Logan - Finite Element Method

1. 0/30 points | Previous Answers LoganFinElem6 5.2.002. My Notes

Question Part

Points

Submissions Used

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

0/1

–/1

1/100

0/100

Total

0/30

This exercise will let students review and reference question concepts.
Read It links are available as a learning tool under each question so students can quickly jump to the corresponding section of the eTextbook.
Students get just-in-time learning support with Watch It videos that contain narrated and closed-captioned videos walking students through the proper steps to solve a similar problem.

For all elements in the rigid frame shown below, let E = 30 ✕ 10⁶ psi,

A = 14 in²,

and

I = 515 in⁴.

(Due to the nature of this problem, do not use rounded intermediate values in your calculations—including answers submitted in WebAssign.)

A rigid frame is composed of three elements of length 30 ft.

Element extends upward from a fixed support at node 1 to its top end at node 2.

Element extends to the right from node 2 to node 3.

Element extends downward from its top end at node 3 to a fixed support at node 4.

A horizontal point load of 4,300 lb acts to the right at node 2.

circled 1 — A rigid frame is composed of three elements of length 30 ft.

Element extends upward from a fixed support at node 1 to its top end at node 2.

Element extends to the right from node 2 to node 3.

Element extends downward from its top end at node 3 to a fixed support at node 4.

A horizontal point load of 4,300 lb acts to the right at node 2.

(a)

Determine the nodal displacement components and rotations in each element. (Enter displacements in inches and rotations in radians. Use the sign conventions for nodal forces and equilibrium.)

Location	u (inches)	v (inches)	ϕ (radians)
Node 1	0	0	0
Node 2	Your response differs from the correct answer by more than 10%. Double check your calculations.	Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully.	The response you submitted has the wrong sign.
Node 3	Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error.	The response you submitted has the wrong sign.	The response you submitted has the wrong sign.
Node 4	0	0	0

(b)

Determine the support reactions. (Enter forces in lb and moments in lb · in. Assume the positive directions are to the right, upward, and counterclockwise. Include the signs of the values in your answers.)

Support Location	F_x (lb)	F_y (lb)	M (lb · in.)
Node 1
Node 4

(c)

Determine the forces in each element. (Enter forces in lb and moments in lb · in. Use the nodal and beam theory sign conventions for shear forces and bending moments.)

Element

f ′_1x⁽¹⁾ = lb f ′_1y⁽¹⁾ = lb m′₁⁽¹⁾ = lb · in. f ′_2x⁽¹⁾ = lb f ′_2y⁽¹⁾ = lb m′₂⁽¹⁾ = lb · in.

Element

f ′_2x⁽²⁾ = lb f ′_2y⁽²⁾ = lb m′₂⁽²⁾ = lb · in. f ′_3x⁽²⁾ = lb f ′_3y⁽²⁾ = lb m′₃⁽²⁾ = lb · in.

Element

f ′_3x⁽³⁾ = lb f ′_3y⁽³⁾ = lb m′₃⁽³⁾ = lb · in. f ′_4x⁽³⁾ = lb f ′_4y⁽³⁾ = lb m′₄⁽³⁾ = lb · in.

Need Help?

Viewing Saved Work Revert to Last Response

2. –/30 points LoganFinElem6 5.2.007. My Notes

Question Part

Points

Submissions Used

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

–/1

0/100

Total

–/30

This exercise will let students review and reference question concepts.
Read It links are available as a learning tool under each question so students can quickly jump to the corresponding section of the eTextbook.

For the rigid frame shown in the figure below, the values of E, A, and I to be used are listed next to the figure. (Due to the nature of this problem, do not use rounded intermediate values in your calculations—including answers submitted in WebAssign.)

A rigid frame is composed of the following three elements, acted upon by the following two point loads.

Element extends up 8 m from a fixed support at node 1 to node 2. Point load 30 kN acts to the right at the midpoint.

Element is horizontal, and goes 8 m to the right from node 2 to a fixed support at node 3. Point load 60 kN acts downward at the midpoint.

Element extends down and to the right from node 2 to a fixed support at node 4. A horizontal line connects nodes 1 and 4, while a vertical line connects nodes 3 and 4.

The frame has properties E = 210 GPa, A = 1.2 ✕ 10⁻² m², and I = 1.2 ✕ 10⁻⁴ m⁴.

(a)

Determine the displacements and rotations of the nodes. (Enter displacements in meters and rotations in radians. Assume the positive directions are to the right, upward, and counterclockwise. Include the signs of the values in your answers.)

Location	u (m)	v (m)	ϕ (radians)
Node 1	0	0	0
Node 2
Node 3	0	0	0
Node 4	0	0	0

(b)

Determine the element forces. (Enter forces in kN and moments in kN · m. Use the nodal and beam theory sign conventions for shear forces and bending moments.)

Element

from node 1 to node 2

f ′_1x⁽¹⁾ = kN f ′_1y⁽¹⁾ = kN m₁⁽¹⁾ = kN · m f ′_2x⁽¹⁾ = kN f ′_2y⁽¹⁾ = kN m₂⁽¹⁾ = kN · m

Element

from node 2 to node 3

f ′_2x⁽²⁾ = kN f ′_2y⁽²⁾ = kN m₂⁽²⁾ = kN · m f ′_3x⁽²⁾ = kN f ′_3y⁽²⁾ = kN m₃⁽²⁾ = kN · m

Element

from node 2 to node 4

f ′_2x⁽³⁾ = kN f ′_2y⁽³⁾ = kN m₂⁽³⁾ = kN · m f ′_4x⁽³⁾ = kN f ′_4y⁽³⁾ = kN m₄⁽³⁾ = kN · m

(c)

Determine the reactions. (Enter forces in kN and moments in kN · m. Assume the positive directions are to the right, upward, and counterclockwise. Include the signs of the values in your answers.)

Support Location	F_x (kN)	F_y (kN)	M (kN · m)
Node 1
Node 3
Node 4

Viewing Saved Work Revert to Last Response

3. –/21 points LoganFinElem6 5.2.009. My Notes

Question Part

Points

Submissions Used

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

–/1

0/100

Total

–/21

This exercise will let students review and reference question concepts.
Read It links are available as a learning tool under each question so students can quickly jump to the corresponding section of the eTextbook.
Students get just-in-time learning support with Watch It videos that contain narrated and closed-captioned videos walking students through the proper steps to solve a similar problem.

Consider the rigid frame shown in the figure below. The values of E, A, and I are listed next to the figure. (Due to the nature of this problem, do not use rounded intermediate values in your calculations—including answers submitted in WebAssign.)

A rigid frame is composed of two elements.

Element has length 3.7 m and extends upward from a fixed support at node 1 to node 2.

Element has length 4.7 m and extends to the right from node 2 to a fixed support at node 3.

Moment 590 kN · m acts counterclockwise at node 2. The frame has properties E = 210 GPa, A = 2.1 ✕ 10⁻² m², and I = 2.8 ✕ 10⁻⁴ m⁴.

(a)

Determine the displacements and rotations of the nodes. (Enter displacements in meters and rotations in radians. Assume the positive directions are to the right, upward, and counterclockwise. Include the signs of the values in your answers.)

Location	u (m)	v (m)	θ (radians)
Node 1	0	0	0
Node 2
Node 3	0	0	0

(b)

Determine the element forces. (Enter forces in kN and moments in kN · m. Use the nodal and beam theory sign conventions for shear forces and bending moments.)

Element

from node 1 to node 2

f ′_1x⁽¹⁾ = kN f ′_1y⁽¹⁾ = kN m₁⁽¹⁾ = kN · m f ′_2x⁽¹⁾ = kN f ′_2y⁽¹⁾ = kN m₂⁽¹⁾ = kN · m

Element

from node 2 to node 3

f ′_2x⁽²⁾ = kN f ′_2y⁽²⁾ = kN m₂⁽²⁾ = kN · m f ′_3x⁽²⁾ = kN f ′_3y⁽²⁾ = kN m₃⁽²⁾ = kN · m

(c)

Determine the reactions. (Enter forces in kN and moments in kN · m. Assume the positive directions are to the right, upward, and counterclockwise. Include the signs of the values in your answers.)

Support Location	F_x (kN)	F_y (kN)	M (kN · m)
Node 1
Node 3

Need Help?

Viewing Saved Work Revert to Last Response

4. –/21 points LoganFinElem6 5.2.014. My Notes

Question Part

Points

Submissions Used

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

–/1

0/100

Total

–/21

This exercise will let students review and reference question concepts.
Read It links are available as a learning tool under each question so students can quickly jump to the corresponding section of the eTextbook.

For the rigid frame shown in the figure below, the values of E, A, and I to be used are listed next to the figure. (Due to the nature of this problem, do not use rounded intermediate values in your calculations—including answers submitted in WebAssign.)

A rigid frame is composed of two elements.

Element extends up and to the right at a 45° angle from the horizontal. The element goes from a fixed support at node 1 to node 2, at a height of 8 ft.

Element is horizontal, and goes 16 ft to the right from node 2 to a fixed support at node 3.

A uniformly distributed load of constant intensity 3,750 lb/ft acts downward across the full length of element . The frame has properties E = 30 ✕ 10⁶ psi, A = 5 in.², and I = 220 in.⁴.

(a)

Determine the displacements and rotations of the nodes. (Enter displacements in inches and rotations in radians. Use the sign conventions for nodal forces and equilibrium.)

Location	u (inches)	v (inches)	ϕ (radians)
Node 1	0	0	0
Node 2
Node 3	0	0	0

(b)

Determine the element forces. (Enter forces in kip and moments in kip · in. Use the nodal and beam theory sign conventions for shear forces and bending moments.)

Element

f ′_1x⁽¹⁾ = kip f ′_1y⁽¹⁾ = kip m₁⁽¹⁾ = kip · in. f ′_2x⁽¹⁾ = kip f ′_2y⁽¹⁾ = kip m₂⁽¹⁾ = kip · in.

Element

f ′_2x⁽²⁾ = kip f ′_2y⁽²⁾ = kip m₂⁽²⁾ = kip · in. f ′_3x⁽²⁾ = kip f ′_3y⁽²⁾ = kip m₃⁽²⁾ = kip · in.

(c)

Determine the reactions. (Enter forces in kip and moments in kip · in. Assume the positive directions are to the right, upward, and counterclockwise. Include the signs of the values in your answers.)

Support Location	F_x (kip)	F_y (kip)	M (kip · in.)
Node 1
Node 3

Viewing Saved Work Revert to Last Response

5. –/4 points LoganFinElem6 5.6.016. My Notes

This exercise will let students review and reference question concepts.
Read It links are available as a learning tool under each question so students can quickly jump to the corresponding section of the eTextbook.
Students get just-in-time learning support with Watch It videos that contain narrated and closed-captioned videos walking students through the proper steps to solve a similar problem.

Solve the structure in the figure below by using substructuring.

A structure is composed of nine elements. The nodes are arranged in two rows of three nodes each, as follows from left to right.

Top row: Nodes 1, 3, 5

Bottom row: Nodes 2, 4, 6

The nine elements are arranged, as follows:

horizontally 1 to 3,

horizontally 3 to 5,

horizontally 2 to 4,

horizontally 4 to 6,

vertically 1 to 2,

vertically 3 to 4,

vertically 5 to 6,

diagonally 2 to 3, and

diagonally 4 to 5.

The horizontal and vertical elements have equal length. The diagonal elements are slightly longer. The structure is supported by pin supports at nodes 1 and 2. A point load P = 23 kN acts downward at node 5. The elements have properties E = 200 GPa and A = 1 ✕ 10⁻² m².

By using substructuring, determine the horizontal and vertical displacements (in m) at nodes 3 and 4. (Assume each horizontal and vertical element in the structure has length L = 1 m. Use the sign conventions for nodal forces and equilibrium.)

u₃ = m v₃ = m u₄ = m v₄ = m

Need Help?

Viewing Saved Work Revert to Last Response

6. –/1 points LoganFinElem6 5.2.023. My Notes

This exercise will let students review and reference question concepts.
Read It links are available as a learning tool under each question so students can quickly jump to the corresponding section of the eTextbook.
Students get just-in-time learning support with Watch It videos that contain narrated and closed-captioned videos walking students through the proper steps to solve a similar problem.

Solve by using a computer program. For the rigid frame shown in the figure below, the values of E, A, and I to be used are listed next to the figure. (Submit a file with a maximum size of 1 MB. Your submission will be graded by your instructor after the due date based on its accuracy. Your grade may change.)

A rigid frame is composed of many elements, all of which have modulus of elasticity E = 30 ✕ 10⁶ psi.

Element extends up 15 ft from a pin support at node 1 to node 3.

Element extends up 15 ft from a pin support at node 2 to node 4.

Element extends up 15 ft from node 3 to node 5.

Element extends up 15 ft from node 4 to node 6.

An element extends to the right from node 5 to node 6, with area A = 12 in², moment of inertia I = 200 in⁴, and length 25 ft.

An element extends to the right from node 3 to node 4, with area A = 12 in², moment of inertia I = 300 in⁴, and length 25 ft.

Four diagonally aligned elements connect other pairs, each with area A = 2.0 in² and moment of inertia I = 1.0 in⁴. :

An element extends from node 2 to node 3,

an element extends from node 1 to node 4,

an element extends from node 4 to node 5, and

an element extends from node 3 to node 6.

Elements , , , and have area A = 10 in² and moment of inertia I = 150 in⁴.

Two point loads and a moment are applied to the frame.

Point load 2,400 lb acts to the right at node 3.

Point load 1,200 lb acts to the right at node 5.

Moment 3,000 lb-ft acts counterclockwise at node 3.

(a)

Determine the displacements and rotations at the nodes.

(b)

Determine the element forces for the vertical and horizontal elements.

(c)

Determine the reactions.

This answer has not been graded yet.

Need Help?

Viewing Saved Work Revert to Last Response

7. 0/42 points | Previous Answers LoganFinElem6 5.2.034. My Notes

Question Part

Points

Submissions Used

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40

41

42

0/1

–/1

1/100

0/100

Total

0/42

This exercise will let students review and reference question concepts.
Read It links are available as a learning tool under each question so students can quickly jump to the corresponding section of the eTextbook.

For the rigid beam shown in the figure below, solve the following by using a computer program.

A horizontal beam is composed of three elements of length 6.5 m.

Element extends to the right from a pin support at node 1 to a pin support at node 2.

Element extends to the right from node 2 to a roller support at node 3.

Element extends to the right from node 3 to a roller support at node 4.

A uniformly distributed load of constant intensity 7,400 N/m acts downward across the full length of the beam. The beam has properties E = 210 GPa, I = 1.8 ✕ 10⁻⁴ m⁴, and A = 1.8 ✕ 10⁻² m².

(a)

Determine the displacements and rotations at the nodes. (Enter displacements in meters and rotations in radians. Use the sign conventions for nodal forces and equilibrium.)

Location	u (m)	v (m)	θ (radians)
Node 1	Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully.
Node 2
Node 3
Node 4

(b)

Determine the element forces. (Enter forces in kN and moments in kN · m. Use the nodal and beam theory sign conventions for shear forces and bending moments.)

Element

f ′_1x⁽¹⁾ = kN f ′_1y⁽¹⁾ = kN m₁⁽¹⁾ = kN · m f ′_2x⁽¹⁾ = kN f ′_2y⁽¹⁾ = kN m₂⁽¹⁾ = kN · m

Element

f ′_2x⁽²⁾ = kN f ′_2y⁽²⁾ = kN m₂⁽²⁾ = kN · m f ′_3x⁽²⁾ = kN f ′_3y⁽²⁾ = kN m₃⁽²⁾ = kN · m

Element

f ′_3x⁽³⁾ = kN f ′_3y⁽³⁾ = kN m₃⁽³⁾ = kN · m f ′_4x⁽³⁾ = kN f ′_4y⁽³⁾ = kN m₄⁽³⁾ = kN · m

(c)

Determine the reactions. (Enter forces in kN and moments in kN · m. Assume the positive directions are to the right, upward, and counterclockwise. Include the signs of the values in your answers.)

Support Location	F_x (kN)	F_y (kN)	M (kN · m)
Node 1
Node 2
Node 3
Node 4

Viewing Saved Work Revert to Last Response

8. –/1 points LoganFinElem6 5.2.038. My Notes

This exercise will let students review and reference question concepts.
Read It links are available as a learning tool under each question so students can quickly jump to the corresponding section of the eTextbook.

Solve by using a computer program. For the rigid frame shown in the figure below, the values of E, A, and I to be used are listed next to the figure. (Submit a file with a maximum size of 1 MB. Your submission will be graded by your instructor after the due date based on its accuracy. Your grade may change.)

A rigid frame is composed of six elements, all with properties E = 210 GPa, I = 2.0 ✕ 10⁻⁴ m⁴, and A = 1.0 ✕ 10⁻² m².

Element of length 8 m extends up from a fixed support at node 1 to node 2.

Element of length 7 m extends to the right from node 2 to node 3.

Element of length 8 m extends down from node 3 to a fixed support at node 7.

Element of length 3.5 m extends to the right from node 3 to node 4.

Element of length 3.5 m extends to the right from node 4 to node 5.

Element ⑥ of length 8 m extends down from node 5 to a fixed support at node 6.

A point load and two distributed loads are applied to the frame.

A triangularly distributed load acts to the right on element . It has maximum intensity 300 kN/m at node 1 and minimum intensity 0 at node 2.

A uniformly distributed load of constant intensity 15 kN/m acts downward across the full length of element .

Point load 80 kN acts downward at node 4.

(a)

Determine the displacements and rotations at the nodes.

(b)

Determine the element forces for the vertical elements.

(c)

Determine the reactions.

This answer has not been graded yet.

Need Help?

Viewing Saved Work Revert to Last Response

9. 0/2 points | Previous Answers LoganFinElem6 5.4.046. My Notes

This exercise will let students review and reference question concepts.
Read It links are available as a learning tool under each question so students can quickly jump to the corresponding section of the eTextbook.

For the nonprismatic torsion bar shown in the figure below, the radius of the shaft is given by

r = r₀ +

x

L

r₀,

where r₀ is the radius at x = 0.

Derive the stiffness matrix for the torsion bar. (For the scalar factor, use the following as necessary: length L, shear modulus of elasticity G, and torsional constant J₀ at x = 0. For the elements of the matrix, use either 1, 0, or −1.)

[K] =

iκ

Viewing Saved Work Revert to Last Response

10. 0/3 points | Previous Answers LoganFinElem6 5.4.047. My Notes

This exercise will let students review and reference question concepts.
Read It links are available as a learning tool under each question so students can quickly jump to the corresponding section of the eTextbook.
Students get just-in-time learning support with Watch It videos that contain narrated and closed-captioned videos walking students through the proper steps to solve a similar problem.

A prismatic circular cross-section torsion bar is shown in the figure below.

A torsion bar spans length L from its left end "1" to its right end "2". Torque per unit length t (in lb · in.⁄in.) acts counterclockwise on the bar as viewed from the right end.

Derive the total potential energy for the bar subjected to uniform torque per unit length (lb · in./in.). Let G be the shear modulus and J be the polar moment of inertia of the bar. (Use the following as necessary: L, G, J, and torque per unit length t.)

U =

9

Also, determine the equivalent nodal torques for the bar. (Enter the magnitudes. Use the following as necessary: L, G, J, and torque per unit length t.)

M_1x

=

3

M_2x

=

4

Viewing Saved Work Revert to Last Response

Logan - Finite Element Method - 6/e (Homework)

Current Score : 0 / 155

Due : Sunday, January 27, 2030 12:00 EST

Last Saved : n/a Saving... ()

Instructions

Assignment Submission

Assignment Scoring

WebAssign

Welcome, demo@demo

Logan - Finite Element Method - 6/e (Homework)

Current Score : 0 / 155

Due : Sunday, January 27, 2030 12:00 EST

Last Saved : n/a Saving... ()

Instructions

Assignment Submission

Assignment Scoring