김경종
17 KiB
d _ { 2 z } = 1 5 mm. Determine the displacement along the local x^ axis at node 2 of the elements. The coordinates, in meters, are shown in the figures.
3.40–3.41 For the space trusses shown in Figures P3–40 and P3–41, determine the nodal displacements and the stresses in each element. Let E = 2 1 0 \mathrm { G P a } and A = 1 0 \times 1 0 ^ { - 4 } \mathrm { m } ^ { 2 } for all elements. Verify force equilibrium at node 1. The coordinates of each node, in meters, are shown in the figure. All supports are ball-and-socket joints.
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y (0, 4, 0) 2 ① ② 1 (4, 4, 3) 3 (0, 4, 6) ③ ④ x 10 kN 4 (4, 0, 3) z (8, -1, 1)
Figure P3–40
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(0, 0, 0) 2 x ① ③ 4 (14, 6, 0) (12, -3, -4) 1 20 kN (in the x direction) ② 3 (12, -3, -7)
Figure P3–41
3.42 For the space truss subjected to a 1000-lb load in the x direction, as shown in Figure P3–42, determine the displacement of node 5. Also determine the stresses in each element. Let A = 4 ~ \mathrm { i n } ^ { 2 } and E = 3 0 \times 1 0 ^ { 6 } psi for all elements. The coordinates of each
node, in inches, are shown in the figure. Nodes 1–4 are supported by ball-and-socket joints (fixed supports).
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(0, 0, 72) 1000 lb 5 ④ ③ 4 (0, 36, 0) (−72, 36, 0) 3 ① y z x ② 1 (−72, −36, 0) 2 (0, −36, 0)
Figure P3–42
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(0, 0, 144) 4000 lb (0, 0, 144) 4 2 (-144, 72, 0) ② ① ③ y z x 1 (-36, 0, 0) (-144, -72, 0) 3
Figure P3–43
3.43 For the space truss subjected to the 4000-lb load acting as shown in Figure P3–43, determine the displacement of node 4. Also determine the stresses in each element. Let
A = 6 ~ \mathrm { i n } ^ { 2 } and E = 3 0 \times 1 0 ^ { 6 } psi for all elements. The coordinates of each node, in inches, are shown in the figure. Nodes 1–3 are supported by ball-and-socket joints (fixed supports).
3.44 Verify Eq. (3.7.9) for k by first expanding \underline { { \boldsymbol { T } } } ^ { * } , given by Eq. (3.7.7), to a 6 \times 6 square matrix in a manner similar to that done in Section 3.4 for the two-dimensional case. Then expand \underline { { \hat { k } } } to a 6 \times 6 matrix by adding appropriate rows and columns of zeros (for the \hat { d } _ { z } terms) to Eq. (3.4.17). Finally, perform the matrix triple product \underline { { k } } = \underline { { T } } ^ { T } \hat { \underline { { k } } } \underline { { T } } .
3.45 Derive Eq. (3.7.21) for stress in space truss elements by a process similar to that used to derive Eq. (3.5.6) for stress in a plane truss element.
3.46 For the truss shown in Figure P3–46, use symmetry to determine the displacements of the nodes and the stresses in each element. All elements have E = 3 0 \times 1 0 ^ { 6 } psi. Elements 1, 2, 4, and 5 have A = 1 0 ~ \mathrm { i n } ^ { 2 } and element 3 has A = 2 0 \mathrm { i n } ^ { 2 } .
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3 ① ② ③ ④ ⑤ 1 2 4 6 ft 8 ft 8 ft 20,000 lb
Figure P3–46
3.47 All elements of the structure in Figure P3–47 have the same A E except element 1, which has an axial stiffness of 2 A E . . Find the displacements of the nodes and the stresses in elements 2, 3, and 4 by using symmetry. Check equilibrium at node 4. You might want to use the results obtained from the stiffness matrix of Problem 3.24.
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y P ② P 3 ⑦ P 4 5 15 ft ③ ④ ⑤ ⑨ ⑧ ⑩ ⑥ ① 2 20 ft 20 ft x 1 2 6 6 x
Figure P3–47
3.48 For the roof truss shown in Figure P3–48, use symmetry to determine the displacements of the nodes and the stresses in each element. All elements have E = 2 1 0 \ \mathrm { G P a } and A = 1 0 \times 1 0 ^ { - 4 } ~ \mathrm { m } ^ { 2 } .
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20 kN 20 kN 3 20 kN 1 2 3 4 5 6 7 8 9 4 m 1 2 3 4 5 6 8 m 8 m
Figure P3–48
3.49–3.51 For the plane trusses with inclined supports shown in Figures P3–49—P3–51, solve for the nodal displacements and element stresses in the bars. Let A = 2 \mathrm { i n } ^ { 2 } , E = 3 0 \times 1 0 ^ { 6 } ~ \mathrm { p s i } , and L = 3 0 in. for each truss.
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45° L 1 L 2 3 2000 lb
Figure P3–49
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3 L 2 L 2000 lb 1
Figure P3–50
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45° L L 3 2000 lb 45° 1 2 L
Figure P3–51
3.52 Use the principle of minimum potential energy developed in Section 3.10 to solve the bar problems shown in Figure P3–52. That is, plot the total potential energy for variations in the displacement of the free end of the bar to determine the minimum potential energy. Observe that the displacement that yields the minimum potential energy also yields the stable equilibrium position. Use displacement increments of
0.002 in., beginning with x = - 0 . 0 0 4 . . Let E = 3 0 \times 1 0 ^ { 6 } psi and A = 2 ~ \mathrm { i n } ^ { 2 } for the bars.
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20,000 lb 30 in.
(a)
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50 in. 10,000 lb
(b)
Figure P3–52
3.53 Derive the stiffness matrix for the nonprismatic bar shown in Figure P3–53 using the principle of minimum potential energy. Let E be constant.
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A(x) = A₀ + A₀ x/L x L
Figure P3–53
3.54 For the bar subjected to the linear varying axial load shown in Figure P3–54, determine the nodal displacements and axial stress distribution using (a) two equal-length elements and (b) four equal-length elements. Let A = 2 ~ \mathrm { i n } . ^ { 2 } and E = 3 0 \times 1 0 ^ { 6 } psi. Compare the finite element solution with an exact solution.
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Tx = 10x lb/in. 60 in.
Figure P3–54
3.55 For the bar subjected to the uniform line load in the axial direction shown in Figure P3–55, determine the nodal displacements and axial stress distribution using (a) two equal-length elements and (b) four equal-length elements. Compare the finite element results with an exact solution. Let \overset { \cdot } { A } \overset { = } 2 \mathrm { i n } ^ { 2 } and E = 3 0 \times 1 0 ^ { 6 } psi.
3.56 For the bar fixed at both ends and subjected to the uniformly distributed loading shown in Figure P3–56, determine the displacement at the middle of the bar and the stress in the bar. Let A = 2 \mathrm { i n } ^ { 2 } and E = 3 0 \times 1 0 ^ { 6 } psi.
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300 lb/in. 1 3 2 30 in. 30 in.
Figure P3–55
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Tₓ = 100 lb/in. 30 in. 30 in.
Figure P3–56
3.57 For the bar hanging under its own weight shown in Figure P3–57, determine the nodal displacements using (a) two equal-length elements and (b) four equal-length elements. Let A = 2 \ \mathrm { i n } ^ { 2 } , \stackrel { \cdot } { E } = 3 0 \times \mathrm { i 0 } ^ { 6 } psi, and weight density \rho _ { w } = 0 . 2 8 3 ~ \mathrm { 1 b } / \mathrm { i n } ^ { 3 } . (Hint: The internal force is a function of x. Use the potential energy approach.)
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60 in. Figure P3-57 x
3.58 Determine the energy equivalent nodal forces for the axial distributed loading shown acting on the bar elements in Figure P3–58.
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Tₓ = 100 + 5x lb/in. 1 → x → 2 10 in.
(a)
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T_x = 5x^2 kN/m 1 → x → 2 4 m
(b)
Figure P3–58
3.59 Solve problem 3.55 for the axial displacement in the bar using collocation, subdomain, least squares, and Galerkin’s methods. Choose a quadratic polynomial u ( x ) = c _ { 1 } x + c _ { 2 } x ^ { 2 } in each method. Compare these weighted residual method solutions to the exact solution.
3.60 For the tapered bar shown with cross sectional areas A _ { 1 } = 2 ~ \mathrm { i n } . ^ { 2 } and A _ { 2 } = 1 ~ \mathrm { i n } . ^ { 2 } at each end, use the collocation, subdomain, least squares, and Galerkin’s methods to obtain the displacement in the bar. Compare these weighted residual solutions to the exact solution. Choose a cubic polynomial u ( x ) = c _ { 1 } x + c _ { 2 } x ^ { 2 } + c _ { 3 } x ^ { 3 } .
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A₁ → x → A₂ P = 1000 lb E = 10 × 10⁶ psi L = 20 in.
Figure P3–60
3.61 For the bar shown in Figure P 3–61 subjected to the linear varying axial load, determine the displacements and stresses using (a) one and then two finite element models and (b) the collocation, subdomain, least squares, and Galerkin’s methods assuming a cubic polynomial of the form u ( x ) = c _ { 1 } x + c _ { 2 } x ^ { 2 } + c _ { 3 } x ^ { 3 } .
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T(x) = 10x kN/m AE = 2 × 10⁴ kN 3.0 m x
Figure P3–61
3.62–3.67
Use a computer program to solve the truss design problems shown in Figures P3. 62–3.67. Determine the single most critical cross-sectional area based on maximum allowable yield strength or buckling strength (based on either Euler’s or Johnson’s formula as relevant) using a factor of safety (FS) listed next to each truss. Recommend a common structural shape and size for each truss. List the largest three nodal displacements and their locations. Also include a plot of the deflected shape of the truss and a principal stress plot.
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25' 10' 10' 25' F = 20 kip
Figure P3–62 Derrick truss (FS ¼ 4:0Þ
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4000 lb 16,000 lb 15' 18' 3'
Figure P3–63 Truss bridge (FS ¼ 3:0Þ
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60 kN 50 kN 100 kN 100 kN 100 kN 4 m 60 kN 4 m 4 m 4 m 4 m 4 m 4 m 4 m 4 m 4 m 4 m 4 m 4 m 4 m 4 m 4 m 4 m 4 m 4 m 4 m 4 m 4 m 4 m 4 m 4 m 4 m 4 m 4 m 4 m 4 m 4 m 4 m 4 m 4 m 4 m
Figure P3–64 Tower (FS ¼ 2:5Þ
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3' 3' 3' 3' A B C 3' 3' 3' G D E F 40 kip
Figure P3–65 Boxcar lift (FS ¼ 3:0Þ
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2.0 kip 2.0 kip 2.0 kip 1.0 kip 2.0 kip 2.0 kip 1.0 kip 6 ft 4.5 ft 8 ft 8 ft 8 ft 8 ft 8 ft 8 ft A B C D E F G H I J K L
Figure P3–66 Howe scissors roof truss (FS ¼ 2:0Þ
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1 kip 2 kip 2 kip 2 kip 9 ft 31 ft E F G H 12 ft 14 ft 14 ft 15.5 ft K L 8 ft 8 ft
Figure P3–67 Stadium roof truss (FS ¼ 3:0)
Development of Beam Equations
Introduction
We begin this chapter by developing the stiffness matrix for the bending of a beam element, the most common of all structural elements as evidenced by its prominence in buildings, bridges, towers, and many other structures. The beam element is considered to be straight and to have constant cross-sectional area. We will first derive the beam element stiffness matrix by using the principles developed for simple beam theory.
We will then present simple examples to illustrate the assemblage of beam element stiffness matrices and the solution of beam problems by the direct stiffness method presented in Chapter 2. The solution of a beam problem illustrates that the degrees of freedom associated with a node are a transverse displacement and a rotation. We will include the nodal shear forces and bending moments and the resulting shear force and bending moment diagrams as part of the total solution.
Next, we will discuss procedures for handling distributed loading, because beams and frames are often subjected to distributed loading as well as concentrated nodal loading. We will follow the discussion with solutions of beams subjected to distributed loading and compare a finite element solution to an exact solution for a beam subjected to a distributed loading.
We will then develop the beam element stiffness matrix for a beam element with a nodal hinge and illustrate the solution of a beam with an internal hinge.
To further acquaint you with the potential energy approach for developing stiffness matrices and equations, we will again develop the beam bending element equations using this approach. We hope to increase your confidence in this approach. It will be used throughout much of this text to develop stiffness matrices and equations for more complex elements, such as two-dimensional (plane) stress, axisymmetric, and three-dimensional stress.
Finally, the Galerkin residual method is applied to derive the beam element equations.
The concepts presented in this chapter are prerequisite to understanding the concepts for frame analysis presented in Chapter 5.
4.1 Beam Stiffness
In this section, we will derive the stiffness matrix for a simple beam element. A beam is a long, slender structural member generally subjected to transverse loading that produces significant bending effects as opposed to twisting or axial effects. This bending deformation is measured as a transverse displacement and a rotation. Hence, the degrees of freedom considered per node are a transverse displacement and a rotation (as opposed to only an axial displacement for the bar element of Chapter 3).
Consider the beam element shown in Figure 4–1. The beam is of length L with axial local coordinate x^ and transverse local coordinate y^. The local transverse nodal displacements are given by \hat { d } _ { i y } { \bf ' s } and the rotations by \hat { \phi _ { i } } \mathrm { \widetilde s } . The local nodal forces are given by \hat { f } _ { i y } ^ { \mathrm { ~ \tiny ~ s ~ } } and the bending moments by \hat { m _ { i } } \mathrm { ' s } as shown. We initially neglect all axial effects.
At all nodes, the following sign conventions are used:
- Moments are positive in the counterclockwise direction.
- Rotations are positive in the counterclockwise direction.
- Forces are positive in the positive y^ direction.
- Displacements are positive in the positive y^ direction.
Figure 4–2 indicates the sign conventions used in simple beam theory for positive shear forces \hat { V } and bending moments m^.
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ŷ, v̂ φ̂₁, m̂₁ 1 x̂ L f̂₁y, d̂₁y 2 m̂₂, φ̂₂ f̂₂y, d̂₂y
Figure 4–1 Beam element with positive nodal displacements, rotations, forces, and moments
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m̂ L m̂ V̂ V̂
Figure 4–2 Beam theory sign conventions for shear forces and bending moments