Analysis of beams with one or more internal hinges
When designing structures it is often necessary to join structural members along their lengths.
It is generally useful to classify such a connection,
also called a splice,
as either a moment connection or a shear connection.
A moment splice rigidly connects the two beam segments,
such that an internal bending moment, as well as a shear force, can develop at the joint.
In a shear splice, however, no internal moment can develop at the joint.
The joint acts as an internal hinge.
It can only support a shear force.
In this section we're going to focus on the analysis of beams that embody one or more internal hinges.
Consider this beam:
It is fixed at the left end and rests on a roller support at the right end.
The beam has an internal hinge at point C.
The structure has four unknown support reactions;
three at the fixed end and one at the roller end.
Although we can write three equilibrium equations for the entire structure
but we cannot solve them for the unknowns,
because we have only three equations but four unknowns.
Does this mean the structure is statically indeterminate?
This would have been the case if the beam did not have an internal hinge.
But the hinge at point C does make the structure statically determinate,
that is, we should be able to determine the support reactions solely using the static equilibrium equations.
Here is how.
The analysis of a beam with an internal hinge involves separating the beam segments at the hinge,
formulating three equilibrium equations for each segment,
then solving them simultaneously for the unknown reaction forces
and the internal shear force at the hinge.
Here is the free-body diagram for each beam segment:
Observe that the internal shear force at the hinge is drawn on each beam segment.
On the left segment the force is shown to act downward.
On the right segment the force is drawn upward.
These two arrows must always be drawn in opposing directions.
Also note that I have placed an internal axial force at point C.
Although most beams do not carry such a force,
for completeness we need to show the force on the free-body diagram.
Similar to the internal shear force, the internal axial force at point C
must appear in a pair acting in opposite directions.
The equilibrium equations for the left beam segment are:
The equilibrium equations for the right beam segment are:
Therefore we have a total of six equations and six unknowns.
The unknowns are: A_x, A_y, M, V_c, C_x and B_y.
Solving the equations for the unknowns, we get:
A_x=0
A_y=2 kN,
M=-16 kN
B_y=2 kN,
C_x=0
V_c=2 kN.
What happens if the beam has multiple internal hinges?
How do we analyze such a beam?
Here is a beam with two internal hinges.
The beam rests on four supports, a pin support and three roller supports.
The number of support reactions, therefore, is five;
we have two reactions at the pin support and one reaction at each roller support.
The two hinges divide the beam into three segments.
When we cut the beam segments at the hinges,
in effect we are introducing four more unknowns.
We are introducing a shear force and an axial force at each cut.
Now we have a total of nine unknowns.
There are five support reactions and four internal forces.
And since we have three beam segments,
and for each segment we can write three equilibrium equations,
there are nine equations.
So we can solve the equations for the unknowns:
Here are the equilibrium equations for AE:
Here are the equations for segment EF:
And here are the equations for segment DE:
Solving these equations simultaneously, we get:
A_x=0
A_y=2.14 kN
B_y=-7.47 kN
E_x=0
V_e=5.33
C_y=21.33 kN
F_x=0
V_f=0
And D_y=0.