03-01ChapGere.0013 - : U NITS : Newtons, Meters ( p )(30 3...

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Solution 3.3-13 Vertical pole SECTION 3.3 Circular Bars and Tubes 193 P 5 1100 lb c 5 5.0 in. t allow 5 4500 psi Find d min T ORSION FORMULA T 5 P (2 c 1 d ) Ê I P 5 p d 4 32 t max 5 Tr I P 5 Td 2 I P ( pt max ) d 3 2 (16 P ) d 2 32 Pc 5 0 S UBSTITUTE NUMERICAL VALUES : U NITS : Pounds, Inches ( p )(4500) d 3 2 (16)(1100) d 2 32(1100)(5.0) 5 0 or d 3 2 1.24495 d 2 12.4495 5 0 Solve numerically: d 5 2.496 in. d min 5 2.50 in. Ê t max 5 P (2 c 1 d ) d p d 4 / 16 5 16 P (2 c 1 d ) p d 3 d c A P B P c Problem 3.3-14 Solve the preceding problem if the horizontal forces have magnitude P 5 5.0 kN, the distance c 5 125 mm, and the allowable shear stress is 30 MPa. Solution 3.3-14 Vertical pole T ORSION FORMULA ( pt max ) d 3 2 (16 P ) d 2 32 Pc 5 0 S UBSTITUTE NUMERICAL VALUES
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Unformatted text preview: : U NITS : Newtons, Meters ( p )(30 3 10 6 ) d 3 2 (16)(5000) d 2 32(5000)(0.125) 5 or d 3 2 848.826 3 10 2 6 d 2 212.207 3 10 2 6 5 Solve numerically: d 5 0.06438 m d min 5 64.4 mm t max 5 P (2 c 1 d ) d p d 4 / 16 5 16 P (2 c 1 d ) p d 3 T 5 P (2 c 1 d ) I P 5 p d 4 32 t max 5 Tr I P 5 Td 2 I P d c A P B P c P 5 5.0 kN c 5 125 mm t allow 5 30 MPa Find d min A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
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This note was uploaded on 09/20/2009 for the course COE 3001 taught by Professor Armanios during the Spring '08 term at Georgia Institute of Technology.

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