Find the strain energy of the prismatic beam AB .

Question

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Answer 1

Question

Chapter 11.3, Problem 40P

To determine

Find the strain energy of the prismatic beam AB.

Expert Solution & Answer

Answer to Problem 40P

The strain energy of the prismatic beam AB is $U=2M02LEbd3{1+3Ed210GL2}_$ .

Explanation of Solution

Given information:

Taking into account the effect of both normal and shearing stresses.

Calculation:

Calculate the moment of inertia (I) for the rectangular cross section as shown below.

$I=bd312$

Here, b is the width of the cross section and d is the depth of the cross section.

Calculate the area of the cross section (A) as shown below.

$A=bd$

Calculate the centroid (c) as shown below.

$c=d2$

Calculate the reactions as shown below.

Take moment about B is Equal to zero.

$∑MB=0RAL+M0=0RAL=−M0RA=−M0L$

Summation of forces along vertical direction is Equal to zero.

$∑Fy=0RB−M0L=0RB=M0L$

Calculate the shear force as shown below.

Shear force at A is $SF@ A=−M0L$ .

Shear force at B is $SF@ B=−M0L+M0L=0$

Calculate the bending moment as shown below.

Bending moment at A is $BM@ A=M0$ .

Bending moment at B is $BM@ B=M0−M0L×L=0$

Sketch the shear force and bending moment diagram as shown in Figure 1.

Mechanics of Materials, 7th Edition, Chapter 11.3, Problem 40P

Refer to Figure 1.

Maximum shear force $V=−M0L$ .

Maximum bending moment $Mmax=M0$ .

Bending moment at a distance $v$ from B $M=M0Lv$

Calculate the strain energy due to bending $(U1)$ as shown below.

$U1=∫abM22EIdv$

Substitute $M=M0Lv$ for M and apply the limits.

$U1=∫0L(M0Lv)22EIdv=12EI∫0LM02v2L2dv=M022EIL2(v33)0L=M026EIL2×L3$

$=M02L6EI$

Substitute $bd312$ for I.

$U1=M02L6E(bd312)=2M02Lbd3E$

Calculate the shear stress $(τxy)$ as shown below.

$τxy=32VA(1−y2c2)$

Calculate the strain energy density $(u)$ as shown below.

$u=τxy22G$

Here, G is the modulus of rigidity.

Substitute $32VA(1−y2c2)$ for $τxy$ .

$u=(32VA(1−y2c2))22G=9V28GA2(1+y4c4−2y2c2)$

Substitute $bd$ for A.

$u=9V28G(bd)2(1+y4c4−2y2c2)=9V28Gb2d2(1+y4c4−2y2c2)$

The value of $v=bdy$ .

Differentiate both sides of the Equation as shown below.

$dv=bdy dx$

Calculate the strain energy due to shear as shown below.

$U2=∫u dv$

Substitute $9V28Gb2d2(1+y4c4−2y2c2)$ for u, $bdy dx$ for $dv$ , and apply the limits.

$U2=∫0L∫−cc9V28Gb2d2(1+y4c4−2y2c2)bdy dx=9V28Gbd2∫0L∫−cc(1+y4c4−2y2c2)dy dx=9V28Gbd2∫0L(y+y55c4−2y33c2)−ccdx=9V28Gbd2{(c+c55c4−2c33c2)−((−c)+(−c)55c4−2(−c)33c2)}(x)0L$

$=9V28Gbd2{c+c5−2c3+c+c5−2c3}L=9V2cL8Gbd2×(15+3−10+15+3−10)15=9V2cL8Gbd2×1615=6V2cL5Gbd2$

Substitute $−M0L$ for V and $d2$ for c.

$U2=6(−M0L)2(d2)L5Gbd2=3M025GbdL$

Calculate the total strain energy as shown below.

$U=U1+U2$

Substitute $2M02Lbd3E$ for $U1$ and $3M025GbdL$ for $U2$ .

$U=2M02Lbd3E+3M025GbdL=2M02LEbd3(1+3Ed210GL2)$

Hence, the strain energy of the prismatic beam AB is $U=2M02LEbd3{1+3Ed210GL2}_$ .

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Answer 2

Question

Chapter 11.3, Problem 40P

To determine

Find the strain energy of the prismatic beam AB.

Expert Solution & Answer

Answer to Problem 40P

The strain energy of the prismatic beam AB is $U=2M02LEbd3{1+3Ed210GL2}_$ .

Explanation of Solution

Given information:

Taking into account the effect of both normal and shearing stresses.

Calculation:

Calculate the moment of inertia (I) for the rectangular cross section as shown below.

$I=bd312$

Here, b is the width of the cross section and d is the depth of the cross section.

Calculate the area of the cross section (A) as shown below.

$A=bd$

Calculate the centroid (c) as shown below.

$c=d2$

Calculate the reactions as shown below.

Take moment about B is Equal to zero.

$∑MB=0RAL+M0=0RAL=−M0RA=−M0L$

Summation of forces along vertical direction is Equal to zero.

$∑Fy=0RB−M0L=0RB=M0L$

Calculate the shear force as shown below.

Shear force at A is $SF@ A=−M0L$ .

Shear force at B is $SF@ B=−M0L+M0L=0$

Calculate the bending moment as shown below.

Bending moment at A is $BM@ A=M0$ .

Bending moment at B is $BM@ B=M0−M0L×L=0$

Sketch the shear force and bending moment diagram as shown in Figure 1.

Mechanics of Materials, 7th Edition, Chapter 11.3, Problem 40P

Refer to Figure 1.

Maximum shear force $V=−M0L$ .

Maximum bending moment $Mmax=M0$ .

Bending moment at a distance $v$ from B $M=M0Lv$

Calculate the strain energy due to bending $(U1)$ as shown below.

$U1=∫abM22EIdv$

Substitute $M=M0Lv$ for M and apply the limits.

$U1=∫0L(M0Lv)22EIdv=12EI∫0LM02v2L2dv=M022EIL2(v33)0L=M026EIL2×L3$

$=M02L6EI$

Substitute $bd312$ for I.

$U1=M02L6E(bd312)=2M02Lbd3E$

Calculate the shear stress $(τxy)$ as shown below.

$τxy=32VA(1−y2c2)$

Calculate the strain energy density $(u)$ as shown below.

$u=τxy22G$

Here, G is the modulus of rigidity.

Substitute $32VA(1−y2c2)$ for $τxy$ .

$u=(32VA(1−y2c2))22G=9V28GA2(1+y4c4−2y2c2)$

Substitute $bd$ for A.

$u=9V28G(bd)2(1+y4c4−2y2c2)=9V28Gb2d2(1+y4c4−2y2c2)$

The value of $v=bdy$ .

Differentiate both sides of the Equation as shown below.

$dv=bdy dx$

Calculate the strain energy due to shear as shown below.

$U2=∫u dv$

Substitute $9V28Gb2d2(1+y4c4−2y2c2)$ for u, $bdy dx$ for $dv$ , and apply the limits.

$U2=∫0L∫−cc9V28Gb2d2(1+y4c4−2y2c2)bdy dx=9V28Gbd2∫0L∫−cc(1+y4c4−2y2c2)dy dx=9V28Gbd2∫0L(y+y55c4−2y33c2)−ccdx=9V28Gbd2{(c+c55c4−2c33c2)−((−c)+(−c)55c4−2(−c)33c2)}(x)0L$

$=9V28Gbd2{c+c5−2c3+c+c5−2c3}L=9V2cL8Gbd2×(15+3−10+15+3−10)15=9V2cL8Gbd2×1615=6V2cL5Gbd2$

Substitute $−M0L$ for V and $d2$ for c.

$U2=6(−M0L)2(d2)L5Gbd2=3M025GbdL$

Calculate the total strain energy as shown below.

$U=U1+U2$

Substitute $2M02Lbd3E$ for $U1$ and $3M025GbdL$ for $U2$ .

$U=2M02Lbd3E+3M025GbdL=2M02LEbd3(1+3Ed210GL2)$

Hence, the strain energy of the prismatic beam AB is $U=2M02LEbd3{1+3Ed210GL2}_$ .

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Answer 3

Question

Chapter 11.3, Problem 40P

To determine

Find the strain energy of the prismatic beam AB.

Expert Solution & Answer

Answer to Problem 40P

The strain energy of the prismatic beam AB is $U=2M02LEbd3{1+3Ed210GL2}_$ .

Explanation of Solution

Given information:

Taking into account the effect of both normal and shearing stresses.

Calculation:

Calculate the moment of inertia (I) for the rectangular cross section as shown below.

$I=bd312$

Here, b is the width of the cross section and d is the depth of the cross section.

Calculate the area of the cross section (A) as shown below.

$A=bd$

Calculate the centroid (c) as shown below.

$c=d2$

Calculate the reactions as shown below.

Take moment about B is Equal to zero.

$∑MB=0RAL+M0=0RAL=−M0RA=−M0L$

Summation of forces along vertical direction is Equal to zero.

$∑Fy=0RB−M0L=0RB=M0L$

Calculate the shear force as shown below.

Shear force at A is $SF@ A=−M0L$ .

Shear force at B is $SF@ B=−M0L+M0L=0$

Calculate the bending moment as shown below.

Bending moment at A is $BM@ A=M0$ .

Bending moment at B is $BM@ B=M0−M0L×L=0$

Sketch the shear force and bending moment diagram as shown in Figure 1.

Mechanics of Materials, 7th Edition, Chapter 11.3, Problem 40P

Refer to Figure 1.

Maximum shear force $V=−M0L$ .

Maximum bending moment $Mmax=M0$ .

Bending moment at a distance $v$ from B $M=M0Lv$

Calculate the strain energy due to bending $(U1)$ as shown below.

$U1=∫abM22EIdv$

Substitute $M=M0Lv$ for M and apply the limits.

$U1=∫0L(M0Lv)22EIdv=12EI∫0LM02v2L2dv=M022EIL2(v33)0L=M026EIL2×L3$

$=M02L6EI$

Substitute $bd312$ for I.

$U1=M02L6E(bd312)=2M02Lbd3E$

Calculate the shear stress $(τxy)$ as shown below.

$τxy=32VA(1−y2c2)$

Calculate the strain energy density $(u)$ as shown below.

$u=τxy22G$

Here, G is the modulus of rigidity.

Substitute $32VA(1−y2c2)$ for $τxy$ .

$u=(32VA(1−y2c2))22G=9V28GA2(1+y4c4−2y2c2)$

Substitute $bd$ for A.

$u=9V28G(bd)2(1+y4c4−2y2c2)=9V28Gb2d2(1+y4c4−2y2c2)$

The value of $v=bdy$ .

Differentiate both sides of the Equation as shown below.

$dv=bdy dx$

Calculate the strain energy due to shear as shown below.

$U2=∫u dv$

Substitute $9V28Gb2d2(1+y4c4−2y2c2)$ for u, $bdy dx$ for $dv$ , and apply the limits.

$U2=∫0L∫−cc9V28Gb2d2(1+y4c4−2y2c2)bdy dx=9V28Gbd2∫0L∫−cc(1+y4c4−2y2c2)dy dx=9V28Gbd2∫0L(y+y55c4−2y33c2)−ccdx=9V28Gbd2{(c+c55c4−2c33c2)−((−c)+(−c)55c4−2(−c)33c2)}(x)0L$

$=9V28Gbd2{c+c5−2c3+c+c5−2c3}L=9V2cL8Gbd2×(15+3−10+15+3−10)15=9V2cL8Gbd2×1615=6V2cL5Gbd2$

Substitute $−M0L$ for V and $d2$ for c.

$U2=6(−M0L)2(d2)L5Gbd2=3M025GbdL$

Calculate the total strain energy as shown below.

$U=U1+U2$

Substitute $2M02Lbd3E$ for $U1$ and $3M025GbdL$ for $U2$ .

$U=2M02Lbd3E+3M025GbdL=2M02LEbd3(1+3Ed210GL2)$

Hence, the strain energy of the prismatic beam AB is $U=2M02LEbd3{1+3Ed210GL2}_$ .

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Answer 4

Question

Chapter 11.3, Problem 40P

To determine

Find the strain energy of the prismatic beam AB.

Expert Solution & Answer

Answer to Problem 40P

The strain energy of the prismatic beam AB is $U=2M02LEbd3{1+3Ed210GL2}_$ .

Explanation of Solution

Given information:

Taking into account the effect of both normal and shearing stresses.

Calculation:

Calculate the moment of inertia (I) for the rectangular cross section as shown below.

$I=bd312$

Here, b is the width of the cross section and d is the depth of the cross section.

Calculate the area of the cross section (A) as shown below.

$A=bd$

Calculate the centroid (c) as shown below.

$c=d2$

Calculate the reactions as shown below.

Take moment about B is Equal to zero.

$∑MB=0RAL+M0=0RAL=−M0RA=−M0L$

Summation of forces along vertical direction is Equal to zero.

$∑Fy=0RB−M0L=0RB=M0L$

Calculate the shear force as shown below.

Shear force at A is $SF@ A=−M0L$ .

Shear force at B is $SF@ B=−M0L+M0L=0$

Calculate the bending moment as shown below.

Bending moment at A is $BM@ A=M0$ .

Bending moment at B is $BM@ B=M0−M0L×L=0$

Sketch the shear force and bending moment diagram as shown in Figure 1.

Mechanics of Materials, 7th Edition, Chapter 11.3, Problem 40P

Refer to Figure 1.

Maximum shear force $V=−M0L$ .

Maximum bending moment $Mmax=M0$ .

Bending moment at a distance $v$ from B $M=M0Lv$

Calculate the strain energy due to bending $(U1)$ as shown below.

$U1=∫abM22EIdv$

Substitute $M=M0Lv$ for M and apply the limits.

$U1=∫0L(M0Lv)22EIdv=12EI∫0LM02v2L2dv=M022EIL2(v33)0L=M026EIL2×L3$

$=M02L6EI$

Substitute $bd312$ for I.

$U1=M02L6E(bd312)=2M02Lbd3E$

Calculate the shear stress $(τxy)$ as shown below.

$τxy=32VA(1−y2c2)$

Calculate the strain energy density $(u)$ as shown below.

$u=τxy22G$

Here, G is the modulus of rigidity.

Substitute $32VA(1−y2c2)$ for $τxy$ .

$u=(32VA(1−y2c2))22G=9V28GA2(1+y4c4−2y2c2)$

Substitute $bd$ for A.

$u=9V28G(bd)2(1+y4c4−2y2c2)=9V28Gb2d2(1+y4c4−2y2c2)$

The value of $v=bdy$ .

Differentiate both sides of the Equation as shown below.

$dv=bdy dx$

Calculate the strain energy due to shear as shown below.

$U2=∫u dv$

Substitute $9V28Gb2d2(1+y4c4−2y2c2)$ for u, $bdy dx$ for $dv$ , and apply the limits.

$U2=∫0L∫−cc9V28Gb2d2(1+y4c4−2y2c2)bdy dx=9V28Gbd2∫0L∫−cc(1+y4c4−2y2c2)dy dx=9V28Gbd2∫0L(y+y55c4−2y33c2)−ccdx=9V28Gbd2{(c+c55c4−2c33c2)−((−c)+(−c)55c4−2(−c)33c2)}(x)0L$

$=9V28Gbd2{c+c5−2c3+c+c5−2c3}L=9V2cL8Gbd2×(15+3−10+15+3−10)15=9V2cL8Gbd2×1615=6V2cL5Gbd2$

Substitute $−M0L$ for V and $d2$ for c.

$U2=6(−M0L)2(d2)L5Gbd2=3M025GbdL$

Calculate the total strain energy as shown below.

$U=U1+U2$

Substitute $2M02Lbd3E$ for $U1$ and $3M025GbdL$ for $U2$ .

$U=2M02Lbd3E+3M025GbdL=2M02LEbd3(1+3Ed210GL2)$

Hence, the strain energy of the prismatic beam AB is $U=2M02LEbd3{1+3Ed210GL2}_$ .

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Answer 5

Chapter 11.3, Problem 40P

To determine

Find the strain energy of the prismatic beam AB.

Expert Solution & Answer

Answer to Problem 40P

The strain energy of the prismatic beam AB is $U=2M02LEbd3{1+3Ed210GL2}_$ .

Explanation of Solution

Given information:

Taking into account the effect of both normal and shearing stresses.

Calculation:

Calculate the moment of inertia (I) for the rectangular cross section as shown below.

$I=bd312$

Here, b is the width of the cross section and d is the depth of the cross section.

Calculate the area of the cross section (A) as shown below.

$A=bd$

Calculate the centroid (c) as shown below.

$c=d2$

Calculate the reactions as shown below.

Take moment about B is Equal to zero.

$∑MB=0RAL+M0=0RAL=−M0RA=−M0L$

Summation of forces along vertical direction is Equal to zero.

$∑Fy=0RB−M0L=0RB=M0L$

Calculate the shear force as shown below.

Shear force at A is $SF@ A=−M0L$ .

Shear force at B is $SF@ B=−M0L+M0L=0$

Calculate the bending moment as shown below.

Bending moment at A is $BM@ A=M0$ .

Bending moment at B is $BM@ B=M0−M0L×L=0$

Sketch the shear force and bending moment diagram as shown in Figure 1.

Mechanics of Materials, 7th Edition, Chapter 11.3, Problem 40P

Refer to Figure 1.

Maximum shear force $V=−M0L$ .

Maximum bending moment $Mmax=M0$ .

Bending moment at a distance $v$ from B $M=M0Lv$

Calculate the strain energy due to bending $(U1)$ as shown below.

$U1=∫abM22EIdv$

Substitute $M=M0Lv$ for M and apply the limits.

$U1=∫0L(M0Lv)22EIdv=12EI∫0LM02v2L2dv=M022EIL2(v33)0L=M026EIL2×L3$

$=M02L6EI$

Substitute $bd312$ for I.

$U1=M02L6E(bd312)=2M02Lbd3E$

Calculate the shear stress $(τxy)$ as shown below.

$τxy=32VA(1−y2c2)$

Calculate the strain energy density $(u)$ as shown below.

$u=τxy22G$

Here, G is the modulus of rigidity.

Substitute $32VA(1−y2c2)$ for $τxy$ .

$u=(32VA(1−y2c2))22G=9V28GA2(1+y4c4−2y2c2)$

Substitute $bd$ for A.

$u=9V28G(bd)2(1+y4c4−2y2c2)=9V28Gb2d2(1+y4c4−2y2c2)$

The value of $v=bdy$ .

Differentiate both sides of the Equation as shown below.

$dv=bdy dx$

Calculate the strain energy due to shear as shown below.

$U2=∫u dv$

Substitute $9V28Gb2d2(1+y4c4−2y2c2)$ for u, $bdy dx$ for $dv$ , and apply the limits.

$U2=∫0L∫−cc9V28Gb2d2(1+y4c4−2y2c2)bdy dx=9V28Gbd2∫0L∫−cc(1+y4c4−2y2c2)dy dx=9V28Gbd2∫0L(y+y55c4−2y33c2)−ccdx=9V28Gbd2{(c+c55c4−2c33c2)−((−c)+(−c)55c4−2(−c)33c2)}(x)0L$

$=9V28Gbd2{c+c5−2c3+c+c5−2c3}L=9V2cL8Gbd2×(15+3−10+15+3−10)15=9V2cL8Gbd2×1615=6V2cL5Gbd2$

Substitute $−M0L$ for V and $d2$ for c.

$U2=6(−M0L)2(d2)L5Gbd2=3M025GbdL$

Calculate the total strain energy as shown below.

$U=U1+U2$

Substitute $2M02Lbd3E$ for $U1$ and $3M025GbdL$ for $U2$ .

$U=2M02Lbd3E+3M025GbdL=2M02LEbd3(1+3Ed210GL2)$

Hence, the strain energy of the prismatic beam AB is $U=2M02LEbd3{1+3Ed210GL2}_$ .

Answer 6

Chapter 11.3, Problem 40P

To determine

Find the strain energy of the prismatic beam AB.

Expert Solution & Answer

Answer to Problem 40P

The strain energy of the prismatic beam AB is $U=2M02LEbd3{1+3Ed210GL2}_$ .

Explanation of Solution

Given information:

Taking into account the effect of both normal and shearing stresses.

Calculation:

Calculate the moment of inertia (I) for the rectangular cross section as shown below.

$I=bd312$

Here, b is the width of the cross section and d is the depth of the cross section.

Calculate the area of the cross section (A) as shown below.

$A=bd$

Calculate the centroid (c) as shown below.

$c=d2$

Calculate the reactions as shown below.

Take moment about B is Equal to zero.

$∑MB=0RAL+M0=0RAL=−M0RA=−M0L$

Summation of forces along vertical direction is Equal to zero.

$∑Fy=0RB−M0L=0RB=M0L$

Calculate the shear force as shown below.

Shear force at A is $SF@ A=−M0L$ .

Shear force at B is $SF@ B=−M0L+M0L=0$

Calculate the bending moment as shown below.

Bending moment at A is $BM@ A=M0$ .

Bending moment at B is $BM@ B=M0−M0L×L=0$

Sketch the shear force and bending moment diagram as shown in Figure 1.

Mechanics of Materials, 7th Edition, Chapter 11.3, Problem 40P

Refer to Figure 1.

Maximum shear force $V=−M0L$ .

Maximum bending moment $Mmax=M0$ .

Bending moment at a distance $v$ from B $M=M0Lv$

Calculate the strain energy due to bending $(U1)$ as shown below.

$U1=∫abM22EIdv$

Substitute $M=M0Lv$ for M and apply the limits.

$U1=∫0L(M0Lv)22EIdv=12EI∫0LM02v2L2dv=M022EIL2(v33)0L=M026EIL2×L3$

$=M02L6EI$

Substitute $bd312$ for I.

$U1=M02L6E(bd312)=2M02Lbd3E$

Calculate the shear stress $(τxy)$ as shown below.

$τxy=32VA(1−y2c2)$

Calculate the strain energy density $(u)$ as shown below.

$u=τxy22G$

Here, G is the modulus of rigidity.

Substitute $32VA(1−y2c2)$ for $τxy$ .

$u=(32VA(1−y2c2))22G=9V28GA2(1+y4c4−2y2c2)$

Substitute $bd$ for A.

$u=9V28G(bd)2(1+y4c4−2y2c2)=9V28Gb2d2(1+y4c4−2y2c2)$

The value of $v=bdy$ .

Differentiate both sides of the Equation as shown below.

$dv=bdy dx$

Calculate the strain energy due to shear as shown below.

$U2=∫u dv$

Substitute $9V28Gb2d2(1+y4c4−2y2c2)$ for u, $bdy dx$ for $dv$ , and apply the limits.

$U2=∫0L∫−cc9V28Gb2d2(1+y4c4−2y2c2)bdy dx=9V28Gbd2∫0L∫−cc(1+y4c4−2y2c2)dy dx=9V28Gbd2∫0L(y+y55c4−2y33c2)−ccdx=9V28Gbd2{(c+c55c4−2c33c2)−((−c)+(−c)55c4−2(−c)33c2)}(x)0L$

$=9V28Gbd2{c+c5−2c3+c+c5−2c3}L=9V2cL8Gbd2×(15+3−10+15+3−10)15=9V2cL8Gbd2×1615=6V2cL5Gbd2$

Substitute $−M0L$ for V and $d2$ for c.

$U2=6(−M0L)2(d2)L5Gbd2=3M025GbdL$

Calculate the total strain energy as shown below.

$U=U1+U2$

Substitute $2M02Lbd3E$ for $U1$ and $3M025GbdL$ for $U2$ .

$U=2M02Lbd3E+3M025GbdL=2M02LEbd3(1+3Ed210GL2)$

Hence, the strain energy of the prismatic beam AB is $U=2M02LEbd3{1+3Ed210GL2}_$ .

Answer 7

Chapter 11.3, Problem 40P

To determine

Find the strain energy of the prismatic beam AB.

Answer 8

Expert Solution & Answer

Answer to Problem 40P

The strain energy of the prismatic beam AB is $U=2M02LEbd3{1+3Ed210GL2}_$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given information:

Taking into account the effect of both normal and shearing stresses.

Calculation:

Calculate the moment of inertia (I) for the rectangular cross section as shown below.

$I=bd312$

Here, b is the width of the cross section and d is the depth of the cross section.

Calculate the area of the cross section (A) as shown below.

$A=bd$

Calculate the centroid (c) as shown below.

$c=d2$

Calculate the reactions as shown below.

Take moment about B is Equal to zero.

$∑MB=0RAL+M0=0RAL=−M0RA=−M0L$

Summation of forces along vertical direction is Equal to zero.

$∑Fy=0RB−M0L=0RB=M0L$

Calculate the shear force as shown below.

Shear force at A is $SF@ A=−M0L$ .

Shear force at B is $SF@ B=−M0L+M0L=0$

Calculate the bending moment as shown below.

Bending moment at A is $BM@ A=M0$ .

Bending moment at B is $BM@ B=M0−M0L×L=0$

Sketch the shear force and bending moment diagram as shown in Figure 1.

Mechanics of Materials, 7th Edition, Chapter 11.3, Problem 40P

Refer to Figure 1.

Maximum shear force $V=−M0L$ .

Maximum bending moment $Mmax=M0$ .

Bending moment at a distance $v$ from B $M=M0Lv$

Calculate the strain energy due to bending $(U1)$ as shown below.

$U1=∫abM22EIdv$

Substitute $M=M0Lv$ for M and apply the limits.

$U1=∫0L(M0Lv)22EIdv=12EI∫0LM02v2L2dv=M022EIL2(v33)0L=M026EIL2×L3$

$=M02L6EI$

Substitute $bd312$ for I.

$U1=M02L6E(bd312)=2M02Lbd3E$

Calculate the shear stress $(τxy)$ as shown below.

$τxy=32VA(1−y2c2)$

Calculate the strain energy density $(u)$ as shown below.

$u=τxy22G$

Here, G is the modulus of rigidity.

Substitute $32VA(1−y2c2)$ for $τxy$ .

$u=(32VA(1−y2c2))22G=9V28GA2(1+y4c4−2y2c2)$

Substitute $bd$ for A.

$u=9V28G(bd)2(1+y4c4−2y2c2)=9V28Gb2d2(1+y4c4−2y2c2)$

The value of $v=bdy$ .

Differentiate both sides of the Equation as shown below.

$dv=bdy dx$

Calculate the strain energy due to shear as shown below.

$U2=∫u dv$

Substitute $9V28Gb2d2(1+y4c4−2y2c2)$ for u, $bdy dx$ for $dv$ , and apply the limits.

$U2=∫0L∫−cc9V28Gb2d2(1+y4c4−2y2c2)bdy dx=9V28Gbd2∫0L∫−cc(1+y4c4−2y2c2)dy dx=9V28Gbd2∫0L(y+y55c4−2y33c2)−ccdx=9V28Gbd2{(c+c55c4−2c33c2)−((−c)+(−c)55c4−2(−c)33c2)}(x)0L$