Statically indeterminate beams and shafts.docx

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1. Statically Indeterminate Beams and Shafts
Alright, so today we are focusing on statically indeterminate beams and shafts.
We’ve already touched on statically determinate systems in earlier sections, but
now we are going to deal with structures where the unknown reactions are more
than the available equilibrium equations.
What does "statically indeterminate" mean?
It means that for a structure to be stable, we have more support reactions than we
can solve using just the equations of equilibrium: ∑F
x
=0, ∑F
y
=0, and ∑M=0.
When that happens, we call the system statically indeterminate.
Take a look at Figure 12-32:
In Figure 12-32a, we have a beam with a load P and supports at points A
and B.
There are four unknown reactions: the vertical reactions at A (A
y), B
y, the
horizontal reaction at A (A
x), and the moment M
A at support A.
However, we only have three equilibrium equations. That means one of
these reactions is redundant, making the system indeterminate to the first
degree. We call this extra reaction a redundant.
For this beam, any one of the reactions could be chosen as the redundant reaction.
If we removed, say, A
y, the beam would still be in equilibrium, but we need to use
additional methods (like compatibility conditions) to solve for it.

2. Now, in Figure 12-33, we have a more complex beam that is continuous and spans
multiple supports:
There are five unknown reactions at supports A, B, C, and D for vertical
forces and at A
x for the horizontal force.
But again, we only have three equilibrium equations, so this beam is
indeterminate to the second degree. This means two reactions are redundant,
and we need extra conditions to solve for them.
In this case, we can choose any two reactions—like B
y and C
y—as redundant.
What do we do next?
To solve for the unknown reactions, we need more than just equilibrium. We have
to introduce additional relationships, called compatibility conditions, that relate
how the beam deforms. We will cover these methods—like integration and
superposition—in the following sections.
Does that make sense so far? Let me know if you need me to clarify any part of the
diagrams or the concepts.
That’s a simplified explanation. We used Figures 12-32 and 12-33 to explain first
and second-degree indeterminate beams, and I highlighted the key points on
redundants and how we need more than just equilibrium equations to solve these
problems.
Let me now explain the next part using your style and for a lecture environment.
We are going to focus on Method of Integration for solving statically
indeterminate beams and shafts.
Statically Indeterminate Beams and Shafts—Method
of Integration
Now that we know what statically indeterminate beams are, let’s discuss how we
actually solve for the unknown reactions. One common method is called the
method of integration.

3. Here’s the basic idea: when we have a statically indeterminate beam, we can’t just
use equilibrium equations like we normally would. We need to consider how the
beam deforms, and for that, we use the differential equation of the elastic curve.
It’s written like this:
d
2
v
dx
2
=
M
EI
where:
v is the deflection,
M is the internal bending moment at a point,
EI is the flexural rigidity (E is the modulus of elasticity, I is the moment of
inertia).
Now, here’s where the method of integration comes in. We take this equation and
integrate it twice. Why? Because integrating the second derivative of deflection
will give us the slope, and then integrating again will give us the deflection itself.
However, if the beam is statically indeterminate, the moment M will depend on the
unknown redundants—the extra reactions we talked about earlier. So, we’ll have to
express M in terms of those unknowns, and that’s what makes the process more
complex.
After we integrate, we end up with two constants of integration, and those
constants, along with the redundants, are determined by applying boundary
conditions (like how the beam is supported) and compatibility conditions (how the
beam deforms).
Let me give you an example of how this works. Imagine a beam supporting a
bridge deck, like in the picture below the text. This is a classic case of a statically
indeterminate beam, because the deck is supported at multiple points. We can’t just
use equilibrium equations to find the reactions at each support. We need to use the
method of integration to calculate how the beam deflects under load, and from
there, determine the reactions.
In the next sections, we’ll walk through detailed examples using this method step
by step, so you can see how the integrations are carried out and how we apply the
boundary and compatibility conditions to solve for the unknowns.

4. So, the method of integration is a systematic way to solve these more complex
problems by focusing on the beam’s deformation, not just the forces acting on it.
The key takeaway is that we use the elastic curve equation, integrate it twice, and
then apply additional conditions to solve for everything.
Let’s now discuss another important method for solving statically indeterminate
beams—the Method of Superposition. This method is particularly useful for
beams and shafts subjected to redundant loading, such as multiple supports.
Statically Indeterminate Beams and Shafts—Method of Superposition
The method of superposition is based on the idea of breaking down the original
problem into simpler, more manageable problems. To apply it, we first remove the
redundant support reactions (the extra reactions beyond those solvable by
equilibrium equations). Once we do that, we treat the beam as a simpler structure
—called the primary beam—which is statically determinate.
Take a look at Figure 12-41(a). Here, we have an actual beam loaded by a force P,
with two supports at points A and B. This is
a statically indeterminate beam, so we
can’t solve it directly using equilibrium
alone.
1.Removing the redundant
reaction: Let’s say we treat the
vertical reaction at B
y as redundant,
and we remove it from the beam.
Now, we have a statically
determinate beam, shown in Figure
12-41(b). This is the primary beam.
2.Adding the redundant force back:
After solving for the primary beam,
we then bring back the redundant
reaction by adding the same
redundant force B
y but applying it
separately to the beam, as shown in
Figure 12-41(c). The beam will
deform due to this added force, and

5. this deformation must be compatible with the actual conditions at the
supports.
The important step here is to ensure that the displacements due to the redundant
force match with the original system’s conditions. In this case, the displacement at
point B must be zero in the final beam configuration. We write a compatibility
equation for this condition:
0=−v
B
1
+v
B
r
Here, v
B
1 is the displacement at point B from the primary beam, and v
B
r is the
displacement due to the redundant reaction B
y.
From Appendix C, we can directly find the displacements:
v
B
1
=
5PL
3
48EI
,v
B
r
=
B
yL
3
3EI
By substituting these into the compatibility equation, we solve for the unknown
redundant reaction B
y:
0=
5PL
3
48EI
+
B
yL
3
3EI
Solving for B
y, we get:
B
y=
5
16
P
Now that we know B
y, we can determine the remaining reactions at support A
using the equilibrium equations. From the free-body diagram in Figure 12-41(d),
we get:
A
y=
11
16
P,M
A=
3
16
PL
So, the method of superposition allows us to break down
the problem by removing redundant forces and then
solving the beam in two steps: first by analyzing the
primary beam and then by applying the redundant force.
It’s a powerful method because it simplifies complex
systems into solvable parts while ensuring that the
conditions of the original beam are met.
Any questions about how this method works? Would you
like to dive into another example or a different method?

6. Let’s continue explaining Figure 12-42 and this next step in the method of
superposition, while
keeping things simple for a lecture-style explanation.
Statically Indeterminate Beams and Shafts—Method
of Superposition
In the previous example, we removed a vertical reaction as a redundant force.
Now, in Figure 12-42, we are removing a moment at point A as a redundant force.
This gives us another way to handle statically indeterminate systems by treating
moments as redundants.
1.Figure 12-42(a): This is the actual beam, with load P applied at the
midpoint and supports at points A and B. The beam is statically
indeterminate because we have both the vertical reactions and the moment at
A that we need to solve for.
2.Figure 12-42(b): Here, we have removed the moment M
A, treating it as the
redundant. With the moment removed, the beam becomes statically
determinate, which allows us to solve for the reactions and internal moments
based only on the load P.
Now, after we’ve solved for the primary beam, we need to reintroduce the moment
M
A. This is shown in Figure 12-42(c). When we apply the moment M
A, the beam
will rotate at point A, and we need to ensure that this rotation matches the original
conditions. In this case, we know that the slope at point A should be zero (the
beam is fixed at that point).
The compatibility equation we write here is based on the slope at point A. We want
the total slope (rotation) at A, caused by both the load P and the moment M
A, to
equal zero. Mathematically, this is expressed as:
0=θ
A
+θ
A
r
Where:
θ
A is the slope due to the load P,

7. θ
A
r is the slope caused by the redundant moment M
A.
Using the equations from Appendix C, we can directly substitute the slopes:
θ
A
=
PL
2
16EI
,θ
A
r
=
M
A
L
3EI
Substituting these values into the compatibility equation:
0=
PL
2
16EI
+
M
AL
3EI
Now we solve for the redundant moment M
A:
M
A=
−3
16
PL
What does this result mean? The negative sign indicates that the moment M
A acts
in the opposite direction to the one assumed initially in the diagram. So, in Figure
12-42(c), M
A will act to resist the rotation caused by the load P, bringing the slope
at A back to zero.
This is essentially how we apply the method of superposition when dealing with
moments as redundant forces. We first remove the redundant moment, solve the
primary beam, then add back the moment and ensure compatibility by checking
that the slope or deflection conditions are met.
Let's now go over Figure 12-43, which gives us
another example using the method of
superposition for a beam that is indeterminate to
the second degree. This example shows how we
can solve a system where we need to use two
redundant forces and two compatibility
conditions.
In this case, the beam shown in Figure 12-43(a)
is indeterminate to the second degree, meaning
we have two unknowns that can't be solved using
equilibrium equations alone. The load P is
distributed at three points: P
1, P
2, and P
3, with roller supports at points B and C.

8. 1.Removing the redundants: To simplify this problem, we treat the vertical
reactions at B
y and C
y as redundant. We remove these reactions, as shown in
Figure 12-43(b), and what remains is the primary beam.
2.Primary beam deformation: Without the redundant reactions, the primary
beam will deform due to the external loads, causing vertical displacements at
points B and C. The displacements at these points are denoted as v
B
1 and v
C
1.
3.Adding back redundant forces: Now, we add back the redundant reactions
B
y and C
y, one at a time. In Figure 12-43(c), we see the effect of applying
the redundant force B
y alone. This force causes additional displacement at
point B, denoted as v
B
r.
Similarly, in Figure 12-43(d), we add the redundant force C
y, which causes
additional displacement at point C, denoted as v
C
r.
Compatibility Conditions
The goal is to ensure that the total displacements at points B and C in the actual
beam configuration are zero, which leads us to the compatibility equations. We
express these conditions mathematically by summing the displacements from the
primary beam and the displacements caused by the redundant forces:
At point B:
0=v
B
1
+v
B
r
(12-20)
At point C:
0=v
C
1
+v
C
r
(12-21)
Here, v
B
1 and v
C
1 are the displacements of the primary beam due to the external
loads, and v
B
r and v
C
r are the displacements caused by the redundant reactions at B
and C, respectively.
Solving the System
The displacements v
B
r and v
C
r depend on the unknown redundant forces B
y and C
y.
By substituting the known values from the primary beam and expressing the
displacements in terms of B
y and C
y, we can solve these two equations (Equations
12-20 and 12-21) simultaneously to find the values of the redundant reactions.

9. Once we have solved for B
y and C
y, we can apply these values to determine all
other reactions and internal forces in the beam using the equilibrium equations.
Summary
In this method, we systematically:
Remove the redundant reactions to create a statically determinate primary
beam.
Add the redundant forces one at a time, ensuring the total displacement at
the points of interest is zero by applying compatibility conditions.
Solve the system of equations to find the redundant forces.
This approach simplifies complex problems into solvable parts while ensuring that
the conditions of the original beam are met.