Direct Stiffness Method for a distributed load

[lawa44]

I'm working on a question which asks to determine the deflection, curvature, forces and moments of a simply supported beam with a distributed load. diagram shown in here http://imgur.com/a/wpI4kInitially, I've done the calculation with 1 plane beam element with 2 nodes. At L = 0 and L = 3. But that doesn't give me the deflection of the beam at L = x/2.

So I'm trying to use 2 elements with 3 nodes as shown here.

I've reduced the matrix down to a 4 x 4 with {theta_1; v_2; theta_2; theta_3} as unknowns.

The result i got were:

• theta_1 = -0.0462
• v_2 = -0.0865
• theta_2 = 0
• theta_3 = 0.0462

The results from the 2 node analysis were:

• Q_1 = 3N
• theta_1 = -0.0231
• Q_2 = 3N
• theta_2 = 0.0231

Which do not agree with the results from the 3 node analysis.

Moreover, from using the Euler beam equation I calculated a deflection result of -21.6mm which does not agree with v_2 either.

Because of the difference in values, I got from the 3 node analysis, I have yet to substitute the values back into calculate the other nodal values.

Did I make a mistake?

Is it correct to use L/2 instead of L for my three node equivalent loads?

i.e. -w(L/2)/2 instead of -wL/2 as was the case in the 2 node analysis.

 

[Future Bruno]

Yes, it is correct to use L/2 instead of L for your three node equivalent loads. The reason for this is because the three node analysis takes into account the effect of the distributed load being split over two nodes instead of being concentrated at one node. In other words, the distributed load is being shared between the two nodes on either side of the midpoint. In addition, you may have made a mistake in your calculations when solving for the nodal values. Make sure you are using the correct equations and boundary conditions when solving for each of the unknowns.