Ask Question
9 September, 10:57

A bar having a length of 5 in. and cross-sectional area of 0.7 i n. 2 is subjected to an axial force of 8000 lb. If the bar stretches 0.002 in., determine the modulus of elasticity of the material. The material has linear elastic behavior.

Answers (1)
  1. 9 September, 11:24
    0
    E=1.969 * 10¹¹ Pa

    Explanation:

    The formula to apply is;

    E=F*L/A*ΔL

    where

    E=Young modulus of elasticity

    F=Force in newtons

    L=Original length in meters, m

    A=area in square meters m²

    ΔL = Change in length in meters, m

    Given

    F = 8000 lb = 8000*4.448 = 35584 N

    L = 5 in = 0.127 m

    A = 0.7 in² = 0.0004516 m²

    ΔL = 0.002 in = 5.08e-5 m

    Applying the formula

    E = (35584 * 0.127) / (0.0004516*5.08e-5)

    E=1.969 * 10¹¹ Pa
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “A bar having a length of 5 in. and cross-sectional area of 0.7 i n. 2 is subjected to an axial force of 8000 lb. If the bar stretches 0.002 ...” in 📗 Engineering if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers
Sign In
Ask Question