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177_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

# 177_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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SECTION 2.5 Thermal Effects 171 Problem 2.5-22 Consider the sleeve made from two copper tubes joined by tin-lead solder over distance s . The sleeve has brass caps at both ends, which are held in place by a steel bolt and washer with the nut turned just snug at the outset. Then, two “loadings” are applied: n 1/2 turn applied to the nut; at the same time the internal temperature is raised by T 30°C. (a) Find the forces in the sleeve and bolt, P s and P B , due to both the prestress in the bolt and the temperature increase. For copper, use E c 120 GPa and c 17 10 6 /°C; for steel, use E s 200 GPa and s 12 10 6 /°C. The pitch of the bolt threads is p 1.0 mm. Assume s 26 mm and bolt diameter d b 5 mm. (b) Find the required length of the solder joint, s , if shear stress in the sweated joint cannot exceed the allowable shear stress aj 18.5 MPa. (c) What is the final elongation of the entire assemblage due to both tempera- ture change T and the initial prestress in the bolt? Brass cap d = np Steel bolt L 1 = 40 mm, d 1 = 25 mm, t 1 = 4 mm L 2 = 50 mm, d 2 = 17 mm, t 2 = 3 mm T T
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Unformatted text preview: ; s p ± P p A p ± 2 np E s A s E p L ( E p A p + 2 E s A s ) P p ± 2 np E s A s E p A p L ( E p A p + 2 E s A s ) S UBSTITUTE NUMERICAL VALUES : N ± E s A s E p ± 900 ´ 10 9 lb 2 /in. 2 D ± E p A p · 2 E s A s ± 4350 ´ 10 3 lb ± 2400 psi ; s p ± 2 np L a N D b ± 2(1)(0.058in.) 10 in. a N D b Solution 2.5-22 The figure shows a section through the sleeve, cap and bolt N UMERICAL PROPERTIES (SI UNITS ) p ± 1.0 mm ² T ± 30°C E c ± 120 GPa c ± 17 ´ (10 µ 6 )/°C E s ± 200 GPa s ± 12 ´ (10 µ 6 )/°C aj ± 18.5 MPa s ± 26 mm d b ± 5 mm L 1 ± 40 mm t 1 ± 4 mm L 2 ± 50 mm t 2 ± 3 mm d 1 ± 25 mm d 1 µ 2t 1 ± 17 mm d 2 ± 17 mm n ± 1 2 A b ± 19.635 mm 2 A 1 ± 263.894 mm 2 A 2 ± 131.947 mm 2 (a) F ORCES IN SLEEVE & BOLT one-degree stat-indet - cut bolt & use force in bolt (P B ) as redundant (see sketches below) ¸ B1 ± µ np · s ² T(L 1 · L 2 µ s) A 2 ± p 4 [ d 2 2 µ 1 d 2 µ 2t 2 2 2 ] A 1 ± p 4 [d 1 2 µ 1 d 1 µ 2 t 1 2 2 ] A b ± p 4 d b 2 Sec_2.5.qxd 9/25/08 3:00 PM Page 171...
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