김경종
30 KiB
Answers to Selected Problems
Chapter 2
2.1
\mathbf {a}. \underline {{{K}}} = \left[ \begin{array}{c c c c} k _ {1} & 0 & - k _ {1} & 0 \\ 0 & k _ {3} & 0 & - k _ {3} \\ - k _ {1} & 0 & k _ {1} + k _ {2} & - k _ {2} \\ 0 & - k _ {3} & - k _ {2} & k _ {2} + k _ {3} \end{array} \right]
\mathbf {b}. d _ {3 x} = \frac {k _ {2} P}{k _ {1} k _ {2} + k _ {1} k _ {3} + k _ {2} k _ {3}}, d _ {4 x} = \frac {(k _ {1} + k _ {2}) P}{k _ {1} k _ {2} + k _ {1} k _ {3} + k _ {2} k _ {3}}
\mathbf {c}. F _ {1 x} = \frac {- k _ {1} k _ {2} P}{k _ {1} k _ {2} + k _ {1} k _ {3} + k _ {2} k _ {3}}, \quad F _ {2 x} = \frac {- k _ {3} (k _ {1} + k _ {2}) P}{k _ {1} k _ {2} + k _ {1} k _ {3} + k _ {2} k _ {3}}
2. 2 \quad d _ {2 x} = 0. 5 \text { in. }, \quad F _ {3 x} = 2 5 0 \text { lb }, \quad \hat {f} _ {1 x} ^ {(1)} = - \hat {f} _ {2 x} ^ {(1)} = - 2 5 0 \text { lb }, \quad \hat {f} _ {2 x} ^ {(2)} = - \hat {f} _ {3 x} ^ {(2)} = - 2 5 0 \text { lb }
\mathbf {2}. 3 \quad \mathbf {a}. \underline {{\boldsymbol {K}}} = \left[ \begin{array}{c c c c c} k & - k & 0 & 0 & 0 \\ - k & 2 k & - k & 0 & 0 \\ 0 & - k & 2 k & - k & 0 \\ 0 & 0 & - k & 2 k & - k \\ 0 & 0 & 0 & - k & k \end{array} \right]
\mathbf {b}. d _ {2 x} = \frac {P}{2 k}, \quad d _ {3 x} = \frac {P}{k}, \quad d _ {4 x} = \frac {P}{2 k} \quad \mathbf {c}. F _ {1 x} = - \frac {P}{2}, \quad F _ {5 x} = - \frac {P}{2}
\mathbf {2 . 4} \quad \mathbf {a .} \underline {{\boldsymbol {K}}} \text { same as 2.3a. } \quad \mathbf {b .} d _ {2 x} = \frac {\delta}{4}, \quad d _ {3 x} = \frac {\delta}{2}, \quad d _ {4 x} = \frac {3 \delta}{4} \quad \mathbf {c .} F _ {1 x} = \frac {- k \delta}{4}, \quad F _ {5 x} = \frac {k \delta}{4}
\mathbf {2 . 5} \quad \underline {{\boldsymbol {K}}} = \left[ \begin{array}{r r r r} 1 & - 1 & 0 & 0 \\ - 1 & 1 0 & 0 & - 9 \\ 0 & 0 & 5 & - 5 \\ 0 & - 9 & - 5 & 1 4 \end{array} \right]
2.6 d_{2x}=0.4746 in
2.7 d_{2x}=1 in., d_{3x}=2 in.
\hat {f} _ {1 x} ^ {(1)} = - \hat {f} _ {2 x} ^ {(1)} = - 5 0 0 \mathrm{lb}, \quad \hat {f} _ {2 x} ^ {(2)} = - \hat {f} _ {3 x} ^ {(2)} = - 5 0 0 \mathrm{lb}, \quad F _ {1 x} = - 5 0 0 \mathrm{lb}
2.8 d_{1x}=0,\quad d_{2x}=3\ in.,\quad d_{3x}=7\ in.,\quad d_{4x}=11\ in.
\hat {f} _ {1 x} ^ {(1)} = - \hat {f} _ {2 x} ^ {(1)} = - 3 0 0 0 \mathrm{lb}, \quad \hat {f} _ {2 x} ^ {(2)} = - \hat {f} _ {3 x} ^ {(2)} = - 4 0 0 0 \mathrm{lb}
\hat {f} _ {3 x} ^ {(3)} = - \hat {f} _ {4 x} ^ {(3)} = - 4 0 0 0 \mathrm{lb}, \quad F _ {1 x} = - 3 0 0 0 \mathrm{lb}
2.9 d_{2x} = -2 in.
\hat {f} _ {1 x} ^ {(1)} = - \hat {f} _ {2 x} ^ {(1)} = 2 0 0 0 \mathrm{lb}, \quad \hat {f} _ {2 x} ^ {(2)} = - \hat {f} _ {3 x} ^ {(2)} = - 1 0 0 0 \mathrm{lb}
\hat {f} _ {2 x} ^ {(3)} = - \hat {f} _ {4 x} ^ {(3)} = - 1 0 0 0 \mathrm{lb}, \quad F _ {1 x} = 2 0 0 0 \mathrm{lb}, \quad F _ {3 x} = F _ {4 x} = 1 0 0 0 \mathrm{lb}
2.10 d_{2x}=0.01\ m,\quad\hat{f}_{1x}^{(1)}=-\hat{f}_{2x}^{(1)}=-20\ N
\hat {f} _ {2 x} ^ {(2)} = - \hat {f} _ {3 x} ^ {(2)} = - 2 0 \mathrm{N}, \quad F _ {1 x} = - 2 0 \mathrm{N}
2.11 d_{2x}=0.027\ m,\quad d_{3x}=0.018\ m
\hat {f} _ {1 x} ^ {(1)} = - \hat {f} _ {2 x} ^ {(1)} = - 2 7 0 \mathrm{N}, \quad \hat {f} _ {2 x} ^ {(2)} = - \hat {f} _ {3 x} ^ {(2)} = 1 8 0 \mathrm{N}
\hat {f} _ {3 x} ^ {(3)} = - \hat {f} _ {4 x} ^ {(3)} = 1 8 0 \mathrm{N}, \quad F _ {1 x} = - 2 7 0 \mathrm{N}, \quad F _ {4 x} = - 1 8 0 \mathrm{N}
2.12 d_{2x}=0.125\ m,\quad d_{3x}=0.25\ m,\quad d_{4x}=0.125\ m
\hat {f} _ {1 x} ^ {(1)} = - \hat {f} _ {2 x} ^ {(1)} = - 2. 5 \mathrm{kN}, \quad \hat {f} _ {2 x} ^ {(2)} = - \hat {f} _ {3 x} ^ {(2)} = - 2. 5 \mathrm{kN}
\hat {f} _ {3 x} ^ {(3)} = - \hat {f} _ {4 x} ^ {(3)} = 2. 5 \mathrm{kN}, \quad \hat {f} _ {4 x} ^ {(4)} = - \hat {f} _ {5 x} ^ {(4)} = 2. 5 \mathrm{kN}
F _ {1 x} = - 2. 5 \mathrm{kN}, \quad F _ {5 x} = - 2. 5 \mathrm{kN}
2.13 d_{2x} = -0.25 \, m,\quad d_{3x} = -0.75 \, m
\hat {f} _ {1 x} ^ {(1)} = - \hat {f} _ {2 x} ^ {(1)} = 1 0 0 \mathrm{N}, \quad \hat {f} _ {2 x} ^ {(2)} = - \hat {f} _ {3 x} ^ {(2)} = 2 0 0 \mathrm{N}
F _ {1 x} = 1 0 0 \mathrm{N}
2.14 d_{3x} = 0.001\mathrm{m},\hat{f}_{1x}^{(1)} = -\hat{f}_{3x}^{(1)} = -0.5\mathrm{kN}
\hat {f} _ {2 x} ^ {(2)} = - \hat {f} _ {3 x} ^ {(2)} = - 0. 5 \mathrm{kN}, \quad \hat {f} _ {3 x} ^ {(3)} = - \hat {f} _ {4 x} ^ {(3)} = 1 \mathrm{kN}
F _ {1 x} = - 0. 5 \mathrm{kN}, \quad F _ {2 x} = - 0. 5 \mathrm{kN}, \quad F _ {4 x} = - 1 \mathrm{kN}
2.15 d_{2x}=1/3 in., d_{3x}=-1/3 in.
2.16 a. x = 0.5 in. \downarrow , \pi_{p_{\min}} = -125 lb-in.
b. x = 2.0 \, in. \leftarrow, \quad \pi_{p_{\min}} = -1000 \, lb-in.
c. x = 1.962 \, mm \downarrow , \pi_{p_{\min}} = -3849 \, N \cdot mm
d. x = 2.4525 \, mm \rightarrow, \quad \pi_{p_{\min}} = -1203 \, N \cdot mm
2.17 x = 2.0 in. \uparrow
2.18 x = 0.707 in. \leftarrow , \pi_{p_{\min}} = -235.7 in-lb
2.19 Same as 2.10
2.20 Same as 2.15
Chapter 3
3.1
\mathbf {a}. \underline {{{\boldsymbol {K}}}} = \left[ \begin{array}{c c c c} \frac {A _ {1} E _ {1}}{L _ {1}} & \frac {- A _ {1} E _ {1}}{L _ {1}} & 0 & 0 \\ \frac {- A _ {1} E _ {1}}{L _ {1}} & \frac {A _ {1} E _ {1}}{L _ {1}} + \frac {A _ {2} E _ {2}}{L _ {2}} & \frac {- A _ {2} E _ {2}}{L _ {2}} & 0 \\ 0 & \frac {- A _ {2} E _ {2}}{L _ {2}} & \frac {A _ {2} E _ {2}}{L ^ {2}} + \frac {A _ {3} E _ {3}}{L _ {3}} & \frac {- A _ {3} E _ {3}}{L _ {3}} \\ 0 & 0 & \frac {- A _ {3} E _ {3}}{L _ {3}} & \frac {A _ {3} E _ {3}}{L _ {3}} \end{array} \right]
b. d_{2x} = \frac{PL}{3AE}, d_{3x} = \frac{2PL}{3AE}
c. i. d_{2x}=3.33\times10^{-4} in., d_{3x}=6.67\times10^{-4} in.
ii. F_{1x} = -333 lb, F_{4x} = -667 lb
iii. \sigma^{(1)} = 333 psi (T), \sigma^{(2)} = 333 psi (T), \sigma^{(3)} = -667 psi (C)
3.2 d_{2x} = -0.595 \times 10^{-4} \, m,\quad d_{3x} = -1.19 \times 10^{-4} \, m,\quad F_{1x} = 5 \, kN
\hat {f} _ {1 x} ^ {(1)} = - \hat {f} _ {2 x} ^ {(1)} = 5 \mathrm{kN}, \quad \hat {f} _ {2 x} ^ {(2)} = - \hat {f} _ {3 x} ^ {(2)} = 5 \mathrm{kN}
3.3 d_{2x}=1.91\times10^{-3} in., F_{1x}=-5715 lb, F_{3x}=-2286 lb
\hat {f} _ {1 x} ^ {(1)} = - \hat {f} _ {2 x} ^ {(1)} = - 5 7 1 5 \mathrm{lb}, \quad \hat {f} _ {2 x} ^ {(2)} = - \hat {f} _ {3 x} ^ {(2)} = 2 2 8 6 \mathrm{lb}
3.4 d_{2x} = -1.66 \times 10^{-4} in., d_{3x} = -1.33 \times 10^{-3} in.
F _ {1 x} = 6 6 7 \mathrm{lb}, \quad F _ {4 x} = 5 3 3 3 \mathrm{lb}
\hat {f} _ {1 x} ^ {(1)} = - \hat {f} _ {2 x} ^ {(1)} = 6 6 7 \mathrm{lb}, \quad \hat {f} _ {2 x} ^ {(2)} = - \hat {f} _ {3 x} ^ {(2)} = 4 6 6 7 \mathrm{lb}
\hat {f} _ {3 x} ^ {(3)} = - \hat {f} _ {4 x} ^ {(3)} = - 5 3 3 3 \mathrm{lb}
3.5 d_{2x} = 0.003 in., d_{3x} = 0.009 in., F_{1x} = -15000 lb
\hat {f} _ {1 x} ^ {(1)} = - \hat {f} _ {2 x} ^ {(1)} = \hat {f} _ {2 x} ^ {(2)} = - \hat {f} _ {3 x} ^ {(2)} = - 1 5 0 0 0 \mathrm{lb}
3.6 d_{2x} = 3.16 \times 10^{-3} in., F_{1x} = -3790 lb, F_{3x} = F_{4x} = -2105 lb
\hat {f} _ {1 x} ^ {(1)} = - \hat {f} _ {2 x} ^ {(1)} = - 3 7 9 0 \mathrm{lb}, \quad \hat {f} _ {2 x} ^ {(2)} = - \hat {f} _ {3 x} ^ {(2)} = \hat {f} _ {2 x} ^ {(3)} = - \hat {f} _ {4 x} ^ {(3)} = 2 1 0 5 \mathrm{lb}
3.7 d_{2x} = 2.21 \times 10^{-5} in., d_{3x} = 6.65 \times 10^{-3} in.
F _ {1 x} = - 3 3. 1 5 \mathrm{lb}, \quad F _ {4 x} = - 9 9 7 5 \mathrm{lb}
\hat {f} _ {1 x} ^ {(1)} = - \hat {f} _ {2 x} ^ {(1)} = \hat {f} _ {2 x} ^ {(2)} = - \hat {f} _ {3 x} ^ {(2)} = - 3 3. 1 5 \mathrm{lb}, \quad \hat {f} _ {3 x} ^ {(3)} = - \hat {f} _ {4 x} ^ {(3)} = 9 9 7 5 \mathrm{lb}
3.8 d_{2x} = -0.250 \, mm,\quad d_{3x} = -1.678 \, mm,\quad F_{1x} = 20 \, kN
3.9 d_{2x}=0.01238\ m,\quad F_{1x}=-520\ kN,\quad F_{3x}=530\ kN
\hat {f} _ {1 x} ^ {(1)} = - \hat {f} _ {2 x} ^ {(1)} = - 5 2 0 \mathrm{kN}, \quad \hat {f} _ {2 x} ^ {(2)} = - \hat {f} _ {3 x} ^ {(2)} = - 5 3 0 \mathrm{kN}
3.10 d_{2x}=0.935\times10^{-3} m, d_{3x}=0.727\times10^{-3} m
F _ {1 x} = - 6. 5 4 6 \mathrm{kN}, \quad F _ {4 x} = - 1. 4 5 5 \mathrm{kN}
\hat {f} _ {1 x} ^ {(1)} = - \hat {f} _ {2 x} ^ {(1)} = - 6. 5 4 6 \mathrm{kN}, \quad \hat {f} _ {2 x} ^ {(2)} = - \hat {f} _ {3 x} ^ {(2)} = 1. 4 5 5 \mathrm{kN},
\hat {f} _ {3 x} ^ {(3)} = - \hat {f} _ {4 x} ^ {(3)} = 1. 4 5 5 \mathrm{kN}
3.11 d_{2x}=3.572\times10^{-4} m, F_{1x}=-7.50 kN, F_{3x}=F_{4x}=F_{5x}=-7.50 kN
\hat {f} _ {1 x} ^ {(1)} = - \hat {f} _ {2 x} ^ {(1)} = - 7. 5 0 \mathrm{kN},
\hat {f} _ {2 x} ^ {(2)} = - \hat {f} _ {3 x} ^ {(2)} = \hat {f} _ {2 x} ^ {(3)} = - \hat {f} _ {4 x} ^ {(3)} = \hat {f} _ {2 x} ^ {(4)} = - \hat {f} _ {5 x} ^ {(4)} = 7. 5 0 \mathrm{kN}
3.12 two-element solution, d_{1x} = -0.686 \times 10^{-3} in.
one-element solution, d_{1x} = -0.667 \times 10^{-3} in.
3.13 B=\left[-\frac{1}{L}+\frac{4x}{L^{2}}\quad\frac{-8x}{L^{2}}\quad\frac{1}{L}+\frac{4x}{L^{2}}\right],\quad\underline{k}=A\int_{-L/2}^{L/2}\underline{B}^{T}E\underline{B}dx
3.15 a. \underline{k} = 2.25 \times 10^{6}\left[ \begin{array}{rrrr} 1 & 1 & -1 & -1 \\ 1 & 1 & -1 & -1 \\ -1 & -1 & 1 & 1 \\ -1 & -1 & 1 & 1 \end{array} \right] lb/in.
b. \underline{k}=\frac{10^{6}}{4}\begin{bmatrix}1&-\sqrt{3}&-1&\sqrt{3}\\ -\sqrt{3}&3&\sqrt{3}&-3\\ -1&\sqrt{3}&1&-\sqrt{3}\\ \sqrt{3}&-3&-\sqrt{3}&3\end{bmatrix} lb/in.
c. \underline{k} = 7000\left[ \begin{array}{cccc}3 & -\sqrt{3} & -3 & \sqrt{3}\\ -\sqrt{3} & 1 & \sqrt{3} & -1\\ -3 & \sqrt{3} & 3 & -\sqrt{3}\\ \sqrt{3} & -1 & -\sqrt{3} & 1 \end{array} \right]\mathrm{kN / m}
\mathbf{d}.\underline{\boldsymbol{k}} = 1.4\times 10^{4}\left[ \begin{array}{cccc}0.883 & 0.321 & -0.883 & -0.321\\ 0.321 & 0.117 & -0.321 & -0.117\\ -0.883 & -0.321 & 0.883 & 0.321\\ -0.321 & -0.117 & 0.321 & 0.117 \end{array} \right]\mathrm{kN / m}
3.16 a. \hat{d}_{1x} = 0.433 in., \hat{d}_{2x} = 0.592 in.
b. \hat{d}_{1x}=0.433 in., \hat{d}_{2x}=-0.1585 in.
3.17 a. \hat{d}_{1x} = 2.165\mathrm{mm},\quad \hat{d}_{1y} = -1.25\mathrm{mm},
\hat {d} _ {2 x} = 0. 0 9 8 \mathrm{mm}, \quad \hat {d} _ {2 y} = - 5. 8 3 \mathrm{mm}
b. \hat{d}_{1x} = -1.25 \, mm,\quad \hat{d}_{1y} = 2.165 \, mm,
\hat {d} _ {2 x} = 3. 0 3 \mathrm{mm}, \quad \hat {d} _ {2 y} = 5. 0 9 8 \mathrm{mm}
3.18 a. \sigma = 10,600 psi, b. 45.47 MPa
3.19
\mathbf {a}. \underline {{\boldsymbol {K}}} = k \left[ \begin{array}{c c c c c c c c} 2 & 0 & - \frac {1}{2} & \frac {1}{2} & - 1 & 0 & - \frac {1}{2} & - \frac {1}{2} \\ 0 & 1 & \frac {1}{2} & - \frac {1}{2} & 0 & 0 & - \frac {1}{2} & - \frac {1}{2} \\ - \frac {1}{2} & \frac {1}{2} & \frac {1}{2} & - \frac {1}{2} & 0 & 0 & 0 & 0 \\ \frac {1}{2} & - \frac {1}{2} & - \frac {1}{2} & \frac {1}{2} & 0 & 0 & 0 & 0 \\ - 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ - \frac {1}{2} & - \frac {1}{2} & 0 & 0 & 0 & 0 & \frac {1}{2} & \frac {1}{2} \\ - \frac {1}{2} & - \frac {1}{2} & 0 & 0 & 0 & 0 & \frac {1}{2} & \frac {1}{2} \end{array} \right]
\mathbf {b}. d _ {1 x} = 0, \quad d _ {1 y} = \frac {- 1 0}{k}
3.20 d _ { 2 x } = 0 , d _ { 2 y } = 0 . 1 4 2 \ \mathrm { i n . , } \sigma ^ { \left( 1 \right) } = \sigma ^ { \left( 2 \right) } = 7 0 7 \ \mathrm { p s i \left( T \right) }
3.21 d _ { 1 x } = \frac { 2 3 1 L } { A E } , d _ { 1 y } = \frac { 4 3 . 5 L } { A E } AE d1y ¼
3.22 d1x ¼ d _ { 1 x } = { \frac { 4 2 2 L } { A E } } , d _ { 1 y } = { \frac { 1 5 7 0 L } { A E } } AE
\sigma^ {(1)} = \frac {5 7 4}{A} (\mathrm{C}), \quad \sigma^ {(2)} = \frac {4 2 2}{A} (\mathrm{T}), \quad \sigma^ {(3)} = \frac {9 9 6}{A} (\mathrm{T})
3.23 d _ { 1 x } = 0 . 2 4 \ \mathrm { i n . } , d _ { 1 \nu } = 0 , \sigma ^ { ( 1 ) } = 1 2 0 0 0 \ \mathrm { p s i }
26;675 105;021 -26;675 105;021 3.24 d2x ¼ AE ， d2y 二 AE ， d3x ¼ AE ， d3y ¼ AE
\hat {f} _ {1 x} ^ {(1)} = - \hat {f} _ {2 x} ^ {(1)} = - 1 3 3 3 \mathrm{lb}, \quad \hat {f} _ {1 x} ^ {(2)} = - \hat {f} _ {3 x} ^ {(2)} = - 1 6 6 7 \mathrm{lb}
\hat {f} _ {2 x} ^ {(3)} = - \hat {f} _ {4 x} ^ {(3)} = 1 6 6 7 \mathrm{lb}, \quad \hat {f} _ {2 x} ^ {(4)} = - \hat {f} _ {3 x} ^ {(4)} = 0
\hat {f} _ {3 x} ^ {(5)} = - \hat {f} _ {4 x} ^ {(5)} = 1 3 3 3 \mathrm{lb}, \quad \hat {f} _ {1 x} ^ {(6)} = - \hat {f} _ {4 x} ^ {(6)} = 0
3 . 2 5 d _ { 2 x } = 0 , d _ { 2 y } = \frac { 2 2 5 , 0 0 0 } { A E } , d _ { 3 x } = \frac { - 5 3 , 3 4 0 } { A E } , d _ { 3 y } = \frac { 2 1 0 , 0 0 0 } { A E } d3y ¼
\hat {f} _ {1 x} ^ {(1)} = - \hat {f} _ {2 x} ^ {(1)} = 0, \quad \hat {f} _ {1 x} ^ {(2)} = - \hat {f} _ {3 x} ^ {(2)} = - 3 3 3 3 \mathrm{lb}
\hat {f} _ {2 x} ^ {(4)} = - \hat {f} _ {3 x} ^ {(4)} = 1 0 0 0 \mathrm{lb}, \quad \hat {f} _ {3 x} ^ {(5)} = - \hat {f} _ {4 x} ^ {(5)} = 2 6 6 7 \mathrm{lb}
\hat {f} _ {1 x} ^ {(6)} = - \hat {f} _ {4 x} ^ {(6)} = 0
3.26 \mathrm { N o } , the truss is unstable, | \underline { { K } } | = 0 .
3.27 d _ { 3 x } = 0 . 0 4 6 3 in., d _ { 3 \nu } = - 0 . 0 1 7 6 in.
\hat {f} _ {1 x} ^ {(1)} = - \hat {f} _ {3 x} ^ {(1)} = - 2. 0 5 5 \text { kip }, \quad \hat {f} _ {2 x} ^ {(2)} = - \hat {f} _ {3 x} ^ {(2)} = 6. 2 7 9 \text { kip }
\hat {f} _ {3 x} ^ {(3)} = - \hat {f} _ {4 x} ^ {(3)} = - 6. 6 \text { kip }
\begin{array} { r } { \underline { { T } } ^ { T } = \left[ \begin{array} { c c c c } { C } & { - S } & { 0 } & { 0 } \\ { S } & { C } & { 0 } & { 0 } \\ { 0 } & { 0 } & { C } & { - S } \\ { 0 } & { 0 } & { S } & { C } \end{array} \right] } \end{array} \begin{array} { r } { \underline { { T } } \underline { { T } } ^ { T } = \left[ \begin{array} { l l l l } { 1 } & { 0 } & { 0 } & { 0 } \\ { 0 } & { 1 } & { 0 } & { 0 } \\ { 0 } & { 0 } & { 1 } & { 0 } \\ { 0 } & { 0 } & { 0 } & { 1 } \end{array} \right] } \end{array} 12 3.28 and 0
\therefore \underline {{T}} ^ {T} = \underline {{T}} ^ {- 1}
3.29 d _ { 1 x } = - 0 . 8 9 3 \times 1 0 ^ { - 4 } \mathrm { m } , d _ { 1 \nu } = - 4 . 4 6 \times 1 0 ^ { - 4 } \mathrm { m }
\sigma^ {(1)} = 3 1. 2 \mathrm{MPa(T)}, \quad \sigma^ {(2)} = 2 6. 5 \mathrm{MPa(T)}, \quad \sigma^ {(3)} = 6. 2 5 \mathrm{MPa(T)}
3.30 d_{1x} = 1.71 \times 10^{-4} \mathrm{~m}, d_{1y} = -7.55 \times 10^{-4} \mathrm{~m}
\sigma^ {(1)} = 7 9. 2 8 \mathrm{MPa(T)}, \quad \sigma^ {(2)} = 1 1. 9 7 \mathrm{MPa(T)}, \quad \sigma^ {(3)} = - 2 3. 8 7 \mathrm{MPa(C)}
3.31 d_{1x} = 8.25 \times 10^{-4} \mathrm{~m}, d_{1y} = -3.65 \times 10^{-3} \mathrm{~m}
\sigma^ {(2)} = 5 7. 7 4 \mathrm{MPa(T)}, \quad \sigma^ {(3)} = - 1 1 5. 5 \mathrm{MPa(C)}
3.32 d_{2x}=0.135\times10^{-2}\ m,\quad d_{2y}=-0.850\times10^{-2}\ m,
d _ {3 y} = - 0. 1 3 7 \times 1 0 ^ {- 1} \mathrm{m}, \quad d _ {4 y} = - 0. 1 6 4 \times 1 0 ^ {- 1} \mathrm{m},
\sigma^ {(1)} = - 1 9 8 \mathrm{MPa(C)}, \quad \sigma^ {(2)} = 0, \quad \sigma^ {(3)} = 4 4. 6 \mathrm{MPa(T)}
\sigma^ {(4)} = - 3 1. 6 \mathrm{MPa(C)}, \quad \sigma^ {(5)} = - 1 9 1 \mathrm{MPa(C)},
\sigma^ {(6)} = - 6 3. 1 \mathrm{MPa(C)}
3.33 a. d_{1x} = -3.448 \times 10^{-3} \mathrm{~m}, d_{1y} = -6.896 \times 10^{-3} \mathrm{~m}
\sigma^ {(1)} = 1 0 2. 4 \mathrm{MPa(T)}, \quad \sigma^ {(2)} = - 7 2. 4 \mathrm{MPa(C)}
3.34 d_{4x} = 9.93 \times 10^{-3} in., d_{4y} = -2.46 \times 10^{-3} in.
\sigma^ {(1)} = 3 1. 2 5 \text { ksi (T) }, \quad \sigma^ {(2)} = 3. 4 5 9 \text { ksi (T) }, \quad \sigma^ {(3)} = - 1. 5 3 8 \text { ksi (C) }
\sigma^ {(4)} = - 3. 1 0 3 \mathrm{ksi} (\mathrm{C}), \quad \sigma^ {(5)} = 0
3.35 d_{1y} = -0.5 \times 10^{-3} in., \sigma^{(1)} = 250 psi (T)
3.36 \hat{d}_{1x}=0.212 in.
3.37 \hat{d}_{1x} = 0.0397 in.
3.38 \hat{d}_{2x} = 16.98\mathrm{mm}
3.39 \hat{d}_{2x}=1.71\ mm
3.40 d_{1x} = -3.018 \times 10^{-5} \, m,\quad d_{1y} = -1.517 \times 10^{-5} \, m,
d _ {1 z} = 2. 6 8 4 \times 1 0 ^ {- 5} \mathrm{m}, \quad \sigma^ {(1)} = - 3 3 8 \mathrm{kN} / \mathrm{m} ^ {2} (\mathrm{C}),
\sigma^ {(2)} = - 1 6 9 0 \mathrm{kN} / \mathrm{m} ^ {2} (\mathrm{C}), \quad \sigma^ {(3)} = - 7 9 6 5 \mathrm{kN} / \mathrm{m} ^ {2} (\mathrm{C})
\sigma^ {(4)} = - 2 7 2 6 \mathrm{kN} / \mathrm{m} ^ {2} (\mathrm{C})
3.41 d_{1x} = 1.383 \times 10^{-3} \mathrm{~m}, d_{1y} = -5.119 \times 10^{-5} \mathrm{~m}
d _ {1 x} = 6. 0 1 5 \times 1 0 ^ {- 5} \mathrm{m}, \quad \sigma^ {(1)} = 2 0. 5 1 \mathrm{MPa(T)},
\sigma^ {(2)} = 4. 2 1 \mathrm{MPa(T)}, \quad \sigma^ {(3)} = - 5. 2 9 \mathrm{MPa(C)}
3.42 d_{5x} = 0.0014 in., d_{5y} = 0 , d_{5z} = -0.00042 in.
\sigma^ {(1)} = \sigma^ {(4)} = 1 8 0 \text { psi (T) }, \quad \sigma^ {(2)} = \sigma^ {(3)} = 1 4 0 \text { psi (C) }
3.43 d_{4x} = 0.00863 in., d_{4y} = 0 , d_{4z} = -0.00683 in.
\sigma^ {(1)} = - 9 1 6 \text { psi (C) }
3.46 d_{2y} = -0.0192 in., d_{3y} = -0.0168 in.
\sigma^ {(1)} = - 1 6 6 8 \text { psi (C) }, \quad \sigma^ {(2)} = 1 3 3 2 \text { psi (T) }, \quad \sigma^ {(3)} = 1 0 0 0 \text { psi (T) }
3 . 4 7 \ : \ : \ : d _ { 1 x } = \frac { - 1 1 0 P } { A E } \mathrm { i n . , } \ : \ : d _ { 1 y } = 0 , \ : \ : \ : d _ { 2 x } = 0 , \ : \ : \ : d _ { 2 y } = \frac { - 4 0 5 P } { A E } \mathrm { i n . , } in., in.,
d _ {3 x} = 0, \quad d _ {3 y} = \frac {- 4 3 3 P}{A E} \text {in.}, \quad d _ {4 x} = \frac {5 0 P}{A E} \text {in.}, \quad d _ {4 y} = \frac {- 2 0 8 P}{A E} \text {in.}
\sigma^ {(1)} = - 0. 1 5 6 \frac {P}{A}, \quad \sigma^ {(2)} = - 0. 2 0 8 \frac {P}{A}, \quad \sigma^ {(3)} = - 1. 1 6 \frac {P}{A}
\sigma^ {(4)} = 0. 2 6 0 \frac {P}{A}, \quad \sigma^ {(5)} = - 0. 5 7 3 \frac {P}{A}, \quad \sigma^ {(6)} = 0. 4 5 8 \frac {P}{A}
3.48 d _ { 2 y } = - 0 . 9 5 5 \times 1 0 ^ { - 2 } \mathrm { m } , d _ { 4 y } = - 1 . 0 3 \times 1 0 ^ { - 2 } \mathrm { m } ,
\sigma^ {(1)} = 6 7. 1 \mathrm{MPa(C)}, \quad \sigma^ {(2)} = 6 0. 0 \mathrm{MPa(T)}, \quad \sigma^ {(3)} = 2 2. 4 \mathrm{MPa(C)}
\sigma^ {(4)} = 4 4. 7 \mathrm{MPa(C)}, \quad \sigma^ {(5)} = 2 0. 0 \mathrm{MPa(T)}
3.49 d 0 ¼ 0, d2y ¼ -0:00283 in., F2x ¼ 2000 lb
\sigma^ {(1)} = 0, \quad \sigma^ {(2)} = 1 4 1 4 \text { psi (T) }, \quad \sigma^ {(3)} = 0
3.50 d _ { 2 \nu } = - 0 . 0 0 2 8 3 \mathrm { i n } .
3.51 d _ { 2 x } ^ { \prime } = 0 . 0 0 2 ~ \mathrm { i n . , } ~ f _ { 1 x } ^ { \prime } = - 2 8 0 0 ~ \mathrm { l b . , } ~ f _ { 2 x } ^ { \prime } = - 2 0 0 0 ~ \mathrm { l b }
F _ {2 y} ^ {\prime} = - 2 8 2 8 \mathrm{lb}
3.52 a. d _ { 1 x } = 0 . 0 1 0 in. #, \pi _ { p _ { \mathrm { m i n } } } = - 1 0 0 ~ \mathrm { l b \cdot i n }
\mathbf {b}. d _ {1 x} = 0. 0 0 8 3 3 \text { in. } \rightarrow , \quad \pi_ {p _ {\min}} = - 4 1. 6 7 \text { lb - in. }
3.53 \underline { { { k } } } = \frac { 3 A _ { 0 } E } { 2 L } \left[ \begin{array} { r r } { 1 } & { - 1 } \\ { - 1 } & { 1 } \end{array} \right]
3.54 two-element solution: d _ { 2 x } = 0 . 0 0 8 2 5 \mathrm { i r } ., d3x ¼ 0:012 in., \sigma ^ { ( 1 ) } = 8 2 5 0 \mathrm { p s i ( T ) } , \sigma ^ { ( 2 ) } = 3 7 5 0 \mathrm { p s i ( T ) } ,
3.55 two-element solution: d _ { 2 x } = 6 . 7 5 \times 1 0 ^ { - 3 } \ \mathrm { i n . , } \quad d _ { 3 x } = 0 . 0 0 9 \ \mathrm { i t } n.
\sigma^ {(1)} = 6 7 5 0 \text { psi (T) }, \quad \sigma^ {(2)} = 2 2 5 0 \text { psi (T) }
3.56 d _ { 2 x } = 0 . 7 5 \times 1 0 ^ { - 3 } \ \mathrm { i n . , } \quad \sigma ^ { ( 1 ) } = 7 5 0 \ \mathrm { p s i ( T ) }
3.57 d _ { 1 x } = \gamma L ^ { 2 } / ( 2 E ) , d _ { 2 x } = 3 \gamma L ^ { 2 } / ( 8 E ) , \sigma ^ { ( 1 ) } = \gamma L / 8 , \sigma ^ { ( 2 ) } = 3 \gamma L / 8
3.58 a. f _ { 1 x } = 5 8 3 . 3 1 \mathrm { k \Omega } , f _ { 2 x } = 6 6 6 . 7 1 \mathrm { { t } } b
\mathbf {b}. f _ {1 x} = 2 6. 7 \mathrm{kN}, f _ {2 x} = 8 0 \mathrm{kN}
Chapter 4
{ \bf 4 . 3 } d _ { 2 y } = { \frac { - 7 P L ^ { 3 } } { 7 6 8 E I } } , \phi _ { 1 } = { \frac { - P L ^ { 2 } } { 3 2 E I } } , \phi _ { 2 } = { \frac { P L ^ { 2 } } { 1 2 8 E I } } -PL2
F _ {1 y} = \frac {5 P}{1 6}, \quad M _ {1} = 0, \quad F _ {3 y} = \frac {1 1 P}{1 6}, \quad M _ {3} = \frac {- 3 P L}{1 6}
4 . 4 d _ { 1 y } = \frac { - P L ^ { 3 } } { 3 E I } , \phi _ { 1 } = \frac { P L ^ { 3 } } { 2 E I } , F _ { 2 y } = P , M _ { 2 } = - P L -PL3
4.5 d_{1y} = -2.688 in., \phi_{1} = 0.0144 rad, \phi_{2} = 0.0048 rad
4.6 d_{3y} = -3.94 in.
4.7 d_{2y} = -0.105 in., \phi_2 = -0.003 rad, d_{3y} = -0.345 in., \phi_3 = -0.0045 rad
4.8 d_{2y} = -1.34 \times 10^{-4} \mathrm{~m}, \quad \phi_{2} = 8.93 \times 10^{-5} \mathrm{rad}
4.9 d_{3y} = -7.619 \times 10^{-4} \, m,\quad \phi_{2} = -3.809 \times 10^{-4} \, rad,\quad \phi_{1} = 1.904 \times 10^{-4} \, rad
F _ {2 y} = 2. 5 \mathrm{kip}, \quad F _ {3 y} = - 1. 5 \mathrm{kip}, \quad M _ {3} = 1 0. 0 \mathrm{k-ft}
F _ {1 y} = 1 0 \mathrm{kN}, \quad M _ {1} = 1 2. 5 \mathrm{kN} \cdot \mathrm{m}, \quad F _ {3 y} = 1. 8 7 \mathrm{N}, \quad M _ {3} = - 2. 5 \mathrm{kN} \cdot \mathrm{m}
F _ {1 y} = - 0. 8 8 9 \mathrm{kN}, \quad F _ {2 y} = 4. 8 8 9 \mathrm{kN}
4.10 d_{2y} = -0.886 in., \phi_{2} = -0.00554 rad
F _ {1 y} = 1 1 1 5 \mathrm{lb}, \quad M _ {1} = - 2 6 7 \mathrm{k-in}.
4.11 d_{2y} = -7.934 \times 10^{-3} \mathrm{~m}, \quad \phi_{1} = -2.975 \times 10^{-3} \mathrm{rad}
F _ {1 y} = 5. 2 0 8 \mathrm{kN}, \quad F _ {3 y} = 5. 2 0 8 \mathrm{kN}
F _ {\text { spring }} = 1. 5 8 7 \mathrm{kN}
4.12 d_{2y} = d_{4y} = \frac{-1wL^4}{607.5EI}, d_{3y} = \frac{-wL^4}{507EI}
\phi_ {2} = \frac {- 1 w L ^ {3}}{2 7 0 E I}, \quad \phi_ {4} = - \phi_ {2}
F _ {1 y} = \frac {w L}{2}, \quad M _ {1} = \frac {w L ^ {2}}{1 2}
4.13 d_{2y} = \frac{-wL^4}{384EI}, F_{1y} = \frac{wL}{2}, M_1 = \frac{wL^2}{12}
4.14 d_{2y} = \frac{-5wL^4}{384EI},\quad \phi_1 = -\phi_3 = \frac{-wL^3}{24EI},\quad F_{1y} = \frac{wL}{2}
4.15 d_{3y} = \frac{-wL^4}{4EI},\quad \phi_2 = \frac{-wL^3}{8EI},\quad \phi_3 = \frac{-7wL^3}{24EI}
F _ {1 y} = \frac {- 3 w L}{4}, \quad M _ {1} = \frac {- w L ^ {2}}{4}, \quad F _ {2 y} = \frac {7 w L}{4}
4.16 \hat{f}_{1y} = \frac{-3wL}{20},\quad \hat{m}_1 = \frac{-wL^2}{30},\quad \hat{f}_{2y} = \frac{-7wL}{20},\quad \hat{m}_2 = \frac{wL^2}{20}
4.17 F_{1y} = \frac{3wL}{20}, M_1 = \frac{wL^2}{30}, F_{3y} = \frac{7wL}{20}, M_3 = \frac{-wL^2}{20}
4.18 \phi_{2} = \frac{wL^{3}}{80EI},\quad F_{1y} = \frac{9wL}{40},\quad M_{1} = \frac{7wL^{2}}{120},\quad F_{2y} = \frac{11wL}{40}
4.19 d_{3y} = -0.0244 \, m,\quad \phi_{3} = -0.0071 \, rad,\quad \phi_{2} = -0.00305 \, rad
F _ {1 y} = - 2 4 \mathrm{kN}, \quad M _ {1} = - 3 2 \mathrm{kN} \cdot \mathrm{m}, \quad F _ {2 y} = 5 6 \mathrm{kN}
\hat {f} _ {1 y} ^ {(1)} = - \hat {f} _ {2 y} ^ {(1)} = - 2 4 \mathrm{kN}, \quad \hat {m} _ {1} ^ {(1)} = - 3 2 \mathrm{kN} \cdot \mathrm{m}, \quad \hat {m} _ {2} ^ {(1)} = - 6 4 \mathrm{kN} \cdot \mathrm{m}
\hat {f} _ {2 y} ^ {(2)} = 3 2 \mathrm{kN}, \quad \hat {m} _ {2} ^ {(2)} = 6 4 \mathrm{kN} \cdot \mathrm{m}, \quad \hat {f} _ {3 y} ^ {(2)} = 0, \quad \hat {m} _ {3} ^ {(2)} = 0
4.20 f1 ¼ -0:0032 rad, d2y ¼ -0:0115 m, f3 ¼ 0:0032 rad
F _ {1 y} = 2 9. 9 4 \mathrm{kN}, \quad F _ {2 y} = 0. 1 1 5 2 \mathrm{kN}, \quad F _ {3 y} = 2 9. 9 4 \mathrm{kN}
\hat {f} _ {1 y} ^ {(1)} = 2 9. 9 4 \mathrm{kN}, \quad \hat {m} _ {1} ^ {(1)} = 0, \quad \hat {f} _ {2 y} ^ {(1)} = 0. 0 5 8 \mathrm{kN}, \quad \hat {m} _ {2} ^ {(1)} = 5 9. 6 5 \mathrm{kN} \cdot \mathrm{m}
4.21 d2y ¼ -2:514 in., f2 ¼ -0:00698 rad, f3 ¼ 0:0279 rad
F _ {1 y} = 3 7. 5 \mathrm{kip}, \quad M _ {1} = 2 2 5 \mathrm{k-in.}, \quad F _ {3 y} = 2 2. 5 \mathrm{kip}
4 . 2 2 \quad d _ { 3 y } = - 3 . 2 7 7 \mathrm { i n . , } \quad \phi _ { 3 } = - 0 . 0 3 2 3 \mathrm { r a d , } \quad \phi _ { 2 } = - 0 . 0 1 3 0 \mathrm { r a d }
F _ {1 y} = - 2 0. 5 \mathrm{kip}, \quad M _ {1} = - 7 1. 6 7 \mathrm{k-ft}, \quad F _ {2 y} = 6 0. 5 \mathrm{kip}
4.23 d _ { 2 y } = - 2 . 3 4 \mathrm { i n . , } \quad F _ { 1 y } = 5 3 2 5 \mathrm { l b } = F _ { 3 y } , \quad M _ { 1 } = 1 9 , 9 0 0 \mathrm { l b } \cdot \mathrm { f t } = - M _ { 3 }
4.24 \phi _ { 1 } = - 3 . 5 9 6 \times 1 0 ^ { - 4 } ~ \mathrm { r a d } , \quad \phi _ { 2 } = 9 . 9 2 \times 1 0 ^ { - 5 } ~ \mathrm { r a d } , \quad \phi _ { 3 } = 1 . 0 9 1 \times 1 0 ^ { - 4 } ~ \mathrm { r a d }
F _ {1 y} = 9 8 7 5 \mathrm{N}, \quad F _ {2 y} = 2 8, 4 0 6 \mathrm{N}, \quad F _ {3 y} = 6 7 1 9 \mathrm{N}
4.25 \mathrm { d } _ { \mathrm { m a x } } = - 0 . 0 0 0 7 5 6 \mathrm { m } at midspan of AB and BC
\sigma_ {\max} = 3 4. 3 \mathrm{MPa} \quad \text { at midspan of AB and BC }
\sigma_ {\mathrm{min}} = - 5 1. 0 \mathrm{MPa} \quad \text { at B }
4.26 \mathrm { d } _ { \mathrm { m a x } } = - 0 . 1 9 5 3 \mathrm { m } at midspan of BC
\sigma_ {\mathrm{min}} = - 4 6 9 \mathrm{MPa}
4.27 \mathrm { d } _ { \mathrm { m a x } } = - 1 . 0 2 8 \mathrm { i n } . under 7.5 kip load at B
\sigma_ {\mathrm{max}} = 3 4 0 0 0 \mathrm{psi}
\sigma_ {\mathrm{min}} = - 6 5 8 0 0 \mathrm{psi}
4 . 2 8 \mathrm { \textrm { d } } \mathrm { d } _ { \mathrm { m a x } } = - 0 . 0 4 1 9 \mathrm { m } \mathrm { a t } \mathrm { C }
\sigma_ {\max} = 6 6. 9 7 \mathrm{MPa} \quad \text { at fixed end A }
\sigma_ {\mathrm{min}} = - 1 3 3. 9 \mathrm{MPa} \quad \text { at B }
4.29 \mathrm { d } _ { \mathrm { m a x } } = - 0 . 4 9 5 \mathrm { i n } . at C
\sigma_ {\max} = 5 6 2 5 \mathrm{psi} \quad \text { at A }
\sigma_ {\min} = - 2 2 5 0 0 \mathrm{psi} \quad \text { at B }
4.30 dmax ¼ -0:087 m at C
\sigma_ {\max} = 2 5 7 \mathrm{MPa} \quad \text { at B }
4.37 d2y ¼ d _ { 2 y } = \frac { - P L ^ { 3 } } { 1 9 2 E I } - \frac { w L ^ { 4 } } { 3 8 4 E I } , \quad F _ { 1 y } = \frac { P + w L } { 2 } , \quad M _ { 1 } = \frac { P L } { 8 } + \frac { w L ^ { 2 } } { 1 2 } F1y ¼ wL2
4.38 d2y ¼ d _ { 2 y } = { \frac { - 5 P L ^ { 3 } } { 6 4 8 E I } }
4.39 d2y ¼ -ð25P þ 22wLÞL d _ { 2 y } = \frac { - ( 2 5 P + 2 2 w L ) L ^ { 3 } } { 2 4 0 E I } , \phi _ { 2 } = \frac { - ( P L ^ { 2 } + w L ^ { 3 } ) } { 8 E I } 3 , f ¼ -ðPL2 þ wL3Þ
F _ {1 y} = P + \frac {w L}{2}, \quad M _ {1} = \frac {P L}{2} + \frac {w L ^ {2}}{3}
4.40 d _ { 2 y } = - 1 . 5 7 \times 1 0 ^ { - 4 } ~ \mathrm { m } , \quad \phi _ { 2 } = 1 . 1 9 \times 1 0 ^ { - 4 } ~ \mathrm { r a d }
4.41 d_{2y} = -3.18 \times 10^{-4} \, m,\quad \phi_{2} = 1.58 \times 10^{-4} \, rad,\quad \phi_{3} = 1.58 \times 10^{-4} \, rad
4.42 d_{3y} = -2.13 \times 10^{-5} \, m,\quad \phi_{2} = -1.28 \times 10^{-5} \, rad,\quad \phi_{3} = 2.69 \times 10^{-5} \, rad
4.44 \underline{k}=\frac{GA_{W}}{L}\begin{bmatrix}1&-1\\ -1&1\end{bmatrix}
4.47 \hat{\underline{k}} = EI\int_{0}^{L}[B]^{T}[B]d\hat{x} +k_{f}\int_{0}^{L}[N]^{T}[N]d\hat{x}
Chapter 5
5.1 d_{2x} = 0.0278 in., d_{2y} = 0 , \phi_2 = -0.555 \times 10^{-4} rad
\hat {f} _ {1 x} ^ {(1)} = - \hat {f} _ {2 x} ^ {(1)} = - 8 3 0 0 \mathrm{lb}, \quad \hat {f} _ {1 y} ^ {(1)} = - \hat {f} _ {2 y} ^ {(1)} = 4. 6 \mathrm{lb}
\hat {m} _ {1} ^ {(1)} = 2 7 7 5 \mathrm{lb-in.}, \quad \hat {m} _ {2} ^ {(1)} = 0
5.2 d_{2x}=d_{3x}=0.688 in., d_{2y}=-d_{3y}=0.00171 in.
\phi_ {2} = - \phi_ {3} = - 0. 0 0 1 7 3 \mathrm{rad}
\hat {f} _ {1 x} ^ {(1)} = - \hat {f} _ {2 x} ^ {(1)} = - 2 1 4 0 \mathrm{lb}, \quad \hat {f} _ {1 y} ^ {(1)} = - \hat {f} _ {2 y} ^ {(1)} = - 2 5 0 3 \mathrm{lb}
\hat {m} _ {1} ^ {(1)} = 3 4 3, 6 0 0 \mathrm{lb-in.}, \quad \hat {m} _ {2} ^ {(1)} = 2 5 7, 0 0 0 \mathrm{lb-in.}
\hat {f} _ {2 x} ^ {(2)} = - \hat {f} _ {3 x} ^ {(2)} = 2 4 9 7 \mathrm{lb}, \quad \hat {f} _ {2 y} ^ {(2)} = - \hat {f} _ {3 y} ^ {(2)} = - 2 1 4 0 \mathrm{lb}
\hat {m} _ {2} ^ {(2)} = - 2 5 7, 0 0 0 \mathrm{lb-in.}, \quad \hat {m} _ {3} ^ {(2)} = - 2 5 6, 6 0 0 \mathrm{lb-in.}
\hat {f} _ {3 x} ^ {(3)} = - \hat {f} _ {4 x} ^ {(3)} = 2 1 4 0 \mathrm{lb}, \quad \hat {f} _ {3 y} ^ {(3)} = - \hat {f} _ {4 y} ^ {(3)} = 2 4 9 7 \mathrm{lb}
\hat {m} _ {3} ^ {(3)} = 2 5 6, 6 0 0 \mathrm{lb-in.}, \quad \hat {m} _ {4} ^ {(3)} = 3 4 2, 7 0 0 \mathrm{lb-in.}
F _ {1 x} = F _ {4 x} = - 2 5 0 3 \mathrm{lb}, \quad F _ {1 y} = - F _ {4 y} = - 2 1 4 0 \mathrm{lb}
M _ {1} = 3 4 3, 6 0 0 \mathrm{lb-in.}, \quad M _ {4} = 3 4 2, 7 0 0 \mathrm{lb-in.}
5.3 Channel section 6 \times 8.2 based on M_{max} = 106,900 lb-in.
5.4 d_{4x}=0.00445 in., d_{4y}=-0.0123 in., \phi_{4}=-0.00290 rad
\hat {f} _ {1 x} ^ {(1)} = - \hat {f} _ {4 x} ^ {(1)} = 4. 0 4 \text { kip }, \quad \hat {f} _ {1 y} ^ {(1)} = - \hat {f} _ {4 y} ^ {(1)} = - 1. 4 3 \text { kip }
\hat {m} _ {1} ^ {(1)} = - 2 5 4 \mathrm{k-in.}, \quad \hat {m} _ {4} ^ {(1)} = - 5 1 3 \mathrm{k-in.}
\hat {f} _ {2 x} ^ {(2)} = - \hat {f} _ {4 x} ^ {(2)} = 5. 8 2 \text { kip }, \quad \hat {f} _ {2 y} ^ {(2)} = - \hat {f} _ {4 y} ^ {(2)} = - 1. 4 5 \text { kip }
\hat {m} _ {2} ^ {(2)} = - 2 6 0 \mathrm{k-in.}, \quad \hat {m} _ {4} ^ {(2)} = - 5 1 9 \mathrm{k-in.}
F _ {1 x} = 3. 1 \mathrm{kip}, \quad F _ {1 y} = 2. 9 6 \mathrm{kip}, \quad M _ {1} = - 2 5 4 \mathrm{k-in}.
F _ {2 x} = - 1. 3 1 \mathrm{kip}, \quad F _ {2 y} = 5. 8 6 \mathrm{kip}, \quad M _ {2} = - 2 6 0 \mathrm{k-in}.
F _ {3 x} = - 1. 7 8 \mathrm{kip}, \quad F _ {3 y} = 1 1. 1 7 \mathrm{kip}, \quad M _ {3} = - 1 7 3 6 \mathrm{k-in}.
5.5 d_{2x}=0.05618 in., d_{2y}=-0.1792 in., \phi_{2}=-0.00965 rad
\hat {f} _ {1 x} ^ {(1)} = 9 0. 0 7 \mathrm{kip}, \quad \hat {f} _ {1 y} ^ {(1)} = 3. 8 3 \mathrm{kip}, \quad \hat {m} _ {1} ^ {(1)} = 3 6 1 \mathrm{k-in}.
\hat {f} _ {2 x} ^ {(1)} = - 7 3. 4 3 \text { kip }, \quad \hat {f} _ {2 y} ^ {(1)} = 7. 2 7 \text { kip }, \quad \hat {m} _ {2} ^ {(1)} = - 1 1 0 6 \text { k - in }.
\hat {f} _ {2 x} ^ {(2)} = - \hat {f} _ {3 x} ^ {(2)} = 4 6. 8 \text { kip }, \quad \hat {f} _ {2 y} ^ {(2)} = 1 7. 0 5 \text { kip }, \quad \hat {m} _ {2} ^ {(2)} = 1 1 0 7 \text { k - in }.