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\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*}

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