急問!!Calculation about stress and strain
A 0.505-cm-diameter aluminium(Al) alloy test bar is subjected to a load of 2000N.If the diameter of the bar is 0.490cm under this load, determine (a)the stress and (b)strain under this load.
Assume no volume change of the bar during extension.
- 1 十年前最愛解答
a) Stress = Force (N) / Area (cm2)
= 2000 / 3.14 x 0.505x 0.505
= 2497.6 N/cm2
= 24.98 N / mm2
b) (i)Young modulus =(ii)Stress / (iii) Strain
(i) Young modulus of (Al) , MUST has been given by the question.
(ii) Stress = 2000 / 3.14 x 0.49 x 0.49
= 2652.8 N/cm2
=26.53 N / mm2
So, (iii) = 2652.8 N/cm2 / Young modulus pf Al
2006-12-11 17:58:50 補充：
Sorry , my mistake about the diameter. Please re-calculate the Area(s) . Area = 3.14 x r2