Design of Compression Members

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1 Design of Compression Members

2 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 strength of the material. 4- Stress distribution across the width of the plate element. There are basically four different cross-section classes and they are defined as: 1- Class 1 cross-sections: are those which can form a plastic hinge with the rotation capacity required from plastic analysis without reduction of the resistance. 2- Class 2 cross-sections: are those which can develop their plastic moment resistance, but have limited rotation capacity because of local buckling. 3- Class3 cross-sections: are those in which the stress in the extreme compression fiber of the steel member assuming an elastic distribution of stresses can reach the yield strength, but local buckling is liable to prevent development of the plastic moment resistance. 4- Class 4 cross-sections: are those in which local buckling will occur before the attainment of yield stress in one or more parts of the cross section. Page(2) Dr.Mamoun Alqedra Eng.Mohammed AbuRahma Eng. Haya Baker

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6 2.2 Support Conditions Note: 1- If buckling occurs, it will take place in a plane perpendicular to the crossponding principal axis of inertia. 2- The slenderness to be used, λ, is the larger of those calculated in the y and z directions, that is λ = max (λy, λz). For example, it is possible to make reference to the compression member in the below Figure, restrained in different ways in the x y and x z planes, where the x-axis is the one along the length of the member. The effective length in the x y plane has to be taken as L/4 (i.e. LZ = 2.25 m), whereas in the x z plane it is 0.7L(Ly = 6.3 m). Page(6) Dr.Mamoun Alqedra Eng.Mohammed AbuRahma Eng. Haya Baker

7 2.3 Design for compression 1- Plastic Resistance: The design values of the compression force N Ed at each cross-section shall satisfy: N Ed N c,rd 1.0 The design resistance of the cross-section for uniform compression N c,rd should be determined as follows: N c,rd = A f y γ M0 for class 1,2 or 3 cross-sections N c,rd = A eff f y γ M0 For class 4 cross-sections 2- Buckling Resistance: A compression member should be verified against buckling as follows: N Ed N b,rd 1.0 The design buckling resistance of a compression member should be taken as: N c,rd = χ A f y γ M1 for class 1,2 or 3 cross-sections N c,rd = χ A eff f y γ M1 for class 4 cross-sections Where χ is the reduction factor for the relevant buckling mode. Page(7) Dr.Mamoun Alqedra Eng.Mohammed AbuRahma Eng. Haya Baker

2.4 Buckling curves For axial compression in members the value of χ for the appropriate non-dimensional slenderness λ should be determined from the relevant buckling curve according to: 1 χ =, but χ

8 2.4 Buckling curves For axial compression in members the value of χ for the appropriate non-dimensional slenderness λ should be determined from the relevant buckling curve according to: 1 χ =, but χ λ 2 where = 0.5[1 + α(λ 0.2) + λ 2 ] λ = A f y N cr = L cr λ = A eff f y N cr i 1 λ 1 = L cr A eff /A i λ 1 for class 1,2 or 3 cross-sections for class 4 cross-sections α is an imperfection factor. N cr is the elastic critical force for the relevant buckling mode based on the gross cross-sectional properties. The imperfection factor α corresponding to the appropriate buckling curve should be obtained from Table 6.1 and Table 6.2. Page(8) Dr.Mamoun Alqedra Eng.Mohammed AbuRahma Eng. Haya Baker

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10 Values of the reduction factor χ for the appropriate non-dimensional slenderness λ may be obtained from Figure 6.4. For slenderness λ 0.2 or for N Ed N cr 0.04 the buckling effects may be ignored and only cross sectional checks apply. Page(10) Dr.Mamoun Alqedra Eng.Mohammed AbuRahma Eng. Haya Baker

11 2.5 Solved Problems Problem (1) Design a lighting column subjected to an axial compression force 30 KN, using a CHS (circular hollow section) cross section in S 275 steel, according to EC The column is fixed at the base and free from the other end. With length of 3 m. Solution: Preliminary design Assuming class 1, 2 or 3 cross sections, yields: N Ed = 30 KN N c,rd = Af y γ M0 = A /1.0 A m 2 = 1.09 cm 2 Use CHS section with A=2.38 cm 2, d/t=8.41, I = 1.7 cm 4, i= cm. Classification of the section d/t=8.41,ε = 235/275 = 0.92 d t < 50ε < 50(0.92) < Class 1 Buckling lengths According to the support conditions, the buckling lengths are equal in both planes, given by: Buckling in the plane of the structure - L E = 2 3 = 6 m Determination of the slenderness coefficients λ 1 = π = λ = L i = = λ = λ = = 8.16 > 1 Long column λ Calculation of the reduction factor x curve a α = x =, but x λ 2 = 0.5[1 + α(λ 0.2) + λ 2 ] = 0.5[ ( ) ] =34.62 x = = Page(11) Dr.Mamoun Alqedra Eng.Mohammed AbuRahma Eng. Haya Baker

12 -Safety verification N b,rd = xaf y γ M1 = = KN As NEd = 30 kn > N b,rd = kn Safety is not verified Try heavier section such CHS section with A=81.1 cm 2, d/t=17.5, I = 4350 cm 4, i= 7.32 cm. Classification of the section d/t=17.5,ε = 235/275 = 0.92 d t < 50ε < 50(0.92) < Class 1 Buckling lengths According to the support conditions, the buckling lengths are equal in both planes, given by: Buckling in the plane of the structure - L E = 2 3 = 6 m Determination of the slenderness coefficients λ 1 = π = λ = L i = = λ = λ = = < 1 short column λ Calculation of the reduction factor x curve a α = 0.21 = 0.5[ ( ) ] =1.02 x = = Safety verification N b,rd = xaf y γ M1 = = KN As NEd = 30 kn < N b,rd = kn Safety is verified Page(12) Dr.Mamoun Alqedra Eng.Mohammed AbuRahma Eng. Haya Baker

13 Problem (2) Check a column subjected to an axial compression force 6000 KN, using a UC (universal column) cross section in S 355 steel, according to EC The column is supported as shown in the figure. With length of 5 m. buckling in X Y plan buckling in X Z plane Solution: Section with A=213 cm 2, (c/t) flange=3.48, (c/t) web=10.4, Iy-y = cm 4, Iz-z = 9870 cm 4, iy-y = 11.9 cm, iz-z = 6.81 cm Classification of the section Flange: (c/t) =3.48, ε = 235 = 0.81 c < 9ε 3.48 < t 3.48 < 7.29 Class 1 Web: (c/t) =10.4, ε = 235 = 0.81 c < 33ε 10.4 < t 10.4 < Class 1 The cross section is Class 1 N c,rd = Af y γ M0 = = > 6000 Safe 1.0 Buckling lengths According to the support conditions, the buckling lengths are equal in both planes, given by: Buckling in the plane of the structure (plane x-z) - L E y = = 5.0 m Buckling in the plane of the structure (plane x-y) - L E z = = 3.0 m Page(13) Dr.Mamoun Alqedra Eng.Mohammed AbuRahma Eng. Haya Baker

14 Determination of the slenderness coefficients λ 1 = π = 76.4 λ y = L Ey = = 42.01, λ i y 11.9 y = λ y = = 0.55 < 1 Short column λ λ z = L Ez = = 44.05, λ i z 6.81 z = λ z = = 0.57 < 1 Short column λ Calculation of the reduction factor x h/b = 289.1/265.2 = 1.09 < 1.2, tf=31.7 mm < 100 mm bending around z curve c α = 0.49 = 0.5[ ( ) ] = 0.75 x = = Safety verification N b,rd = xaf y γ M1 = = KN As NEd = 6000 kn < N b,rd = kn Safety is verified. Page(14) Dr.Mamoun Alqedra Eng.Mohammed AbuRahma Eng. Haya Baker

15 Problem (3) Check a column subjected to an axial compression force 2500 KN, using a UB (universal beam) cross section in S 275 steel, according to EC The column is supported as shown in the figure. With length of 6 m. Solution: Section with A=105 cm 2, (c/t) flange=6.58, (c/t) web=49.6, Iy-y = cm 4, Iz-z = 2010 cm 4, iy-y = 21.3 cm, iz-z = 4.38 cm Classification of the section Flange: (c/t) =6.58, ε = 235 = 0.92 c < 9ε 6.58 < t 6.58 < 8.28 Class 1 Web: (c/t) =49.6, ε = 235 = 0.92 c < 42ε 49.6 < t 49.6 < Class 4 From bluebook A eff = 96.4 cm 2 N c,rd = A eff f y γ M0 = = 2651 > 2500 Safe 1.0 Buckling lengths According to the support conditions, the buckling lengths are equal in both planes, given by: Buckling in the plane of the structure (plane x-z) - L E y = = 6.0 m Buckling in the plane of the structure (plane x-y) - L E z = = 2.0 m Page(15) Dr.Mamoun Alqedra Eng.Mohammed AbuRahma Eng. Haya Baker

16 Determination of the slenderness coefficients λ 1 = π = = L cr λ y λ z i y = L cr i z A eff /A λ 1 A eff /A λ 1 = / = / Calculation of the reduction factor x = 0.31 < 1 Short column = 0.50 < 1 Short column h/b = 528.3/208.8 = 2.53 > 1.2, tf=13.2 mm < 40 mm bending around z curve b α = 0.34 = 0.5[ ( ) ] = x = = Safety verification N b,rd = x A eff f y γ M1 = = KN As NEd = 2500 kn > N b,rd = kn Safety is not verified. Page(16) Dr.Mamoun Alqedra Eng.Mohammed AbuRahma Eng. Haya Baker

17 Problem (4) The following truss design the upper cord members compressed members, considering the same type of cross section, that is: Square hollow sections (SHS), with welded connections between the members of the structure. Solution: Based on the axial force diagrams represented in Figure 3.53, the most compressed chord member is under an axial force of kn and it is simultaneously one of the longest members, with L = 3.00 m; For the definition of the buckling lengths of the members, it is assumed that all the nodes of the truss are braced in the direction perpendicular to the plane of the structure. Preliminary design Assuming class 1, 2 or 3 cross sections, yields: Upper cord: N Ed = KN N c,rd = Af y γ M0 = A /1.0 A m 2 = 27 cm 2 Use SHS for upper cord with A=35.5 cm 2, I=738cm 4, i=4.56cm Classification of the section Upper cord: c/t=12,ε = 235/275 = 0.92 c < 33ε 12 < 33(0.92) 12 < t 30.4 Class 1 Determination of the slenderness coefficients λ 1 = π = 86.81, LE=3.0 m λ = L E = = λ = λ = = < 1 Short column i 4.56 λ Calculation of the reduction factor x curve a α = 0.21 Page(17) Dr.Mamoun Alqedra Eng.Mohammed AbuRahma Eng. Haya Baker

18 = 0.5[ ( ) ] = x = = Safety verification N b,rd = xaf y γ M1 = = KN As NEd = kn < N b,rd = kn Safety is verified Page(18) Dr.Mamoun Alqedra Eng.Mohammed AbuRahma Eng. Haya Baker

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