In the previous lectures,
we showed how the slope-deflection method can be used to analyze continuous beams.
In this lecture,
we're going to examine the use of the method for analyzing frame structures.
Consider a two-story frame structure fixed at the base.
The frame has six joints.
So there are six joint rotations.
Two of these rotations are known.
theta_A and theta_B are zero, since the frame is fixed at A and B.
However, joint rotation at C, D, E and F are unknown.
Also, we consider moment reaction at A and B as unknown.
Let's refer to them as m_A and m_B.
Further, under the applied loads, the frame is going to deform, like this:
Let's refer to the lateral displacement at the top of the first story as Delta_1,
and we use Delta_2 for the horizontal displacement
at the top of the second story relative to the first story.
These displacements are also unknown.
Therefore, we have a total of eight unknowns.
There are four unknown slopes.
Two unknown displacements, and two unknown moment reactions.
We need eight equations for determining these unknowns.
Six of these equations come from the moment equilibrium condition for the joints. They are:
The remaining two equations come from the rigid-body equilibrium condition
applied to two distinct parts of the frame.
More specifically, one of the equations is obtained by cutting the frame at the first level, like this:
And summing the forces in the x-direction, like this:
The other equation comes from cutting the frame at the second level, like this:
Then, summing the forces in the x-direction gives:
Since we can express these shear forces in terms of member-end moments,
all eight equations can be written in terms of the unknown slopes and deflections.
We can then solve the equilibrium equations simultaneously for the unknowns.
Once the slopes and deflections are determined,
we can calculate the member-end moment and shear forces, then determine the support reactions.
Let's go through an example to see this process in action.
Here is our two-story frame.
For the sake of simplicity, we assume a constant EI for the entire structure.
Since the frame has a side sway,
the slope-deflection equations for the vertical frame members embody the non-zero deflection term.
So, for member AC, we have:
For BD, we get:
For CE, we can write:
And for DF, we have:
The deflection term however, is zero in the slope-deflection equations
associated with the horizontal members.
So for CD, we can write:
And for EF, we get:
Here are our six joint equilibrium equations in the expanded form:
We can use the slope-deflection equations to expand the equilibrium equations:
To rewrite them in terms of the unknown slopes and deflections:
As for the remaining two equations,
let's replace each shear force with its corresponding member-end moments.
But how?
How do we express shear in terms of moments?
For a typical vertical frame member, we can draw its free-body diagram like this:
Taking the sum of the moments about point J, we get:
Or:
Then, V_AC becomes:
V_BD is:
V_CE can be written as:
And here is V_DF:
So the last two equilibrium equations can be written this way:
Solving all eight equilibrium equations simultaneously, we get:
When we substitute these slopes and deflections in the slope-deflection equations,
we get the following values for member-end moments:
Now we can calculate member- end shear forces, like this:
Now, we transfer member-end shear forces to the joints,
and apply the equilibrium conditions to each joint until support reactions are determined.
But since moment equilibrium has already been maintained,
at this point, we only need to consider force equilibrium at each joint.
Let's start by transferring member- end shear forces to the joints.
Here is the transfer for member EF:
For member CE:
For DF:
For CD:
For AC:
And for BD:
Now we enforce the equilibrium conditions.
For joint E, we get:
For joint F, we have:
Now we transfer the axial force in member CE, applied to joint E, to joint C.
Then, we balance the forces at joint C.
Transferring the axial force in member DF, applied to joint F, to joint D.
And balance the forces at joint D.
Now let's transfer the axial force in member AC, applied to joint C, to joint A.
And do the same for member BD.
Finally, apply the equilibrium conditions to joint A,
to determine the support reactions at A.
Do the same at B in order to determine the reaction forces at B.
We've already determined moment reactions at A and B.
Here are the remaining support reactions:
Now it's your turn.