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

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).

v+dv

+d

d

v

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

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

B

w

A

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.

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,

B

w

A

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.

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.

Figure 20: Free-body diagram showing reactions, shear force (V) and bending moment (M)

From equilibrium of forces and moments,

A

RA

MA V

M

x

B P

A

w A

RA

MA V

M

x

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.

Figure 21: Cantilever beam with linearly varying distributed load

For this beam, let us evaluate the deflection by integrating where w is the distributed load. In

this example,

o

Substituting for q,

n in g ng

B

wo

A

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.

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: slo

in the above four equations. We get

B w

A

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.

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).

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

P w

+

B

w

A

P

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.

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).

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.

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

B A

P P P

L/3 L/3 L/3

B A + B

(w2 - w1)

A

w1

B

w2

A

w1

Example 12

Evaluate the deflection and rotation at the free end for the cantilever beam in Figure 28.

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.

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.

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.

B

w

A

L b b

w

2L/3

w

L/3

_

B

w

A

L/3 L/3 L/3

Figure 31: Beam for Example 13

Example 14

Evaluate the rotation at A and deflection at B for the beam in Figure 32.

Figure 32: Beam for Example 14

Drawing the free-body diagrams as in Figure 33 and then finding the support reaction at A, =10/3

kN.

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

w=4kN/m P=10kN

F F

w=4kN/m

A 2 m 4 m

B

P=10kN

1 m

w

MA MB MA

MB

Example 15

Evaluate the deflection at mid-span and at the end of the overhanging portion for the beam in Figure 34.

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

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.

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 3b b

P

b b

B C D

w

MB MB

B A

L a

C D

w

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.

A P

B C D B C

P/2 P/2

P/2 P/2