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\documentclass{article} |
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\usepackage{amsmath} |
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\usepackage{amssymb} |
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\usepackage{siunitx} |
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\begin{document} |
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\title{Engineering Mechanics} |
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\author{Ahmad Saalim Lone, 2019BCSE017} |
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\date{} |
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\maketitle |
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\section*{Question 1} |
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Calculate Reactions: |
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\begin{align*} |
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\sum M_J &= 0 \\ |
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(12 kN)(4.8) + (12 kN)(2.4) - B(9.6) &= 0 \\ |
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B &= 9 kN |
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\end{align*} |
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\begin{align*} |
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\sum F_y &= 0 \\ |
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9 kN - 12 kN - 12 kN + J &= 0 \\ |
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J &= 15kN |
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\end{align*} |
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Member CD:\@ |
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\begin{align*} |
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\sum F_y &= 0 \\ |
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9 kN + F_{CD} &= 0 \\ |
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F_{CD} &= 9 kN \to \text{compression} |
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\end{align*} |
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Member DF:\@ |
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\begin{align*} |
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\sum M_c &= 0 \\ |
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F_{DF}1.8 m - 9kN \times 2.4m &= 0 \\ |
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F_{DF} &= 12kN \to \text{Tension} |
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\end{align*} |
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\section*{Question 2} |
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Reactions:\@ |
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\[ |
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A = N = 0 |
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\] |
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DF member:\@ |
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\begin{align*} |
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\sum M_E &= 0 \\ |
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(16 kN)(6m) - \frac{3}{5} F_{DF} (4m) &= 0 \\ |
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F_{DF} &= 40 kN \to \text{Tension} |
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\end{align*} |
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|
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EF member:\@ |
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\begin{align*} |
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\sum F&= 0 \\ |
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16 kN \sin{\beta} - F_{EF} \cos{\beta} &= 0 \\ |
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F_{EF} &= 16 \tan{\beta} \\ |
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&= 12 kN \to \text{Tension} |
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\end{align*} |
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|
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EG member:\@ |
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\begin{align*} |
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\sum M_F &= 0 \\ |
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16kN \times 9m + \frac{4}{5}F_{EG} \times 3m &= 0 \\ |
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F_{EG} &= -60 kN \to \text{Compression} |
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\end{align*} |
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\section*{Question 3} |
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Reactions |
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\begin{align*} |
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\sum M_k &= 0 \\ |
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36\times 2.4 - B \times 13.5 + 20 \times 9 + 20 \times 4.5 &= 0 \\ |
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B &= 26.4kN \\ |
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\end{align*} |
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\begin{align*} |
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\sum F_x &= 0 \\ |
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K_x &= 36 \\ |
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\end{align*} |
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\begin{align*} |
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\sum F_y &= 0 \\ |
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26.4 - 20 -20 + K_y &= 0 \\ |
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K_y &= 13.6 kN \uparrow \\ |
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\end{align*} |
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\begin{align*} |
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\sum M_C &= 0 \\ |
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36 \times 1.2 - 26.4 \times 2.25 - F_{AD} \times 1.2 &= 0 \\ |
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F_{AD} &= 13.5 kN \to \text{compression} \\ |
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\end{align*} |
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\begin{align*} |
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\sum M_A &= 0 \\ |
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\left( \frac{8}{17}F_{CD}\right)(4.5) &= 0 \\ |
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F_{CD} &= 0 \\ |
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\end{align*} |
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\begin{align*} |
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\sum M_D &= 9 \\ |
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\frac{15}{17} \times F_{CE} \times 2.4 - 26.4 \times 4.5 &= 0 \\ |
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F_{CE} &= 56.1 kN \to \text{Tension} |
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\end{align*} |
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\section*{Question 4} |
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|
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Support reactions |
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\begin{align*} |
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\sum M_I &= 0 \\ |
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2\times 12 + 5\times 8 3\times 6 + 2\times 4 - A_y \times 16 &= 0 \\ |
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A_y &= 5.625 kN |
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\end{align*} |
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\begin{align*} |
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\sum A_x &= 0 \\ |
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A_x &= 0 |
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\end{align*} |
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Method of joints: By inspection, members BN, NC, DO, OC, HJ, LE \& JG are zero force members \\ |
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Method of sections: |
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\begin{align*} |
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\sum M_M &= 0 \\ |
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4F_{CD} - 5.625 \times 4 &= 0 \\ |
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F_{CD} &= 5.625 kN \to \text{Tension} |
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\end{align*} |
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\begin{align*} |
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\sum M_A &= 0 \\ |
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4F_{CM} - 2\times 4 &= 0 \\ |
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F_{CM} &= 2 kN \to \text{Tension} |
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\end{align*} |
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\end{document} |
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\documentclass{article} |
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\usepackage{amsmath} |
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\usepackage{amssymb} |
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\usepackage{siunitx} |
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\begin{document} |
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\title{Engineering Mechanics} |
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\author{Ahmad Saalim Lone, 2019BCSE017} |
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\date{} |
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\maketitle |
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\section*{Question 1} |
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Applying the equations of the equilibrium to the FBD of the entire truss, we have |
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\begin{align*} |
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\sum M_A &= 0 \\ |
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N_C (2 + 2) - 4(2) - 3(1.5) &= 0 \\ |
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N_C &= 3.125 kN |
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\end{align*} |
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\begin{align*} |
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\sum F_x &= 0 \\ |
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3 - A_x &= 0 \\ |
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A_x &= 3 kN |
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\end{align*} |
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\begin{align*} |
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\sum F_y &= 0 \\ |
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A_y + 3.125 - 4 &= 0 \\ |
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A_y &= 0.875 kN |
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\end{align*} |
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Method of joints \\ |
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Joint C:\@ Just assume it to be in equilibrium |
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\begin{align*} |
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\sum F_y &= 0 \\ |
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3.125 - F_{CD} \frac{3}{5} &= 0 \\ |
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F_{CD} &= 5.21 kN \to \text{Compression} |
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\end{align*} |
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\begin{align*} |
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\sum F_x &= 0 \\ |
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5.208 \times \frac{4}{5} - F_{CB} &= 0 \\ |
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F_{CB} &= 4.17kN \to \text{Tension} |
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\end{align*} |
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Joint A:\@ |
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\begin{align*} |
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\sum F_y &= 0 \\ |
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0.875 - F_{AD} \times \frac{3}{5} &= 0 \\ |
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F_{AD} &= 1.46 kN \to \text{Compression} |
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\end{align*} |
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|
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\begin{align*} |
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\sum F_x &= 0 \\ |
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F_{AB} - 3 - 1.458 \times \frac{4}{5} &= 0 \\ |
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F_{AB} &= 4.167 kN \to \text{Tension} |
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\end{align*} |
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Joint B |
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\begin{align*} |
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\sum F_y &= 0 \\ |
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F_{BD} &= 4 kN |
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\end{align*} |
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\begin{align*} |
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\sum F_x &= 0 \\ |
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4.167 - 4.167 &= 0 |
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\end{align*} |
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\section*{Question 2} |
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Analyze equilibrium of joint D, C \& E. |
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Joint D |
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\begin{align*} |
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\sum F_x &= 0 \\ |
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F_{DE} \times \frac{3}{5} - 600 &= 0 \\ |
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F_{DE} &= 1 kN \to \text{Compression} |
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\end{align*} |
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\begin{align*} |
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\sum F_y &= 0 \\ |
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1000 \times \frac{4}{5} - F_{DC} &= 0 \\ |
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F_{DC} &= 800 N \to \text{Tension} |
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\end{align*} |
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Joint C |
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\begin{align*} |
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\sum F_x &= 0 \\ |
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F_{CE} &= 900 N \to \text{Compression} |
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\end{align*} |
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\begin{align*} |
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\sum F_y &= 0 \\ |
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F_{CB} &= 800 N \to \text{Tension} |
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\end{align*} |
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Joint E |
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\begin{align*} |
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\sum F_x &= 0 \\ |
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F_{EB} &= 750 N \to \text{Tension} |
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\end{align*} |
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\begin{align*} |
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\sum F_y &= 0 \\ |
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F_{BA} &= 1.75 kN \to \text{Compression} |
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\end{align*} |
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\section*{Question 3} |
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Joint A |
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\begin{align*} |
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\sum F_y &= 0 \\ |
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F_{AL} &= 28.28 kN \to \text{Compression} |
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\end{align*} |
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\begin{align*} |
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\sum F_x &= 0 \\ |
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F_{AB} &= 20 kN \to \text{Tension} |
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\end{align*} |
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Joint B |
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\begin{align*} |
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\sum F_x &= 0 \\ |
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F_{BC} &= 20 kN \to \text{Tension} |
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\end{align*} |
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\begin{align*} |
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\sum F_y &= 0 \\ |
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F_{BL} &= 0 |
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\end{align*} |
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Joint L |
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\begin{align*} |
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\sum F_x &= 0 \\ |
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F_{LC} &= 0 |
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\end{align*} |
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\begin{align*} |
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\sum F_y &= 0 \\ |
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F_{LK} &= 28.28 kN \to \text{Compression} |
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\end{align*} |
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Joint C |
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\begin{align*} |
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\sum F_x &= 0 \\ |
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F_{CD} &= 20 kN \to \text{Tension} |
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\end{align*} |
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\begin{align*} |
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\sum F_y &= 0 \\ |
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F_{CK} &= 10 kN \to \text{Tension} |
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\end{align*} |
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Joint K |
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\begin{align*} |
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\sum F_x &= 0 \\ |
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F_{KD} &= 7.454 kN |
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\end{align*} |
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\begin{align*} |
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\sum F_y &= 0 \\ |
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F_{KJ} &= 23.57 kN \to \text{Compression} |
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\end{align*} |
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Joint J |
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\begin{align*} |
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\sum F_x &= 0 \\ |
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F_{IJ} &= 23.57 kN |
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\end{align*} |
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\begin{align*} |
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\sum F_y &= 0 \\ |
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F_{JD} &= 33.3 kN \to \text{Tension} |
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\end{align*} |
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Now we know that there exists symmetry, |
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\begin{gather*} |
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F_{AL} = F_{GH} = F{LK} = F{HI} = 28.3 kN \\ |
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F_{AB} = F_{GF} = F_{BC} = F_{FE} = F_{CD} = F_{ED} = 20 kN \\ |
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F_{BL} = F_{FH} = F_{LC} = F_{HE} = 0 \\ |
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F_{CK} = F_{EI} = 10 kN \\ |
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F_{KJ} = F_{IJ} = 23.6 kN \\ |
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F_{KD} = F_{ID} = 7.45 kN |
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\end{gather*} |
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\section*{Question 4} |
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To evaluate support reactions |
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\begin{align*} |
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\sum M_E &= 0 \\ |
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A_y &= \frac{4}{3} P |
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\end{align*} |
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\begin{align*} |
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\sum F_y &= 0 \\ |
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E_y &= \frac{4}{3} P |
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\end{align*} |
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\begin{align*} |
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\sum F_x &= 0 \\ |
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E_x &= P |
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\end{align*} |
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Methods of joints: By inspecting joint C, members CB \& CD are zero force members. Hence |
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\[ |
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F_{CB} = F_{CD} = 0 |
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\] |
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Joint A |
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\begin{align*} |
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\sum F_y &= 0 \\ |
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F_{AB} &= 2.40 P \to \text{Compression} |
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\end{align*} |
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\begin{align*} |
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\sum F_x &= 0 \\ |
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F_{AF} &= 2P \to \text{Tension} |
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\end{align*} |
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Joint B |
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\begin{align*} |
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\sum F_x &= 0 \\ |
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2.404P \times \frac{1.5}{\sqrt{3.25}} - P - F_{BF} \times \frac{0.5}{\sqrt{1.25}} - F_{BD} \times \frac{0.5}{\sqrt{1.25}} &= 0 \\ |
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P - 0.447 F_{BF} - 0.447 F_{BD} &= 0 |
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\end{align*} |
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\begin{align*} |
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\sum F_y &= 0 \\ |
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2.404P \times \frac{1}{\sqrt{3.25}} - F_{BF} \times \frac{1}{\sqrt{1.25}} + F_{BD} \times \frac{1}{\sqrt{1.25}} &= 0 \\ |
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1.33P - 0.8944 F_{BF} + 0.8944 F_{BD} &= 0 |
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\end{align*} |
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Solving the above equations, we get |
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\begin{align*} |
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F_{BD} &= 0.3727 P \to \text{Compression}\\ |
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F_{BF} &= 1.863 P \to \text{Tension} |
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\end{align*} |
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Joint D |
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\begin{align*} |
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\sum F_y &= 0 \\ |
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F_{DE} &= 0.3727 P \to \text{Compression} |
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\end{align*} |
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\section*{Question 5} |
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FBD of Joint A |
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\begin{gather*} |
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\frac{F_{AB}}{2.29} = \frac{F_{AC}}{2.29} = \frac{1.2}{1.2} kN \\ |
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F_{AB} = 2.29 kN \to \text{Tension} \\ |
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F_{AC} = 2.29 kN \to \text{Compression} |
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\end{gather*} |
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FBD of Joint F |
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\begin{gather*} |
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\frac{F_{DF}}{2.29} = \frac{F_{EF}}{2.29} = \frac{1.2}{1.2} kN \\ |
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F_{DF} = 2.29 kN \to \text{Tension} \\ |
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F_{EF} = 2.29 kN \to \text{Compression} |
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\end{gather*} |
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FBD of Joint D |
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\begin{gather*} |
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\frac{F_{BD}}{2.21} = \frac{F_{DE}}{0.6} = \frac{2.29}{2.29} kN \\ |
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F_{DE} = 0.6 kN \to \text{Compression}\\ |
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F_{EF} = 2.21 kN \to \text{Tension} |
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\end{gather*} |
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FBD of Joint C |
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\begin{align*} |
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\sum F_x &= 0 \\ |
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F_{CE} &= 2.21 kN \to \text{Compression} |
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\end{align*} |
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\begin{align*} |
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\sum F_y &= 0 \\ |
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F_{CH} &= 1.2 kN \to \text{Compression} |
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\end{align*} |
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FBD of Joint E |
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\begin{align*} |
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\sum F_x &= 0 \\ |
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F_{BH} &= 0 |
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\end{align*} |
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\begin{align*} |
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\sum F_y &= 0 \\ |
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F_{EJ} &= 1.2 kN \to \text{Compression} |
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\end{align*} |
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\section*{Question 6} |
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Zero Force Members |
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Analyzing joint F:\@ Note that $DF$ and $EF$ are zero force members. |
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\[ |
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F_{DF} = F_{EF} = 0 |
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\] |
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Analyzing joint D:\@ Note that $BD$ and $DE$ are zero force members. |
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\[ |
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F_{BD} = F_{DE} = 0 |
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\] |
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FBD of joint A |
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\begin{gather*} |
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\frac{F_{AB}}{2.29} = \frac{F_{AC}}{2.29} = \frac{1.2}{1.2} kN \\ |
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F_{AB} = 2.29 kN \to \text{Tension}\\ |
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F_{AC} = 2.29 kN \to \text{Compression} |
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\end{gather*} |
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FBD of joint B |
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\begin{align*} |
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\sum F_x &= 0 \\ |
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F_{BE} &= 2.7625 kN \to \text{Tension} |
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\end{align*} |
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\begin{align*} |
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\sum F_y &= 0 \\ |
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F_{BC} &= 2.25 kN \to \text{Compression} |
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\end{align*} |
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FBD of joint C |
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\begin{align*} |
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\sum F_x &= 0 \\ |
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F_{CE} &= 2.21 kN \to \text{Compression} |
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\end{align*} |
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\begin{align*} |
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\sum F_y &= 0 \\ |
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F_{CH} &= 2.86 kN \to \text{Compression} |
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\end{align*} |
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FBD of joint E |
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\begin{align*} |
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\sum F_x &= 0 \\ |
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F_{EH} &= 0 |
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\end{align*} |
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\begin{align*} |
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\sum F_y &= 0 \\ |
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f_{EJ} &= 1.657 kN \to \text{Tension} |
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\end{align*} |
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|
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\end{document} |
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