35_ch 04 Mechanical Design budynas_SM_ch04

# 35_ch 04 Mechanical Design budynas_SM_ch04 - = F R 2 [ (1...

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104 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design ( δ A ) y =− U P =− ± 1 EI ² π/ 2 0 P ( R sin θ ) 2 Rd θ + 1 GJ ² π/ 2 0 P [ R (1 cos θ )] 2 Rd θ ¾ Integrating and substituting J = 2 I and G = E / [2(1 + ν )] ( δ A ) y =− PR 3 EI ³ π 4 + (1 + ν ) ´ 3 π 4 2 µ¶ =− [4 π 8 + (3 π 8) ν ] PR 3 4 EI =− [4 π 8 + (3 π 8)(0 . 29)] (200)(100) 3 4(200)(10 3 )( π/ 64)(5) 4 =− 40 . 6mm 4-68 Consider the horizontal reaction, to be applied at B, subject to the constraint ( δ B ) H = 0 . (a) ( δ B ) H = U H = 0 Due to symmetry, consider half of the structure. F does not deﬂect horizontally. M = FR 2 (1 cos θ ) HR sin θ , M H =− R sin θ , 0 <θ< π 2 U H = 1 EI ² π/ 2 0 ³ FR 2 (1 cos θ ) HR sin θ ( R sin θ ) Rd θ = 0 F 2 + F 4 + H π 4 = 0 H = F π Ans. Reaction at A is the same where H goes to the left (b) For 0 <θ< π 2 , M = FR 2 (1 cos θ ) FR π sin θ M
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Unformatted text preview: = F R 2 [ (1 cos ) 2 sin ] Ans. Due to symmetry, the solution for the left side is identical. (c) M F = R 2 [ (1 cos ) 2 sin ] F = U F = 2 E I / 2 F R 2 4 2 [ (1 cos ) 2 sin ] 2 R d = F R 3 2 2 E I / 2 ( 2 + 2 cos 2 + 4 sin 2 2 2 cos 4 sin + 4 sin cos ) d = F R 3 2 2 E I 2 2 + 2 4 + 4 4 2 2 4 + 2 = (3 2 8 4) 8 F R 3 E I Ans. F A B H R F 2...
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## This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

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