Calculate the support reactions for the given structure using method of consistent deformation. Sketch the shear and bending moment diagrams for the given structure.

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Answer 1

Question

Chapter 13, Problem 38P

To determine

Calculate the support reactions for the given structure using method of consistent deformation.

Sketch the shear and bending moment diagrams for the given structure.

Expert Solution & Answer

Answer to Problem 38P

The vertical reaction at B is $By=25.5 k↑_$ .

The vertical reaction at D is $Dy=41.2 k↑_$ .

The vertical reaction at E is $Ey=18.3 k↑_$ .

The moment at E is $ME=56 k-ft (↻)_$ .

Explanation of Solution

Given information:

The structure is given in the Figure.

Apply the sign conventions for calculating reactions, forces and moments using the three equations of equilibrium as shown below.

For summation of forces along x-direction is equal to zero $(∑Fx=0)$ , consider the forces acting towards right side as positive $(→+)$ and the forces acting towards left side as negative $(←−)$ .
For summation of forces along y-direction is equal to zero $(∑Fy=0)$ , consider the upward force as positive $(↑+)$ and the downward force as negative $(↓−)$ .
For summation of moment about a point is equal to zero $(∑Mat a point=0)$ , consider the clockwise moment as positive when moment taken from left and the counter clockwise moment as positive when moment taken from right.
Consider the positive sign indicates the clockwise moment the negative sign indicates the counterclockwise moment.

Calculation:

Sketch the free body diagram of the beam as shown in Figure 1.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 1

Refer Figure 1.

Find the degree of indeterminacy of the structure:

Degree of determinacy of the beam is equal to the number of unknown reactions minus the number of equilibrium equations.

The beam is supported by 4 support reactions and the number of equilibrium equations is 2.

Therefore, the degree of indeterminacy of the beam is $i=2$ .

Select the vertical reaction $By$ and $Dy$ at the supports B and D as redundant.

Release the support B and D and consider the notation of moments due to external load as $MO$ .

Sketch the free body diagram of primary beam subjected to external loading without support B and D as shown in Figure 2.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 2

Refer Figure 2.

Find the reactions at the supports without considering support B and D using equilibrium equations:

Summation of moments of all forces about E is equal to 0.

$∑ME=0MEO−1(10)(102+40)−35(30)−2(20)(202)=0MEO=1,900 k-ft (↻)$

Summation of forces along y-direction is equal to 0.

$+↑∑Fy=0EyO−1(10)−35−2(20)=0EyO=85 k (↑)$

For unit value of the unknown redundant $Dy$ :

Consider the notation of moments due to external load as $mD$ .

Sketch the free body diagram of primary beam Subjected to unit value of redundant $Dy$ as shown in Figure 3.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 3

Refer Figure 3.

Find the support reaction and moment at E when 1 k vertical load applied at B.

Summation of moments of all forces about A is equal to 0.

$∑ME=0MEB=1(40)MEB=40 k-ft (↺)$

Summation of forces along y-direction is equal to 0.

$+↑∑Fy=0−EyB+1=0EyB=1 k (↓)$

For unit value of the unknown redundant $By$ :

Consider the notation of moments due to external load as $mB$ .

Sketch the free body diagram of primary beam Subjected to unit value of redundant $By$ as shown in Figure 4.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 4

Refer Figure 4.

Find the support reaction and moment at E when 1 k vertical load applied at D.

Summation of moments of all forces about E is equal to 0.

$∑MD=0MED=1(20)MED=20 k-ft (↺)$

Summation of forces along y-direction is equal to 0.

$+↑∑Fy=0−EyD+1=0EyD=1 k (↓)$

Find the moment equation of the beam for different sections on the beam.

Consider a section XX in the portion AB of the primary structure at a distance of x from A.

Refer Figure 1.

Draw the primary structure with section XX as shown in Figure 5.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 5

Refer Figure 5.

Portion AB.

Consider origin as A. $(0≤x≤10 ft)$ .

Find the moment at section XX in the portion AB as shown in Figure 5.

$MO=1(x)(x2)=x22$

Similarly calculate the moment of the remaining section in the external loading and redundant loading structures.

Tabulate the moment equation of different segment of beam as in Table 1.

Segment	x-coordinate		$MO$ $(k-ft)$	$mBY$ $(k-ft/k)$	$mDY$ $(k-ft/k)$
Segment	Origin	Limits (ft)	$MO$ $(k-ft)$	$mBY$ $(k-ft/k)$	$mDY$ $(k-ft/k)$
AB	A	$0−10$	$x22$	$0$	0
BC	E	$10−20$	$1(10)(x−102)$	$−1(x−10)$	0
CD	G	$20−30$	$1(10)(x−5)+35(x−20)$	$−1(x−10)$	0
DE	D	$30−50$	$[1(10)(x−5)+35(x−20)+(x−30)2]$	$−1(x−10)$	$−1(x−30)$

The vertical deflection at point B due to external loading is $ΔBO$ , vertical deflection at point D due to external loading is $ΔDO$ , and the flexibility coefficient representing the deflection at B due to unit value of redundant $By$ is $fBB$ , the redundant $Dy$ is $fDD$ , and the both redundant $By$ and $Dy$ is $fBD=fDB$ .

Calculate the value of $ΔBO$ using the equation as follows:

$ΔBO=Σ∫MOmBYEIdx=1EI{∫010(x22)(0)dx+∫1020[1(10)(x−102)][−1(x−10)]dx+∫2030[1(10)(x−5)+35(x−20)][−1(x−10)]dx+∫3050[1(10)(x−5)+35(x−20)+(x−30)2][−1(x−10)]dx}=1EI[0−5,833.33−60,000−753,333.33]=−819,166.67EI k-ft3$

Calculate the value of $ΔDO$ using the equation as follows:

$ΔDO=Σ∫MOmDYEIdx=1EI{∫010(0)dx+∫1020(0)dx+∫2030(0)dx+∫3050[1(10)(x−5)+35(x−20)+(x−30)2][−1(x−30)]dx}=1EI[0+0+0−280,000]=−280,000EI k-ft3$

Calculate the value of $fBB$ using the equation as follows:

$fBB=Σ∫mBY2EIdx=1EI{∫010(0)2dx+∫1020[−1(x−10)]2dx+∫2030[−1(x−10)]2dx+∫3050[−1(x−10)]2dx}=1EI[0+333.33+2,333.33+18,666.67]=−21,333.33EI k-ft3/k$

Calculate the value of $fDD$ using the equation as follows:

$fDD=Σ∫mDY2EIdx=1EI{∫010(0)2dx+∫1020(0)2dx+∫2030(0)2dx+∫3050[−1(x−30)]2dx}=1EI[0+0+0+2,666.67]=2,666.67EI k-ft3/k$

Calculate the value of $fFG$ using the equation as follows:

$fBD=fDB=Σ∫mBYmBYEIdx=1EI{∫010(0)dx+∫1020(0)dx+∫2030(0)dx+∫3050[−1(x−10)][−1(x−30)]dx}=1EI[0+0+0+,666.67]=6,666.67EI k-ft3/k$

Find the reactions and moment:

Find the horizontal and vertical reaction at F and G.

Show the first compatibility Equation as follows:

$ΔBO+fBBBy+fBDDy=0$

Substitute $−819,166.67EI k-ft3$ for $ΔBO$ , $21,333.33EI k-ft3$ for $fBB$ , and $6,666.67EI k-ft3/k$ for $fBD$ .

$−819,166.67EI +(21,333.33EI)By+(6,666.67EI)Dy=021,333.33By+6,666.67Dy=819,166.67$ (1)

Show the second compatibility Equation as follows:

$ΔDO+fDBBy+fDDDy=0$

Substitute $−280,000EI k-ft3$ for $ΔDO$ , $6,666.67EI k-ft3/k$ for $fDB$ , and $2,666.67EI k-ft3/k$ for $fDD$ .

$−280,000EI+(6,666.67EI)By+(2,666.67EI)Dy=06,666.67By+2,666.67Dy=280,000$ (2)

Solve Equation (1) and (2).

$By=25.5 k (↑)Dy=41.2 k (↑)$

Find the vertical reaction at E.

Summation of forces along y-direction is equal to 0.

$+↑∑Fy=0By+Dy+Ey=1(10)+35+(2)(20)25.5+41.2+Ey=85Ey=18.3 k (↑)$

Find the moment at E.

Summation of moments of all forces about E is equal to 0.

$∑ME=0ME−1(10)(102+40)−35(30)−2(20)(202)+(25.5×40)+(41.2×20)=0ME=56 k-ft (↻)$

Therefore, the vertical reaction at B is $By=25.5 k↑_$ .

Therefore, the vertical reaction at D is $Dy=41.2 k↑_$ .

Therefore, the vertical reaction at E is $Ey=18.3 k↑_$ .

Therefore, the moment at E is $ME=56 k-ft (↻)_$ .

Sketch the free body diagram with support reactions of the beam as shown in Figure 6.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 6

Refer Figure 6,

Find the shear force $(S)$ for the given beam:

For span AB,

At point A.

$SA=0$

At point B, (negative side):

$SB−=−1(10)=−10 k$

For span BD,

At point B, (positive side):

$SB+=−10+25.5=15.5 k$

At point C, (positive side):

$SC+=15.5 k$

At point C, (negative side):

$SC−=15.5−35=−19.5 k$

At point D, (negative side):

$SD−=−19.5 k$

For span DF,

At point D, (positive side):

$SD+=−19.5+41.2=21.7 k$

At point E, (negative side):

$SE−=21.7−2(20)=−18.3 k$

At point E, (positive side):

$SE+=−18.3+18.3=0$

Sketch the shear diagram for the given beam as shown in Figure 7.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 7

Refer Figure 7.

The points of zero shear as F.

Find the point of zero shear force from E.

$18.3−2(x1)=SF018.3−2(x1)=02x1=18.3x1=9.15 ft$

Refer Figure 6.

Find the bending moment $(M)$ for the beam:

For span AB,

At point A, (free end)

$MA=0$

At point B,

$MB=−1(10)(102)=−50 k-ft$

At point C,

$MC=−1(10)(15)+25.5(10)=105 k-ft$

At point D,

$MD=−1(10)(25)+25.5(20)−35(10)=−90 k-ft$

At point F,

$MF=2(9.15)(9.152)−56=27.7 k-ft$

At point E,

$ME=56 k-ft$

Sketch the bending moment diagram for the given beam as shown in Figure 8.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 8

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Answer 2

Question

Chapter 13, Problem 38P

To determine

Calculate the support reactions for the given structure using method of consistent deformation.

Sketch the shear and bending moment diagrams for the given structure.

Expert Solution & Answer

Answer to Problem 38P

The vertical reaction at B is $By=25.5 k↑_$ .

The vertical reaction at D is $Dy=41.2 k↑_$ .

The vertical reaction at E is $Ey=18.3 k↑_$ .

The moment at E is $ME=56 k-ft (↻)_$ .

Explanation of Solution

Given information:

The structure is given in the Figure.

Apply the sign conventions for calculating reactions, forces and moments using the three equations of equilibrium as shown below.

For summation of forces along x-direction is equal to zero $(∑Fx=0)$ , consider the forces acting towards right side as positive $(→+)$ and the forces acting towards left side as negative $(←−)$ .
For summation of forces along y-direction is equal to zero $(∑Fy=0)$ , consider the upward force as positive $(↑+)$ and the downward force as negative $(↓−)$ .
For summation of moment about a point is equal to zero $(∑Mat a point=0)$ , consider the clockwise moment as positive when moment taken from left and the counter clockwise moment as positive when moment taken from right.
Consider the positive sign indicates the clockwise moment the negative sign indicates the counterclockwise moment.

Calculation:

Sketch the free body diagram of the beam as shown in Figure 1.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 1

Refer Figure 1.

Find the degree of indeterminacy of the structure:

Degree of determinacy of the beam is equal to the number of unknown reactions minus the number of equilibrium equations.

The beam is supported by 4 support reactions and the number of equilibrium equations is 2.

Therefore, the degree of indeterminacy of the beam is $i=2$ .

Select the vertical reaction $By$ and $Dy$ at the supports B and D as redundant.

Release the support B and D and consider the notation of moments due to external load as $MO$ .

Sketch the free body diagram of primary beam subjected to external loading without support B and D as shown in Figure 2.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 2

Refer Figure 2.

Find the reactions at the supports without considering support B and D using equilibrium equations:

Summation of moments of all forces about E is equal to 0.

$∑ME=0MEO−1(10)(102+40)−35(30)−2(20)(202)=0MEO=1,900 k-ft (↻)$

Summation of forces along y-direction is equal to 0.

$+↑∑Fy=0EyO−1(10)−35−2(20)=0EyO=85 k (↑)$

For unit value of the unknown redundant $Dy$ :

Consider the notation of moments due to external load as $mD$ .

Sketch the free body diagram of primary beam Subjected to unit value of redundant $Dy$ as shown in Figure 3.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 3

Refer Figure 3.

Find the support reaction and moment at E when 1 k vertical load applied at B.

Summation of moments of all forces about A is equal to 0.

$∑ME=0MEB=1(40)MEB=40 k-ft (↺)$

Summation of forces along y-direction is equal to 0.

$+↑∑Fy=0−EyB+1=0EyB=1 k (↓)$

For unit value of the unknown redundant $By$ :

Consider the notation of moments due to external load as $mB$ .

Sketch the free body diagram of primary beam Subjected to unit value of redundant $By$ as shown in Figure 4.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 4

Refer Figure 4.

Find the support reaction and moment at E when 1 k vertical load applied at D.

Summation of moments of all forces about E is equal to 0.

$∑MD=0MED=1(20)MED=20 k-ft (↺)$

Summation of forces along y-direction is equal to 0.

$+↑∑Fy=0−EyD+1=0EyD=1 k (↓)$

Find the moment equation of the beam for different sections on the beam.

Consider a section XX in the portion AB of the primary structure at a distance of x from A.

Refer Figure 1.

Draw the primary structure with section XX as shown in Figure 5.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 5

Refer Figure 5.

Portion AB.

Consider origin as A. $(0≤x≤10 ft)$ .

Find the moment at section XX in the portion AB as shown in Figure 5.

$MO=1(x)(x2)=x22$

Similarly calculate the moment of the remaining section in the external loading and redundant loading structures.

Tabulate the moment equation of different segment of beam as in Table 1.

Segment	x-coordinate		$MO$ $(k-ft)$	$mBY$ $(k-ft/k)$	$mDY$ $(k-ft/k)$
Segment	Origin	Limits (ft)	$MO$ $(k-ft)$	$mBY$ $(k-ft/k)$	$mDY$ $(k-ft/k)$
AB	A	$0−10$	$x22$	$0$	0
BC	E	$10−20$	$1(10)(x−102)$	$−1(x−10)$	0
CD	G	$20−30$	$1(10)(x−5)+35(x−20)$	$−1(x−10)$	0
DE	D	$30−50$	$[1(10)(x−5)+35(x−20)+(x−30)2]$	$−1(x−10)$	$−1(x−30)$

The vertical deflection at point B due to external loading is $ΔBO$ , vertical deflection at point D due to external loading is $ΔDO$ , and the flexibility coefficient representing the deflection at B due to unit value of redundant $By$ is $fBB$ , the redundant $Dy$ is $fDD$ , and the both redundant $By$ and $Dy$ is $fBD=fDB$ .

Calculate the value of $ΔBO$ using the equation as follows:

$ΔBO=Σ∫MOmBYEIdx=1EI{∫010(x22)(0)dx+∫1020[1(10)(x−102)][−1(x−10)]dx+∫2030[1(10)(x−5)+35(x−20)][−1(x−10)]dx+∫3050[1(10)(x−5)+35(x−20)+(x−30)2][−1(x−10)]dx}=1EI[0−5,833.33−60,000−753,333.33]=−819,166.67EI k-ft3$

Calculate the value of $ΔDO$ using the equation as follows:

$ΔDO=Σ∫MOmDYEIdx=1EI{∫010(0)dx+∫1020(0)dx+∫2030(0)dx+∫3050[1(10)(x−5)+35(x−20)+(x−30)2][−1(x−30)]dx}=1EI[0+0+0−280,000]=−280,000EI k-ft3$

Calculate the value of $fBB$ using the equation as follows:

$fBB=Σ∫mBY2EIdx=1EI{∫010(0)2dx+∫1020[−1(x−10)]2dx+∫2030[−1(x−10)]2dx+∫3050[−1(x−10)]2dx}=1EI[0+333.33+2,333.33+18,666.67]=−21,333.33EI k-ft3/k$

Calculate the value of $fDD$ using the equation as follows:

$fDD=Σ∫mDY2EIdx=1EI{∫010(0)2dx+∫1020(0)2dx+∫2030(0)2dx+∫3050[−1(x−30)]2dx}=1EI[0+0+0+2,666.67]=2,666.67EI k-ft3/k$

Calculate the value of $fFG$ using the equation as follows:

$fBD=fDB=Σ∫mBYmBYEIdx=1EI{∫010(0)dx+∫1020(0)dx+∫2030(0)dx+∫3050[−1(x−10)][−1(x−30)]dx}=1EI[0+0+0+,666.67]=6,666.67EI k-ft3/k$

Find the reactions and moment:

Find the horizontal and vertical reaction at F and G.

Show the first compatibility Equation as follows:

$ΔBO+fBBBy+fBDDy=0$

Substitute $−819,166.67EI k-ft3$ for $ΔBO$ , $21,333.33EI k-ft3$ for $fBB$ , and $6,666.67EI k-ft3/k$ for $fBD$ .

$−819,166.67EI +(21,333.33EI)By+(6,666.67EI)Dy=021,333.33By+6,666.67Dy=819,166.67$ (1)

Show the second compatibility Equation as follows:

$ΔDO+fDBBy+fDDDy=0$

Substitute $−280,000EI k-ft3$ for $ΔDO$ , $6,666.67EI k-ft3/k$ for $fDB$ , and $2,666.67EI k-ft3/k$ for $fDD$ .

$−280,000EI+(6,666.67EI)By+(2,666.67EI)Dy=06,666.67By+2,666.67Dy=280,000$ (2)

Solve Equation (1) and (2).

$By=25.5 k (↑)Dy=41.2 k (↑)$

Find the vertical reaction at E.

Summation of forces along y-direction is equal to 0.

$+↑∑Fy=0By+Dy+Ey=1(10)+35+(2)(20)25.5+41.2+Ey=85Ey=18.3 k (↑)$

Find the moment at E.

Summation of moments of all forces about E is equal to 0.

$∑ME=0ME−1(10)(102+40)−35(30)−2(20)(202)+(25.5×40)+(41.2×20)=0ME=56 k-ft (↻)$

Therefore, the vertical reaction at B is $By=25.5 k↑_$ .

Therefore, the vertical reaction at D is $Dy=41.2 k↑_$ .

Therefore, the vertical reaction at E is $Ey=18.3 k↑_$ .

Therefore, the moment at E is $ME=56 k-ft (↻)_$ .

Sketch the free body diagram with support reactions of the beam as shown in Figure 6.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 6

Refer Figure 6,

Find the shear force $(S)$ for the given beam:

For span AB,

At point A.

$SA=0$

At point B, (negative side):

$SB−=−1(10)=−10 k$

For span BD,

At point B, (positive side):

$SB+=−10+25.5=15.5 k$

At point C, (positive side):

$SC+=15.5 k$

At point C, (negative side):

$SC−=15.5−35=−19.5 k$

At point D, (negative side):

$SD−=−19.5 k$

For span DF,

At point D, (positive side):

$SD+=−19.5+41.2=21.7 k$

At point E, (negative side):

$SE−=21.7−2(20)=−18.3 k$

At point E, (positive side):

$SE+=−18.3+18.3=0$

Sketch the shear diagram for the given beam as shown in Figure 7.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 7

Refer Figure 7.

The points of zero shear as F.

Find the point of zero shear force from E.

$18.3−2(x1)=SF018.3−2(x1)=02x1=18.3x1=9.15 ft$

Refer Figure 6.

Find the bending moment $(M)$ for the beam:

For span AB,

At point A, (free end)

$MA=0$

At point B,

$MB=−1(10)(102)=−50 k-ft$

At point C,

$MC=−1(10)(15)+25.5(10)=105 k-ft$

At point D,

$MD=−1(10)(25)+25.5(20)−35(10)=−90 k-ft$

At point F,

$MF=2(9.15)(9.152)−56=27.7 k-ft$

At point E,

$ME=56 k-ft$

Sketch the bending moment diagram for the given beam as shown in Figure 8.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 8

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Answer 3

Question

Chapter 13, Problem 38P

To determine

Calculate the support reactions for the given structure using method of consistent deformation.

Sketch the shear and bending moment diagrams for the given structure.

Expert Solution & Answer

Answer to Problem 38P

The vertical reaction at B is $By=25.5 k↑_$ .

The vertical reaction at D is $Dy=41.2 k↑_$ .

The vertical reaction at E is $Ey=18.3 k↑_$ .

The moment at E is $ME=56 k-ft (↻)_$ .

Explanation of Solution

Given information:

The structure is given in the Figure.

Apply the sign conventions for calculating reactions, forces and moments using the three equations of equilibrium as shown below.

For summation of forces along x-direction is equal to zero $(∑Fx=0)$ , consider the forces acting towards right side as positive $(→+)$ and the forces acting towards left side as negative $(←−)$ .
For summation of forces along y-direction is equal to zero $(∑Fy=0)$ , consider the upward force as positive $(↑+)$ and the downward force as negative $(↓−)$ .
For summation of moment about a point is equal to zero $(∑Mat a point=0)$ , consider the clockwise moment as positive when moment taken from left and the counter clockwise moment as positive when moment taken from right.
Consider the positive sign indicates the clockwise moment the negative sign indicates the counterclockwise moment.

Calculation:

Sketch the free body diagram of the beam as shown in Figure 1.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 1

Refer Figure 1.

Find the degree of indeterminacy of the structure:

Degree of determinacy of the beam is equal to the number of unknown reactions minus the number of equilibrium equations.

The beam is supported by 4 support reactions and the number of equilibrium equations is 2.

Therefore, the degree of indeterminacy of the beam is $i=2$ .

Select the vertical reaction $By$ and $Dy$ at the supports B and D as redundant.

Release the support B and D and consider the notation of moments due to external load as $MO$ .

Sketch the free body diagram of primary beam subjected to external loading without support B and D as shown in Figure 2.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 2

Refer Figure 2.

Find the reactions at the supports without considering support B and D using equilibrium equations:

Summation of moments of all forces about E is equal to 0.

$∑ME=0MEO−1(10)(102+40)−35(30)−2(20)(202)=0MEO=1,900 k-ft (↻)$

Summation of forces along y-direction is equal to 0.

$+↑∑Fy=0EyO−1(10)−35−2(20)=0EyO=85 k (↑)$

For unit value of the unknown redundant $Dy$ :

Consider the notation of moments due to external load as $mD$ .

Sketch the free body diagram of primary beam Subjected to unit value of redundant $Dy$ as shown in Figure 3.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 3

Refer Figure 3.

Find the support reaction and moment at E when 1 k vertical load applied at B.

Summation of moments of all forces about A is equal to 0.

$∑ME=0MEB=1(40)MEB=40 k-ft (↺)$

Summation of forces along y-direction is equal to 0.

$+↑∑Fy=0−EyB+1=0EyB=1 k (↓)$

For unit value of the unknown redundant $By$ :

Consider the notation of moments due to external load as $mB$ .

Sketch the free body diagram of primary beam Subjected to unit value of redundant $By$ as shown in Figure 4.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 4

Refer Figure 4.

Find the support reaction and moment at E when 1 k vertical load applied at D.

Summation of moments of all forces about E is equal to 0.

$∑MD=0MED=1(20)MED=20 k-ft (↺)$

Summation of forces along y-direction is equal to 0.

$+↑∑Fy=0−EyD+1=0EyD=1 k (↓)$

Find the moment equation of the beam for different sections on the beam.

Consider a section XX in the portion AB of the primary structure at a distance of x from A.

Refer Figure 1.

Draw the primary structure with section XX as shown in Figure 5.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 5

Refer Figure 5.

Portion AB.

Consider origin as A. $(0≤x≤10 ft)$ .

Find the moment at section XX in the portion AB as shown in Figure 5.

$MO=1(x)(x2)=x22$

Similarly calculate the moment of the remaining section in the external loading and redundant loading structures.

Tabulate the moment equation of different segment of beam as in Table 1.

Segment	x-coordinate		$MO$ $(k-ft)$	$mBY$ $(k-ft/k)$	$mDY$ $(k-ft/k)$
Segment	Origin	Limits (ft)	$MO$ $(k-ft)$	$mBY$ $(k-ft/k)$	$mDY$ $(k-ft/k)$
AB	A	$0−10$	$x22$	$0$	0
BC	E	$10−20$	$1(10)(x−102)$	$−1(x−10)$	0
CD	G	$20−30$	$1(10)(x−5)+35(x−20)$	$−1(x−10)$	0
DE	D	$30−50$	$[1(10)(x−5)+35(x−20)+(x−30)2]$	$−1(x−10)$	$−1(x−30)$

The vertical deflection at point B due to external loading is $ΔBO$ , vertical deflection at point D due to external loading is $ΔDO$ , and the flexibility coefficient representing the deflection at B due to unit value of redundant $By$ is $fBB$ , the redundant $Dy$ is $fDD$ , and the both redundant $By$ and $Dy$ is $fBD=fDB$ .

Calculate the value of $ΔBO$ using the equation as follows:

$ΔBO=Σ∫MOmBYEIdx=1EI{∫010(x22)(0)dx+∫1020[1(10)(x−102)][−1(x−10)]dx+∫2030[1(10)(x−5)+35(x−20)][−1(x−10)]dx+∫3050[1(10)(x−5)+35(x−20)+(x−30)2][−1(x−10)]dx}=1EI[0−5,833.33−60,000−753,333.33]=−819,166.67EI k-ft3$

Calculate the value of $ΔDO$ using the equation as follows:

$ΔDO=Σ∫MOmDYEIdx=1EI{∫010(0)dx+∫1020(0)dx+∫2030(0)dx+∫3050[1(10)(x−5)+35(x−20)+(x−30)2][−1(x−30)]dx}=1EI[0+0+0−280,000]=−280,000EI k-ft3$

Calculate the value of $fBB$ using the equation as follows:

$fBB=Σ∫mBY2EIdx=1EI{∫010(0)2dx+∫1020[−1(x−10)]2dx+∫2030[−1(x−10)]2dx+∫3050[−1(x−10)]2dx}=1EI[0+333.33+2,333.33+18,666.67]=−21,333.33EI k-ft3/k$

Calculate the value of $fDD$ using the equation as follows:

$fDD=Σ∫mDY2EIdx=1EI{∫010(0)2dx+∫1020(0)2dx+∫2030(0)2dx+∫3050[−1(x−30)]2dx}=1EI[0+0+0+2,666.67]=2,666.67EI k-ft3/k$

Calculate the value of $fFG$ using the equation as follows:

$fBD=fDB=Σ∫mBYmBYEIdx=1EI{∫010(0)dx+∫1020(0)dx+∫2030(0)dx+∫3050[−1(x−10)][−1(x−30)]dx}=1EI[0+0+0+,666.67]=6,666.67EI k-ft3/k$

Find the reactions and moment:

Find the horizontal and vertical reaction at F and G.

Show the first compatibility Equation as follows:

$ΔBO+fBBBy+fBDDy=0$

Substitute $−819,166.67EI k-ft3$ for $ΔBO$ , $21,333.33EI k-ft3$ for $fBB$ , and $6,666.67EI k-ft3/k$ for $fBD$ .

$−819,166.67EI +(21,333.33EI)By+(6,666.67EI)Dy=021,333.33By+6,666.67Dy=819,166.67$ (1)

Show the second compatibility Equation as follows:

$ΔDO+fDBBy+fDDDy=0$

Substitute $−280,000EI k-ft3$ for $ΔDO$ , $6,666.67EI k-ft3/k$ for $fDB$ , and $2,666.67EI k-ft3/k$ for $fDD$ .

$−280,000EI+(6,666.67EI)By+(2,666.67EI)Dy=06,666.67By+2,666.67Dy=280,000$ (2)

Solve Equation (1) and (2).

$By=25.5 k (↑)Dy=41.2 k (↑)$

Find the vertical reaction at E.

Summation of forces along y-direction is equal to 0.

$+↑∑Fy=0By+Dy+Ey=1(10)+35+(2)(20)25.5+41.2+Ey=85Ey=18.3 k (↑)$

Find the moment at E.

Summation of moments of all forces about E is equal to 0.

$∑ME=0ME−1(10)(102+40)−35(30)−2(20)(202)+(25.5×40)+(41.2×20)=0ME=56 k-ft (↻)$

Therefore, the vertical reaction at B is $By=25.5 k↑_$ .

Therefore, the vertical reaction at D is $Dy=41.2 k↑_$ .

Therefore, the vertical reaction at E is $Ey=18.3 k↑_$ .

Therefore, the moment at E is $ME=56 k-ft (↻)_$ .

Sketch the free body diagram with support reactions of the beam as shown in Figure 6.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 6

Refer Figure 6,

Find the shear force $(S)$ for the given beam:

For span AB,

At point A.

$SA=0$

At point B, (negative side):

$SB−=−1(10)=−10 k$

For span BD,

At point B, (positive side):

$SB+=−10+25.5=15.5 k$

At point C, (positive side):

$SC+=15.5 k$

At point C, (negative side):

$SC−=15.5−35=−19.5 k$

At point D, (negative side):

$SD−=−19.5 k$

For span DF,

At point D, (positive side):

$SD+=−19.5+41.2=21.7 k$

At point E, (negative side):

$SE−=21.7−2(20)=−18.3 k$

At point E, (positive side):

$SE+=−18.3+18.3=0$

Sketch the shear diagram for the given beam as shown in Figure 7.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 7

Refer Figure 7.

The points of zero shear as F.

Find the point of zero shear force from E.

$18.3−2(x1)=SF018.3−2(x1)=02x1=18.3x1=9.15 ft$

Refer Figure 6.

Find the bending moment $(M)$ for the beam:

For span AB,

At point A, (free end)

$MA=0$

At point B,

$MB=−1(10)(102)=−50 k-ft$

At point C,

$MC=−1(10)(15)+25.5(10)=105 k-ft$

At point D,

$MD=−1(10)(25)+25.5(20)−35(10)=−90 k-ft$

At point F,

$MF=2(9.15)(9.152)−56=27.7 k-ft$

At point E,

$ME=56 k-ft$

Sketch the bending moment diagram for the given beam as shown in Figure 8.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 8

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Answer 4

Question

Chapter 13, Problem 38P

To determine

Calculate the support reactions for the given structure using method of consistent deformation.

Sketch the shear and bending moment diagrams for the given structure.

Expert Solution & Answer

Answer to Problem 38P

The vertical reaction at B is $By=25.5 k↑_$ .

The vertical reaction at D is $Dy=41.2 k↑_$ .

The vertical reaction at E is $Ey=18.3 k↑_$ .

The moment at E is $ME=56 k-ft (↻)_$ .

Explanation of Solution

Given information:

The structure is given in the Figure.

Apply the sign conventions for calculating reactions, forces and moments using the three equations of equilibrium as shown below.

For summation of forces along x-direction is equal to zero $(∑Fx=0)$ , consider the forces acting towards right side as positive $(→+)$ and the forces acting towards left side as negative $(←−)$ .
For summation of forces along y-direction is equal to zero $(∑Fy=0)$ , consider the upward force as positive $(↑+)$ and the downward force as negative $(↓−)$ .
For summation of moment about a point is equal to zero $(∑Mat a point=0)$ , consider the clockwise moment as positive when moment taken from left and the counter clockwise moment as positive when moment taken from right.
Consider the positive sign indicates the clockwise moment the negative sign indicates the counterclockwise moment.

Calculation:

Sketch the free body diagram of the beam as shown in Figure 1.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 1

Refer Figure 1.

Find the degree of indeterminacy of the structure:

Degree of determinacy of the beam is equal to the number of unknown reactions minus the number of equilibrium equations.

The beam is supported by 4 support reactions and the number of equilibrium equations is 2.

Therefore, the degree of indeterminacy of the beam is $i=2$ .

Select the vertical reaction $By$ and $Dy$ at the supports B and D as redundant.

Release the support B and D and consider the notation of moments due to external load as $MO$ .

Sketch the free body diagram of primary beam subjected to external loading without support B and D as shown in Figure 2.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 2

Refer Figure 2.

Find the reactions at the supports without considering support B and D using equilibrium equations:

Summation of moments of all forces about E is equal to 0.

$∑ME=0MEO−1(10)(102+40)−35(30)−2(20)(202)=0MEO=1,900 k-ft (↻)$

Summation of forces along y-direction is equal to 0.

$+↑∑Fy=0EyO−1(10)−35−2(20)=0EyO=85 k (↑)$

For unit value of the unknown redundant $Dy$ :

Consider the notation of moments due to external load as $mD$ .

Sketch the free body diagram of primary beam Subjected to unit value of redundant $Dy$ as shown in Figure 3.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 3

Refer Figure 3.

Find the support reaction and moment at E when 1 k vertical load applied at B.

Summation of moments of all forces about A is equal to 0.

$∑ME=0MEB=1(40)MEB=40 k-ft (↺)$

Summation of forces along y-direction is equal to 0.

$+↑∑Fy=0−EyB+1=0EyB=1 k (↓)$

For unit value of the unknown redundant $By$ :

Consider the notation of moments due to external load as $mB$ .

Sketch the free body diagram of primary beam Subjected to unit value of redundant $By$ as shown in Figure 4.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 4

Refer Figure 4.

Find the support reaction and moment at E when 1 k vertical load applied at D.

Summation of moments of all forces about E is equal to 0.

$∑MD=0MED=1(20)MED=20 k-ft (↺)$

Summation of forces along y-direction is equal to 0.

$+↑∑Fy=0−EyD+1=0EyD=1 k (↓)$

Find the moment equation of the beam for different sections on the beam.

Consider a section XX in the portion AB of the primary structure at a distance of x from A.

Refer Figure 1.

Draw the primary structure with section XX as shown in Figure 5.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 5

Refer Figure 5.

Portion AB.

Consider origin as A. $(0≤x≤10 ft)$ .

Find the moment at section XX in the portion AB as shown in Figure 5.

$MO=1(x)(x2)=x22$

Similarly calculate the moment of the remaining section in the external loading and redundant loading structures.

Tabulate the moment equation of different segment of beam as in Table 1.

Segment	x-coordinate		$MO$ $(k-ft)$	$mBY$ $(k-ft/k)$	$mDY$ $(k-ft/k)$
Segment	Origin	Limits (ft)	$MO$ $(k-ft)$	$mBY$ $(k-ft/k)$	$mDY$ $(k-ft/k)$
AB	A	$0−10$	$x22$	$0$	0
BC	E	$10−20$	$1(10)(x−102)$	$−1(x−10)$	0
CD	G	$20−30$	$1(10)(x−5)+35(x−20)$	$−1(x−10)$	0
DE	D	$30−50$	$[1(10)(x−5)+35(x−20)+(x−30)2]$	$−1(x−10)$	$−1(x−30)$

The vertical deflection at point B due to external loading is $ΔBO$ , vertical deflection at point D due to external loading is $ΔDO$ , and the flexibility coefficient representing the deflection at B due to unit value of redundant $By$ is $fBB$ , the redundant $Dy$ is $fDD$ , and the both redundant $By$ and $Dy$ is $fBD=fDB$ .

Calculate the value of $ΔBO$ using the equation as follows:

$ΔBO=Σ∫MOmBYEIdx=1EI{∫010(x22)(0)dx+∫1020[1(10)(x−102)][−1(x−10)]dx+∫2030[1(10)(x−5)+35(x−20)][−1(x−10)]dx+∫3050[1(10)(x−5)+35(x−20)+(x−30)2][−1(x−10)]dx}=1EI[0−5,833.33−60,000−753,333.33]=−819,166.67EI k-ft3$

Calculate the value of $ΔDO$ using the equation as follows:

$ΔDO=Σ∫MOmDYEIdx=1EI{∫010(0)dx+∫1020(0)dx+∫2030(0)dx+∫3050[1(10)(x−5)+35(x−20)+(x−30)2][−1(x−30)]dx}=1EI[0+0+0−280,000]=−280,000EI k-ft3$

Calculate the value of $fBB$ using the equation as follows:

$fBB=Σ∫mBY2EIdx=1EI{∫010(0)2dx+∫1020[−1(x−10)]2dx+∫2030[−1(x−10)]2dx+∫3050[−1(x−10)]2dx}=1EI[0+333.33+2,333.33+18,666.67]=−21,333.33EI k-ft3/k$

Calculate the value of $fDD$ using the equation as follows:

$fDD=Σ∫mDY2EIdx=1EI{∫010(0)2dx+∫1020(0)2dx+∫2030(0)2dx+∫3050[−1(x−30)]2dx}=1EI[0+0+0+2,666.67]=2,666.67EI k-ft3/k$

Calculate the value of $fFG$ using the equation as follows:

$fBD=fDB=Σ∫mBYmBYEIdx=1EI{∫010(0)dx+∫1020(0)dx+∫2030(0)dx+∫3050[−1(x−10)][−1(x−30)]dx}=1EI[0+0+0+,666.67]=6,666.67EI k-ft3/k$

Find the reactions and moment:

Find the horizontal and vertical reaction at F and G.

Show the first compatibility Equation as follows:

$ΔBO+fBBBy+fBDDy=0$

Substitute $−819,166.67EI k-ft3$ for $ΔBO$ , $21,333.33EI k-ft3$ for $fBB$ , and $6,666.67EI k-ft3/k$ for $fBD$ .

$−819,166.67EI +(21,333.33EI)By+(6,666.67EI)Dy=021,333.33By+6,666.67Dy=819,166.67$ (1)

Show the second compatibility Equation as follows:

$ΔDO+fDBBy+fDDDy=0$

Substitute $−280,000EI k-ft3$ for $ΔDO$ , $6,666.67EI k-ft3/k$ for $fDB$ , and $2,666.67EI k-ft3/k$ for $fDD$ .

$−280,000EI+(6,666.67EI)By+(2,666.67EI)Dy=06,666.67By+2,666.67Dy=280,000$ (2)

Solve Equation (1) and (2).

$By=25.5 k (↑)Dy=41.2 k (↑)$

Find the vertical reaction at E.

Summation of forces along y-direction is equal to 0.

$+↑∑Fy=0By+Dy+Ey=1(10)+35+(2)(20)25.5+41.2+Ey=85Ey=18.3 k (↑)$

Find the moment at E.

Summation of moments of all forces about E is equal to 0.

$∑ME=0ME−1(10)(102+40)−35(30)−2(20)(202)+(25.5×40)+(41.2×20)=0ME=56 k-ft (↻)$

Therefore, the vertical reaction at B is $By=25.5 k↑_$ .

Therefore, the vertical reaction at D is $Dy=41.2 k↑_$ .

Therefore, the vertical reaction at E is $Ey=18.3 k↑_$ .

Therefore, the moment at E is $ME=56 k-ft (↻)_$ .

Sketch the free body diagram with support reactions of the beam as shown in Figure 6.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 6

Refer Figure 6,

Find the shear force $(S)$ for the given beam:

For span AB,

At point A.

$SA=0$

At point B, (negative side):

$SB−=−1(10)=−10 k$

For span BD,

At point B, (positive side):

$SB+=−10+25.5=15.5 k$

At point C, (positive side):

$SC+=15.5 k$

At point C, (negative side):

$SC−=15.5−35=−19.5 k$

At point D, (negative side):

$SD−=−19.5 k$

For span DF,

At point D, (positive side):

$SD+=−19.5+41.2=21.7 k$

At point E, (negative side):

$SE−=21.7−2(20)=−18.3 k$

At point E, (positive side):

$SE+=−18.3+18.3=0$

Sketch the shear diagram for the given beam as shown in Figure 7.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 7

Refer Figure 7.

The points of zero shear as F.

Find the point of zero shear force from E.

$18.3−2(x1)=SF018.3−2(x1)=02x1=18.3x1=9.15 ft$

Refer Figure 6.

Find the bending moment $(M)$ for the beam:

For span AB,

At point A, (free end)

$MA=0$

At point B,

$MB=−1(10)(102)=−50 k-ft$

At point C,

$MC=−1(10)(15)+25.5(10)=105 k-ft$

At point D,

$MD=−1(10)(25)+25.5(20)−35(10)=−90 k-ft$

At point F,

$MF=2(9.15)(9.152)−56=27.7 k-ft$

At point E,

$ME=56 k-ft$

Sketch the bending moment diagram for the given beam as shown in Figure 8.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 8

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Answer 5

Chapter 13, Problem 38P

To determine

Calculate the support reactions for the given structure using method of consistent deformation.

Sketch the shear and bending moment diagrams for the given structure.

Expert Solution & Answer

Answer to Problem 38P

The vertical reaction at B is $By=25.5 k↑_$ .

The vertical reaction at D is $Dy=41.2 k↑_$ .

The vertical reaction at E is $Ey=18.3 k↑_$ .

The moment at E is $ME=56 k-ft (↻)_$ .

Explanation of Solution

Given information:

The structure is given in the Figure.

Apply the sign conventions for calculating reactions, forces and moments using the three equations of equilibrium as shown below.

For summation of forces along x-direction is equal to zero $(∑Fx=0)$ , consider the forces acting towards right side as positive $(→+)$ and the forces acting towards left side as negative $(←−)$ .
For summation of forces along y-direction is equal to zero $(∑Fy=0)$ , consider the upward force as positive $(↑+)$ and the downward force as negative $(↓−)$ .
For summation of moment about a point is equal to zero $(∑Mat a point=0)$ , consider the clockwise moment as positive when moment taken from left and the counter clockwise moment as positive when moment taken from right.
Consider the positive sign indicates the clockwise moment the negative sign indicates the counterclockwise moment.

Calculation:

Sketch the free body diagram of the beam as shown in Figure 1.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 1

Refer Figure 1.

Find the degree of indeterminacy of the structure:

Degree of determinacy of the beam is equal to the number of unknown reactions minus the number of equilibrium equations.

The beam is supported by 4 support reactions and the number of equilibrium equations is 2.

Therefore, the degree of indeterminacy of the beam is $i=2$ .

Select the vertical reaction $By$ and $Dy$ at the supports B and D as redundant.

Release the support B and D and consider the notation of moments due to external load as $MO$ .

Sketch the free body diagram of primary beam subjected to external loading without support B and D as shown in Figure 2.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 2

Refer Figure 2.

Find the reactions at the supports without considering support B and D using equilibrium equations:

Summation of moments of all forces about E is equal to 0.

$∑ME=0MEO−1(10)(102+40)−35(30)−2(20)(202)=0MEO=1,900 k-ft (↻)$

Summation of forces along y-direction is equal to 0.

$+↑∑Fy=0EyO−1(10)−35−2(20)=0EyO=85 k (↑)$

For unit value of the unknown redundant $Dy$ :

Consider the notation of moments due to external load as $mD$ .

Sketch the free body diagram of primary beam Subjected to unit value of redundant $Dy$ as shown in Figure 3.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 3

Refer Figure 3.

Find the support reaction and moment at E when 1 k vertical load applied at B.

Summation of moments of all forces about A is equal to 0.

$∑ME=0MEB=1(40)MEB=40 k-ft (↺)$

Summation of forces along y-direction is equal to 0.

$+↑∑Fy=0−EyB+1=0EyB=1 k (↓)$

For unit value of the unknown redundant $By$ :

Consider the notation of moments due to external load as $mB$ .

Sketch the free body diagram of primary beam Subjected to unit value of redundant $By$ as shown in Figure 4.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 4

Refer Figure 4.

Find the support reaction and moment at E when 1 k vertical load applied at D.

Summation of moments of all forces about E is equal to 0.

$∑MD=0MED=1(20)MED=20 k-ft (↺)$

Summation of forces along y-direction is equal to 0.

$+↑∑Fy=0−EyD+1=0EyD=1 k (↓)$

Find the moment equation of the beam for different sections on the beam.

Consider a section XX in the portion AB of the primary structure at a distance of x from A.

Refer Figure 1.

Draw the primary structure with section XX as shown in Figure 5.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 5

Refer Figure 5.

Portion AB.

Consider origin as A. $(0≤x≤10 ft)$ .

Find the moment at section XX in the portion AB as shown in Figure 5.

$MO=1(x)(x2)=x22$

Similarly calculate the moment of the remaining section in the external loading and redundant loading structures.

Tabulate the moment equation of different segment of beam as in Table 1.

Segment	x-coordinate		$MO$ $(k-ft)$	$mBY$ $(k-ft/k)$	$mDY$ $(k-ft/k)$
Segment	Origin	Limits (ft)	$MO$ $(k-ft)$	$mBY$ $(k-ft/k)$	$mDY$ $(k-ft/k)$
AB	A	$0−10$	$x22$	$0$	0
BC	E	$10−20$	$1(10)(x−102)$	$−1(x−10)$	0
CD	G	$20−30$	$1(10)(x−5)+35(x−20)$	$−1(x−10)$	0
DE	D	$30−50$	$[1(10)(x−5)+35(x−20)+(x−30)2]$	$−1(x−10)$	$−1(x−30)$

The vertical deflection at point B due to external loading is $ΔBO$ , vertical deflection at point D due to external loading is $ΔDO$ , and the flexibility coefficient representing the deflection at B due to unit value of redundant $By$ is $fBB$ , the redundant $Dy$ is $fDD$ , and the both redundant $By$ and $Dy$ is $fBD=fDB$ .

Calculate the value of $ΔBO$ using the equation as follows:

$ΔBO=Σ∫MOmBYEIdx=1EI{∫010(x22)(0)dx+∫1020[1(10)(x−102)][−1(x−10)]dx+∫2030[1(10)(x−5)+35(x−20)][−1(x−10)]dx+∫3050[1(10)(x−5)+35(x−20)+(x−30)2][−1(x−10)]dx}=1EI[0−5,833.33−60,000−753,333.33]=−819,166.67EI k-ft3$

Calculate the value of $ΔDO$ using the equation as follows:

$ΔDO=Σ∫MOmDYEIdx=1EI{∫010(0)dx+∫1020(0)dx+∫2030(0)dx+∫3050[1(10)(x−5)+35(x−20)+(x−30)2][−1(x−30)]dx}=1EI[0+0+0−280,000]=−280,000EI k-ft3$

Calculate the value of $fBB$ using the equation as follows:

$fBB=Σ∫mBY2EIdx=1EI{∫010(0)2dx+∫1020[−1(x−10)]2dx+∫2030[−1(x−10)]2dx+∫3050[−1(x−10)]2dx}=1EI[0+333.33+2,333.33+18,666.67]=−21,333.33EI k-ft3/k$

Calculate the value of $fDD$ using the equation as follows:

$fDD=Σ∫mDY2EIdx=1EI{∫010(0)2dx+∫1020(0)2dx+∫2030(0)2dx+∫3050[−1(x−30)]2dx}=1EI[0+0+0+2,666.67]=2,666.67EI k-ft3/k$

Calculate the value of $fFG$ using the equation as follows:

$fBD=fDB=Σ∫mBYmBYEIdx=1EI{∫010(0)dx+∫1020(0)dx+∫2030(0)dx+∫3050[−1(x−10)][−1(x−30)]dx}=1EI[0+0+0+,666.67]=6,666.67EI k-ft3/k$

Find the reactions and moment:

Find the horizontal and vertical reaction at F and G.

Show the first compatibility Equation as follows:

$ΔBO+fBBBy+fBDDy=0$

Substitute $−819,166.67EI k-ft3$ for $ΔBO$ , $21,333.33EI k-ft3$ for $fBB$ , and $6,666.67EI k-ft3/k$ for $fBD$ .

$−819,166.67EI +(21,333.33EI)By+(6,666.67EI)Dy=021,333.33By+6,666.67Dy=819,166.67$ (1)

Show the second compatibility Equation as follows:

$ΔDO+fDBBy+fDDDy=0$

Substitute $−280,000EI k-ft3$ for $ΔDO$ , $6,666.67EI k-ft3/k$ for $fDB$ , and $2,666.67EI k-ft3/k$ for $fDD$ .

$−280,000EI+(6,666.67EI)By+(2,666.67EI)Dy=06,666.67By+2,666.67Dy=280,000$ (2)

Solve Equation (1) and (2).

$By=25.5 k (↑)Dy=41.2 k (↑)$

Find the vertical reaction at E.

Summation of forces along y-direction is equal to 0.

$+↑∑Fy=0By+Dy+Ey=1(10)+35+(2)(20)25.5+41.2+Ey=85Ey=18.3 k (↑)$

Find the moment at E.

Summation of moments of all forces about E is equal to 0.

$∑ME=0ME−1(10)(102+40)−35(30)−2(20)(202)+(25.5×40)+(41.2×20)=0ME=56 k-ft (↻)$

Therefore, the vertical reaction at B is $By=25.5 k↑_$ .

Therefore, the vertical reaction at D is $Dy=41.2 k↑_$ .

Therefore, the vertical reaction at E is $Ey=18.3 k↑_$ .

Therefore, the moment at E is $ME=56 k-ft (↻)_$ .

Sketch the free body diagram with support reactions of the beam as shown in Figure 6.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 6

Refer Figure 6,

Find the shear force $(S)$ for the given beam:

For span AB,

At point A.

$SA=0$

At point B, (negative side):

$SB−=−1(10)=−10 k$

For span BD,

At point B, (positive side):

$SB+=−10+25.5=15.5 k$

At point C, (positive side):

$SC+=15.5 k$

At point C, (negative side):

$SC−=15.5−35=−19.5 k$

At point D, (negative side):

$SD−=−19.5 k$

For span DF,

At point D, (positive side):

$SD+=−19.5+41.2=21.7 k$

At point E, (negative side):

$SE−=21.7−2(20)=−18.3 k$

At point E, (positive side):

$SE+=−18.3+18.3=0$

Sketch the shear diagram for the given beam as shown in Figure 7.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 7

Refer Figure 7.

The points of zero shear as F.

Find the point of zero shear force from E.

$18.3−2(x1)=SF018.3−2(x1)=02x1=18.3x1=9.15 ft$

Refer Figure 6.

Find the bending moment $(M)$ for the beam:

For span AB,

At point A, (free end)

$MA=0$

At point B,

$MB=−1(10)(102)=−50 k-ft$

At point C,

$MC=−1(10)(15)+25.5(10)=105 k-ft$

At point D,

$MD=−1(10)(25)+25.5(20)−35(10)=−90 k-ft$

At point F,

$MF=2(9.15)(9.152)−56=27.7 k-ft$

At point E,

$ME=56 k-ft$

Sketch the bending moment diagram for the given beam as shown in Figure 8.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 8

Answer 6

Chapter 13, Problem 38P

To determine

Calculate the support reactions for the given structure using method of consistent deformation.

Sketch the shear and bending moment diagrams for the given structure.

Expert Solution & Answer

Answer to Problem 38P

The vertical reaction at B is $By=25.5 k↑_$ .

The vertical reaction at D is $Dy=41.2 k↑_$ .

The vertical reaction at E is $Ey=18.3 k↑_$ .

The moment at E is $ME=56 k-ft (↻)_$ .

Explanation of Solution

Given information:

The structure is given in the Figure.

Apply the sign conventions for calculating reactions, forces and moments using the three equations of equilibrium as shown below.

For summation of forces along x-direction is equal to zero $(∑Fx=0)$ , consider the forces acting towards right side as positive $(→+)$ and the forces acting towards left side as negative $(←−)$ .
For summation of forces along y-direction is equal to zero $(∑Fy=0)$ , consider the upward force as positive $(↑+)$ and the downward force as negative $(↓−)$ .
For summation of moment about a point is equal to zero $(∑Mat a point=0)$ , consider the clockwise moment as positive when moment taken from left and the counter clockwise moment as positive when moment taken from right.
Consider the positive sign indicates the clockwise moment the negative sign indicates the counterclockwise moment.

Calculation:

Sketch the free body diagram of the beam as shown in Figure 1.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 1

Refer Figure 1.

Find the degree of indeterminacy of the structure:

Degree of determinacy of the beam is equal to the number of unknown reactions minus the number of equilibrium equations.

The beam is supported by 4 support reactions and the number of equilibrium equations is 2.

Therefore, the degree of indeterminacy of the beam is $i=2$ .

Select the vertical reaction $By$ and $Dy$ at the supports B and D as redundant.

Release the support B and D and consider the notation of moments due to external load as $MO$ .

Sketch the free body diagram of primary beam subjected to external loading without support B and D as shown in Figure 2.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 2

Refer Figure 2.

Find the reactions at the supports without considering support B and D using equilibrium equations:

Summation of moments of all forces about E is equal to 0.

$∑ME=0MEO−1(10)(102+40)−35(30)−2(20)(202)=0MEO=1,900 k-ft (↻)$

Summation of forces along y-direction is equal to 0.

$+↑∑Fy=0EyO−1(10)−35−2(20)=0EyO=85 k (↑)$

For unit value of the unknown redundant $Dy$ :

Consider the notation of moments due to external load as $mD$ .

Sketch the free body diagram of primary beam Subjected to unit value of redundant $Dy$ as shown in Figure 3.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 3

Refer Figure 3.

Find the support reaction and moment at E when 1 k vertical load applied at B.

Summation of moments of all forces about A is equal to 0.

$∑ME=0MEB=1(40)MEB=40 k-ft (↺)$

Summation of forces along y-direction is equal to 0.

$+↑∑Fy=0−EyB+1=0EyB=1 k (↓)$

For unit value of the unknown redundant $By$ :

Consider the notation of moments due to external load as $mB$ .

Sketch the free body diagram of primary beam Subjected to unit value of redundant $By$ as shown in Figure 4.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 4

Refer Figure 4.

Find the support reaction and moment at E when 1 k vertical load applied at D.

Summation of moments of all forces about E is equal to 0.

$∑MD=0MED=1(20)MED=20 k-ft (↺)$

Summation of forces along y-direction is equal to 0.

$+↑∑Fy=0−EyD+1=0EyD=1 k (↓)$

Find the moment equation of the beam for different sections on the beam.

Consider a section XX in the portion AB of the primary structure at a distance of x from A.

Refer Figure 1.

Draw the primary structure with section XX as shown in Figure 5.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 5

Refer Figure 5.

Portion AB.

Consider origin as A. $(0≤x≤10 ft)$ .

Find the moment at section XX in the portion AB as shown in Figure 5.

$MO=1(x)(x2)=x22$

Similarly calculate the moment of the remaining section in the external loading and redundant loading structures.

Tabulate the moment equation of different segment of beam as in Table 1.

Segment	x-coordinate		$MO$ $(k-ft)$	$mBY$ $(k-ft/k)$	$mDY$ $(k-ft/k)$
Segment	Origin	Limits (ft)	$MO$ $(k-ft)$	$mBY$ $(k-ft/k)$	$mDY$ $(k-ft/k)$
AB	A	$0−10$	$x22$	$0$	0
BC	E	$10−20$	$1(10)(x−102)$	$−1(x−10)$	0
CD	G	$20−30$	$1(10)(x−5)+35(x−20)$	$−1(x−10)$	0
DE	D	$30−50$	$[1(10)(x−5)+35(x−20)+(x−30)2]$	$−1(x−10)$	$−1(x−30)$

The vertical deflection at point B due to external loading is $ΔBO$ , vertical deflection at point D due to external loading is $ΔDO$ , and the flexibility coefficient representing the deflection at B due to unit value of redundant $By$ is $fBB$ , the redundant $Dy$ is $fDD$ , and the both redundant $By$ and $Dy$ is $fBD=fDB$ .

Calculate the value of $ΔBO$ using the equation as follows:

$ΔBO=Σ∫MOmBYEIdx=1EI{∫010(x22)(0)dx+∫1020[1(10)(x−102)][−1(x−10)]dx+∫2030[1(10)(x−5)+35(x−20)][−1(x−10)]dx+∫3050[1(10)(x−5)+35(x−20)+(x−30)2][−1(x−10)]dx}=1EI[0−5,833.33−60,000−753,333.33]=−819,166.67EI k-ft3$

Calculate the value of $ΔDO$ using the equation as follows:

$ΔDO=Σ∫MOmDYEIdx=1EI{∫010(0)dx+∫1020(0)dx+∫2030(0)dx+∫3050[1(10)(x−5)+35(x−20)+(x−30)2][−1(x−30)]dx}=1EI[0+0+0−280,000]=−280,000EI k-ft3$

Calculate the value of $fBB$ using the equation as follows:

$fBB=Σ∫mBY2EIdx=1EI{∫010(0)2dx+∫1020[−1(x−10)]2dx+∫2030[−1(x−10)]2dx+∫3050[−1(x−10)]2dx}=1EI[0+333.33+2,333.33+18,666.67]=−21,333.33EI k-ft3/k$

Calculate the value of $fDD$ using the equation as follows:

$fDD=Σ∫mDY2EIdx=1EI{∫010(0)2dx+∫1020(0)2dx+∫2030(0)2dx+∫3050[−1(x−30)]2dx}=1EI[0+0+0+2,666.67]=2,666.67EI k-ft3/k$

Calculate the value of $fFG$ using the equation as follows:

$fBD=fDB=Σ∫mBYmBYEIdx=1EI{∫010(0)dx+∫1020(0)dx+∫2030(0)dx+∫3050[−1(x−10)][−1(x−30)]dx}=1EI[0+0+0+,666.67]=6,666.67EI k-ft3/k$

Find the reactions and moment:

Find the horizontal and vertical reaction at F and G.

Show the first compatibility Equation as follows:

$ΔBO+fBBBy+fBDDy=0$

Substitute $−819,166.67EI k-ft3$ for $ΔBO$ , $21,333.33EI k-ft3$ for $fBB$ , and $6,666.67EI k-ft3/k$ for $fBD$ .

$−819,166.67EI +(21,333.33EI)By+(6,666.67EI)Dy=021,333.33By+6,666.67Dy=819,166.67$ (1)

Show the second compatibility Equation as follows:

$ΔDO+fDBBy+fDDDy=0$

Substitute $−280,000EI k-ft3$ for $ΔDO$ , $6,666.67EI k-ft3/k$ for $fDB$ , and $2,666.67EI k-ft3/k$ for $fDD$ .

$−280,000EI+(6,666.67EI)By+(2,666.67EI)Dy=06,666.67By+2,666.67Dy=280,000$ (2)

Solve Equation (1) and (2).

$By=25.5 k (↑)Dy=41.2 k (↑)$

Find the vertical reaction at E.

Summation of forces along y-direction is equal to 0.

$+↑∑Fy=0By+Dy+Ey=1(10)+35+(2)(20)25.5+41.2+Ey=85Ey=18.3 k (↑)$

Find the moment at E.

Summation of moments of all forces about E is equal to 0.

$∑ME=0ME−1(10)(102+40)−35(30)−2(20)(202)+(25.5×40)+(41.2×20)=0ME=56 k-ft (↻)$

Therefore, the vertical reaction at B is $By=25.5 k↑_$ .

Therefore, the vertical reaction at D is $Dy=41.2 k↑_$ .

Therefore, the vertical reaction at E is $Ey=18.3 k↑_$ .

Therefore, the moment at E is $ME=56 k-ft (↻)_$ .

Sketch the free body diagram with support reactions of the beam as shown in Figure 6.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 6

Refer Figure 6,

Find the shear force $(S)$ for the given beam:

For span AB,

At point A.

$SA=0$

At point B, (negative side):

$SB−=−1(10)=−10 k$

For span BD,

At point B, (positive side):

$SB+=−10+25.5=15.5 k$

At point C, (positive side):

$SC+=15.5 k$

At point C, (negative side):

$SC−=15.5−35=−19.5 k$

At point D, (negative side):

$SD−=−19.5 k$

For span DF,

At point D, (positive side):

$SD+=−19.5+41.2=21.7 k$

At point E, (negative side):

$SE−=21.7−2(20)=−18.3 k$

At point E, (positive side):

$SE+=−18.3+18.3=0$

Sketch the shear diagram for the given beam as shown in Figure 7.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 7

Refer Figure 7.

The points of zero shear as F.

Find the point of zero shear force from E.

$18.3−2(x1)=SF018.3−2(x1)=02x1=18.3x1=9.15 ft$

Refer Figure 6.

Find the bending moment $(M)$ for the beam:

For span AB,

At point A, (free end)

$MA=0$

At point B,

$MB=−1(10)(102)=−50 k-ft$

At point C,

$MC=−1(10)(15)+25.5(10)=105 k-ft$

At point D,

$MD=−1(10)(25)+25.5(20)−35(10)=−90 k-ft$

At point F,

$MF=2(9.15)(9.152)−56=27.7 k-ft$

At point E,

$ME=56 k-ft$

Sketch the bending moment diagram for the given beam as shown in Figure 8.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 8

Answer 7

Chapter 13, Problem 38P

To determine

Calculate the support reactions for the given structure using method of consistent deformation.

Sketch the shear and bending moment diagrams for the given structure.

Answer 8

Expert Solution & Answer

Answer to Problem 38P

The vertical reaction at B is $By=25.5 k↑_$ .

The vertical reaction at D is $Dy=41.2 k↑_$ .

The vertical reaction at E is $Ey=18.3 k↑_$ .

The moment at E is $ME=56 k-ft (↻)_$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given information:

The structure is given in the Figure.

Apply the sign conventions for calculating reactions, forces and moments using the three equations of equilibrium as shown below.

For summation of forces along x-direction is equal to zero $(∑Fx=0)$ , consider the forces acting towards right side as positive $(→+)$ and the forces acting towards left side as negative $(←−)$ .
For summation of forces along y-direction is equal to zero $(∑Fy=0)$ , consider the upward force as positive $(↑+)$ and the downward force as negative $(↓−)$ .
For summation of moment about a point is equal to zero $(∑Mat a point=0)$ , consider the clockwise moment as positive when moment taken from left and the counter clockwise moment as positive when moment taken from right.
Consider the positive sign indicates the clockwise moment the negative sign indicates the counterclockwise moment.

Calculation:

Sketch the free body diagram of the beam as shown in Figure 1.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 1

Refer Figure 1.

Find the degree of indeterminacy of the structure:

Degree of determinacy of the beam is equal to the number of unknown reactions minus the number of equilibrium equations.

The beam is supported by 4 support reactions and the number of equilibrium equations is 2.

Therefore, the degree of indeterminacy of the beam is $i=2$ .

Select the vertical reaction $By$ and $Dy$ at the supports B and D as redundant.

Release the support B and D and consider the notation of moments due to external load as $MO$ .

Sketch the free body diagram of primary beam subjected to external loading without support B and D as shown in Figure 2.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 2

Refer Figure 2.

Find the reactions at the supports without considering support B and D using equilibrium equations:

Summation of moments of all forces about E is equal to 0.

$∑ME=0MEO−1(10)(102+40)−35(30)−2(20)(202)=0MEO=1,900 k-ft (↻)$

Summation of forces along y-direction is equal to 0.

$+↑∑Fy=0EyO−1(10)−35−2(20)=0EyO=85 k (↑)$

For unit value of the unknown redundant $Dy$ :

Consider the notation of moments due to external load as $mD$ .

Sketch the free body diagram of primary beam Subjected to unit value of redundant $Dy$ as shown in Figure 3.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 3

Refer Figure 3.

Find the support reaction and moment at E when 1 k vertical load applied at B.

Summation of moments of all forces about A is equal to 0.

$∑ME=0MEB=1(40)MEB=40 k-ft (↺)$

Summation of forces along y-direction is equal to 0.

$+↑∑Fy=0−EyB+1=0EyB=1 k (↓)$

For unit value of the unknown redundant $By$ :

Consider the notation of moments due to external load as $mB$ .

Sketch the free body diagram of primary beam Subjected to unit value of redundant $By$ as shown in Figure 4.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 4

Refer Figure 4.

Find the support reaction and moment at E when 1 k vertical load applied at D.

Summation of moments of all forces about E is equal to 0.

$∑MD=0MED=1(20)MED=20 k-ft (↺)$

Summation of forces along y-direction is equal to 0.

$+↑∑Fy=0−EyD+1=0EyD=1 k (↓)$

Find the moment equation of the beam for different sections on the beam.

Consider a section XX in the portion AB of the primary structure at a distance of x from A.

Refer Figure 1.

Draw the primary structure with section XX as shown in Figure 5.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 5

Refer Figure 5.

Portion AB.

Consider origin as A. $(0≤x≤10 ft)$ .

Find the moment at section XX in the portion AB as shown in Figure 5.

$MO=1(x)(x2)=x22$

Similarly calculate the moment of the remaining section in the external loading and redundant loading structures.

Tabulate the moment equation of different segment of beam as in Table 1.

Segment	x-coordinate		$MO$ $(k-ft)$	$mBY$ $(k-ft/k)$	$mDY$ $(k-ft/k)$
Segment	Origin	Limits (ft)	$MO$ $(k-ft)$	$mBY$ $(k-ft/k)$	$mDY$ $(k-ft/k)$
AB	A	$0−10$	$x22$	$0$	0
BC	E	$10−20$	$1(10)(x−102)$	$−1(x−10)$	0
CD	G	$20−30$	$1(10)(x−5)+35(x−20)$	$−1(x−10)$	0
DE	D	$30−50$	$[1(10)(x−5)+35(x−20)+(x−30)2]$	$−1(x−10)$	$−1(x−30)$

The vertical deflection at point B due to external loading is $ΔBO$ , vertical deflection at point D due to external loading is $ΔDO$ , and the flexibility coefficient representing the deflection at B due to unit value of redundant $By$ is $fBB$ , the redundant $Dy$ is $fDD$ , and the both redundant $By$ and $Dy$ is $fBD=fDB$ .

Calculate the value of $ΔBO$ using the equation as follows:

$ΔBO=Σ∫MOmBYEIdx=1EI{∫010(x22)(0)dx+∫1020[1(10)(x−102)][−1(x−10)]dx+∫2030[1(10)(x−5)+35(x−20)][−1(x−10)]dx+∫3050[1(10)(x−5)+35(x−20)+(x−30)2][−1(x−10)]dx}=1EI[0−5,833.33−60,000−753,333.33]=−819,166.67EI k-ft3$

Calculate the value of $ΔDO$ using the equation as follows:

$ΔDO=Σ∫MOmDYEIdx=1EI{∫010(0)dx+∫1020(0)dx+∫2030(0)dx+∫3050[1(10)(x−5)+35(x−20)+(x−30)2][−1(x−30)]dx}=1EI[0+0+0−280,000]=−280,000EI k-ft3$

Calculate the value of $fBB$ using the equation as follows:

$fBB=Σ∫mBY2EIdx=1EI{∫010(0)2dx+∫1020[−1(x−10)]2dx+∫2030[−1(x−10)]2dx+∫3050[−1(x−10)]2dx}=1EI[0+333.33+2,333.33+18,666.67]=−21,333.33EI k-ft3/k$

Calculate the value of $fDD$ using the equation as follows:

$fDD=Σ∫mDY2EIdx=1EI{∫010(0)2dx+∫1020(0)2dx+∫2030(0)2dx+∫3050[−1(x−30)]2dx}=1EI[0+0+0+2,666.67]=2,666.67EI k-ft3/k$

Calculate the value of $fFG$ using the equation as follows:

$fBD=fDB=Σ∫mBYmBYEIdx=1EI{∫010(0)dx+∫1020(0)dx+∫2030(0)dx+∫3050[−1(x−10)][−1(x−30)]dx}=1EI[0+0+0+,666.67]=6,666.67EI k-ft3/k$

Find the reactions and moment:

Find the horizontal and vertical reaction at F and G.

Show the first compatibility Equation as follows:

$ΔBO+fBBBy+fBDDy=0$

Substitute $−819,166.67EI k-ft3$ for $ΔBO$ , $21,333.33EI k-ft3$ for $fBB$ , and $6,666.67EI k-ft3/k$ for $fBD$ .

$−819,166.67EI +(21,333.33EI)By+(6,666.67EI)Dy=021,333.33By+6,666.67Dy=819,166.67$ (1)

Show the second compatibility Equation as follows:

$ΔDO+fDBBy+fDDDy=0$

Substitute $−280,000EI k-ft3$ for $ΔDO$ , $6,666.67EI k-ft3/k$ for $fDB$ , and $2,666.67EI k-ft3/k$ for $fDD$ .

$−280,000EI+(6,666.67EI)By+(2,666.67EI)Dy=06,666.67By+2,666.67Dy=280,000$ (2)

Solve Equation (1) and (2).

$By=25.5 k (↑)Dy=41.2 k (↑)$

Find the vertical reaction at E.

Summation of forces along y-direction is equal to 0.

$+↑∑Fy=0By+Dy+Ey=1(10)+35+(2)(20)25.5+41.2+Ey=85Ey=18.3 k (↑)$

Find the moment at E.

Summation of moments of all forces about E is equal to 0.

$∑ME=0ME−1(10)(102+40)−35(30)−2(20)(202)+(25.5×40)+(41.2×20)=0ME=56 k-ft (↻)$

Therefore, the vertical reaction at B is $By=25.5 k↑_$ .

Therefore, the vertical reaction at D is $Dy=41.2 k↑_$ .

Therefore, the vertical reaction at E is $Ey=18.3 k↑_$ .

Therefore, the moment at E is $ME=56 k-ft (↻)_$ .

Sketch the free body diagram with support reactions of the beam as shown in Figure 6.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 6

Refer Figure 6,

Find the shear force $(S)$ for the given beam:

For span AB,

At point A.

$SA=0$

At point B, (negative side):

$SB−=−1(10)=−10 k$

For span BD,

At point B, (positive side):

$SB+=−10+25.5=15.5 k$

At point C, (positive side):

$SC+=15.5 k$

At point C, (negative side):

$SC−=15.5−35=−19.5 k$

At point D, (negative side):

$SD−=−19.5 k$

For span DF,

At point D, (positive side):

$SD+=−19.5+41.2=21.7 k$

At point E, (negative side):

$SE−=21.7−2(20)=−18.3 k$

At point E, (positive side):

$SE+=−18.3+18.3=0$

Sketch the shear diagram for the given beam as shown in Figure 7.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 7

Refer Figure 7.

The points of zero shear as F.

Find the point of zero shear force from E.

$18.3−2(x1)=SF018.3−2(x1)=02x1=18.3x1=9.15 ft$

Refer Figure 6.

Find the bending moment $(M)$ for the beam:

For span AB,

At point A, (free end)

$MA=0$

At point B,

$MB=−1(10)(102)=−50 k-ft$

At point C,

$MC=−1(10)(15)+25.5(10)=105 k-ft$

At point D,

$MD=−1(10)(25)+25.5(20)−35(10)=−90 k-ft$

At point F,

$MF=2(9.15)(9.152)−56=27.7 k-ft$

At point E,

$ME=56 k-ft$

Sketch the bending moment diagram for the given beam as shown in Figure 8.

Structural Analysis, SI Edition, Chapter 13, Problem 38P , additional homework tip 8

Answer 12

Ch. 13 - Prob. 1P Ch. 13 - Prob. 2P Ch. 13 - Prob. 3P Ch. 13 - Prob. 4P Ch. 13 - Prob. 5P Ch. 13 - Prob. 6P Ch. 13 - Prob. 7P Ch. 13 - Prob. 8P Ch. 13 - Prob. 9P Ch. 13 - Prob. 10P

Calculate the support reactions for the given structure using method of consistent deformation. Sketch the shear and bending moment diagrams for the given structure.

Concept explainers

Calculate the support reactions for the given structure using method of consistent deformation.

Sketch the shear and bending moment diagrams for the given structure.

Answer to Problem 38P

Explanation of Solution

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