2nd Degree Inderteminacy for Structure Using Force Method

[Tygra]

Homework Statement

Calculating Deflections

Relevant Equations

In question

Dear all

I am trying to find the support reactions on the following structure:

The frame member in the horizontal is 8m long and in the vertical the member is 5m.

To do this I am using the force method (or unit load method or virtual work method).

Firstly, I removing the redundants at the pinned support to make a statically determinant frame as shown below:

Next, I apply units loads in place of the pinned support that looks like this:

Firstly, lets consider the horizontal unit load. I need the displacement in the parallel and perpendicular directions as a result of this horizontal unit load.

Calculating the horizontal displacement as a result of this load is no problem. It is simply the sumof the integration of the bending moments.

So, the moment functions are: Mx = -1*x and Mx = 5. Hence, the integration to compute the delection in the horizontal direction is

The area where I am a little stuck is computing the vertical deflection as a result of the horizontal unit load.

If you see here from the software the vertical displacement is 4.074 mm.

So my question is: how do I calculate this displacement using the force method?


Many thanks in advance.

 

[Lnewqban]

I can follow your description, but it is impossible to see details of the posted diagrams.

 

[Tygra]

Hi Lnewqban,

Sorry for this.

Is this any better?

Frame:

Statically Determinant Frame

Application of Unit Loads:

Bending Moment diagram for horizontal unit load:

Displacement for horizontal unit load:

Bending moment diagram for vertical unit load:

Displacement for vertical unit load:

 

[Lnewqban]

The numbers can't be clearly seen still.
I see that the diagrams shown the deformation do not match how the structure would really deform.
Hinge 3 can't move horizontally and joint 2 can't move upwards.
The beam located next to 1 can't remain horizontal due to the distributed load.
The angle of joint 2 should remain more or less 90°.