23_ch 03 Mechanical Design budynas_SM_ch03

# 23_ch 03 Mechanical Design budynas_SM_ch03 -...

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36 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 3-23 (a) A a = A b = 0 . 25(1 . 5) = 0 . 375 in 2 A = 3(0 . 375) = 1 . 125 in 2 ¯ y = 2(0 . 375)(0 . 75) + 0 . 375(0 . 5) 1 . 125 = 0 . 667 in I a = 0 . 25(1 . 5) 3 12 = 0 . 0703 in 4 I b = 1 . 5(0 . 25) 3 12 = 0 . 001 95 in 4 I 1 = 2[0 . 0703 + 0 . 375(0 . 083) 2 ] + [0 . 001 95 + 0 . 375(0 . 167) 2 ] = 0 . 158 in 4 Ans. σ A = 10 000(0 . 667) 0 . 158 = 42(10) 3 psi Ans. σ B = 10 000(0 . 667 0 . 375) 0 . 158 = 18 . 5(10) 3 psi Ans. σ C = 10 000(0 . 167 0 . 125) 0 . 158 = 2 . 7(10) 3 psi Ans. σ D =− 10 000(0 . 833) 0 . 158 52 . 7(10) 3 psi Ans. (b) Here we treat the hole as a negative area.
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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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