Support reactions of a rigid, rectangular steel frame

Tags:
1. May 17, 2017

InYoung

1. The problem statement, all variables and given/known data
Solve for support reactions of a rigid, rectangular steel frame sitting on rubber pads at the four corners (See attached), when a shear force and a moment is applied at some point on the frame. All the known variables are shown in black while the unknown are in red. It is a two-dimensional problem.

2. Relevant equations
Force balances : Sum of X = T & Sum of Y = 0
Moment balance about point 1 (upper left hand corner) : (X3+X4)W + (Y2+Y3)L - M = 0

3. The attempt at a solution
There are eight unknown reaction forces and three equilibrium equations. It seems like a statically indeterminate structure to the degree of 5 (8 unknowns - 3 equations). Could anyone offer help on how to find the appropriate compatibility equations? Thank you!

Attached Files:

• Question.png
File size:
15 KB
Views:
47
2. May 18, 2017

haruspex

The four pads will deform, and they are identical.
Each extent of deformation relates to the shear force.
What can you say about the relationships between the different extents of deformation?

3. May 19, 2017

InYoung

Umm, can I say that the x-direction deformation of pad #1 and that of pad #2 have to be the same given that the steel frame is rigid? Also for pads #3 and 4. As for the y-direction deformations would be equal for pads 1 and 4 (and for pads 2 and 3). Are these correct assumptions to make?

4. May 19, 2017

haruspex

For small deformations, yes.
You can also get a relationship between the two different x deformations and the two different y deformations. Think about how the frame as a whole can move. How many degrees of freedom?

5. May 19, 2017

InYoung

Three degrees of freedom: x and y translations and rotation about z-axis. In the case of translations, the two different x deformations must be identical and the two different y deformations also must be identical. In the case of rotation about z-axis, the two different x deformations must be equal and opposite (also true for the two different y deformations). Is this correct?

Thank you for the helpful pointers.

6. May 19, 2017

haruspex

For a pure rotation, yes, but in general it will not be.
More interestingly, there is a relationship between the x deformations and the y deformations.

7. May 19, 2017

InYoung

I'm afraid I can't see what that relationship is. Could you give me some pointers? Thank you!

8. May 19, 2017

haruspex

Work in terms of the displacement of the frame. If its centre is displaced (Δx, Δy), and its orientation displaced Δθ, you can compute all the deformations from those.

9. May 19, 2017

InYoung

So for a general frame displacement, if I look at the pure rotational deformations, these would be due to the differences in the shear forces (Y2 - Y1) and (X2 - X1). These deformations would be 'constrained' by the given geometry of the frame, as shown in the attached photo. (Again assuming small displacements). Is this correct?

10. May 19, 2017

haruspex

Yes.
Relate tan theta to the given dimensions of the frame.
You can represent the whole movement of the frame as a rotational displacement, plus an x displacement, plus a y displacement.

11. May 21, 2017

InYoung

Thank you very much for your help.