427_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

# 427_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: 05Ch05.qxd 9/24/08 4:59 AM Page 421 SECTION 5.6 Design of Beams 421 Solution 5.6-12 Cantilever beam MINIMUM DIAMETER Mmax sallow S L 450 mm P 400 N sallow 60 MPa g weight density of steel DATA PL + 3 77.0 kN/m ga sallow a pd 3 b 32 Rearrange the equation: WEIGHT OF BEAM PER UNIT LENGTH q pgd 2L2 8 sallow d 3 pd 2 b 4 4gL2 d 2 32 PL p 0 (Cubic equation with diameter d as unknown.) MAXIMUM BENDING MOMENT Mmax q L2 PL + 2 SECTION MODULUS Substitute numerical values (d pgd3L2 PL + 8 S 6 2 3 4(77,000 N/m3)(0.45 m)2d2 (60 * 10 N/m )d 32 (400 N)(0.45 m) p pd 3 32 60,000d 3 meters): 62.37d 2 0 1.833465 0 Solve the equation numerically: d 0.031614 m 31.61 mm ; q Problem 5.6-13 A compound beam ABCD (see figure) is supported at points A, B, and D and has a splice at point C. The distance a 6.25 ft, and the beam is a S 18 70 wide-flange shape with an allowable bending stress of 12,800 psi. dmin A B C D Splice (a) If the splice is a moment release, find the allowable 4a uniform load qallow that may be placed on top of the beam, taking into account the weight of the beam itself. [See figure part (a).] (b) Repeat assuming now that the splice is a shear release, as in figure part (b). a 4a (a) (b) Moment Shear release release Solution 5.6-13 NUMERICAL DATA lb S 103 in.3 w 70 ft a 6.25 ft sa 12800 psi MZ 2.000E+00 @ 2.000E+00 MZ 9.453E–01 @ 1.375E+00 × × (a) MOMENT RELEASE AT C-GIVES MAX. MOMENT AT B 2.5 q a2 (SEE MOMENT DIAGRAM) Mmax S and Mmax sa w a Mmax [1qallow + w2 a2 (2.5)] saS lb S 103 in.3 ft sa 12800 psi 6.25 ft 70 MZ –2.500E+00 @ 4.000E+00 × ...
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## This note was uploaded on 12/22/2011 for the course MEEG 310 taught by Professor Staff during the Fall '11 term at University of Delaware.

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