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35_ch 04 Mechanical Design budynas_SM_ch04

# 35_ch 04 Mechanical Design budynas_SM_ch04 - = F R 2 π...

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104 Solutions Manual Instructor’s Solution Manual to Accompany Mechanical Engineering Design ( δ A ) y = − U P = − 1 E I π/ 2 0 P ( R sin θ ) 2 R d θ + 1 G J π/ 2 0 P [ R (1 cos θ )] 2 R d θ Integrating and substituting J = 2 I and G = E / [2(1 + ν )] ( δ A ) y = − P R 3 E I π 4 + (1 + ν ) 3 π 4 2 = − [4 π 8 + (3 π 8) ν ] P R 3 4 E I = − [4 π 8 + (3 π 8)(0 . 29)] (200)(100) 3 4(200)(10 3 )( π/ 64)(5) 4 = − 40 . 6 mm 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 = F R 2 (1 cos θ ) H R sin θ , M H = − R sin θ , 0 < θ < π 2 U H = 1 E I π/ 2 0 F R 2 (1 cos θ ) H R sin θ ( R sin θ ) R d 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 = F R 2 (1 cos θ ) F R
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Unformatted text preview: = F R 2 π [ π (1 − cos θ ) − 2 sin θ ] Ans. Due to symmetry, the solution for the left side is identical. (c) ∂ M ∂ F = R 2 π [ π (1 − cos θ ) − 2 sin θ ] δ F = ∂ U ∂ F = 2 E I ² π/ 2 F R 2 4 π 2 [ π (1 − cos θ ) − 2 sin θ ] 2 R d θ = F R 3 2 π 2 E I ² π/ 2 ( π 2 + π 2 cos 2 θ + 4 sin 2 θ − 2 π 2 cos θ − 4 π sin θ + 4 π sin θ cos θ ) d θ = F R 3 2 π 2 E I ³ π 2 ´ π 2 µ + π 2 ´ π 4 µ + 4 ´ π 4 µ − 2 π 2 − 4 π + 2 π ¶ = (3 π 2 − 8 π − 4) 8 π F R 3 E I Ans. F A B H R F 2 ±...
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