Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN: 9781337093347
Author: Barry J. Goodno, James M. Gere
Publisher: Cengage Learning
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Chapter 7, Problem 7.7.18P

(a)

To determine

The strains for an element oriented at θ=50°.

(a)

Expert Solution
Check Mark

Answer to Problem 7.7.18P

The normal strain in x direction is 14.69×104.

The normal strain along y direction is 9.07×104.

The shear strain is 7.18×104in x-y plane.

Explanation of Solution

Given:

Stress along x direction is 150MPaand stress along y direction is 210MPaand shear stress along x-y plane is 16MPa.

Write the Equation for strain along x axis

εx=1E(σxνσy)(I)

Here, the normal strain along x direction is εx, direct stress along x direction is σxand direct stress along y direction is σyPoissons ratio is νand Young modulus is E.

Write the Equation for strain along y axis

εy=1E(σyνσx)(II)

Here, the normal strain along the y direction is εy.

Write the expression for the shear strain ix-y plane

γxy=2τxy(1+ν)E(III)

Here, modulus of rigidity is Gand the shear strain along x-y plane is γxy.

Write the expression for strain along x direction

εx1=εx+εy2+εxεy2cos2θ+γxy2sin2θ(IV)

Write the expression for strain along y direction

εy1=εx+εyεx1(V)

Write the expression for shear strain

γx1y12=εxεy2sin2θ+γxy2cos2θ(VI)

Here, the shear strain along x-y plane is γxy.

Calculation:

Substitute 150MPafor σx, 210MPafor σyand 0.34for νand 100GPafor Ein Equation (I).

εx=[(150MPa)((0.34)×(210MPa))]100GPa=(150MPa)((0.34)×(210MPa))100GPa(1000MPa1GPa)=1105MPa[(150MPa)((0.34)×(210MPa))]=7.86×104

Substitute 150MPafor σx, 210MPafor σyand 0.34for νan 100GPafor Ein Equation (II).

εy=(210MPa)((0.34)×(150MPa))100GPa=1100GPa(1000MPa1GPa)[(210MPa)((0.34)×(150MPa))]=159MPa105MPa=15.9×104

Substitute 16MPafor τxy, 0.34for νand 100GPafor Ein Equation (III).

γxy=2(16MPa)(1+0.34)100GPa=2(16MPa)(1+0.34)100GPa(1000MPa1GPa)=42.88MPa105MPa=4.29×104

Substitute 7.86×104for εx, 15.9×104for εyand 4.29×104for γxyin Equation (IV).

εx1=((7.86×10415.9×104)2+(7.86×104+15.9×104)2cos(2×(50°))+(4.29×104)2sin(2×(50°)))=(11.88×1040.7×1042.11×104)=14.69×104

Substitute 7.86×104for εx, 15.9×104for εyand 14.69×104for εx1in Equation (V).

εy1=7.86×10415.9×104+14.69×104=23.76×104+14.69×104=9.07×104

Substitute 7.86×104for εx, 15.9×104for εyand 4.29×104for γxyin Equation (VI).

γx1y12=[((7.86×104+15.9×104)2sin(2×50°))+((4.29×1042)cos(2×50°))]γx1y12=3.59×104γx1y1=7.18×104

In below figure the normal strains are shown

:

    Mechanics of Materials (MindTap Course List), Chapter 7, Problem 7.7.18P , additional homework tip  1

    Figure (1)

Conclusion:

The normal strain along x direction is 14.69×104.

The normal strain along y direction is 9.07×104.

The shear strain along x-y plane is 7.18×104.

(b)

To determine

The principal strains.

(b)

Expert Solution
Check Mark

Answer to Problem 7.7.18P

The maximum principal strain is 7.32×104, and the minimum principal strain is 16.44×104.

Explanation of Solution

Write expression for the principal strains.

ε1,2=(εx+εy2)+(εxεy2)2+(γxy2)2(VII)

Here, the maximum principal strain is ε1and the minimum principal strain is ε2.

Write the expression for the principal angle.

tan2θp=γxyεxεy(VIII)

Here, θpis the principal angle.

Calculation:

Substitute 7.86×104for εx, 15.9×104for εyand 4.29×104for γxyin Equation (VII).

ε1,2=((7.86×10415.9×104)2)±((7.86×104+15.9×104)2)2+(4.29×1042)2=11.88×104±(16.16×108)+(4.60×108)=11.88×104±4.556×104

While taking positive sign you get maximum principal strain:

ε1=(11.88+4.556)×104=7.32×104

While taking negative sign you get minimum principal strain.

ε1=(11.884.556)×104=16.44×104

Substitute 7.86×104for εx, 15.9×104for εyand 4.29×104for γxyin Equation (VIII).

tan2θp=4.29×104(7.86×104+15.9×104)tan2θp=0.53358

1802θp=28.1°2θp=151.9°θp=75.95°

In below figure principal strains and principal angle are shown:

    Mechanics of Materials (MindTap Course List), Chapter 7, Problem 7.7.18P , additional homework tip  2

    Figure (2)

Conclusion:

The maximum principal strain is 7.32×104and minimum principal strain is 16.44×104.

(c)

To determine

The maximum shear strain.

(c)

Expert Solution
Check Mark

Answer to Problem 7.7.18P

The maximum shear strain is 9.12×104.

Explanation of Solution

Write the expression for maximum shear strain,

γmax2=(εxεy2)2+(γxy2)2(IX)

Here, γmaxis the maximum shear strain.

Write expression for first shear angle

θs1=θp145° (X)

Here, θs1is the first shear angle.

Write expression for second shear angle

θs2=θs145° (XI)

Here, θs2is the second shear angle.

Write the Equation for average strain

εavg=εx+εy2(XII)

Write expression for maximum shear strain.

γmin=γmax (XIII)

Calculation:

Substitute 7.86×104for εx, 15.9×104for εyand 4.29×104for γxyin Equation (IX).

γmax2=((7.86×104+15.9×104)2)2+(4.29×1042)2γmax2=((7.86+15.9)×1042)2+(2.145×104)2γmax2=(4.56×104)γmax=9.12×104

Substitute 165.95°for θp1in Equation (X).

θs1=165.95°45°=120.95°

Substitute 120.95°for θs1in Equation (XI)

θs2=120.95°90°=30.95°

Substitute 7.86×104for εx, 15.9×104for εyin Equation (XII).

εavg=7.86×10415.9×1042=23.76×1042=11.88×104

Substitute 9.12×104for γmax.in Equation (XIII).

γmin=γmax=9.12×104

In below figure average minimum strain and shear angle are shown:

    Mechanics of Materials (MindTap Course List), Chapter 7, Problem 7.7.18P , additional homework tip  3

    Figure (3)

Conclusion:

The maximum shear strain is 9.12×104.

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Chapter 7 Solutions

Mechanics of Materials (MindTap Course List)

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