Partially Fixed Cantilever

[egc]

I need to aproximate the deflection stiffness of an "L" shape frame (one with a vertical section supporting a shorter horizontal section - see attached *.png). The existing steel supporting the vertical member (not shown in picture) is assumed to provide complete fixity.

I am going to assume small rotations and deformations to keep it all in the elastic range. For discussion purposes, assume the point load at the end of the horizontal member. I'm using Stadd Pro to validate my equations and I'm clearly missing something; I'm assuming its the partial fixity of the cantilevered horizontal member.

Where:
Member S1 has length = L1 and IZ = I1
Member S2 has length = L2 and IZ = I2

Then the vertical deflection at point P2 can be approximated as:
[Part 1] [Part 2]
Delta = (P*L2^3 / (3*E*I2)) + (2 * L2 * sin(P * L2 * L1 / (E * I1)))

Part 1 is simply the fixed cantilever deflection w/ a point load at the end.

Part 2 is the deflection caused by rotation at P1 from the moment applied at P1. The 2 at the front of part 2 is my assumption to account for the partially fixity of member S1.

Where am I going wrong?

 

[bradleyelwood]

egc,

The first part is right, just a cantilever beam call it Delta1.

The second part should be Delta2 = Theta (L2) where Theta = (Moment (L1))/EI1

And the Moment at P1 is just P(L2)

Then Delta2 = (P(L2)(L2)(L1))/EI1

Total vertical deflection a P2 is Delta1 + Delta2

 

[egc]

So your saying that since the angle is small to just use the fact that sin(theta) = theta (Theta is in radians)?

 

[bradleyelwood]

Yes,

Unless you have deflections which are large compared to the beam lengths then sine Theta = Theta, and Theta is in radians is a good assumption.

If you really think about it, all of our BASIC assumptions of beam deflection theory are that loading remains orthogonal to beam centerlines.

 

[egc]

Well I found my biggest issue as I was using the sin(rads) when the code I'm using expects angles in degs. I fixed that and then compared to sin(tehta) = theta and the results are the same (surprise!).

But my calculations and Staad still don't agree. P1 actually deflects horizontally a small amount. This would add an additional rigid body rotation to the P2 deflection so I'm going to add that in and see what happens.

 

[BAretired]

You asked where you were going wrong.

Delta = (P*L2^3 / (3*E*I2)) + (2 * L2 * sin(P * L2 * L1 / (E * I1)))

This should read:
Delta = (P*L2^3 / (3*E*I2)) + (L2 * sin(P * L2 * L1 / (E * I1)))
which for a small angle is approximately equal to:
Delta = (P*L2^3 / (3*E*I2)) + (L2 * P * L2 * L1 / (E * I1))

You had a misplaced factor of 2 in the second expression.

BA

 

[egc]

Thanks but I had already removed the 2.

 

[BAretired]

The angle has to be in radians, not degrees. If the angle is in degrees, sin(theta) is not equal to theta.

P1 deflects horizontally by M1*L1^2/2E*I1 where M1 = P*L2. The horizontal deflection of P1 does not add a "rigid body rotation to the P2 deflection" as you stated earlier. It adds a horizontal rigid body translation to point P2 but has no effect on the vertical deflection of point P2.

BA

 

[egc]

I forgot that we also have the lengthing of the vertical member so the equation is actually:
Delta = P*L2^3 / (3*E*I2) + L2*P*L2*L1/(E*I1) + P*L1/(A1 *E)

Now I'm within 5% of Staad. Why I can't get closer I don't know.

 

[BAretired]

My guess would be that Staad is including shear deformation.

BA

 

[egc]

Yes it was. Once I turned it off, we're exactly the same. Thanks BA and Brad.

 

[rb1957]

"partially fixed" cantilever always sounds to me like "partially pregnant"