ECM2109: Structures Deflection of beams We have previously shown that the curvature is proportional to the rate of change of , i.e.,
Figure 15 below shows the deflected shape of a cantilever. v represents the deflection of the beam. Then the slope of the deflection curve can be given as
d
+d v
v+dv
Figure 15: Deflections in a cantilever beam
Assuming that deflections and curvatures are very small (and it holds true for many structures in practice), we can introduce the approximation for simplifying the analysis of beams. Therefore
Relating to curvature,
For a linear- l s ic m
i l following Hook ’s l w
This differential equation defines the deflected shape of the beam. It can be integrated twice to obtain the deflection v knowing the bending moment variation M(x) and the flexural rigidity (EI). We have also previously seen the following two relations between the load (w), shear force (V) and bending moment (M).
Therefore, substituting from above and assuming prismatic beams, i.e., the moment of inertia I does not vary along the length of the beam, n The derivatives of the deflection v are often written in the following simple form.
Therefore the above equations between M, V, EI and v are often represented as follows.
The deflection of a beam is evaluated by integrating the above expressions and then applying boundary conditions (e.g., v=0 at a support) to evaluate any constants of integration. Example 4 Evaluate the deflected shape and the maximum deflection for the beam in Figure 16 w A
B
Figure 16: Simply-supported beam with udl w
We evaluated the bending moment and shear force distribution for this beam in a previous lecture (see Example 3). From Figure 10 (in an earlier handout),
To find deflections,
Integrating the above expression
where A and B are constants from the integration. Applying the boundary conditions that v=0 @ x=0, we find that B=0. Similarly, v=0 @ x=L implies
fo The maximum deflection will be at mid-span due to symmetry. It can be evaluated by substituting x=L/2 in above expression for v.
We can also evaluate the slope of the beam at any point along its length. For instance, at the left support, x=0,
Similarly at the right support,
Slopes at the two supports are of equal magnitudes but with opposite signs (as would be expected from symmetry). Example 5 Evaluate the deflection at the free end of the cantilever of length L shown in Figure 17. w A
B Figure 17: Cantilever beam with udl w
The support reaction can be evaluated from vertical equilibrium as
Moment at the fixed end is also obtained from moment equilibrium as
Figure 18 shows a free-body diagram of a portion of the beam. From equilibrium of forces and moments,
w A
M
MA
x
V
RA Figure 18: Free-body diagram showing reactions, shear force (V) and bending moment (M)
Substituting for M in
and integrating,
Applying boundary conditions, v=0 @ x=0 => B=0. Then using v’=0 @ x=0 gives A=0. Therefore expression for deflection is given by
Maximum deflection is at the free end (x=L) and is given by
Example 6 Evaluate the deflection at the free end of the cantilever of length L shown in Figure 19. P A
B Figure 19: Cantilever beam with concentrated load P at free end
The support reaction and moment at the fixed end can be evaluated as for the previous example.
Figure 20 shows a free-body diagram of a portion of the beam. A
M
MA
x
V
RA Figure 20: Free-body diagram showing reactions, shear force (V) and bending moment (M)
From equilibrium of forces and moments,
Substituting for M in
and integrating,
Applying boundary conditions, v=0 @ x=0 => B=0. Then using v’=0 @ x=0 gives A=0. Therefore expression for deflection is given by
Maximum deflection is at the free end (x=L) and is given by
Example 7 Evaluate the deflection at the free end of the cantilever of length L with a linearly varying distributed load as shown in Figure 21. wo A
B Figure 21: Cantilever beam with linearly varying distributed load
For this beam, let us evaluate the deflection by integrating this example,
where w is the distributed load. In
o Substituting for q, n in g
ng
Shear force (V) should be zero at the free end, i.e., @x=L. Use this boundary condition in
Similarly, bending moment (M) should be zero at the free end. Enforce this condition in
We know that the slope is zero at the fixed end, i.e., v’=0 @ x=0. Using this in the expression for we get C=0. Next, since v=0 @ x=0, we get D=0. Therefore, the expression for deflection can be given as
Maximum deflection is at the free end,
Example 8 Evaluate the maximum deflection for the beam of length L in Figure 22. w
B
A
Figure 22: Simply-supported beam with udl w
The beam has a pin support at the end A. At end B, it has no restraint against translation in the vertical direction. However, there is a restraint against rotation. Again, integrating
Again using boundary conditions: in the above four equations. We get
slo
Therefore the deflection can be expressed as
Maximum deflection is at x=L
Method of Superposition When evaluating deflections for beams with multiple loads, it could be easier to use the superposition principle, in particular, when the deflections are known for the individual loads. The principle is applicable if the following conditions are satisfied. 1. Linear-elastic behaviour 2. Small deformations The idea is to superpose the deflections due to the individual loads to obtain the net deflection of all the loads. Example 9 Evaluate the deflection at the free end for the cantilever beam in Figure 23. w A
P B
Figure 23: Cantilever beam with udl w and a concentrated load P
The deflection at B could be evaluated as the net deflection due to that of a udl and that of a concentrated load (see Figure 24). P
w
+ Figure 24: Superposition applied to beam in Figure 23
Deflection at the free end of a cantilever v1 due to a udl w is given as (see Table G-1 cases 1 and 4)
Note that the –ve sign is due to the followed sign convention for deflections. Deflection is positive upward and downward negative. Similarly deflection due to the concentrated load P is given as
Therefore deflection at B for the total load is
Example 10 Evaluate the deflection at the free end for the cantilever beam in Figure 25.
w2
w1
B
A
Figure 25: Cantilever beam with linearly varying distributed load
Deflection could be evaluated as the sum of the deflection due to the following two load cases (see Figure 26). w1
(w2 - w1)
+
B
A
B
A
Figure 26: Superposition applied to beam in Figure 25
Net deflection at B is given by (see Table G-1, cases 1 and 8)
Example 11 Evaluate the deflection and rotation at the free end for the cantilever beam in Figure 27. P
P
P
A
B L/3
L/3
L/3
Figure 27: Cantilever beam with three concentrated loads
Using the superposition principle, the deflection at the free end B is (see Table G-1 case 5)
Evaluating rotations in a similar manner (and assuming counter-clockwise rotations are positive), we find
Example 12 Evaluate the deflection and rotation at the free end for the cantilever beam in Figure 28. w A
B L/3
L/3
L/3
Figure 28: Cantilever beam for Example 12
This load case could be represented as the superposition of the two load cases shown in Figure 29. w
_
w L/3
2L/3
Figure 29: Superposition applied to beam in Figure 28
Deflection could be evaluated using case 2 in Table G-1.
Similarly finding the slope at end of the beam
Example 13 Evaluate the deflection at the middle of the center-span for the beam in Figure 30. w A b
B L
b
Figure 30: Beam for Example 13 This beam could be analyzed as a simply-supported beam. The effect of the loading on the overhang portions could be taken into account as moments applied over the supports as shown in Figure 31.
Apply the superposition principle with cases 1 and 10 in Table G-2 to find the maximum deflection in the center-span.
w
MA
MB MB
MA Figure 31: Beam for Example 13
Example 14 Evaluate the rotation at A and deflection at B for the beam in Figure 32. w=4kN/m
P=10kN B
A 2m
1m
4m
Figure 32: Beam for Example 14
Drawing the free-body diagrams as in Figure 33 and then finding the support reaction at A, kN.
=10/3
w=4kN/m
P=10kN F
F
Figure 33: Free-body diagrams for beam in Figure 32
From the free-body diagram of the cantilever portion in Figure 33, deflection at B can be calculated as the superposition of the deflections due to a concentrated load and a udl.
The rotation at support A could be considered as the sum of two cases: 1. Rotation solely due to the downward displacement at the hinge ( ). This can be evaluated as the deflection at B divided by the length of the portion AB, and 2. Rotation due to the bending caused by the load P over AB ( . This could be evaluated from case 5 in Table G-2 Therefore, net rotation at A is given by
Example 15 Evaluate the deflection at mid-span and at the end of the overhanging portion for the beam in Figure 34. w A
D
C
B
L
a
Figure 34: Beam for Example 15
Let us obtain the free-body diagrams for the overhanging portion and the simply-supported portion (see Figure 35). Therefore w MB MB Figure 35: Free-body diagrams for Example 15
Deflection at mid-span could be evaluated as the sum of the deflection due to the udl over the span and that due to the moment MB at the right support. Use cases 1 and 7 in Table G-2.
Deflection at C could be calculated as the sum of deflections due to (i) the net rotation at support B of a simply-supported portion with udl w and a moment MB and (ii) deflection at the end of a cantilever supported of length a.For (i), see cases 1 and 7 in Table G-2.
Example 16 Evaluate the deflections at B and C for the beam in Figure 34. P
B
D
C
A 3b
b
b
b
Figure 36: Beam for Example 16
The free-body diagrams for the three portions and the corresponding reaction forces at the hinges are shown in Figure 37. The deflection at B can therefore be evaluated as that at the free end of a cantilever with a concentrated load of P/2.
A
B
P
P/2 B P/2
P/2
C
C
D
P/2
Figure 37: Free-body diagram for beam in Figure 36
The maximum deflection in the beam is under the load. This is evaluated as the total of the deflection at mid-span of a simply-supported beam of length 2b and the average of the deflections at the two supports B and C.
