Analysis the gain truss for nodal displacement element stresses and strain. Take $P_1 = 5KN, P_2 = 2KN, E = 180GPa, A = 6cm^2$ for all element

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written 6.6 years ago by

teamques10 ★ 68k

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a) Number of elements and nodes,

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b) Elemental table,

Ele. No	Nodes(i)	Nodes (j)	$x_i$	$y_i$	$x_j$	$y_j$	Length L	$l=\frac{x_j-x_i}{L}$	$m=\frac{y_j-y_i}{l}$	$l^2$	$m^2$	$lm$
1	1	2	0	0	0.6	0	0.6	1	0	0	0	0
2	2	3	0.6	0	0.231	0.4	0.544	-0.678	0.735	0.46	0.54	-0.5
3	3	1	0.321	0.4	0	0	0.462	-0.5	-0.866	0.25	0.75	0.433

c) Element matrix equation.
$\begin{bmatrix} K \end{bmatrix}^e=\frac{AE}{L} \begin{bmatrix} l^2&lm&-l^2&-lm \\ lm&m^2&-lm&-m^2 \\ -l^2&-lm&l^2&lm \\ -lm&-m^2&lm&m^2 \end{bmatrix}$

For element 1, $A=600\,mm^2\,\,\,E=1.8*10^5\,\,\,L=600mm$
$\begin{bmatrix} K \end{bmatrix}^1=\frac{600*1.8*10^5}{600} \begin{bmatrix} 1&0-1&0 \\ 0&0&0&0 \\ -1&0&1&0 \\ 0&0&0&0 \end{bmatrix} $
$\hspace{2cm} \therefore = 10^4 \begin{bmatrix} 18&0-18&0 \\ 0&0&0&0 \\ -18&0&18&0 \\ 0&0&0&0 \end{bmatrix} $

For element 2, $A=600\,mm^2\,\,\,E=1.8*10^5\,\,\,L=544mm$
$\begin{bmatrix} K \end{bmatrix}^2=\frac{600*1.8*10^5}{544} \begin{bmatrix} 0.46&-0.5&-0.46&0.5 \\ -0.5&054&0.5&-0.54 \\ -0.46&0.5&0.46&-0.5 \\ 0.5&-0.54&-0.5&0.54 \end{bmatrix} $
$\hspace{2cm} \therefore = 10^4 \begin{bmatrix} 9.132&-9.927&-9.132&9.927 \\ -9.927&10.72&9.92&-10.72 \\ -9.132&9.927&9.132&-9.927 \\ 9.927&-10.72&-9.92&10.72 \end{bmatrix} $

For element 3, $A=600\,mm^2\,\,\,E=1.8*10^5\,\,\,L=462mm$
$\begin{bmatrix} K \end{bmatrix}^3=\frac{600*1.8*10^5}{462} \begin{bmatrix} 0.25&0.433&-0.25&-0.433 \\ 0.433&0.75&-0.433&-0.75 \\ -0.25&-0.433&0.25&0.433 \\ -0.433&-0.75&0.433&0.75 \end{bmatrix} $
$\hspace{2cm} \therefore = 10^4 \begin{bmatrix} 5.844&10.122&-5.844&-10.122 \\ 10.122&17.532&-10.122&-17.532 \\ -5.844&-10.122&5.844&10.122 \\ -10.122&-17.532&10.122&17.532 \end{bmatrix} $

Elemental matrix,
$\begin{bmatrix} K \end{bmatrix}= 10^4\begin{bmatrix} 18+5.844&10.122&-18&0&-5.844&-10.122 \\ 10.122&17.532&0&0&-10.122&-17.532 \\ -18&0&18+9.132&-9.927&-9.132&9.927 \\ 0&0&-9.927&10.72&9.927&-10.72 \\ -5.844&-10.122&-9.132&9.927&9.132+5.844&-9.927+10.122 \\ -10.122&-17.532&9.927&-10.72&-9.927+10.122&10.72+17.532 \end{bmatrix}$

Global martix equation,
$\begin{bmatrix} K \end{bmatrix}\begin{Bmatrix} u \end{Bmatrix}=\begin{Bmatrix} P \end{Bmatrix}$
$ 10^4\begin{bmatrix} 23.844&10.122&-18&0&-5.844&-10.122 \\ 10.122&17.532&0&0&-10.122&-17.532 \\ -18&0&27.1322&-9.927&-9.132&9.927 \\ 0&0&-9.927&10.72&9.927&-10.72 \\ -5.844&-10.122&-9.132&9.927&14.976&0.195 \\ -10.122&-17.532&9.927&-10.72&0.195&28.252 \end{bmatrix}\begin{Bmatrix} u_1\\v_1\\u_2\\v_2\\u_3\\v_3 \end{Bmatrix}=\begin{Bmatrix} P_{1x}\\P_{1y}\\P_{2x}\\P_{2y}\\P_{3x}\\P_{3y} \end{Bmatrix}$

Imposing boundary conditions,
$u_1=0\,\,\,v_1=0\hspace{4cm}$ -both direction constrained.
$v_2=0\hspace{5cm}$ -constrained only in vertices direction.
$P_{3x}=-2000N\,\,\,\,P_{3y}=-5000 $
$P_{2x}=0\hspace{6cm}$ -No horizontal reaction.

Frame the equation,
$10^4[-18\,u_2-5.844\,u_3-10.122\,v_3]=P_{1x}\hspace{3.7cm}$ -(1)
$10^4[-10.122\,u_3-17.532\,v_3]=P_{1y} \hspace{5cm}$ -(2)
$10^4[27.132\,u_2-9.927*0-9.132\,u_3+9.927\,v_3]=0\hspace{2cm}$ -(3)
$10^4[-9.927\,u_2+9.927\,u_3-10.72\,v_3]=P_{2y} \hspace{3.5cm}$ -(4)
$10^4[-9.132\,u_2+14.976\,u_3+0.195\,v_3]=-2000 \hspace{2.7cm}$ -(5)
$10^4[9.927\,u_2+0.195\,u_3+28.252\,v_3]=-5000 \hspace{3cm}$ -(6)

By solving equation (3),(5),(6), we get,
$\therefore u_2=30.53*10^{-4}mm$
$\therefore u_3=-112.5*10^{-4}mm$
$\therefore v_3=-186.93*10^{-4}mm$

Substituting above values in equation (1),(2),(4),we get,
$P_{1x}=2KN$
$P_{1y}=4.416KN$
$P_{2y}=0.584KN$

Now,
$\sum P_x=P_{1x}+P_{2x}+P_{3x}= 2+0-2=0$
$\sum P_y=P_{1y}+P_{2y}+P_{3y}= 4.416+0.584-5=0 \hspace{3cm}$ -Hence Verified.

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