Answer:
σ = 1.09 mm
Step-by-step 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