Question

Two dogbone specimens of identical geometry but made of two different materials: steel and aluminum are tested under tension at a fixed load rate. If the displacement rate needed to achieve the specified load rate for aluminum specimen is 0.001 inch/min, determine the displacement rate that must be applied on the steel specimen to achieve the specified load rate. Given, Young's modulus (E) of steel is 29,000 ksi and that of aluminum is 10,000 ksi.

Answer 1

vsteel = 0.00034483 inch/min

Step-by-step explanation:

Given

vAl = 0.001 inch/min

EAl = 10,000 ksi

Esteel = 29,000 ksi

P is the same value for the dogbone specimens

A is the same value for the dogbone specimens

L is the same value for the dogbone specimens

We can apply

ΔL = P*L/(A*E)

⇒ ΔL*E = P*L/A

then

ΔLAl*EAl = ΔLsteel*Esteel

ΔLAl*10,000 ksi = ΔLsteel*29,000 ksi

⇒ ΔLAl = 2.9*ΔLsteel

If ΔLAl = 2.9*ΔLsteel we have

vAl = 2.9*vsteel

0.001 inch/min = 2.9*vsteel

⇒ vsteel = 0.00034483 inch/min

Answer 2

`\dot L_(steel) = 3.448* 10^(-4)\,(in)/(min)`

Step-by-step explanation:

The Young's module is:

`E = (\sigma)/((\Delta L)/(L_(o)) )`

`E = (\sigma\cdot L_(o))/(\dot L \cdot \Delta t)`

Let assume that both specimens have the same geometry and load rate. Then:

`E_(aluminium) \cdot \dot L_(aluminium) = E_(steel) \cdot \dot L_(steel)`

The displacement rate for steel is:

`\dot L_(steel) = (E_(aluminium))/(E_(steel))\cdot \dot L_(aluminium)`

`\dot L_(steel) = \left((10000\,ksi)/(29000\,ksi)\right)\cdot (0.001\,(in)/(min) )`

`\dot L_(steel) = 3.448* 10^(-4)\,(in)/(min)`