Question

13. A dumbbell test specimen with a gauge length of 100 mm and a circular cross-section with a diameter of 10 mm is loaded in tension in a tensile testing machine. Upon application of a tensile force of 1000 N, the gage length extends to 100.5 mm. Calculate the stress and strain in the gage section. If the deformation is all within the linear elastic range, what is the elastic modulus of the material?

Answer 1

ε = 0.005

σ = 12.73 MPa

E= 2.546 GPa

Step-by-step 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

`\varepsilon =(0.5)/(100)`

ε = 0.005

We know that stress given as

`\sigma  =(P)/(A)`

`A=(\pi d^2)/(4)`

`A=(\pi * 10^2)/(4)\ mm^2`

`A=78.53\ mm^2`

`\sigma  =(1000)/(78.53)\ MPa`

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)