Question If the 20-mm-diameter rod is made of A-36 steel and the stiffness of the spring is k = 61 MN/m , determine the displacement of end A when the 60-kN force is applied. Take E = 200 GPa.

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}K= L A∗E L = \frac{A*E}{K}L= K A∗E A=\frac{\pi d^{2}}{4}A= 4 πd 2 d = 20-mm = 0.02 m A=\frac{\pi (0.02)^{2}}{4}A= 4 π(0.02) 2 A = 0.0003 m² L = \frac{A*E}{K}L= K A∗E L = \frac{(0.0003142)*(200X10^9)}{55X10^6}L= 55X10 6 (0.0003142)∗(200X10 9 ) L = 1.14 m Step 3: calculate the displacement of the rod, σ \sigma = \frac{F*L}{A*E}σ= A∗E F∗L \sigma = \frac{(60X10^3)*(1.14)}{(0.0003142)*(200X10^9)}σ= (0.0003142)∗(200X10 9 ) (60X10 3 )∗(1.14) σ = 0.00109 m σ = 1.09 mm Therefore, the displacement at the end of A is 1.09 mm Log in to Reply

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}K=

L

A∗E

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

K

A∗E

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

4

πd

2

d = 20-mm = 0.02 m

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

4

π(0.02)

2

A = 0.0003 m²

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

K

A∗E

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

55X10

6

(0.0003142)∗(200X10

9

)

L = 1.14 m

Step 3: calculate the displacement of the rod, σ

\sigma = \frac{F*L}{A*E}σ=

A∗E

F∗L

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

(0.0003142)∗(200X10

9

)

(60X10

3

)∗(1.14)

σ = 0.00109 m

σ = 1.09 mm

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