408_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

408_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: 05Ch05.qxd 9/24/08 402 4:59 AM Page 402 CHAPTER 5 Stresses in Beams (Basic Topics) Solution 5.5-14 Circular core From Appendix D, Cases 9 and 15: 4 Iy b radians a 4 Iy pd 64 4 d 2 r aa 2 pr 4 r 4 a dp a 32 2 ab pd 4 64 d4 p a 32 2 b d4 p a 32 2 d4 (4 b 128 b+ 1 sin 4 b b 4 sin4 b ) MAXIMUM BENDING STRESS r cos b b sin b cos b + 2 sin b cos3 b b b dp a 32 2 r4 pd 4 64 b 4 pd 64 2ab + r2 r sin b radians a 4 p 2 3 b smax 2 (sin b cos b )(1 b 2 cos b ) b smax 1 a sin 2 b b ( 2 cos 2 b ) b For b smax Mc Iy c 64M sin b 3 d (4 b ; sin 4 b ) (8p 13 + 9)d 60° p/3 rad: 576M 10.96 3 P Problem 5.5-15 A simple beam AB of span length L 24 ft is subjected to two wheel loads acting at distance d = 5 ft apart (see figure). Each wheel transmits a load P = 3.0 k, and the carriage may occupy any position on the beam. Determine the maximum bending stress smax due to the wheel loads if the beam is an I-beam having section modulus S 16.2 in.3 d sin b 2 r sin b d M ; d3 P A B C L Solution 5.5-15 Wheel loads on a beam Substitute x into the equation for M: Mmax L 24 ft d 5 ft P 3k S 288 in. 60 in. P L L x+ M RA x P (2L x L dM dx P (2L L d P (L L x dx 4x) d2 b 2 MAXIMUM BENDING STRESS smax 3 16.2 in. Mmax S P aL 2LS d2 b 2 ; Substitute numerical values: MAXIMUM BENDING MOMENT RA P aL 2L d) P (2L L 2x) smax 3k 2(288 in.)(16.2 in.3) 21.4 ksi 2x2) 0x d L 2 d 4 ; (288 in. 30 in.)2 ...
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This note was uploaded on 12/22/2011 for the course MEEG 310 taught by Professor Staff during the Fall '11 term at University of Delaware.

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