Question

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.

Answer 1

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 $10.2 \times 10^6 psi = 10.2 \times 10^3 kpsi$

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

$\Delta L = 0.01569 ft$

$\Delta L = 0.1883 inch$