161_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

161_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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SECTION 2.5 Thermal Effects 155 Problem 2.5-6 A plastic bar ACB having two different solid circular cross sections is held between rigid supports as shown in the figure. The diameters in the left- and right-hand parts are 50 mm and 75 mm, respectively. The corresponding lengths are 225 mm and 300 mm. Also, the modulus of elasticity E is 6.0 GPa, and the coeffi- cient of thermal expansion ± is 100 ² 10 ³ 6 /°C. The bar is subjected to a uniform temperature increase of 30°C. (a) Calculate the following quantities: (1) the compressive force N in the bar; (2) the maximum compressive stress ´ c ; and (3) the displacement µ C of point C . (b) Repeat (a) if the rigid support at A is replaced by an elastic support having spring constant k 50 MN/m (see figure part b; assume that only the bar ACB is subject to the temperature increase). (a) 300 mm 75 mm 225 mm A B C 50 mm (b) 300 mm 75 mm 225 mm A B C 50 mm k Solution N UMERICAL DATA d 1 50 mm d 2 75 mm L 1 225 mm L 2 300 mm E 6.0 GPa 100 ² (10 ³ 6 /°C · T 30°C k 50 MN/m (a) C OMPRESSIVE FORCE N, MAX . COMPRESSIVE STRESS & DISPL . OF PT . C one-degree stat-indet - use R B as redundant B1 · T(L 1 ¸ L 2 ) compatibility: B1 B2 , solve for R
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Unformatted text preview: B N R B N 51.8 kN max. compressive stress in AC since it has the smaller area (A 1 A 2 ) cmax 26.4 MPa displacement C of point C superposition of dis-placements in two released structures at C s cmax N A 1 ; R B a T( L 1 + L 2 ) L 1 E A 1 + L 2 E A 2 d B2 R B a L 1 E A 1 + L 2 E A 2 b A 2 p 4 d 2 2 A 1 p 4 d 1 2 C 0.314 mm ( ) sign means jt C moves left (b) C OMPRESSIVE FORCE N, MAX . COMPRESSIVE STRESS & DISPL . OF PT . C FOR ELASTIC SUPPORT CASE Use R B as redundant as in (a) B1 T(L 1 L 2 ) ^ now add effect of elastic support; equate B1 and B2 then solve for R B N R B N 31.2 kN cmax 15.91 MPa super position C 0.546 mm ( ) sign means jt C moves left ; d C a T( L 1 ) R B a L 1 E A 1 + 1 k b ; s cmax N A 1 ; R B a T 1 L 1 + L 2 2 L 1 E A 1 + L 2 EA 2 + 1 k d B 2 R B a L 1 E A 1 + L 2 E A 2 + 1 k b ; d C a T( L 1 ) R B L 1 E A 1 Sec_2.5.qxd 9/25/08 3:00 PM Page 155...
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