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11_ch 04 Mechanical Design budynas_SM_ch04

# 11_ch 04 Mechanical Design budynas_SM_ch04 - Ans 4-22(a l =...

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80 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Adding the two deﬂections, y C =− 0 . 057 52 0 . 1923 =− 0 . 2498 in Ans. (b) At O : Due to 450 lbf: dy dx ± ± ± ± x = 0 = Fa 6 EIl ( l 2 3 x 2 ) ± ± ± ± x = 0 = Fal 6 EI θ O =− 720(11)(0 + 11 2 400) 6(30)(10 6 )(0 . 1198)(20) + 450(12)(20) 6(30)(10 6 )(0 . 1198) = 0 . 010 13 rad = 0 . 5805 At B : θ B =− 4 . 793(10 3 ) + 450(12) 6(30)(10 6 )(0 . 1198)(20) [20 2 3(20 2 )] =− 0 . 014 81 rad = 0 . 8485 I = 0 . 1198 ² 0 . 8485 0 . 06 ³ = 1 . 694 in 4 d = ² 64 I π ³ 1 / 4 = ´ 64(1 . 694) π µ 1 / 4 = 2 . 424 in Use d = 2 . 5in Ans. I = π 64 ( 2 . 5 4 ) = 1 . 917 in 4 y C =− 0 . 2498 ² 0 . 1198 1 . 917 ³ =− 0 . 015 61 in
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Unformatted text preview: Ans. 4-22 (a) l = 36(12) = 432 in y max = − 5 w l 4 384 E I = − 5(5000 / 12)(432) 4 384(30)(10 6 )(5450) = − 1 . 16 in The frame is bowed up 1.16 in with respect to the bolsters. It is fabricated upside down and then inverted. Ans. (b) The equation in xy-coordinates is for the center sill neutral surface y = w x 24 E I (2 lx 2 − x 3 − l 3 ) Ans. y x l...
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