439_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

439_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: 05Ch05.qxd 9/24/08 4:59 AM Page 433 433 SECTION 5.7 Nonprismatic Beams Solution 5.7-2 (a) FIND MAX. BENDING STRESS FOR TAPERED CANTILEVER d Aa1 + d(x) P 8m t 2x b L 2ptdA2 smax sB pd(x) t 4 S(x) PL a smax sB 10 mm dA 90 mm dB 4P L b 9 pt dA2 9 8 2.4 kN L 2 ; 270 mm Evaluate using numerical data M(x) S(x) s(x) 4P pt s(x) J 4P xL2 d c2 pt dA (L + 2x)2 s(x) d s(x) dx c dA a 1 + or so xmax smax smax smax ptdA2 (L + 2x)3 d L2 aL + 2 b 2 Px a L+x b L Px a 0 pt 2x 2 c dA a 1 + bd 4 L x) L ptdA2 (L 2x)2 c 0 then solve for xmax 4Px ( 4P ( 4Px L xmax L ptdA2 (L 2x)2 d 0 L ptdA2 (L + 2x)2 L ptdA2 (L L OR simplifying so xmax x) L + x) 2ptdA2 s(L) L L+x b L tension on top, compression on bottom of beam d s(x) dx ¥ s(x) 4Px ( 16Px ( PL 4P L 9 pt dA2 Px xb L2 2 M(x) S(x) d c dx L2 L 2 dA2 a Px s(x) ; 4m Stress at support sB sB 0 L sa b 2 4P ≥ pt (b) REPEAT (A) BUT NOW ADD DISTRIBUTED LOAD s(x) L + 2x L 2 smax 0 xL2 P d pt dA2 (L + 2x)3 c 4PL2 2p (0.010) (0.090)2 37.7 MPa ; M(x) L2 P c4 2 pt dA (L + 2x)2 16 2x bd L K 2 M(x) 0 then solve for xmax xL2 d 4P c c2 dd dx pt dA (L + 2x)2 (2400) (8) smax x L 4 2m + 2x)2 x) c ; L ptdA2 (L 4PL2 2x)3 d 0 L + 4x ptdA2 (L + 2x)3 d 0 ...
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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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