Answer:
ε = 0.005
σ = 12.73 MPa
E= 2.546 GPa
Explanation:
Given that
Li= 100 mm
d = 10 mm
P= 1000 N
Lf= 100.5 mm
ΔL = 100.5 - 100 mm
ΔL = 0.5 mm
We know that strain,ε
ε = ΔL / L
[tex]\varepsilon =\dfrac{0.5}{100}[/tex]
ε = 0.005
We know that stress given as
[tex]\sigma =\dfrac{P}{A}[/tex]
[tex]A=\dfrac{\pi d^2}{4}[/tex]
[tex]A=\dfrac{\pi \times 10^2}{4}\ mm^2[/tex]
[tex]A=78.53\ mm^2[/tex]
[tex]\sigma =\dfrac{1000}{78.53}\ MPa[/tex]
σ = 12.73 MPa
We know that with in the elastic limit
σ = ε .E
E Modulus of elasticity
σ = ε .E
12.73 = 0.005 x E
E= 2546 MPa
E= 2.546 GPa ( 1 GPa = 1000 MPa)
