FM_TOC 46060 6/22/10 11:26 AM Page iii CONTENTS To the Instructor iv 1 Stress 1 2 Strain 73 3 Mechanical Properties of Materials 92 4 Axial Load 122 5 Torsion 214 6 Bending 329 7 Transverse Shear 472 8 Combined Loadings 532 9 Stress Transformation 619 10 Strain Transformation 738 11 Design of Beams and Shafts 830 12 Deflection of Beams and Shafts 883 13 Buckling of Columns 1038 14 Energy Methods 1159 13 Solutions 46060 6/11/10 11:56 AM Page 1038 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •13–1. Determine the critical buckling load for the column. The material can be assumed rigid. P L 2 k Equilibrium: The disturbing force F can be determined by summing moments about point A. a + ©MA = 0; P(Lu) - F a L b = 0 2 A F = 2Pu Spring Formula: The restoring spring force F1 can be determine using spring formula Fs = kx. Fs = ka L kLu ub = 2 2 Critical Buckling Load: For the mechanism to be on the verge of buckling, the disturbing force F must be equal to the restoring spring force F1. 2Pcr u = Pcr = L 2 kLu 2 kL 4 Ans. 1038 13 Solutions 46060 6/11/10 11:56 AM Page 1039 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–2. Determine the critical load Pcr for the rigid bar and spring system. Each spring has a stiffness k. P Equilibrium: The disturbing forces F1 and F2 can be related to P by writing the moment equation of equlibrium about point A. Using small angle ananlysis, where cos u ⬵ 1 and sin u = u, + ©MA = 0; F2 a L 3 k L 2 b + F1 a L b - PLu = 01 3 3 L 3 F2 + 2F1 = 3Pu k (1) Spring Force. The restoring spring force A Fsp B 1 and A Fsp B 2 can be determined using the spring formula, 2 1 Lu and x2 = Lu, Fig. b. Thus, 3 3 2 2 = kx1 = ka Lu b = kLu 3 3 L 3 A Fsp = kx, where x1 = A Fsp B 1 A Fsp B 2 = kx2 = ka Lu b = 1 3 Critical Buckling Load. When the mechanism is on the verge of buckling the disturbing force F must be equal to the restoring force of the spring Fsp. Thus, F1 = A Fsp B 1 = 2 kLu 3 F2 = A Fsp B 2 = 1 kLu 3 Substituting this result into Eq. (1), 2 1 kLu + 2 a kLu b = 3Pcr u 3 3 Pcr = 5 kL 9 Ans. 1039 1 kLu 3 13 Solutions 46060 6/11/10 11:56 AM Page 1040 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–3. The leg in (a) acts as a column and can be modeled (b) by the two pin-connected members that are attached to a torsional spring having a stiffness k (torque兾rad). Determine the critical buckling load. Assume the bone material is rigid. a + ©MA = 0; - P(u)a P L b + 2ku = 0 2 L — 2 Require: k Pcr = 4k L Ans. L — 2 (a) *13–4. Rigid bars AB and BC are pin connected at B. If the spring at D has a stiffness k, determine the critical load Pcr for the system. (b) P A Equilibrium. The disturbing force F can be related P by considering the equilibrium of joint A and then the equilibrium of member BC, a B Joint A (Fig. b) + c ©Fy = 0; FAB cos f - P = 0 FAB = a P cos f k D Member BC (Fig. c) a ©MC = 0; F(a cos u) - P P cos f (2a sin u) sin f(2a cos u) = 0 cos f cos f C F = 2P(tan u + tan f) Since u and f are small, tan u ⬵ u and tan f ⬵ f. Thus, F = 2P(u + f) (1) Also, from the geometry shown in Fig. a, 2au = af f = 2u Thus Eq. (1) becomes F = 2P(u + 2u) = 6Pu Spring Force. The restoring spring force Fsp can be determined using the spring formula, Fsp = kx, where x = au, Fig. a. Thus, Fsp = kx = kau 1040 13 Solutions 46060 6/11/10 11:56 AM Page 1041 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–4. Continued Critical Buckling Load. When the mechanism is on the verge of buckling the disturbing force F must be equal to the restoring spring force Fsp. F = Fsp 6Pcru = kau Pcr = ka 6 Ans. 1041 13 Solutions 46060 6/11/10 11:56 AM Page 1042 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •13–5. An A-36 steel column has a length of 4 m and is pinned at both ends. If the cross sectional area has the dimensions shown, determine the critical load. 25 mm Section Properties: A = 0.01(0.06) + 0..05(0.01) = 1.10 A 10 - 3 B m2 Ix = Iy = 10 mm 1 1 (0.01) A 0.063 B + (0.05) A 0.013 B = 0.184167 A 10 - 6 B m4 12 12 25 mm Critical Buckling Load: K = 1 for pin supported ends column. Applying Euler’s formula, Pcr = 25 mm 25 mm 10 mm p2EI (KL)2 p2 (200)(109)(0.184167)(10 - 6) = [1(4)]2 = 22720.65 N = 22.7 kN Ans. Critical Stress: Euler’s formula is only valid if scr 6 sg. scr = Pcr 22720.65 = 20.66 MPa 6 sg = 250 MPa = A 1.10(10 - 3) O.K. 13–6. Solve Prob. 13–5 if the column is fixed at its bottom and pinned at its top. 25 mm Section Properties: A = 0.01(0.06) + 0.05(0.01) = 1.10 A 10 - 3 B m2 10 mm 1 1 Ix = Iy = (0.01) A 0.063 B + (0.05) A 0.013 B = 0.184167 A 10 - 6 B m4 12 12 25 mm Critical Buckling Load: K = 0.7 for one end fixed and the other end pinned column. Applying Euler’s formula, Pcr = p EI (EL)2 p2 (200)(109)(0.184167)(10 - 6) = [0.7(4)]2 = 46368.68 N = 46.4 kN Ans. Critical Stress: Euler’s formula is only valid if scr 6 sg. scr = 25 mm 25 mm 10 mm 2 Pcr 46368.68 = 42.15 MPa 6 sg = 250 MPa = A 1.10(10 - 3) 1042 O.K. 13 Solutions 46060 6/11/10 11:56 AM Page 1043 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–7. A column is made of A-36 steel, has a length of 20 ft, and is pinned at both ends. If the cross-sectional area has the dimensions shown, determine the critical load. 6 in. 0.25 in. The cross sectional area and moment of inertia of the square tube is 5.5 in. A = 6(6) - 5.5(5.5) = 5.75 in2 I = 0.25 in. 1 1 (6)(63) (5.5)(5.53) = 31.74 in4 12 12 0.25 in. 0.25 in. The column is pinned at both of its end, k = 1. For A36 steel, E = 29.0(103) ksi and sg = 36 ksi (table in appendix). Applying Euler’s formula, Pcr = p2 C 29.0(103) D (31.74) p2EI = (KL)2 C 1(20)(12) D 2 Ans. = 157.74 kip = 158 Critical Stress. Euler’s formula is valid only if scr 6 sg. Pcr 157.74 = = 27.4 ksi 6 sg = 36 ksi A 5.75 scr = O.K. *13–8. A column is made of 2014-T6 aluminum, has a length of 30 ft, and is fixed at its bottom and pinned at its top. If the cross-sectional area has the dimensions shown, determine the critical load. 6 in. 0.25 in. 5.5 in. The cross-sectional area and moment of inertia of the square tube is 0.25 in. A = 6(6) - 5.5(5.5) = 5.75 in2 0.25 in. 1 1 I = (6)(63) (5.5)(5.53) = 31.74 in4 12 12 The column is fixed at one end, K = 0.7. For 2014–76 aluminium, E = 10.6(103) ksi and sg = 60 ksi (table in appendix). Applying Euler’s formula, Pcr = p2 C 10.6(103) D (31.74) p2EI = (KL)2 C 0.7(30)(12) D 2 Ans. = 52.29 kip = 52.3 kip Critical Stress. Euler’s formula is valid only if scr 6 sg. scr = Pcr 52.3 = = 9.10 ksi 6 sg = 60 ksi A 5.75 O.K. 1043 0.25 in. 13 Solutions 46060 6/11/10 11:56 AM Page 1044 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •13–9. The W14 * 38 column is made of A-36 steel and is fixed supported at its base. If it is subjected to an axial load of P = 15 kip, determine the factor of safety with respect to buckling. P From the table in appendix, the cross-sectional area and moment of inertia about weak axis (y-axis) for W14 * 38 are 20 ft A = 11.2 in2 Iy = 26.7 in4 The column is fixed at its base and free at top, k = 2. Here, the column will buckle about the weak axis (y axis). For A36 steel, E = 29.0(103) ksi and sy = 36 ksi. Applying Euler’s formula, p2 C 29.0(103) D (26.7) p2EIy Pcr = C 2 (20)(12) D 2 = (KL)2 = 33.17 kip Thus, the factor of safety with respect to buckling is F.S = Pcr 33.17 = = 2.21 P 15 Ans. The Euler’s formula is valid only if scr 6 sg. scr = Pcr 33.17 = = 2.96 ksi 6 sg = 36 ksi A 11.2 O.K. 13–10. The W14 * 38 column is made of A-36 steel. Determine the critical load if its bottom end is fixed supported and its top is free to move about the strong axis and is pinned about the weak axis. P From the table in appendix, the cross-sectional area and moment of inertia about weak axis (y-axis) for W14 * 38 are A = 11.2 in2 Ix = 385 in4 Iy = 26.7 in4 The column is fixed at its base and free at top about strong axis. Thus, kx = 2. For A36 steel, E = 29.0(103) ksi and sg = 36 ksi. Pcr = p2EIx (KxLx) 2 = p2 C 29.0(103) D (385) C 2 (20)(12) D 2 = 478.28 kip The column is fixed at its base and pinned at top about weak axis. Thus, ky = 0.7. Pcr = p2EIy 2 (KyLy) = p2 C 29.0(103) D (26.7) C 0.7(20)(12) D 2 Ans. = 270.76 kip = 271 kip (Control) The Euler’s formula is valid only if scr 6 sg. scr = Pcr 270.76 = = 24.17 ksi 6 sg = 36 ksi A 11.2 O.K. 1044 20 ft 13 Solutions 46060 6/11/10 11:56 AM Page 1045 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–11. The A-36 steel angle has a cross-sectional area of A = 2.48 in2 and a radius of gyration about the x axis of rx = 1.26 in. and about the y axis of ry = 0.879 in. The smallest radius of gyration occurs about the z axis and is rz = 0.644 in. If the angle is to be used as a pin-connected 10-ft-long column, determine the largest axial load that can be applied through its centroid C without causing it to buckle. y z C x x z y The least radius of gyration: r2 = 0.644 in. scr = p2E 2 A KL r B controls. K = 1.0 ; p2 (29)(103) (120) 2 C 1.00.644 D = = 8.243 ksi 6 sg O.K. Pcr = scr A = 8.243 (2.48) = 20.4 kip Ans. *13–12. An A-36 steel column has a length of 15 ft and is pinned at both ends. If the cross-sectional area has the dimensions shown, determine the critical load. 8 in. 0.5 in. 0.5 in. 6 in. 0.5 in. Ix = 1 1 (8)(73) (7.5)(63) = 93.67 in4 12 12 Iy = 2 a Pcr = 1 1 b (0.5)(83) + (6)(0.53) = 42.729 in4 (controls) 12 12 p2(29)(103)(42.729) p2EI = 2 (EL) [(1.0)(15)(12)]2 = 377 kip Ans. Check: A = (2)(8)(0.5) + 6(0.5) = 11 in2 scr = Pcr 377 = = 34.3 ksi 6 sg A 11 Therefore, Euler’s formula is valid 1045 13 Solutions 46060 6/11/10 11:56 AM Page 1046 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •13–13. An A-36 steel column has a length of 5 m and is fixed at both ends. If the cross-sectional area has the dimensions shown, determine the critical load. I = 10 mm 10 mm 50 mm 1 1 (0.1)(0.053) (0.08)(0.033) = 0.86167 (10 - 6) m4 12 12 Pcr = 100 mm p2(200)(109)(0.86167)(10 - 6) p2EI = 2 (KL) [(0.5)(5)]2 = 272 138 N = 272 kN scr = = Pcr ; A Ans. A = (0.1)(0.05) - (0.08)(0.03) = 2.6(10 - 3) m2 272 138 = 105 MPa 6 sg 2.6 (10 - 3) Therefore, Euler’s formula is valid. 13–14. The two steel channels are to be laced together to form a 30-ft-long bridge column assumed to be pin connected at its ends. Each channel has a cross-sectional area of A = 3.10 in2 and moments of inertia Ix = 55.4 in4, Iy = 0.382 in4. The centroid C of its area is located in the figure. Determine the proper distance d between the centroids of the channels so that buckling occurs about the x–x and y¿ – y¿ axes due to the same load. What is the value of this critical load? Neglect the effect of the lacing. Est = 2911032 ksi, sY = 50 ksi. y 0.269 in. C d y In order for the column to buckle about x - x and y - y at the same time, Iy must be equal to Ix Iy = Ix 0.764 + 1.55 d2 = 110.8 d = 8.43 in. Ans. Check: d 7 2(1.231) = 2.462 in. O.K. 3 p (29)(10 )(110.8) p2 EI = 2 (KL) [1.0(360)]2 Ans. = 245 kip Check stress: scr = x C d 2 Iy = 2(0.382) + 2 (3.10) a b = 0.764 + 1.55 d2 2 Pcr = 1.231 in. x Ix = 2(55.4) = 110.8 in.4 2 y¿ Pcr 245 = = 39.5 ksi 6 sg A 2(3.10) Therefore, Euler’s formula is valid. 1046 y¿ 13 Solutions 46060 6/11/10 11:56 AM Page 1047 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–15. An A-36-steel W8 * 24 column is fixed at one end and free at its other end. If it is subjected to an axial load of 20 kip, determine the maximum allowable length of the column if F.S. = 2 against buckling is desired. Section Properties. From the table listed in the appendix, the cross-sectional area and moment of inertia about the y axis for a W8 * 24 are A = 7.08 in2 Iy = 18.3 in4 Critical Buckling Load. The critical buckling load is Pcr = Pallow (F.S) = 20(2) = 40 kip Applying Euler’s formula, p2 EIy Pcr = 40 = (KL)2 p2 C 29 A 103 B D (18.3) (2L)2 L = 180.93 in = 15.08 ft = 15.1 ft Ans. Critical Stress. Euler’s formula is valid only if scr 6 sY. scr = Pcr 40 = = 5.65 ksi 6 sY = 36 ksi A 7.08 O.K. *13–16. An A-36-steel W8 * 24 column is fixed at one end and pinned at the other end. If it is subjected to an axial load of 60 kip, determine the maximum allowable length of the column if F.S. = 2 against buckling is desired. Section Properties. From the table listed in the appendix, the cross-sectional area and moment of inertia about the y axis for a W8 * 24 are A = 7.08 in2 Iy = 18.3 in4 Critical Buckling Load. The critical buckling load is Pcr = Pallow (F.S.) = 60(2) = 120 kip Applying Euler’s formula, Pcr = 120 = p2EIy (KL)2 p2 C 24 A 103 B D (18.3) (0.7L)2 L = 298.46 in = 24.87 ft = 24.9 ft Ans. Critical Stress. Euler’s formula is valid only if scr 6 sY. scr = Pcr 120 = = 16.95 ksi 6 sY = 36 ksi A 7.08 O.K. 1047 13 Solutions 46060 6/11/10 11:56 AM Page 1048 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •13–17. The 10-ft wooden rectangular column has the dimensions shown. Determine the critical load if the ends are assumed to be pin connected. Ew = 1.611032 ksi, sY = 5 ksi. Section Properties: 10 ft A = 4(2) = 8.00 in2 4 in. Ix = 1 (2) A 43 B = 10.667 in4 12 Iy = 1 (4) A 23 B = 2.6667 in4 (Controls !) 12 2 in. Critical Buckling Load: K = 1 for pin supported ends column. Applying Euler’s formula,. Pcr = p2EI (KL)2 p2(1.6)(103)(2.6667) = [1(10)(12)]2 Ans. = 2.924 kip = 2.92 kip Critical Stress: Euler’s formula is only valid if scr 6 sg. scr = Pcr 2.924 = = 0.3655 ksi 6 sg = 5 ksi A 8.00 O.K. 13–18. The 10-ft column has the dimensions shown. Determine the critical load if the bottom is fixed and the top is pinned. Ew = 1.611032 ksi, sY = 5 ksi. Section Properties: A = 4(2) = 8.00 in2 10 ft 1 (2) A 43 B = 10.667 in4 Ix = 12 4 in. 2 in. 1 Iy = (4) A 23 B = 2.6667 in4 (Controls!) 12 Critical Buckling Load: K = 0.7 for column with one end fixed and the other end pinned. Applying Euler’s formula. Pcr = p2EI (KL)2 p2 (1.6)(103)(2.6667) = [0.7(10)(12)]2 Ans. = 5.968 kip = 5.97 kip Critical Stress: Euler’s formula is only valid if scr 6 sg. scr = Pcr 5.968 = = 0.7460 ksi 6 sg = 5 ksi A 8.00 O.K. 1048 13 Solutions 46060 6/11/10 11:56 AM Page 1049 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–19. Determine the maximum force P that can be applied to the handle so that the A-36 steel control rod BC does not buckle. The rod has a diameter of 25 mm. P 350 mm A 250 mm 45⬚ Support Reactions: a + ©MA = 0; P(0.35) - FBC sin 45°(0.25) = 0 FBC = 1.9799P Section Properties: A = p A 0.0252 B = 0.15625 A 10 - 3 B p m2 4 I = p A 0.01254 B = 19.17476 A 10 - 9 B m4 4 Critical Buckling Load: K = 1 for a column with both ends pinned. Appyling Euler’s formula, Pcr = FBC = 1.9799P = p2EI (KLBC)2 p2(200)(109) C 19.17476(10 - 9) D [1(0.8)]2 P = 29 870 N = 29.9 kN Ans. Critical Stress: Euler’s formula is only valid if scr 6 sg. scr = 1.9799(29 870) Pcr = 120.5 MPa 6 sg = 250 MPa = A 0.15625(10 - 3)p 1049 O.K. C B 800 mm 13 Solutions 46060 6/11/10 11:56 AM Page 1050 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *13–20. The W10 * 45 is made of A-36 steel and is used as a column that has a length of 15 ft. If its ends are assumed pin supported, and it is subjected to an axial load of 100 kip, determine the factor of safety with respect to buckling. P Critical Buckling Load: Iy = 53.4 in4 for a W10 * 45 wide flange section and K = 1 for pin supported ends column. Applying Euler’s formula, Pcr = 15 ft p2EI (KL)2 p2 (29)(103)(53.4) = [1(15)(12)]2 P = 471.73 kip Critical Stress: Euler’s formula is only valid if scr 6 sg. A = 13.3 in2 for the W10 * 45 wide-flange section. scr = Pcr 471.73 = = 35.47 ksi 6 sg = 36 ksi A 13.3 O.K. Pcr 471.73 = = 4.72 P 100 Ans. Factor of Safety: F.S = The W10 * 45 is made of A-36 steel and is used as a column that has a length of 15 ft. If the ends of the column are fixed supported, can the column support the critical load without yielding? •13–21. P Critical Buckling Load: Iy = 53.4 in4 for W10 * 45 wide flange section and K = 0.5 for fixed ends support column. Applying Euler’s formula, Pcr = 15 ft p2EI (KL)2 p2 (29)(103)(53.4) = [0.5(15)(12)]2 P = 1886.92 kip Critical Stress: Euler’s formula is only valid if scr 6 sg. A = 13.3 in2 for W10 * 45 wide flange section. scr = Pcr 1886.92 = = 141.87 ksi 7 sg = 36 ksi (No!) A 13.3 Ans. The column will yield before the axial force achieves the critical load Pcr and so Euler’s formula is not valid. 1050 13 Solutions 46060 6/11/10 11:56 AM Page 1051 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–22. The W12 * 87 structural A-36 steel column has a length of 12 ft. If its bottom end is fixed supported while its top is free, and it is subjected to an axial load of P = 380 kip, determine the factor of safety with respect to buckling. W 12 * 87 A = 25.6 in2 Ix = 740 in4 P Iy = 241 in4 (controls) 12 ft K = 2.0 Pcr = p2(29)(103)(241) p2EI = = 831.63 kip 2 (KL) [(2.0)(12)(12)]2 Pcr 831.63 = = 2.19 P 380 F.S. = Ans. Check: scr = = Pcr A 831.63 = 32.5 ksi 6 sg 25.6 O.K. 13–23. The W12 * 87 structural A-36 steel column has a length of 12 ft. If its bottom end is fixed supported while its top is free, determine the largest axial load it can support. Use a factor of safety with respect to buckling of 1.75. W 12 * 87 A = 25.6 in2 Ix = 740 in4 P Iy = 241 in4 (controls) K = 2.0 12 ft Pcr p2(29)(103)(241) p2EI = = = 831.63 kip 2 (KL) (2.0(12)(12))2 P = Pcr 831.63 = = 475 ksi F.S 1.75 Ans. Check: scr = P 831.63 = = 32.5 ksi 6 sg A 25.6 O.K. 1051 13 Solutions 46060 6/11/10 11:56 AM Page 1052 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *13–24. An L-2 tool steel link in a forging machine is pin connected to the forks at its ends as shown. Determine the maximum load P it can carry without buckling. Use a factor of safety with respect to buckling of F.S. = 1.75. Note from the figure on the left that the ends are pinned for buckling, whereas from the figure on the right the ends are fixed. P P 1.5 in. 0.5 in. 24 in. Section Properties: A = 1.5(0.5) = 0.750 in2 Ix = 1 (0.5) A 1.53 B = 0.140625 in4 12 Iy = 1 (1.5) A 0.53 B = 0.015625 in4 12 P Critical Buckling Load: With respect to the x - x axis, K = 1 (column with both ends pinned). Applying Euler’s formula, Pcr = p2EI (KL)2 p2(29.0)(103)(0.140625) = [1(24)]2 = 69.88 kip With respect to the y - y axis, K = 0.5 (column with both ends fixed). Pcr = p2EI (KL)2 p2(29.0)(103)(0.015625) = [0.5(24)]2 = 31.06 kip (Controls!) Critical Stress: Euler’s formula is only valid if scr 6 sg. scr = Pcr 31.06 = = 41.41 ksi 6 sg = 102 ksi A 0.75 O.K. Factor of Safety: F.S = 1.75 = Pcr P 31.06 P P = 17.7 kip Ans. 1052 P 13 Solutions 46060 6/11/10 11:56 AM Page 1053 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. The W14 * 30 is used as a structural A-36 steel column that can be assumed pinned at both of its ends. Determine the largest axial force P that can be applied without causing it to buckle. •13–25. P From the table in appendix, the cross-sectional area and the moment of inertia about weak axis (y-axis) for W14 * 30 are A = 8.85 in2 Iy = 19.6 in4 25 ft Critical Buckling Load: Since the column is pinned at its base and top, K = 1. For A36 steel, E = 29.0(103) ksi and sg = 36 ksi. Here, the buckling occurs about the weak axis (y-axis). P = Pcr = p2EIy (KL)2 = p2 C 29.0(103) D (19.6) C 1(25)(12) D 2 Ans. = 62.33 kip = 62.3 kip Euler’s formula is valid only if scr 6 sg. scr = Pcr 62.33 = = 7.04 ksi 6 sg = 36 ksi A 8.85 O.K. 13–26. The A-36 steel bar AB has a square cross section. If it is pin connected at its ends, determine the maximum allowable load P that can be applied to the frame. Use a factor of safety with respect to buckling of 2. a + ©MA = 0; C FBC sin 30°(10) - P(10) = 0 FBC = 2 P + : ©Fx = 0; A 1.5 in. 30⬚ B 1.5 in. FA - 2P cos 30° = 0 1.5 in. 10 ft FA = 1.732 P P Buckling load: Pcr = FA(F.S.) = 1.732 P(2) = 3.464 P L = 10(12) = 120 in. I = 1 (1.5)(1.5)3 = 0.421875 in4 12 Pcr = p2 EI (KL)2 3.464 P = p2 (29)(103)(0.421875) [(1.0)(120)]2 P = 2.42 kip Ans. Pcr = FA(F.S.) = 1.732(2.42)(2) = 8.38 kip Check: scr = Pcr 8.38 = = 3.72 ksi 6 sg A 1.5 (1.5) O.K. 1053 13 Solutions 46060 6/11/10 11:56 AM Page 1054 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–27. Determine the maximum allowable intensity w of the distributed load that can be applied to member BC without causing member AB to buckle. Assume that AB is made of steel and is pinned at its ends for x–x axis buckling and fixed at its ends for y–y axis buckling. Use a factor of safety with respect to buckling of 3. Est = 200 GPa, sY = 360 MPa. w C 1.5 m B 0.5 m 2m 30 mm x Ix = 1 (0.02)(0.033) = 45.0(10 - 9)m4 12 Iy = 1 (0.03)(0.023) = 20(10 - 9) m4 12 x x-x axis: Pcr = FAB (F.S.) = 1.333w(3) = 4.0 w K = 1.0, Pcr = L = 2m p2EI (KL)2 4.0w = p2(200)(109)(45.0)(10 - 9) [(1.0)(2)]2 w = 5552 N>m = 5.55 kN>m Ans. (controls) y -y axis K = 0.5, 4.0w = L = 2m p2 (200)(109)(20)(10 - 9) [(0.5)(2)]2 w = 9870 N>m = 9.87 kN>m Check: scr = 20 mm y y Moment of inertia: 4(5552) Pcr = = 37.0 MPa 6 sg A (0.02)(0.03) O.K. 1054 30 mm A 13 Solutions 46060 6/11/10 11:56 AM Page 1055 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *13–28. Determine if the frame can support a load of w = 6 kN>m if the factor of safety with respect to buckling of member AB is 3. Assume that AB is made of steel and is pinned at its ends for x–x axis buckling and fixed at its ends for y–y axis buckling. Est = 200 GPa, sY = 360 MPa. w C B 1.5 m 0.5 m Check x -x axis buckling: Ix = 1 (0.02)(0.03)3 = 45.0(10 - 9) m4 12 K = 1.0 Pcr 2m 30 mm x 20 mm y y L = 2m p2(200)(109)(45.0)(10 - 9) p2EI = = 2 (KL) ((1.0)(2))2 x A 30 mm Pcr = 22.2 kN a + ©MC = 0; FAB(1.5) - 6(2)(1) = 0 FAB = 8 kN Preq’d = 8(3) = 24 kN 7 22.2 kN No, AB will fail. Ans. The beam supports the load of P = 6 kip. As a result, the A-36 steel member BC is subjected to a compressive load. Due to the forked ends on the member, consider the supports at B and C to act as pins for x–x axis buckling and as fixed supports for y–y axis buckling. Determine the factor of safety with respect to buckling about each of these axes. •13–29. a + ©MA = 0; P 4 ft A 3 ft 1 )(1)(3)3 p2(29)(103)(12 p2EI = = 178.9 kip 2 (KL) (1.0(5)(12))2 178.9 = 8.94 20 Ans. y - y axis buckling: Pcr = F.S. = 3 in. y x -x axis buckling: F.S. = B C x 3 FBC a b (4) - 6000(8) = 0 5 FBC = 20 kip Pcr = 4 ft 1 )(3)(1)3 p2 (29)(103)(12 p2EI = = 79.51 2 (KL) (0.5(5)(12))2 79.51 = 3.98 20 Ans. 1055 1 in. x y 13 Solutions 46060 6/11/10 11:56 AM Page 1056 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–30. Determine the greatest load P the frame will support without causing the A-36 steel member BC to buckle. Due to the forked ends on the member, consider the supports at B and C to act as pins for x–x axis buckling and as fixed supports for y–y axis buckling. P 4 ft A 3 ft 3 FBC a b (4) - P(8) = 0 5 a + ©MA = 0; 4 ft B y 3 in. C x 1 in. FBC = 3.33 P y x -x axis buckling: Pcr = x 1 )(1)(3)3 p2(29)(103)(12 p2EI = = 178.9 kip (KL)2 (1.0(5)(12))2 y - y axis buckling: Pcr = 1 )(3)(1)3 p2(29)(103)(12 p2EI = = 79.51 kip (KL)2 (0.5(5)(12))2 Thus, 3.33 P = 79.51 P = 23.9 kip Ans. 13–31. Determine the maximum distributed load that can be applied to the bar so that the A-36 steel strut AB does not buckle. The strut has a diameter of 2 in. It is pin connected at its ends. w C A 2 ft The compressive force developed in member AB can be determined by writing the moment equation of equilibrium about C. a + ©MC = 0; FAB(2) - w(2)(3) = 0 A = p(12) = p in2 I = FAB = 3w 4 ft p 4 p (1 ) = in4 4 4 Since member AB is pinned at both ends, K = 1. For A36 steel, E = 29.0(103) ksi and sg = 36 ksi. Pcr = p EI ; (KL)2 p C 29.0(10 ) D (p>4) 2 2 3w = 3 C 1(4)(12) D 2 Ans. w = 32.52 kip>ft = 32.5 kip>ft The Euler’s formula is valid only if scr 6 sg. scr = 3(32.52) Pcr = = 31.06 ksi 6 sg = 36 ksi p A O.K. 1056 B 2 ft 13 Solutions 46060 6/11/10 11:56 AM Page 1057 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *13–32. The members of the truss are assumed to be pin connected. If member AC is an A-36 steel rod of 2 in. diameter, determine the maximum load P that can be supported by the truss without causing the member to buckle. P C B 4 ft D A 3 ft Section the truss through a -a, the FBD of the top cut segment is shown in Fig. a. The compressive force developed in member AC can be determined directly by writing the force equation of equilibrium along x axis. + : ©Fx = 0; 3 FAC a b - P = 0 5 A = p(12) = p in2 I = FAC = 5 P (C) 3 p 4 p (1 ) = in4 4 4 Since both ends of member AC are pinned, K = 1. For A-36 steel, E = 29.0(103) ksi and sg = 36 ksi. The length of member AC is LAC = 232 + 42 = 5 ft. Pcr = p2EI ; (KL)2 p2 C 29.0(103) D (p>4) 5 P = 3 C 1(5)(12) D 2 P = 37.47 kip = 37.5 kip Ans. Euler’s formula is valid only if scr 6 sg. scr 5 (37.47) Pcr 3 = = = 19.88 ksi 6 sg = 36 ksi p A O.K. 1057 13 Solutions 46060 6/11/10 11:56 AM Page 1058 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •13–33. The steel bar AB of the frame is assumed to be pin connected at its ends for y–y axis buckling. If w = 3 kN>m, determine the factor of safety with respect to buckling about the y–y axis due to the applied loading. Est = 200 GPa, sY = 360 MPa. 6m w B C 40 mm 40 mm 3m 40 mm y x A 4m The force with reference to the FBD shown in Fig. a. a + ©MC = 0; 3 3(6)(3) - FAB a b (6) = 0 5 A = 0.04(0.08) = 3.2(10 - 3) m2 Iy = FAB = 15 kN 1 (0.08)(0.043) = 0.4267(10 - 6)m4 12 The length of member AB is L = 232 + 42 = 5m. Here, buckling will occur about the weak axis, (y-axis). Since both ends of the member are pinned, Ky = 1. Pcr = p2EIy (KyLy)2 = p2 C 200(109) D C 0.4267(10 - 6) D C 1.0(5) D 2 = 33.69 kN Euler’s formula is valid only if scr 6 sg. scr = 33.69(103) Pcr = 10.53(106)Pa = 10.53 MPa 6 sg = 360 MPa = A 3.2(10 - 3) O.K. Thus, the factor of safety against buckling is F.S = Pcr 33.69 = = 2.25 FAB 15 Ans. 1058 13 Solutions 46060 6/11/10 11:56 AM Page 1059 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–34. The members of the truss are assumed to be pin connected. If member AB is an A-36 steel rod of 40 mm diameter, determine the maximum force P that can be supported by the truss without causing the member to buckle. 2m C E D 1.5 m B A 2m P By inspecting the equilibrium of joint E, FAB = 0. Then, the compressive force developed in member AB can be determined by analysing the equilibrium of joint A, Fig. a. + c ©Fy = 0; 3 FAC a b - P = 0 5 + : ©Fx = 0; 5 4 P a b - FAB = 0 3 5 A = p(0.022) = 0.4(10 - 3)p m2 I = FAC = 5 P (T) 3 FAB = 4 P(c) 3 p (0.024) = 40(10 - 9) p m4 4 Since both ends of member AB are pinned, K = 1. For A36 steel, E = 200 GPa and sg = 250 MPa. Pcr = p2EI ; (KL)2 p2 C 200(109) D C 40(10 - 9)p D 4 P = 3 C 1(2) D 2 P = 46.51(103) N = 46.5 kN Ans. The Euler’s formula is valid only if scr 6 sg. scr 4 (46.51)(103) Pcr 3 = 49.35(106) Pa = 49.35 MPa 6 sg = 250 MPa O.K. = = A 0.4(10 - 3)p 1059 13 Solutions 46060 6/11/10 11:56 AM Page 1060 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–35. The members of the truss are assumed to be pin connected. If member CB is an A-36 steel rod of 40 mm diameter, determine the maximum load P that can be supported by the truss without causing the member to buckle. 2m C E D 1.5 m B A 2m P Section the truss through a–a, the FBD of the left cut segment is shown in Fig. a. The compressive force developed in member CB can be obtained directly by writing the force equation of equilibrium along y axis. + c ©Fy = 0; FCB - P = 0 A = p(0.022) = 0.4(10 - 3)p m2 FCB = P (C) I = p (0.024) = 40(10 - 9)p m4 4 Since both ends of member CB are pinned, K = 1. For A36 steel, E = 200 GPa and sg = 250 MPa. Pcr = p2EI ; (KL)2 P = p2 C 200(109) D C 40(10 - 9)p D C 1(1.5) D 2 = 110.24(103) N = 110 kN Ans. The Euler’s formula is valid only if scr 6 sg. scr = 110.24(103) Pcr = 87.73(106) Pa = 87.73 MPa 6 sg = 250 MPa = A 0.4(10 - 3)p 1060 O.K. 13 Solutions 46060 6/11/10 11:56 AM Page 1061 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *13–36. If load C has a mass of 500 kg, determine the required minimum diameter of the solid L2-steel rod AB to the nearest mm so that it will not buckle. Use F.S. = 2 against buckling. A 45° 4m D Equilibriun. The compressive force developed in rod AB can be determined by analyzing the equilibrium of joint A, Fig. a. ©Fy¿ = 0; FAB sin 15° - 500(9.81) cos 45° = 0 FAB = 13 400.71 N Section Properties. The cross-sectional area and moment of inertia of the solid rod are A = p 2 d 4 I = p d 4 p 4 a b = d 4 2 64 Critical Buckling Load. Since the rod is pinned at both of its ends, K = 1. Here, Pcr = FAB (F.S.) = 13400.71(2) = 26801.42 N. Applying Euler’s formula, Pcr = p2EIy (KL)2 26801.42 = p2 C 200 A 109 B D c p 4 d d 64 [1(4)]2 d = 0.04587 m = 45.87 mm Use d = 46 mm Ans. Critical Stress. Euler’s formula is valid only if scr 6 sY. scr = Pcr 26801.42 = = 16.13 MPa 6 sY = 703 MPa p A 2 A 0.046 B 4 1061 O.K. 60° B C 13 Solutions 46060 6/11/10 11:56 AM Page 1062 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •13–37. If the diameter of the solid L2-steel rod AB is 50 mm, determine the maximum mass C that the rod can support without buckling. Use F.S. = 2 against buckling. A 45° 4m D Equilibrium. The compressive force developed in rod AB can be determined by analyzing the equilibrium of joint A, Fig. a. ©Fy¿ = 0; FAB sin 15° - m(9.81) cos 45° = 0 FAB = 26.8014m B Section Properties. The cross-sectional area and moment of inertia of the rod are A = I = p A 0.052 B = 0.625 A 10 - 3 B pm2 4 p A 0.0254 B = 97.65625 A 10 - 9 B pm4 4 Critical Buckling Load. Since the rod is pinned at both of its ends, K = 1. Here, Pcr = FAB (F.S.) = 26.8014m(2) = 53.6028m. Applying Euler’s formula, Pcr = p2EIy (KL)2 53.6028m = p2 c 200 A 109 B d c 97.65625 A 10 - 9 B p d [1(4)]2 m = 706.11 kg = 7.06 kg Ans. Critical Stress. Euler’s formula is valid only if scr 6 sY. scr = 53.6028(706.11) Pcr = = 19.28 MPa 6 sY = 703 MPa A p 0.625 A 10 - 3 B 1062 60° O.K. C 13 Solutions 46060 6/11/10 11:56 AM Page 1063 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–38. The members of the truss are assumed to be pin connected. If member GF is an A-36 steel rod having a diameter of 2 in., determine the greatest magnitude of load P that can be supported by the truss without causing this member to buckle. H G 16 ft P Support Reactions: As shown on FBD(a). Member Forces: Use the method of sections [FBD(b)]. FGF = 1.3333P (C) Section Properties: A = I = p 2 A 2 B = p in2 4 p 4 A 1 B = 0.250p in4 4 Critical Buckling Load: K = 1 for a column with both ends pinned. Applying Euler’s formula, Pcr = FGF = 1.3333P = p2EI (KLGF)2 p2 (29)(103)(0.250p) [1(16)(12)]2 P = 4.573 kip = 4.57 kip Ans. Critical Stress: Euler’s formula is only valid if scr 6 sg. scr = 1.3333(4.573) Pcr = = 1.94 ksi 6 sg = 36 ksi p A 1063 O.K. D C B 16 ft FGF (12) - P(16) = 0 E 12 ft A + ©MB = 0; F 16 ft P 13 Solutions 46060 6/11/10 11:56 AM Page 1064 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–39. The members of the truss are assumed to be pin connected. If member AG is an A-36 steel rod having a diameter of 2 in., determine the greatest magnitude of load P that can be supported by the truss without causing this member to buckle. H G 16 ft P Support Reactions: As shown on FBD(a). Member Forces: Use the method of joints [FBD(b)]. 3 = 0 F 5 AG FAG = 1.6667P (C) Section Properties: LAG = 2162 + 122 = 20.0 ft A = I = p 2 A 2 B = p in2 4 p 4 A 1 B = 0.250p in4 4 Critical Buckling Load: K = 1 for a column with both ends pinned. Applying Euler’s formula, Pcr = FGF = 1.6667P = p2EI (KLGF)2 p2 (29)(103)(0.250p) [1(20)(12)]2 P = 2.342 kip = 2.34 kip Ans. Critical Stress: Euler’s formula is only valid if scr = sg. scr = 1.6667(2.342) Pcr = = 1.24 ksi 6 sg = 36 ksi p A 1064 O.K. D C B 16 ft P - E 12 ft A + c ©Fy = 0; F 16 ft P 13 Solutions 46060 6/11/10 11:56 AM Page 1065 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *13–40. The column is supported at B by a support that does not permit rotation but allows vertical deflection. Determine the critical load Pcr . EI is constant. L B Pcr A Elastic curve: EI d2y = M = -P y dx2 P d2y + y = 0 EI dx2 y = C1 sin c P P x d + C2 cos c xd A EI A EI Boundry conditions: At x = 0; 0 = 0 + C2; At x = L; y = 0 C2 = 0 dv = 0 dx P P dv = C1 cos c L] d = 0; dx A EI A EI cos c P L d = 0; A EI For n = 1 ; Pcr = C1 P p L = na b A EI 2 P Z 0 A EI n = 1, 3, 5 p2 P = EI 4L2 p2EI 4L2 Ans. 1065 13 Solutions 46060 6/11/10 11:56 AM Page 1066 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. w •13–41. The ideal column has a weight w (force兾length) and rests in the horizontal position when it is subjected to the axial load P. Determine the maximum moment in the column at midspan. EI is constant. Hint: Establish the differential equation for deflection, Eq. 13–1, with the origin at the mid span. The general solution is v = C1 sin kx + C2 cos kx + 1w>12P22x2 - 1wL>12P22x - 1wEI>P22 where k2 = P>EI. P L Moment Functions: FBD(b). a + ©Mo = 0; M(x) = wL x wx a b - M(x) - a bx - Pv = 0 2 2 w 2 A x - Lx B - Pv 2 [1] Differential Equation of The Elastic Curve: EI d2y = M(x) dx2 EI d2y w 2 = A x - Lx B - Py 2 dx2 w d2y P y = + A x2 - Lx B EI 2EI dx2 The solution of the above differential equation is of the form v = C1 sin a P P w 2 wL wEI x b + C2 cos ¢ xb + x x A EI A EI 2P 2P P2 [2] dv P P P P w wL = C1 cos ¢ x ≤ - C2 sin ¢ x≤ + xdx A EI A EI A EI A EI P 2P [3] and The integration constants can be determined from the boundary conditions. Boundary Condition: At x = 0, y = 0. From Eq. [2], 0 = C2 - wEI P2 C2 = wEI P2 At x = L dy = 0. From Eq.[3], , 2 dx 0 = C1 P wEI P w L P L P L wL cos ¢ sin ¢ ≤ ≤ + a b A EI A EI 2 A EI 2 P 2 2P P2 A EI C1 = wEI P L tan ¢ ≤ A EI 2 P2 1066 13 Solutions 46060 6/11/10 11:56 AM Page 1067 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–41. Continued Elastic Curve: y = w EI P L P EI P x2 L EI x≤ + x≤ + tan ¢ cos ¢ - x B ≤ sin ¢ R P P A EI 2 A EI P A EI 2 2 P However, y = ymax at x = ymax = = L . Then, 2 P L P L P L EI w EI EI L2 tan ¢ cos ¢ B ≤ sin ¢ ≤ + ≤ R P P A EI 2 A EI 2 P A EI 2 8 P wEI P L PL2 sec - 1R B ¢ ≤ A EI 2 8EI P2 Maximum Moment: The maximum moment occurs at x = Mmax = L . From, Eq.[1], 2 w L2 L - L a b R - Pymax B 2 4 2 = - wL2 wEI P L PL2 - P b 2 B sec ¢ - 1R r ≤ 8 A EI 2 8EI P = - PL wEI B sec ¢ ≤ - 1R P A EI 2 Ans. 13–42. The ideal column is subjected to the force F at its midpoint and the axial load P. Determine the maximum moment in the column at midspan. EI is constant. Hint: Establish the differential equation for deflection, Eq. 13–1. The general solution is v = C1 sin kx + C2 cos kx - c2x>k2, where c2 = F>2EI, k2 = P>EI. F P L 2 Moment Functions: FBD(b). a + ©Mo = 0; M(x) + F x + P(v) = 0 2 M(x) = - F x - Pv 2 [1] Differential Equation of The Elastic Curve: EI d2y = M(x) dx2 EI F d2y = - x - Py 2 2 dx d2y F P y = x + EI 2EI dx2 The solution of the above differential equation is of the form, v = C1 sin a P P F x b + C2 cos ¢ xb x A EI A EI 2P 1067 [2] L 2 13 Solutions 46060 6/11/10 11:56 AM Page 1068 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–42. Continued and dv P P P P F = C1 cos ¢ x ≤ - C2 sin ¢ x≤ dx A EI A EI A EI A EI 2P [3] The integration constants can be determined from the boundary conditions. Boundary Conditions: At x = 0, y = 0. From Eq.[2], C2 = 0 At x = L dy = 0. From Eq.[3], , 2 dx 0 = C1 C1 = P L P F cos ¢ ≤ A EI A EI 2 2P F EI P L sec ¢ ≤ 2P A P A EI 2 Elastic Curve: y = F EI P L P F sec ¢ x≤ x ≤ sin ¢ 2P A P A EI 2 A EI 2P = F EI P L P sec ¢ x≤ - xR B ≤ sin ¢ 2P A P A EI 2 A EI However, y = ymax at x = ymax = = L . Then, 2 F EI P L P L L sec ¢ B ≤ sin ¢ ≤ - R 2P A P A EI 2 A EI 2 2 F EI P L L tan ¢ B ≤ - R 2P A P A EI 2 2 Maximum Moment: The maximum moment occurs at x = Mmax = - L . From Eq.[1], 2 F L a b - Pymax 2 2 = - FL F EI P L L - Pb tan ¢ B ≤ - Rr 4 2P A P A EI 2 2 = - P L F EI tan ¢ ≤ 2 AP A EI 2 Ans. 1068 13 Solutions 46060 6/11/10 11:56 AM Page 1069 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–43. The column with constant EI has the end constraints shown. Determine the critical load for the column. P L Moment Function. Referring to the free-body diagram of the upper part of the deflected column, Fig. a, a + ©MO = 0; M + Pv = 0 M = - Pv Differential Equation of the Elastic Curve. EI d2v = M dx2 EI d2v = - Pv dx2 d2v P + v = 0 2 EI dx The solution is in the form of v = C1 sin a P P x b + C2 cos ¢ xb A EI A EI (1) dv P P P P = C1 cos ¢ x ≤ - C2 sin ¢ x≤ dx A EI A EI A EI A EI (2) Boundary Conditions. At x = 0, v = 0. Then Eq. (1) gives 0 = 0 + C2 At x = L, C2 = 0 dv = 0. Then Eq. (2) gives dx 0 = C1 P P cos ¢ L≤ A EI A EI C1 = 0 is the trivial solution, where v = 0. This means that the column will remain straight and buckling will not occur regardless of the load P. Another possible solution is cos ¢ P L≤ = 0 A EI P np L = A EI 2 n = 1, 3, 5 The smallest critical load occurs when n = 1, then p Pcr L = A EI 2 Pcr = p2EI 4L2 Ans. 1069 13 Solutions 46060 6/11/10 11:56 AM Page 1070 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *13–44. Consider an ideal column as in Fig. 13–10c, having both ends fixed. Show that the critical load on the column is given by Pcr = 4p2EI>L2. Hint: Due to the vertical deflection of the top of the column, a constant moment M¿ will be developed at the supports. Show that d2v>dx2 + 1P>EI2v = M¿>EI. The solution is of the form v = C1 sin1 1P>EIx2 + C2 cos1 1P>EIx2 + M¿>P. Moment Functions: M(x) = M¿ - Py Differential Equation of The Elastic Curve: EI d2y = M(x) dx2 d2y = M¿ - Py dx2 EI M¿ d2y P + y = 2 EI EI dx (Q.E.D.) The solution of the above differential equation is of the form v = C1 sin a P P M¿ x b + C2 cos ¢ xb + A EI P A EI [1] and dv P P P P = C1 cos ¢ x ≤ - C2 sin ¢ x≤ dx A EI A EI A EI A EI [2] The integration constants can be determined from the boundary conditions. Boundary Conditions: At x = 0, y = 0. From Eq.[1], C2 = At x = 0, M¿ P dy = 0. From Eq.[2], C1 = 0 dx Elastic Curve: y = M¿ P x≤ R B 1 - cos ¢ P A EI and M¿ P P dy = sin ¢ x≤ dx P A EI A EI However, due to symmetry sin B L dy = 0 at x = . Then, dx 2 P L a bR = 0 A EI 2 or P L a b = np A EI 2 where n = 1, 2, 3,... The smallest critical load occurs when n = 1. Pce = 4p2EI L2 (Q.E.D.) 1070 13 Solutions 46060 6/11/10 11:56 AM Page 1071 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •13–45. Consider an ideal column as in Fig. 13–10d, having one end fixed and the other pinned. Show that the critical load on the column is given by Pcr = 20.19EI>L2. Hint: Due to the vertical deflection at the top of the column,a constant moment M¿ will be developed at the fixed support and horizontal reactive forces R¿ will be developed at both supports. Show that d2v>dx2 + 1P>EI2v = 1R¿>EI21L - x2. The solution is of the form v = C1 sin 11P>EIx2 + C2 cos 11P>EIx2 + 1R¿>P21L - x2. After application of the boundary conditions show that tan 1 1P>EIL2 = 1P>EI L. Solve by trial and error for the smallest nonzero root. Equilibrium. FBD(a). Moment Functions: FBD(b). M(x) = R¿(L - x) - Py Differential Equation of The Elastic Curve: EI d2y = M(x) dx2 EI d2y = R¿(L - x) - Py dx2 d2y P R¿ + y = (L - x) 2 EI EI dx (Q.E.D.) The solution of the above differential equation is of the form v = C1 sin a P P R¿ (L - x) x b + C2 cos ¢ xb + A EI P A EI [1] and dv P P P P R¿ = C1 cos ¢ x ≤ - C2 sin ¢ x≤ dx A EI A EI A EI A EI P The integration constants can be determined from the boundary conditions. Boundary Conditions: At x = 0, y = 0. From Eq.[1], C2 = - At x = 0, R¿L P dy R¿ EI = 0. From Eq.[2], C1 = dx P AP Elastic Curve: y = = R¿ EI P R¿L P R¿ sin ¢ x≤ cos ¢ x≤ + (L - x) P AP A EI P A EI P EI P P R¿ sin ¢ x ≤ - L cos ¢ x ≤ + (Lx) R B P AP A EI A EI 1071 [2] 13 Solutions 46060 6/11/10 11:56 AM Page 1072 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–45. Continued However, y = 0 at x = L. Then, 0 = P P EI sin ¢ L ≤ - L cos ¢ L≤ A EI A EI AP tan ¢ P P L≤ = L A EI A EI (Q.E.D.) By trial and error and choosing the smallest root, we have P L = 4.49341 A EI Then, Pcr = 20.19EI L2 (Q.E.D.) 13–46. Determine the load P required to cause the A-36 steel W8 * 15 column to fail either by buckling or by yielding. The column is fixed at its base and free at its top. 1 in. P Section properties for W8 * 15: A = 4.44 in2 Ix = 48.0 in4 rx = 3.29 in. d = 8.11 in. Iy = 3.41 in4 8 ft Buckling about y - y axis: K = 2.0 P = Pcr = L = 8(12) = 96 in. p2EIy (KL)2 Check: scr = p2(29)(103)(3.41) = [(2.0)(96)]2 = 26.5 kip (controls) Pcr 26.5 = = 5.96 ksi 6 sg A 4.44 Ans. O.K. Check yielding about x - x axis: smax = P ec KL P c 1 + 2 sec a bd A 2r A EA r 26.5 P = = 5.963 ksi A 4.44 (1) A 8.11 ec 2 B = = 0.37463 2 r (3.29)2 2.0(96) P 26.5 KL = = 0.4184 2r A EA 2(3.29) A 29(103)(4.44) smax = 5.963[1 + 0.37463 sec (0.4184)] = 8.41 ksi 6 sg = 36 ksi 1072 O.K. 13 Solutions 46060 6/11/10 11:56 AM Page 1073 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–47. The hollow red brass C83400 copper alloy shaft is fixed at one end but free at the other end. Determine the maximum eccentric force P the shaft can support without causing it to buckle or yield. Also, find the corresponding maximum deflection of the shaft. 2m a a P 150 mm 30 mm Section Properties. A = p A 0.032 - 0.022 B = 0.5 A 10 - 3 B p m2 I = 20 mm Section a – a p A 0.034 - 0.024 B = 0.1625 A 10 - 6 B p m4 4 0.1625 A 10 B p I = = 0.01803 m C 0.5 A 10 - 3 B p AA -6 r = e = 0.15 m c = 0.03 m For a column that is fixed at one end and free at the other, K = 2. Thus, KL = 2(2) = 4 m Yielding. In this case, yielding will occur before buckling. Applying the secant formula, smax = P ec KL P B 1 + 2 sec ¢ ≤R A 2rx A EA rx 70.0 A 106 B = 70.0 A 106 B = P 0.5 A 10 -3 Bp P 0.5 A 10 - 3 B p D1 + 0.15(0.03) 0.018032 secC P 4 ST 2(0.01803)A 101 A 109 B C 0.5 A 10 - 3 B p D a 1 + 13.846 sec 8.8078 A 10 - 3 B 2Pb Solving by trial and error, P = 5.8697 kN = 5.87 kN Ans. Maximum Deflection. vmax = e B sec ¢ P KL ≤ - 1R A EI 2 = 0.15 D sec C 5.8697 A 103 B 4 a b S - 1T C 101 A 109 B C 0.1625 A 10 - 6 B p D 2 = 0.04210 m = 42.1 mm Ans. 1073 13 Solutions 46060 6/11/10 11:56 AM Page 1074 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *13–48. The hollow red brass C83400 copper alloy shaft is fixed at one end but free at the other end. If the eccentric force P = 5 kN is applied to the shaft as shown, determine the maximum normal stress and the maximum deflection. 2m a a P 150 mm 30 mm Section Properties. A = p A 0.032 - 0.022 B = 0.5 A 10 - 3 B p m2 I = 20 mm Section a – a p A 0.034 - 0.024 B = 0.1625 A 10 - 6 B p m4 4 0.1625 A 10 - 6 B p I = 0.01803 m = r = C 0.5 A 10 - 3 B p AA e = 0.15 m c = 0.03 m For a column that is fixed at one end and free at the other, K = 2. Thus, KL = 2(2) = 4 m Yielding. Applying the secant formula, smax = P ec KL P B 1 + 2 sec ¢ ≤R A 2r A EA r 5 A 103 B 5 A 103 B 4 ST = secC D1 + 2(0.01803) C 101 A 109 B C 0.5 A 10 - 3 B p D 0.018032 0.5 A 10 - 3 B p 0.15(0.03) = 57.44 MPa = 57.4 MPa Ans. Since smax 6 sY = 70 MPa, the shaft does not yield. Maximum Deflection. vmax = e B sec ¢ P KL ≤ - 1R A EI 2 = 0.15D sec C 5 A 103 B 4 a b S - 1T C 101 A 109 B C 0.1625 A 10 - 6 B p D 2 = 0.03467 m = 34.7 mm Ans. 1074 13 Solutions 46060 6/11/10 11:56 AM Page 1075 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •13–49. The tube is made of copper and has an outer diameter of 35 mm and a wall thickness of 7 mm. Using a factor of safety with respect to buckling and yielding of F.S. = 2.5, determine the allowable eccentric load P. The tube is pin supported at its ends. Ecu = 120 GPa, sY = 750 MPa. 2m P 14 mm Section Properties: A = p (0.0352 - 0.0212) = 0.61575(10 - 3) m2 4 I = p (0.01754 - 0.01054) = 64.1152(10 - 9) m4 4 r = I 64.1152(10 - 9) = = 0.010204 m AA A 0.61575(10 - 3) For a column pinned at both ends, K = 1. Then KL = 1(2) = 2 m. Buckling: Applying Euler’s formula, Pmax = Pcr = p2 (120)(109) C 64.1152(10 - 9) D p2EI = = 18983.7 N = 18.98 kN (KL)2 22 Critical Stress: Euler’s formula is only valid if scr 6 sg. scr = Pcr 18983.7 = 30.83 MPa 6 sg = 750 MPa = A 0.61575(10 - 3) O.K. Yielding: Applying the secant formula, smax = (KL) Pmax Pmax ec B 1 + 2 sec ¢ ≤R A 2r A EA r 750 A 106 B = 0.61575(10 - 3) 750 A 106 B = 0.61575(10 - 3) Pmax Pmax B1 + 0.014(0.0175) 0.0102042 sec ¢ Pmax 2 ≤R 2(0.010204)A 120(109)[0.61575(10 - 3)] A 1 + 2.35294 sec 0.0114006 2Pmax B Solving by trial and error, Pmax = 16 885 N = 16.885 kN (Controls!) Factor of Safety: P = Pmax 16.885 = = 6.75 kN F.S. 2.5 Ans. 1075 P 13 Solutions 46060 6/11/10 11:56 AM Page 1076 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–50. The tube is made of copper and has an outer diameter of 35 mm and a wall thickness of 7 mm. Using a factor of safety with respect to buckling and yielding of F.S. = 2.5, determine the allowable eccentric load P that it can support without failure. The tube is fixed supported at its ends. Ecu = 120 GPa, sY = 750 MPa. 2m P 14 mm Section Properties: A = p A 0.0352 - 0.0212 B = 0.61575 A 10 - 3 B m2 4 I = p A 0.01754 - 0.01054 B = 64.1152 A 10 - 9 B m4 4 r = I 64.1152(10 - 9) = = 0.010204 ms AA A 0.61575(10 - 3) For a column fixed at both ends, K = 0.5. Then KL = 0.5(2) = 1 m. Buckling: Applying Euler’s formula, Pmax = Pcr = p2(120)(109) C 64.1152(10 - 9) D p2EI = = 75 935.0 N = 75.93 kN (KL)2 12 Critical Stress: Euler’s formula is only valid if scr 6 sg. scr = Pcr 75 935.0 = 123.3 MPa 6 sg = 750 MPa = A 0.61575(10 - 3) O. K. Yielding: Applying the secant formula, smax = (KL) Pmax Pmax ec B 1 + 2 sec ¢ ≤R A 2r A EA r 750 A 106 B = 0.61575(10 - 3) 750 A 106 B = 0.61575(10 - 3) Pmax Pmax B1 + 0.014(0.0175) 0.0102042 sec ¢ 2 Pmax ≤R 2(0.010204)A 120(109)[0.61575(10 - 3)] A 1 + 2.35294 sec 5.70032 A 10 - 3 B 2P B Solving by trial and error, Pmax = 50 325 N = 50.325 kN (Controls!) Factor of Safety: P = Pmax 50.325 = = 20.1 kN F.S. 2.5 Ans. 1076 P 13 Solutions 46060 6/11/10 11:56 AM Page 1077 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–51. The wood column is fixed at its base and can be assumed pin connected at its top. Determine the maximum eccentric load P that can be applied without causing the column to buckle or yield. Ew = 1.811032 ksi, sY = 8 ksi. P y 4 in. x x P y 10 in. 10 ft Section Properties: A = 10(4) = 40 in2 ry = Iy = 1 (4)(103) = 333.33 in4 12 Ix = 1 (10)(43) = 53.33 in4 12 Ix 333.33 = = 2.8868 in. AA A 40 Buckling about x -x axis: P = Pcr = p2(1.8)(103)(53.33) p2EI = = 134 kip (KL)2 [(0.7)(10)(12)]2 Check: scr = Pcr 134 = = 3.36 ksi 6 sg A 40 O.K. Yielding about y- y axis: smax = P ec KL P a 1 + 2 sec a b b A 2r A EA r 5(5) ec = = 3.0 r2 2.88682 a 0.7(10)(12) P P KL b = = 0.0542212P 2r A EA 2(2.8868) A 1.8(103)(40) 8(40) = P[1 + 3.0 sec (0.0542212P)] By trial and error: P = 73.5 kip Ans. (controls) 1077 13 Solutions 46060 6/11/10 11:56 AM Page 1078 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *13–52. The wood column is fixed at its base and can be assumed fixed connected at its top. Determine the maximum eccentric load P that can be applied without causing the column to buckle or yield. Ew = 1.811032 ksi, sY = 8 ksi. P y 4 in. x x P y 10 in. 10 ft Section Properties: A = 10(4) = 40 in2 ry = Iy = 1 (4)(103) = 333.33 in4 12 Ix = 1 (10)(43) = 53.33 in4 12 Iy 333.33 = = 2.8868 in. AA A 40 Buckling about x - x axis: P = Pcr = p2(1.8)(103)(53.33) p2EI = = 263 kip 2 (KL) [(0.5)(10)(12)]2 Check: scr = Pcr 263 = = 6.58 ksi 6 sg A 40 O.K. Yielding about y -y axis: smax = P ec KL P a1 + 2 sec a bb A 2r A EA r 5(5) ec = = 3.0 r2 2.88682 a 0.5(10)(12) P P KL b = = 0.0387292P 2r A EA 2(2.8868) A 1.8(103)(40) 8(40) = P[1 + 3.0 sec (0.0387292P)] By trial and error: P = 76.5 kip Ans. (controls) 1078 13 Solutions 46060 6/11/10 11:56 AM Page 1079 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. The W200 * 22 A-36-steel column is fixed at its base. Its top is constrained to rotate about the y–y axis and free to move along the y–y axis. Also, the column is braced along the x–x axis at its mid-height. Determine the allowable eccentric force P that can be applied without causing the column either to buckle or yield. Use F.S. = 2 against buckling and F.S. = 1.5 against yielding. •13–53. 100 mm y A = 2860 mm2 = 2.86 A 10 - 3 B m2 ry = 22.3 mm = 0.0223 m Ix = 20.0 A 106 B mm4 = 20.0 A 10 - 6 B m4 c = bf 2 = 102 = 51 mm = 0.051 m 2 e = 0.1m Buckling About the Strong Axis. Since the column is fixed at the base and free at the top, Kx = 2. Applying Euler’s formula, Pcr = p2EIx (KL)x 2 = p2 c 200 A 109 B d c 20.0 A 10 - 6 B d [2(10)]2 = 98.70kN Euler’s formula is valid if scr 6 sY. scr = 98.70 A 103 B Pcr = = 34.51 MPa 6 sY = 250MPa A 2.86 A 10 - 3 B O.K. Then, Pallow = Pcr 98.70 = = 49.35 kN F.S. 2 Yielding About Weak Axis. Since the support provided by the bracing can be considered a pin connection, the upper portion of the column is pinned at both of its ends. Then Ky = 1 and L = 5 m. Applying the secant formula, smax = A KL B y Pmax Pmax ec C 1 + 2 sec B RS A 2ry A EA ry 250 A 106 B = 250 A 106 B = Pmax 2.86 A 10 -3 Pmax B 2.86 A 10 - 3 B D1 + 0.1(0.051) 0.02232 sec C 1(5) Pmax ST 2(0.0223)A 200 A 109 B C 2.86 A 10 - 3 B D c 1 + 10.2556 sec 4.6875 A 10 - 3 B 2Pmax d Solving by trial and error, Pmax = 39.376 kN Then, Pallow = Pmax 39.376 = = 26.3 kN (controls) 1.5 1.5 Ans. 1079 5m y x 5m Section Properties. From the table listed in the appendix, the necessary section properties for a W200 * 22 are P x 13 Solutions 46060 6/11/10 11:56 AM Page 1080 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–54. The W200 * 22 A-36-steel column is fixed at its base. Its top is constrained to rotate about the y–y axis and free to move along the y–y axis. Also, the column is braced along the x–x axis at its mid-height. If P = 25 kN, determine the maximum normal stress developed in the column. 100 mm y Section Properties. From the table listed in the appendix, necessary section properties for a W200 * 22 are A = 2860 mm2 = 2.86 A 10 - 3 B m2 ry = 22.3 mm = 0.0223 m Ix = 20.0 A 106 B mm4 = 20.0 A 10 - 6 B m4 c = bf 2 = 102 = 51 mm = 0.051 m 2 e = 0.1m Buckling About the Strong Axis. Since the column is fixed at the base and free at the top, Kx = 2. Applying Euler’s formula, Pcr = p2EIx (KL)x 2 = p2 c 200 A 109 B d c 20.0 A 10 - 6 B d [2(10)]2 = 98.70kN Euler’s formula is valid only if scr 6 sY. scr = 98.70 A 103 B Pcr = = 34.51 MPa 6 sY = 250 MPa A 2.86 A 10 - 3 B O.K. Since P = 25 kN 6 Pcr, the column does not buckle. Yielding About Weak Axis. Since the support provided by the bracing can be considered a pin connection, the upper portion of the column is pinned at both of its ends. Then Ky = 1 and L = 5 m. Applying the secant formula, smax = = (KL) P P ec C 1 + 2 sec B RS A 2ry A EA ry 2.5 A 103 B 2.86 A 10 - 3 B D1 + 0.1(0.051) 0.02232 secC 25 A 103 B 1(5) ST 2(0.0223) C 200 A 109 B C 2.86 A 10 - 3 B D = 130.26 MPa = 130 MPa Ans. Since smax 6 sY = 250 MPa, the column does not yield. 1080 y x 5m 5m P x 13 Solutions 46060 6/11/10 11:56 AM Page 1081 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–55. The wood column is fixed at its base, and its top can be considered pinned. If the eccentric force P = 10 kN is applied to the column, investigate whether the column is adequate to support this loading without buckling or yielding. Take E = 10 GPa and sY = 15 MPa. x P 150 mm 25 mm yx 25 mm 75 mm 5m Section Properties. A = 0.05(0.15) = 7.5 A 10 - 3 B m2 Ix = rx = 1 (0.05) A 0.153 B = 14.0625 A 10 - 6 B m4 12 14.0625 A 10 - 6 B Ix = 0.04330 m = AA C 7.5 A 10 - 3 B 1 (0.15) A 0.053 B = 1.5625 A 10 - 6 B m4 12 e = 0.15 m c = 0.075 m Iy = For a column that is fixed at one end and pinned at the other K = 0.7. Then, (KL)x = (KL)y = 0.7(5) = 3.5 m Buckling About the Weak Axis. Applying Euler’s formula, Pcr = p2EIy (KL)y 2 = p2 C 10 A 109 B D C 1.5625 A 10 - 6 B D 3.52 = 12.59 kN Euler’s formula is valid if scr 6 sY. scr = 12.59 A 103 B Pcr = = 1.68 MPa 6 sY = 15 MPa A 7.5 A 10 - 3 B O.K. Since Pcr 7 P = 10 kN, the column will not buckle. Yielding About Strong Axis. Applying the secant formula. smax = = (KL)x P ec P C 1 + 2 sec B RS A 2rx A EA rx 10 A 103 B 7.5 A 10 - 3 B D1 + 0.15(0.075) 2 0.04330 secC 10 A 103 B 3.5 ST 2(0.04330) C 10 A 109 B C 7.5 A 10 - 3 B D = 10.29 MPa Since smax 6 sY = 15 MPa , the column will not yield. Ans. 1081 75 mm 13 Solutions 46060 6/11/10 11:56 AM Page 1082 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *13–56. The wood column is fixed at its base, and its top can be considered pinned. Determine the maximum eccentric force P the column can support without causing it to either buckle or yield. Take E = 10 GPa and sY = 15 MPa. x P 150 mm 25 mm yx 25 mm 75 mm 5m Section Properties. A = 0.05(0.15) = 7.5 A 10 - 3 B m2 Ix = 1 (0.05) A 0.153 B = 14.0625 A 10 - 6 B m4 12 14.0625 A 10 Ix = C 7.5 A 10 - 3 B AA -6 rx = B = 0.04330 m 1 (0.15) A 0.053 B = 1.5625 A 10 - 6 B m4 12 e = 0.15 m c = 0.075 m Iy = For a column that is fixed at one end and pinned at the other K = 0.7. Then, (KL)x = (KL)y = 0.7(5) = 3.5 m Buckling About the Weak Axis. Applying Euler’s formula, Pcr = p2EIy (KL)y 2 = p2 C 10 A 109 B D C 1.5625 A 10 - 6 B D 3.52 = 12.59 kN = 12.6 kN Ans. Euler’s formula is valid if scr 6 sY. scr 12.59 A 103 B Pcr = = = 1.68 MPa 6 sY = 15 MPa A 7.5 A 10 - 3 B O.K. Yielding About Strong Axis. Applying the secant formula with P = Pcr = 12.59 kN, smax = = (KL)x P P ec C B 1 + 2 sec B RS A 2rx A EA rx 12.59 A 103 B 7.5 A 10 - 3 B D1 + 0.15(0.075) 0.043302 secC 12.59 A 103 B B 3.5 ST 2(0.04330) C 10 A 109 B C 7.5 A 10 - 3 B D = 13.31 MPa 6 sY = 15 MPa O.K. 1082 75 mm 13 Solutions 46060 6/11/10 11:56 AM Page 1083 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. The W250 * 28 A-36-steel column is fixed at its base. Its top is constrained to rotate about the y–y axis and free to move along the y–y axis. If e = 350 mm, determine the allowable eccentric force P that can be applied without causing the column either to buckle or yield. Use F.S. = 2 against buckling and F.S. = 1.5 against yielding. •13–57. P e x x y 6m Section Properties. From the table listed in the appendix, necessary section properties for a W250 * 28 are A = 3620 mm2 = 3.62 A 10 - 3 B m2 rx = 105 mm = 0.105 m Iy = 1.78 A 106 B mm4 = 1.78 A 10 - 6 B m4 c = 260 d = = 130 mm = 0.13 m 2 2 e = 0.35 m Buckling About the Strong Axis. Since the column is fixed at the base and pinned at the top, Kx = 0.7. Applying Euler’s formula, Pcr = p2EIy (KL)y 2 = p2 C 200 A 109 B D C 1.78 A 10 - 6 B D [0.7(6)]2 = 199.18 kN Euler’s formula is valid only if scr 6 sY. scr = 199.18 A 103 B Pcr = = 55.02 MPa 6 sY = 250 MPa A 3.62 A 10 - 3 B O.K. Thus, Pallow = Pcr 199.18 = = 99.59 kN F.S. 2 Yielding About Strong Axis. Since the column is fixed at its base and free at its top, Kx = 2. Applying the secant formula, smax = (KL)x Pmax Pmax ec C 1 + 2 sec B RS A 2rx A EA rx 250 A 106 B = 250 A 106 B = Pmax 3.62 A 10 -3 Pmax B 3.62 A 10 - 3 B D1 + 0.35(0.13) 0.1052 sec C 2(6) Pmax ST 2(0.105)A 200 A 109 B C 3.62 A 10 - 3 B D A 1 + 4.1270 sec (0.0021237)2Pmax B Solving by trial and error, Pmax = 133.45 kN Then, Pallow = y Pmax 133.45 = = 88.97 kN = 89.0 kN (controls) 1.5 1.5 Ans. 1083 13 Solutions 46060 6/11/10 11:56 AM Page 1084 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–58. The W250 * 28 A-36-steel column is fixed at its base. Its top is constrained to rotate about the y–y axis and free to move along the y–y axis. Determine the force P and its eccentricity e so that the column will yield and buckle simultaneously. P e x x y 6m Section Properties. From the table listed in the appendix, the necessary section properties for a W250 * 28 are A = 3620 mm2 = 3.62 A 10 - 3 B m2 rx = 105 mm = 0.105 m Iy = 1.78 A 106 B mm4 = 1.78 A 10 - 6 B m4 c = d 260 = = 130 mm = 0.13 m 2 2 Buckling About the Weak Axis. Since the column is fixed at the base and pinned at its top, Kx = 0.7. Applying Euler’s formula, Pcr = p2EIy (KL)y 2 = p2 C 200 A 109 B D C 1.78 A 10 - 6 B D [0.7(6)]2 = 199.18 kN = 199 kN Ans. Euler’s formula is valid only if scr 6 sY. scr = 199.18 A 103 B Pcr = = 55.02 MPa 6 sY = 250 MPa A 3.62 A 10 - 3 B O.K. Yielding About Strong Axis. Since the column is fixed at its base and free at its top, Kx = 2. Applying the secant formula with P = Pcr = 199.18 kN, smax = (KL)x P P ec C 1 + 2 sec B RS A 2rx A EA rx 250 A 106 B = 199.18 A 103 B 3.62 A 10 - 3 B D1 + e(0.13) 0.1052 secC y 199.18 A 103 B 2(6) ST 2(0.105) C 200 A 109 B C 3.62 A 10 - 3 B D e = 0.1753 m = 175 mm Ans. 1084 13 Solutions 46060 6/11/10 11:56 AM Page 1085 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–59. The steel column supports the two eccentric loadings. If it is assumed to be pinned at its top, fixed at the bottom, and fully braced against buckling about the y–y axis, determine the maximum deflection of the column and the maximum stress in the column. Est = 200 GPa, sY = 360 MPa. 130 kN 50 kN 80 mm 120 mm 100 mm 10 mm 6m A = 0.12(0.1) - (0.1)(0.09) = 3.00 A 10 - 3 B m2 Ix = 1 1 (0.1) A 0.123 B (0.09) A 0.13 B = 6.90 A 10 - 6 B m4 12 12 rx = Ix 6.90(10 - 6) = = 0.047958 m AA A 3.00(10 - 3) For a column fixed at one end and pinned at the other end, K = 0.7. (KL)x = 0.7(6) = 4.2 m The eccentricity of the two applied loads is, e = 130(0.12) - 50(0.08) = 0.06444 m 180 Yielding About x–x Axis: Applying the secant formula, (KL)x P P ec B 1 + 2 sec ¢ ≤R A 2rx A EA rx 180(103) = -3 3.00(10 ) y B1 + 0.06444(0.06) 0.047958 2 sec ¢ 180(103) 4.2 ≤R 2(0.047958)A 200(109)(3.00)(10 - 3) = 199 MPa Ans. Since smax 6 sg = 360 MPa, the column does not yield. Maximum Displacement: ymax = e B sec ¢ P KL ≤ - 1R A EI 2 = 0.06444 B sec ¢ 4.2 180(103) a b ≤ - 1R A 200(109)[6.90(10 - 6)] 2 = 0.02433 m = 24.3 mm Ans. 1085 10 mm 100 mm x Section Properties: smax = x 10 mm y 13 Solutions 46060 6/11/10 11:56 AM Page 1086 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *13–60. The steel column supports the two eccentric loadings. If it is assumed to be fixed at its top and bottom, and braced against buckling about the y–y axis, determine the maximum deflection of the column and the maximum stress in the column. Est = 200 GPa, sY = 360 MPa. 130 kN 50 kN 80 mm 120 mm 100 mm 10 mm 6m A = 0.12(0.1) - (0.1)(0.09) = 3.00 A 10 - 3 B m2 Ix = 1 1 (0.1) A 0.123 B (0.09) A 0.013 B = 6.90 A 10 - 6 B m4 12 12 rx = 6.90(10 - 6) Ix = = 0.047958 m AA A 3.00(10 - 3) For a column fixed at both ends, K = 0.5. (KL)x = 0.5(6) = 3.00 m The eccentricity of the two applied loads is, e = 130(0.12) - 50(0.08) = 0.06444 m 180 Yielding About x–x Axis: Applying the secant formula, (KL)x P P ec B 1 + 2 sec ¢ ≤R A 2rx A EA rx 180(103) = -3 3.00(10 ) y B1 + 0.06444(0.06) 0.047958 2 sec ¢ 180(103) 3.00 ≤R 2(0.047958)A 200(109)(3.00)(10 - 3) = 178 MPa Ans. Since smax 6 sg = 360 MPa, the column does not yield. Maximum Displacement: ymax = e B sec ¢ P KL ≤ - 1R A EI 2 = 0.06444 B sec ¢ 3 180(103) a b ≤ - 1R 9 6 A 200(10 )[6.90(10 )] 2 = 0.01077 m = 10.8 mm Ans. 1086 10 mm 100 mm x Section Properties: smax = x 10 mm y 13 Solutions 46060 6/11/10 11:56 AM Page 1087 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–61. The W250 * 45 A-36-steel column is pinned at its top and fixed at its base. Also, the column is braced along its weak axis at mid-height. If P = 250 kN, investigate whether the column is adequate to support this loading. Use F.S. = 2 against buckling and F.S. = 1.5 against yielding. P 4 250 mm P 250 mm 4m Section Properties. From the table listed in the appendix, the necessary section properties for a W250 * 45 are A = 5700 mm = 5.70 A 10 2 -3 Bm 2 rx = 112 mm = 0.112 m Iy = 7.03 A 106 B mm4 = 7.03 A 10 - 6 B m4 c = 266 d = = 133 mm = 0.133 m 2 2 The eccentricity of the equivalent force P¿ = 250 + 250 = 312.5 kN is 4 250 (0.25) 4 = 0.15 m 250 250 + 4 250(0.25) e = Buckling About the Weak Axis. The column is braced along the weak axis at midheight and the support provided by the bracing can be considered as a pin. The top portion of the column is critical is since the top is pinned so Ky = 1 and L = 4 m Applying Euler’s formula, Pcr = p2EIy (KL)y 2 = p2 C 200 A 109 B D C 7.03 A 10 - 6 B D [1(4)]2 = 867.29 kN Euler’s equation is valid only if scr 6 sY. scr = 867.29 A 103 B Pcr = = 152.16 MPa 6 sY = 250 MPa A 5.70 A 10 - 3 B O.K. Then, œ Pallow = Pcr 867.29 = = 433.65 kN F.S. 2 œ Since Pallow 7 P¿ , the column does not buckle. Yielding About Strong Axis. Since the column is fixed at its base and pinned at its top, Kx = 0.7 and L = 8 m. Applying the secant formula with œ = P¿(F.S.) = 312.5(1.5) = 468.75 kN Pmax smax = = œ œ (KL)x Pmax Pmax ec C 1 + 2 sec B RS A 2rx A EA rx 468.75 A 103 B 5.70 A 10 - 3 B C1 + 0.15(0.133) 0.112 2 sec B 468.75 A 103 B 0.7(8) RS 2(0.112) C 200 A 109 B C 5.70 A 10 - 3 B D = 231.84 MPa Since smax 6 sY = 250 MPa, the column does not yield. 1087 4m 13 Solutions 46060 6/11/10 11:56 AM Page 1088 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •13–62. The W250 * 45 A-36-steel column is pinned at its top and fixed at its base. Also, the column is braced along its weak axis at mid-height. Determine the allowable force P that the column can support without causing it either to buckle or yield. Use F.S. = 2 against buckling and F.S. = 1.5 against yielding. P 4 250 mm P 250 mm 4m Section Properties. From the table listed in the appendix, the necessary section properties for a W250 * 45 are A = 5700 mm2 = 5.70 A 10 - 3 B m2 rx = 112 mm = 0.112 m Iy = 7.03 A 106 B mm4 = 7.03 A 10 - 6 B m4 c = The eccentricity of the equivalent force P¿ = P + d 266 = = 133 mm = 0.133 m 2 2 P = 1.25P is 4 P (0.25) 4 = 0.15 m P P + 4 P(0.25) e = Buckling About the Weak Axis. The column is braced along the weak axis at midheight and the support provided by the bracing can be considered as a pin. The top portion of the column is critical is since the top is pinned so Ky = 1 and L = 4 m. Applying Euler’s formula, Pcr = p2EIy (KL)y 2 = p2 C 200 A 109 B D C 7.03 A 10 - 6 B D [1(4)]2 = 867.29 kN Euler’s equation is valid only if scr 6 sY. scr 867.29 A 103 B Pcr = = = 152.16 MPa 6 sY = 250 MPa A 5.70 A 10 - 3 B O.K. Then, œ = Pallow Pcr F.S. 867.29 2 = 346.92 kN 1.25Pallow = Pallow Yielding About Strong Axis. Since the column is fixed at its base and pinned at its top, Kx = 0.7 and L = 8 m. Applying the secant formula, smax = œ œ (KL)x Pmax Pmax ec C 1 + 2 sec B RS A 2rx A EA rx 250 A 106 B = 250 A 106 B = 1.25Pmax 5.70 A 10 -3 1.25Pmax B 5.70 A 10 - 3 B C1 + 0.15(0.133) 0.1122 sec B 0.7(8) 1.25Pmax RS 2(0.112) A 200 A 109 B C 5.70 A 10 - 3 B D A 1 + 1.5904 sec (0.00082783) 2Pmax B Solving by trial and error, Pmax = 401.75 kN Then, Pallow = 401.75 = 267.83 kN = 268 kN (controls) 1.5 Ans. 1088 4m 13 Solutions 46060 6/11/10 11:56 AM Page 1089 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–63. The W14 * 26 structural A-36 steel member is used as a 20-ft-long column that is assumed to be fixed at its top and fixed at its bottom. If the 15-kip load is applied at an eccentric distance of 10 in., determine the maximum stress in the column. 15 kip Section Properties for W 14 * 26 A = 7.69 in2 d = 13.91 in. 10 in. 20 ft rx = 5.65 in. Yielding about x - x axis: smax = P ec KL P b d; c 1 + 2 sec a A 2 r AE A r 15 P = = 1.9506 ksi; A 7.69 K = 0.5 10 A 13.91 ec 2 B = = 2.178714 r2 (5.65)2 0.5 (20)(12) 15 KL P = = 0.087094 2 r A EA 2(5.65) A 29 (103)(7.69) smax = 1.9506[1 + 2.178714 sec (0.087094)] = 6.22 ksi 6 sg = 36 ksi O.K. Ans. *13–64. The W14 * 26 structural A-36 steel member is used as a column that is assumed to be fixed at its top and pinned at its bottom. If the 15-kip load is applied at an eccentric distance of 10 in., determine the maximum stress in the column. 15 kip Section Properties for W 14 * 26 A = 7.69 in2 d = 13.91 in. 20 ft rx = 5.65 in. Yielding about x - x axis: smax = P ec KL P c 1 + 2 sec a b d; A 2 r AE A r 15 P = = 1.9506 ksi ; A 7.69 K = 0.7 10 A 13.91 ec 2 B = = 2.178714 2 r (5.65)2 0.7 (20)(12) 15 KL P = = 0.121931 2 r A EA 2(5.65) A 29 (103)(7.69) smax = 1.9506[1 + 2.178714 sec (0.121931)] = 6.24 ksi 6 sg = 36 ksi 10 in. O.K. Ans. 1089 13 Solutions 46060 6/11/10 11:56 AM Page 1090 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •13–65. Determine the maximum eccentric load P the 2014-T6-aluminum-alloy strut can support without causing it either to buckle or yield. The ends of the strut are pin-connected. P 150 mm 100 mm a a 3m 50 mm 100 mm Section a – a Section Properties. The necessary section properties are A = 0.05(0.1) = 5 A 10 - 3 B m2 Iy = 1 (0.1) A 0.053 B = 1.04167 A 10 - 6 B m4 12 4.1667 A 10 Ix = C 5 A 10 - 3 B AA -6 rx = B = 0.02887 m For a column that is pinned at both of its ends K = 1. Thus, (KL)x = (KL)y = 1(3) = 3 m Buckling About the Weak Axis. Applying Euler’s formula, Pcr = p2EIy (KL)y 2 = p2 C 73.1 A 109 B D C 1.04167 A 10 - 6 B D 32 = 83.50 kN = 83.5 kN Ans. Critical Stress: Euler’s formula is valid only if scr 6 sY. scr = 83.50 A 103 B Pcr = = 16.70 MPa 6 sY = 414 MPa A 5 A 10 - 3 B O.K. Yielding About Strong Axis. Applying the secant formula, smax = = (KL)x P P ec C 1 + 2 sec B RS A 2rx A EA rx 83.50 A 103 B 5 A 10 - 3 B D1 + 0.15(0.05) 0.028872 83.50 A 10 B 3 ST 2(0.02887) C 73.1 A 109 B C 5 A 10 - 3 B D 3 secC = 229.27 MPa 6 sY = 414 MPa O.K. 1090 P 150 mm 13 Solutions 46060 6/11/10 11:56 AM Page 1091 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–66. The W8 * 48 structural A-36 steel column is fixed at its bottom and free at its top. If it is subjected to the eccentric load of 75 kip, determine the factor of safety with respect to either the initiation of buckling or yielding. 75 kip y 12 ft Section Properties: For a wide flange section W8 * 48, A = 14.1 in2 rx = 3.61 in. Iy = 60.9 in4 d = 8.50 in. For a column fixed at one end and free at the other and, K = 2. (KL)y = (KL)x = 2(12)(12) = 288 in. Buckling About y–y Axis: Applying Euler’s formula, P = Pcr = p2EIy (KL)2y p2 (29.0)(103)(60.9) = 2882 = 210.15 kip Critical Stress: Euler’s formula is only valid if scr 6 sg. scr = Pcr 210.15 = = 14.90 ksi 6 sg = 36 ksi A 14.1 O. K. Yielding About x–x Axis: Applying the secant formula, smax = 36 = (KL)x Pmax Pmax ec B 1 + 2 sec ¢ ≤R A 2rx A EA rx 8 A 1.50 Pmax Pmax 288 2 B sec ¢ B1 + ≤R 14.1 2(3.61)A 29.0(103)(14.1) 3.612 36(14.1) = Pmax A 1 + 2.608943 sec 0.06238022Pmax B Solving by trial and error, Pmax = 117.0 kip (Controls!) Factor of Safety: F.S. = Pmax 117.0 = = 1.56 P 75 Ans. 1091 8 in. y x 13 Solutions 46060 6/11/10 11:56 AM Page 1092 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–67. The W8 * 48 structural A-36 steel column is fixed at its bottom and pinned at its top. If it is subjected to the eccentric load of 75 kip, determine if the column fails by yielding. The column is braced so that it does not buckle about the y–y axis. 75 kip y 12 ft Section Properties: For a wide flange section W8 * 48, A = 14.1 in2 rx = 3.61 in. d = 8.50 in. For a column fixed at one end and pinned at the other end, K = 0.7. (KL)x = 0.7(12)(12) = 100.8 in. Yielding About x–x Axis: Applying the secant formula, smax = = (KL)x P P ec B 1 + 2 sec ¢ ≤R A 2rx A EA rx 8 A 1.50 75 100.8 75 2 B sec ¢ B1 + ≤R 2 14.1 2(3.61)A 3.61 29.0(103)(14.1) = 19.45 ksi 6 sg = 36 ksi O.K. Hence, the column does not fail by yielding. Ans. 1092 8 in. y x 13 Solutions 46060 6/11/10 11:56 AM Page 1093 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *13–68. Determine the load P required to cause the steel W12 * 50 structural A-36 steel column to fail either by buckling or by yielding. The column is fixed at its bottom and the cables at its top act as a pin to hold it. 2 in. P 25 ft Section Properties: For a wide flange section W12 * 50, A = 14.7 in2 rx = 5.18 in. Iy = 56.3 in4 d = 12.19 in. For a column fixed at one end and pinned at the other end, K = 0.7. (KL)y = (KL)x = 0.7(25)(12) = 210 in. Buckling About y–y Axis: Applying Euler’s formula, P = Pcr = p2EIy (KL)2y p2 (29.0)(103)(56.3) = 2102 = 365.40 kip Critical Stress: Euler’s formula is only valid if scr 6 sg. scr = Pcr 365.4 = = 24.86 ksi 6 sg = 36 ksi A 14.7 O. K. Yielding About x–x Axis: Applying the secant formula, smax = 36 = (KL)x P P ec B 1 + 2 sec ¢ ≤R A 2rx A EA rx 2 A 12.19 P 210 P 2 B sec ¢ B1 + ≤R 14.7 2(5.18)A 29.0(103)(14.7) 5.182 36(14.7) = P A 1 + 0.454302 sec 0.03104572P B Solving by trial and error, Pmax = 343.3 kip = 343 kip (Controls!) Ans. 1093 13 Solutions 46060 6/11/10 11:56 AM Page 1094 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •13–69. Solve Prob. 13–68 if the column is an A-36 steel W12 * 16 section. 2 in. P 25 ft Section Properties: For a wide flange section W12 * 16, A = 4.71 in2 rx = 4.67 in. Iy = 2.82 in4 d = 11.99 in. For a column fixed at one end and pinned at the other end, K = 0.7. (KL)y = (KL)x = 0.7(25)(12) = 210 in. Buckling About y–y Axis: Applying Euler’s formula, P = Pcr = p 2EIy (KL)2y p2 (29.0)(103)(2.82) = 2102 = 18.30 kip = 18.3 kip‚ Ans. Critical Stress: Euler’s formula is only valid if scr 6 sg. scr = Pcr 18.30 = = 3.89 ksi 6 sg = 36 ksi A 4.71 Yielding About x–x Axis: Applying the secant formula, smax = = (KL)x P P ec B 1 + 2 sec ¢ ≤R A 2rx A EA rx 2 A 11.99 18.30 210 18.30 2 B seca bR B1 + 4.71 2(4.67)A 29.0(103)(4.71) 4.672 = 6.10 ksi 6 sg = 36 ksi O.K. 1094 13 Solutions 46060 6/11/10 11:56 AM Page 1095 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–70. A column of intermediate length buckles when the compressive stress is 40 ksi. If the slenderness ratio is 60, determine the tangent modulus. p2 Et A KrL B 2 scr = 40 = ; a KL b = 60 r p2 Et (60)2 Et = 14590 ksi = 14.6 (103) ksi ‚ Ans s(ksi) 13–71. The 6-ft-long column has the cross section shown and is made of material which has a stress-strain diagram that can be approximated as shown. If the column is pinned at both ends, determine the critical load Pcr for the column. 0.5 in. 55 0.5 in. 5 in. 25 0.5 in. 3 in. Section Properties: The neccessary section properties are A = 2[0.5(3)] + 5(0.5) = 5.5 in2 P (in./in.) 0.001 I = 2B ry = 1 1 (0.5) A 33 B R + (5) A 0.53 B = 2.3021 in4 12 12 2.3021 Iy = = 0.6470 in. AA A 5.5 For the column pinned at both of its ends, K = 1. Thus, 1(6)(12) KL = 111.29 = ry 0.6470 Critical Stress. Applying Engesser’s equation, scr = p2Et a KL r b = p2Et 111.292 = 0.7969 A 10 - 3 B Et (1) From the stress - strain diagram, the tangent moduli are (Et)1 = 25 ksi = 25 A 103 B ksi 0.001 (Et)2 = (55 - 25) ksi = 10 A 103 B (ksi) 0.004 - 0.001 0 … s 6 25 ksi 25ksi 6 s … 40 ksi Substituting (Et)1 = 25 A 103 B into Eq. (1), scr = 0.7969 A 10 - 3 B c 25 A 103 B d = 19.92 ksi Since scr 6 sY = 25 ksi, elastic buckling occurs. Thus, Pcr = scr A = 19.92(5.5) = 109.57 kip = 110 kip‚ Ans. 1095 0.004 13 Solutions 46060 6/11/10 11:56 AM Page 1096 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. s(ksi) *13–72. The 6-ft-long column has the cross section shown and is made of material which has a stress-strain diagram that can be approximated as shown. If the column is fixed at both ends, determine the critical load Pcr for the column. 0.5 in. 55 0.5 in. 5 in. 25 0.5 in. 3 in. Section Properties. The neccessary section properties are A = 2[0.5(3)] + 5(0.5) = 5.5 in2 P (in./in.) 0.001 I = 2B ry = 1 1 (0.5) A 33 B R + (5) A 0.53 B = 2.3021 in4 12 12 Iy 2.3021 = = 0.6470 in. AA A 5.5 For the column fixed at its ends, K = 0.5. Thus, 0.5(6)(12) KL = 55.64 = ry 0.6470 Critical Stress. Applying Engesser’s equation, From the stress - strain diagram, the tangent moduli are (Et)1 = 25 ksi = 25 A 103 B ksi 0.001 (Et)2 = (55 - 25)ksi = 10 A 103 B ksi 0.004 - 0.001 0 … s 6 25 ksi 25 ksi 6 s … 40 ksi Substituting (Et)1 = 25 A 103 B ksi into Eq. (1), scr = 3.1875 A 10 - 3 B c 25 A 103 B d = 79.69 ksi Since scr 7 sY = 25 ksi, the inelastic buckling occurs. Substituting (Et)2 into Eq. (1), scr = 3.1875 A 10 - 3 B c 10 A 103 B d = 31.88 ksi Since 25 ksi 6 scr 6 55 ksi, this result can be used to calculate the critical load. Pcr = scr A = 31.88(5.5) = 175.31 kip = 175 kip Ans. 1096 0.004 13 Solutions 46060 6/11/10 11:56 AM Page 1097 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. s (MPa) •13–73. The stress-strain diagram of the material of a column can be approximated as shown. Plot P兾A vs. KL兾r for the column. 350 200 Tangent Moduli. From the stress - strain diagram, (Et)1 = (Et)2 = 200 A 106 B 0.001 0 … s 6 200MPa = 200 GPa (350 - 200) A 106 B 0.004 - 0.001 = 50 GPa 0 P (in./in.) 0.001 0.004 200MPa 6 s … 350 MPa Critical Stress. Applying Engesser’s equation, scr P = A p2Et (1) KL 2 a b r If Et = (Et)1 = 200 GPa, Eq. (1) becomes p2 C 200 A 109 B D 1.974 A 106 B P = = MPa A KL 2 KL 2 a b a b r r when scr = P = sY = 200 MPa, this equation becomes A 200 A 106 B = p2 C 200 A 109 B D a KL 2 b r KL = 99.346 = 99.3 r If Et = (Et)2 = 50 GPa, Eq. (1) becomes P = A p2 c 50 A 109 B d 0.4935 A 106 B MPa KL 2 KL 2 ¢ ≤ ¢ ≤ r r P = sY = 200 MPa, this when scr = A equation gives 200 A 106 B = = p2 C 50 A 109 B D a KL 2 b r KL = 49.67 = 49.7 r Using these results, the graphs of KL P vs. is shown in Fig. a can be plotted. r A 1097 13 Solutions 46060 6/11/10 11:56 AM Page 1098 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. s (MPa) 13–74. Construct the buckling curve, P兾A versus L兾r, for a column that has a bilinear stress–strain curve in compression as shown. The column is pinned at its ends. 260 140 0.001 Tangent modulus: From the stress–strain diagram, (Et)1 = 140(106) = 140 GPa 0.001 (Et)2 = (260 - 140)(106) = 40 GPa 0.004 - 0.001 Critical Stress: Applying Engesser’s equation, scr p2Et P = A L 2 a b r [1] Substituting (Et)1 = 140 GPa into Eq. [1], we have p2 C 140(109) D P = A ALB2 r P = A When 1.38(106) A Lr B 2 MPa L P = 140 MPa, = 99.3 r A Substitute (Et)2 = 40 GPa into Eq. [1], we have p2 C 40(109) D P = A ALB2 r P = A When 0.395(106) A Lr B 2 MPa L P = 140 MPa, = 53.1 r A 1098 0.004 P (mm/mm) 13 Solutions 46060 6/11/10 11:56 AM Page 1099 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–75. The stress-strain diagram for a material can be approximated by the two line segments shown. If a bar having a diameter of 80 mm and a length of 1.5 m is made from this material, determine the critical load provided the ends are pinned. Assume that the load acts through the axis of the bar. Use Engesser’s equation. s (MPa) 1100 200 0.001 E1 = 200 (106) = 200 GPa 0.001 E2 = 1100 (106) - 200 (106) = 150 GPa 0.007 - 0.001 Section properties: I = p 4 c; 4 r = p 4 I 0.04 c 4 c = = = = 0.02 m AA C p c2 2 2 A = pc2 Engesser’s equation: 1.0(1.5) KL = = 75 r 0.02 scr = p2 Et A B KL 2 r = p2 Et (75)2 = 1.7546(10 - 3) Et Assume Et = E1 = 200 GPa scr = 1.7546 (10 - 3)(200)(109) = 351 MPa 7 200 MPa Therefore, inelastic buckling occurs: Assume Et = E2 = 150 GPa scr = 1.7546 (10 - 3)(150)(109) = 263.2 MPa 200 MPa 6 scr 6 1100 MPa O.K. Critical load: Pcr = scr A = 263.2 (106)(p)(0.042) = 1323 kN‚ Ans 1099 0.007 P (mm/mm) 13 Solutions 46060 6/11/10 11:56 AM Page 1100 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *13–76. The stress-strain diagram for a material can be approximated by the two line segments shown. If a bar having a diameter of 80 mm and a length of 1.5 m is made from this material, determine the critical load provided the ends are fixed. Assume that the load acts through the axis of the bar. Use Engesser’s equation. s (MPa) 1100 200 0.001 E1 = 200 (106) = 200 GPa 0.001 E2 = 1100 (106) - 200 (106) = 150 GPa 0.007 - 0.001 Section properties: I = p 4 c; 4 r = p 4 I 0.04 c 4 c = = = = 0.02 m AA C p c2 2 2 A = pc2 Engesser’s equation: 0.5 (1.5) KL = = 37.5 r 0.02 scr = p2 Et A B KL 2 r = p2 Et (37.5)2 = 7.018385(10 - 3) Et Assume Et = E1 = 200 GPa scr = 7.018385 (10 - 3)(200)(109) = 1403.7 MPa 7 200 MPa NG Assume Et = E2 = 150 GPa scr = 7.018385 (10 - 3)(150)(109) = 1052.8 MPa 200 MPa 6 scr 6 1100 MPa O.K. Critical load: Pcr = scr A = 1052.8 (106)(p)(0.042) = 5292 kN Ans. 1100 0.007 P (mm/mm) 13 Solutions 46060 6/11/10 11:56 AM Page 1101 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •13–77. The stress-strain diagram for a material can be approximated by the two line segments shown. If a bar having a diameter of 80 mm and length of 1.5 m is made from this material, determine the critical load provided one end is pinned and the other is fixed. Assume that the load acts through the axis of the bar. Use Engesser’s equation. s (MPa) 1100 200 0.001 E1 = 200 (106) = 200 GPa 0.001 E2 = 1100 (106) - 200 (106) = 150 GPa 0.007 - 0.001 Section properties: I = p 4 c ; 4 r = p 4 I 0.04 c 4 c = = = = 0.02 m AA C pc2 2 2 A = pc2 Engesser’s equation: 0.7 (1.5) KL = = 52.5 r 0.02 scr = p2 Et A B KL 2 r = p2Et (52.5)2 = 3.58081 (10 - 3) Et Assume Et = E1 = 200 GPa scr = 3.58081 (10 - 3)(200)(109) = 716.2 MPa 7 200 MPa NG Assume Et = E2 = 150 GPa scr = 3.58081 (10 - 3)(150)(109) = 537.1 MPa 200 MPa 6 scr 6 1100 MPa O.K. Critical load: Pcr = scr A = 537.1 (106)(p)(0.042) = 2700 kN Ans. 1101 0.007 P (mm/mm) 13 Solutions 46060 6/11/10 11:56 AM Page 1102 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–78. Determine the largest length of a structural A-36 steel rod if it is fixed supported and subjected to an axial load of 100 kN. The rod has a diameter of 50 mm. Use the AISC equations. Section Properties: A = p A 0.0252 B = 0.625 A 10 - 3 B p m2 I = p A 0.0254 B = 97.65625 A 10 - 9 B p m4 4 r = 97.65625(10 - 9)p I = = 0.0125 m AA A 0.625(10 - 3)p Slenderness Ratio: For a column fixed at both ends, K = 0.5. Thus, 0.5L KL = = 40.0L r 0.0125 AISC Column Formula: Assume a long column. sallow = 100(103) 0.625(10 - 3)p = 12p2E 2 23 A KL r B 12p2 C 200(109) D 23(40.0L)3 L = 3.555 m KL KL 2p2E = 40.0(3.555) = 142.2 and for A–36 steel, a b = r r e A sg KL KL 2p2[200(109)] = = 125.7. Since a b … … 200, the assumption is correct. r e r A 250(106) Here, Thus, L = 3.56 m Ans. 1102 13 Solutions 46060 6/11/10 11:56 AM Page 1103 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–79. Determine the largest length of a W10 * 45 structural steel column if it is pin supported and subjected to an axial load of 290 kip. Est = 29(103) ksi, sY = 50 ksi. Use the AISC equations. Section Properties: For a W10 * 45 wide flange section, A = 13.3 in2 ry = 2.01 in Slenderness Ratio: For a column pinned at both ends, K = 1. Thus, a 1(L) KL b = = 0.49751L r y 2.01 AISC Column Formula: Assume a long column, sallow = 12p2E 2 23 A KL r B 12p2 C 29(103) D 290 = 13.3 23(0.49751L)2 L = 166.3 in. KL KL 2p2E = 0.49751 (166.3) = 82.76 and for grade 50 steel, a b = r r c A sg KL KL 2p2[29(103)] 6 a b , the assumption is not correct. = = 107.0. Since r r c 50 A Thus, the column is an intermediate column. Here, Applying Eq. 13–23, B1 sallow = (KL>r)2 2(KL>r)2c R sg 3(KL>r) (KL>r)3 5 + 3 8(KL>r)c 8(KL>r)3c B1 - (0.49751L)2 R (50) 2(107.02) 290 = 13.3 3(0.49751L) (0.49751L)3 5 + 3 8(107.0) 8(107.03) 0 = 12.565658 A 10 - 9 B L3 - 24.788132 A 10 - 6 B L2 - 1.743638 A 10 - 3 B L + 0.626437 Solving by trial and error, L = 131.12 in. = 10.9 ft Ans. 1103 13 Solutions 46060 6/11/10 11:56 AM Page 1104 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *13–80. Determine the largest length of a W10 * 12 structural A-36 steel section if it is pin supported and is subjected to an axial load of 28 kip. Use the AISC equations. For a W 10 * 12, A = 3.54 in2 ry = 0.785 in. s = 28 P = = 7.91 ksi A 3.54 Assume a long column: sallow = a a 12p2E 23(KL>r)2 KL 12p2(29)(103) 12p2E = = 137.4 b = r A 23(7.91) A 23sallow KL 2p2(29)(103) 2p2E = = 126.1, b = r c A sg A 36 KL KL 7 a b r r c Long column. KL = 137.4 r L = 137.4 ¢ r 0.785 ≤ = 137.4 ¢ ≤ = 107.86 in. K 1 = 8.99 ft Ans. •13–81. Using the AISC equations, select from Appendix B the lightest-weight structural A-36 steel column that is 14 ft long and supports an axial load of 40 kip.The ends are pinned. Take sY = 50 ksi. Try, W6 * 15 (A = 4.43 in2 ry = 1.46 in.) a KL 2p2(29)(103) 2p2E b = = = 107 r c A sY 50 A a (1.0)(14)(12) KL b = = 115.1, ry 1.46 a KL KL b 7 a b ry r c Long column sallow = 12p2(29)(103) 12 p2E = = 11.28 ksi 2 23(KL>r) 23(115.1)2 Pallow = sallowA = 11.28(4.43) = 50.0 kip 7 40 kip O.K. Use W6 * 15 Ans. 1104 13 Solutions 46060 6/11/10 11:56 AM Page 1105 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–82. Using the AISC equations, select from Appendix B the lightest-weight structural A-36 steel column that is 12 ft long and supports an axial load of 40 kip. The ends are fixed. Take sY = 50 ksi. A = 2.68 in2 Try W6 * 9 a ry = 0.905 in. KL 2p2(29)(103) 2p2E b = = = 107 r c A sY A 50 0.5(12)(12) KL = = 79.56 ry 0.905 KL KL 6 a b ry r c Intermediate column sallow = KL>r 2 C 1 - 12 A (KL>r)c B D sg KL>r KL>r 3 C 53 + 38 A (KL>r)c B - 18 A (KL>r)c B D = 2 C 1 - 12 A 79.56 126.1 B D 36 ksi 1 79.56 3 C 53 + 38 A 79.56 126.1 B - 8 A 12.61 B D = 15.40 ksi Pallow = sallowA = 15.40(2.68) = 41.3 kip 7 40 kip O.K. Use W6 * 9 Ans. 1105 13 Solutions 46060 6/11/10 11:56 AM Page 1106 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–83. Using the AISC equations, select from Appendix B the lightest-weight structural A-36 steel column that is 24 ft long and supports an axial load of 100 kip.The ends are fixed. Section Properties: Try a W8 * 24 wide flange section, A = 7.08 in2 ry = 1.61 in Slenderness Ratio: For a column fixed at both ends, K = 0.5. Thus, a AISC Column 0.5(24)(12) KL b = = 89.44 r y 1.61 Formula: For A–36 steel, a KL 2p2E b = r c A sg KL KL 2p2[29(103)] 6 a b , the column is an intermediate = 126.1. Since r r c A 36 column. Applying Eq. 13–23, = (KL>r)2 B1 sallow = R sg 3(KL>r) (KL>r)3 5 + 3 8(KL>r)c 8(KL>r)3c B1 = 2(KL>r)2c (89.442) 2(126.12) R (36) 3(89.44) (89.443) 5 + 3 8(126.1) 8(126.13) = 14.271 ksi The allowable load is Pallow = sallowA = 14.271(7.08) = 101 kip 7 P = 100 kip O.K. W8 * 24 Ans. Thus, Use 1106 13 Solutions 46060 6/11/10 11:56 AM Page 1107 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *13–84. Using the AISC equations, select from Appendix B the lightest-weight structural A-36 steel column that is 30 ft long and supports an axial load of 200 kip.The ends are fixed. a A = 14.1 in2 ry = 2.08 in. Try W8 * 48 KL 2 p2 (29)(103) 2 p2E b = = = 126.1 r c A sg A 36 0.5 (30)(12) KL = = 86.54 ry 2.08 a KL KL b 6 a b intermediate column. ry r c b 1 - 12 B sallow = b 53 + 38 B KL r A KL r Bc e1 = e 53 + 3 8 C 1 2 KL r A KL r Bc 2 R r sg R - 18 B KL r A KL r Bc 2 C 86.54 126.1 D f36 86.54 126.1 D - C D 1 86.54 3 f 8 126.1 3 R r = 14.611 ksi Pallow = sallow A = 14.611 (14.1) = 206 kip 7 P = 200 kip Use O.K. W 8 * 48 Ans. 1107 13 Solutions 46060 6/11/10 11:56 AM Page 1108 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. A W8 * 24 A-36-steel column of 30-ft length is pinned at both ends and braced against its weak axis at midheight. Determine the allowable axial force P that can be safely supported by the column. Use the AISC column design formulas. •13–85. Section Properties. From the table listed in the appendix, the necessary section properties for a W8 * 24 are A = 7.08 in2 rx = 3.42 in. ry = 1.61 in. Slenderness Ratio. For a column pinned at both of its ends, K = 1. Here, Lx = 30(12) = 360 in. and Ly = 15(12) = 180 in. Thus, ¢ 1(360) KL = 105.26 ≤ = r x 3.42 ¢ 1(180) KL = 111.80 (controls) ≤ = r y 1.61 AISC = C Column 2p2 C 29 A 103 B D 36 Formulas. For = 126.10. Since ¢ A-36 steel ¢ KL KL ≤ 6 ¢ ≤ , the r y r c KL 2p2E ≤ = r c A sY column is an intermediate column. (KL>r)2 B1 sallow = R sY 3(KL>r) (KL>r)3 5 + 3 8(KL>r)c 8(KL>r)c 3 C1 - = 2(KL>r)c 2 111.802 2 A 126.102 B S(36) 3(111.80) 5 111.803 + 3 8(126.10) 8 A 126.103 B = 11.428 ksi Thus, the allowable force is Pallow = sallowA = 11.428(7.08) = 80.91 kip = 80.9 kip 1108 Ans. 13 Solutions 46060 6/11/10 11:56 AM Page 1109 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–86. Check if a W10 * 39 column can safely support an axial force of P = 250 kip. The column is 20 ft long and is pinned at both ends and braced against its weak axis at mid-height. It is made of steel having E = 29(103) ksi and sY = 50 ksi. Use the AISC column design formulas. Section Properties. From the table listed in the appendix, the necessary section properties for a W10 * 39 are A = 11.5 in2 rx = 4.27 in. ry = 1.98 in. Slenderness Ratio. For a column pinned at both of its ends, K = 1. Here, Lx = 20(12) = 240 in. and Ly = 10(12) = 120 in. Thus, ¢ 1(240) KL = 56.21 ≤ = r x 4.27 ¢ 1(120) KL = 60.606 (controls) ≤ = r y 1.98 AISC = Column 2p2 c 29 A 103 B d S 50 Formulas. For A-36 = 107.00 . Since ¢ steel ¢ KL 2p2E ≤ = r c A sY KL KL ≤ 6 ¢ ≤ , the column is an r y r c intermediate column. B1 sallow = 2(KL>r)c 2 R sY 3(KL>r) (KL>r)3 5 + 3 8(KL>r)c 8(KL>r)c 3 C1 - = (KL>r)2 60.6062 2 A 107.002 B S(50) 3(60.606) 5 60.6063 + 3 8(107.00) 8 A 107.003 B = 22.614 ksi Thus, the allowable force is Pallow = sallowA = 22.614(11.5) = 260.06 kip 7 P = 250 kip Thus, a W10 * 39 column is adequate. 1109 O.K. 13 Solutions 46060 6/11/10 11:56 AM Page 1110 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–87. A 5-ft-long rod is used in a machine to transmit an axial compressive load of 3 kip. Determine its smallest diameter if it is pin connected at its ends and is made of a 2014-T6 aluminum alloy. Section properties: A = p 2 d ; 4 I = p d 4 pd4 a b = 4 2 64 pd4 r = I d 64 = = AA C p4 d2 4 sallow = P = A p 4 3 3.820 = d2 d2 Assume long column: 1.0 (5)(12) 240 KL = = d r d 4 sallow = 54 000 A B KL 2 r ; 3.820 54000 = d2 C 240 D 2 d d = 1.42 in. Ans. KL 240 = = 169 7 55 r 1.42 O.K. 1110 13 Solutions 46060 6/11/10 11:56 AM Page 1111 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *13–88. Check if a W10 * 45 column can safely support an axial force of P = 200 kip. The column is 15 ft long and is pinned at both of its ends. It is made of steel having E = 29(103) ksi and sY = 50 ksi. Use the AISC column design formulas. Section Properties. Try W10 * 45. From the table listed in the appendix, the necessary section properties are A = 13.3 in2 ry = 2.01 in. Slenderness Ratio. For a column pinned at both of its ends, K = 1. Thus, a 1(15)(12) KL b = = 89.552 r y 2.01 KL 2p2E AISC Column Formulas. Here, a b = = r c A sY S KL KL Since a b 6 a b , the r y r c 2p2 c29 A 103 B d 50 = 107.00. column is an intermediate column. (KL>r)2 B1 sallow = R sY 3(KL>r) (KL>r)3 5 + 3 8(KL>r)c 8(KL>r)c 3 C1 - = 2(KL>r)c 2 89.5522 2 A 107.002 B S(50) 3(89.552) 5 89.5523 + 3 8(107.00) 8 A 107.003 B = 17.034 ksi Thus, the allowable force is Pallow = sallowA = 17.034(13.3) = 226.55 kip 7 P = 200 kip O.K. Thus, A W10 * 45 can be used Ans. 1111 13 Solutions 46060 6/11/10 11:56 AM Page 1112 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •13–89. Using the AISC equations, check if a column having the cross section shown can support an axial force of 1500 kN. The column has a length of 4 m, is made from A-36 steel, and its ends are pinned. 20 mm 350 mm 300 mm 10 mm Section Properties: A = 0.3(0.35) - 0.29(0.31) = 0.0151 m2 Iy = 1 1 (0.04) A 0.33 B + (0.31) A 0.013 B = 90.025833 A 10 - 6 B m4 12 12 ry = Iy 90.02583(10 - 6) = = 0.077214 m AA A 0.0151 Slenderness Ratio: For a column pinned at both ends, K = 1. Thus, a AISC = Column 2p2[200(109)] 1(4) KL b = = 51.80 r y 0.077214 Formula: = 125.7. Since A 250(10 ) column. Applying Eq. 13–23, 6 For a KL 2p2E b = r c A sg (KL>r)2 2(KL>r)2c R sg 3(KL>r) (KL>r)3 5 + 3 8(KL>r)c 8(KL>r)3c B1 = steel, KL KL 6 a b , the column is an intermediate r r c B1 sallow = A–36 (51.802) 2(125.72) 20 mm R (250)(106) 3(51.80) (51.803) 5 + 3 8(125.7) 8(125.73) = 126.2 MPa The allowable load is Pallow = sallowA = 126.2 A 106 B (0.0151) = 1906 kN 7 P = 1500 kN O.K. Thus, the column is adequate. Ans. 1112 13 Solutions 46060 6/11/10 11:56 AM Page 1113 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–90. The A-36-steel tube is pinned at both ends. If it is subjected to an axial force of 150 kN, determine the maximum length that the tube can safely support using the AISC column design formulas. 100 mm 80 mm Section Properties. A = p A 0.052 - 0.042 B = 0.9 A 10 - 3 B p m2 I = r = p A 0.054 - 0.044 B = 0.9225 A 10 - 6 B p m4 4 0.9225 A 10 - 6 B p I = = 0.03202 m AA C 0.9 A 10 - 3 B p Slenderness Ratio. For a column pinned at both of its ends, K = 1. Thus, 1(L) KL = = 31.23L r 0.03202 AISC Column Formulas. sallow = 12p2E 23(KL>r)2 150 A 103 B .9 A 10 - 3 B p = 12p2 C 200 A 109 B D 23(31.23L)2 L = 4.4607 m = 4.46 m Here, KL = 31.23(4.4607) = 139.33. r 2p2 C 200 A 109 B D = 125.66. Since a 250 A 10 B long column is correct. = C 6 Ans. For A-36 steel a KL 2p2E b = r c A sY KL KL b 6 6 200, the assumption of a r c r 1113 13 Solutions 46060 6/11/10 11:56 AM Page 1114 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–91. The bar is made of a 2014-T6 aluminum alloy. Determine its smallest thickness b if its width is 5b. Assume that it is pin connected at its ends. 600 lb b 5b 8 ft Section Properties: A = b(5b) = 5b2 Iy = 1 5 4 (5b) A b3 B = b 12 12 ry = 5 4 Iy 23 12 b = = b AA C 5b2 6 600 lb Slenderness Ratio: For a column pinned at both ends, K = 1. Thus, a 1(8)(12) 332.55 KL = b = 23 b r y 6 b Aluminum (2014 - T6 alloy) Column Formulas: Assume a long column and apply Eq. 13–26. sallow = 54 000 (KL>r)2 0.600 54 000 = 2 5b A 332.55 B 2 b b = 0.7041 in. Here, KL KL 332.55 = 7 55, the assumption is correct. Thus, = 472.3. Since r r 0.7041 b = 0.704 in. Ans. 1114 13 Solutions 46060 6/11/10 11:56 AM Page 1115 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *13–92. The bar is made of a 2014-T6 aluminum alloy. Determine its smallest thickness b if its width is 5b. Assume that it is fixed connected at its ends. 600 lb b 5b 8 ft Section Properties: A = b(5b) = 5b2 Iy = 1 5 4 (5b) A b3 B = b 12 12 ry = Iy 23 12 b = = b AA C 5b2 6 5 600 lb 4 Slenderness Ratio: For a column fixed at both ends, K = 0.5. Thus, a 0.5(8)(12) 166.28 KL = b = r y b 23 6 b Aluminum (2014 - T6 alloy) Column Formulas: Assume a long column and apply Eq. 13–26. sallow = 54 000 (KL>r)2 0.600 54 000 = 5b2 A 166.28 B 2 b b = 0.4979 in. Here, KL KL 166.28 = 334.0. Since = 7 55, the assumption is correct. r r 0.4979 Thus, b = 0.498 in. Ans. 1115 13 Solutions 46060 6/11/10 11:56 AM Page 1116 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •13–93. The 2014-T6 aluminum column of 3-m length has the cross section shown. If the column is pinned at both ends and braced against the weak axis at its mid-height, determine the allowable axial force P that can be safely supported by the column. 15 mm 170 mm 15 mm 15 mm 100 mm Section Properties. A = 0.1(0.2) - 0.085(0.17) = 5.55 A 10 - 3 B m2 Ix = 1 1 (0.1) A 0.23 B (0.085) A 0.173 B = 31.86625 A 10 - 6 B m4 12 12 Iy = 2 c rx = ry = 1 1 (0.015) A 0.13 B d + (0.17) A 0.0153 B = 2.5478 A 10 - 6 B m4 12 12 31.86625 A 10 - 6 B Ix = = 0.07577 AA C 5.55 A 10 - 3 B 2.5478 A 10 - 6 B Iy = = 0.02143 m AA C 5.55 A 10 - 3 B Slenderness Ratio. For a column pinned at both of its ends, K = 1. Here, Lx = 3 m and Ly = 1.5 m. Thus, a (1)(3) KL b = = 39.592 r x 0.07577 a (1)(1.5) KL b = = 70.009 (controls) r y 0.02143 2014-T6 Alumimum Alloy Column Formulas. Since a KL b 7 55, the column can r y be classified a long column, sallow = D 373 A 103 B T Mpa = C 373 A 103 B S MPa a KL 2 b r 70.0092 = 76.103 MPa Thus, the allowed force is Pallow = sallowA = 76.103 A 106 B C 5.55 A 10 - 3 B D = 422.37 kN = 422 kN 1116 Ans. 13 Solutions 46060 6/11/10 11:56 AM Page 1117 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–94. The 2014-T6 aluminum column has the cross section shown. If the column is pinned at both ends and subjected to an axial force P = 100 kN, determine the maximum length the column can have to safely support the loading. 15 mm 170 mm 15 mm 15 mm 100 mm Section Properties. A = 0.1(0.2) - 0.085(0.17) = 5.55 A 10 - 3 B m2 Iy = 2 c ry = 1 1 (0.015) A 0.13 B d + (0.17) A 0.0153 B = 2.5478 A 10 - 6 B m4 12 12 2.5478 A 10 - 6 B Iy = = 0.02143 m AA C 5.55 A 10 - 3 B Slenderness Ratio. For a column pinned at both of its ends, K = 1. Then, a 1(L) KL b = = 46.6727L r y 0.02143 2014-T6 Alumimum Alloy Column Formulas. Assuming a long column, sallow = D 100 A 103 B 373 A 103 B 5.55 A 10 - 3 B a KL 2 b r = C T MPa 373 A 103 B (46.672L)2 S A 106 B Pa L = 3.083 m = 3.08 m Since a Ans. KL b = 46.6727(3.083) = 143.88 7 55, the assumption is correct. r y 1117 13 Solutions 46060 6/11/10 11:56 AM Page 1118 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–95. The 2014-T6 aluminum hollow section has the cross section shown. If the column is 10 ft long and is fixed at both ends, determine the allowable axial force P that can be safely supported by the column. 4 in. 3 in. Section Properties. A = p A 22 - 1.52 B = 1.75p in2 r = I = p 4 A 2 - 1.54 B = 2.734375p in4 4 I 2.734375p = = 1.25 in. AA A 1.75p Slenderness Ratio. For a column fixed at both of its ends, K = 0.5. Thus, 0.5(10)(12) KL = = 48 r 1.25 2014-T6 Aluminum Alloy Column Formulas. Since 12 6 KL 6 55, the column can r be classified as an intermediate column. sallow = c 30.7 - 0.23a KL b d ksi r = [30.7 - 0.23(48)] ksi = 19.66 ksi Thus, the allowable load is Pallow = sallowA = 19.66 A 106 B (1.75p) = 108.09 kip = 108 kip 1118 Ans. 13 Solutions 46060 6/11/10 11:56 AM Page 1119 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *13–96. The 2014-T6 aluminum hollow section has the cross section shown. If the column is fixed at its base and pinned at its top, and is subjected to the axial force P = 100 kip, determine the maximum length of the column for it to safely support the load. 4 in. 3 in. Section Properties. A = p A 22 - 1.52 B = 1.75p in2 r = I = p 4 A 2 - 1.54 B = 2.734375p in4 4 I 2.734375p = = 1.25 in. AA A 1.75p Slenderness Ratio. For a column fixed at its base and pinned at its top, K = 0.7. Thus, 0.7(L) KL = = 0.56L r 1.25 2014-T6 Aluminum Alloy Column Formulas. Assuming an intermediate column, sallow = c 30.7 - 0.23a KL b d ksi r 100 = 30.7 - 0.23(0.56L) 1.75p L = 97.13 in. = 8.09 ft Ans. KL = 0.56(97.13) = 54.39 6 55, the assumption of an intermediate column r is correct. Since 1119 13 Solutions 46060 6/11/10 11:56 AM Page 1120 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •13–97. The tube is 0.25 in. thick, is made of a 2014-T6 aluminum alloy, and is fixed at its bottom and pinned at its top. Determine the largest axial load that it can support. P x y 6 in. x 6 in. y 10 ft Section Properties: P A = 6(6) - 5.5(5.5) = 5.75 in2 I = 1 1 (6) A 63 B (5.5) A 5.53 B = 31.7448 in4 12 12 r = I 31.7448 = = 2.3496 in. A 5.75 AA Slenderness Ratio: For a column fixed at one end and pinned at the other end, K = 0.7. Thus, 0.7(10)(12) KL = = 35.75 r 2.3496 Aluminium (2014 –∑ T6 alloy) Column Formulas: Since 12 6 KL 6 55, the r column is classified as an intermediate column. Applying Eq. 13–25, sallow = c 30.7 - 0.23a KL b d ksi r = [30.7 - 0.23(33.75)] = 24.48 ksi The allowable load is Pallow = sallowA = 22.48(5.75) = 129 kip Ans. 1120 13 Solutions 46060 6/11/10 11:56 AM Page 1121 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–98. The tube is 0.25 in. thick, is made of a 2014-T6 aluminum alloy, and is fixed connected at its ends. Determine the largest axial load that it can support. P x y 6 in. x 6 in. y 10 ft Section Properties: P A = 6(6) - 5.5(5.5) = 5.75 in2 I = 1 1 (6) A 63 B (5.5) A 5.53 B = 31.7448 in4 12 12 r = I 31.7448 = = 2.3496 in. AA A 5.75 Slenderness Ratio: For column fixed at both ends, K = 0.5. Thus, 0.5(10)(12) KL = = 25.54 r 2.3496 Aluminium (2014 – T6 alloy) Column Formulas: Since 12 6 KL 6 55, the r column is classified as an intermediate column. Applying Eq. 13–25, sallow = c 30.7 - 0.23 a KL b d ksi r = [30.7 - 0.23(25.54)] = 24.83 ksi The allowable load is Pallow = sallowA = 24.83(5.75) = 143 kip Ans. 1121 13 Solutions 46060 6/11/10 11:56 AM Page 1122 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–99. The tube is 0.25 in. thick, is made of 2014-T6 aluminum alloy and is pin connected at its ends. Determine the largest axial load it can support. P x y 6 in. x 6 in. y Section Properties: A = 6(6) - 5.5(5.5) = 5.75 in2 I = 1 1 (6) A 63 B (5.5) A 5.53 B = 31.7448 in4 12 12 r = I 31.7448 = = 2.3496 in. AA A 5.75 10 ft P Slenderness Ratio: For a column pinned as both ends, K = 1. Thus, 1(10)(12) KL = = 51.07 r 2.3496 Aluminum (2014 – T6 alloy) Column Formulas: Since 12 6 KL 6 55, the r column is classified as an intermediate column. Applying Eq. 13–25, sallow = c 30.7 - 0.23 a KL b d ksi r = [30.7 - 0.23(51.07)] = 18.95 ksi The allowable load is Pallow = sallowA = 18.95(5.75) = 109 kip Ans. *13–100. A rectangular wooden column has the cross section shown. If the column is 6 ft long and subjected to an axial force of P = 15 kip, determine the required minimum 1 dimension a of its cross-sectional area to the nearest 16 in. so that the column can safely support the loading. The column is pinned at both ends. a 2a Slenderness Ratio. For a column pinned at both of its ends, K = 1. Then, (1)(6)(12) KL 72 = = a a d NFPA Timber Column Formula. Assuming an intermediate column, sallow = 1.20 c 1 - 1 KL>d 2 a b d ksi 3 26.0 15 1 72>a 2 = 1.20 c 1 - a b d 2a(a) 3 26.0 a = 2.968 in. Use a = 3 in. Ans. KL 72 KL = = 24. Since 11 6 6 26, the assumption is correct. d 3 d 1122 13 Solutions 46060 6/11/10 11:56 AM Page 1123 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •13–101. A rectangular wooden column has the cross section shown. If a = 3 in. and the column is 12 ft long, determine the allowable axial force P that can be safely supported by the column if it is pinned at its top and fixed at its base. a 2a Slenderness Ratio. For a column fixed at its base and pinned at its top K = 0.7. Then, 0.7(12)(12) KL = = 33.6 d 3 NFPA Timer Column Formula. Since 26 6 KL 6 50, the column can be classified d as a long column. sallow = 540 ksi 540 = = 0.4783 ksi 2 (KL>d) 33.62 The allowable force is Pallow = sallowA = 0.4783(3)(6) = 8.61 kip Ans. 13–102. A rectangular wooden column has the cross section shown. If a = 3 in. and the column is subjected to an axial force of P = 15 kip, determine the maximum length the column can have to safely support the load. The column is pinned at its top and fixed at its base. a 2a Slenderness Ratio. For a column fixed at its base and pinned at its top, K = 0.7. Then, KL 0.7L = = 0.2333L d 3 NFPA Timber Column Formula. Assuming an intermediate column, sallow = 1.20 c 1 - 1 KL>d 2 a b d ksi 3 26.0 15 1 0.2333L 2 = 1.20 c 1 - a b d 3(6) 3 26.0 L = 106.68 in. = 8.89 ft Ans. KL KL = 0.2333(106.68) = 24.89. Since 11 6 6 26, the assumption is d d correct. Here, 1123 13 Solutions 46060 6/11/10 11:56 AM Page 1124 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–103. The timber column has a square cross section and is assumed to be pin connected at its top and bottom. If it supports an axial load of 50 kip, determine its smallest side dimension a to the nearest 12 in. Use the NFPA formulas. 14 ft a Section properties: A = a2 sallow = s = 50 P = 2 A a Assume long column: sallow = 50 = a2 C 540 2 A KL d B 540 (1.0)(14)(12) a D 2 a = 7.15 in. (1.0)(14)(12) KL KL = = 23.5, 6 26 d 7.15 d Assumption NG Assume intermediate column: sallow = 1.20 B 1 - 1 KL>d 2 a b R 3 26.0 2 50 1 a b R = 1.20 B 1 - a 2 a 26.0 3 1.0(14)(12) a = 7.46 in. 1.0(14)(12) KL KL = = 22.53, 11 6 6 26 d 7.46 d Assumption O.K. 1 Use a = 7 in. 2 Ans. 1124 13 Solutions 46060 6/11/10 11:56 AM Page 1125 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *13–104. The wooden column shown is formed by gluing together the 6 in. * 0.5 in. boards. If the column is pinned at both ends and is subjected to an axial load P = 20 kip, determine the required number of boards needed to form the column in order to safely support the loading. P 6 in. 0.5 in. 9 ft Slenderness Ratio. For a column pinned at both of its ends, K = 1. If the number of the boards required is n and assuming that n(0.5) 6 6 in. Then, d = n(0.5). Thus, (1)(9)(12) KL 216 = = n d n(0.5) P NFPA Timber Column Formula. Assuming an intermediate column, sallow = 1.20 B 1 - 1 KL>d 2 a b R ksi 3 26.0 1 216>n 2 20 = 1.20 B 1 - a b R [n(0.5)](6) 3 26.0 n2 - 5.5556n - 23.01 = 0 Solving for the positive root, n = 8.32 Use n = 9 Ans. KL KL 216 = = 24. Since n(0.5) = 9(0.5) = 4.5 in. 6 6 in. and 11 6 6 26, d 9 d the assumptions made are correct. Here, 1125 13 Solutions 46060 6/11/10 11:56 AM Page 1126 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •13–105. The column is made of wood. It is fixed at its bottom and free at its top. Use the NFPA formulas to determine its greatest allowable length if it supports an axial load of P = 2 kip. P 2 in. y y x x 4 in. Slenderness Ratio: For a column fixed at one end and free at the other end, K = 2. Thus, L 2(L) KL = = 1.00L d 2 NFPA Timber Column Formulas: Assume a long column. Apply Eq. 13–29, sallow = 540 ksi (KL>d)2 2 540 = 2(4) (1.00L)2 L = 46.48 in Here, KL KL = 1.00(46.48) = 46.48. Since 26 6 6 50, the assumption is correct. d d Thus, L = 46.48 in. = 3.87 ft Ans. 13–106. The column is made of wood. It is fixed at its bottom and free at its top. Use the NFPA formulas to determine the largest allowable axial load P that it can support if it has a length L = 4 ft. P 2 in. y x y x 4 in. Slenderness Ratio: For a column fixed at one end and free at the other end, K = 2. Thus, L 2(4)(12) KL = = 48.0 d 2 NFPA Timber Column Formulas: Since 26 6 KL 6 50, it is a long column. Apply d Eq. 13–29, sallow = = 540 ksi (KL>d)2 540 48.02 = 0.234375 ksi The allowable axial force is Pallow = sallowA = 0.234375[2(4)] = 1.875 kip Ans. 1126 13 Solutions 46060 6/11/10 11:56 AM Page 1127 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–107. The W14 * 53 structural A-36 steel column supports an axial load of 80 kip in addition to an eccentric load P. Determine the maximum allowable value of P based on the AISC equations of Sec. 13.6 and Eq. 13–30. Assume the column is fixed at its base, and at its top it is free to sway in the x–z plane while it is pinned in the y–z plane. z 80 kip x y 12 ft Section Properties: For a W14 * 53 wide flange section. A = 15.6 in2 Ix = 541 in4 d = 13.92 in. rx = 5.89 in. ry = 1.92 in. Slenderness Ratio: By observation, the largest slenderness ratio is about y -y axis. For a column fixed at one end and free at the other end, K = 2. Thus, a 2(12)(12) KL b = = 150 r y 1.92 Allowable Stress: The allowable stress can be determined using AISC Column 2p2[29(103)] 2p2E KL b = = = 126.1. Since Formulas. For A–36 steel, a r c B sY B 36 KL KL b … … 200, the column is a long column. Applying Eq. 13–21, a r c r sallow = 12p2E 23(KL>r)2 12p2(29.0)(103) = 23(1502) = 6.637 ksi Maximum Stress: Bending is about x - x axis. Applying we have smax = sallow = 6.637 = Mc P + A I P(10) A 13.92 P + 80 2 B + 15.6 541 P = 7.83 kip Ans. 1127 P y x 10 in. 13 Solutions 46060 6/11/10 11:56 AM Page 1128 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *13–108. The W12 * 45 structural A-36 steel column supports an axial load of 80 kip in addition to an eccentric load of P = 60 kip. Determine if the column fails based on the AISC equations of Sec. 13.6 and Eq. 13–30. Assume that the column is fixed at its base, and at its top it is free to sway in the x–z plane while it is pinned in the y–z plane. z 80 kip x y 12 ft Section Properties: For a W12 * 45 wide flange section, A = 13.2 in2 d2 = 12.06 in. Ix = 350 in4 rx = 5.15 in. ry = 1.94 in. Slenderness Ratio: By observation, the largest slenderness ratio is about y- y axis. For a column fixed at one end and free at the other end, K = 2. Thus, a 2(12)(12) KL b = = 148.45 r y 1.94 Allowable Stress: The allowable stress can be determined using AISC Column 2p2[29(103)] 2p2E KL b = = = 126.1. Since Formulas. For A–36 steel, a r c B sY B 36 KL KL b … … 200, the column is a long column. Applying Eq. 13–21, a r c r sallow = 12p2E 23(KL>r)2 12p2(29.0)(103) = 23(148.452) = 6.776 ksi Maximum Stress: Bending is about x - x axis. Applying Eq. 1 we have smax = = Mc P + A I 60(10) A 12.06 140 2 B + 13.2 350 = 20.94 ksi Since smax 7 sallow, the column is not adequate. 1128 P y x 10 in. 13 Solutions 46060 6/11/10 11:56 AM Page 1129 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •13–109. The W14 * 22 structural A-36 steel column is fixed at its top and bottom. If a horizontal load (not shown) causes it to support end moments of M = 10 kip # ft, determine the maximum allowable axial force P that can be applied. Bending is about the x–x axis. Use the AISC equations of Sec. 13.6 and Eq. 13–30. P x M y y x 12 ft Section properties for W14 * 22: A = 6.49 in2 d = 13.74 in2 Ix = 199 in4 ry = 1.04 in. M Allowable stress method: P 0.5(12)(12) KL = = 69.231 ry 1.04 a KL KL 2p2E KL 2p2(29)(103) b = = = 126.1, 6 a b r c r r c B sY B 36 y Hence, B1 (sa)allow = B 53 + 3 8 smax = (sa)allow = 16.510 = 1 2 ¢ KL r 2 A KL r B 2 A KL r Bc A KL r Bc - c1 - ≤ R sY 1 8 3 A KL r B 3 A KL r Bc = R c 53 + 3 8 1 2 2 A 69.231 126.1 B d36 1 69.231 3 A 69.231 126.1 B - 8 A 126.1 B d = 16.510 ksi My c P + A Ix 10(12)(13.74 P 2 ) + 6.49 199 P = 80.3 kip Ans. 1129 13 Solutions 46060 6/11/10 11:56 AM Page 1130 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–110. The W14 * 22 column is fixed at its top and bottom. If a horizontal load (not shown) causes it to support end moments of M = 15 kip # ft, determine the maximum allowable axial force P that can be applied. Bending is about the x–x axis. Use the interaction formula with 1sb2allow = 24 ksi. P x M y y x 12 ft Section Properties for W 14 * 22: A = 6.49 in2 d = 13.74 in2 Ix = 199 in4 ry = 1.04 in. M P Interaction method: 0.5(12)(12) KL = = 69.231 ry 1.04 a KL 2p2E KL KL 2p2(29)(103) b = = = 126.1, 6 a b r c ry r c B sY B 36 Hence, B1 (sa)allow = B 53 sa = + 3 8 1 2 ¢ KL r 2 A KL r B 2 A KL r Bc A KL r Bc - c1 - ≤ R sY 1 8 P P = = 0.15408 P A 6.49 3 A KL r B 3 A KL r Bc = R c 53 + 3 8 A 1 2 2 A 69.231 126.1 B d36 69.231 126.1 B - 1 8 A B 69.231 3 d 126.1 = 16.510 ksi 15(12) A 13.74 Mxc 2 B = = 6.214 ksi sb = Ix 199 sb sa + = 1.0 (sa)allow (sb)allow 6.2141 0.15408 P + = 1.0 16.510 24 P = 79.4 kip Ans. 1130 13 Solutions 46060 6/11/10 11:56 AM Page 1131 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–111. The W14 * 43 structural A-36 steel column is fixed at its bottom and free at its top. Determine the greatest eccentric load P that can be applied using Eq. 13–30 and the AISC equations of Sec. 13.6. 40 kip 16 in. 10 ft Section properties for W14 * 43: A = 12.6 in2 d = 13.66 in. Iy = 45.2 in4 ry = 1.89 in. b = 7.995 Allowable stress method: 2(10)(12) KL = = 126.98 ry 1.89 a 2p2E KL KL KL 2p2 (29)(103) b = = = 126.1, 200 7 7 a b r c ry r c B sY B 36 (sa)allow = 12p2(29)(103) 12p2E = = 9.26 ksi 2 23(KL>r) 23(126.98)2 smax = (sa)allow = P My c P + A Iy P(16) A 7.995 P + 40 A B 9.26 = + 12.6 45.2 P = 4.07 kip Ans. 1131 13 Solutions 46060 6/11/10 11:56 AM Page 1132 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *13–112. The W10 * 45 structural A-36 steel column is fixed at its bottom and free at its top. If it is subjected to a load of P = 2 kip, determine if it is safe based on the AISC equations of Sec. 13.6 and Eq. 13–30. 40 kip 16 in. 10 ft Section Properties for W10 * 45: A = 13.3 in2 d = 10.10 in. Iy = 53.4 in4 ry = 2.01 in. b = 8.020 in. Allowable stress method: 2.0(10)(12) KL = = 119.4 ry 2.01 a KL 2p2E 2p2(29)(103) b = = = 126.1 r c B sY B 36 KL KL 6 a b r r c (sa)allow Ú (sa)allow = 10.37 Ú c1 5 3 + A 1 (KL>r) 2 (KL>r)3 d sY 3 KL>r 8 KL>rc My c P + A Iy 2 C 1 - 12 A 119.4 126.1 B D 36 2 B - c 3 1 (KL>r) 8 (KL>rc)3 = 5 3 + 3 8 1 119.4 3 A 119.4 136.1 B - 8 A 126.1 B P = 10.37 ksi 2(16) A 8.020 42 2 B + 13.3 53.4 10.37 Ú 5.56 O.K. Column is safe. Yes. Ans. 1132 13 Solutions 46060 6/11/10 11:56 AM Page 1133 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. The A-36-steel W10 * 45 column is fixed at its base. Its top is constrained to move along the x–x axis but free to rotate about and move along the y–y axis. Determine the maximum eccentric force P that can be safely supported by the column using the allowable stress method. •13–113. 12 in. y Section Properties. From the table listed in the appendix, the section properties for a W10 * 45 are bf = 8.02 in. rx = 4.32 in. Iy = 53.4 in4 ry = 2.01 in. Slenderness Ratio. Here, Lx = 24(12) = 288 in. and for a column fixed at its base and free at its top, Kx = 2. Thus, a 2(288) KL b = = 133.33 (controls) r x 4.32 Since the column is fixed at its base and pinned at its top, Ky = 0.7 and Ly = 24(12) = 288 in. Then, a 0.7(288) KL b = = 100.30 r y 2.01 Allowable Stress. The allowable stress will be determined using the AISC column formulas. For A-36 steel, 2p2 C 29 A 103 B D KL 2p2E KL KL b = = = 126.10. Since a b 6 a b 6 200, r c r c r x B sY C 36 the column is classified as a long column. a sallow = = 12p2E 23(KL>r)2 12p2 C 29 A 103 B D 23(133.332) = 8.400 ksi Maximum Stress. Bending is about the weak axis. Since M = P(12) and bf 8.02 = = 4.01 in, c = 2 2 sallow = Mc P + A I 8.400 = [P(12)](4.01) P + 13.3 53.4 y x 24 ft A = 13.3 in2 P P = 8.604 kip = 8.60 kip Ans. 1133 x 13 Solutions 46060 6/11/10 11:56 AM Page 1134 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–114. The A-36-steel W10 * 45 column is fixed at its base. Its top is constrained to move along the x–x axis but free to rotate about and move along the y–y axis. Determine the maximum eccentric force P that can be safely supported by the column using an interaction formula. The allowable bending stress is (sb)allow = 15 ksi. 12 in. bf = 8.02 in. rx = 4.32 in. y 24 ft Iy = 53.4 in4 ry = 2.01 in. Slenderness Ratio. Here, Lx = 24(12) = 288 in and for a column fixed at its base and free at its top, Kx = 2. Thus, a 2(288) KL b = = 133.33 (controls) r x 4.32 Since the column is fixed at its base and pinned at its top, Ky = 0.7 and Ly = 24(12) = 288 in. Then, a 0.7(288) KL b = = 100.30 r y 2.01 Allowable Stress. The allowable stress will be determined using the AISC column 2p2 C 29 A 103 B D 2p2E KL = 126.10. Since formulas. For A-36 steel, a b = = C r c B sY 36 a KL KL b 6 a b 6 200, the column is classified as a long column. r c r x sallow = = 12p2E 23(KL>r)2 12p2 C 29 A 103 B D 23 A 133.332 B = 8.400 ksi Interaction Formula. Bending is about the weak axis. Here, M = P(12) and bf 8.02 = = 4.01 in. c = 2 2 P>A Mc>Ar2 + = 1 (sa)allow (sb)allow P>13.3 + 8.400 P(12)(4.01) n C 13.3 A 2.012 B D 15 y x Section Properties. From the table listed in the appendix, the section properties for a W10 * 45 are A = 13.3 in2 P = 1 P = 14.57 kip = 14.6 kip Ans. 14.57>13.3 sa = = 0.1304 6 0.15 (sa)allow 8.400 O.K. 1134 x 13 Solutions 46060 6/11/10 11:56 AM Page 1135 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–115. The A-36-steel W12 * 50 column is fixed at its base. Its top is constrained to move along the x–x axis but free to rotate about and move along the y–y axis. If the eccentric force P = 15 kip is applied to the column, investigate if the column is adequate to support the loading. Use the allowable stress method. 12 in. bf = 8.08 in. rx = 5.18 in. y 24 ft Iy = 56.3 in4 ry = 1.96 in. Slenderness Ratio. Here, Lx = 24(12) = 288 in. and for a column fixed at its base and pinned at its top, Kx = 2. Thus, a 2(288) KL b = = 111.20 (controls) r x 5.18 Since the column is fixed at its base and pinned at its top, Ky = 0.7 and Ly = 24(12) = 288 in. Thus, a 0.7(288) KL b = = 102.86 r y 1.96 Allowable Stress. The allowable stress will be determined using the AISC column formulas. For A-36 steel, a a 2p2 C 29 A 103 B D KL 2p2E = 126.10. Since b = = C r c A sY 36 KL KL b 6 a b , the column can be classified as an intermediate column. r x r c B1 sallow = 2(KL>r)C 2 R sY 3(KL>r) (KL>r)3 5 + 3 8(KL>r)C 8(KL>r)C 3 C1 - = (KL>r)2 111.202 2 A 126.102 B S(36) 3(111.20) 5 111.203 + 3 8(126.10) 8 A 126.103 B = 11.51 ksi Maximum Stress. Bending is about the weak axis. Since, M = 15(12) = 180 kip # in. bf 8.08 = = 4.04 in., and c = 2 2 smax = y x Section Properties. From the table listed in the appendix, the section properties for a W12 * 50 are A = 14.7 in2 P 180(4.04) P Mc 15 + = + = 13.94 ksi A I 14.7 56.3 Since smax 7 sallow, the W12 * 50 column is inadequate according to the allowable stress method. 1135 x 13 Solutions 46060 6/11/10 11:56 AM Page 1136 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *13–116. The A-36-steel W12 * 50 column is fixed at its base. Its top is constrained to move along the x–x axis but free to rotate about and move along the y–y axis. If the eccentric force P = 15 kip is applied to the column, investigate if the column is adequate to support the loading. Use the interaction formula. The allowable bending stress is (sb)allow = 15 ksi. 12 in. bf = 8.08 in. rx = 5.18 in. y 24 ft ry = 1.96 in. Slenderness Ratio. Here, Lx = 24(12) = 288 in and for a column fixed at its base and pinned at its top, Kx = 2. Thus, a 2(288) KL b = = 111.20 (controls) r x 5.18 Since the column is fixed at its base and pinned at its top, Ky = 0.7 and Ly = 24(12) = 288 in. Then, a Allowable = C Axial 2p2 C 29 A 103 B D 36 0.7(288) KL b = = 102.86 r y 1.96 Stress. For = 126.10. Since a A-36 steel, a KL 2p2E b = r c A sY KL KL b 6 a b , the column can be r x r c classified as an intermediate column. C1 sallow = 2(KL>r)c 2 SsY 3(KL>r) (KL>r)3 5 + 3 8(KL>r)c 8(KL>r)c 3 C1 - = (KL>r)2 111.202 2 A 126.102 B S(36) 3(111.20) 5 111.203 + 3 8(126.10) 8 A 126.103 B = 11.51 ksi Interaction Formula. Bending is about the weak axis. Here, M = 15(12) bf 8.08 = 180 kip # in. and c = = = 4.04 in. 2 2 P>A Mc>Ar2 15>14.7 + + = (sa)allow (sb)allow 11.51 180(4.04) n C 14.7 A 1.962 B D 15 y x Section Properties. From the table listed in the appendix, the section properties for a W12 * 50 are A = 14.7 in2 P = 0.9471 6 1 P = 14.57 kip = 14.6 kip Ans. 15>14.7 sa = = 0.089 6 0.15 (sa)allow 11.51 O.K. Thus, a W12 * 50 column is adequate according to the interaction formula. 1136 x 13 Solutions 46060 6/11/10 11:56 AM Page 1137 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •13–117. A 16-ft-long column is made of aluminum alloy 2014-T6. If it is fixed at its top and bottom, and a compressive load P is applied at point A, determine the maximum allowable magnitude of P using the equations of Sec. 13.6 and Eq. 13–30. P A 4.25 in. x 0.5 in. y 8 in. Section properties: A = 2(0.5)(8) + 8(0.5) = 12 in2 Ix = 1 1 (8)(93) (7.5)(83) = 166 in4 12 12 Iy = 2 a ry = 1 1 b (0.5)(83) + (8)(0.53) = 42.75 in4 12 12 Iy 42.75 = = 1.8875 in. AA A 12 Allowable stress method: 0.5(16)(12) KL KL = 50.86, 12 6 = 6 55 ry ry 1.8875 sallow = c 30.7 - 0.23a KL bd r = [30.7 - 0.23(50.86)] = 19.00 ksi smax = sallow = 19.00 = Mx c P + A Ix P(4.25)(4.5) P + 12 166 P = 95.7 kip Ans. 1137 y x 8 in. 0.5 in. 0.5 in. 13 Solutions 46060 6/11/10 11:56 AM Page 1138 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–118. A 16-ft-long column is made of aluminum alloy 2014-T6. If it is fixed at its top and bottom, and a compressive load P is applied at point A, determine the maximum allowable magnitude of P using the equations of Sec. 13.6 and the interaction formula with 1sb2allow = 20 ksi. P A 4.25 in. x 0.5 in. y 8 in. Section Properties: A = 2(0.5)(8) + 8(0.5) = 12 in2 Ix = 1 1 (8)(93) (7.5)(83) = 166 in4 12 12 Iy = 2 a ry = 1 1 b (0.5)(83) + (8)(0.53) = 42.75 in4 12 12 Iy 42.75 = = 1.8875 in. AA A 12 Interaction method: 0.5(16)(12) KL KL = 50.86, 12 6 = 6 55 ry ry 1.8875 sallow = c 30.7 - 0.23a KL bd r = [30.7 - 0.23(50.86)] = 19.00 ksi sa = P P = = 0.08333P A 12 sb = P(4.25)(4.50) Mc = = 0.1152P Ix 166 sb sa + = 1.0 (sa)allow (sb)allow 0.08333P 0.1152P + = 1 19.00 20 P = 98.6 kip Ans. 1138 y x 8 in. 0.5 in. 0.5 in. 13 Solutions 46060 6/11/10 11:56 AM Page 1139 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–119. The 2014-T6 hollow column is fixed at its base and free at its top. Determine the maximum eccentric force P that can be safely supported by the column. Use the allowable stress method. The thickness of the wall for the section is t = 0.5 in. 6 in. P 3 in. 6 in. 8 ft Section Properties. A = 6(3) - 5(2) = 8 in2 Ix = 1 1 (3) A 63 B (2) A 53 B = 33.1667 in4 12 12 rx = 33.1667 Ix = = 2.036 in. AA A 8 Iy = 1 1 (6) A 33 B (5) A 23 B = 10.1667 in4 12 12 ry = Iy 10.1667 = = 1.127 in. AA A 8 Slenderness Ratio. For a column fixed at its base and free at its top, K = 2. Thus, a 2(8)(12) KL b = = 170.32 r y 1.127 Allowable Stress. Since a KL b 7 55, the column can be classified as a long r y column. sallow = 54 000 ksi 54 000 ksi = = 1.862 ksi (KL>r)2 170.312 Maximum Stress. Bending occurs about the strong axis so that M = P(6) and 6 c = = 3 in. 2 sallow = 1.862 = Mc P + A I C P(6) D (3) P + 8 33.1667 P = 2.788 kip = 2.79 kip Ans. 1139 13 Solutions 46060 6/11/10 11:56 AM Page 1140 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *13–120. The 2014-T6 hollow column is fixed at its base and free at its top. Determine the maximum eccentric force P that can be safely supported by the column. Use the interaction formula. The allowable bending stress is (sb)allow = 30 ksi. The thickness of the wall for the section is t = 0.5 in. 6 in. P 3 in. 6 in. 8 ft Section Properties. A = 6(3) - 5(2) = 8 in2 Ix = 1 1 (3) A 63 B (2) A 53 B = 33.1667 in4 12 12 rx = 33.1667 Ix = = 2.036 in. AA A 8 Iy = 1 1 (6) A 33 B (5) A 23 B = 10.1667 in4 12 12 ry = Iy 10.1667 = = 1.127 in. AA A 8 Slenderness Ratio. For a column fixed at its base and pinned at its top, K = 2. Thus, a 2(8)(12) KL b = = 170.32 r y 1.127 KL b 7 55, the column can be classified as the column is r y classified as a long column. Allowable Stress. Since a sallow = 54000 ksi 54000 ksi = = 1.862 ksi (KL>r)2 170.312 Interaction Formula. Bending is about the strong axis. Since M = P(6) and 6 = 3 in, c = 2 P>A Mc>Ar2 + = 1 (sa)allow (sb)allow P>8 + 1.862 [P(6)](3) n C 8 A 2.0362 B D 30 = 1 P = 11.73 kip = 11.7 kip Ans. 1140 13 Solutions 46060 6/11/10 11:56 AM Page 1141 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •13–121. The 10-ft-long bar is made of aluminum alloy 2014-T6. If it is fixed at its bottom and pinned at the top, determine the maximum allowable eccentric load P that can be applied using the formulas in Sec. 13.6 and Eq. 13–30. P x 1.5 in. 1.5 in. x Section Properties: A = 6(4) = 24.0 in2 Ix = 1 (4) A 63 B = 72.0 in4 12 Iy = 1 (6) A 43 B = 32.0 in4 12 ry = Iy 32.0 = = 1.155 in. AA A 24 Slenderness Ratio: The largest slenderness ratio is about y- y axis. For a column pinned at one end fixed at the other end, K = 0.7. Thus, a 0.7(10)(12) KL b = = 72.75 r y 1.155 Allowable Stress: The allowable stress can be determined using aluminum KL (2014 –T6 alloy) column formulas. Since 7 55, the column is classified as a long r column. Applying Eq. 13–26, sallow = c = 54 000 d ksi (KL>r)2 54 000 72.752 = 10.204 ksi Maximum Stress: Bending is about x - x axis. Applying Eq. 13–30, we have smax = sallow = 10.204 = P Mc + A I P(1.5)(3) P + 24.0 72.0 P = 98.0 kip Ans. 1141 y 3 in. 2 in. y 2 in. 13 Solutions 46060 6/11/10 11:56 AM Page 1142 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–122. The 10-ft-long bar is made of aluminum alloy 2014-T6. If it is fixed at its bottom and pinned at the top, determine the maximum allowable eccentric load P that can be applied using the equations of Sec. 13.6 and the interaction formula with 1sb2allow = 18 ksi. P x 1.5 in. 1.5 in. x Section Properties: A = 6(4) = 24.0 in2 Ix = 1 (4) A 63 B = 72.0 in4 12 Iy = 1 (6) A 43 B = 32.0 in4 12 rx = Ix 72.0 = = 1.732 in. AA A 24.0 ry = Iy 32.0 = = 1.155 in. AA A 24.0 Slenderness Ratio: The largest slenderness radio is about y -y axis. For a column pinned at one end and fixed at the other end, K = 0.7. Thus a 0.7(10)(12) KL b = = 72.75 r y 1.155 Allowable Stress: The allowable stress can be determined using aluminum KL 7 55, the column is classified as a long (2014 –T6 alloy) column formulas. Since r column. Applying Eq. 13–26, (sa)allow = c = 54 000 d ksi (KL>r)2 54 000 72.752 = 10.204 ksi Interaction Formula: Bending is about x - x axis. Applying Eq. 13–31, we have Mc>Ar2 P>A + = 1 (sa)allow (sb)allow P(1.5)(3)>24.0(1.7322) P>24.0 + = 1 10.204 18 P = 132 kip Ans. 1142 y 3 in. 2 in. y 2 in. 13 Solutions 46060 6/11/10 11:56 AM Page 1143 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–123. The rectangular wooden column can be considered fixed at its base and pinned at its top. Also, the column is braced at its mid-height against the weak axis. Determine the maximum eccentric force P that can be safely supported by the column using the allowable stress method. 6 in. 6 in. Section Properties. Ix = dx = 6 in. 5 ft 1 (3) A 63 B = 54 in4 12 dy = 3 in. Slenderness Ratio. Here, Lx = 10(12) = 120 in. and for a column fixed at its base and pinned at its top, K = 0.7. Thus, a 0.7(120) KL b = = 14 d x 6 Since the bracing provides support equivalent to a pin, Ky = 1 and Ly = 5(12) = 60 in. Then a 1(60) KL b = = 20 (controls) d y 3 KL 6 26, the column can be classified as the column d is classified as an intermediate column. Allowable Stress. Since 11 6 sallow = 1.20c 1 = 1.20 c 1 - 1 KL>d 2 a b d ksi 3 26.0 1 20 2 a b d ksi = 0.9633 ksi 3 26.0 Maximum Stress. Bending occurs about the strong axis. Here, M = P(6) and 6 c = = 3 in. 2 sallow = 0.9633 = 6 in. 3 in. 5 ft A = 6(3) = 18 in2 P P Mc + A I [P(6)](3) P + 18 54 P = 2.477 kip = 2.48 kip Ans. 1143 13 Solutions 46060 6/11/10 11:56 AM Page 1144 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *13–124. The rectangular wooden column can be considered fixed at its base and pinned at its top. Also, the column is braced at its mid-height against the weak axis. Determine the maximum eccentric force P that can be safely supported by the column using the interaction formula. The allowable bending stress is (sb)allow = 1.5 ksi. 6 in. 6 in. 5 ft Section Properties. rx = Ix = 1 (3) A 63 B = 54 in4 12 Ix 54 = = 1.732 in. AA A 18 dx = 6 in. dy = 3 in. Slenderness Ratio. Here, Lx = 10(12) = 120 in. and for a column fixed at its base pinned at its top, K = 0.7. Thus, a 0.7(120) KL b = = 14 d x 6 Since the bracing provides support equivalent to a pin, Ky = 1 and Ly = 5(12) = 60 in. Then a 1(60) KL b = = 20 (controls) d y 3 KL 6 26, the column can be classified as the d column is classified as an intermediate column. Allowable Axial Stress. Since 11 6 sallow = 1.20c 1 = 1.20 c 1 - 1 KL>d 2 a b d ksi 3 26.0 1 20 2 a b d ksi = 0.9633 ksi 3 26.0 Interaction Formula. Bending occurs about the strong axis. Since M = P(6) and 6 c = = 3 in. 2 P>A Mc>Ar2 + = 1 (sa)allow (sb)allow P>18 + 0.9633 [P(6)](3) n C 18 A 1.7322 B D 1.5 6 in. 3 in. 5 ft A = 6(3) = 18 in2 P = 1 P = 3.573 kip = 3.57 kip Ans. 1144 13 Solutions 46060 6/11/10 11:56 AM Page 1145 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •13–125. The 10-in.-diameter utility pole supports the transformer that has a weight of 600 lb and center of gravity at G. If the pole is fixed to the ground and free at its top, determine if it is adequate according to the NFPA equations of Sec. 13.6 and Eq. 13–30. G 15 in. 18 ft 2(18)(12) KL = = 43.2 in. d 10 26 6 43.2 … 50 Use Eq. 13–29, sallow = 540 540 = = 0.2894 ksi (KL>d) (43.2)2 smax = Mc P + A I smax = (600)(15)(5) 600 + 2 p (5) A p4 B (5)4 smax = 99.31 psi 6 0.289 ksi O.K. Yes. Ans. 1145 13 Solutions 46060 6/11/10 11:56 AM Page 1146 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–126. Using the NFPA equations of Sec. 13.6 and Eq. 13–30, determine the maximum allowable eccentric load P that can be applied to the wood column. Assume that the column is pinned at both its top and bottom. P 0.75 in. 6 in. 3 in. 12 ft Section Properties: A = 6(3) = 18.0 in2 Iy = 1 (6) A 33 B = 13.5 in4 12 Slenderness Ratio: For a column pinned at both ends, K = 1.0. Thus, a 1.0(12)(12) KL b = = 48.0 d y 3 Allowable Stress: The allowable stress can be determined using NFPA timber KL column formulas. Since 26 6 6 50, it is a long column. Applying Eq. 13–29, d sallow = = 540 ksi (KL>d)2 540 = 0.234375 ksi 48.02 Maximum Stress: Bending is about y -y axis. Applying Eq. 13–30, we have smax = sallow = 0.234375 = P Mc + A I P(0.75)(1.5) P + 18.0 13.5 P = 1.69 kip Ans. 1146 13 Solutions 46060 6/11/10 11:56 AM Page 1147 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–127. Using the NFPA equations of Sec. 13.6 and Eq. 13–30, determine the maximum allowable eccentric load P that can be applied to the wood column. Assume that the column is pinned at the top and fixed at the bottom. P 0.75 in. 6 in. 3 in. 12 ft Section Properties: A = 6(3) = 18.0 in2 Iy = 1 (6) A 33 B = 13.5 in4 12 Slenderness Ratio: For a column pinned at one end and fixed at the other end, K = 0.7. Thus, a 0.7(12)(12) KL b = = 33.6 d y 3 Allowable Stress: The allowable stress can be determined using NFPA timber KL column formulas. Since 26 6 6 50, it is a long column. Applying Eq. 13–29, d sallow = = 540 ksi (KL>d)2 540 = 0.4783 ksi 33.62 Maximum Stress: Bending is about y -y axis. Applying Eq. 13–30, we have smax = sallow = 0.4783 = Mc P + A I P(0.75)(1.5) P + 18.0 13.5 P = 3.44 kip Ans. 1147 13 Solutions 46060 6/11/10 11:56 AM Page 1148 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *13–128. The wood column is 4 m long and is required to support the axial load of 25 kN. If the cross section is square, determine the dimension a of each of its sides using a factor of safety against buckling of F.S. = 2.5. The column is assumed to be pinned at its top and bottom. Use the Euler equation. Ew = 11 GPa, and sY = 10 MPa. 25 kN a 4m a 1 a4 (a) A a3 B = , P = (2.5)25 = 62.5 kN and K = 1 12 12 cr for pin supported ends column. Applying Euler’s formula, Critical Buckling Load: I = Pcr = 62.5 A 10 3 B = p2EI (KL)2 a p2(11)(109) A 12 B 4 [1(4)]2 a = 0.1025 m = 103 mm Ans. Critical Stress: Euler’s formula is only valid if scr 6 sY. scr = 62.5(103) Pcr = = 5.94 MPa 6 s Y = 10 MPa A 0.1025(0.1025) 1148 O.K. 13 Solutions 46060 6/11/10 11:56 AM Page 1149 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •13–129. If the torsional springs attached to ends A and C of the rigid members AB and BC have a stiffness k, determine the critical load Pcr. P k A L 2 B Equilibrium. When the system is given a slight lateral disturbance, the configuration shown in Fig. a is formed. The couple moment M can be related to P by considering the equilibrium of members AB and BC. Member AB + c ©Fy = 0; a + ©MA = 0; By - P = 0 (1) By a (2) L L sin u b + Bx a cos u b - M = 0 2 2 Member BC a + ©MC = 0; - By a L L sin u b + Bx a cos u b + M = 0 2 2 (3) Solving Eqs. (1), (2), and (3), we obtain Bx = 0 By = 2M L sin u M = PL sin u 2 Since u is very small, the small angle analysis gives sin u ⬵ u. Thus, M = PL u 2 (4) Torslonal Spring Moment. The restoring couple moment Msp can be determined using the torsional spring formula, M = ku. Thus, Msp = ku Critical Buckling Load. When the mechanism is on the verge of bucklling M must equal Msp. M = Msp Pcr L u = ku 2 Pcr = 2k L Ans. 1149 L 2 k C 13 Solutions 46060 6/11/10 11:56 AM Page 1150 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–129. Continued 1150 13 Solutions 46060 6/11/10 11:56 AM Page 1151 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–130. Determine the maximum intensity w of the uniform distributed load that can be applied on the beam without causing the compressive members of the supporting truss to buckle. The members of the truss are made from A-36-steel rods having a 60-mm diameter. Use F.S. = 2 against buckling. w B A C 2m Equilibrium. The force developed in member BC can be determined by considering the equilibrium of the free-body diagram of the beam AB, Fig. a. 3 w(5.6)(2.8) - FBC a b(5.6) = 0 FBC = 4.6667w 5 a + ©MA = 0; The Force developed in member CD can be obtained by analyzing the equilibrium of joint C, Fig. b, + c ©Fy = 0; FAC a + : ©Fx = 0; 4 12 4.6667w a b + 7.28a b w -FCD = 0 5 13 5 3 b - 4.6667w a b = 0 13 5 FAC = 7.28w (T) FCD = 10.4533w (C) Section Properties. The cross-sectional area and moment of inertia of the solid circular rod CD are A = p A 0.032 B = 0.9 A 10 - 3 B p m2 I = p A 0.034 B = 0.2025 A 10 - 6 B p m4 4 Critical Buckling Load. Since both ends of member CD are pinned, K = 1. The critical buckling load is Pcr = FCD (F.S.) = 10.4533w(2) = 20.9067w Applying Euler’s formula, Pcr = p2EI (KL)2 20.9067w = p2 C 200 A 109 B D C 0.2025 A 10 - 6 B p D [1(3.6)]2 w = 4634.63 N>m = 4.63 kN>m Ans. Critical Stress: Euler’s formula is valid only if scr 6 sY. scr = 20.907(4634.63) Pcr = = 34.27 MPa 6 sY = 250 MPa A 0.9 A 10 - 3 B p 1151 O.K. 3.6 m D 1.5 m 13 Solutions 46060 6/11/10 11:56 AM Page 1152 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–131. The W10 * 45 steel column supports an axial load of 60 kip in addition to an eccentric load P. Determine the maximum allowable value of P based on the AISC equations of Sec. 13.6 and Eq. 13–30. Assume that in the x–z plane Kx = 1.0 and in the y–z plane Ky = 2.0. Est = 2911032 ksi, sY = 50 ksi. z P 60 kip x y 10 ft Section properties for W 10 * 45: A = 13.3 in2 d = 10.10 in. rx = 4.32 in. ry = 2.01 in. Ix = 248 in4 Allowable stress method: a 1.0(10)(12) KL b = = 27.8 r x 4.32 a 2.0(10)(12) KL b = = 119.4 r y 2.01 a 2p2(29)(103) KL 2p2E b = = = 107 r c B sg B 50 (controls) KL KL 7 a b r r c (sa)allow = 12p2(29)(103) 12p2E = = 10.47 ksi 2 23(KL>r) 23(119.4)4 smax = (sa)allow = Mc P + A I P(8) A 10.10 P + 60 2 B + 10.47 = 13.3 248 P = 25.0 kip Ans. 1152 y x 8 in. 13 Solutions 46060 6/11/10 11:56 AM Page 1153 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *13–132. The A-36-steel column can be considered pinned at its top and fixed at its base. Also, it is braced at its mid-height along the weak axis. Investigate whether a W250 * 45 section can safely support the loading shown. Use the allowable stress method. 600 mm 10 kN 4.5 m Section Properties. From the table listed in the appendix, the necessary section properties for a W250 * 45 are A = 5700 mm2 = 5.70 A 10 - 3 B m2 d = 266 mm = 0.266 m Ix = 71.1 A 106 B mm4 = 71.1 A 10 - 6 B m4 rx = 112 mm = 0.112 m ry = 35.1 mm = 0.0351 mm Slenderness Ratio. Here, Lx = 9 m and for a column fixed at its base and pinned at its top, Kx = 0.7. Thus, ¢ 0.7(9) KL = 56.25 ≤ = r x 0.112 Since the bracing provides support equivalent to a pin, Ky = 1 and Ly = 4.5 m. Then, ¢ 1(4.5) KL = 128.21 (controls) ≤ = r y 0.0351 Allowable Stress. For A-36 steel, a Since a 2p2 C 200 A 109 B D KL 2p2E b = = = 125.66. r c C 250 A 106 B B sY KL KL b 6 a b 6 200, the column can be classified as a long column. r c r y sallow = 12p2 C 200 A 109 B D 12p2E = = 62.657 MPa 2 23(KL>r) 23(128.21)2 Maximum Stress. Bending occurs about the strong axis. Here, P = 10 + 40 0.266 d = 50 kN, M = 40(0.6) = 24 kN # m and c = = = 0.133 m, 2 2 smax = 50 A 103 B 24 A 103 B (0.133) P Mc + = + = 53.67 MPa A I 5.70 A 10 - 3 B 71.1 A 10 - 6 B Since smax 6 sallow, the column is adequate according to the allowable stress method. 1153 4.5 m 40 kN 13 Solutions 46060 6/11/10 11:56 AM Page 1154 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •13–133. The A-36-steel column can be considered pinned at its top and fixed at its base. Also, it is braced at its mid-height along the weak axis. Investigate whether a W250 * 45 section can safely support the loading shown. Use the interaction formula. The allowable bending stress is (sb)allow = 100 MPa. 600 mm 10 kN 4.5 m Section Properties. From the table listed in the appendix, the necessary section properties for a W250 * 45 are A = 5700 mm2 = 5.70 A 10 - 3 B m2 Ix = 71.1 A 10 B mm = 71.1 A 10 6 4 -6 d = 266 mm = 0.266 m 4.5 m Bm 4 rx = 112 mm = 0.112 m ry = 35.1 mm = 0.0351 mm Slenderness Ratio. Here, Lx = 9 m and for a column fixed at its base and pinned at its top, Kx = 0.7. Thus, a 0.7(9) KL b = = 56.25 r x 0.112 Since the bracing provides support equivalent to a pin, Ky = 1 and Ly = 4..5 m. Then, a Allowable = C 1(4.5) KL b = = 128.21 (controls) r y 0.0351 Axial 2p2 C 200 A 109 B D 250 A 10 6 B Stress. For = 125.66. Since a A–36 steel, a KL 2p2E b = r c B sY KL KL b 6 a b 6 200, the column can be r c r y classified as a long column. sallow = 12p2 C 200 A 109 B D 12p2E = = 62.657 MPa 23(KL>r)2 23(128.21)2 Interaction Formula. Bending is about the strong axis. Here, P = 10 + 40 = 50 kN, 0.266 d M = 40(0.6) = 24 kN # m and c = = = 0.133 m, 2 2 Mc>Ar2 P>A + = (sa)allow (sb)allow 50 A 103 B n 5.70 A 10 - 3 B 62.657 A 106 B + 24 A 103 B (0.133) n C 5.70 A 10 - 3 B A 0.1122 B D 100 A 106 B = 0.5864 6 1 sa = (sa)allow 50 A 103 B n 5.7 A 10 - 3 B 62.657 A 106 B O.K. = 0.140 6 0.15 O.K. Thus, a W250 * 45 column is adequate according to the interaction formula. 1154 40 kN 13 Solutions 46060 6/11/10 11:56 AM Page 1155 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–134. The member has a symmetric cross section. If it is pin connected at its ends, determine the largest force it can support. It is made of 2014-T6 aluminum alloy. 0.5 in. 2 in. P 5 ft Section properties: A = 4.5(0.5) + 4(0.5) = 4.25 in2 P 1 1 (0.5)(4.53) + (4)(0.5)3 = 3.839 in4 I = 12 12 r = I 3.839 = = 0.9504 in. AA A 4.25 Allowable stress: 1.0(5)(12) KL = = 63.13 r 0.9504 KL 7 55 r Long column sallow = 54000 54000 = = 13.55 ksi (KL>r)2 63.132 Pallow = sallowA = 13.55(4.25) = 57.6 kip Ans. 1155 13 Solutions 46060 6/11/10 11:56 AM Page 1156 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13–135. The W200 * 46 A-36-steel column can be considered pinned at its top and fixed at its base. Also, the column is braced at its mid-height against the weak axis. Determine the maximum axial load the column can support without causing it to buckle. 6m Section Properties. From the table listed in the appendix, the section properties for a W200 * 46 are A = 5890 mm2 = 5.89 A 10 - 3 B m2 Iy = 15.3 A 106 B mm4 = 15.3 A 10 - 6 B m4 Ix = 45.5 A 106 B mm4 = 45.5 A 10 - 6 B m4 Critical Buckling Load. For buckling about the strong axis, Kx = 0.7 and Lx = 12 m. Since the column is fixed at its base and pinned at its top, Pcr = p2EIx (KL)x 2 = p2 c 200 A 109 B d c 45.5 A 10 - 6 B d [0.7(12)]2 = 1.273 A 106 B N = 1.27 MN For buckling about the weak axis, Ky = 1 and Ly = 6 m since the bracing provides a support equivalent to a pin. Applying Euler’s formula, Pcr = p2EIy (KL)y 2 = p2 c 200 A 109 B d c 15.3 A 10 - 6 B d [1(6)]2 = 838.92 kN = 839 kN (controls)Ans. Critical Stress. Euler’s formula is valid only if scr 6 sY. scr = 838.92 A 103 B Pcr = = 142.43 MPa 6 sY = 250 MPa A 5.89 A 10 - 3 B 1156 O.K. 6m 13 Solutions 46060 6/11/10 11:56 AM Page 1157 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *13–136. The structural A-36 steel column has the cross section shown. If it is fixed at the bottom and free at the top, determine the maximum force P that can be applied at A without causing it to buckle or yield. Use a factor of safety of 3 with respect to buckling and yielding. P 20 mm A 10 mm Section properties: 100 mm -3 2 4m ©A = 0.2(0.01) + 0.15 (0.01) + 0.1(0.01) = 4.5(10 ) m 150 mm A 100 mm 10 mm 0.005 (0.2)(0.01) + 0.085 (0.15)(0.01) + 0.165 (0.1)(0.01) ©xA x = = ©A 4.5(10 - 3) = 0.06722 m 1 (0.2)(0.013) + 0.2 (0.01)(0.06722 - 0.005)2 12 Iy = + 1 (0.01)(0.153) + 0.01 (0.15)(0.085 - 0.06722)2 12 + 1 (0.1)(0.013) + 0.1 (0.01)(0.165 - 0.06722)2 12 = 20.615278 (10 - 6) m4 1 1 1 (0.01)(0.23) + (0.15)(0.013) + (0.01)(0.13) 12 12 12 Ix = = 7.5125 (10 - 6) m4 ry = Iy 20.615278(10 - 6) = = 0.0676844 AA A 4.5 (10 - 3) Buckling about x - x axis: Pcr = p2(200)(109)(7.5125)(10 - 6) p2 EI = 2 (KL) [2.0(4)]2 = 231.70 kN scr = (controls) 231.7 (103) Pcr = 51.5 MPa 6 sg = 250 MPa = A 4.5 (10 - 3) Yielding about y- y axis: smax = P ec KL P c 1 + 2 sec a b d; A 2r A EA r e = 0.06722 - 0.02 = 0.04722 m 0.04722 (0.06722) ec = = 0.692919 0.0676844 r2 2.0 (4) P P KL = = 1.96992 P (10 - 3) 2P 2r A EA 2(0.0676844) A 200 (109)(4.5)(10 - 3) 250(106)(4.5)(10 - 3) = P[1 + 0.692919 sec (1.96992P (10 - 3) 2P)] By trial and error: P = 378.45 kN Hence, Pallow = 231.70 = 77.2 kN 3 Ans. 1157 10 mm 100 mm 13 Solutions 46060 6/11/10 11:56 AM Page 1158 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •13–137. The structural A-36 steel column has the cross section shown. If it is fixed at the bottom and free at the top, determine if the column will buckle or yield when the load P = 10 kN. Use a factor of safety of 3 with respect to buckling and yielding. P 20 mm A 10 mm 100 mm Section properties: 4m ©A = 0.2 (0.01) + 0.15 (0.01) + 0.1 (0.01) = 4.5 (10 - 3) m2 0.005 (0.2)(0.01) + 0.085 (0.15)(0.01) + 0.165 (0.1)(0.01) ©xA = 0.06722 m = ©A 4.5 (10 - 3) Iy = 1 (0.2)(0.013) + 0.2 (0.01)(0.06722 - 0.005)2 12 ry = + 1 (0.01)(0.153) + 0.01 (0.15)(0.085 - 0.06722)2 12 + 1 (0.1)(0.013) + 0.1 (0.01)(0.165 - 0.06722)2 = 20.615278 (10 - 6) m4 12 1 1 1 (0.01)(0.23) + (0.15)(0.013) + (0.01)(0.13) = 7.5125 (10 - 6) m4 12 12 12 Iy BA 20.615278 (10 - 6) = B 4.5 (10 - 3) = 0.067843648 m Buckling about x -x axis: Pcr = p2(200)(109)(7.5125)(10 - 6) p2 EI = = 231.70 kN 2 (KL) [2.0(4)]2 scr = 231.7 (103) Pcr = 51.5 MPa 6 sg = 250 MPa = A 4.5 (10 - 3) Pallow = O.K. Pcr 231.7 = = 77.2 kN 7 P = 10 kN FS 3 Hence the column does not buckle. Yielding about y- y axis: smax = P = P KL ec P bd c 1 + 2 sec a A 2r A EA r A 100 mm 10 mm x = Ix = 150 mm e = 0.06722 - 0.02 = 0.04722 m 10 = 3.333 kN 3 3.333 (103) P = 0.7407 MPa = A 4.5 (10 - 3) 0.04722 (0.06722) ec = 0.689815 = (0.067844) r2 2.0 (4) P KL 3.333 (103) = = 0.1134788 2 r AE A 2(0.06783648) A 200 (109)(4.5)(10 - 3) smax = 0.7407 [1 + 0.692919 sec (0.1134788)] = 1.25 MPa 6 sg = 250 MPa Hence the column does not yield! No. Ans. 1158 10 mm 100 mm