Question

A nickel alloy rod having an elastic modulus of 200 GPa and an original diameter of 10 mm experiences only elastic deformation when a tensile load of 80000 N is applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is 0.25 mm.

Answer 1

Answer:

50 mm

Explanation:

$200 GPa = 200\times10^9 Pa$

10 mm in diameter = 5 mm in radius or 0.005 m

So the cross-section area of the rod is

$A = \pi r^2 = \pi 0.005^2 =  0.00008 m^2$

The stress on the rod is the ratio of force and cross-section area

$\sigma = F / A = 80000 / 0.00008 = 10^9 Pa$

So the maximum strain when this stress is applied is

$\epsilon = \sigma / E = 10^9/(200\times10^9 Pa) = 1/200 = 0.005$

If the maximum allowable deformation is 0.25 mm, then the maximum original length is

$\epsilon = \Delta L / L$

$L = \Delta L / \epsilon = 0.25 / 0.005 = 50 mm$

Answer 2

Answer:

$l_{o}=0.05m$

Explanation:

Given data

Diameter d=10 mm

Force F=80000 N

Elongation Δl=0.25 mm

Elastic E=200 GPa

We want the length l₀ before deformation.The modulus of elasticity is property of material and could be calculated by Hooks Law which is expresses a relationship between the stress and strain for elastic deformation as:

E=σ/ε

Where ε is strain resulted from applied force and represent as:

ε=Δl/l₀

l₀=Δl/ε.......(i)

As the stress σ  the force per unit area so we can find its value first

σ=F/A

σ=(80000N)/π*r²

σ=(80000N)/π*(5×10⁻³m)²

σ=1018591636 Pa

By using the expressing ε from eq(i)

l₀=Δl/ε

l₀=(Δl×E)/σ

Substitute the given values

So

l₀=(Δl×E)/σ

$=\frac{0.25*10^{-3}m(200*10^{9}Pa)}{1018591636Pa}\\ =0.05m\\l_{o}=0.05m$