Add assignment three and four

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  1. 115
      engg_mech/assignment/t_four.tex
  2. 284
      engg_mech/assignment/t_three.tex

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\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{siunitx}
\begin{document}
\title{Engineering Mechanics}
\author{Ahmad Saalim Lone, 2019BCSE017}
\date{}
\maketitle
\section*{Question 1}
Calculate Reactions:
\begin{align*}
\sum M_J &= 0 \\
(12 kN)(4.8) + (12 kN)(2.4) - B(9.6) &= 0 \\
B &= 9 kN
\end{align*}
\begin{align*}
\sum F_y &= 0 \\
9 kN - 12 kN - 12 kN + J &= 0 \\
J &= 15kN
\end{align*}
Member CD:\@
\begin{align*}
\sum F_y &= 0 \\
9 kN + F_{CD} &= 0 \\
F_{CD} &= 9 kN \to \text{compression}
\end{align*}
Member DF:\@
\begin{align*}
\sum M_c &= 0 \\
F_{DF}1.8 m - 9kN \times 2.4m &= 0 \\
F_{DF} &= 12kN \to \text{Tension}
\end{align*}
\section*{Question 2}
Reactions:\@
\[
A = N = 0
\]
DF member:\@
\begin{align*}
\sum M_E &= 0 \\
(16 kN)(6m) - \frac{3}{5} F_{DF} (4m) &= 0 \\
F_{DF} &= 40 kN \to \text{Tension}
\end{align*}
EF member:\@
\begin{align*}
\sum F&= 0 \\
16 kN \sin{\beta} - F_{EF} \cos{\beta} &= 0 \\
F_{EF} &= 16 \tan{\beta} \\
&= 12 kN \to \text{Tension}
\end{align*}
EG member:\@
\begin{align*}
\sum M_F &= 0 \\
16kN \times 9m + \frac{4}{5}F_{EG} \times 3m &= 0 \\
F_{EG} &= -60 kN \to \text{Compression}
\end{align*}
\section*{Question 3}
Reactions
\begin{align*}
\sum M_k &= 0 \\
36\times 2.4 - B \times 13.5 + 20 \times 9 + 20 \times 4.5 &= 0 \\
B &= 26.4kN \\
\end{align*}
\begin{align*}
\sum F_x &= 0 \\
K_x &= 36 \\
\end{align*}
\begin{align*}
\sum F_y &= 0 \\
26.4 - 20 -20 + K_y &= 0 \\
K_y &= 13.6 kN \uparrow \\
\end{align*}
\begin{align*}
\sum M_C &= 0 \\
36 \times 1.2 - 26.4 \times 2.25 - F_{AD} \times 1.2 &= 0 \\
F_{AD} &= 13.5 kN \to \text{compression} \\
\end{align*}
\begin{align*}
\sum M_A &= 0 \\
\left( \frac{8}{17}F_{CD}\right)(4.5) &= 0 \\
F_{CD} &= 0 \\
\end{align*}
\begin{align*}
\sum M_D &= 9 \\
\frac{15}{17} \times F_{CE} \times 2.4 - 26.4 \times 4.5 &= 0 \\
F_{CE} &= 56.1 kN \to \text{Tension}
\end{align*}
\section*{Question 4}
Support reactions
\begin{align*}
\sum M_I &= 0 \\
2\times 12 + 5\times 8 3\times 6 + 2\times 4 - A_y \times 16 &= 0 \\
A_y &= 5.625 kN
\end{align*}
\begin{align*}
\sum A_x &= 0 \\
A_x &= 0
\end{align*}
Method of joints: By inspection, members BN, NC, DO, OC, HJ, LE \& JG are zero force members \\
Method of sections:
\begin{align*}
\sum M_M &= 0 \\
4F_{CD} - 5.625 \times 4 &= 0 \\
F_{CD} &= 5.625 kN \to \text{Tension}
\end{align*}
\begin{align*}
\sum M_A &= 0 \\
4F_{CM} - 2\times 4 &= 0 \\
F_{CM} &= 2 kN \to \text{Tension}
\end{align*}
\end{document}

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\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{siunitx}
\begin{document}
\title{Engineering Mechanics}
\author{Ahmad Saalim Lone, 2019BCSE017}
\date{}
\maketitle
\section*{Question 1}
Applying the equations of the equilibrium to the FBD of the entire truss, we have
\begin{align*}
\sum M_A &= 0 \\
N_C (2 + 2) - 4(2) - 3(1.5) &= 0 \\
N_C &= 3.125 kN
\end{align*}
\begin{align*}
\sum F_x &= 0 \\
3 - A_x &= 0 \\
A_x &= 3 kN
\end{align*}
\begin{align*}
\sum F_y &= 0 \\
A_y + 3.125 - 4 &= 0 \\
A_y &= 0.875 kN
\end{align*}
Method of joints \\
Joint C:\@ Just assume it to be in equilibrium
\begin{align*}
\sum F_y &= 0 \\
3.125 - F_{CD} \frac{3}{5} &= 0 \\
F_{CD} &= 5.21 kN \to \text{Compression}
\end{align*}
\begin{align*}
\sum F_x &= 0 \\
5.208 \times \frac{4}{5} - F_{CB} &= 0 \\
F_{CB} &= 4.17kN \to \text{Tension}
\end{align*}
Joint A:\@
\begin{align*}
\sum F_y &= 0 \\
0.875 - F_{AD} \times \frac{3}{5} &= 0 \\
F_{AD} &= 1.46 kN \to \text{Compression}
\end{align*}
\begin{align*}
\sum F_x &= 0 \\
F_{AB} - 3 - 1.458 \times \frac{4}{5} &= 0 \\
F_{AB} &= 4.167 kN \to \text{Tension}
\end{align*}
Joint B
\begin{align*}
\sum F_y &= 0 \\
F_{BD} &= 4 kN
\end{align*}
\begin{align*}
\sum F_x &= 0 \\
4.167 - 4.167 &= 0
\end{align*}
\section*{Question 2}
Analyze equilibrium of joint D, C \& E.
Joint D
\begin{align*}
\sum F_x &= 0 \\
F_{DE} \times \frac{3}{5} - 600 &= 0 \\
F_{DE} &= 1 kN \to \text{Compression}
\end{align*}
\begin{align*}
\sum F_y &= 0 \\
1000 \times \frac{4}{5} - F_{DC} &= 0 \\
F_{DC} &= 800 N \to \text{Tension}
\end{align*}
Joint C
\begin{align*}
\sum F_x &= 0 \\
F_{CE} &= 900 N \to \text{Compression}
\end{align*}
\begin{align*}
\sum F_y &= 0 \\
F_{CB} &= 800 N \to \text{Tension}
\end{align*}
Joint E
\begin{align*}
\sum F_x &= 0 \\
F_{EB} &= 750 N \to \text{Tension}
\end{align*}
\begin{align*}
\sum F_y &= 0 \\
F_{BA} &= 1.75 kN \to \text{Compression}
\end{align*}
\section*{Question 3}
Joint A
\begin{align*}
\sum F_y &= 0 \\
F_{AL} &= 28.28 kN \to \text{Compression}
\end{align*}
\begin{align*}
\sum F_x &= 0 \\
F_{AB} &= 20 kN \to \text{Tension}
\end{align*}
Joint B
\begin{align*}
\sum F_x &= 0 \\
F_{BC} &= 20 kN \to \text{Tension}
\end{align*}
\begin{align*}
\sum F_y &= 0 \\
F_{BL} &= 0
\end{align*}
Joint L
\begin{align*}
\sum F_x &= 0 \\
F_{LC} &= 0
\end{align*}
\begin{align*}
\sum F_y &= 0 \\
F_{LK} &= 28.28 kN \to \text{Compression}
\end{align*}
Joint C
\begin{align*}
\sum F_x &= 0 \\
F_{CD} &= 20 kN \to \text{Tension}
\end{align*}
\begin{align*}
\sum F_y &= 0 \\
F_{CK} &= 10 kN \to \text{Tension}
\end{align*}
Joint K
\begin{align*}
\sum F_x &= 0 \\
F_{KD} &= 7.454 kN
\end{align*}
\begin{align*}
\sum F_y &= 0 \\
F_{KJ} &= 23.57 kN \to \text{Compression}
\end{align*}
Joint J
\begin{align*}
\sum F_x &= 0 \\
F_{IJ} &= 23.57 kN
\end{align*}
\begin{align*}
\sum F_y &= 0 \\
F_{JD} &= 33.3 kN \to \text{Tension}
\end{align*}
Now we know that there exists symmetry,
\begin{gather*}
F_{AL} = F_{GH} = F{LK} = F{HI} = 28.3 kN \\
F_{AB} = F_{GF} = F_{BC} = F_{FE} = F_{CD} = F_{ED} = 20 kN \\
F_{BL} = F_{FH} = F_{LC} = F_{HE} = 0 \\
F_{CK} = F_{EI} = 10 kN \\
F_{KJ} = F_{IJ} = 23.6 kN \\
F_{KD} = F_{ID} = 7.45 kN
\end{gather*}
\section*{Question 4}
To evaluate support reactions
\begin{align*}
\sum M_E &= 0 \\
A_y &= \frac{4}{3} P
\end{align*}
\begin{align*}
\sum F_y &= 0 \\
E_y &= \frac{4}{3} P
\end{align*}
\begin{align*}
\sum F_x &= 0 \\
E_x &= P
\end{align*}
Methods of joints: By inspecting joint C, members CB \& CD are zero force members. Hence
\[
F_{CB} = F_{CD} = 0
\]
Joint A
\begin{align*}
\sum F_y &= 0 \\
F_{AB} &= 2.40 P \to \text{Compression}
\end{align*}
\begin{align*}
\sum F_x &= 0 \\
F_{AF} &= 2P \to \text{Tension}
\end{align*}
Joint B
\begin{align*}
\sum F_x &= 0 \\
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 \\
P - 0.447 F_{BF} - 0.447 F_{BD} &= 0
\end{align*}
\begin{align*}
\sum F_y &= 0 \\
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 \\
1.33P - 0.8944 F_{BF} + 0.8944 F_{BD} &= 0
\end{align*}
Solving the above equations, we get
\begin{align*}
F_{BD} &= 0.3727 P \to \text{Compression}\\
F_{BF} &= 1.863 P \to \text{Tension}
\end{align*}
Joint D
\begin{align*}
\sum F_y &= 0 \\
F_{DE} &= 0.3727 P \to \text{Compression}
\end{align*}
\section*{Question 5}
FBD of Joint A
\begin{gather*}
\frac{F_{AB}}{2.29} = \frac{F_{AC}}{2.29} = \frac{1.2}{1.2} kN \\
F_{AB} = 2.29 kN \to \text{Tension} \\
F_{AC} = 2.29 kN \to \text{Compression}
\end{gather*}
FBD of Joint F
\begin{gather*}
\frac{F_{DF}}{2.29} = \frac{F_{EF}}{2.29} = \frac{1.2}{1.2} kN \\
F_{DF} = 2.29 kN \to \text{Tension} \\
F_{EF} = 2.29 kN \to \text{Compression}
\end{gather*}
FBD of Joint D
\begin{gather*}
\frac{F_{BD}}{2.21} = \frac{F_{DE}}{0.6} = \frac{2.29}{2.29} kN \\
F_{DE} = 0.6 kN \to \text{Compression}\\
F_{EF} = 2.21 kN \to \text{Tension}
\end{gather*}
FBD of Joint C
\begin{align*}
\sum F_x &= 0 \\
F_{CE} &= 2.21 kN \to \text{Compression}
\end{align*}
\begin{align*}
\sum F_y &= 0 \\
F_{CH} &= 1.2 kN \to \text{Compression}
\end{align*}
FBD of Joint E
\begin{align*}
\sum F_x &= 0 \\
F_{BH} &= 0
\end{align*}
\begin{align*}
\sum F_y &= 0 \\
F_{EJ} &= 1.2 kN \to \text{Compression}
\end{align*}
\section*{Question 6}
Zero Force Members
Analyzing joint F:\@ Note that $DF$ and $EF$ are zero force members.
\[
F_{DF} = F_{EF} = 0
\]
Analyzing joint D:\@ Note that $BD$ and $DE$ are zero force members.
\[
F_{BD} = F_{DE} = 0
\]
FBD of joint A
\begin{gather*}
\frac{F_{AB}}{2.29} = \frac{F_{AC}}{2.29} = \frac{1.2}{1.2} kN \\
F_{AB} = 2.29 kN \to \text{Tension}\\
F_{AC} = 2.29 kN \to \text{Compression}
\end{gather*}
FBD of joint B
\begin{align*}
\sum F_x &= 0 \\
F_{BE} &= 2.7625 kN \to \text{Tension}
\end{align*}
\begin{align*}
\sum F_y &= 0 \\
F_{BC} &= 2.25 kN \to \text{Compression}
\end{align*}
FBD of joint C
\begin{align*}
\sum F_x &= 0 \\
F_{CE} &= 2.21 kN \to \text{Compression}
\end{align*}
\begin{align*}
\sum F_y &= 0 \\
F_{CH} &= 2.86 kN \to \text{Compression}
\end{align*}
FBD of joint E
\begin{align*}
\sum F_x &= 0 \\
F_{EH} &= 0
\end{align*}
\begin{align*}
\sum F_y &= 0 \\
f_{EJ} &= 1.657 kN \to \text{Tension}
\end{align*}
\end{document}
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