163_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

163_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: Sec_2.5.qxd 9/25/08 3:00 PM Page 157 SECTION 2.5 Thermal Effects Solution 2.5-8 Brass sleeve fitted over a Steel bolt sS ¢T ES(aS aB) a1 + ES AS b EB AB ; SUBSTITUTE NUMERICAL VALUES: 25 MPa S d2 ES Subscript S means “sleeve”. 36 mm 100 GPa Subscript B means “bolt”. 21 S Use the results of Example 2-8. S EQUATION (2-20A): aB)( ¢ T)ES EB AB sS (Compression) ES AS + EB AB SOLVE FOR T: ¢T sS(ES AS + EB AB) (aS aB)ES EB AB EB 10 /°C AS p2 (d 42 p (dB)2 4 (aS 26 mm d2) 1 B 10 p (620 mm2) 4 ES AS p (625 mm2) 1 + 4 EB AB (100 GPa)(11 * 10 34°C 6 /°C) ; (Increase in temperature) or Problem 2.5-9 Rectangular bars of copper and aluminum are held by pins at their ends, as shown in the figure. Thin spacers provide a separation between the bars. The copper bars have cross-sectional dimensions 0.5 in. 2.0 in., and the aluminum bar has dimensions 1.0 in. 2.0 in. Determine the shear stress in the 7/16 in. diameter pins if the temperature is raised by 100°F. (For copper, Ec 18,000 ksi and c 9.5 10 6/°F; for aluminum, Ea 10,000 ksi and a 13 10 6/°F.) Suggestion: Use the results of Example 2-8. Solution 2.5-9 Rectangular bars held by pins Diameter of pin: dP Area of pin: AP 7 in. 16 p2 d 4P 0.4375 in. 0.15033 in.2 Area of two copper bars: Ac Area of aluminum bar: Aa T 100°F 25 mm 10 6/°C 25 MPa (1.496) ¢T T dB 200 GPa 6 AB compressive force in sleeve d1 2.0 in.2 2.0 in.2 1.496 157 ...
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