Beam Problem 2 Solution first created 28/03/08 - last modified 28/03/08 Page Author: Ty Harness


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Figure 1 - Beam split in 2 where V and M hold each side in equillibrium



$EI (d^2y)/(dx^2) = M$

$EI (d^2y)/(dx^2) = w(L-x) (L-x)/2 = (w/2) {L^2 - 2Lx + x^2} $


$EI (dy)/(dx) = (w/2) {L^2 x - Lx^2 + (x^3)/3} + A $


$EI y = (w/2) {(L^2 x^2)/2 - (Lx^3)/3 + (x^4)/12} + Ax + B $


y= 0 at x = 0 therefore B=0

dy/dy = 0 at x = 0 therefore A = 0

$EI y = (w x^2)/24 {6*L^2 - 4*L*x + L^2} $


The deflection when x=L will be the maximum
$y(L) = w/(24*E*I) {6L^2 - 4*L^2 + L^2} = (wL^4)/(8*E*I) $
Working in Newtons and mm

We need the second moment of area,I about the neutral axis:

$I = (b*d^3)/12 = 3.333*10^7 mm^4$

$E = 200 000 (MN)/m^2 = 200 000 N/(mm^2)$
We need to find the load per metre, w [N/mm] for the self weight of the beam.
$v = 50*mm * 200 mm = 10000 mm^2$
$rho = 8000 (kg)/m^3 = (8000 kg * m^3 ) / (m^3 * 1e9 (mm)^3)$

$rho = 8*10^-6 * (kg)/(mm^3)$

$w = rho*v*g = 8*10^-6 * (kg)/(mm^3) *10000*mm^2 * 10 m/s^2 = 0.8 N/(mm) $

$y(L) = (0.8*1000^4)/(8*2*10^5*3.333*10^7) = (8*10^11)/(5.3333*10^13) = 0.015 (mm) $
I've chosen to use Prof. Rieg's Z88 Finite Elements Program because it's free and unlimited and ideal for student use. The theory behind FEM is beyond the scope of this tutorial but you can learn a great deal from the Z88 manual.

Using a Beam Element No. 13 we create a

General Structure Data File, Z88i1.txt which basically consists of : 2D Analysis, 10 Nodes, 9 Elements, 3DOF per node 30 in total.
Nodal co-ordinates, elements defined and adjacent elements. All elements 1 to 9 have the same properties Young's Modulus but this time working in metres = 2e11 N/m^2. Poisson's Ratio = 0.3 from tables, Area = 1e-2 m^2, I = 3.33e-5 m^4
2 10 9 30 1 0 1 0 Z88I1.TXT,typed in by Ty
1 3 +0.00000E+000 +0.00000E+000 +0.00000E+000 node #1
2 3 +1.11111E+002 +0.00000E+000 +0.00000E+000 node #2
3 3 +2.22222E+002 +0.00000E+000 +0.00000E+000 node #3
4 3 +3.33333E+002 +0.00000E+000 +0.00000E+000 node #4
5 3 +4.44444E+002 +0.00000E+000 +0.00000E+000 node #5
6 3 +5.55556E+002 +0.00000E+000 +0.00000E+000 node #6
7 3 +6.66667E+002 +0.00000E+000 +0.00000E+000 node #7
8 3 +7.77778E+002 +0.00000E+000 +0.00000E+000 node #8
9 3 +8.88889E+002 +0.00000E+000 +0.00000E+000 node #9
10 3 +1.00000E+003 +0.00000E+000 +0.00000E+000 node #10
1 13 element #1
1 2
2 13 element #2
2 3
3 13 element #3
3 4
4 13 element #4
4 5
5 13 element #5
5 6
6 13 element #6
6 7
7 13 element #7
7 8
8 13 element #8
8 9
9 13 element #9
9 10
1 9 +2.00000E+005 +3.00000E-001 1 +1.00000E+004 0 0 +3.33333E+007 +1.00000E+002 0 0
13 , typed in by ty
1 1 2 0.00000E+000
1 2 2 0.00000E+000
1 3 2 0.00000E+000
1 2 1 -8.00000E+001
2 2 1 -8.00000E+001
3 2 1 -8.00000E+001
4 2 1 -8.00000E+001
5 2 1 -8.00000E+001
6 2 1 -8.00000E+001
7 2 1 -8.00000E+001
8 2 1 -8.00000E+001
9 2 1 -8.00000E+001
10 2 1 -8.00000E+001


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Figure 3 - Z88 plot of the deflection for bp2


output file Z88O2.TXT : displacements, computed by Z88F V10
*************

Knoten U(1) U(2) U(3) U(4) U(5) U(6)

1 +0.0000000E+000 +0.0000000E+000 +0.0000000E+000
2 +0.0000000E+000 -3.4567869E-004 -6.0000007E-006
3 +0.0000000E+000 -1.2866930E-003 -1.0740743E-005
4 +0.0000000E+000 -2.6913560E-003 -1.4370376E-005
5 +0.0000000E+000 -4.4444415E-003 -1.7037047E-005
6 +0.0000000E+000 -6.4472032E-003 -1.8888917E-005
7 +0.0000000E+000 -8.6173000E-003 -2.0074101E-005
8 +0.0000000E+000 -1.0888906E-002 -2.0740767E-005
9 +0.0000000E+000 -1.3212637E-002 -2.1037063E-005
10 +0.0000000E+000 -1.5555573E-002 -2.1111137E-005
The end deflection is U(2) on node 10 = 0.0156 mm

There is a small difference - maybe you could repeat the analysis with 20 and 30 nodes to see if approximating the UDL by nodal forces converges to the theoretical answer of 0.015.



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