29. During laboratory testing of an 8-inch aluminum-alloy rod with a cross sectional area of 0.32in², a tensile load of 40,000 lbs.is applied. If the croSs sectional area decreases to 0.3188in? and lengthens to 8.021 inches, what is the Modulus of Elasticity and Poisson's Ratio? E= 6/8 v =-e(lateral)/e(axial)

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 1.5.5P: The results of a tensile test are shown in Table 1.5.2. The test was performed on a metal specimen...
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During laboratory testing of an 8-inch aluminum-alloy rod with a cross sectional
area of 0.32in, a tensile load of 40,000 lbs.is applied. If the cross sectional area
decreases to 0.3188in and lengthens to 8.021 inches, what is the Modulus of
Elasticity and Poisson's Ratio?
29.
E= 6/8
v =-ɛ(lateral)/E(axial)
Transcribed Image Text:During laboratory testing of an 8-inch aluminum-alloy rod with a cross sectional area of 0.32in, a tensile load of 40,000 lbs.is applied. If the cross sectional area decreases to 0.3188in and lengthens to 8.021 inches, what is the Modulus of Elasticity and Poisson's Ratio? 29. E= 6/8 v =-ɛ(lateral)/E(axial)
Expert Solution
Step 1

Given:-

The length of the aluminum-alloy rod during testing = 8 inch 

The cross-sectional area of the rod = 0.32 in2 

The magnitude of the load applied = 40,000 lbs

The decreased cross-sectional area = 0.3188 in2 

The changed length = 8.021 inches

To determine:-

The modulus of elasticity and Poisson's ratio 

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