285 lines
6.3 KiB
TeX
285 lines
6.3 KiB
TeX
\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}
|