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02-03ChapGere.0002 - 8 C(15 mm 2 ¢ T 5 2 t d B 2 E a d R 2...

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SECTION 2.5 Thermal Effects 107 Problem 2.5-4 A steel rod of diameter 15 mm is held snugly (but without any initial stresses) between rigid walls by the arrangement shown in the figure. Calculate the temperature drop T (degrees Celsius) at which the average shear stress in the 12-mm diameter bolt becomes 45 MPa. (For the steel rod, use 12 10 6 /°C and E 200 GPa.) Solution 2.5-4 Steel rod with bolted connection 15 mm 12 mm diameter bolt R rod B bolt P tensile force in steel rod due to temperature drop T A R cross-sectional area of steel rod From Eq. (2-17) of Example 2-7: P EA R ( T ) Bolt is in double shear. V shear force acting over one cross section of the bolt average shear stress on cross section of the bolt A B cross-sectional area of bolt t V A B EA R ( ¢ T ) 2 A B V P 2 1 2 EA R ( ¢ T ) S UBSTITUTE NUMERICAL VALUES : 45 MPa d B 12 mm d R 15 mm 12 10 6 / C E 200 GPa ¢ T 24 C ¢ T 2(45 MPa)(12
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Unformatted text preview: / 8 C)(15 mm) 2 ¢ T 5 2 t d B 2 E a d R 2 A R 5 p d R 2 4 Ê where d R 5 diameter of steel rod A B 5 p d B 2 4 Ê where d B 5 diameter of bolt Solve for ¢ T : ¢ T 5 2 t A B EA R a 15 mm 12 mm diameter bolt B R Problem 2.5-5 A bar AB of length L is held between rigid supports and heated nonuniformly in such a manner that the temperature increase D T at distance x from end A is given by the expression D T 5 D T B x 3 / L 3 , where D T B is the increase in temperature at end B of the bar (see figure). Derive a formula for the compressive stress s c in the bar. (Assume that the material has modulus of elasticity E and coefficient of thermal expansion a .) L A ∆ T ∆ T B B x A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
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