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

James Finch

Engineering, section 1, Fall 2019

Instructor: Dr. Friendly

Current Score : 0 / 155

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

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1. 0/2 points  |  Previous Answers LoganFinElem6SI 5.4.046. My Notes
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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.
A nonprismatic torsion bar of length L starts at x = 0 at its left end with radius r0 and extends to the right. The radius increases linearly with the length.
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] =
QE
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2. /30 points LoganFinElem6SI 5.2.002. My Notes
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/30
 
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For all elements in the rigid frame shown below, let E = 210 GPa, A = 7.00 103 m2, and I = 8.2 105 m4. Use the finite element stiffness method. (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 5 m.
  • Element circled 1 extends upward from a fixed support at node 1 to its top end at node 2.
  • Element circled 2 extends to the right from node 2 to node 3.
  • Element circled 3 extends downward from its top end at node 3 to a fixed support at node 4.
A horizontal point load of 15 kN acts to the right at node 2.
(a)
Determine the nodal displacement components and rotations. (Enter displacements in meters and rotations in radians. Assume the positive directions are to the right, upward, and counterclockwise. Indicate the direction with the signs of your answers.)
Location u (m) v (m) ϕ (radians)
Node 1 0 0 0
Node 2
Node 3
Node 4 0 0 0
(b)
Determine the support reactions. (Enter forces in kN and moments in kN · m. Assume the positive directions are to the right, upward, and counterclockwise. Indicate the direction with the signs of your answers.)
Support Location Fx (kN) Fy (kN) M (kN · m)
Node 1
Node 4
(c)
Determine the forces in each element. (Enter forces in kN and moments in kN · m. Use the nodal and beam theory sign conventions for shear forces and bending moments.)
Element circled 1 from node 1 to node 2
f1x(1) = kN f1y(1) = kN m1(1) = kN · m f2x(1) = kN f2y(1) = kN m2(1) = kN · m
Element circled 2 from node 2 to node 3
f2x(2) = kN f2y(2) = kN m2(2) = kN · m f3x(2) = kN f3y(2) = kN m3(2) = kN · m
Element circled 3 from node 3 to node 4
f3x(3) = kN f3y(3) = kN m3(3) = kN · m f4x(3) = kN f4y(3) = kN m4(3) = kN · m

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3. /30 points LoganFinElem6SI 5.2.007. My Notes
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Total
/30
 
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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 circled 1 extends up 8 m from a fixed support at node 1 to node 2. Point load 36 kN acts to the right at the midpoint.
  • Element circled 2 is horizontal, and goes 8 m to the right from node 2 to a fixed support at node 3. Point load 72 kN acts downward at the midpoint.
  • Element circled 3 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 = 1.8 102 m2, and I = 1.4 104 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 circled 1 from node 1 to node 2
f1x(1) = kN f1y(1) = kN m1(1) = kN · m f2x(1) = kN f2y(1) = kN m2(1) = kN · m
Element circled 2 from node 2 to node 3
f2x(2) = kN f2y(2) = kN m2(2) = kN · m f3x(2) = kN f3y(2) = kN m3(2) = kN · m
Element circled 3 from node 2 to node 4
f2x(3) = kN f2y(3) = kN m2(3) = kN · m f4x(3) = kN f4y(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 Fx (kN) Fy (kN) M (kN · m)
Node 1
Node 3
Node 4
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4. /21 points LoganFinElem6SI 5.2.009. My Notes
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Total
/21
 
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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 circled 1 has length 3.8 m and extends upward from a fixed support at node 1 to node 2.
  • Element circled 2 has length 4.8 m and extends to the right from node 2 to a fixed support at node 3.
Moment 660 kN · m acts counterclockwise at node 2. The frame has properties E = 210 GPa, A = 2.6 102 m2, and I = 2.1 104 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 circled 1 from node 1 to node 2
f1x(1) = kN f1y(1) = kN m1(1) = kN · m f2x(1) = kN f2y(1) = kN m2(1) = kN · m
Element circled 2 from node 2 to node 3
f2x(2) = kN f2y(2) = kN m2(2) = kN · m f3x(2) = kN f3y(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 Fx (kN) Fy (kN) M (kN · m)
Node 1
Node 3

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5. /21 points LoganFinElem6SI 5.2.014. My Notes
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/1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1
0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100
Total
/21
 
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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. For the following, use the finite element stiffness method. (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 circled 1 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 5 m.
  • Element circled 2 is horizontal, and goes 7.5 m to the right from node 2 to a fixed support at node 3.
A uniformly distributed load of constant intensity 24 kN/m acts downward across the full length of element circled 2. The frame has properties E = 210 GPa, A = 4 103 m2, and I = 8 105 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. Indicate the direction with the signs of your answers.)
Location u (m) v (m) ϕ (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 circled 1 from node 1 to node 2
f1x(1) = kN f1y(1) = kN m1(1) = kN · m f2x(1) = kN f2y(1) = kN m2(1) = kN · m
Element circled 2 from node 2 to node 3
f2x(2) = kN f2y(2) = kN m2(2) = kN · m f3x(2) = kN f3y(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. Indicate the direction with the signs of your answers.)
Support Location Fx (kN) Fy (kN) M (kN · m)
Node 1
Node 3
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6. /1 points LoganFinElem6SI 5.2.023. My Notes
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/1
0/100
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/1
 
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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 = 210 GPa.
  • Element circled 1 extends up 4.5 m from a pin support at node 1 to node 3.
  • Element circled 2 extends up 4.5 m from a pin support at node 2 to node 4.
  • Element circled 3 extends up 4.5 m from node 3 to node 5.
  • Element circled 4 extends up 4.5 m from node 4 to node 6.
  • An element element extends to the right from node 5 to node 6, with area A = 7.5 103 m2, moment of inertia I = 8 105 m4, and length 7.5 m.
  • An element extends to the right from node 3 to node 4, with area A = 7.5 103 m2, and moment of inertia I = 12 105 m4, and length 7.5 m.
  • Four diagonally aligned elements connect other pairs, each with area A = 12.5 103 m2 and moment of inertia I = 4 107 m4. :
    • 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 circled 1, circled 2, circled 3, and circled 4 have area A = 6 103 m2 and moment of inertia I = 6 105 m4.
Two point loads and a moment are applied to the frame.
  • Point load 70 kN acts to the right at node 3.
  • Point load 5 kN acts to the right at node 5.
  • Moment 4 kN · m acts counterclockwise at node 5.
(a)
Determine the displacements and rotations at the nodes.
(b)
Determine the element forces for the vertical elements.
(c)
Determine the element forces for the horizontal elements.

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7. /42 points LoganFinElem6SI 5.2.034. My Notes
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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
/1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1 /1
0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100
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/42
 
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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 7 m.
  • Element circled 1 extends to the right from a pin support at node 1 to a pin support at node 2.
  • Element circled 2 extends to the right from node 2 to a roller support at node 3.
  • Element circled 3 extends to the right from node 3 to a roller support at node 4.
A uniformly distributed load of constant intensity 7,000 N/m acts downward across the full length of the beam. The beam has properties E = 210 GPa, I = 1.0 104 m4, and A = 1.0 102 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
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 circled 1
f1x(1) = kN f1y(1) = kN m1(1) = kN · m f2x(1) = kN f2y(1) = kN m2(1) = kN · m
Element circled 2
f2x(2) = kN f2y(2) = kN m2(2) = kN · m f3x(2) = kN f3y(2) = kN m3(2) = kN · m
Element circled 3
f3x(3) = kN f3y(3) = kN m3(3) = kN · m f4x(3) = kN f4y(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 Fx (kN) Fy (kN) M (kN · m)
Node 1
Node 2
Node 3
Node 4
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8. /1 points LoganFinElem6SI 5.2.038. My Notes
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Points
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/1
0/100
Total
/1
 
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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 = 2.0 104 m4, and A = 1.0 102 m2.
  • Element circled 1 of length 8 m extends up from a fixed support at node 1 to node 2.
  • Element circled 2 of length 7 m extends to the right from node 2 to node 3.
  • Element circled 3 of length 8 m extends up from a fixed support at node 7 to node 3.
  • Element circled 4 of length 3.5 m extends to the right from node 3 to node 4.
  • Element circled 5 of length 3.5 m extends to the right from node 4 to node 5.
  • Element 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 circled 1. 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 circled 2.
  • 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 and horizontal elements.

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9. /3 points LoganFinElem6SI 5.4.047. My Notes
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/1 /1 /1
0/100 0/100 0/100
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/3
 
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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 kN · m/m) 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 (kN · m/m). 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 =
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.)
M1x
=
M2x
=
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10. /4 points LoganFinElem6SI 5.2.016. My Notes
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/4
 
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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 = 12 kN acts downward at node 5. The elements have properties E = 200 GPa and A = 1 102 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
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