH of the truss shown. D : SOLID MECHANICS Q. 1 Q. 9 carry one mark each. Q.1 Find the force (in kn) in the member BH of the truss shown. Q.2 Consider the forces of magnitude F acting on the sides of the regular hexagon having

More information

two structural analysis (statics & mechanics) APPLIED ACHITECTURAL STRUCTURES: DR. ANNE NICHOLS SPRING 2017 lecture STRUCTURAL ANALYSIS AND SYSTEMS

two structural analysis (statics & mechanics) APPLIED ACHITECTURAL STRUCTURES: DR. ANNE NICHOLS SPRING 2017 lecture STRUCTURAL ANALYSIS AND SYSTEMS APPLIED ACHITECTURAL STRUCTURES: STRUCTURAL ANALYSIS AND SYSTEMS DR. ANNE NICHOLS SPRING 2017 lecture two structural analysis (statics & mechanics) Analysis 1 Structural Requirements strength serviceability

More information