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?

Stress will be same for both specimen.

Titanium alloy have more strain as compare to stainless steel.

Stainless steel is more stronger .

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

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)

Specimen 2:

E = 200 GPa

σ =  ε .E

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

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 .

energy can cross the boundaries of a closed system in the form of heat and mass

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