Topic: Strength of Materials
An aluminum tensile test specimen has a diameter, do = 25 mm and a gauge length of Lo = 250 mm. If a force of 175 kN elongates the gauge length by 1.25 mm, the modulus of elasticity of the material is nearly
- A. 71 GPa
- B. 71 MPa
- C. 142 GPa
- D. 142 MPa
Answer: A. 71 GPa
Explanation: The cross-sectional area A = (pi/4) * do² = (pi/4) * 25^2 = 490.87 mm². Using the elongation formula delta_L = (P * L) / (A * E), we can rearrange for Modulus of Elasticity E = (P * L) / (A * delta_L). Substituting values gives E = (175000 N * 250 mm) / (490.87 mm² * 1.25 mm) = 71301 N/mm² (or MPa). Converting to GPa gives approximately 71 GPa.