201_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

201_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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SECTION 2.6 Stresses on Inclined Sections 195 Solution 2.6-19 (a) S TAT - INDET NONPRISMATIC BAR WITH LOAD P AT jt 2 apply load P and temp. change ± T - use R 3 as ² 3a ³ Pf 12 ´ ( µ 1 b 1 ´ 2 b 2 ) ± T 3b ³ R 3 (f 12 ´ f 23 ) compatibility: 3a ´ 3b ³ 0 compression at Location B due to both P and temp. increase statics: R 1 ³· P · R 3 compression due to temp. increase, tension due to P, at Location A ¸ a1 ³ a Numerical data a ³ 11 ksi A ³ 6in 2 P ³ 12 kips E ³ 30000 ksi ³ 6.5 ¹ (10 · 6 )/°F steel (1) check normal and shear stresses at element A location & solve for ± T max using a1 & º a1 ³ s a A ± b E A + 3 4 b 4 3 E A 2 ≤¥ + ± P 3 4 b 4 3 E A 2 a a b + 4 5 a 3 4 b b ¢ T max f 23 ³ b 2 E 2 A 2 f 12 ³ b 1 E 1 A 1 ¢ T max ³ [ s a1 A 1 ( f 12 + f 23 )] + P f 23 ( a 1 b 1 + a 2 b 2 ) s xA ³ R 1 A 1 t a2 ³ 9 20 s a t a1 ³ 2 5 s a s a2 ³ 3 4 s a R 1 ³ · P f 23 + 1 a 1 b 1 + a 2 b 2 2 ¢ T f 12 + f 23 R 3 ³ · P f 12 · 1 a 1 b 1 + a 2 b 2 2 ¢ T f 12 + f 23 f 23 ³ b 2 E 2 A 2 f 12 ³ b 1 E 1 A 1 ± T max ³ 82.1°F compression due to temp. rise but tension due to P ¢ T max º max A ³ s xA 2 ¢ T max ³ 85 s a A + 45 P 64EA a ± T max ³ 67.1°F shear controls for Location A where temp. rise causes compressive stress but load P causes tensile stress
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Unformatted text preview: (2) check normal and shear stresses at element B location & solve for ± T max using a2 & a2 ¶ compression due to both temp. rise & load P normal stress controls for Location B where temp. rise & load P both cause compressive stress; as a result, permissible temp. rise is reduced at B compared to Location A where temp. rise effect is offset by load P effect ; ¢ T max ³ 21.7°F ¢ T max ³ 255 s a A · 320 P 512EA a ³ ≥ 3 4 s a A 2 ± b EA + 3 4 b 4 3 E A 2 ≤¥ · P b E A a a b + 4 5 a 3 4 b b ¢ T max ¢ T max ³ [ s a2 A 2 ( f 12 + f 23 )] + P f 12 ( a 1 b 1 + a 2 b 2 ) s xB ³ R 3 A 2 ; ¢ T max ³ 68 s a A + 45P 64EA a ³ 2 a 2 5 s a b A ± b E A + 3 4 b 4 3 E A 2 ≤ + P 3 4 b 4 3 E A 2 a a b + 4 5 a 3 4 b b Sec_2.6.qxd 9/25/08 11:40 AM Page 195...
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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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