Find the orientation and magnitude of the principal strains in the plane.

Question

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Answer 1

Question

Chapter 7.9, Problem 149P

(a)

To determine

Find the orientation and magnitude of the principal strains in the plane.

(a)

Expert Solution

Answer to Problem 149P

The orientation of the principal strains is $θp=30°and 120°_$ .

The in-plane maximum principal strain is $εmax=560×10−6 in./in._$ .

The in-plane minimum principal strain is $εmin=−140×10−6 in./in._$ .

Explanation of Solution

Given information:

The normal strain in Rosette 1 is $ε1=−93.1×10−6 in./in.$ .

The Rosette 1 makes an angle with the x-axis is $θ1=−75°$ .

The normal strain in Rosette 2 is $ε2=+385×10−6 in./in.$ .

The Rosette 2 makes an angle with the x-axis is $θ2=0°$ .

The normal strain in Rosette 3 is $ε3=+210×10−6 in./in.$ .

The Rosette 3 makes an angle with the x-axis is $θ3=75°$ .

Calculation:

Write the equation for the normal strain 1 as follows;

$ε1=εxcos2θ1+εysin2θ1+γxysinθ1cosθ1$

Substitute $−93.1×10−6 in./in.$ for $ε1$ and $−75°$ for $θ1$ .

$−93.1×10−6=εxcos2(−75°)+εysin2(−75°)+γxysin(−75°)cos(−75°)$ (1)

Write the equation for the normal strain 2 as follows;

$ε2=εxcos2θ2+εysin2θ2+γxysinθ2cosθ2$

Substitute $+385×10−6 in./in.$ for $ε2$ and $0°$ for $θ2$ .

$385×10−6=εxcos2(0°)+εysin2(0°)+γxysin(0°)cos(0°)=εx+0+0εx=385×10−6 in./in.$

Write the equation for the normal strain 3 as follows;

$ε3=εxcos2θ3+εysin2θ3+γxysinθ3cosθ3$

Substitute $+210×10−6 in./in.$ for $ε3$ and $75°$ for $θ3$ .

$+210×10−6=εxcos2(75°)+εysin2(75°)+γxysin(75°)cos(75°)$ (2)

Add equation (1) and (2).

$−93.1×10−6+210×10−6=(εxcos2(−75°)+εysin2(−75°)+γxysin(−75°)cos(−75°)+εxcos2(75°)+εysin2(75°)+γxysin(75°)cos(75°))116.9×10−6=(εxcos2(−75°)+εysin2(−75°)−0.25γxy+εxcos2(75°)+εysin2(75°)+0.25γxy)=εxcos2(−75°)+εysin2(−75°)+εxcos2(75°)+εysin2(75°)$

Substitute $385×10−6 in./in.$ for $εx$ .

$116.9×10−6=((385×10−6)cos2(−75°)+εysin2(−75°)+(385×10−6)cos2(75°)+εysin2(75°))=5.158×10−5+1.866εyεy=35×10−6 in./in.$

Substitute $385×10−6 in./in.$ for $εx$ and $35×10−6 in./in.$ for $εy$ in Equation (2).

$+210×10−6=(385×10−6)cos2(75°)+(35×10−6)sin2(75°)+γxysin(75°)cos(75°)=2.579×10−5+3.266×10−5+0.25γxyγxy=606.2×10−6 in./in.$

Find the orientation $(θp)$ of the principal strains using the relation.

$tan2θp=γxyεx−εy$

Substitute $606.2×10−6 in./in.$ for $γxy$ , $385×10−6 in./in.$ for $εx$ , and $35×10−6 in./in.$ for $εy$

$tan2θp=606.2×10−6385×10−6−35×10−6θp=30°and 120°$

Therefore, the orientation of the principal strains is $θp=30°and 120°_$ .

Find the average normal strain $(εave)$ using the relation.

$εave=εx+εy2$

Substitute $385×10−6 in./in.$ for $εx$ and $35×10−6 in./in.$ for $εy$ .

$εave=385×10−6+35×10−62=210×10−6 in./in.$

Find the radius of the Mohr’s circle (R) using the equation.

$R=(εx−εy2)2+(γxy2)2$

Substitute $385×10−6 in./in.$ for $εx$ , $35×10−6 in./in.$ for $εy$ , and $606.2×10−6 in./in.$ for $γxy$ .

$R=(385×10−6−35×10−62)2+(606.2×10−62)2=350×10−6 in./in.$

Find the in-plane maximum principal strain $(εmax)$ using the relation.

$εmax=εave+R$

Substitute $210×10−6 in./in.$ for $εave$ and $350×10−6 in./in.$ for R.

$εmax=210×10−6+350×10−6=560×10−6 in./in.$

Find the in-plane minimum principal strain $(εmin)$ using the relation.

$εmin=εave−R$

Substitute $210×10−6 in./in.$ for $εave$ and $350×10−6 in./in.$ for R.

$εmin=210×10−6−350×10−6=−140×10−6 in./in.$

Therefore,

The in-plane maximum principal strain is $εmax=560×10−6 in./in._$ .

The in-plane minimum principal strain is $εmin=−140×10−6 in./in._$ .

(b)

To determine

Find the in-plane maximum shearing strain.

(b)

Expert Solution

Answer to Problem 149P

The in-plane maximum shearing strain is $(γmax)in-plane=700×10−6 in./in._$ .

Explanation of Solution

Refer to part (a);

The radius of the Mohr’s circle is $R=350×10−6 in./in.$

Find the in-plane maximum shearing strain $(γmax)in-plane$ using the relation.

$(γmax)in-plane=2R$

Substitute $350×10−6 in./in.$ for R.

$(γmax)in-plane=2×350×10−6=700×10−6 in./in.$

Therefore, the in-plane maximum shearing strain is $(γmax)in-plane=700×10−6 in./in._$ .

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Answer 2

Question

Chapter 7.9, Problem 149P

(a)

To determine

Find the orientation and magnitude of the principal strains in the plane.

(a)

Expert Solution

Answer to Problem 149P

The orientation of the principal strains is $θp=30°and 120°_$ .

The in-plane maximum principal strain is $εmax=560×10−6 in./in._$ .

The in-plane minimum principal strain is $εmin=−140×10−6 in./in._$ .

Explanation of Solution

Given information:

The normal strain in Rosette 1 is $ε1=−93.1×10−6 in./in.$ .

The Rosette 1 makes an angle with the x-axis is $θ1=−75°$ .

The normal strain in Rosette 2 is $ε2=+385×10−6 in./in.$ .

The Rosette 2 makes an angle with the x-axis is $θ2=0°$ .

The normal strain in Rosette 3 is $ε3=+210×10−6 in./in.$ .

The Rosette 3 makes an angle with the x-axis is $θ3=75°$ .

Calculation:

Write the equation for the normal strain 1 as follows;

$ε1=εxcos2θ1+εysin2θ1+γxysinθ1cosθ1$

Substitute $−93.1×10−6 in./in.$ for $ε1$ and $−75°$ for $θ1$ .

$−93.1×10−6=εxcos2(−75°)+εysin2(−75°)+γxysin(−75°)cos(−75°)$ (1)

Write the equation for the normal strain 2 as follows;

$ε2=εxcos2θ2+εysin2θ2+γxysinθ2cosθ2$

Substitute $+385×10−6 in./in.$ for $ε2$ and $0°$ for $θ2$ .

$385×10−6=εxcos2(0°)+εysin2(0°)+γxysin(0°)cos(0°)=εx+0+0εx=385×10−6 in./in.$

Write the equation for the normal strain 3 as follows;

$ε3=εxcos2θ3+εysin2θ3+γxysinθ3cosθ3$

Substitute $+210×10−6 in./in.$ for $ε3$ and $75°$ for $θ3$ .

$+210×10−6=εxcos2(75°)+εysin2(75°)+γxysin(75°)cos(75°)$ (2)

Add equation (1) and (2).

$−93.1×10−6+210×10−6=(εxcos2(−75°)+εysin2(−75°)+γxysin(−75°)cos(−75°)+εxcos2(75°)+εysin2(75°)+γxysin(75°)cos(75°))116.9×10−6=(εxcos2(−75°)+εysin2(−75°)−0.25γxy+εxcos2(75°)+εysin2(75°)+0.25γxy)=εxcos2(−75°)+εysin2(−75°)+εxcos2(75°)+εysin2(75°)$

Substitute $385×10−6 in./in.$ for $εx$ .

$116.9×10−6=((385×10−6)cos2(−75°)+εysin2(−75°)+(385×10−6)cos2(75°)+εysin2(75°))=5.158×10−5+1.866εyεy=35×10−6 in./in.$

Substitute $385×10−6 in./in.$ for $εx$ and $35×10−6 in./in.$ for $εy$ in Equation (2).

$+210×10−6=(385×10−6)cos2(75°)+(35×10−6)sin2(75°)+γxysin(75°)cos(75°)=2.579×10−5+3.266×10−5+0.25γxyγxy=606.2×10−6 in./in.$

Find the orientation $(θp)$ of the principal strains using the relation.

$tan2θp=γxyεx−εy$

Substitute $606.2×10−6 in./in.$ for $γxy$ , $385×10−6 in./in.$ for $εx$ , and $35×10−6 in./in.$ for $εy$

$tan2θp=606.2×10−6385×10−6−35×10−6θp=30°and 120°$

Therefore, the orientation of the principal strains is $θp=30°and 120°_$ .

Find the average normal strain $(εave)$ using the relation.

$εave=εx+εy2$

Substitute $385×10−6 in./in.$ for $εx$ and $35×10−6 in./in.$ for $εy$ .

$εave=385×10−6+35×10−62=210×10−6 in./in.$

Find the radius of the Mohr’s circle (R) using the equation.

$R=(εx−εy2)2+(γxy2)2$

Substitute $385×10−6 in./in.$ for $εx$ , $35×10−6 in./in.$ for $εy$ , and $606.2×10−6 in./in.$ for $γxy$ .

$R=(385×10−6−35×10−62)2+(606.2×10−62)2=350×10−6 in./in.$

Find the in-plane maximum principal strain $(εmax)$ using the relation.

$εmax=εave+R$

Substitute $210×10−6 in./in.$ for $εave$ and $350×10−6 in./in.$ for R.

$εmax=210×10−6+350×10−6=560×10−6 in./in.$

Find the in-plane minimum principal strain $(εmin)$ using the relation.

$εmin=εave−R$

Substitute $210×10−6 in./in.$ for $εave$ and $350×10−6 in./in.$ for R.

$εmin=210×10−6−350×10−6=−140×10−6 in./in.$

Therefore,

The in-plane maximum principal strain is $εmax=560×10−6 in./in._$ .

The in-plane minimum principal strain is $εmin=−140×10−6 in./in._$ .

(b)

To determine

Find the in-plane maximum shearing strain.

(b)

Expert Solution

Answer to Problem 149P

The in-plane maximum shearing strain is $(γmax)in-plane=700×10−6 in./in._$ .

Explanation of Solution

Refer to part (a);

The radius of the Mohr’s circle is $R=350×10−6 in./in.$

Find the in-plane maximum shearing strain $(γmax)in-plane$ using the relation.

$(γmax)in-plane=2R$

Substitute $350×10−6 in./in.$ for R.

$(γmax)in-plane=2×350×10−6=700×10−6 in./in.$

Therefore, the in-plane maximum shearing strain is $(γmax)in-plane=700×10−6 in./in._$ .

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Answer 3

Question

Chapter 7.9, Problem 149P

(a)

To determine

Find the orientation and magnitude of the principal strains in the plane.

(a)

Expert Solution

Answer to Problem 149P

The orientation of the principal strains is $θp=30°and 120°_$ .

The in-plane maximum principal strain is $εmax=560×10−6 in./in._$ .

The in-plane minimum principal strain is $εmin=−140×10−6 in./in._$ .

Explanation of Solution

Given information:

The normal strain in Rosette 1 is $ε1=−93.1×10−6 in./in.$ .

The Rosette 1 makes an angle with the x-axis is $θ1=−75°$ .

The normal strain in Rosette 2 is $ε2=+385×10−6 in./in.$ .

The Rosette 2 makes an angle with the x-axis is $θ2=0°$ .

The normal strain in Rosette 3 is $ε3=+210×10−6 in./in.$ .

The Rosette 3 makes an angle with the x-axis is $θ3=75°$ .

Calculation:

Write the equation for the normal strain 1 as follows;

$ε1=εxcos2θ1+εysin2θ1+γxysinθ1cosθ1$

Substitute $−93.1×10−6 in./in.$ for $ε1$ and $−75°$ for $θ1$ .

$−93.1×10−6=εxcos2(−75°)+εysin2(−75°)+γxysin(−75°)cos(−75°)$ (1)

Write the equation for the normal strain 2 as follows;

$ε2=εxcos2θ2+εysin2θ2+γxysinθ2cosθ2$

Substitute $+385×10−6 in./in.$ for $ε2$ and $0°$ for $θ2$ .

$385×10−6=εxcos2(0°)+εysin2(0°)+γxysin(0°)cos(0°)=εx+0+0εx=385×10−6 in./in.$

Write the equation for the normal strain 3 as follows;

$ε3=εxcos2θ3+εysin2θ3+γxysinθ3cosθ3$

Substitute $+210×10−6 in./in.$ for $ε3$ and $75°$ for $θ3$ .

$+210×10−6=εxcos2(75°)+εysin2(75°)+γxysin(75°)cos(75°)$ (2)

Add equation (1) and (2).

$−93.1×10−6+210×10−6=(εxcos2(−75°)+εysin2(−75°)+γxysin(−75°)cos(−75°)+εxcos2(75°)+εysin2(75°)+γxysin(75°)cos(75°))116.9×10−6=(εxcos2(−75°)+εysin2(−75°)−0.25γxy+εxcos2(75°)+εysin2(75°)+0.25γxy)=εxcos2(−75°)+εysin2(−75°)+εxcos2(75°)+εysin2(75°)$

Substitute $385×10−6 in./in.$ for $εx$ .

$116.9×10−6=((385×10−6)cos2(−75°)+εysin2(−75°)+(385×10−6)cos2(75°)+εysin2(75°))=5.158×10−5+1.866εyεy=35×10−6 in./in.$

Substitute $385×10−6 in./in.$ for $εx$ and $35×10−6 in./in.$ for $εy$ in Equation (2).

$+210×10−6=(385×10−6)cos2(75°)+(35×10−6)sin2(75°)+γxysin(75°)cos(75°)=2.579×10−5+3.266×10−5+0.25γxyγxy=606.2×10−6 in./in.$

Find the orientation $(θp)$ of the principal strains using the relation.

$tan2θp=γxyεx−εy$

Substitute $606.2×10−6 in./in.$ for $γxy$ , $385×10−6 in./in.$ for $εx$ , and $35×10−6 in./in.$ for $εy$

$tan2θp=606.2×10−6385×10−6−35×10−6θp=30°and 120°$

Therefore, the orientation of the principal strains is $θp=30°and 120°_$ .

Find the average normal strain $(εave)$ using the relation.

$εave=εx+εy2$

Substitute $385×10−6 in./in.$ for $εx$ and $35×10−6 in./in.$ for $εy$ .

$εave=385×10−6+35×10−62=210×10−6 in./in.$

Find the radius of the Mohr’s circle (R) using the equation.

$R=(εx−εy2)2+(γxy2)2$

Substitute $385×10−6 in./in.$ for $εx$ , $35×10−6 in./in.$ for $εy$ , and $606.2×10−6 in./in.$ for $γxy$ .

$R=(385×10−6−35×10−62)2+(606.2×10−62)2=350×10−6 in./in.$

Find the in-plane maximum principal strain $(εmax)$ using the relation.

$εmax=εave+R$

Substitute $210×10−6 in./in.$ for $εave$ and $350×10−6 in./in.$ for R.

$εmax=210×10−6+350×10−6=560×10−6 in./in.$

Find the in-plane minimum principal strain $(εmin)$ using the relation.

$εmin=εave−R$

Substitute $210×10−6 in./in.$ for $εave$ and $350×10−6 in./in.$ for R.

$εmin=210×10−6−350×10−6=−140×10−6 in./in.$

Therefore,

The in-plane maximum principal strain is $εmax=560×10−6 in./in._$ .

The in-plane minimum principal strain is $εmin=−140×10−6 in./in._$ .

(b)

To determine

Find the in-plane maximum shearing strain.

(b)

Expert Solution

Answer to Problem 149P

The in-plane maximum shearing strain is $(γmax)in-plane=700×10−6 in./in._$ .

Explanation of Solution

Refer to part (a);

The radius of the Mohr’s circle is $R=350×10−6 in./in.$

Find the in-plane maximum shearing strain $(γmax)in-plane$ using the relation.

$(γmax)in-plane=2R$

Substitute $350×10−6 in./in.$ for R.

$(γmax)in-plane=2×350×10−6=700×10−6 in./in.$

Therefore, the in-plane maximum shearing strain is $(γmax)in-plane=700×10−6 in./in._$ .

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Answer 4

Question

Chapter 7.9, Problem 149P

(a)

To determine

Find the orientation and magnitude of the principal strains in the plane.

(a)

Expert Solution

Answer to Problem 149P

The orientation of the principal strains is $θp=30°and 120°_$ .

The in-plane maximum principal strain is $εmax=560×10−6 in./in._$ .

The in-plane minimum principal strain is $εmin=−140×10−6 in./in._$ .

Explanation of Solution

Given information:

The normal strain in Rosette 1 is $ε1=−93.1×10−6 in./in.$ .

The Rosette 1 makes an angle with the x-axis is $θ1=−75°$ .

The normal strain in Rosette 2 is $ε2=+385×10−6 in./in.$ .

The Rosette 2 makes an angle with the x-axis is $θ2=0°$ .

The normal strain in Rosette 3 is $ε3=+210×10−6 in./in.$ .

The Rosette 3 makes an angle with the x-axis is $θ3=75°$ .

Calculation:

Write the equation for the normal strain 1 as follows;

$ε1=εxcos2θ1+εysin2θ1+γxysinθ1cosθ1$

Substitute $−93.1×10−6 in./in.$ for $ε1$ and $−75°$ for $θ1$ .

$−93.1×10−6=εxcos2(−75°)+εysin2(−75°)+γxysin(−75°)cos(−75°)$ (1)

Write the equation for the normal strain 2 as follows;

$ε2=εxcos2θ2+εysin2θ2+γxysinθ2cosθ2$

Substitute $+385×10−6 in./in.$ for $ε2$ and $0°$ for $θ2$ .

$385×10−6=εxcos2(0°)+εysin2(0°)+γxysin(0°)cos(0°)=εx+0+0εx=385×10−6 in./in.$

Write the equation for the normal strain 3 as follows;

$ε3=εxcos2θ3+εysin2θ3+γxysinθ3cosθ3$

Substitute $+210×10−6 in./in.$ for $ε3$ and $75°$ for $θ3$ .

$+210×10−6=εxcos2(75°)+εysin2(75°)+γxysin(75°)cos(75°)$ (2)

Add equation (1) and (2).

$−93.1×10−6+210×10−6=(εxcos2(−75°)+εysin2(−75°)+γxysin(−75°)cos(−75°)+εxcos2(75°)+εysin2(75°)+γxysin(75°)cos(75°))116.9×10−6=(εxcos2(−75°)+εysin2(−75°)−0.25γxy+εxcos2(75°)+εysin2(75°)+0.25γxy)=εxcos2(−75°)+εysin2(−75°)+εxcos2(75°)+εysin2(75°)$

Substitute $385×10−6 in./in.$ for $εx$ .

$116.9×10−6=((385×10−6)cos2(−75°)+εysin2(−75°)+(385×10−6)cos2(75°)+εysin2(75°))=5.158×10−5+1.866εyεy=35×10−6 in./in.$

Substitute $385×10−6 in./in.$ for $εx$ and $35×10−6 in./in.$ for $εy$ in Equation (2).

$+210×10−6=(385×10−6)cos2(75°)+(35×10−6)sin2(75°)+γxysin(75°)cos(75°)=2.579×10−5+3.266×10−5+0.25γxyγxy=606.2×10−6 in./in.$

Find the orientation $(θp)$ of the principal strains using the relation.

$tan2θp=γxyεx−εy$

Substitute $606.2×10−6 in./in.$ for $γxy$ , $385×10−6 in./in.$ for $εx$ , and $35×10−6 in./in.$ for $εy$

$tan2θp=606.2×10−6385×10−6−35×10−6θp=30°and 120°$

Therefore, the orientation of the principal strains is $θp=30°and 120°_$ .

Find the average normal strain $(εave)$ using the relation.

$εave=εx+εy2$

Substitute $385×10−6 in./in.$ for $εx$ and $35×10−6 in./in.$ for $εy$ .

$εave=385×10−6+35×10−62=210×10−6 in./in.$

Find the radius of the Mohr’s circle (R) using the equation.

$R=(εx−εy2)2+(γxy2)2$

Substitute $385×10−6 in./in.$ for $εx$ , $35×10−6 in./in.$ for $εy$ , and $606.2×10−6 in./in.$ for $γxy$ .

$R=(385×10−6−35×10−62)2+(606.2×10−62)2=350×10−6 in./in.$

Find the in-plane maximum principal strain $(εmax)$ using the relation.

$εmax=εave+R$

Substitute $210×10−6 in./in.$ for $εave$ and $350×10−6 in./in.$ for R.

$εmax=210×10−6+350×10−6=560×10−6 in./in.$

Find the in-plane minimum principal strain $(εmin)$ using the relation.

$εmin=εave−R$

Substitute $210×10−6 in./in.$ for $εave$ and $350×10−6 in./in.$ for R.

$εmin=210×10−6−350×10−6=−140×10−6 in./in.$

Therefore,

The in-plane maximum principal strain is $εmax=560×10−6 in./in._$ .

The in-plane minimum principal strain is $εmin=−140×10−6 in./in._$ .

(b)

To determine

Find the in-plane maximum shearing strain.

(b)

Expert Solution

Answer to Problem 149P

The in-plane maximum shearing strain is $(γmax)in-plane=700×10−6 in./in._$ .

Explanation of Solution

Refer to part (a);

The radius of the Mohr’s circle is $R=350×10−6 in./in.$

Find the in-plane maximum shearing strain $(γmax)in-plane$ using the relation.

$(γmax)in-plane=2R$

Substitute $350×10−6 in./in.$ for R.

$(γmax)in-plane=2×350×10−6=700×10−6 in./in.$

Therefore, the in-plane maximum shearing strain is $(γmax)in-plane=700×10−6 in./in._$ .

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Answer 5

Chapter 7.9, Problem 149P

(a)

To determine

Find the orientation and magnitude of the principal strains in the plane.

(a)

Expert Solution

Answer to Problem 149P

The orientation of the principal strains is $θp=30°and 120°_$ .

The in-plane maximum principal strain is $εmax=560×10−6 in./in._$ .

The in-plane minimum principal strain is $εmin=−140×10−6 in./in._$ .

Explanation of Solution

Given information:

The normal strain in Rosette 1 is $ε1=−93.1×10−6 in./in.$ .

The Rosette 1 makes an angle with the x-axis is $θ1=−75°$ .

The normal strain in Rosette 2 is $ε2=+385×10−6 in./in.$ .

The Rosette 2 makes an angle with the x-axis is $θ2=0°$ .

The normal strain in Rosette 3 is $ε3=+210×10−6 in./in.$ .

The Rosette 3 makes an angle with the x-axis is $θ3=75°$ .

Calculation:

Write the equation for the normal strain 1 as follows;

$ε1=εxcos2θ1+εysin2θ1+γxysinθ1cosθ1$

Substitute $−93.1×10−6 in./in.$ for $ε1$ and $−75°$ for $θ1$ .

$−93.1×10−6=εxcos2(−75°)+εysin2(−75°)+γxysin(−75°)cos(−75°)$ (1)

Write the equation for the normal strain 2 as follows;

$ε2=εxcos2θ2+εysin2θ2+γxysinθ2cosθ2$

Substitute $+385×10−6 in./in.$ for $ε2$ and $0°$ for $θ2$ .

$385×10−6=εxcos2(0°)+εysin2(0°)+γxysin(0°)cos(0°)=εx+0+0εx=385×10−6 in./in.$

Write the equation for the normal strain 3 as follows;

$ε3=εxcos2θ3+εysin2θ3+γxysinθ3cosθ3$

Substitute $+210×10−6 in./in.$ for $ε3$ and $75°$ for $θ3$ .

$+210×10−6=εxcos2(75°)+εysin2(75°)+γxysin(75°)cos(75°)$ (2)

Add equation (1) and (2).

$−93.1×10−6+210×10−6=(εxcos2(−75°)+εysin2(−75°)+γxysin(−75°)cos(−75°)+εxcos2(75°)+εysin2(75°)+γxysin(75°)cos(75°))116.9×10−6=(εxcos2(−75°)+εysin2(−75°)−0.25γxy+εxcos2(75°)+εysin2(75°)+0.25γxy)=εxcos2(−75°)+εysin2(−75°)+εxcos2(75°)+εysin2(75°)$

Substitute $385×10−6 in./in.$ for $εx$ .

$116.9×10−6=((385×10−6)cos2(−75°)+εysin2(−75°)+(385×10−6)cos2(75°)+εysin2(75°))=5.158×10−5+1.866εyεy=35×10−6 in./in.$

Substitute $385×10−6 in./in.$ for $εx$ and $35×10−6 in./in.$ for $εy$ in Equation (2).

$+210×10−6=(385×10−6)cos2(75°)+(35×10−6)sin2(75°)+γxysin(75°)cos(75°)=2.579×10−5+3.266×10−5+0.25γxyγxy=606.2×10−6 in./in.$

Find the orientation $(θp)$ of the principal strains using the relation.

$tan2θp=γxyεx−εy$

Substitute $606.2×10−6 in./in.$ for $γxy$ , $385×10−6 in./in.$ for $εx$ , and $35×10−6 in./in.$ for $εy$

$tan2θp=606.2×10−6385×10−6−35×10−6θp=30°and 120°$

Therefore, the orientation of the principal strains is $θp=30°and 120°_$ .

Find the average normal strain $(εave)$ using the relation.

$εave=εx+εy2$

Substitute $385×10−6 in./in.$ for $εx$ and $35×10−6 in./in.$ for $εy$ .

$εave=385×10−6+35×10−62=210×10−6 in./in.$

Find the radius of the Mohr’s circle (R) using the equation.

$R=(εx−εy2)2+(γxy2)2$

Substitute $385×10−6 in./in.$ for $εx$ , $35×10−6 in./in.$ for $εy$ , and $606.2×10−6 in./in.$ for $γxy$ .

$R=(385×10−6−35×10−62)2+(606.2×10−62)2=350×10−6 in./in.$

Find the in-plane maximum principal strain $(εmax)$ using the relation.

$εmax=εave+R$

Substitute $210×10−6 in./in.$ for $εave$ and $350×10−6 in./in.$ for R.

$εmax=210×10−6+350×10−6=560×10−6 in./in.$

Find the in-plane minimum principal strain $(εmin)$ using the relation.

$εmin=εave−R$

Substitute $210×10−6 in./in.$ for $εave$ and $350×10−6 in./in.$ for R.

$εmin=210×10−6−350×10−6=−140×10−6 in./in.$

Therefore,

The in-plane maximum principal strain is $εmax=560×10−6 in./in._$ .

The in-plane minimum principal strain is $εmin=−140×10−6 in./in._$ .

(b)

To determine

Find the in-plane maximum shearing strain.

(b)

Expert Solution

Answer to Problem 149P

The in-plane maximum shearing strain is $(γmax)in-plane=700×10−6 in./in._$ .

Explanation of Solution

Refer to part (a);

The radius of the Mohr’s circle is $R=350×10−6 in./in.$

Find the in-plane maximum shearing strain $(γmax)in-plane$ using the relation.

$(γmax)in-plane=2R$

Substitute $350×10−6 in./in.$ for R.

$(γmax)in-plane=2×350×10−6=700×10−6 in./in.$

Therefore, the in-plane maximum shearing strain is $(γmax)in-plane=700×10−6 in./in._$ .

Answer 6

Chapter 7.9, Problem 149P

(a)

To determine

Find the orientation and magnitude of the principal strains in the plane.

(a)

Expert Solution

Answer to Problem 149P

The orientation of the principal strains is $θp=30°and 120°_$ .

The in-plane maximum principal strain is $εmax=560×10−6 in./in._$ .

The in-plane minimum principal strain is $εmin=−140×10−6 in./in._$ .

Explanation of Solution

Given information:

The normal strain in Rosette 1 is $ε1=−93.1×10−6 in./in.$ .

The Rosette 1 makes an angle with the x-axis is $θ1=−75°$ .

The normal strain in Rosette 2 is $ε2=+385×10−6 in./in.$ .

The Rosette 2 makes an angle with the x-axis is $θ2=0°$ .

The normal strain in Rosette 3 is $ε3=+210×10−6 in./in.$ .

The Rosette 3 makes an angle with the x-axis is $θ3=75°$ .

Calculation:

Write the equation for the normal strain 1 as follows;

$ε1=εxcos2θ1+εysin2θ1+γxysinθ1cosθ1$

Substitute $−93.1×10−6 in./in.$ for $ε1$ and $−75°$ for $θ1$ .

$−93.1×10−6=εxcos2(−75°)+εysin2(−75°)+γxysin(−75°)cos(−75°)$ (1)

Write the equation for the normal strain 2 as follows;

$ε2=εxcos2θ2+εysin2θ2+γxysinθ2cosθ2$

Substitute $+385×10−6 in./in.$ for $ε2$ and $0°$ for $θ2$ .

$385×10−6=εxcos2(0°)+εysin2(0°)+γxysin(0°)cos(0°)=εx+0+0εx=385×10−6 in./in.$

Write the equation for the normal strain 3 as follows;

$ε3=εxcos2θ3+εysin2θ3+γxysinθ3cosθ3$

Substitute $+210×10−6 in./in.$ for $ε3$ and $75°$ for $θ3$ .

$+210×10−6=εxcos2(75°)+εysin2(75°)+γxysin(75°)cos(75°)$ (2)

Add equation (1) and (2).

$−93.1×10−6+210×10−6=(εxcos2(−75°)+εysin2(−75°)+γxysin(−75°)cos(−75°)+εxcos2(75°)+εysin2(75°)+γxysin(75°)cos(75°))116.9×10−6=(εxcos2(−75°)+εysin2(−75°)−0.25γxy+εxcos2(75°)+εysin2(75°)+0.25γxy)=εxcos2(−75°)+εysin2(−75°)+εxcos2(75°)+εysin2(75°)$

Substitute $385×10−6 in./in.$ for $εx$ .

$116.9×10−6=((385×10−6)cos2(−75°)+εysin2(−75°)+(385×10−6)cos2(75°)+εysin2(75°))=5.158×10−5+1.866εyεy=35×10−6 in./in.$

Substitute $385×10−6 in./in.$ for $εx$ and $35×10−6 in./in.$ for $εy$ in Equation (2).

$+210×10−6=(385×10−6)cos2(75°)+(35×10−6)sin2(75°)+γxysin(75°)cos(75°)=2.579×10−5+3.266×10−5+0.25γxyγxy=606.2×10−6 in./in.$

Find the orientation $(θp)$ of the principal strains using the relation.

$tan2θp=γxyεx−εy$

Substitute $606.2×10−6 in./in.$ for $γxy$ , $385×10−6 in./in.$ for $εx$ , and $35×10−6 in./in.$ for $εy$

$tan2θp=606.2×10−6385×10−6−35×10−6θp=30°and 120°$

Therefore, the orientation of the principal strains is $θp=30°and 120°_$ .

Find the average normal strain $(εave)$ using the relation.

$εave=εx+εy2$

Substitute $385×10−6 in./in.$ for $εx$ and $35×10−6 in./in.$ for $εy$ .

$εave=385×10−6+35×10−62=210×10−6 in./in.$

Find the radius of the Mohr’s circle (R) using the equation.

$R=(εx−εy2)2+(γxy2)2$

Substitute $385×10−6 in./in.$ for $εx$ , $35×10−6 in./in.$ for $εy$ , and $606.2×10−6 in./in.$ for $γxy$ .

$R=(385×10−6−35×10−62)2+(606.2×10−62)2=350×10−6 in./in.$

Find the in-plane maximum principal strain $(εmax)$ using the relation.

$εmax=εave+R$

Substitute $210×10−6 in./in.$ for $εave$ and $350×10−6 in./in.$ for R.

$εmax=210×10−6+350×10−6=560×10−6 in./in.$

Find the in-plane minimum principal strain $(εmin)$ using the relation.

$εmin=εave−R$

Substitute $210×10−6 in./in.$ for $εave$ and $350×10−6 in./in.$ for R.

$εmin=210×10−6−350×10−6=−140×10−6 in./in.$

Therefore,

The in-plane maximum principal strain is $εmax=560×10−6 in./in._$ .

The in-plane minimum principal strain is $εmin=−140×10−6 in./in._$ .

Answer 7

Chapter 7.9, Problem 149P

(a)

To determine

Find the orientation and magnitude of the principal strains in the plane.

Answer 8

(a)

Expert Solution

Answer to Problem 149P

The orientation of the principal strains is $θp=30°and 120°_$ .

The in-plane maximum principal strain is $εmax=560×10−6 in./in._$ .

The in-plane minimum principal strain is $εmin=−140×10−6 in./in._$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given information:

The normal strain in Rosette 1 is $ε1=−93.1×10−6 in./in.$ .

The Rosette 1 makes an angle with the x-axis is $θ1=−75°$ .

The normal strain in Rosette 2 is $ε2=+385×10−6 in./in.$ .

The Rosette 2 makes an angle with the x-axis is $θ2=0°$ .

The normal strain in Rosette 3 is $ε3=+210×10−6 in./in.$ .

The Rosette 3 makes an angle with the x-axis is $θ3=75°$ .

Calculation:

Write the equation for the normal strain 1 as follows;

$ε1=εxcos2θ1+εysin2θ1+γxysinθ1cosθ1$

Substitute $−93.1×10−6 in./in.$ for $ε1$ and $−75°$ for $θ1$ .

$−93.1×10−6=εxcos2(−75°)+εysin2(−75°)+γxysin(−75°)cos(−75°)$ (1)

Write the equation for the normal strain 2 as follows;

$ε2=εxcos2θ2+εysin2θ2+γxysinθ2cosθ2$

Substitute $+385×10−6 in./in.$ for $ε2$ and $0°$ for $θ2$ .

$385×10−6=εxcos2(0°)+εysin2(0°)+γxysin(0°)cos(0°)=εx+0+0εx=385×10−6 in./in.$

Write the equation for the normal strain 3 as follows;

$ε3=εxcos2θ3+εysin2θ3+γxysinθ3cosθ3$

Substitute $+210×10−6 in./in.$ for $ε3$ and $75°$ for $θ3$ .

$+210×10−6=εxcos2(75°)+εysin2(75°)+γxysin(75°)cos(75°)$ (2)

Add equation (1) and (2).

$−93.1×10−6+210×10−6=(εxcos2(−75°)+εysin2(−75°)+γxysin(−75°)cos(−75°)+εxcos2(75°)+εysin2(75°)+γxysin(75°)cos(75°))116.9×10−6=(εxcos2(−75°)+εysin2(−75°)−0.25γxy+εxcos2(75°)+εysin2(75°)+0.25γxy)=εxcos2(−75°)+εysin2(−75°)+εxcos2(75°)+εysin2(75°)$

Substitute $385×10−6 in./in.$ for $εx$ .

$116.9×10−6=((385×10−6)cos2(−75°)+εysin2(−75°)+(385×10−6)cos2(75°)+εysin2(75°))=5.158×10−5+1.866εyεy=35×10−6 in./in.$

Substitute $385×10−6 in./in.$ for $εx$ and $35×10−6 in./in.$ for $εy$ in Equation (2).

$+210×10−6=(385×10−6)cos2(75°)+(35×10−6)sin2(75°)+γxysin(75°)cos(75°)=2.579×10−5+3.266×10−5+0.25γxyγxy=606.2×10−6 in./in.$

Find the orientation $(θp)$ of the principal strains using the relation.

$tan2θp=γxyεx−εy$

Substitute $606.2×10−6 in./in.$ for $γxy$ , $385×10−6 in./in.$ for $εx$ , and $35×10−6 in./in.$ for $εy$

$tan2θp=606.2×10−6385×10−6−35×10−6θp=30°and 120°$

Therefore, the orientation of the principal strains is $θp=30°and 120°_$ .

Find the average normal strain $(εave)$ using the relation.

$εave=εx+εy2$

Substitute $385×10−6 in./in.$ for $εx$ and $35×10−6 in./in.$ for $εy$ .

$εave=385×10−6+35×10−62=210×10−6 in./in.$

Find the radius of the Mohr’s circle (R) using the equation.

$R=(εx−εy2)2+(γxy2)2$

Substitute $385×10−6 in./in.$ for $εx$ , $35×10−6 in./in.$ for $εy$ , and $606.2×10−6 in./in.$ for $γxy$ .

$R=(385×10−6−35×10−62)2+(606.2×10−62)2=350×10−6 in./in.$

Find the in-plane maximum principal strain $(εmax)$ using the relation.

$εmax=εave+R$

Substitute $210×10−6 in./in.$ for $εave$ and $350×10−6 in./in.$ for R.

$εmax=210×10−6+350×10−6=560×10−6 in./in.$

Find the in-plane minimum principal strain $(εmin)$ using the relation.

$εmin=εave−R$

Substitute $210×10−6 in./in.$ for $εave$ and $350×10−6 in./in.$ for R.

$εmin=210×10−6−350×10−6=−140×10−6 in./in.$

Therefore,

The in-plane maximum principal strain is $εmax=560×10−6 in./in._$ .

The in-plane minimum principal strain is $εmin=−140×10−6 in./in._$ .

Answer 13

(b)

To determine

Find the in-plane maximum shearing strain.

(b)

Expert Solution

Answer to Problem 149P

The in-plane maximum shearing strain is $(γmax)in-plane=700×10−6 in./in._$ .

Explanation of Solution

Refer to part (a);

The radius of the Mohr’s circle is $R=350×10−6 in./in.$

Find the in-plane maximum shearing strain $(γmax)in-plane$ using the relation.

$(γmax)in-plane=2R$

Substitute $350×10−6 in./in.$ for R.

$(γmax)in-plane=2×350×10−6=700×10−6 in./in.$

Therefore, the in-plane maximum shearing strain is $(γmax)in-plane=700×10−6 in./in._$ .

Answer 14

(b)

To determine

Find the in-plane maximum shearing strain.

Answer 15

(b)

Expert Solution

Answer to Problem 149P

The in-plane maximum shearing strain is $(γmax)in-plane=700×10−6 in./in._$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Refer to part (a);

The radius of the Mohr’s circle is $R=350×10−6 in./in.$

Find the in-plane maximum shearing strain $(γmax)in-plane$ using the relation.

$(γmax)in-plane=2R$

Substitute $350×10−6 in./in.$ for R.

$(γmax)in-plane=2×350×10−6=700×10−6 in./in.$

Therefore, the in-plane maximum shearing strain is $(γmax)in-plane=700×10−6 in./in._$ .

Answer 20

Find the orientation and magnitude of the principal strains in the plane.

Videos

Find the orientation and magnitude of the principal strains in the plane.

Answer to Problem 149P

Explanation of Solution

Find the in-plane maximum shearing strain.

Answer to Problem 149P

Explanation of Solution

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