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

`\Delta L = 0.1883 inch`

Step-by-step 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 = (PL)/(AE)`

`= (P)/(A)* (L)/(E)`

`= \sigma_(allow) * (L)/(E)`

modulus of elasticity E for aluminium alloy is
`10.2 * 10^6 psi = 10.2 * 10^3 kpsi`

`\Delta L = (20 * 8)/(10.2* 10^3)`

`\Delta L = 0.01569 ft`

`\Delta L = 0.1883 inch`