MATERIALS SCI + ENGR: INT W/ACCESS
MATERIALS SCI + ENGR: INT W/ACCESS
10th Edition
ISBN: 9781119808084
Author: Callister
Publisher: WILEY
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Chapter 6, Problem 1QAP
To determine

To derive:

The Equation 6.4a and 6.4b using mechanics of materials principles.

Expert Solution & Answer
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Answer to Problem 1QAP

The Equation 6.4a and 6.4b using mechanics of materials principles is shown in conclusion.

Explanation of Solution

Write the expression for engineering stress.

σ=PA (I)

Here, the tensile force is P and cross-sectional area is A.

Write the expression for shear stress.

τ=VA (II)

Here, the normal force is P and shear forces is V.

Conclusion:

Show the diagram of a block element of material of cross-sectional area A that is subjected to a tensile force P, the orientations of the applied stress, the normal stress to this plane, as well as the shear stress taken parallel to this inclined plane.

MATERIALS SCI + ENGR: INT W/ACCESS, Chapter 6, Problem 1QAP , additional homework tip  1

MATERIALS SCI + ENGR: INT W/ACCESS, Chapter 6, Problem 1QAP , additional homework tip  2

MATERIALS SCI + ENGR: INT W/ACCESS, Chapter 6, Problem 1QAP , additional homework tip  3

For static equilibrium in the x-direction.

Fx=0

Which means that

PPcosθ=0P=Pcosθ

Write the expression for the stress σ in terms of P and A using the above expression and the relationship between A and A .

σ=PA=PcosθAcosθ=PAcos2θ=σcos2θ

Write the expression for τ .

τ=VA=PsinθAcosθ=PAsinθcosθ=σsinθcosθ

Thus, the τ value is σsinθcosθ_ .

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