25_ch 03 Mechanical Design budynas_SM_ch03

25_ch 03 Mechanical Design budynas_SM_ch03 - = 21 . 33 in 4...

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38 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design I 1 = [21 . 333 + 16(0 . 292) 2 ] [6 . 75 + 9(0 . 292) 2 ] [0 . 020 83 + 1(2 . 292 0 . 25) 2 ] = 10 . 99 in 4 Ans . σ A = 10 000(2 . 292) 10 . 99 = 2086 psi Ans. σ B = 10 000(2 . 292 0 . 5) 10 . 99 = 1631 psi Ans. σ C =− 10 000(1 . 708 0 . 5) 10 . 99 =− 1099 psi Ans. σ D =− 10 000(1 . 708) 10 . 99 =− 1554 psi Ans. (d) Use a as a negative area. A a = 6 . 928 in 2 , A b = 16 in 2 , A = 9 . 072 in 2 ; ¯ y a = 1 . 155 in, ¯ y b = 2in ¯ y = 2(16) 1 . 155(6 . 928) 9 . 072 = 2 . 645 in Ans. c 1 = 4 2 . 645 = 1 . 355 in I a = bh 3 36 = 4(3 . 464) 3 36 = 4 . 618 in 4 I b = 4(4) 3 12
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Unformatted text preview: = 21 . 33 in 4 I 1 = [21 . 33 + 16(0 . 645) 2 ] [4 . 618 + 6 . 928(1 . 490) 2 ] = 7 . 99 in 4 Ans. A = 10 000(2 . 645) 7 . 99 = 3310 psi Ans. B = 10 000(3 . 464 2 . 645) 7 . 99 = 1025 psi Ans. C = 10 000(1 . 355) 7 . 99 = 1696 psi Ans. 3.464" 1 1 G a B b a C A c 1 1.355" c 2 2.645" 1.490" 1.155"...
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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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