Question

A bar having a length of 5 in. and cross-sectional area of0.7 m' is subjected to an axial force of 8000N. If the bar stretches 0.002m., determine the modulus of elasticity of the material. The material has linear- elastic behavior.​

Answer 1

Step-by-step explanation:

he modulus of elasticity (E) can be calculated using the formula:Stress = Force / AreaStrain = Change in length / Initial lengthModulus of Elasticity (E) = Stress / StrainWe have the Force = 8000 N, Area = 0.7 m^2, Change in length = 0.002 m and initial length = 5 in = 0.127 mStress = Force / Area = 8000 N / 0.7 m^2 = 11428.57 N/m^2Strain = Change in length / Initial length = 0.002 m / 0.127 m = 0.0157Modulus of elasticity (E) = Stress / Strain = 11428.57 N/m^2 / 0.0157 = 727,279.9 N/m^2So the modulus of elasticity for the material of the bar is 727,279.9 N/m^2This is the ratio of the applied stress to the corresponding strain within the elastic limit, which is a measure of a material's resistance to deformation.