Question

A diagonal aluminum alloy tension rod of diameter d and initial length l is used in a rectangularframe 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 thisallowable stress.

Answer 1

The stretched length is 0.18816 in.

Step-by-step explanation:

Given that,

Distance = 0.5 in

length = 8 ft

Tensile stress = 20 kpsi

We need to calculate the stretched length

Using formula of length

`\Delta L=(PL)/(AE)`

`\Delta L=\sigma*(L)/(E)`

Where,
`\sigma` = Tensile stress

L = length

E = young's modulus of aluminum

Put the value into the formula

`\Delta L= 20*10^3(8)/(10.2*10^(6))`

`\Delta L= 0.01568\ ft`

`\Delta L=0.01568*12`

`\Delta L=0.18816\ in`

Hence, The stretched length is 0.18816 in.