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3 years ago

13

A diagonal aluminum alloy tension rod of diameter d and initial length l is used in a rectangular frame to prevent collapse. The

rod can safely support a tensile stress of sallow. If d 5 0.5 in, l 5 8 ft, and sallow 5 20 kpsi, determine how much the rod must be stretched to develop this allowable stress.

1 answer:

Ksenya-84 [330]3 years ago

6 0

Answer:

$\Delta L = 0.1883 inch$

Explanation:

Given data:

d = 0.5 inch

$L_i = 8 ft$

$\sigma_{allow} = 20 kpsi$

we know that change in length is given as

$\Delta L = \frac{PL}{AE}$

$= \frac{P}{A}\times \frac{L}{E}$

$= \sigma_{allow} \times \frac{L}{E}$

modulus of elasticity E for aluminium alloy is $10.2 \times 10^6 psi = 10.2 \times 10^3 kpsi$

$\Delta L = \frac{20 \times 8}{10.2\times 10^3}$

$\Delta L = 0.01569 ft$

$\Delta L = 0.1883 inch$

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