05-02 Chap Gere

05-02 Chap Gere - 304 CHAPTER 5 Stresses in Beams Design of...

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Design of Beams Problem 5.6-1 The cross section of a narrow-gage railway bridge is shown in part (a) of the figure. The bridge is constructed with longitudinal steel girders that support the wood cross ties. The girders are restrained against lateral buckling by diagonal bracing, as indicated by the dashed lines. The spacing of the girders is s 1 5 50 in. and the spacing of the rails is s 2 5 30 in. The load transmitted by each rail to a single tie is P 5 1500 lb. The cross section of a tie, shown in part (b) of the figure, has width b 5 5.0 in. and depth d . Determine the minimum value of d based upon an allowable bending stress of 1125 psi in the wood tie. (Disregard the weight of the tie itself.) Solution 5.6-1 Railway cross tie 304 CHAPTER 5 Stresses in Beams s 1 d s 2 PP Wood tie Steel girder Steel rail (a) (b) b s 1 5 50 in. b 5 5.0 in. s 2 5 30 in. d 5 depth of tie P 5 1500 lb s allow 5 1125 psi S 5 bd 2 6 5 1 6 (50 in.)( d 2 ) 5 5 d 2 6 Ê d 5 inches M max 5 P ( s 1 2 s 2 ) 2 5 15,000 lb-in. Solving, d 2 5 16.0 in. d min 5 4.0 in. Note: Symbolic solution: d 2 5 3 P ( s 1 2 s 2 ) b s allow M max 5 s allow S Ê 15,000 5 (1125) ¢ 5 d 2 6 s 1 d s 2 Wood tie Steel rail (a) (b) b Problem 5.6-2 A fiberglass bracket ABCD of solid circular cross section has the shape and dimensions shown in the figure. A vertical load P 5 36 N acts at the free end D . Determine the minimum permissible diameter d min of the bracket if the allowable bending stress in the material is 30 MPa and b 5 35 mm. (Disregard the weight of the bracket itself.) Solution 5.6-2 Fiberglass bracket 5 b 2 b AB DC P 2 b D ATA P 5 36 N s allow 5 30 MPa b 5 35 mm C ROSS SECTION M AXIMUM BENDING MOMENT M max 5 P (3 b ) M AXIMUM BENDING STRESS s allow 5 3 Pbd 2 I 5 96 Pb p d 3 s max 5 M max c I Ê c 5 d 2 I 5 p d 4 64 d = diameter M INIMUM DIAMETER 5 1,283.4 mm 3 d min 5 10.9 mm d 3 5 96 Pb p s allow 5 (96)(36 N)(35 mm) p (30 MPa)

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Problem 5.6-3 A cantilever beam of length L 5 6 ft supports a uniform load of intensity q 5 200 lb/ft and a concentrated load P 5 2500 lb (see figure). Calculate the required section modulus S if s allow 5 15,000 psi. Then select a suitable wide-flange beam ( W shape) from Table E-1, Appendix E, and recalculate S taking into account the weight of the beam. Select a new beam size if necessary. Solution 5.6-3 Cantilever beam SECTION 5.6 Design of Beams 305 L = 6 ft q 5 200 lb/ft P 5 2500 lb P 5 2500 lb q 5 200 lb/ft L 5 6 ft s allow 5 15,000 psi R EQUIRED SECTION MODULUS 5 18,600 lb-ft 5 223,200 lb-in. S 5 M max s allow 5 223,200 lb-in. 15,000 psi 5 14.88 in. 3 M max 5 PL 1 qL 2 2 5 15,000 lb-ft 1 3,600 lb-ft T RIAL SECTION W 8 3 21 S 5 18.2 in. 3 q 0 5 21 lb/ft M max 5 223,200 1 4,536 5 227,700 lb-in. Required 15.2 in. 3 , 18.2 in. 3 [ Beam is satisfactory. Use W 8 3 21 S 5 M max s allow 5 227,700 lb-in. 15,000 psi 5 15.2 in. 3 M 0 5 q 0 L 2 2 5 378 lb-ft 5 4536 lb-in. Problem 5.6-4 A simple beam of length L 5 15 ft carries a uniform load of intensity q 5 400 lb/ft and a concentrated load P 5 4000 lb (see figure).
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This note was uploaded on 03/19/2008 for the course E M 316 taught by Professor Korkolis during the Spring '08 term at University of Texas.

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05-02 Chap Gere - 304 CHAPTER 5 Stresses in Beams Design of...

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