192_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

192_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: Sec_2.6.qxd 186 9/25/08 11:40 AM Page 186 CHAPTER 2 Axially Loaded Members using u acos a A sx s pq s pq cos1u22 )2 xcos( b from (a) (for temperature increase T): sx RR1 xsin( ) cos( ) pq 2 pq xcos( ) sx stresses at sy 1154 psi ; 1700 psi pq p2 b 2 y set max 784 psi max psi si 0 p θ = 34.2° 170 sx 3P tmax 4A a & solve for Pmax1 3300 psi set a 3300 a 3400 psi max P tmax 4A a & solve for Pmax2 P 4 RL2 3 4 Pmax2 4A( 2 14688 lb less than a so shear controls stat-indet analysis for P at L/4 gives, for reactions: RR2 a E a¢ T Pmax2 (c) MAX. LOAD AT QUARTER POINT 2 sx 2 Stresses in bar (L/4 to L) sx 1650 psi 3Pmax1 4A less than Ea ¢ T + x Par a T 3 Pmax1 E a¢T + 2 8A 1650 psi check tmax psi ) t (b Pmax1 Pmax1 4 115 EA E a ¢T 3P + 2 8A 4A 12ta + Ea¢ T2 3 34704 lb ta (b) STRESS ELEMENT FOR PLANE PQ 784 RL1 Ea ¢ T + /2 (y face) s xcos a u + T Stresses in bar (0 to L/4) 34.2° now with , can find shear stress on plane pq pq EA P (tension for 0 to L/4 & compression for rest of bar) tmax sx E a¢ T 2 Ea ¢ T a E T) ; Pmax2 8A Pmax2 4A sx 2 shear in segment (L/4 to L) controls max x 1650 psi 3300 psi ...
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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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