Question: A mild steel link as shown in Figure No. 2, by full lines transmits a pull of 80 kN; find the dimensions b and t if b = 3t. Assume the permissible stress as 70 N/mm2.
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modified 8 days ago  • written 8 days ago by gravatar for bharathchippa49 bharathchippa490
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Given:

For M.S link, P = 80 kN = $80 \times 10^{3}$ , b = 3t , $\sigma = 70 N/mm^{2}$

Solution:

$\sigma = \frac{P}{A}$ $\therefore A = \frac{P}{\sigma} = \frac{80 \times 10^{3}}{70}$

$\therefore A = 1142.86 mm^{2}$

$\therefore A = 1142.86$

$\therefore 3t \times t = 1142.86$

$\therefore t = 19.52 mm$

& b = 3t = $3 \times 19.52 = 58.56 mm$

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written 8 days ago by gravatar for bharathchippa49 bharathchippa490
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