Question

Gauge length prior to starting =50mm

Length after fracture =66.4mm

0.1% proof load=18.5kn

Maximum load=26.6kn

Original diameter =10mm

Calculate using the above data the

2. 0.1 proof stress

3. Tensile strength

Answer 1

1.

0.1% proof stress means 0.1% of E (Young's modulus of elasticity):

$E=\sigma/\varepsilon$

stress:

$\sigma=F/A$

where F is 0.1% proof load,

A is cross-sectional area

$\sigma=\frac{18.5\cdot10^3}{\pi(10\cdot10^{-3})^2}=5.89\cdot10^7$ Pa

stain:

$\varepsilon=x/L$

where x is extension,

L is length

$\varepsilon=\frac{66.4-50}{50}=0.328$

$E=5.89\cdot10^7/0.328=1.796\cdot10^8$ Pa

2.

Tensile strength:

$F_{max}/A=\frac{26.6\cdot10^3}{\pi(10\cdot10^{-3})^2}=8.47\cdot10^7$ Pa