Frames are structural systems consisting of two or more elements.
The type of loads applied to frames often subject the structural elements to axial force,
as well as shear force and bending moment.
Therefore a frame member should be viewed as a beam and a truss element at the same time.
When calculating frame displacements,
it may be necessary to consider both axial deformation and bending.
I say maybe here, because for many frames, displacement due to axial member forces
is significantly less than the displacement due to bending.
Consequently, in practice it is not uncommon to ignore axial deformation
when calculating displacements in frames.
In this lecture however, for the sake of completeness,
I am going to include both axial deformation and bending in dealing with frames.
This lecture draws upon a basic understanding of the virtual work method,
as was presented in lectures SA21 and SA22.
Please review those lectures before continuing with this one.
Consider this simple frame structure.
It consists of three members.
The frame is subjected to a horizontal concentrated load,
and a uniformly distributed load, as shown.
We would like to determine the horizontal displacement at C.
According to the method of virtual work, we can solve for the unknown displacement
by equating the external virtual work to the internal virtual work,
done by a virtual unit load placed at C.
Let W*_e stand for external virtual work,
and W*_i stand for internal virtual work.
Then we can write:
W*_e = W*_i
The expression for external virtual work is easy to obtain.
It equals to 1* times Delta,
where 1* is the magnitude of the virtual load,
and Delta is the horizontal displacement at C.
The expression for internal virtual work however is a bit more involved.
The structure's internal virtual work
equal to the sum of the internal virtual work for the individual members.
So we can write:
Further, the internal virtual work for each member consists of two parts.
A part due to axial deformation, and a part due to bending.
So let's rewrite the equation this way:
Where W*^a stands for virtual work due to axial deformation,
and W*^b stands for virtual work due to bending.
To calculate the individual terms in this equation,
we need to know the member forces.
To get these forces, we need to analyze the structure.
This can be done easily since the frame is statically determinate.
All right, we are almost ready to start calculating the internal virtual work.
But before we do so,
let's determine the member forces when the frame is subjected to the unit virtual load at C.
Here's the analysis results:
Now we're ready to calculate the internal virtual work done in each member.
For member AB,
the internal virtual work has two parts:
A part due to axial deformation,
and a part due to bending.
To determine virtual work due to axial deformation,
we treat the member as a truss element, so we get:
W*^a=f*_ab times delta_ab,
where f*_ab is the member's axial force due to the virtual unit load,
and delta_ab is the member's elongation due to the real loads.
Here is f*_ab,
it equals two.
To determine delta_ab, we use Hooke's law.
Here, F_ab is the axial force in the member due to the real loads.
L is the member's length,
E is the modulus of elasticity of the material,
and A is the member's cross-sectional area.
So we can write:
Then the member's internal virtual work due to axial deformation becomes:
Now let's write the expression for internal virtual work due to bending.
Here's the free-body diagram for the member as a beam.
Using the diagram, we can write the bending moment equation for the member as:
M(x)=Px
For m*, we have this free-body diagram:
So, m*(x)=x
Then, internal virtual work due to bending equals:
Assuming E and I are constant, we get:
Therefore total internal virtual work for member AB can be written as:
Note that this term is due to axial deformation,
and this is due to bending:
Now let's repeat those steps for member BC.
Here's the free-body diagram for the member under the applied loads.
And here is the member's free-body diagram when the virtual unit load is placed at C.
Internal virtual work for the member is:
Since the member does not carry any axial force, its axial deformation is zero.
So, this term is zero.
To determine the internal virtual work due to bending,
we need to have M and m*.
Here is M(x),
and here is m*(x).
So the internal virtual work for BC is:
Assuming all the members have the same E, A and I, we can write:
Finally for member CD, we have the following free body diagrams:
Since there is no bending moment in the member,
internal virtual work is going to be due to axial deformation only.
Axial deformation, delta, equals to:
Therefore internal virtual work equals:
Now we can express the structure's internal virtual work
as the sum of the three internal virtual work expressions.
For member AB, we have:
For BC we have:
And for CD we have:
Adding the three expressions we get:
Then according to the virtual work principle we can write:
Hence the horizontal displacement at C is:
Let's wrap up this lecture with a numeric example.
Here's a simple frame:
Both members have the same cross- sectional area and modulus of elasticity.
We want to determine the horizontal displacement at A, and the joint rotation at B.
We shall start by analyzing the frame,
and determining the member forces under the applied loads.
So axial force in AB is -200 N,
and bending moment in the member is zero.
Axial force in BC is -200 N,
and bending moment in the member is defined using two equations.
To determine the horizontal displacement at A,
we place a virtual unit load at A.
The direction of the load indicates the assumed direction for the displacement.
Here I am assuming the roller displaces to the right.
To determine the member forces, the frame needs to be analyzed.
Here are the member forces in AB:
And here are the member forces in BC:
Then the principle of virtual work can be written as:
This term is zero, since M_ab is zero.
Simplifying the equation, we get:
We assumed that A displaced to the right.
The negative sign however indicates that the direction of the displacement is opposite to the assumed one.
So the joint displaces to the left by 2.88 mm.
To determine the rotation at joint B,
place a virtual unit moment at B like this:
Then analyze the frame in order to determine the member forces:
The equation representing the principle of virtual work can be written as:
Therefore the rotation of joint B is:
Here are couple of practice problems: