425_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

# 425_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 419 419 SECTION 5.6 Design of Beams Solution 5.6-10 Trapeze bar (regular octagon) P 1.2 kN L 2.1 m sallow 200 MPa b 0.41421h 1.85948(0.41421h)4 ‹ Ic Determine minimum height h. SECTION MODULUS MAXIMUM BENDING MOMENT S 0.054738h4 PL 4 Mmax (1.2 kN)(2.1 m) 4 630 N # m 0.054738h4 h/2 Ic h/2 MINIMUM HEIGHT h 8 M S b length of one side 0.109476h3 b 360° n PROPERTIES OF THE CROSS SECTION s Use Appendix D, Case 25, with n 360° 8 45° b tan 2 b (from triangle) h b cot 2 h b h3 For b 45° tan 2 45° cot 2 M s 630 N # m 200 MPa S 28.7735 * 10 ‹ hmin 6 3.15 * 10 m3 h ( 12 Ic a S a M 0.41421 2.41421 PL 4 11 + 8 12 4 bb 12 b 4 12 6 m3 0.030643 m 1)h h b h3 8) 12 ( 12 + 1)b 45° tan 5 6 ; 30.6 mm ALTERNATIVE SOLUTION (n b b 45°: h h b 0.109476h3 b 2 a 4 12 12 h3 1 cot 2(4 12 5 b 2 12 b h4 3PL 5)sallow ; 30.643 mm ; Substitute numerical values: h3 MOMENT OF INERTIA 4 Ic b b nb a cot b a 3 cot2 192 2 2 1b Ic 8b4 (2.41421)[3(2.41421)2 192 1] 1.85948b4 28.7735 * 10 6 m3 hmin 1 ...
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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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