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A 30 mm diameter rod is bent to form an offset link as shown in Figure No. 1, if permissible tensile stress is 80 N/mm2, find the maximum value of P.
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written 4.7 years ago by
teamques10
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• modified 4.7 years ago |
Given: For offset link, d = 30 mm , $\sigma_{Max} = 80 N/mm^{2}$
Eccentricity = e = 40 + $\frac{d}{2}$ = 40 + $\frac{30}{2}$ = 55mm.
Solution:
$c/s Area = A = \frac{\pi}{4} \times d^{2} = \frac{\pi}{4} \times 30^{2} = 706.86 mm^{2}$
$M.I = I = \frac{\pi}{64} \times d^{4} = …
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