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Question

A bar AB of diameter 40 mm and 4 m long is rigidly fixed at its ends. A torque 600 N-m is applied at a section of the bar, 1 m from end A. The fixing couples TA and TB at the supports A and B, respectively are

A
200 N-m and and 400 N-m
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B
300 N-m and 150 N-m
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C
450 N-m and 150 N-m
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D
300 N-m and 100 N-m
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Solution

The correct option is C 450 N-m and 150 N-m

The fixing couple TA and TB at A and B respectively is given by

TA=T0×ba+b

=600×31+3=450Nm

TB=T0×aa+b

=600×14=150 Nm

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