Question

by finite elemnt method

4. (25 points) Using symmetry on the frame shown below, solve for the unknown displacements and rotations. When setting up the stiffness matrix you may neglect terms that would be used to find the reactions. Use a consistent unit system from the table below. Note, the connections at nodes 2 and 4 are pins that do not allow X and Y displacements but do allow rotations. 60 KN 2 100 KN m 3 3 100 kNm E 210 GPS 11.0 10m A10 X 10m

Answer 1

Symmetry 高 s(v) - - 0 (1) -se) - s (r) | tsp [-- - s( 一)- (sa-ori) - - s( 一) - -

From Figure

Element	Anlge	cos(theta)	sin(theta)	cos(theta)^2	sin(theta)^2	sin(theta)*cos(theta)
1	90	0	1	0	1	0
2	0	1	0	1	0	0
3	0	1	0	1	0	0
4	-90	0	-1	0	1	0

and as per given data

E	2.1*1011	Pa
L	6	m
A	0.01	m2
I	0.0001	m4
12*I/L2	3.33333*10-05	m2
6*I/L	0.0001	m3
E/L	35000000000	Pa/m

So from data the K matrix are,

11.67 о к(1) __ 105 | -35 -11.67 о | —35 1 3500 o о к(2) = 105 —3500 0 o 3500 | o к(3) __ 105 | о |-3500 о 0 [11.67 o 51 35 К = 10 11 67 о 35 o 3500 0 o — 3500 o o o о o -11.67 o o o о o -11,67 0 o o о o —3500 o —35 -11.67 o -35 o o — 3500 о 140 35 o 70 35 11.67 o 35 o o 3500 о 7035 о 140 o — 3500 о о1 o o —11.67 о 140 o o 70 o 3500 о о o o 11.67 о 70 o o 140 o — 3500 о ој o o –11.67 о 140 o o 70 o 3500 о о o o 11.67 о 70 o o 1401 35 -11.67 o 35 o o — 3500 о 140 — 35 o 70 — 35 11.67 o -35 o o 3500 о 70 — 35 о 140

  

F= R3y M3 and D= RAI

as per figure node 1 and node 4 are fixed joint and node 2 and node 3 are pin joint so

dir = dy = 0, = d2r = d2y = der = d4y = ds = dsy = 05 = 0

As per given force and moment

M2 = -10 Nm R3r = ON R3y = -60000N M4 = 10 Nm M3 = ONm

Rut Rly M R2x Ray - 100000 F= -60000 and D= RAI R4y 100000 R 5 R5y M5

Global K matrix

о o о — 11.67 о о 70 о о —3500 11.67 о о К= 10° 11.67 o — 35 — 11.67 o — 35 о o o o o o о 3500 o o — 3500 o o o o o o o —35 о 140 35 o 70 о o o o o o — 11.67 o 35 3511.67 o 35 — 3500 o o o oo — 3500 о 3500 o o ооо оо — 35 o 70 35 0 280 0 o o o o o o o — 3500 o o 7000 ооо oo оооо —11.67 o o — 11.67 o o o o о о 70 o o o o o o o o o — 3500 — 3500 о 35 -11.67 о 35 ооо о 11.67 o o — 3500 о ооо о o o o o 280 —35 o 70 o o o o o o o о —35 11.67 o — 35 o o o o o o o — 3500 o o 3500 о o o o o o o o о 70 — 35 о 140 о 280 о o o 3511.67 — 11.67 o o o o o o 70 o o o 35 -11.67 | o 35

For finding the deflection and rotation we have to terminate row and column except 6,7,8,9,12

1280 o o 70 о] о 7000 o o o К = 10° | о о 11.67 o o | 70 o o 280 70 | o o o 70 280 | [ 382.65 o o o 14.29 o К-1-10-10 | o o 8571.43 -102.04 o o | 25.51 o o 25.51 1 o — 102.04 o о 408.16 –102.04 –102.04 382.65

AS we know

F = KD

— 100000 (280 о — 60000 | = 10° | о | 70 100000 | o o 7000 о o o o o 11.67 o o 70 o o | o o 280 70 70 280 | |ө
$\begin{bmatrix} \theta_2 \\ d_{3x} \\ d_{3y} \\ \theta_3 \\ \theta_4 \end{bmatrix}= 10^{-10}\begin{bmatrix} 382.65 & 0 & 0 & -102.04 & 25.51 \\ 0 & 14.29 & 0 & 0 & 0 \\ 0 & 0 & 8571.43 & 0 & 0 \\ -102.04 & 0 & 0 & 408.16 & -102.04 \\ 25.51 & 0 & 0 & -102.04 & 382.65 \\ \end{bmatrix} \begin{bmatrix} -100000 \\ 0 \\ -60000 \\ 0 \\ 100000 \end{bmatrix}$

T 02 1-0.003571429 rad dar от -0.051428571 m O rad 0.003571429 rad 04

so,

Reaction force at other node are given by,

F = KD

Rix Rig All R28 Rzy —100000 —60000 = RAT RAY 100000 RSS RSY Mus
$10^5\begin{bmatrix} 11.67& 0& -35& -11.67& 0& -35& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 3500& 0& 0& -3500& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ -35& 0& 140& 35& 0& 70& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ -11.67& 0& 35& 3511.67& 0& 35& -3500& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& -3500& 0& 0& 3500& 0& 0& -11.67& 0& 0& 0& 0& 0& 0& 0\\ -35& 0& 70& 35& 0& 280& 0& 0& 70& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& -3500& 0& 0& 7000& 0& 0& -3500& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& -11.67& 0& 0& 11.67& 0& 0& -11.67& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 70& 0& 0& 280& 0& 0& 70& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& -3500& 0& 0& 3511.67& 0& 35& -11.67& 0& 35\\ 0& 0& 0& 0& 0& 0& 0& -11.67& 0& 0& 11.67& 0& 0& -3500& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 70& 35& 0& 280& -35& 0& 70\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& -11.67& 0& -35& 11.67& 0& -35\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& -3500& 0& 0& 3500& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 35& 0& 70& -35& 0& 140\\ \end{bmatrix}$$*\begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ -0.0035714 \\ 0 \\ -0.0514286 \\ 0 \\ 0 \\ 0 \\ 0.0035714 \\ 0 \\ 0 \\ 0 \\ \end{bmatrix}$

R3r Rзу 12500N ON - 25000 Nm -12500N 60000N -100000Nm ON - 60000N ONm 12500N 60000N 100000 Nm - 12500N ON 25000Nm

other method to find the solution is

the strucher is symetric so just perfom it for half strucher so,

Element	Anlge	cos(theta)	sin(theta)	cos(theta)^2	sin(theta)^2	sin(theta)*cos(theta)
1	90	0	1	0	1	0
2	0	1	0	1	0	0
3	0	1	0	1	0	0
4	-90	0	-1	0	1	0

and as per given data

E	2.1*1011	Pa
L	6	m
A	0.01	m2
I	0.0001	m4
12*I/L2	3.33333*10-05	m2
6*I/L	0.0001	m3
E/L	35000000000	Pa/m

So from data the K matrix are,

11.67 о о 3500 – 350 к(1) = 10° | 1-11.67 0 о — 3500 | — 35 o 1 3500 o o o к(2) = 10° | о о —3500 o 0 -11.67 0 0 — 35 -11.67 o o — 3500 о 140 35 o 70 35 11.67 o 35 o o 35000 7035 о 140 | o — 3500 о о ] o o –11.67 о 140 o o 70 o 3500 о о o o 11.67 о 70 o o 1401

$\\K= 10^5\begin{bmatrix} 11.67& 0& -35& -11.67& 0& -35& 0& 0& 0\\ 0& 3500& 0& 0& -3500& 0& 0& 0& 0\\ -35& 0& 140& 35& 0& 70& 0& 0& 0\\ -11.67& 0& 35& 3511.67& 0& 35& -3500& 0& 0\\ 0& -3500& 0& 0& 3500& 0& 0& -11.67& 0\\ -35& 0& 70& 35& 0& 280& 0& 0& 70\\ 0& 0& 0& -3500& 0& 0& 3500& 0& 0\\ 0& 0& 0& 0& -11.67& 0& 0& 11.67& 0\\ 0& 0& 0& 0& 0& 70& 0& 0& 140\\ \end{bmatrix}$

as per figure node 1 and node 4 are fixed joint and node 2 and node 3 are pin joint so

$d_{1x}= d_{1y}= \theta_1= d_{2x}= d_{2y}= 0$

As per given force and moment

M2 = -10 Nm R3r = ON R3y = -60000/2 = -30000N M3 = 0Nm

RE Riy M R2 F= R24 and D= -100000 -30000

For finding the deflection and rotation we have to terminate row and column and row 1,2,3,4,5,from K,F,D matrix

$\\K=10^5 \begin{bmatrix} 280 & 0 & 0 & 70 \\ 0 & 3500 & 0 & 0 \\ 0 & 0 & 11.67 & 0 \\ 70 & 0 & 0 & 140 \\ \end{bmatrix} \\K^{-1}= 10^{-10}\begin{bmatrix} 408.16 & 0 & 0 & -204.08 \\ 0 & 28.57 & 0 & 0 \\ 0 & 0 & 8571.43 & 0 \\ -204.08 & 0 & 0 & 816.33 \\ \end{bmatrix}$

F = KD

D] = [K-F

$\begin{bmatrix} \theta_2 \\ d_{3x} \\ d_{3y} \\ \theta_3 \end{bmatrix}= 10^{-10}\begin{bmatrix} 408.16 & 0 & 0 & -204.08 \\ 0 & 28.57 & 0 & 0 \\ 0 & 0 & 8571.43 & 0 \\ -204.08 & 0 & 0 & 816.33 \\ \end{bmatrix} \begin{bmatrix} -100000 \\ 0 \\ -30000 \\0 \end{bmatrix}= \begin{bmatrix} -0.004081633 rad\\ 0 m\\ -0.025714286 m\\ 0.002040816 rad\\ \end{bmatrix}$

Reaction forces:

$F=\begin{bmatrix} R_{1x} \\ R_{1y} \\ M_1 \\ R_{2x} \\ R_{2y} \\ M_2 \\ R_{3x} \\ R_{3y} \\ M_3 \\ \end{bmatrix}=\begin{bmatrix} 12500N \\ 0 N \\ -25000N-m \\ -12500N\\ 30000 N\\ -100000 N-m\\ 0 N-m\\ -30000 N\\ 0 N-m\\ \end{bmatrix}$

Same will be act on other halft portion

$F=\begin{bmatrix} R_{3x} \\ R_{3y} \\ M_3 \\ R_{4x} \\ R_{4y} \\ M_4 \\ R_{5x} \\ R_{5y} \\ M_5 \\ \end{bmatrix}=\begin{bmatrix} 0 N\\ -30000 N\\ 0 Nm\\ -10000 N\\ 30000 N\\ -12500 Nm\\ -25000 N\\ 0 N\\ 12500 Nm\\ \end{bmatrix}$

Total F matrix

12500N ON - 25000 Nm - 12500N 30000N -100000 Nm ON - 60000N ONm 12500N 30000N 100000 Nm - 12500N ON 25000Nm