   Chapter 1, Problem 1.5.4P ### Steel Design (Activate Learning wi...

6th Edition
Segui + 1 other
ISBN: 9781337094740

#### Solutions

Chapter
Section ### Steel Design (Activate Learning wi...

6th Edition
Segui + 1 other
ISBN: 9781337094740
Textbook Problem

# A tensile test was performed on a metal specimen with a diameter of 1 2 inch and a gage length (the length over which the elongation is measured) of 4 inches. The dam were plotted on a load-displacement graph. P vs. Δ L . A best-fit line was drawn through the points, and the slope of the straight-line portion was calculated to be P / Δ L = 1392 kips/in. What is the modulus of elasticity?

To determine

The modulus of elasticity of the metal specimen.

Explanation

Given:

The diameter of metal specimen is 12inch.

The gage length is 4inches.

The slope of straight line is 1392kips/in.

Concept Used:

Write the expression for deformation.

ΔL=PLAE

Rearrange the equation to calculate the young’s modulus.

E=PLA×ΔL      ...... (I)

Here, the modulus of elasticity is E, the slope of straight line from load-displacement graph is PΔL, the gage length is L and the cross-sectional area is A.

Calculation:

Calculate the cross-sectional area of metal specimen.

A=π4×d2      ...... (II)

Here, diameter of the metal specimen is d.

Substitute 12inch for d in Equation (II).

A=π4×(0

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