Question

An 60-m long wire of 5-mm diameter is made of steel with E = 200 GPa and ultimate tensile strength of 400 MPa. If a factor of safety of 3.2 is desired, determine (a) the allowable tension in the wire (b) the corresponding elongation of the wire

Answer 1

a) 2.45 KN

b) 0.0375 m

Step-by-step explanation:

`(a) \quad \sigma_(v)=400 * 10^(6) \mathrm{Pa} \quad A=(\pi)/(4) d^(2)=(\pi)/(4)(5)^(2)=19.635 \mathrm{mm}^(2)=19.635 * 10^(-6) \mathrm{m}^(2)`

`P_(U)=\sigma_(U) A=\left(400 * 10^(6)\right)\left(19.635 * 10^(-6)\right)=7854 \mathrm{N}`

`P_{\text {al }}=(P_(U))/(F S)=(7854)/(3.2)=2454 \mathrm{N}`

(b)
`\quad \delta=(P L)/(A E)=((2454)(60))/(\left(19.635 * 10^(-6)\right)\left(200 * 10^(9)\right))=37.5 * 10^(-3) \mathrm{m}`