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

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Answer 1

Question

Chapter 13, Problem 54P

To determine

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

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

Expert Solution & Answer

Answer to Problem 54P

The horizontal reaction at A is $Ax=0_$ .

The vertical reaction at A is $Ay=45.88 kN(↑)_$ .

The vertical reaction at C is $Cy=100.48 kN(↑)_$ .

The vertical reaction at E is $Ey=198.23 kN(↑)_$ .

The vertical reaction at G is $Gy=45.41 kN(↑)_$ .

Explanation of Solution

Given information:

The settlement at A $(ΔA)$ is $10 mm=0.01 m$ .

The settlement at C $(ΔC)$ is $65 mm=0.065 m$ .

The settlement at E $(ΔE)$ is $40 mm=0.04 m$ .

The settlement at G $(ΔG)$ is $25 mm=0.025 m$ .

The moment of inertia $(I)$ is $500×106 mm4$ .

The modulus of elasticity $(E)$ is 200 GPa.

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 negative and the counter clockwise moment as positive.

Calculation:

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

Structural Analysis, Si Edition (mindtap Course List), Chapter 13, Problem 54P , additional homework tip 1

Calculate 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 5 support reactions and the number of equilibrium equations is 3.

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

Select the bending moments at the supports E and C as redundant.

Let $θCL$ and $θCR$ are the slopes at the ends C of the left and right spans of the primary beam due to external loading.

Let $θEL$ and $θER$ are the slopes at the ends E of the left and right spans of the primary beam due to external loading.

Let $fCCL$ and $fCCR$ are the flexibility coefficient representing the slopes at C of the left and right spans due to unit value of redundant $MC$ .

Let $fEEL$ and $fEER$ are the flexibility coefficient representing the slopes at E of the left and right spans due to unit value of redundant $ME$ .

Let $fCE$ be the flexibility coefficient representing the slope at point C of the primary beam due to unit value at point E.

Use beam deflection formulas:

Calculate the value of $θCL$ using the relation:

$θCL=Pa(L2−a2)6EIL$

Substitute $120 kN$ for $P$ , $6 m$ for a, and $10 m$ for L.

$θCL=120×6(102−62)6EI(10)=768 kN⋅m2EI$

Substitute $500×106 mm4$ for E and 200 GPa for E.

$θCL=768 kN⋅m2200 GPa×106 kN/m21 GPa×500×106 mm4×(1 m1,000 mm)4=0.00768 rad$

Calculate the value of $θCR$ using the relation:

$θCR=Pb(L2−b2)6EIL$

Substitute $120 kN$ for $P$ , $4 m$ for b, $2I$ for $I$ , and $10 m$ for L.

$θCR=120×4(102−42)6E(2I)(10)=336 kN⋅m2EI$

Substitute $500×106 mm4$ for E and 200 GPa for E.

$θCR=336 kN⋅m2200 GPa×106 kN/m21 GPa×500×106 mm4×(1 m1,000 mm)4=0.00336 rad$

Calculate the value of $θEL$ using the relation:

$θEL=Pa(L2−a2)6EIL$

Substitute $120 kN$ for $P$ , $6 m$ for a, $2I$ for $I$ , and $10 m$ for L.

$θEL=120×6(102−62)6E(2I)(10)=384 kN⋅m2EI$

Substitute $500×106 mm4$ for E and 200 GPa for E.

$θEL=384 kN⋅m2200 GPa×106 kN/m21 GPa×500×106 mm4×(1 m1,000 mm)4=0.00384 rad$

Calculate the value of $θER$ using the relation:

$θER=PL216EI$

Substitute $150 kN$ for $P$ and $8 m$ for L.

$θER=150(8)216EI=600 kN⋅m2EI$

Substitute $500×106 mm4$ for E and 200 GPa for E.

$θER=600 kN⋅m2200 GPa×106 kN/m21 GPa×500×106 mm4×(1 m1,000 mm)4=0.006 rad$

Calculate the value of $fCCL$ using the relation:

$fCCL=ML3EI$

Substitute $1 kN⋅m$ for M and 10 m for L.

$fCCL=1(10)3EI=10 kN⋅m2/kN⋅m3EI$

Substitute $500×106 mm4$ for E and 200 GPa for E.

$fCCL=10 kN⋅m2/kN⋅m3×200 GPa×106 kN/m21 GPa×500×106 mm4×(1 m1,000 mm)4=0.0000333 rad/kN⋅m$

Calculate the value of $fCCR$ using the relation:

$fCCR=ML6EI$

Substitute $1 kN⋅m$ for M and 10 m for L.

$fCCL=1(10)6EI=10 kN⋅m2/kN⋅m6EI$

Substitute $500×106 mm4$ for E and 200 GPa for E.

$fCCL=10 kN⋅m2/kN⋅m6×200 GPa×106 kN/m21 GPa×500×106 mm4×(1 m1,000 mm)4=0.0000167 rad/kN⋅m$

Calculate the value of $fCE$ and $fEC$ using the relation:

$fCE=fEC=ML6EI$

Substitute $1 kN⋅m$ for M, $2I$ for $I$ , and 10 m for L.

$fCE=fEC=1(10)6E(2I)=10 kN⋅m2/kN⋅m12EI$

Substitute $500×106 mm4$ for E and 200 GPa for E.

$fCE=fEC=10 kN⋅m2/kN⋅m12×200 GPa×106 kN/m21 GPa×500×106 mm4×(1 m1,000 mm)4=0.00000833 rad/kN⋅m$

Calculate the value of $fEEL$ using the relation:

$fEEL=ML6EI$

Substitute $1 kN⋅m$ for M and 10 m for L.

$fEEL=1(10)6EI=10 kN⋅m2/kN⋅m6EI$

Substitute $500×106 mm4$ for E and 200 GPa for E.

$fEEL=10 kN⋅m2/kN⋅m6×200 GPa×106 kN/m21 GPa×500×106 mm4×(1 m1,000 mm)4=0.0000166 rad/kN⋅m$

Calculate the value of $fEER$ using the relation:

$fEER=ML3EI$

Substitute $1 kN⋅m$ for M and 8 m for L.

$fEER=1(8)3EI=8 kN⋅m2/kN⋅m3EI$

Substitute $500×106 mm4$ for E and 200 GPa for E.

$fEER=8 kN⋅m2/kN⋅m3×200 GPa×106 kN/m21 GPa×500×106 mm4×(1 m1,000 mm)4=0.0000267 rad/kN⋅m$

The change of slope between the two tangents due to external load $(θCO rel.)$ is expressed as,

$θCO rel.=θCL+θCR=0.00768+0.00336=0.01104 rad$

Similarly Calculate the value of $θEO rel.$ as follows:

$θEO rel.=θEL+θER=0.00384+0.006=0.00984 rad$

The flexibility coefficient is expressed as $fCC rel.=fCCL+fCCR$ .

$fCC rel.=0.0000333+0.0000167=0.00005 rad/kN⋅m$

Calculate the value of $fEE rel.$ as follows:

$fEE rel.=fEEL+fEER=0.0000166+0.0000267=0.0000433 rad/kN⋅m$

Sketch the settlement at supports as shown in Figure 2.

Structural Analysis, Si Edition (mindtap Course List), Chapter 13, Problem 54P , additional homework tip 2

Refer to Figure 2.

Calculate the value $θCL$ as shown below.

$θCL=0.065−0.0110=0.0055 rad$

Calculate the value $θCR$ as shown below.

$θCR=0.065−0.0410=0.0025 rad$

Calculate the value $θC$ as shown below.

$θC=θCR+θCL=0.0025+0.0055=0.008 rad$

Calculate the value $θEL$ as shown below.

$θEL=0.04−0.06510=−0.0025 rad$

Calculate the value $θER$ as shown below.

$θER=0.04−0.0258=0.001875 rad$

Calculate the value $θC$ as shown below.

$θE=θER+θEL=0.001875−0.0025=−0.000625 rad$

Calculate the reactions for the given beam:

Show the compatibility Equations of the given beam as follows:

$θCO rel.+fCC rel.MC+fCEME=θCθEO rel.+fECMC+fEE rel.ME=θE$

Substitute $0.001104 rad$ for $θCO rel.$ , $0.00005 rad/kN⋅m$ for $fCC rel.$ , $0.00000833 rad/kN⋅m$ for $fCE$ , $0.008 rad$ for $θC$ , $0.00984 rad$ for $θEO rel.$ , $0.00000833 rad/kN⋅m$ for $fEC$ , $0.0000433 rad/kN⋅m$ for $fEE rel.$ , and $−0.000625 rad$ for $θE$ .

$0.001,104+0.00005MC+0.00000833ME=0.0081,104+5MC+0.833ME=8005MC+0.833ME=−304$ (1)

$0.00984+0.00000833MC+0.0000433ME=−0.000625984+0.833MC+4.33ME=−62.50.833MC+4.33ME=−1,046.5$ (2)

Solve Equation (1) and Equation (2).

$MC=−21.22 kN⋅mME=−237.6 kN⋅m$

Sketch the span end moments and shears for span ABC, CDE, and EFG as shown in Figure 3.

Structural Analysis, Si Edition (mindtap Course List), Chapter 13, Problem 54P , additional homework tip 3

Refer to Figure 3.

Use equilibrium equations:

For span ABC,

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

$∑MA=0Cy(10)−21.22−120(6)=0Cy=74.12 kN$

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

$+↑∑Fy=0Ay−120+Cy=0Ay−120+74.12=0Ay=45.88 kN$

For span CDE,

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

$∑MC=0Ey(10)+21.22−237.6−120(6)=0Ey=93.64 kN$

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

$+↑∑Fy=0Cy−120+Ey=0Cy−120+93.64=0Cy=26.36 kN$

For span EFG,

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

$∑ME=0Gy(8)+236.7−150(4)=0Gy=45.41 kN$

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

$+↑∑Fy=0Ey−150+Gy=0Ey−150+45.41=0Ey=104.59 kN$

Provide the reaction at supports C and E as shown below.

$Cy=74.12+26.36=100.48 kNEy=93.64+104.59=198.23 kN$

Sketch the reactions for the given beam as shown in Figure 4.

Structural Analysis, Si Edition (mindtap Course List), Chapter 13, Problem 54P , additional homework tip 4

Refer to Figure 4.

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

At point A,

$SA=45.88 kN$

At point B,

$SB,L=45.88 kNSB,R=45.88−120=−74.12 kN$

At point C,

$SC,L=45.88−120=−74.12 kNSC,R=45.88−120+100.48=26.36 kN$

At point D,

$SD,L=45.88−120+100.48=26.36 kNSD,R=45.88−120+100.48−120=−93.64 kN$

At point E,

$SE,L=45.88−120+100.48−120=−93.64 kNSE,R=45.88−120+100.48−120+198.23=104.59 kN$

At point F,

$SF,L=45.88−120+100.48−120+198.23=104.59 kNSF,R=45.88−120+100.48−120+198.23−150=−45.41 kN$

At point G,

$SG,L=45.88−120+100.48−120+198.23−150=−45.41 kNSG=45.88−120+100.48−120+198.23−150+45.41=0$

Sketch the shear diagram for the given beam as shown in Figure 5.

Structural Analysis, Si Edition (mindtap Course List), Chapter 13, Problem 54P , additional homework tip 5

Refer to Figure 4.

Calculate the bending moment $(M)$ for the given beam:

At point A,

$MA=0$

At point B,

$MB=45.88(6)=275.28 kN⋅m$

At point C,

$MC=45.88(10)−120(4)=−21.2 kN⋅m$

At point D,

$MD=45.88(16)−120(10)+100.48(6)=136.96 kN⋅m$

At point E,

$ME=45.88(20)−120(14)+100.48(10)−120(4)=−237.6 kN⋅m$

At point F,

$MF=45.88(24)−120(18)+100.48(14)−120(8)+198.23(4)=180.76 kN⋅m$

At point G,

$MG=0$

Sketch the bending moment diagram for the given beam as shown in Figure 6.

Structural Analysis, Si Edition (mindtap Course List), Chapter 13, Problem 54P , additional homework tip 6

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Answer 2

Question

Chapter 13, Problem 54P

To determine

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

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

Expert Solution & Answer

Answer to Problem 54P

The horizontal reaction at A is $Ax=0_$ .

The vertical reaction at A is $Ay=45.88 kN(↑)_$ .

The vertical reaction at C is $Cy=100.48 kN(↑)_$ .

The vertical reaction at E is $Ey=198.23 kN(↑)_$ .

The vertical reaction at G is $Gy=45.41 kN(↑)_$ .

Explanation of Solution

Given information:

The settlement at A $(ΔA)$ is $10 mm=0.01 m$ .

The settlement at C $(ΔC)$ is $65 mm=0.065 m$ .

The settlement at E $(ΔE)$ is $40 mm=0.04 m$ .

The settlement at G $(ΔG)$ is $25 mm=0.025 m$ .

The moment of inertia $(I)$ is $500×106 mm4$ .

The modulus of elasticity $(E)$ is 200 GPa.

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 negative and the counter clockwise moment as positive.