Question

A cylindrical rod of steel (E = 207 GPa, 30 * 106 psi) having a yield strength of 310 MPa (45,000 psi) is to be subjected to a load of 11,100 N (2500 lbf). If the length of the rod is 500 mm (20.0 in.), what must be the diameter to allow an elongation of 0.38 mm (0.015 in.)?

Answer 1

Answer:9.477 mm

Step-by-step explanation:

Given

E=207 GPa

Yield Strength$(s_{yt}) 310 MPa$

load (P)=11,100 N

length of rod(L)=500 mm

$\Delta L=0.38 mm$

we know $\Delta L$ is given by

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

where A= cross-section

$A=\frac{11100\times 500}{0.38\times 207\times 10^9}$

$A=70.556\times 10^{-6} m^2$

$A=\frac{\pi d^2}{2}=70.556\times 10^{-6} m^2$

$d=9.477\times 10^{-3} m$

d=9.477 mm