SolutionsToCh5 - (c) In (a), the moment at the support is...

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Unformatted text preview: (c) In (a), the moment at the support is changed from 5 kN(0.5 m) = 2.5 kN.m to 5(O.5 m) — 10(0.3) = -0.5 kN.m. Hence, stresses are reduced by 80% and reversed in direction. This is true regardless of values of and In (b), the effect of Q would be to reverse the stress and give a magnitude exceeding 20% of the Q = 0 stress. The greater the product E1, the greater the stress. I SOLUTION (5.24) Known: The dimensions of a solid round steel with 8y 2 350 MPa are known. Find: With a safety factor of 4, what axial compressive load can be applied if: (a) Both ends are hinged? (b) Both ends are built in? Schematic and Given Data: Assumptions: 1. The rod does not fail under compression. 2. The rod is straight. 3. The minimum AISC recommended value for Le should be used in the analysis. Analysis: (a) From Appendix B-l, radius of gyration of circle is —Q=E= —4 4 17.5m Le/p = 1000/175 = 57.14 5-35 From Eq. (15.13), Euler and Johnson are tangent at: r.) l»— = 108.05 Le/p : [WEE = [216(207 x 109) 5y 350 x 106 where E = 207 GPa (Appendix C—l). Hence, the Johnson Eq. (5.12) applies: 82 2 scr=sy y — 41t2E P (350 x 106)2 Scr = 350 X 106 - m 411:2(207 x 109) (57.14)2 = 301 X 106 Pa Per = ScrA 2 . PCr = Scr - E3713 = 301 MPa (“(10) mm) = 1158 ><103 N Dividing by a safety factor of 4: P = 290 kN I (b) From Fig. 5.25c, assume Le = .65L = .65(1000) = 650 mm = £— = 37.14. Johnson parabola applies. £2 p 17.5 (350 x 106)2 SCr = 350 X 106 - ——-————————- 41t2(207 x 109) (37.14)2 = 329 X 106 Pa 11:(.70)2 4 PC,=329( )=1266><103 N Dividing by a safety factor of 4: P = 317 kN I 5-36 SOLUTION (5.25) Known: A bar with a known dimension is made of steel and has Sy = 25 ksi. Find: With a safety factor of 4, determine the axial compressive load that can be applied if: (a) Both ends are hinged. (b) One end is built in and the other is unsupported. Schematic and Given Data: Assumptions: 1. The bar does not fail under compression. 2. The bar is straight. 3. The minimum AISC recommended value for Le should be used in the analysis. Analysis: (a) From Appendix B-l, the least radius of gyration p = 0.289 h = 0.289 (1) = 0.289 in. E=__20 = 92 p 0.289 6 ‘ From Eq. (5.13), Euler and Johnson are tangent at: 2 L % = (27st E)2 where E = 30 X 106 psi (Appendix C-l) 5-37 27:2(30 x 106) % Le/p = —____ = 154 25 x 103 Johnson's Eq. (5.12) applies: s = - Syz (1:2)2 “ y 47:25 9 2 scr = 25,000 — 25’000 (692)2 = 22,470 psi 47r2(30 x 106) Pa = ScrA = 22,470 (1)(2) = 44,490 lb. With a safety factor of 4, P = 11,235 lb I (b) From Fig. 5.25(e), assume Le = 2.1 L = 42 in. L 42 J = —— = 145 33 p 0.289 Eq. (5.12) still applies: 2 scr = 25,000 - 25’000 —————— (145.33)2 = 13,854 psi 4n2(30 x 106) Per 2 13,854(1)(2) = 27,708 lb With a safety factor of 4, P = 6927 lb I SOLUTION (5.26) Known: A steel angle iron, loaded in compression, is added to a structure in order to increase its rigidity. The radius of gyration about the centroidal axis parallel to either side is 8 mm, but the minimum radius of gyration is only 5 mm. Find: Determine the compressive load that can be carried with a safety factor of 3. Schematic and Given Data: 5-38 Assumptions: 1. The steel angle iron does not fail under compression. 2. The ends are pinned. 3. The steel angle is straight. Analysis: 1. Using the minimum radius of gyration, E = 1200 = 240 p 5 Euler/Johnson tangent point [Eq. (5.13)] is at 1 E _ 2qu %_an2(207)]5= p '( Sy ) ‘i 0.350 10805 Hence, the Euler Eq. (5.11) applies: Per _ 752E CI=—__ A (Le/P)2 where E = 207 x 109 Pa (Appendix C-l) __ 1:2(207 x 103 MP3) = 35 .47 MPa (240)2 (31' Since the cross-sectional area is not given, the load capacity with SF = 3 can only be given as: P = 35'5 A ; P = 11.8 A where"P" is in Newtons and "A" in m2. 3 I 5-39 SOLUTION (5.27) Known: A 3—in. I-beam is made of steel having Sy = 42 ksi. Find: Determine the safe axial compressive load based on a safety factor of 3 for pinned ends and unsupported lengths of (a) 10 in., (b) 50 in, (c) 100 in., and (d) 200 in. Schematic and Given Data: A: 1.64 in.2 111:2.5in.4 122 = 0.46 in.4 sy = 42 ksi Assumptions: 1. The I-beam does not fail under compression. 2. The I-beam is straight. Analysis: Since I = Apz, pmin = 1&1“. = (11% = 0.53 in. L Euler—Johnson tangent point [EQ- (5.13)] is at If)": = (ZEZE)2 where E = 30 X 106 psi y 1 . L 21:2(30 x1035]? A d C-1 . —e=[— = ( ppen 1x ) p 42 119 (a) 1739: 42- = 189. Johnson equation [Eq. (5.12)] applies: 0.53 S2 2 422 = - Y 1:; =42———(_)— 189 =415k‘ S“ 33’ 41:2E(P) 4n2(30,000)( 'y ‘ S‘ with SF = 3, P = (41,500 psi)(1.64 in.2)/3 = 22,670 lb I (Note that, in this case, column action is almost negligible, and SCI = Sy) 5-40 (b) 1‘5 = i = 94.34. Johnson equation still applies: 9 .53 (42)2 . s = -— 94,342=28.7k Cr 41:2(30000) ( ) 31 P = (28,700) (1.64)/3 = 15,710 lb I = _____ _ —— = 8.32 k C‘ (Lem)2 (188.6202 S] P = (8320)(1.64)/3 = 4550 lb I g- 200 = - . (d) p — 0.53 377.36. Eq. (5.11) applles. 2 Ser = "—6999? = 2.08 ksi (377.36) P = (2080)(1.64)/3 : 1140 lb I SOLUTION (5.28) Known: A 20 mm diameter rod of Sy = 350 MPa is loaded as a column with pinned ends. If sufficiently short, it can carry a limiting load of SyA = 110 kN. Find: Determine the rod length that will carry the following percentage of a 110 kN load: (a) 90%, (b) 50%, (c) 10%, and (d) 2%? Schematic and Given Data: 5—41 Assumptions: 1. The rod does not fail under compression. 2. The rod is straight. Analysis: From Appendix B-l, p = d/4 = 20/4 = 5 mm (a) Assuming Johnson Eq. (5.12) applies: - Syz 1:2)2 y 4an p where E = 207 GPa (Appendix 01). SCI-:8 (350)2 (Le)? cr= SOMP 0.9 =315MP =350______ 6 (3 ax ) a 4n2(207,000) p 2 2 ———(3—§9)——— . Therefore, £6— = 48.32 41t2(207,000) P P From Eq. (5.13), Euler/Johnson tangent is at 35 MPa 2 g— : [2n2(207,000) 1:_e__ 2an F: _( 350 108 P Sy Hence, Johnson does indeed apply and Le = L = 4832(5) = 241.6 m I (b) Assume Euler Eq. (5.11) applies: cr = an (Le/P)2 2 “or = (350 MPa)(0.5) = 175 MPa = M992 (Le/P)2 Therefore, £53 = 108 Hence, both Johnson and Euler apply. and Le = L = 108(5) = 540 m I (c) The Euler equation will definitely apply. 1:2(207,000) 2) 1:9 = G = 350 0.1 =35MPa= 241 or < )< ) (Le/p). p Le=L=241(5)= 1205 m I (d) The Euler equation will apply. 152007.000) L o = 350 0.02 =7MP =————-———=>—9=540 c. < x > a (Le/ml p Le = L = 540(5) = 2700 m I 5-42 SOLUTION (5.29) Known: A boom and tie—rod arrangement is supporting a load of 6 kN. The tie-rod is made of steel having a tensile yield strength of 400 MPa. Find: (a) Determine the safety factor of the tie—rod with respect to static yielding. F (b) Determine the safety factor of the tie-rod if the vertical rod is rotated 180' so that the 6 kN load acts upward. (c) Draw a conclusion with respect to the relative desirability of designing machines with column members loaded in tension vs. loaded in compression. Schematic and Given Data: Assumption: The tie—rod is straight. Analysis: (a) 5—43 6kN(1m) ZMon: F: ij = 8.57 kN Tensile stress in the tie rod is (5 = —P— : ——8§lO—— : 75.8 MPa A 36 11? mm2 = 400 = 3 S 75.8 5' I (b) o = 75.8 MPa in compression Le _ From Appendix B—l, p = S: = 3 mm, L = Le = 700 m z) —p— — 233.3 Euler-Johnson tangent point [Eq. (5.13)] is at pg : 2K2E1§:[2n2(207 x103)]%=101 p sy 400 Where E = 207 X 109 Pa (Appendix C—l). Hence, Euler Eq. (5.11) applies: 2 207 103 Scr = “2E =————————“ X )= 37.5 MPa (Le/p)2 (233.3)2 Since 75.8 > 37.5, the rod will fail in buckling. I (c) Long, slender rods in compression can carry only a small fraction of the load they can carry in tension. 5—44 ...
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This note was uploaded on 07/25/2008 for the course ME 471 taught by Professor Diaz during the Spring '06 term at Michigan State University.

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SolutionsToCh5 - (c) In (a), the moment at the support is...

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