김경종
799 lines
15 KiB
Markdown
799 lines
15 KiB
Markdown
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<details>
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<summary>text_image</summary>
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1
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1 m
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2
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1 m
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3
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5 kN
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E = 210 GPa
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A = 4 × 10⁻⁴ m²
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</details>
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Figure P3–2
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<details>
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<summary>text_image</summary>
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1 ① 2 8000 lb ② 3
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20 in. 50 in. E = 30 × 10⁶ psi
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A = 2.0 in²
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</details>
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Figure P3–3
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<details>
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<summary>text_image</summary>
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1
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①
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2
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4000 lb
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10,000 lb
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3
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②
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③
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4
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30 in.
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30 in.
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30 in.
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E = 30 × 10⁶ psi
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A = 4.0 in²
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</details>
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Figure P3–4
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<details>
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<summary>text_image</summary>
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1 E₁, A 2 E₂, A 3 15,000 lb
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30 in. 30 in. E₁ = 30 × 10⁶ psi
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E₂ = 15 × 10⁶ psi
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A = 5 in²
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</details>
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Figure P3–5
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<details>
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<summary>text_image</summary>
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E1, A
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50 in.
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Rigid bar
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2
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E2, A
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8000 lb
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3
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E1 = 30 × 10^6 psi
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E2 = 10 × 10^6 psi
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A = 2 in^2
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2
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E2, A
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30 in.
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</details>
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Figure P3–6
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<details>
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<summary>text_image</summary>
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1
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①
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30 in.
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2
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k
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3 10,000 lb
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30 in.
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②
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③
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4
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E = 15 × 10^6 psi
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A = 3 in^2
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k = 5000 lb/in.
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</details>
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Figure P3–7
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<details>
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<summary>text_image</summary>
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1 Steel 2 Aluminum 3
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1 m 1 m
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20 kN
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E_st = 200 GPa
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A_st = 4 × 10^-4 m^2
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E_al = 70 GPa
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A_al = 2 × 10^-4 m^2
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</details>
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Figure P3–8
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<!-- source-page: 152 -->
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<details>
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<summary>text_image</summary>
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1 ① 10 kN 2 ② 3 F₃
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2 m 2 m δ δ
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E = 210 GPa
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A = 4 × 10⁻⁴ m²
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δ = 25 mm
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</details>
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Figure P3–9
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<details>
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<summary>text_image</summary>
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1
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①
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2 m
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2
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8 kN
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2 m
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3
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②
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k
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③
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4
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E = 70 GPa
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A = 2 × 10⁻⁴ m²
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k = 2000 kN/m
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</details>
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Figure P3–10
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<details>
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<summary>text_image</summary>
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1
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①
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3 m
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2
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②
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3
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2
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③
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30 kN
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4
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2
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④
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3 m
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5
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E = 210 GPa
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A = 3 × 10⁻⁴ m²
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</details>
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Figure P3–11
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3.12 Solve for the axial displacement and stress in the tapered bar shown in Figure P3–12 using one and then two constant-area elements. Evaluate the area at the center of each element length. Use that area for each element. Let $A _ { 0 } = 2 \mathrm { i n } ^ { 2 } , L = 2 0$ in., $E = 1 0 \times 1 0 ^ { 6 }$ psi, and $P = 1 0 0 0$ lb. Compare your finite element solutions with the exact solution.
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<details>
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<summary>text_image</summary>
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A(x) = A₀(1 + x/L)
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P ← x
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L
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</details>
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Figure P3–12
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3.13 Determine the stiffness matrix for the bar element with end nodes and midlength node shown in Figure P3–13. Let axial displacement $u = a _ { 1 } + a _ { 2 } x + a _ { 3 } x ^ { 2 }$ . (This is a higherorder element in that strain now varies linearly through the element.)
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<details>
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<summary>text_image</summary>
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x,u
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1 3 2
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L
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</details>
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Figure P3–13
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<!-- source-page: 153 -->
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3.14 Consider the following displacement function for the two-noded bar element:
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$$
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u = a + b x ^ {2}
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$$
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Is this a valid displacement function? Discuss why or why not.
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3.15 For each of the bar elements shown in Figure P3–15, evaluate the global x-y stiffness matrix.
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<details>
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<summary>text_image</summary>
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y
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2
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E = 30 × 10⁶ psi
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A = 3 in²
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L = 20 in.
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45°
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1
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x
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(a)
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</details>
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<details>
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<summary>text_image</summary>
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2
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y
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E = 15 × 10⁶ psi
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A = 1 in²
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L = 15 in.
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120°
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x
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(b)
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</details>
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<details>
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<summary>text_image</summary>
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E = 210 GPa
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A = 4 × 10⁻⁴ m²
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L = 3 m
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1
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30°
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2
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(c)
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</details>
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<details>
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<summary>text_image</summary>
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E = 70 GPa
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A = 2 × 10⁻⁴ m²
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L = 1 m
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1
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2
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20°
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x
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y
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(d)
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</details>
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Figure P3–15
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3.16 For the bar elements shown in Figure P3–16, the global displacements have been determined to be $d _ { 1 x } = 0 . 5$ in., $d _ { 1 y } = 0 . 0 , d _ { 2 x } = 0 . 2 5$ in., and $d _ { 2 y } = 0 . 7 5$ in. Determine the local x^ displacements at each end of the bars. Let $E = 1 2 \times 1 0 ^ { 6 } \ \mathrm { p s i } , A = 0 . 5 \ \mathrm { i n } ^ { 2 }$ , and $L = 6 0$ in. for each element.
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<!-- source-page: 154 -->
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<details>
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<summary>text_image</summary>
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y
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2
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x̂
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30°
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1
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x
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(a)
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</details>
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<details>
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<summary>text_image</summary>
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y
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1
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30°
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x
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(b)
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2
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x̂
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</details>
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Figure P3–16
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3.17 For the bar elements shown in Figure P3–17, the global displacements have been determined to be $d _ { 1 x } = 0 . 0 , d _ { 1 y } = 2 . 5$ mm, $d _ { 2 x } = 5 . 0$ mm, and $d _ { 2 y } = 3 . 0$ mm. Determine the local x^ displacements at the ends of each bar. Let E ¼ 210 GPa, $A = 1 0 \times 1 0 ^ { - 4 }$ $\mathbf { m } ^ { 2 } .$ , and $L = 3$ m for each element.
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<details>
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<summary>text_image</summary>
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x̂
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2
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y
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120°
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x
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(a)
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</details>
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<details>
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<summary>text_image</summary>
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y
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1
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30°
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x
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(b)
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</details>
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Figure P3–17
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3.18 Using the method of Section 3.5, determine the axial stress in each of the bar elements shown in Figure P3–18.
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<!-- source-page: 155 -->
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<details>
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<summary>text_image</summary>
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y
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2
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45°
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1
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x
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</details>
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(a)
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$$
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E = 3 0 \times 1 0 ^ {6} \mathrm{psi}
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$$
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$$
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A = 2 \mathrm{in} ^ {2}
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$$
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$$
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L = 6 0 \text { in. }
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$$
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$$
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d _ {1 x} = 0 \quad d _ {1 y} = 0
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$$
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$$
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d _ {2 x} = 0. 0 1 \text { in. } d _ {2 y} = 0. 0 2 \text { in. }
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$$
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<details>
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<summary>text_image</summary>
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y
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2
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1
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30°
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x
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</details>
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(b)
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$$
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E = 2 1 0 \mathrm{GPa}
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$$
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$$
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A = 3 \times 1 0 ^ {- 4} \mathrm{m} ^ {2}
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$$
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$$
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L = 3 \mathrm{m}
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$$
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$$
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d _ {1 x} = 0. 2 5 \mathrm{mm} \quad d _ {1 y} = 0. 0
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$$
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$$
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d _ {2 x} = 1. 0 0 \mathrm{mm} \quad d _ {2 y} = 0. 0
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$$
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Figure P3–18
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3.19 a. Assemble the stiffness matrix for the assemblage shown in Figure P3–19 by superimposing the stiffness matrices of the springs. Here k is the stiffness of each spring.
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b. Find the x and y components of deflection of node 1.
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<details>
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<summary>text_image</summary>
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y
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2
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k
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3
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k
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45°
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45°
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1
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x
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k
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4
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10 lb
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</details>
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Figure P3–19
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<!-- source-page: 156 -->
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3.20 For the plane truss structure shown in Figure P3–20, determine the displacement of node 2 using the stiffness method. Also determine the stress in element 1. Let $A = 5$ $\mathrm { i n } ^ { 2 } , E = 1 \times 1 0 ^ { 6 }$ psi, and $L = 1 0 0$ in.
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<details>
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<summary>text_image</summary>
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L
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L
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10 kip
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2
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①
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②
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1
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45°
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45°
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3
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</details>
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Figure P3–20
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<details>
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<summary>text_image</summary>
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L
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4
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3
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2
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30°
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30°
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141 lb
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45°
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1
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</details>
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Figure P3–21
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3.21 Find the horizontal and vertical displacements of node 1 for the truss shown in Figure P3–21. Assume AE is the same for each element.
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3.22 For the truss shown in Figure P3–22 solve for the horizontal and vertical components of displacement at node 1 and determine the stress in each element. Also verify force equilibrium at node 1. All elements have $A _ { 1 } = 1 ~ \mathrm { i n } . ^ { 2 }$ and $E = 1 0 \times 1 0 ^ { 6 }$ psi. Let $L =$ 100 in.
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<details>
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<summary>text_image</summary>
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2
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①
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1000 lb
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60°
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30°
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1
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1000 lb
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②
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3
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③
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4
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L
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</details>
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Figure P3–22
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<!-- source-page: 157 -->
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3.23 For the truss shown in Figure P3–23, solve for the horizontal and vertical components of displacement at node 1. Also determine the stress in element 1. Let $A = 1 \ \mathrm { i n } ^ { 2 }$ , $E = \bar { 1 0 } . 0 \times 1 0 ^ { 6 }$ psi, and L ¼ 100 in.
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<details>
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<summary>text_image</summary>
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12,000 lb
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①
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②
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60°
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60°
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②
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③
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L
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</details>
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Figure P3–23
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<details>
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<summary>text_image</summary>
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P = 1000 lb
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15 ft
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20 ft
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P = 1000 lb
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</details>
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Figure P3–24
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3.24 Determine the nodal displacements and the element forces for the truss shown in Figure P3–24. Assume all elements have the same AE.
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3.25 Now remove the element connecting nodes 2 and 4 in Figure P3–24. Then determine the nodal displacements and element forces.
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3.26 Now remove both cross elements in Figure P3–24. Can you determine the nodal displacements? If not, why?
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3.27 Determine the displacement components at node 3 and the element forces for the plane truss shown in Figure P3–27. Let $A = 3 ~ \mathrm { i n } ^ { 2 }$ and $E = 3 0 \times 1 0 ^ { 6 }$ psi for all elements. Verify force equilibrium at node 3.
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<details>
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<summary>text_image</summary>
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4
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③
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20 ft
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5 kip
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3
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10 kip
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①
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②
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40 ft
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1
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2
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30 ft
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30 ft
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</details>
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Figure P3–27
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<!-- source-page: 158 -->
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3.28 Show that for the transformation matrix $\underline { T }$ of Eq. (3.4.15), $\underline { T } ^ { T } = \underline { T } ^ { - 1 }$ and hence Eq. (3.4.21) is indeed correct, thus also illustrating that ${ \underline { { k } } } = { \underline { { T } } } ^ { T } { \underline { { \hat { k } } } } { \underline { { T } } }$ is the expression for the global stiffness matrix for an element.
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3.29–3.30 For the plane trusses shown in Figures P3–29 and P3–30, determine the horizontal and vertical displacements of node 1 and the stresses in each element. All elements have E ¼ 210 GPa and $A = 4 . 0 \times 1 0 ^ { - 4 } ~ \mathrm { m } ^ { 2 }$ .
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<details>
|
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<summary>text_image</summary>
|
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2
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3 m
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①
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②
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3 m
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45°
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③
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10 kN
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1
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3 m
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4
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20 kN
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</details>
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Figure P3–29
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<details>
|
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<summary>text_image</summary>
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2
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1
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2 m
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3
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3 m
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1
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2
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60°
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5 m
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3
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4
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40 kN
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</details>
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Figure P3–30
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3.31 Remove element 1 from Figure P3–30 and solve the problem. Compare the displacements and stresses to the results for Problem 3.30.
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3.32 For the plane truss shown in Figure P3–32, determine the nodal displacements, the element forces and stresses, and the support reactions. All elements have E ¼ 70 GPa and $A = 3 . 0 \times 1 0 ^ { - 4 } ~ \mathrm { m } ^ { 2 }$ . Verify force equilibrium at nodes 2 and 4. Use symmetry in your model.
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<details>
|
||
<summary>text_image</summary>
|
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50 kN
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100 kN
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50 kN
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2
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4
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6
|
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3 m
|
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①
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③
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⑤
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⑥
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②
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3 m
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3 m
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1
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5
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1
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</details>
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Figure P3–32
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3.33 For the plane trusses supported by the spring at node 1 in Figure P3–33 (a) and (b), determine the nodal displacements and the stresses in each element. Let $E = 2 1 0 \mathrm { G P a }$ and $A = 5 . 0 \times 1 0 ^ { - 4 } \mathrm { m } ^ { 2 }$ for both truss elements.
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<!-- source-page: 159 -->
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||
|
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<details>
|
||
<summary>text_image</summary>
|
||
|
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50 kN
|
||
5 m
|
||
1
|
||
45°
|
||
10 m
|
||
①
|
||
②
|
||
③
|
||
k = 2000 kN/m
|
||
4
|
||
</details>
|
||
|
||
Figure P3–33(a)
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
100 kN
|
||
5 m
|
||
5 m
|
||
①
|
||
②
|
||
60°
|
||
60°
|
||
③
|
||
k = 4000 N/m
|
||
4
|
||
</details>
|
||
|
||
Figure P3–33(b)
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
4
|
||
③
|
||
3
|
||
④
|
||
⑤
|
||
②
|
||
4 ft
|
||
①
|
||
2
|
||
1
|
||
8 ft
|
||
δ
|
||
</details>
|
||
|
||
Figure P3–34
|
||
|
||
3.34 For the plane truss shown in Figure P3–34, node 2 settles an amount $\delta = 0 . 0 5$ in. Determine the forces and stresses in each element due to this settlement. Let $E =$ $3 0 \times 1 0 ^ { 6 }$ psi and $A = 2 \mathrm { i n } ^ { 2 }$ for each element.
|
||
3.35 For the symmetric plane truss shown in Figure P3–35, determine (a) the deflection of node 1 and (b) the stress in element 1. $A E / L$ for element 3 is twice $A E / L$ for the other
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
y
|
||
4
|
||
3
|
||
2
|
||
3
|
||
5
|
||
2
|
||
L
|
||
30°
|
||
L
|
||
30°
|
||
L
|
||
30°
|
||
L
|
||
30°
|
||
L
|
||
6
|
||
1
|
||
L
|
||
30°
|
||
x
|
||
P = 2000 lb
|
||
</details>
|
||
|
||
Figure P3–35
|
||
|
||
<!-- source-page: 160 -->
|
||
|
||
elements. Let $A E / L = 1 0 ^ { 6 } 1 \mathrm { b / i n }$ . Then let $A = 1 \mathrm { i n } ^ { 2 } , L = 1 0$ in., and $E = 1 0 \times 1 0 ^ { 6 }$ psi to obtain numerical results.
|
||
|
||
3.36–3.37 For the space truss elements shown in Figures P3–36 and P3–37, the global displacements at node 1 have been determined to be $d _ { 1 x } = 0 . 1$ in., $d _ { 1 y } = 0 . 2 \ \mathrm { i n } .$ ., and $d _ { 1 z } =$ 0:15 in. Determine the displacement along the local x^ axis at node 1 of the elements. The coordinates, in inches, are shown in the figures.
|
||
|
||
|
||
Figure P3–36
|
||
Figure P3–37
|
||
|
||
3.38–3.39 For the space truss elements shown in Figures P3–38 and P3–39, the global displacements at node 2 have been determined to be $d _ { 2 x } = 5$ mm, $d _ { 2 y } = 1 0$ mm, and
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
y
|
||
x
|
||
z
|
||
1
|
||
2
|
||
(1, 1.5, 1)
|
||
(0, 0, 0)
|
||
x̂
|
||
</details>
|
||
|
||
Figure P3–38
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
y
|
||
(5, 4, -1)
|
||
2
|
||
x
|
||
1
|
||
(2, 0, 2)
|
||
z
|
||
</details>
|
||
|
||
Figure P3–39
|