Question

At the proportional limit, a 2 in. gage length of a 0.500 in. diameter alloy rod has elongated 0.0035 in. and the diameter has been reduced by 0.0003 in. The total tension force on the rod was 5.45 kips.

Determine the following properties of the material:

(a) the proportional limit.

(b) the modulus of elasticity

(c) Poisson's ratio.

Answer 1

a) The proportional limit is 2.99MPa.

b) The modulus of elasticity is 0.427GPa.

C) The poisson´s ratio is 0.021

Step-by-step explanation:

a) The proportional limit is the maximum stress for wich a tension bar stops acting as a linear material in a stress-strain curve. So this stress can be obtained as:

`\sigma_(p)=(F_(p))/(S_(o)) =(5.45Kips)/(\pi(2in)^2)=2.99MPa`

b) The modulus of elasticity E is the proportion between the strain and the stress in the linear section of the stress-strain curve.

`\sigma=E\epsilon`

The strain in for the proportional limit is:

`\epsilon_0=(\Delta L)/(L)=(0.0035in)/(0.5in)=7 \cdot 10^(-3)`

Therefore:

`E=(\sigma_0)/(\epsilon_0) =(2.99MPa)/(7 \cdot 10^(-3)) =0.427GPa`

c) The Poisson´s ratio is the negative proportion between the transverse strain and axial strain.

In this case, the transverse strain is:

`\epsilon_(tr0)=(\Delta D)/(D)=(0.0003in)/(2in)=1.5 \cdot 10^(-4)`

So the poisson´s ratio is:

`\\u=-(d\epsilon_(tr))/(d\epsilon_(l))\approx-(\epsilon_(tr))/(\epsilon_(l))=-(1.5 \cdot 10^(-4))/(-7 \cdot 10^(-3))=0.021`