Beam analysis due to Support Settlement
This lecture is a continuation of our previous lectures on the slope-deflection method.
Here, we're going to see how this method can be used to calculate internal forces
induced in an indeterminate beam when one or more of its supports settle
Suppose we have a continuous beam acting as a bridge structure.
The beam has three supports, two rollers and a pin.
The left roller rests on solid rock, so does the right roller.
The interior support however is a pin connected to a substructure that rests on soft soil.
The beam is made of structural steel with a modulus of elasticity of 200 GPa.
Assume a uniform cross-section for the entire beam
having a moment of inertia of 56000 cm^4.
After the bridge has been constructed and operational for some time,
the soft soil under the structure settles,
causing a downward displacement of the substructure by 50 mm.
Consequently, point B displaces downward by the same amount.
We wish to determine the effect of this displacement on the beam.
More specifically, we want to compute member forces and support reactions caused by the support settlement.
Once these forces are computed,
we can then assess the impact of the settlement on the overall health of the bridge.
To begin our analysis, let us observe the change in the bridge geometry caused by the settlement.
Note that the downward shift at B forces segment AB to rotate clockwise,
and segment BC to rotate counterclockwise.
According to our previous discussion on the slope-deflection method,
this means AB has a differential vertical displacement of negative 50 mm,
and BC has a differential vertical displacement of positive 50 mm.
To see how the slope-deflection method deals with support settlement,
let's write the slope-deflection equations for each beam segment.
For segment AB, we get:
What is Delta here?
It is the differential vertical displacement between the ends of the segment:
In other words, it is our support settlement,
so Delta equals negative 50 mm.
For the sake of consistency,
I have converted the unit of displacement from millimeter to meter here.
For member BC, we have:
Here, Delta is positive 50 mm.
So when using the slope-deflection method,
dealing with support settlement is rather straightforward.
Delta represents the differential settlement between the ends of the segment.
So if one end of this segment settles by 50 mm relative to the other end,
Delta would be 50 mm.
Here is a question for you.
If the left end of the segment settles by 20 mm,
and the right end settles by 50 mm,
what value should we use for Delta?
Since Delta is differential vertical displacement,
we should use 20-50, or negative 30 for Delta.
Why the negative sign?
Because these end-settlements force a clockwise rotation of the segment.
Back to our problem.
Now we can write the joint equilibrium equations and solve them for the unknown rotations.
The equations are:
When solved for theta_A, theta_B and theta_C, we get:
Now we substitute the rotation values in the slope-deflection equations
in order to determine member-end moments.
They are:
Then, we can easily calculate member-end shear forces for each segment.
For segment AB, we have:
For BC, we have:
Finally, we determine the reaction forces at A, B and C due to the support settlement:
Here is the summary of our analysis.
These are the member forces and support reactions due to a support settlement of 50 mm at B.
Let's examine another case.
Here, we wish to analyze a continuous beam subjected to a uniformly distributed load.
The structure has four supports, a pin support, and three roller supports.
Suppose the beam experiences support settlement at B, C and D.
B settles by 20 millimeters,
C settles by 10 millimeters,
and D settles by 40 millimeters.
Similar to the previous problem,
we start by determining differential vertical displacement for each segment.
Segment AB rotates clockwise due to the settlement at B.
So Delta for this segment is negative 20 mm.
Delta for segment BC is 20-10, or positive 10 mm.
And Delta for segment CD is 10-40, or negative 30 mm.
Now if we write the slope-deflection equations for the three beam segments, we get:
Then, the joint equilibrium equations can be written as:
Solving these for the unknown joint rotations, we get:
When we substitute these values in the slope-deflection equations,
we get the following member-end moments:
Now, we can compute member- end shear forces like this:
And here are the support reactions:
So the applied distributed load,
coupled with the support settlements, result in these member forces and support reactions:
See if you can solve the following problems: