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717_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

# 717_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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SECTION 9.3 Deflection Formulas 711 Problem 9.3-3 What is the span length L of a uniformly loaded simple beam of wide-flange cross section (see figure) if the maximum bending stress is 12,000 psi, the maximum deflection is 0.1 in., the height of the beam is 12 in., and the modulus of elasticity is 30 ± 10 6 psi? (Use the formulas of Example 9-1.) Solution 9.3-3 Simple beam (uniform load) s ² s max ² 12,000 psi d ² d max ² 0.1 in. h ² 12 in. E ² 30 ± 10 6 psi Calculate the span length L . Maximum bending moment: M ² qL 2 8 s ² qL 2 h 16 I (2) Flexure formula: s ² Mc I ² Mh 2 I Eq. (9- 18): d ² d max ² 5 qL 4 384 EI or q ² 384 EI d 5 L 4 (1) (3) Equate (1) and (2) and solve for L : Substitute numerical values: L ² 120 in. ² 10 ft ; L 2 ² 24(30 * 10 6 psi)(12 in.)(0.1 in.) 5(12,000 psi) ² 14,400 in. 2 L 2 ² 24 Eh d 5 s L ² A 24 Eh d 5 s ; Solve Eq. (2) for q : q ² 16 I s L 2 h Problem 9.3-4 Calculate the maximum deflection d max of a uniformly loaded simple beam (see figure) if the span length L ² 2.0 m, the intensity of the uniform load q ² 2.0 kN/m, and the maximum bending stress s ² 60 MPa. The cross section of the beam is square, and the material is aluminum
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Unformatted text preview: having modulus of elasticity E ² 70 GPa. (Use the formulas of Example 9-1.) q = 2.0 kN/m L = 2.0 m Solution 9.3-4 Simple beam (uniform load) L ² 2.0 m q ² 2.0 kN/m s ² s max ² 60 MPa E ² 70 GPa C ROSS SECTION (square; b ² width) (2) (3) Substitute for S : s ² 3 qL 2 4 b 3 Flexure formula with M ² qL 2 8 : s ² M S ² qL 2 8 S Substitute for I : d ² 5 qL 4 32 Eb 4 Maximum deflection (Eq. 9-18): d ² 5 qL 4 384 EI (1) I ² b 4 12 S ² b 3 6 (4) (The term in parentheses is nondimensional.) Substitute numerical values: d max ² 10(80) 1/3 2.8 mm ² 15.4 mm ; a 4 L s 3 q b 1/3 ² c 4(2.0 m)(60 MPa) 3(2000 N/m) d 1/3 ² 10(80) 1/3 5 L s 24 E ² 5(2.0 m)(60 MPa) 24(70 GPa) ² 1 2800 m ² 1 2.8 mm Substitute b into Eq. (2): d max ² 5 L s 24 E a 4 L s 3 q b 1/3 ; Solve for b 3 : b 3 ² 3 qL 2 4 s 09Ch09.qxd 9/27/08 1:31 PM Page 711...
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