1 A specimen of aluminum having a rectangular cross section 10.4 mm x 12.6 mm (0.4094 in. × 0.4961 in.) is pulled in tension with 35900N (8071 Ib;) force, producing only elastic deformation. The elastic modulus for aluminum is 69 GPa (or 10 x 106 psi). Calculate theresulting strain.

So, stress can be defined as ; σ = PA P = Tension pull = 35900 A = 10.4 * 12.6 = 131.04 mm2 σ =…

A specimen of aluminum having a rectangular cross section 10.4 mm × 12.6 mm (0.4094 in. × 0.4961 in.) is pulled in tension with 35900 N (8071 lb;) force, producing only elastic deformation. The elastic modulus for aluminum is 69 GPa (or 10 x 106 psi). Calculate the resulting strain.

A specimen of aluminum having a rectangular cross section 10.4 mm × 12.6 mm (0.4094 in. × 0.4961 in.) is pulled in tension with 35900 N (8071 lb;) force, producing only elastic deformation. The elastic modulus for aluminum is 69 GPa (or 10 x 106 psi). Calculate the resulting strain.

Steel Design (Activate Learning with these NEW titles from Engineering!)

6th Edition

ISBN:9781337094740

Author:Segui, William T.

Publisher:Segui, William T.

Chapter1: Introduction

Section: Chapter Questions

Problem 1.5.2P: A tensile test was performed on a metal specimen having a circular cross section with a diameter of...

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Question

A specimen of aluminum having a rectangular cross section 10.4 mm x 12.6 mm (0.4094 in. × 0.4961 in.) is pulled in tension with 35900
N (8071 Ib;) force, producing only elastic deformation. The elastic modulus for aluminum is 69 GPa (or 10 x 106 psi). Calculate the
resulting strain.

Expert Solution

Step 1 Calculate the amount of Stress developed

So, stress can be defined as ;

$σ = \frac{P}{A}$

P = Tension pull = 35900

A = 10.4 * 12.6 = 131.04 mm2

$σ = \frac{35900}{131.04} = 273.96 M P a$

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