145
6
Beam and Frame Analysis:
Force Method—Part III
Solution
1. Degree of indeterminacy is one. We choose the reaction at b, Rb as the redundant force.
P
a b
Rb
The statically determinate primary structure and the redundant force.
2. Establish the compatibility equation. Comparing the previous two figures, we observe that the combined effect of the load P and the reaction Rb must be such that the total vertical displacement at b is zero, which is dictated by the roller support condition of the original problem. Denoting the total displacement at b as Δb, we can express the compatibility equation as
b= b+Rb bbδ =0
where b is the displacement at b due to the applied load and δbb is the displacement at b due to a unit load at b. Together, Rbδbb
represents the displacement at b due to the reaction Rb. The com- bination of b and Rbδbb is based on the principle of superposi- tion, which states that the displacement of a linear structure due to two loads is the superposition of the displacement due to each of the two loads. This principle is illustrated next.
P
a b
Rb
P
a b
a b
Rb Rbδbb
∆b
=
+
∆b = 0
Compatibility equation based on the principle of superposition.
3. Solve for the redundant force. Clearly, the redundant force is expressed by
Rb = – δbb
∆′b
To find b and δbb, we can use the conjugate beam method for each separately, as shown next.
a b
∆b
a b
PL/2EI
Conjugate Beam
L/2 L/2
L/3
Conjugate beam method to find b.
Use the conjugate beam method to find b. The deflection is computed as the moment at b of the conjugate beam.
∆′b = (ΣMb) = EI
1 ( 2 1
2 PL
2 L)(
3 L+
2 L) =
48EI
5PL3 Downward
a b
δbb 1
a b
L/EI
Conjugate Beam 2L/3
Use the conjugate beam method to find δbb.
δbb = (ΣMb) = EI
1 [(
2 1)(L)(L)] (
3 2L) = L3
3EI Downward Rb = –
δbb
∆′b = – 16
5P Upward
4. Find other reaction forces and draw the shear and moment dia- grams. This is achieved through a series of diagrams.
P
a b
5P/16 11P/16
3PL/16
V
M
∆
5PL/32 11P/16
–5P/16
–3PL/16
Inflection Point
L/2 L/2
Reaction, shear, moment, and deflection diagrams.
Example 6.2
Find all reactions of the same beam as in Example 6.1, but choose a differ- ent redundant force. EI is constant.
L/2 L/2
P
a b
Beam statically indeterminate to the first degree.
Solution
There are different ways of establishing a primary structure. For example, inserting a hinge connection at any point along the beam introduces one condition of construction and renders the resulting structure statically deter- minate. We now choose to put the hinge at the fixed end, effectively select- ing the end moment, Ma, as the redundant force.
L/2 L/2
P
a b
Ma
Primary structure and redundant moment Ma.
The compatibility equation is established from the condition that the total rotation at a of the primary structure due to the combined effect of the applied load and the redundant force Ma must be zero, which is required by the fixed end support.
P
a b
P
a b
a b
=
+
θa
Ma
Maθaa θa = 0
Compatibility condition and principle of superposition.
θa.=.θ′a.+.Maθaa.=.0
The conjugate beam method is used to find θ′a and θaa.
a b
M
PL/4
PL/4EI
PL2/16EI Conjugate Beam
PL2/16EI
PL2/16EI PL2/16EI
Conjugate beam for θa.
From the conjugate beam, the rotation at point a is computed as the shear of the conjugate beam at a.
θ′a = (Va) = – ( PL2 16EI)
To find θaa, the following figure applies.
a b
1
a b
M
1/EI
L/6EI 1
L/3EI θaa
L/2EI
Conjugate Beam
Conjugate beam for θaa.
From the conjugate beam, the rotation at point a is computed as the shear of the conjugate beam at a.
θaa = (Va) = ( L 3EI)
The redundant moment is computed from the compatibility equation as
Ma = – θaa
= – 16 θ′a 3PL
This is the same end moment as obtained in Example 6.1. All reaction forces are shown next.
P
a b
5P/16 11P/16
3PL/16
L/2 Solution showing all reaction forces.
Example 6.3
Analyze the indeterminate beam shown next, and draw the shear, moment, and deflection diagrams. EI is constant.
L L w
Statically indeterminate beam with one redundant force.
Solution
We choose the reaction at the center support as the redundant force. The compatibility condition is that the vertical displacement at the center sup- port be zero. The primary structure, deflections at center due to the load, the redundant force, and so forth, are shown next. The resulting computa- tion is self-evident.
L L
w
w
∆c
c
Rc
Rcδcc
=
+
∆c = 0
Principle of superposition used to find compatibility equation.
The compatibility condition is R
c= c+ δ =c cc 0
For such a simple geometry, we can find the deflections from published deflection formulas.
. = w length = =
EI
w L EI
wL
c 5 ( ) EI
384
5 (2 ) 384
5 24
4 4 4
.
P length EI
P L EI
PL
cc ( ) EI
48
(2 )
48 6
3 3 3
δ = = =
Hence
. Rc 5wL Upward
= − 4 ↑
The reaction, shear, moment, and deflection diagrams are shown next.
L L
w c
5wL/4 3wL/8
3wL/8
3wL/8
–3wL/8 5wL/8
–5wL/8
9wL2/128 M
∆
V 3/8L
3/4L wL2/8
Reaction, shear, moment, and deflection diagrams.
Example 6.4
Outline the formulation of the compatibility equation for the beam shown.
L L
w
L
Statically indeterminate beam with two redundant forces.
Solution
We choose the reaction forces at the two internal supports as the redun- dant forces. As a result, the two conditions of compatibility are the vertical
displacements at the internal support points be zero. The superposition of dis- placements involves three loading conditions as shown in the following figure.
L L
w
L
∆1 ∆2
L L
w
L R1δ11 R1δ21
R1
1 2
1 2
L L
w
L R2δ12 R2 R2δ22
1 2
L L
w
L
+
+
=
∆1 = 0 ∆2 = 0
1 2
Superposition of primary structure solutions.
The two compatibility equations are:
R R
R R
0 0
1 1 1 11 2 12
2 2 1 21 2 22
= + δ + δ =
= + δ + δ =
These two equations can be put in the following matrix form.
.
R R
11 12
21 22
1 2
1
2
δ δ
δ δ = −
.
Note that the square matrix at the left-hand side (LHS) is symmetric because of Maxwell’s reciprocal law. For problems with more than two redundant forces, the same procedures apply and the square matrix is always symmetric.
While.we.have.chosen.support.reactions.as.redundant.forces.in.the.pre- ceding. beam. examples,. it. is. sometimes. advantageous. to. choose. internal.
moments.as.the.redundant.forces.as.shown.in.the.frame.example.next.