From c22f140ad5eeca6522be64c82fe720497043054a Mon Sep 17 00:00:00 2001 From: Ceda EI Date: Thu, 3 Sep 2020 17:26:16 +0530 Subject: [PATCH] Add assignment three and four --- engg_mech/assignment/t_four.tex | 115 +++++++++++++ engg_mech/assignment/t_three.tex | 284 +++++++++++++++++++++++++++++++ 2 files changed, 399 insertions(+) create mode 100644 engg_mech/assignment/t_four.tex create mode 100644 engg_mech/assignment/t_three.tex diff --git a/engg_mech/assignment/t_four.tex b/engg_mech/assignment/t_four.tex new file mode 100644 index 0000000..ec0d254 --- /dev/null +++ b/engg_mech/assignment/t_four.tex @@ -0,0 +1,115 @@ +\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} diff --git a/engg_mech/assignment/t_three.tex b/engg_mech/assignment/t_three.tex new file mode 100644 index 0000000..b8818cf --- /dev/null +++ b/engg_mech/assignment/t_three.tex @@ -0,0 +1,284 @@ +\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}