116 lines
2.5 KiB
TeX
116 lines
2.5 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}
|
|
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}
|