The steel rod fits loosely inside the aluminum sleeve. Both components are attached to a rigid wall A and joined together by a pin at B. Because of a slight misalignment of the pre-drilled holes, th torque To 650 N-m was applied to the steel rod before the pin could be inserted into the hole Determine the torque in each component after To was removed. Use G = 80 GPa for steel and G = 2 GPa for aluminum. A Aluminum 40 mm 50 mm 3 m Pin B To Steel
The steel rod fits loosely inside the aluminum sleeve. Both components are attached to a rigid wall A and joined together by a pin at B. Because of a slight misalignment of the pre-drilled holes, th torque To 650 N-m was applied to the steel rod before the pin could be inserted into the hole Determine the torque in each component after To was removed. Use G = 80 GPa for steel and G = 2 GPa for aluminum. A Aluminum 40 mm 50 mm 3 m Pin B To Steel
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Transcribed Image Text:The steel rod fits loosely inside the aluminum sleeve. Both components are attached to a rigid wall at
A and joined together by a pin at B. Because of a slight misalignment of the pre-drilled holes, the
torque To= 650 N-m was applied to the steel rod before the pin could be inserted into the holes.
Determine the torque in each component after To was removed. Use G = 80 GPa for steel and G = 28
GPa for aluminum.
A
Aluminum
40 mm
50 mm
3 m
Pin
B
-Steel
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