Asked by dmkp
A mild steel bar of width 10mm is subjected to a tensile stress of 3,0x105 Pa. Given that Young’s modulus for mild steel is 200 GPa and Poisson’s ration for mild steel is 0.31, calculate the change in width of the bar. Is the width increased or reduced?
Answers
Answered by
drwls
I suggest you review the definition of Poisson's ratio at
http://www.engineeringtoolbox.com/poissons-ratio-d_1224.html
It is the ratio of the transverse strain to the longitudinal strain, with a minus sign stuck on. If a material gets stretched in uniaxial tension, it simultaneously gets thinner in the two perpendicular directions.
In your case, the strain along the direction of the applied tensile force is
dL/L = Stress/E = 3*10^5/200*10^9
= 1.6^10^-6
For mild steel, the dimensionless strain contraction in width is 0.31 times that number, or 4.65*10^-7.
That gets multiplied by 10 mm for the actual width reduction: 4.65*10^-6 mm
http://www.engineeringtoolbox.com/poissons-ratio-d_1224.html
It is the ratio of the transverse strain to the longitudinal strain, with a minus sign stuck on. If a material gets stretched in uniaxial tension, it simultaneously gets thinner in the two perpendicular directions.
In your case, the strain along the direction of the applied tensile force is
dL/L = Stress/E = 3*10^5/200*10^9
= 1.6^10^-6
For mild steel, the dimensionless strain contraction in width is 0.31 times that number, or 4.65*10^-7.
That gets multiplied by 10 mm for the actual width reduction: 4.65*10^-6 mm
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