Mechanics of Materials, 7th Edition
Mechanics of Materials, 7th Edition
7th Edition
ISBN: 9780073398235
Author: Ferdinand P. Beer, E. Russell Johnston Jr., John T. DeWolf, David F. Mazurek
Publisher: McGraw-Hill Education
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Chapter 2.9, Problem 91P

(a)

To determine

Find the change in length of the cube in the x direction.

(a)

Expert Solution
Check Mark

Answer to Problem 91P

The change in length of the cube in the x direction is 0.0303mm_.

Explanation of Solution

Given information:

The modulus of elasticity in x-direction Ex is 50GPa.

The modulus of elasticity y-direction Ey is 15.2GPa.

The modulus of elasticity in z-direction Ez is 15.2GPa.

The strain along xz axis is 0.254.

The strain along xy axis is 0.254

The strain along zy axis is 0.428.

Calculation:

Write the stress to strain Equation along x direction as follows:

εx=σxExνyxσyEyνzxσzEz (1)

Here, σx,σy,σz is normal stress along x, y, z axis respectively.

Write the stress to strain Equation along y direction as follows:

εy=νxyσxEx+σyEyνzyσzEz (2)

Write the stress to strain Equation along z direction as follows:

εz=νxzσxExνyzσyEy+σzEz (3)

Equate the stress Equation along xy axis.

νxyEx=νyxEy (4)

Equate the stress Equation along yz axis.

νyzEy=νzyEz (5)

Equate the stress Equation along zx axis.

νzxEz=νxzEx (6)

Apply the constraint conditions as follows:

εy=0εz=0

Substitute 0 for εy in Equation (2).

0=νxyσxEx+σyEyνzyσzEz1EyσyνzyEzσz=νxyExσx (7)

Substitute 0 for εz in Equation (3).

0=νxzσxExνyzσyEy+σzEzνyzEyσy+1Ezσz=νxzExσx (8)

Substitute 15.2GPa for Ey, 0.428 for νzy, 15.2GPa for Ez, 0.254 for νxz, and 50GPa for Ex in Equation (7).

115.2σy0.42815.2σz=0.25450σx115.2σy0.42815.2σz=5.08×103σxσy0.428σz=0.077216σx

Substitute 15.2GPa for Ey, 0.428 for vzy, 15.2GPa for Ez, 0.254 for νxy, and 50GPa for Ex in Equation (8).

0.42815.2σy+115.2σz=0.25450σx0.42815.2σy+115.2σz=5.08×103σx0.428σy+σz=0.077216σx

Solving Equation to get values,

σy=σz=0.134993σx

Apply the Equation (4) and (5) in Equation 1 as follows:

εx=1ExσxνxyExσyνxzEσz

Substitute 0.254 for νxy, 0.254 for νxz, 0.134993σx for σy, 0.134993σx for σz.

εx=1Exσx0.254Ex(0.134993σx)0.254E(σz)=1Ex[1(0.254)(0.134993)(0.254)(0.134993)]σx=1Ex[0.93142σx] (9)

Calculate the cross sectional area of cube as follows:

A=a2 (10)

Here, a is the sides.

Substitute 40mm for a in Equation (10).

A=(40)2=1,600mm2(1m103m)2=1,600×106m2

Find the normal stress along x axis as follows:

σx=PA (11)

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

Substitute 65kN for P and 1,600×106m2 for A in Equation (11).

σx=65kN((103N1kN))1,600×106m2=65×1031,600×106=40.625×106Pa

Find the strain along x axis as follows:

Substitute 40.625×106Pa for σx and 50GPa for Ex in Equation (9).

εx=150GPa(109Pa1GPa)[0.93142×40.625×106]=150×109[0.93142×40.625×106]=150×109[37,838,937.5]=756.78×106

Determine the change in length of the cube in the x direction using the relation:

δx=Lxεx (12)

Here, Lx is side length and εx strain along x axis.

Substitute 40mm for Lx and 756.78×106 for εx in Equation (12).

δx=40×756.78×106=0.03027=0.0303mm

Thus, the change in length of the cube in the x direction is 0.0303mm_.

(b)

To determine

The stress values σx,σy,σz.

(b)

Expert Solution
Check Mark

Answer to Problem 91P

The stress value σx is 40.6MPa_

The stress value σy is 5.48MPa_

The stress value σz is 5.48MPa_

Explanation of Solution

Calculation:

Find the normal stress along x axis as follows:

σx=PA (13)

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

Substitute 65kN for P and 1,600×106m2 for A in Equation (13).

σx=65kN((103N1kN))1,600×106m2=65×1031,600×106=40.625×106Pa

Thus, the stress value σx is 40.6MPa_

Find the normal stress along y,z axis as follows:

σy=σz=0.134993σx

Substitute 40.625×106Pa for σx.

σy=σz=0.134993×40.625×106=5,484,090.625Pa(1MPa106Pa)=5.48MPa

Thus, the stress value σy is 5.48MPa_

Thus, the stress value σz is 5.48MPa_.

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Chapter 2 Solutions

Mechanics of Materials, 7th Edition

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