A spring of mean radius 40 mm contains 8 action coils of steel (N = 80000 N/mm2), 4 mm in diameter. The clearance between the coils being 1 mm when unloaded, the minimum compressive load to remove the clearance, is-A-25 NB-30 NC-35 ND-40 N-C-35 N

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C 35 N The question is asked to find the load. Not distance! p= (64W* R^3)/(C*d^4). We find out W, load for 1 clearance, between 2 coils. Between 2 coil, 1 clearance, load= 5N. Between 8 coil, 7 clearance, load= 5x7=35N.

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A spring of mean radius 40 mm contains 8 action coils of steel (N = 80000 N/mm²), 4 mm in diameter. The clearance between the coils being 1 mm when unloaded, the minimum compressive load to remove the clearance, is

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A spring of mean radius 40 mm contains 8 action coils of steel (N = 80000 N/mm2), 4 mm in diameter. The clearance between the coils being 1 mm when unloaded, the minimum compressive load to remove the clearance, is

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A spring of mean radius 40 mm contains 8 action coils of steel (N = 80000 N/mm²), 4 mm in diameter. The clearance between the coils being 1 mm when unloaded, the minimum compressive load to remove the clearance, is

For help Students Orientation
Mcqs Questions