Question

If the 20-mm-diameter rod is made of A-36 steel and the stiffness of the spring is k = 55 MN/m , determine the displacement of end A when the 60-kN force is applied. Take E = 200 GPa.

Answer 1

Answer:

σ = 1.09 mm

Explanation:

Step 1: Identify the given parameters

rod diameter = 20 mm

stiffness constant (k) = 55 MN/m = 55X10⁶N/m

applied force (f) = 60 KN = 60 X 10³N

young modulus (E) = 200 Gpa = 200 X 10⁹pa

Step 2: calculate length of the rod, L

$K = \frac{A*E}{L}$

$L = \frac{A*E}{K}$

$A=\frac{\pi d^{2}}{4}$

d = 20-mm = 0.02 m

$A=\frac{\pi (0.02)^{2}}{4}$

A = 0.0003 m²

$L = \frac{A*E}{K}$

$L = \frac{(0.0003142)*(200X10^9)}{55X10^6}$

L = 1.14 m

Step 3: calculate the displacement of the rod, σ

$\sigma = \frac{F*L}{A*E}$

$\sigma = \frac{(60X10^3)*(1.14)}{(0.0003142)*(200X10^9)}$

σ = 0.00109 m

σ = 1.09 mm

Therefore, the displacement at the end of A is 1.09 mm