02-03ChapGere.0002

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

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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 D T (degrees Celsius) at which the average shear stress in the 12-mm diameter bolt becomes 45 MPa. (For the steel rod, use a 5 12 3 10 2 6 /°C and E 5 200 GPa.) Solution 2.5-4 Steel rod with bolted connection 15 mm 12 mm diameter bolt R 5 rod B 5 bolt P 5 tensile force in steel rod due to temperature drop D T A R 5 cross-sectional area of steel rod From Eq. (2-17) of Example 2-7: P 5 EA R a ( D T ) Bolt is in double shear. V 5 shear force acting over one cross section of the bolt t 5 average shear stress on cross section of the bolt A B 5 cross-sectional area of bolt t 5 V A B 5 EA R a ( ¢ T ) 2 A B V 5 P / 2 5 1 2 EA R a ( ¢ T ) S UBSTITUTE NUMERICAL VALUES : t 5 45 MPa d B 5 12 mm d R 5 15 mm a 5 12 3 10 2 6 / 8 C E 5 200 GPa ¢ T 5 24 8 C Ê ¢ T 5 2(45 MPa)(12 mm) 2 (200 GPa)(12 3 10 2 6
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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 where d R 5 diameter of steel rod A B 5 p d B 2 4 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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## This note was uploaded on 09/20/2009 for the course COE 3001 taught by Professor Armanios during the Spring '08 term at Georgia Institute of Technology.

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