Question

You have two specimens both of which are cylindrical with a diameter of 20 mm and a length of 200 mm. One is made of a titanium alloy (E = 100 GPa) and the other is made out of stainless steel (E = 200 GPa). Both of them are subjected to a tensile force of 500 N. Which of them will have higher stress? Which will develop higher strain? If possible, can you determine which of the specimens will be stronger?

Answer 1

Stress will be same for both specimen.

Titanium alloy have more strain as compare to stainless steel.

Stainless steel is more stronger .

Step-by-step explanation:

d = 20 mm

L= 200 mm

P= 500 N

Specimen 1:

E = 100 GPa

Specimen 2:

E = 200 GPa

We know that stress given as

`\sigma  =(P)/(A)`

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

`A=(\pi * 20^2)/(4)\ mm^2`

`A=314.15\ mm^2`

`\sigma  =(500)/(314.15)\ MPa`

σ = 1.59 MPa

Both specimen have same force and same dimensions so the stress will be same for both specimen.

We know that with in the elastic limit

σ = ε .E

E= Modulus of elasticity

σ= stress

ε = strain

Specimen 1:

E = 100 GPa

σ = ε .E

1.59 = 100 x 1000 x ε ( 1 GPa = 1000 MPa)

`\varepsilon =1.59* 10^(-5)`

Specimen 2:

E = 200 GPa

σ = ε .E

1.59 = 200 x 1000 x ε ( 1 GPa = 1000 MPa)

`\varepsilon =0.795* 10^(-5)`

`\varepsilon_1 =1.59* 10^(-5)`

`\varepsilon_2 =0.795* 10^(-5)`

So we can say that titanium alloy have more strain as compare to stainless steel.

E stainless steel is more then titanium alloy that is why stainless steel is more stronger .