김경종
16 KiB
neutral axis to the top or bottom of the beam cross section, as used in the bending stress formula \sigma = ( M c / I ) .
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1000 lb 4 8 12 2000 lb 3 7 11 2000 lb 2 6 10 1 1 5 9 30 ft 30 ft 10 ft 10 ft 10 ft
Figure P5–22
5.23–5.38 For the rigid frames or beams shown in Figures P5–23—P5–38, determine the displacements and rotations at the nodes, the element forces, and the reactions.
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3000 lb-ft 1200 lb A = 10 in² I = 150 in⁴ (for elements 3 and 4) 2400 lb 3 ③ ④ 300 in⁴ A = 12 in² 15 ft 15 ft 2 ① ② 2 25 ft For cross members: I = 1.0 in⁴ A = 2.0 in² E = 30 × 10⁶ psi (for all members)
Figure P5–23
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300 lb/ft 25 ft 5 ④ 6 ② ⑥ 15 ft 3 ③ 4 ① ⑤ 1 15 ft 2 E = 30 × 10⁶ psi I = 200 in⁴ A = 15 in²
Figure P5–24
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125 lb 2 25 lb 1 3 4 5 6
(a)Design 1
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125 lb 2 25 lb 1 2 3 4 5 6 7
(b）Design 2
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(10, 15) (30, 15) (10, 15) (32, 11.25) (38, 0) (0, 0) (20, 0) (32, 11.25) (38, 0) ① ② ③ ④ ⑤ ⑥ ⑦ ⑧
(a)Design 1
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y 1 2 (10, 15) 3 4 5 6 (30, 15) (32, 11.25) (38, 0) (20, 0) (38, 0) x ① ② ③ ④ ⑤ ⑥ ⑦ ⑧ ⑨
(b）Design 2
\begin{array}{l} \text { Case 1 } \\ E = 3 0 \times 1 0 ^ {6} \mathrm{psi} \\ \text { Case 2 } \\ E = 1 0 \times 1 0 ^ {6} \mathrm{psi} \\ A _ {1} = 0. 1 \text { in } ^ {2} \\ A _ {2} = A _ {3} = A _ {4} = A _ {5} = 0. 1 5 \mathrm{in} ^ {2} \\ A _ {6} = A _ {7} = A _ {8} = 0. 3 \text { in } ^ {2} \\ I _ {1} = 0. 0 1 \mathrm{in} ^ {4} \\ I _ {2} = I _ {3} = I _ {4} = I _ {5} = 0. 0 2 \text { in } ^ {4} \\ I _ {6} = I _ {7} = I _ {8} = 0. 1 \mathrm{in} ^ {4} \\ \end{array}
Figure P5–25 Two bicycle frame models (coordinates shown in inches)
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1000 lb/ft 3 8 ft 2 I₂, A₂ 5 8 ft 6 I₁, A₁ 12 ft I₁, A₁ 1 4 12 ft E = 30 × 10⁶ psi I₁ = 300 in⁴ I₂ = 600 in⁴ A₁ = 15 in² A₂ = 30 in²
Figure P5–26
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60 kN 4 m 4 m 2 ② 3 ③ 4 ① ④ 6 m 1 5 E = 210 GPa I = 1.0 × 10⁻⁴ m⁴ A = 1.0 × 10⁻² m²
Figure P5-27
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10 m 4 4 m 2 4 m 1 300 kN/m 5 E = 210 GPa I = 0.5 × 10⁻⁴ m⁴ A = 0.5 × 10⁻² m²
Figure P5-28
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E = 30 × 10⁶ psi I = 150 in⁴ A = 10 in² 1 ① 2 ② 16 kip 16 kip 4 kip 3 ③ 4 ④ 5 ⑤ 6 30 ft 14 ft 7 ft 7 ft 30 ft
Figure P5–29
\begin{array}{l} E = 3 0 \times 1 0 ^ {6} \mathrm{psi} \\ I = 2 0 0 \mathrm{in} ^ {4} \\ A = 1 2 \mathrm{in} ^ {2} \end{array}
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30 kip 30 kip 15 ft 20 ft 15 ft ② 3 ③ 4 ④ 2 5 10 ft ① ⑤ 30 ft 1 6 50 ft
Figure P5–30
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E = 30 × 10⁶ psi I = 100 in⁴ A = 8 in² ① ② ③ 500 lb/ft 10 ft 15 ft 3 4 1 45°
Figure P5–31
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15 kN 10 kN · m 2 6 m 3 ① ② ③ 6 m 1 4
\begin{array}{l} E = 2 1 0 \mathrm{GPa} \\ I = 2 \times 1 0 ^ {- 4} \mathrm{m} ^ {4} \\ A = 2 \times 1 0 ^ {- 2} \mathrm{m} ^ {2} \\ \end{array}
Figure P5–32
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20 kN 7 60 kN·m I₁, A₁ 8 I₃, A₃ I₁, A₁ I₃, A₃ 20 kN 5 6 I₂, A₂ I₂, A₂ 20 kN 3 I₁, A₁ 4 I₂, A₂ 2 1 2 10 m 3 m 3 m 4 m
\begin{array}{l} E = 2 1 0 \mathrm{GPa} \\ I _ {1} = 2 \times 1 0 ^ {- 4} \mathrm{m} ^ {4} \\ A _ {1} = 2 \times 1 0 ^ {- 2} \mathrm{m} ^ {2} \\ I _ {2} = 1 \times 1 0 ^ {- 4} \mathrm{m} ^ {4} \\ A _ {2} = 1 \times 1 0 ^ {- 2} \mathrm{m} ^ {2} \\ I _ {3} = 0. 5 \times 1 0 ^ {- 4} \mathrm{m} ^ {4} \\ A _ {3} = 0. 5 \times 1 0 ^ {- 2} \mathrm{m} ^ {2} \\ \end{array}
Figure P5-33
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7000 N/m 1 ① 2 ② 3 ③ 4 7 m 7 m 7 m
E = 2 1 0 \mathrm{GPa}
I = 1 \times 1 0 ^ {- 4} \mathrm{m} ^ {4}
A = 1 \times 1 0 ^ {- 2} \mathrm{m} ^ {2}
Figure P5-34
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2800 N · m 4000 N → 9 → 2A₁, 3I₁ 10 A₃, I₃ 4 m 8000 N → 7 → 2A₁, 3I₁ 8 A₃, I₃ 4 m 8000 N → 5 → 2A₁, 3I₁ 6 A₂, I₂ 4 m 12,000 N → 3 → 2A₁, 3I₁ 4 A₁, I₁ 6 m 1 → 2 10 m
E = 2 1 0 \mathrm{GPa}
A _ {1} = 2. 0 \times 1 0 ^ {- 2} \mathrm{m} ^ {2}
I _ {1} = 2. 0 \times 1 0 ^ {- 4} \mathrm{m} ^ {4}
A _ {2} = 1. 5 \times 1 0 ^ {- 2} \mathrm{m} ^ {2}
I _ {2} = 1. 5 \times 1 0 ^ {- 4} \mathrm{m} ^ {4}
A _ {3} = 1. 0 \times 1 0 ^ {- 2} \mathrm{m} ^ {2}
I _ {3} = 1. 0 \times 1 0 ^ {- 4} \mathrm{m} ^ {4}
Figure P5-35
A _ {1} = 2. 0 \times 1 0 ^ {- 2} \mathrm{m} ^ {2}
I _ {1} = 2. 0 \times 1 0 ^ {- 4} \mathrm{m} ^ {4}
A _ {2} = 1. 5 \times 1 0 ^ {- 2} \mathrm{m} ^ {2}
I _ {2} = 1. 5 \times 1 0 ^ {- 4} \mathrm{m} ^ {4}
A _ {3} = 1. 0 \times 1 0 ^ {- 2} \mathrm{m} ^ {2}
I _ {3} = 1. 0 \times 1 0 ^ {- 4} \mathrm{m} ^ {4}
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E = 210 GPa I = 1 × 10⁻⁴ m⁴ A = 1 × 10⁻² m² 10 kN 10 kN 4 6 20 kN 13 0.8m 2.5 m 2 1 m 2 3 5 6 7 8 9 10 11 12 13 14 1 2 3 2 m 2 m
Figure P5–36
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E = 210 GPa I = 4 × 10⁻⁴ m⁴ A = 4 × 10⁻² m² 18 kN 72 kN 72 kN 6 m 4 m 3 m 5 m 6 m 1 3 4 5 6 7 10 m 2 8 10 m
Figure P5–37
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15 kN/m 80 kN 3.5 m E = 210 GPa I = 2.0 × 10⁻⁴ m⁴ A = 1.0 × 10⁻² m² 300 kN/m ① ② ③ ④ ⑤ ⑥ 1 3 7 8 m 7 m 7 m
Figure P5–38
5.39 Consider the plane structure shown in Figure P5–39. First assume the structure to be a plane frame with rigid joints, and analyze using a frame element. Then assume the structure to be pin-jointed and analyze as a plane truss, using a truss element. If the structure is actually a truss, is it appropriate to model it as a rigid frame? How
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10 kN 7 4 m 8 3 m 20 kN 5 6 3 m 20 kN 3 4 20 kN 1 2 3 4 3 m 1 2
Figure P5–39
can you model the truss using the frame (or beam) element? In other words, what idealization could you make in your model to use the beam element to approximate a truss?
5.40 For the two-story, two-bay rigid frame shown, determine (1) the nodal displacement components and (2) the shear force and bending moments in each member. Let E = \mathsf { 2 0 0 ~ G P a } , I = \overset { \cdot } { 2 } \times 1 0 ^ { - 4 } \mathrm { m } ^ { 4 } for each horizontal member and I = 1 . 5 \times 1 0 ^ { - 4 } \mathrm { m } ^ { 4 } for each vertical member.
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12 kN/m G H 12 kN/m I 5 m D E F 5 m A B C 10 m 10 m
Figure P5–40
5.41 For the two-story, three-bay rigid frame shown, determine (1) the nodal displacements and (2) the member end shear forces and bending moments. (3) Draw the shear force and bending moment diagrams for each member. Let E ¼ 200 GPa; I = 1 . 2 9 \times 1 0 ^ { - 4 } \mathrm { m } ^ { 4 } for the beams and I = 0 . 4 6 2 \times 1 0 ^ { - 4 } \mathrm { m } ^ { 4 } for the columns.
The properties for I correspond to a W 6 1 0 \times 1 5 5 and a W 4 1 0 \times 1 1 4 wide-flange section, respectively, in metric units.
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25 kN 50 kN I J K L E F G H A B C D 4 m 6 m 8 m 6 m 8 m
Figure P5–41
5.42 For the rigid frame shown, determine (1) the nodal displacements and rotations and (2) the member shear forces and bending moments. Let E = 2 0 0 GPa, \dot { I } = 0 . 7 9 5 \times 1 0 ^ { - 4 } \mathrm { m } ^ { 4 } for the horizontal members and I = 0 . 3 1 6 \times 1 0 ^ { - 4 } \mathrm { m } ^ { 4 } for the vertical members. These I values correspond to a W 4 6 0 \times 1 5 8 and a W 4 1 0 \times 8 5 wide-flange section, respectively.
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20 kN → G H I 40 kN → D E F A B C 3 m 3 m 5 m 5 m
Figure P5–42
5.43 For the rigid frame shown, determine (1) the nodal displacements and rotations and (2) the shear force and bending moments in each member. Let E = 2 9 \times 1 0 6 psi, I = 3 1 0 0 ~ \mathrm { i n } . ^ { 4 } for the horizontal members and I = 1 1 1 0 ~ \mathrm { i n } . ^ { 4 } for the vertical members. The I values correspond to a { \textsf { W } } 2 4 \times 1 0 4 and a { \bf W } 1 6 \times 7 7 \mathrm { ~ \AA ~ } :
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7.5 kip M N 15 kip I J K L 15 kip E F G H A B C D 30 ft 20 ft 30 ft 15 ft 15 ft 15 ft
Figure P5–43
5.44 A structure is fabricated by welding together three lengths of I-shaped members as shown in Figure P5–44. The yield strength of the members is 36 ksi, E = 2 9 \mathrm { e 6 p s i } , and Poisson’s ratio is 0.3. The members all have cross-section properties corresponding to a W18 by 76. That is, A = 2 2 . 3 \mathrm { i n } ^ { 2 } , depth of section is d = 1 8 . 2 1 in., I _ { x } = 1 3 3 0 \mathrm { i n } ^ { 4 } , S _ { x } = 1 4 6 \mathrm { { i n } } ^ { 3 } , I _ { \nu } = 1 5 2 \mathrm { i n } ^ { 4 } , and S _ { \nu } = 2 7 . 6 \mathrm { i n } ^ { 3 } . Determine whether a load of Q = 1 0 { , } 0 0 0 lb downward is safe against general yielding of the material. The factor of safety against general yielding is to be 2.0. Also, determine the maximum vertical and horizontal deflections of the structure.
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90" 90" Q
Figure P5–44
5.45 For the tapered beam shown in Figure P5–45, determine the maximum deflection using one, two, four, and eight elements. Calculate the moment of inertia at the midlength station for each element. Let E = 3 0 \times 1 0 ^ { 6 } psi, I _ { 0 } = 1 0 0 ~ \mathrm { i n } ^ { 4 } , and L = 1 0 0 in. Run cases where n = 1 , 3 . , and 7. Use a beam element. The analytical solution for n = 7 is given by Reference [7]:
v _ {1} = \frac {P L ^ {3}}{4 9 E I _ {0}} (1 / 7 \ln 8 + 2. 5) = \frac {1}{1 7 . 5 5} \frac {P L ^ {3}}{E I _ {0}}
\theta_ {1} = \frac {P L ^ {2}}{4 9 E I _ {0}} (\ln 8 - 7) = - \frac {1}{9 . 9 5} \frac {P L ^ {2}}{E I _ {0}}
I (x) = I _ {0} \left(1 + n \frac {x}{L}\right)
where n ¼ arbitrary numerical factor and I _ { 0 } = \mathrm { m o m e n t } of inertia of section at x = 0 .
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P = 500 lb 1 A 2 x w h(x) A-A 1 2 y L One-element approximation
Figure P5–45 Tapered cantilever beam
5.46 Derive the stiffness matrix for the nonprismatic torsion bar shown in Figure P5–46. The radius of the shaft is given by
r = r _ {0} + (x / L) r _ {0}, \text { where } r _ {0} \text { is the radius at } x = 0.
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r₀工 x L
Figure P5–46
5.47 Derive the total potential energy for the prismatic circular cross-section torsion bar shown in Figure P5–47. Also determine the equivalent nodal torques 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.
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T (lb-in./in.) 1 2 L
Figure P5–47
5.48 For the grid shown in Figure P5–48, determine the nodal displacements and the local element forces. Let E = 3 0 \times 1 0 ^ { 6 } psi, G ¼ 12 106 psi, I = 2 0 0 \mathrm { i n } ^ { 4 } , and J = 1 0 0 \mathrm { i n } ^ { 4 } for both elements.
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5000 lb 10 ft 10 ft x y z 1 2 3
Figure P5–48
5.49 Resolve Problem 5–48 with an additional nodal moment of 1000 k-in. applied about the x axis at node 2.
5.50–5.51 For the grids shown in Figures P5–50 and P5–51, determine the nodal displacements and the local element forces. Let E ¼ 210 GPa, G ¼ 84 GPa, I = 2 \times 1 0 ^ { - 4 } \mathrm { ~ m } ^ { 4 } , J = 1 \times 1 0 ^ { - 4 } ~ \mathrm { m } ^ { 4 } , and A = 1 \times 1 0 ^ { - 2 } \mathrm { m } ^ { 2 } .