In the previous lecture, we learned that given a beam segment, like this,
we can express its end moments
in terms of the end rotations, like this:
We call these slope-deflection equations.
In this lecture, we're going to come up with the slope-deflection equations
when the beam segment is subjected to a distributed load, like this:
We are going to follow our adopted sign convention,
and show the end rotations and moments in the positive,
counterclockwise direction, like this:
We need to write two slope-deflection equations.
We want one equation for defining M_AB in terms of theta_A, theta_B and w,
And another equation for expressing M_BA in a similar manner.
To get these equations,
we are going to employ the superposition strategy we adopted in the previous lecture.
We start by writing end-rotations for a simply supported beam
under three separate loading cases,
then we superimpose the three cases to arrive at our target equations.
Here are the three loading cases we need to consider.
A simply supported beam subjected to a positive moment at A,
the same beam subjected to a positive moment at B,
and the beam subjected to a uniformly distributed load.
For each case, first we compute member-end rotations.
Then we combine the individual rotations at each end to come up with the total end rotations.
As you'll see in a minute,
the right-hand side of these equations can be written in terms of M_AB, M_BA, and w.
This means we can solve them for the end moments in terms of the end rotations and the load magnitude.
Here are the details. Loading Case 1:
The load is a counterclockwise moment placed at A, like this:
The beam is going to deflect upward, causing a counterclockwise rotation at A,
and the clockwise rotation at B.
We refer to the moment as M_AB,
and denote the end rotations as theta_A1 and theta_B1.
We can use the virtual work method to calculate these rotations. Here they are:
Loading Case 2:
The load is a counterclockwise moment placed at B.
The bending moment is labeled M_BA.
And the end rotations are labeled theta_A2 and theta_B2.
Similar to Case 1, using the virtual work method,
we can easily determine these end rotations.
Loading Case 3:
The load is uniformly distributed over the entire length of the beam, like this:
Similar to the previous cases,
we use the virtual work method to calculate these end rotations. They are:
Here is a summary of the three loading cases:
If we superimpose these diagrams, we get this diagram:
This means, rotation A, here, equals to the sum of these three rotations, or:
Similarly, rotation at B can be written as:
In the expanded form, these equations become:
Let's treat M_AB and M_BA as unknowns and solve for them:
To do so, first arrange the two equations like this:
Then multiply the first equation by two:
And add it to the second equation:
This gives us:
Now solve for M_AB:
Substituting this expression for M_AB in one of these equations, we get:
So for a beam segment under a uniformly distributed load,
the slope-deflection equations are:
Let's compare these equations with the ones we got for a load-free beam segment:
Here are the two sets of equations side by side:
The difference between the two sets is these terms:
They can be viewed as the effect of the applied load on the slope-deflection equations:
We refer to them as fixed-end moments.
For a uniformly distributed load, fixed-end moments are:
positive wL^2/12 and negative wL^2/12.
Obviously different loads give us different fixed-end moments.
For example,
if we subject the beam segment to a downward concentrated load of P at mid-span,
fixed-end moments become:
Generically, we can write these slope-deflection equations as,
where FEM_AB stands for fixed-end moment at end A,
and FEM_BA represents fixed-end moment at end B.
So, for a uniformly distributed load of w,
FEM_AB is positive wL^2/12,
and FEM_BA is negative WL^2/12.
Note that here too, we are using counterclockwise direction as positive.
So this represents a counterclockwise moment, whereas this represents a clockwise moment.
Let's use an example to see these equations in action.
Here is a continuous beam subjected to a uniformly distributed load.
To analyze the beam using the slope-deflection method,
we are going to separate the joints from the members, like this:
For each beam segment, we show the member-end moments in the positive direction, like this:
Then we write the slope-deflection equations for each segment:
For segment AB, we get:
For segment BC, we have:
Now show the member-end moments acting at each joint:
Then write the joint equilibrium equations.
We have three of them:
Substituting the slope-deflection expressions in the above equations, we get:
Solving these equations simultaneously for the unknown rotations, we get:
Now we substitute these values back in the slope-deflection equations
to determine the member-end moments:
Knowing the member-end moments, we can easily determine member-end shear forces.
For segment AB, we have:
For segment BC, we have:
Now we transfer member-end shear forces to the joints.
So the support reactions are:
Or:
In the next lecture, we will discuss the formulation of fixed end-moments
for various load shapes in more details.
Here are couple of exercise problems involving
uniformly distributed loads and joint moments only.
Analyze the following indeterminate structures
using the slope-deflection method.