Plot the shear diagram, bending moment diagram, axial force diagram, and the qualitative deflected shape of the frame.

Question

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Answer 1

Question

Chapter 5, Problem 64P

To determine

Plot the shear diagram, bending moment diagram, axial force diagram, and the qualitative deflected shape of the frame.

Expert Solution & Answer

Explanation of Solution

Write the condition for static instability, determinacy and indeterminacy of plane frames as follows:

$3m+r<3j+ec statically unstable frame$ (1)

$3m+r=3j+ec statically determinate frame$ (2)

$3m+r>3j+ec statically indeterminate frame$ (3)

Here, number of members is m, number of external reactions is r, the number of joints is j, and the number of elastic hinges is $ec$ .

Find the degree of static indeterminacy (i) using the equation;

$i=(3m+r)−(3j+ec)$ (4)

Refer to the Figure in the question;

The number of members (m) is 3.

The number of external reactions (r) is 3.

The number of joints (j) is 4.

The number of elastic hinges $ec$ is 0.

Substitute the values in Equation (2);

$3(3)+3=3(4)+012=12statically determinate frame$

Show the free-body diagram of the entire frame as in Figure 1.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 1

Refer Figure 1,

Find the vertical reaction at point D by taking moment about point A.

$+↺∑MA=0Dy(10)−15(10)(102)+90(4)=010Dy−750+360=0Dy=39 kN↑$

Find the vertical reaction at point A by resolving the vertical component of forces.

$+↑∑Fy=0Ay−15(10)+Dy=0Ay−150+39=0Ay=111 kN↑$

Find the horizontal reaction at point A by resolving the horizontal component of forces.

$+→∑Fx=0Ax−90=0Ax=90 kN→$

Show the free-body diagram of the members and joints of the entire frame as in Figure 2.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 2

Consider point A:

Resolve the vertical component of forces.

$+↑∑FY=0111−AYAB=0AYAB=111 kN$

Resolve the horizontal component of forces.

$+→∑FX=090−AXAB=0AXAB=90 kN$

Consider the member AB:

Resolve the vertical component of forces.

$+↑∑FY=0AYAB−BYAB=0111−BYAB=0BYAB=111 kN$

Resolve the horizontal component of forces.

$+→∑FX=0AXAB−BXAB=090−BXAB=0BXAB=90 kN$

Take moment about the point B.

$+↺∑MBAB=0−MBAB+AXAB(8)=0−MBAB+90(8)=0MBAB=720 kN-m$

Consider the point B:

Resolve the vertical component of forces.

$+↑∑FY=0BYAB−BYBC=0111−BYBC=0BYBC=111 kN$

Resolve the horizontal component of forces.

$+→∑FX=0BXAB−BXBC=090−BXBC=0BXBC=90 kN$

Take moment about the point B.

$+↺∑MB=0MBAB−MBBC=0720−MBBC=0MBBC=720 kN-m$

Consider the member BC:

Resolve the vertical component of forces.

$+↑∑FY=0BYBC−15(10)+CYBC=0111−150+CYBC=0CYBC=39 kN$

Resolve the horizontal component of forces.

$+→∑FX=0−BXBC+CXBC=0−90+CXBC=0CXBC=90 kN$

Take moment about the point C.

$+↺∑MCBC=0−BYBC(10)+MBBC+15(10)(102)−MCBC=0−111(10)+720+750−MCBC=0MCBC=360 kN-m$

Consider the point C:

Resolve the vertical component of forces.

$+↑∑FY=0−CYBC+CYCD=0−39+CYCD=0CYCD=39 kN$

Resolve the horizontal component of forces.

$+→∑FX=0CXBC−CXCD=090−CXCD=0CXCD=90 kN$

Take moment about the point C.

$+↺∑MB=0MCBC−MCCD=0360−MCCD=0MCCD=360 kN-m$

Consider the point D:

Resolve the vertical component of forces.

$+↑∑FY=039−DYCD=0DYCD=39 kN$

Plot the moment end forces of the frame as in Figure 3.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 3

Refer to the moment end force diagram plot the shear diagram, bending moment diagram, and the axial force diagrams.

Refer to the shear force diagram, the maximum bending moment occurs at point F where the shear force changes sign.

Use similar triangle concept for the region BC:

$111x=3910−x1110−111x=39x150x=1110x=7.4 m from pointB$

Plot the shear force diagram as in Figure 4.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 4

Plot the bending moment diagram as in Figure 5.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 5

Plot the axial force diagram as in Figure 6.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 6

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Answer 2

Question

Chapter 5, Problem 64P

To determine

Plot the shear diagram, bending moment diagram, axial force diagram, and the qualitative deflected shape of the frame.

Expert Solution & Answer

Explanation of Solution

Write the condition for static instability, determinacy and indeterminacy of plane frames as follows:

$3m+r<3j+ec statically unstable frame$ (1)

$3m+r=3j+ec statically determinate frame$ (2)

$3m+r>3j+ec statically indeterminate frame$ (3)

Here, number of members is m, number of external reactions is r, the number of joints is j, and the number of elastic hinges is $ec$ .

Find the degree of static indeterminacy (i) using the equation;

$i=(3m+r)−(3j+ec)$ (4)

Refer to the Figure in the question;

The number of members (m) is 3.

The number of external reactions (r) is 3.

The number of joints (j) is 4.

The number of elastic hinges $ec$ is 0.

Substitute the values in Equation (2);

$3(3)+3=3(4)+012=12statically determinate frame$

Show the free-body diagram of the entire frame as in Figure 1.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 1

Refer Figure 1,

Find the vertical reaction at point D by taking moment about point A.

$+↺∑MA=0Dy(10)−15(10)(102)+90(4)=010Dy−750+360=0Dy=39 kN↑$

Find the vertical reaction at point A by resolving the vertical component of forces.

$+↑∑Fy=0Ay−15(10)+Dy=0Ay−150+39=0Ay=111 kN↑$

Find the horizontal reaction at point A by resolving the horizontal component of forces.

$+→∑Fx=0Ax−90=0Ax=90 kN→$

Show the free-body diagram of the members and joints of the entire frame as in Figure 2.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 2

Consider point A:

Resolve the vertical component of forces.

$+↑∑FY=0111−AYAB=0AYAB=111 kN$

Resolve the horizontal component of forces.

$+→∑FX=090−AXAB=0AXAB=90 kN$

Consider the member AB:

Resolve the vertical component of forces.

$+↑∑FY=0AYAB−BYAB=0111−BYAB=0BYAB=111 kN$

Resolve the horizontal component of forces.

$+→∑FX=0AXAB−BXAB=090−BXAB=0BXAB=90 kN$

Take moment about the point B.

$+↺∑MBAB=0−MBAB+AXAB(8)=0−MBAB+90(8)=0MBAB=720 kN-m$

Consider the point B:

Resolve the vertical component of forces.

$+↑∑FY=0BYAB−BYBC=0111−BYBC=0BYBC=111 kN$

Resolve the horizontal component of forces.

$+→∑FX=0BXAB−BXBC=090−BXBC=0BXBC=90 kN$

Take moment about the point B.

$+↺∑MB=0MBAB−MBBC=0720−MBBC=0MBBC=720 kN-m$

Consider the member BC:

Resolve the vertical component of forces.

$+↑∑FY=0BYBC−15(10)+CYBC=0111−150+CYBC=0CYBC=39 kN$

Resolve the horizontal component of forces.

$+→∑FX=0−BXBC+CXBC=0−90+CXBC=0CXBC=90 kN$

Take moment about the point C.

$+↺∑MCBC=0−BYBC(10)+MBBC+15(10)(102)−MCBC=0−111(10)+720+750−MCBC=0MCBC=360 kN-m$

Consider the point C:

Resolve the vertical component of forces.

$+↑∑FY=0−CYBC+CYCD=0−39+CYCD=0CYCD=39 kN$

Resolve the horizontal component of forces.

$+→∑FX=0CXBC−CXCD=090−CXCD=0CXCD=90 kN$

Take moment about the point C.

$+↺∑MB=0MCBC−MCCD=0360−MCCD=0MCCD=360 kN-m$

Consider the point D:

Resolve the vertical component of forces.

$+↑∑FY=039−DYCD=0DYCD=39 kN$

Plot the moment end forces of the frame as in Figure 3.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 3

Refer to the moment end force diagram plot the shear diagram, bending moment diagram, and the axial force diagrams.

Refer to the shear force diagram, the maximum bending moment occurs at point F where the shear force changes sign.

Use similar triangle concept for the region BC:

$111x=3910−x1110−111x=39x150x=1110x=7.4 m from pointB$

Plot the shear force diagram as in Figure 4.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 4

Plot the bending moment diagram as in Figure 5.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 5

Plot the axial force diagram as in Figure 6.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 6

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Answer 3

Question

Chapter 5, Problem 64P

To determine

Plot the shear diagram, bending moment diagram, axial force diagram, and the qualitative deflected shape of the frame.

Expert Solution & Answer

Explanation of Solution

Write the condition for static instability, determinacy and indeterminacy of plane frames as follows:

$3m+r<3j+ec statically unstable frame$ (1)

$3m+r=3j+ec statically determinate frame$ (2)

$3m+r>3j+ec statically indeterminate frame$ (3)

Here, number of members is m, number of external reactions is r, the number of joints is j, and the number of elastic hinges is $ec$ .

Find the degree of static indeterminacy (i) using the equation;

$i=(3m+r)−(3j+ec)$ (4)

Refer to the Figure in the question;

The number of members (m) is 3.

The number of external reactions (r) is 3.

The number of joints (j) is 4.

The number of elastic hinges $ec$ is 0.

Substitute the values in Equation (2);

$3(3)+3=3(4)+012=12statically determinate frame$

Show the free-body diagram of the entire frame as in Figure 1.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 1

Refer Figure 1,

Find the vertical reaction at point D by taking moment about point A.

$+↺∑MA=0Dy(10)−15(10)(102)+90(4)=010Dy−750+360=0Dy=39 kN↑$

Find the vertical reaction at point A by resolving the vertical component of forces.

$+↑∑Fy=0Ay−15(10)+Dy=0Ay−150+39=0Ay=111 kN↑$

Find the horizontal reaction at point A by resolving the horizontal component of forces.

$+→∑Fx=0Ax−90=0Ax=90 kN→$

Show the free-body diagram of the members and joints of the entire frame as in Figure 2.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 2

Consider point A:

Resolve the vertical component of forces.

$+↑∑FY=0111−AYAB=0AYAB=111 kN$

Resolve the horizontal component of forces.

$+→∑FX=090−AXAB=0AXAB=90 kN$

Consider the member AB:

Resolve the vertical component of forces.

$+↑∑FY=0AYAB−BYAB=0111−BYAB=0BYAB=111 kN$

Resolve the horizontal component of forces.

$+→∑FX=0AXAB−BXAB=090−BXAB=0BXAB=90 kN$

Take moment about the point B.

$+↺∑MBAB=0−MBAB+AXAB(8)=0−MBAB+90(8)=0MBAB=720 kN-m$

Consider the point B:

Resolve the vertical component of forces.

$+↑∑FY=0BYAB−BYBC=0111−BYBC=0BYBC=111 kN$

Resolve the horizontal component of forces.

$+→∑FX=0BXAB−BXBC=090−BXBC=0BXBC=90 kN$

Take moment about the point B.

$+↺∑MB=0MBAB−MBBC=0720−MBBC=0MBBC=720 kN-m$

Consider the member BC:

Resolve the vertical component of forces.

$+↑∑FY=0BYBC−15(10)+CYBC=0111−150+CYBC=0CYBC=39 kN$

Resolve the horizontal component of forces.

$+→∑FX=0−BXBC+CXBC=0−90+CXBC=0CXBC=90 kN$

Take moment about the point C.

$+↺∑MCBC=0−BYBC(10)+MBBC+15(10)(102)−MCBC=0−111(10)+720+750−MCBC=0MCBC=360 kN-m$

Consider the point C:

Resolve the vertical component of forces.

$+↑∑FY=0−CYBC+CYCD=0−39+CYCD=0CYCD=39 kN$

Resolve the horizontal component of forces.

$+→∑FX=0CXBC−CXCD=090−CXCD=0CXCD=90 kN$

Take moment about the point C.

$+↺∑MB=0MCBC−MCCD=0360−MCCD=0MCCD=360 kN-m$

Consider the point D:

Resolve the vertical component of forces.

$+↑∑FY=039−DYCD=0DYCD=39 kN$

Plot the moment end forces of the frame as in Figure 3.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 3

Refer to the moment end force diagram plot the shear diagram, bending moment diagram, and the axial force diagrams.

Refer to the shear force diagram, the maximum bending moment occurs at point F where the shear force changes sign.

Use similar triangle concept for the region BC:

$111x=3910−x1110−111x=39x150x=1110x=7.4 m from pointB$

Plot the shear force diagram as in Figure 4.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 4

Plot the bending moment diagram as in Figure 5.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 5

Plot the axial force diagram as in Figure 6.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 6

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Answer 4

Question

Chapter 5, Problem 64P

To determine

Plot the shear diagram, bending moment diagram, axial force diagram, and the qualitative deflected shape of the frame.

Expert Solution & Answer

Explanation of Solution

Write the condition for static instability, determinacy and indeterminacy of plane frames as follows:

$3m+r<3j+ec statically unstable frame$ (1)

$3m+r=3j+ec statically determinate frame$ (2)

$3m+r>3j+ec statically indeterminate frame$ (3)

Here, number of members is m, number of external reactions is r, the number of joints is j, and the number of elastic hinges is $ec$ .

Find the degree of static indeterminacy (i) using the equation;

$i=(3m+r)−(3j+ec)$ (4)

Refer to the Figure in the question;

The number of members (m) is 3.

The number of external reactions (r) is 3.

The number of joints (j) is 4.

The number of elastic hinges $ec$ is 0.

Substitute the values in Equation (2);

$3(3)+3=3(4)+012=12statically determinate frame$

Show the free-body diagram of the entire frame as in Figure 1.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 1

Refer Figure 1,

Find the vertical reaction at point D by taking moment about point A.

$+↺∑MA=0Dy(10)−15(10)(102)+90(4)=010Dy−750+360=0Dy=39 kN↑$

Find the vertical reaction at point A by resolving the vertical component of forces.

$+↑∑Fy=0Ay−15(10)+Dy=0Ay−150+39=0Ay=111 kN↑$

Find the horizontal reaction at point A by resolving the horizontal component of forces.

$+→∑Fx=0Ax−90=0Ax=90 kN→$

Show the free-body diagram of the members and joints of the entire frame as in Figure 2.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 2

Consider point A:

Resolve the vertical component of forces.

$+↑∑FY=0111−AYAB=0AYAB=111 kN$

Resolve the horizontal component of forces.

$+→∑FX=090−AXAB=0AXAB=90 kN$

Consider the member AB:

Resolve the vertical component of forces.

$+↑∑FY=0AYAB−BYAB=0111−BYAB=0BYAB=111 kN$

Resolve the horizontal component of forces.

$+→∑FX=0AXAB−BXAB=090−BXAB=0BXAB=90 kN$

Take moment about the point B.

$+↺∑MBAB=0−MBAB+AXAB(8)=0−MBAB+90(8)=0MBAB=720 kN-m$

Consider the point B:

Resolve the vertical component of forces.

$+↑∑FY=0BYAB−BYBC=0111−BYBC=0BYBC=111 kN$

Resolve the horizontal component of forces.

$+→∑FX=0BXAB−BXBC=090−BXBC=0BXBC=90 kN$

Take moment about the point B.

$+↺∑MB=0MBAB−MBBC=0720−MBBC=0MBBC=720 kN-m$

Consider the member BC:

Resolve the vertical component of forces.

$+↑∑FY=0BYBC−15(10)+CYBC=0111−150+CYBC=0CYBC=39 kN$

Resolve the horizontal component of forces.

$+→∑FX=0−BXBC+CXBC=0−90+CXBC=0CXBC=90 kN$

Take moment about the point C.

$+↺∑MCBC=0−BYBC(10)+MBBC+15(10)(102)−MCBC=0−111(10)+720+750−MCBC=0MCBC=360 kN-m$

Consider the point C:

Resolve the vertical component of forces.

$+↑∑FY=0−CYBC+CYCD=0−39+CYCD=0CYCD=39 kN$

Resolve the horizontal component of forces.

$+→∑FX=0CXBC−CXCD=090−CXCD=0CXCD=90 kN$

Take moment about the point C.

$+↺∑MB=0MCBC−MCCD=0360−MCCD=0MCCD=360 kN-m$

Consider the point D:

Resolve the vertical component of forces.

$+↑∑FY=039−DYCD=0DYCD=39 kN$

Plot the moment end forces of the frame as in Figure 3.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 3

Refer to the moment end force diagram plot the shear diagram, bending moment diagram, and the axial force diagrams.

Refer to the shear force diagram, the maximum bending moment occurs at point F where the shear force changes sign.

Use similar triangle concept for the region BC:

$111x=3910−x1110−111x=39x150x=1110x=7.4 m from pointB$

Plot the shear force diagram as in Figure 4.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 4

Plot the bending moment diagram as in Figure 5.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 5

Plot the axial force diagram as in Figure 6.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 6

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Answer 5

Chapter 5, Problem 64P

To determine

Plot the shear diagram, bending moment diagram, axial force diagram, and the qualitative deflected shape of the frame.

Expert Solution & Answer

Explanation of Solution

Write the condition for static instability, determinacy and indeterminacy of plane frames as follows:

$3m+r<3j+ec statically unstable frame$ (1)

$3m+r=3j+ec statically determinate frame$ (2)

$3m+r>3j+ec statically indeterminate frame$ (3)

Here, number of members is m, number of external reactions is r, the number of joints is j, and the number of elastic hinges is $ec$ .

Find the degree of static indeterminacy (i) using the equation;

$i=(3m+r)−(3j+ec)$ (4)

Refer to the Figure in the question;

The number of members (m) is 3.

The number of external reactions (r) is 3.

The number of joints (j) is 4.

The number of elastic hinges $ec$ is 0.

Substitute the values in Equation (2);

$3(3)+3=3(4)+012=12statically determinate frame$

Show the free-body diagram of the entire frame as in Figure 1.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 1

Refer Figure 1,

Find the vertical reaction at point D by taking moment about point A.

$+↺∑MA=0Dy(10)−15(10)(102)+90(4)=010Dy−750+360=0Dy=39 kN↑$

Find the vertical reaction at point A by resolving the vertical component of forces.

$+↑∑Fy=0Ay−15(10)+Dy=0Ay−150+39=0Ay=111 kN↑$

Find the horizontal reaction at point A by resolving the horizontal component of forces.

$+→∑Fx=0Ax−90=0Ax=90 kN→$

Show the free-body diagram of the members and joints of the entire frame as in Figure 2.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 2

Consider point A:

Resolve the vertical component of forces.

$+↑∑FY=0111−AYAB=0AYAB=111 kN$

Resolve the horizontal component of forces.

$+→∑FX=090−AXAB=0AXAB=90 kN$

Consider the member AB:

Resolve the vertical component of forces.

$+↑∑FY=0AYAB−BYAB=0111−BYAB=0BYAB=111 kN$

Resolve the horizontal component of forces.

$+→∑FX=0AXAB−BXAB=090−BXAB=0BXAB=90 kN$

Take moment about the point B.

$+↺∑MBAB=0−MBAB+AXAB(8)=0−MBAB+90(8)=0MBAB=720 kN-m$

Consider the point B:

Resolve the vertical component of forces.

$+↑∑FY=0BYAB−BYBC=0111−BYBC=0BYBC=111 kN$

Resolve the horizontal component of forces.

$+→∑FX=0BXAB−BXBC=090−BXBC=0BXBC=90 kN$

Take moment about the point B.

$+↺∑MB=0MBAB−MBBC=0720−MBBC=0MBBC=720 kN-m$

Consider the member BC:

Resolve the vertical component of forces.

$+↑∑FY=0BYBC−15(10)+CYBC=0111−150+CYBC=0CYBC=39 kN$

Resolve the horizontal component of forces.

$+→∑FX=0−BXBC+CXBC=0−90+CXBC=0CXBC=90 kN$

Take moment about the point C.

$+↺∑MCBC=0−BYBC(10)+MBBC+15(10)(102)−MCBC=0−111(10)+720+750−MCBC=0MCBC=360 kN-m$

Consider the point C:

Resolve the vertical component of forces.

$+↑∑FY=0−CYBC+CYCD=0−39+CYCD=0CYCD=39 kN$

Resolve the horizontal component of forces.

$+→∑FX=0CXBC−CXCD=090−CXCD=0CXCD=90 kN$

Take moment about the point C.

$+↺∑MB=0MCBC−MCCD=0360−MCCD=0MCCD=360 kN-m$

Consider the point D:

Resolve the vertical component of forces.

$+↑∑FY=039−DYCD=0DYCD=39 kN$

Plot the moment end forces of the frame as in Figure 3.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 3

Refer to the moment end force diagram plot the shear diagram, bending moment diagram, and the axial force diagrams.

Refer to the shear force diagram, the maximum bending moment occurs at point F where the shear force changes sign.

Use similar triangle concept for the region BC:

$111x=3910−x1110−111x=39x150x=1110x=7.4 m from pointB$

Plot the shear force diagram as in Figure 4.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 4

Plot the bending moment diagram as in Figure 5.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 5

Plot the axial force diagram as in Figure 6.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 6

Answer 6

Chapter 5, Problem 64P

To determine

Plot the shear diagram, bending moment diagram, axial force diagram, and the qualitative deflected shape of the frame.

Expert Solution & Answer

Explanation of Solution

Write the condition for static instability, determinacy and indeterminacy of plane frames as follows:

$3m+r<3j+ec statically unstable frame$ (1)

$3m+r=3j+ec statically determinate frame$ (2)

$3m+r>3j+ec statically indeterminate frame$ (3)

Here, number of members is m, number of external reactions is r, the number of joints is j, and the number of elastic hinges is $ec$ .

Find the degree of static indeterminacy (i) using the equation;

$i=(3m+r)−(3j+ec)$ (4)

Refer to the Figure in the question;

The number of members (m) is 3.

The number of external reactions (r) is 3.

The number of joints (j) is 4.

The number of elastic hinges $ec$ is 0.

Substitute the values in Equation (2);

$3(3)+3=3(4)+012=12statically determinate frame$

Show the free-body diagram of the entire frame as in Figure 1.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 1

Refer Figure 1,

Find the vertical reaction at point D by taking moment about point A.

$+↺∑MA=0Dy(10)−15(10)(102)+90(4)=010Dy−750+360=0Dy=39 kN↑$

Find the vertical reaction at point A by resolving the vertical component of forces.

$+↑∑Fy=0Ay−15(10)+Dy=0Ay−150+39=0Ay=111 kN↑$

Find the horizontal reaction at point A by resolving the horizontal component of forces.

$+→∑Fx=0Ax−90=0Ax=90 kN→$

Show the free-body diagram of the members and joints of the entire frame as in Figure 2.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 2

Consider point A:

Resolve the vertical component of forces.

$+↑∑FY=0111−AYAB=0AYAB=111 kN$

Resolve the horizontal component of forces.

$+→∑FX=090−AXAB=0AXAB=90 kN$

Consider the member AB:

Resolve the vertical component of forces.

$+↑∑FY=0AYAB−BYAB=0111−BYAB=0BYAB=111 kN$

Resolve the horizontal component of forces.

$+→∑FX=0AXAB−BXAB=090−BXAB=0BXAB=90 kN$

Take moment about the point B.

$+↺∑MBAB=0−MBAB+AXAB(8)=0−MBAB+90(8)=0MBAB=720 kN-m$

Consider the point B:

Resolve the vertical component of forces.

$+↑∑FY=0BYAB−BYBC=0111−BYBC=0BYBC=111 kN$

Resolve the horizontal component of forces.

$+→∑FX=0BXAB−BXBC=090−BXBC=0BXBC=90 kN$

Take moment about the point B.

$+↺∑MB=0MBAB−MBBC=0720−MBBC=0MBBC=720 kN-m$

Consider the member BC:

Resolve the vertical component of forces.

$+↑∑FY=0BYBC−15(10)+CYBC=0111−150+CYBC=0CYBC=39 kN$

Resolve the horizontal component of forces.

$+→∑FX=0−BXBC+CXBC=0−90+CXBC=0CXBC=90 kN$

Take moment about the point C.

$+↺∑MCBC=0−BYBC(10)+MBBC+15(10)(102)−MCBC=0−111(10)+720+750−MCBC=0MCBC=360 kN-m$

Consider the point C:

Resolve the vertical component of forces.

$+↑∑FY=0−CYBC+CYCD=0−39+CYCD=0CYCD=39 kN$

Resolve the horizontal component of forces.

$+→∑FX=0CXBC−CXCD=090−CXCD=0CXCD=90 kN$

Take moment about the point C.

$+↺∑MB=0MCBC−MCCD=0360−MCCD=0MCCD=360 kN-m$

Consider the point D:

Resolve the vertical component of forces.

$+↑∑FY=039−DYCD=0DYCD=39 kN$

Plot the moment end forces of the frame as in Figure 3.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 3

Refer to the moment end force diagram plot the shear diagram, bending moment diagram, and the axial force diagrams.

Refer to the shear force diagram, the maximum bending moment occurs at point F where the shear force changes sign.

Use similar triangle concept for the region BC:

$111x=3910−x1110−111x=39x150x=1110x=7.4 m from pointB$

Plot the shear force diagram as in Figure 4.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 4

Plot the bending moment diagram as in Figure 5.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 5

Plot the axial force diagram as in Figure 6.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 6

Answer 7

Chapter 5, Problem 64P

To determine

Plot the shear diagram, bending moment diagram, axial force diagram, and the qualitative deflected shape of the frame.

Answer 8

Expert Solution & Answer

Explanation of Solution

Write the condition for static instability, determinacy and indeterminacy of plane frames as follows:

$3m+r<3j+ec statically unstable frame$ (1)

$3m+r=3j+ec statically determinate frame$ (2)

$3m+r>3j+ec statically indeterminate frame$ (3)

Here, number of members is m, number of external reactions is r, the number of joints is j, and the number of elastic hinges is $ec$ .

Find the degree of static indeterminacy (i) using the equation;

$i=(3m+r)−(3j+ec)$ (4)

Refer to the Figure in the question;

The number of members (m) is 3.

The number of external reactions (r) is 3.

The number of joints (j) is 4.

The number of elastic hinges $ec$ is 0.

Substitute the values in Equation (2);

$3(3)+3=3(4)+012=12statically determinate frame$

Show the free-body diagram of the entire frame as in Figure 1.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 1

Refer Figure 1,

Find the vertical reaction at point D by taking moment about point A.

$+↺∑MA=0Dy(10)−15(10)(102)+90(4)=010Dy−750+360=0Dy=39 kN↑$

Find the vertical reaction at point A by resolving the vertical component of forces.

$+↑∑Fy=0Ay−15(10)+Dy=0Ay−150+39=0Ay=111 kN↑$

Find the horizontal reaction at point A by resolving the horizontal component of forces.

$+→∑Fx=0Ax−90=0Ax=90 kN→$

Show the free-body diagram of the members and joints of the entire frame as in Figure 2.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 2

Consider point A:

Resolve the vertical component of forces.

$+↑∑FY=0111−AYAB=0AYAB=111 kN$

Resolve the horizontal component of forces.

$+→∑FX=090−AXAB=0AXAB=90 kN$

Consider the member AB:

Resolve the vertical component of forces.

$+↑∑FY=0AYAB−BYAB=0111−BYAB=0BYAB=111 kN$

Resolve the horizontal component of forces.

$+→∑FX=0AXAB−BXAB=090−BXAB=0BXAB=90 kN$

Take moment about the point B.

$+↺∑MBAB=0−MBAB+AXAB(8)=0−MBAB+90(8)=0MBAB=720 kN-m$

Consider the point B:

Resolve the vertical component of forces.

$+↑∑FY=0BYAB−BYBC=0111−BYBC=0BYBC=111 kN$

Resolve the horizontal component of forces.

$+→∑FX=0BXAB−BXBC=090−BXBC=0BXBC=90 kN$

Take moment about the point B.

$+↺∑MB=0MBAB−MBBC=0720−MBBC=0MBBC=720 kN-m$

Consider the member BC:

Resolve the vertical component of forces.

$+↑∑FY=0BYBC−15(10)+CYBC=0111−150+CYBC=0CYBC=39 kN$

Resolve the horizontal component of forces.

$+→∑FX=0−BXBC+CXBC=0−90+CXBC=0CXBC=90 kN$

Take moment about the point C.

$+↺∑MCBC=0−BYBC(10)+MBBC+15(10)(102)−MCBC=0−111(10)+720+750−MCBC=0MCBC=360 kN-m$

Consider the point C:

Resolve the vertical component of forces.

$+↑∑FY=0−CYBC+CYCD=0−39+CYCD=0CYCD=39 kN$

Resolve the horizontal component of forces.

$+→∑FX=0CXBC−CXCD=090−CXCD=0CXCD=90 kN$

Take moment about the point C.

$+↺∑MB=0MCBC−MCCD=0360−MCCD=0MCCD=360 kN-m$

Consider the point D:

Resolve the vertical component of forces.

$+↑∑FY=039−DYCD=0DYCD=39 kN$

Plot the moment end forces of the frame as in Figure 3.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 3

Refer to the moment end force diagram plot the shear diagram, bending moment diagram, and the axial force diagrams.

Refer to the shear force diagram, the maximum bending moment occurs at point F where the shear force changes sign.

Use similar triangle concept for the region BC:

$111x=3910−x1110−111x=39x150x=1110x=7.4 m from pointB$

Plot the shear force diagram as in Figure 4.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 4

Plot the bending moment diagram as in Figure 5.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 5

Plot the axial force diagram as in Figure 6.

STRUCTURAL ANALYSIS (LL), Chapter 5, Problem 64P , additional homework tip 6

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

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

Plot the shear diagram, bending moment diagram, axial force diagram, and the qualitative deflected shape of the frame.

Concept explainers

Plot the shear diagram, bending moment diagram, axial force diagram, and the qualitative deflected shape of the frame.

Explanation of Solution

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