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09-01ChapGere.0005 - having modulus of elasticity E 5 70...

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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) SECTION 9.3 Deflection Formulas 551 max 12,000 psi max 0.1 in. h 12 in. E 30 10 6 psi Calculate the span length L . Eq. (9-18): or (1) Flexure formula: Maximum bending moment: (2) s qL 2 h 16 I M qL 2 8 s Mc I Mh 2 I q 384 EI 5 L 4 max 5 qL 4 384 EI Solve Eq. (2) for q : (3) Equate (1) and (2) and solve for L : Substitute numerical values: L 120 in. 10 ft L 2 24(30 10 6 psi)(12 in.)(0.1 in.) 5(12,000 psi) 14,400 in. 2 L B 24 Eh 5 s L 2 24 Eh 5 s q 16 I s L 2 h Problem 9.3-4 Calculate the maximum deflection max of a uniformly loaded simple beam (see figure) if the span length L 2.0 m, the intensity of the uniform load q 2.0 kN/m, and the maximum bending stress 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 5 70 GPa. (Use the formulas of Example 9-1.) Solution 9.3-4 Simple beam (uniform load) L 5 2.0 m q 5 2.0 kN / m s 5 s max 5 60 MPa E 5 70 GPa C ROSS SECTION (square; b 5 width) Maximum deflection (Eq. 9-18): (1) Substitute for I : (2) Flexure formula with : Substitute for S : (3) s 5 3 qL 2 4 b 3 s 5 M S 5 qL 2 8 S M 5 qL 2 8 d 5 5 qL 4 32 Eb 4 d 5 5 qL 4 384 EI S 5 b 3 6 I 5 b 4 12 Solve for b 3 : (4) Substitute b into Eq. (2): (The term in parentheses is nondimensional.) Substitute numerical values: d max 5 10(80) 1 / 3 2.8 mm 5 15.4 mm ¢ 4 L s 3 q ≤ 1 / 3 5 B 4(2.0 m)(60 MPa) 3(2000 N / m) R 1 / 3 5 10(80) 1 / 3 5 L s 24 E 5 5(2.0 m)(60 MPa) 24(70 GPa) 5 1 2800 m 5 1 2.8 mm d max 5 5 L s 24 E ¢ 4 L s 3 q ≤ 1 / 3 b 3 5 3 qL 2 4 s q = 2.0 kN/m L = 2.0 m A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
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