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

Answer 2

The displacement at the end will be "1.09 mm".

Force and Displacement

According to the question,

Rod's diameter, d= 20 mm

Stiffness constant, k = 55 M-N/m or,

                                  = 55 × 10⁶ N/m

Force, f = 60 kN or,

             = 60 × 10³ N

Young's modulus, E = 200 Gpa or,

                                 = 200 × 10⁹ pa

We know the formula,

→ K = $\frac{A\times E}{L}$

or,

   L = $\frac{A\times E}{K}$

or,

   A = $\frac{\pi d^2}{4}$

By substituting the values,

      = $\frac{\pi(0.02)^2}{4}$

      = 0.0003 m³

The length, L =  $\frac{A\times E}{K}$

                      = $\frac{0.0003\times 200\times 10^9}{55\times 10^6}$

                      = 1.14 m

hence,

The displacement be:

→ σ = $\frac{F\times L}{A\times E}$

      = $\frac{60\times 10^3\times 1.14}{0.0003\times 200\times 10^9}$

      = 0.00109 m or,

      = 1.09 mm

Thus the answer above is correct.            

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