In: Mechanical Engineering
10 cm of copper with square cross section of 1x1 cm exposed to
200 KN tension force
caused 15% permanent reduction in cross section area , if
the fracture got on this point,
calculate the following:
1 – the ductility
2 – stress
3 – true stress
4- fracture stress
A0 = 1 cm2
L0 = 10 cm
Af = 0.85 cm2
For a tensile specimen volume will be constant.
A0*L0 = Af*Lf
1*10 = 0.85*Lf
Lf = 11.77 cm
1 – the ductility
Ductility is the strain at fracture. or percentage elongation at fracture.
2 –
stress
3 – true
stress
true stress at fracture. area at fracture i considered.
4- fracture
stress
Fracture stress is the stress at which fracture occurs. As from the question fracture happens when 200 kN force is getting applied. So it will be equal to engineering stress at this point. That is force/initial area
= 200000/100 mm2 = 2000 N/mm2 = 2000 MPa
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