CHAPTER

    6
DEFLECTION

6.1 Introduction
Structures may deform and change shape when being subjected to forces. Other
common causes of structure deformation include temperature changes,
construction errors, and support settlements. If the structure regains its original
shape after the deformation-causing actions are removed, then the deformation is
termed “elastic deformation”. In this chapter, the deformation of the beam is
assumed linearly elastic.

Deflection is an important quantity to be calculated in structural analysis and

design. Deflections are determined and used to solve problems in the analysis of
statically indeterminate structures. Deflections are also important in dynamic
analyses where the response of structures subjected to dynamic loadings
(earthquake, wind, or vibration) is determined. In structural design, deflections
are calculated and checked in order to make sure they are within the maximum
allowable limits stipulated in the design codes. Large deflections can cause
cracks in non-structural elements (non-load-bearing walls or ceilings), distortion
of door and window frames, and water ponding of flat slab of buildings or
undesirable vibrations of structures. Excessive deflections can cause structure
unsightly, unoccupied, and possibly unsafe.

Deflections can be computed using (1) geometric methods, and (2) work energy
methods. In this chapter, the elastic beam theory and the direct (double)
integration method are briefly reviewed. Next, the geometric methods are
discussed particularly to determine the slopes and vertical displacement of
statically determinate beams. Geometric methods are based on a consideration of
the geometry of the deflected shapes of structures. There are two approaches in
geometric method, which are moment-area method and conjugate beam method.

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Deflection

6.2 Objectives
At the end of Chapter 6, students should be able to

1. understand the elastic beam theory, and

2. calculate the deflection of simple beam using the moment-area method
and the conjugate beam method.

6.3 Elastic-Beam Theory

To derive the deflection equation, an initially straight beam that is elastically
deformed by arbitrary loads is considered. The loads are applied perpendicular to
the x axis of the beam and lying in x-y axes as shown in Figure 6.1(a). Due to the
loads, the beam deforms into a curve under shear and bending forces. The
longitudinal axis of the beam in the deformed state is known as the elastic curve
of the beam. Consider a differential element dx of the beam, the angle between
the cross-sections and the width of the small element of the deflected beam
shown in Figure 6.1(b) are dθ and dx, respectively.
y

w(x) P
dx
A B
x
θ
x

(a)
O'

dθ
ds
R Ds'
y Neutral
axis y
M M

dx dx

Before deformation After deformation

(b)
Figure 6.1

222
 Deflection

The radius of' curvature, R for this arc is measured from the centre of curvature
O' to the neutral axis of the element. The strain in arc, ds that is located at a
distance of y from the neutral axis is given by:

ds '− ds
ε= . (6.1)
ds

However,

ds
= dx
= Rdθ, (6.2)

and

ds ' =( R − y ) dθ. (6.3)

Substitute equations (6.2) and (6.3) into equation (6.1), the following equation is
obtained.

ε=
( R − y ) dθ − Rdθ ,
Rdθ

which can be further simplified as follows:

1 ε
= − . (6.4)
R y

The material of the beam is assumed to be homogeneous and it behaves in a

linear elastic manner, thus Hooke’s Law applies:

σ
ε= . (6.5)
E

The flexure formula also applies:

My
σ=− . (6.6)
I

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Deflection

Combine equations (6.5) and (6.6), the following equation is obtained:

1 M
= , (6.7)
R EI

where R = the radius of curvature at a specific point on the elastic curve

M = the internal moment in the beam at the point where R is to be
determined
E = the material’s modulus of elasticity
I = the beam’s moment of inertia

The product of EI in the equation is flexural rigidity of the beam.

Since

dx
= Rdθ, (6.8)

Substitute equation (6.7) into equation (6.8), thus

M
dθ =  dx. (6.9)
 EI 

Since θ = dy/dx, equation (6.9) can also be expressed as

d2 y M
= . (6.10)
dx 2 EI

This differential equation for the deflection of beams is also known as the
Bernoulli-Euler beam equation.

6.4 Direct Integration Method

For simple beams subject to simple loading in which M/EI can be expressed as a
single continuous function of x over the entire length of the beams, the direct
(double) integration method is the most simple and convenient approach for
computing slopes and deflections of beams. The direct integration method
essentially involves writing the expression for M/EI in terms of the distance x
along the axis of the beam and integrating this expression successively to obtain
equations for the slope and deflection of the elastic curve. The constants of
integration are determined from the boundary conditions.

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 Deflection

Applying equation (6.10) and integrating in x, the slope of beams is obtained:

dy M
=
θ =
dx ∫ EI dx. (6.11)

Integrating equation (6.11) twice, the vertical deflection of beams is obtained:

M
y= ∫∫ EI
dx dx. (6.12)

When a beam is subjected to several loads or a beam consists of different cross

sections for which M/EI function is not continuous, each discontinuity due to a
change in loading and/or flexural rigidity (EI) introduces two additional constants
of integration in the analysis and the problem becomes complicated.
Superposition method is applied to determine slope and deflection caused by the
combined effect of loads by superimposing (algebraically adding) the slopes and
deflections (calculated from the direct integration method) due to each of the
loads acting individually on the beam. The difficulty can also be simplified by
employing the singularity functions.

Example 6.1

Determine the equations for slope and deflection of the beam shown in Figure
6.2. Also determine the slopes and deflections at points B and C of the beam.
Given that E = 200 GPa and I = 360(106) mm4.
10 kN

A B C

5m
10 m

Figure 6.2

225
Deflection

Solution

1. Draw the FBD of the beam and determine the reaction forces at support A
using static equilibrium equations.

10 kN

A B C
MA = 100 kNm

Ay = 10 kN 5m 5m

2. Determine the equation for bending moment. Make a cut section at a distance
x from support A. Considering the free body to the left of this section, the
equation obtained as follows:
A
MA = 100 kNm Mcut

x
Ay = 10 kN

M cut = A y x − M A = 10x − 100.

3. The EI of the beam is constant. The equation for M/EI can be written as

d2 y M 1
= = (10x − 100 ) .
dx 2 EI EI

4. Integrating the equation for M/EI in x and the equation for slope of the beam
as

dy 1
=
θ =
dx EI
(
5x 2 − 100x + C1 , )
where C1 is constant.

5. Integrating once more, the equation for deflection of the beam is obtained as

226
 Deflection

1  5x 3 
y
=  − 50x 2 + C1x  + C2 ,
EI  3 
where C2 is constant.

6. The constants of integration, C1 and C2 are evaluated by applying the

boundary conditions:

At end A,=x 0, =θ 0; =
C1 0.
At end=C, x 0,= y 0; =
C2 0.

7. The equations for slope and deflection of the beam are

1 1  5x 3 
=θ
EI
(
5x 2 − 100x ) and=y 
EI  3
− 50x 2  .


8. Substitute numerical data for E, I, and x = 5 m and x = 10 m for points B and

C, respectively, the slopes and deflections at ends B and C are obtained as

(5 kN)(5 m) 2 − (100 kNm)(5 m)

θB = =−0.00521 rad,
[200(106 ) kN/m 2 ][360(106 )(10−12 ) m 4 ]

(5 kN)(10 m) 2 − (100 kNm)(10 m)

θC = =−0.00694 rad,
[200(106 ) kN/m 2 ][360(106 )(10−12 ) m 4 ]

[(5 kN)(5 m)3 /3] − (50 kNm)(5 m) 2

y B =6 −0.0145 m =
= 14.5 mm, 
[200(10 ) kN/m 2 ][360(106 )(10−12 ) m 4 ]

[(5 kN)(10 m)3 /3] − (50 kNm)(10 m) 2

yC =6 −0.0463 m =
= 46.3 mm. 
[200(10 ) kN/m 2 ][360(106 )(10−12 ) m 4 ]

The “negative sign” indicates that the angle is measured clockwise from
A and the deflection is downward.

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Deflection

6.5 Moment-Area Method

The moment-area method was developed based on two theorems by Otto Mohr
and later stated formally by Charles E. Greene in 1873. It employs a semi-
graphical technique to determine the slope and vertical displacement of the
elastic curve due to bending. Furthermore, it relates the geometry of the beam’s
elastic curve to its M/EI diagram.

The following procedures are used to determine the displacement and slope at a
point on the elastic curve of a beam using the moment-area theorems.

Step 1: Draw M/EI diagram

The analysis of statically determinate beam is carried out using static equilibrium
equations and from the analysis, the BMD is drawn. To draw the M/EI diagram,
the BMD is divided by the flexural rigidity of the beam. Figure 6.3 shows the
steps in drawing the M/EI diagram. Note that the flexural rigidity throughout the
beam “may not be the same” as shown in Figure 6.3(b).

M
2I I
I
M

BMD

M/EI
M/EI
M/2EI

M/EI diagram

(a) (b)

Figure 6.3

228
 Deflection

Step 2: Draw the elastic curve of the beam under the applied load

Draw an exaggerated view of the elastic curve of the beam. Note that the points
of zero slope occur at fixed supports while zero displacement at fixed, pinned,
and roller supports. The bending moment or M/EI diagram can be referred if the
general shape of the elastic curve is difficult to be drawn. Realize that when the
beam is subjected to a positive moment, the beam bends concave up whereas
negative moment bends the beam concave down. The displacement and slope
which are to be determined should be indicated on the elastic curve.

Step 3: Apply Theorem 1 and Theorem 2 of the moment-area method

Theorem 1

Figures 6.4(b) and 6.4(c) show the elastic curve and M/EI diagram of the beam
under the applied load. By using equation (6.9) and Figure 6.4, dθ on either side
of the element is equal to the shaded area under the M/EI diagram. This equation
forms the basis for the first moment-area theorem.

Theorem 1
The change in slope between any two points on the elastic curve equals to
the area of the M/EI diagram between the two points.

Hence, the area under M/EI diagram between points A and B on the elastic curve
represents the internal angle between the two tangent lines at A and B,
respectively, which can be computed as follows:

B M
θAB = ∫A EI
dx. (6.13)

Theorem 2

The second moment-area theorem is based on the relative derivation of tangents

to the elastic curve. Figures 6.4(c) and 6.4(d) show a greatly exaggerated view of
the vertical deviation, dt of the tangents on each side of the differential element,
dx. Since the slope of elastic curve and its deflection are assumed to be very
small, it is satisfactory to approximate the length of each tangent line by x and
the arc ds' by dt as shown in Figures 6.4(c) and 6.4(d). From the equation of
calculating the arc length, s = rθ, thus

dt = xdθ. (6.14)

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Deflection

Substitute equation (6.9) into equation (6.14),

M
dt = x   dx. (6.15)
 EI 

w(x)

A B

dx
(a)

(M/EI)dx

(MA/EI) (MB/EI)

A B
dx
(b)
Elastic curve

Tangent A Tangent
B
line at A line at B
dθ
tAB θAB tBA
dt

(c)

dt

ds'

(d)

Figure 6.4

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 Deflection

Theorem 2
The vertical deviation of the tangent at a point (A) on the elastic curve with
respect to the tangent extended from another point (B) equals to the
“moment” of the area under the M/EI diagram between the two points
(A and B).

Thus, the vertical deviation of the tangent at A with respect to the tangent at B
can be determined by integration,

B M
t AB = ∫
A
x
EI
dx. (6.16)

Example 6.2

Determine the slope at points B and C of the beam as shown in Figure 6.5. Given
that E = 200 GPa and I = 360(106) mm4.
10 kN

A B C

5m
10 m

Figure 6.5

Solution

1. Draw the FBD of the beam and determine the reaction forces at support A
using static equilibrium equations.
10 kN

A B C
MA = 100 kNm

5m 5m
Ay = 10 kN

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Deflection

2. Draw the BMD and then divide it by the EI of the beam to obtain M/EI
diagram. It is easier to solve the problem in terms of EI.

M/EI

A B C x

–50/EI

–100/EI

3. The elastic curve of the beam subjected to 10 kN concentrated load is shown

in figure below. Draw tangent line at A, B, and C. The tangent line at A is
horizontally straight (no angle of rotation at A since the beam is fixed at A).

θAB tan C
tan A
A θAC
B
θB
θBC tan B

C θC

4. The angle between tan A and tan B, θAB is equivalent to θB.

θB =θAB ; θC =θAC .

5. Applying Theorem 1, θAB is equal to the area under the M/EI diagram
between points A and B.

1  100 kNm 50 kNm 

θAB =−  +  (5 m)
2 EI EI 

375 kNm 2
θAB =− .
EI

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 Deflection

6. Substitute the numerical data for E and I.

375 kNm 2
θB =− =−0.00521 rad.
[200(106 ) kN/m 2 ][360(106 )(10−12 ) m 4 ]

7. The “negative sign” indicates that the angle is measured clockwise from A.

8. In a similar manner, the area under the M/EI diagram between points A and
C (θAC) is equivalent to θC, hence

1  100 kNm  500 kNm 2

θC =θC/A =  −  (10 m) =− .
2 EI  EI

9. Substitute numerical data for E and I.

−500 kNm 2
θC = =−0.00694 rad.
[200(106 ) kN/m 2 ][360(106 )(10−12 ) m 4 ]

Example 6.3

Determine the vertical displacement at points B and C of the beam shown in

Figure 6.6. Values for the moment of inertia of each segment are indicated as
shown below. Given that E = 200 GPa and I = 4(106) mm4.

A B C
500 Nm
2I I

4m 3m

Figure 6.6

233
Deflection

Solution

1. Draw the FBD of the beam and calculate the reaction forces at A.

A B C
MA 500 Nm

Ay 4m 3m

2. Draw the BMD and then divide it by the EI of the beam to obtain M/EI
diagram as shown in figure below. It is easier to solve the problem in terms
of EI. Note that the EI of span AB and span BC are not the same.
500 Nm
BMD

500/EI
M/EI 250/EI
diagram
A B C

3. The couple moment at C causes the beam to deflect as shown in the figure
below. The tangent lines at A, B, and C are indicated.
tan C

tCA
B
A tBA
tan A
tan B

4. Vertical displacement at point A and B can be related directly to the

deviations between the tangent lines. So that from the elastic curve of the
beam above,

∆B = tBA and ∆C = tCA.

234
 Deflection

5. Applying Theorem 2, vertical displacement at B is equivalent to the moment

of the area under the M/EI diagram between A and B which is computed
about B. Hence

∆B =t BA

 250 Nm 
= (4 m)  (2 m)
 EI 
2000 Nm3
=
EI
2000 Nm3
=
[200(109 ) N/m 2 ][4(106 ) mm 4 (1 m 4 /(103 ) 4 mm 4 )]

= 0.0025 m
= 2.5 mm.

6. Likewise for tCA, the vertical displacement at C is equivalent to the moment

of the entire M/EI diagram from A to C which is computed about C.

∆C =t CA

 250 Nm   500 Nm 
=  (4 m)  (5 m) +  (3 m)  (1.5 m)
 EI   EI 

7250 Nm3
=
EI
7250 Nm3
=
[200(109 ) N/m 2 ][4(106 )(10−12 ) m 4 )]

= 0.00906 m
= 9.06 mm.

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Deflection

6.6 Conjugate Beam Method

The conjugate beam method was developed by H. Müller-Breslau in 1865.
Essentially, it requires the same amount of computation as the moment-area
theorems to determine a beam’s slope or deflection. However, this method is
based on the similarity between the relationships for loading and internal shear,
also internal shear and moment as listed below.

The slope and deflection of the elastic curve are related to the internal moment by
the following expressions [equations (6.17) and (6.18)].

dθ M d2 v M
= and = , (6.17)
dx EI dx 2 EI

or

M M
θ= ∫ dx and v = ∫∫ dx dx. (6.18)
EI EI

The following expressions [equations (6.19) and (6.20)] relate the internal shear
and moment to the applied load.

dV d2M
=w and = w, (6.19)
dx dx 2
or
V = ∫ w dx and M = ∫ ∫ w dx dx. (6.20)

Compare the expression for slope, θ and internal shear, V. If the applied load, w
is replaced by the term M/EI, the expression for slope and shear are identical.
The same explanation goes to the expression for displacement, v and internal
moment, M. These comparisons can be considered by having a beam with the
same length as the real beam which referred as the “conjugate beam”. Conjugate
beam is loaded with the M/EI diagram derived from the load, won the real beam.
Therefore, the two theorems that are related to the conjugate beam method can be
written as follows.

Theorem 1
The slope at a point in the real beam is numerically equal to the shear at
the corresponding point in the conjugate beam.

236
 Deflection

Theorem 2
The displacement of a point in the real beam is numerically equal to the
moment at the corresponding point in the conjugate beam.

Figure 6.7 shows two examples of conjugate beam corresponding to the real
beam. Note the sign of loading, w and the M/EI on the conjugate beam.

P
θA = 0 θC ≠ 0
∆A = 0 ∆C ≠ 0 θA ≠ 0 θB ≠ 0 θC ≠ 0
∆A = 0 ∆B = 0 ∆C ≠ 0
M
C C
A 2I B I
A B
Real beam

M/EI
M/2EI

–M/EI
M/EI diagram
-M/EI
B' M/EI C'
VA' = 0 VB' = /
A' M/2EI
MA' = 0 MB' = 0 VC' ≠ 0
MC' ≠ 0
VA' = 0
A' M/EI C'
MA' = 0 VC' ≠ 0
MC' ≠
(a) (b)
Conjugate beam loaded with M/EI diagram

Figure 6.7

6.6.1 Conjugate beam support

When drawing the conjugate beam, it is important that the shear force and
bending moment developed at the supports of conjugate beam correspond to the
slope and displacement of the real beam. For example, a cantilever beams as
shown in Figure 6.7(a). The real beam is fixed at A while free end at C which
provides zero slope and displacement at A, but the beam has a non-zero slope
and displacement at C. Thus, from Theorem 1 and Theorem 2, the conjugate
beam must be fixed supported at C' and free end at A' as the shear force and

237
Deflection

bending moment at the support of the conjugate beam correspond to the slope
and displacement of the real beam. Figure 6.8 shows more examples of the real
beam and the conjugate beam subjected to their support conditions.

6.6.2 Procedure for analysis

The following procedure provides a method that could be used to determine the
displacement and slope at a point on the elastic curve of a beam using the
conjugate beam method.

1. Draw the BMD of the real beam and consequently draw the M/EI
diagram. Note that the EI of the beam which in some cases, they are not
the same throughout the beam.
2. Draw the conjugate beam and it is important to make sure the support
condition of the conjugate beam corresponds to the real beam.
3. The conjugate beam is loaded with M/EI diagram from the real beam.
This loading is assumed to be distributed over the conjugate beam and is
directed upward when M/EI is positive or downward when M/EI is
negative. In other words, the loading always acts away from the beam.
4. Analyse the conjugate-beam using the equations of equilibrium to
determine the reaction forces at the conjugate beam’s supports
5. Section the conjugate beam at the point where the slope θ and
displacement ∆ of the real beam are to be determined. At the section,
show the unknown shear, V' and moment, M' according to their positive
convention.
6. Determine the shear and moment using method of section or moment
equation method. V' and M' in the conjugate beam is equivalent to θ and
∆ in the real beam, respectively.

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 Deflection

REAL BEAM CONJUGATE BEAM

θ≠0 V≠0
∆=0 M=0
(a) Pinned and roller end (a) Pinned and roller end

θ=0 V=0
∆=0 M=0

(b) Fixed end (b) Free end

θ≠0 V≠0
∆≠0 M≠0
(c) Free end (c) Fixed end

θ≠0 V≠0
∆=0 M=0

(d) Internal hinge (d) Internal roller

θ≠0 V≠0
∆≠0 M≠0

(e) Internal hinge (e) Internal roller

Figure 6.8

Example 6.4

Determine the slope and vertical displacement at point C of the steel beam shown
in Figure 6.9. Take E = 200 GPa and I = 475(106) mm4.
20 kN

A B C

5m
10 m

Figure 6.9

239
Deflection

Solution

1. Draw the FBD and compute the reaction forces at support A.

20 kN

A B C
MA = 100 kNm

5m 5m
Ay = 20 kN

2. Draw the BMD and M/EI diagram correspond to the applied load and EI of
the beam, respectively.

A B C

–100 kNm
BMD

A B C

–100/EI
M/EI diagram

240
 Deflection

3. Draw the conjugate beam as shown in figure below. The end condition at A'
and C' in the conjugate beam are corresponding to the support A and the free
end C in the real beam.

A' B' C'

4. Load the conjugate beam with the M/EI diagram from the real beam. The
M/EI diagram is negative, so it acts downwards on the conjugate beam which
is away from the beam as shown in figure below.

A' B' C'

100/EI

5. Draw the FBD of the conjugate beam and calculate the reaction forces at
support C'. As shown in figure below, the slope and vertical displacement at
C in the real beam are equivalent to the reaction force and moment at support
C' in the conjugate beam, respectively.

A' B' C'

MC'

Cy '
100/EI

6. Calculate reaction force at C' in the conjugate beam.

+ ∑ F = 0,
y

241
Deflection

 1  100 
−   −  ( 5) + Cy ' =0
 2  EI 
250
Cy' = −
EI
250 kNm 2
θC =−
[200(106 ) kN/m 2 ][475(106 )(10−12 ) m 4 ]
= −0.00263 rad.

The negative sign indicates that the angle is measured clockwise from A.

7. Calculate moment at C' in the conjugate beam.

+ ∑M C = 0,

 1  100   2  
− M C' −   −  ( 5 )   ( 5 ) + 5  =
0
 2  EI   3  

6250
M C' = −
3EI
6250 kNm
∆ C =−
3 [200(10 ) kN/m 2 ][475(106 )(10−12 ) m 4 ]
6

= −0.0219 m

= −21.9 mm.

The negative sign indicates that the deflection is downward.

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 Deflection

Example 6.5

Determine the maximum vertical displacement of the steel beam shown in Figure
6.10. Take E = 200 GPa and I = 60(106) mm4.

8 kN

A B C
I

9m 3m

Figure 6.10

Solution

1. Draw the FBD of the beam and compute the reaction forces at both supports
A and C as shown in figure below.
8 kN

A B C

9m 3m
Ay = 2kN Cy = 6 kN

2. Draw the BMD and M/EI diagram as shown below.

18/EI

3. Draw the conjugate beam which corresponds to the real beam particularly
end/support conditions. Load the conjugate beam with M/EI diagram from
the real beam as shown in figure below. Since M/EI diagram is positive, it
loads the conjugate beam in upward direction.

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Deflection

18/EI

9m 3m
A' B' C'

a. Draw the FBD of the conjugate beam and calculate the reaction forces at
both supports A' and C' by using static equilibrium equations. The
reaction forces at A' and C' in the conjugate beam are shown in figure
below.
18/EI

9m
B' 3 m
' '
Ay = 45/EI Cy = 63/EI

Maximum displacement of the real beam occurs at the point where the
slope of the beam is zero. Assume this point acts within span A'B'
(0 ≤ x ≤ 9 m) which the origin of x is taken from point A'. Note that the
peak value of the distributed load is determined from proportional
triangles.

 18 
 
w  EI 
=
x 9
M'
2x
x V' = 0 w=
EI
'
A = 45/EI
y

When V' = 0,

+ ∑ F = 0,
y

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 Deflection

45 1  2x 
− +  x = 0
EI 2  EI 
x = 6.71 m.

From the above calculation, it shows that the maximum displacement

occurs at x = 6.71 m from point A' which is within span A'B'. The
maximum displacement in the real beam corresponds to the moment, M'
in the conjugate beam can be calculated as follows:

+ ∑M C' = 0,

 45   1  2x  1 
− M '−   x +    ( x )  ( x )  =
0
 EI   2  EI  3 
45x x 3
− M '− + 0
=
EI 3EI
45x x 3
M' =
− +
EI 3EI

=
 45 
−   ( 6.71) +
( 6.71) 3

 EI  3EI
201.24
= −
EI

−201.2 kNm3
∆C =
[200(106 ) kN/m 2 ][60(106 ) mm 4 (1 m 4 /(103 ) 4 mm 4 )]

= −0.0168 m
= −16.8 mm.

The negative sign indicates the deflection is downward.

245
Deflection

EXERCISES

Exercise 1

Sketch the elastic curve for the beams shown in Figures 6.11(a) and 6.11(b). The
corresponding BMD is shown directly beneath each beam.

P P
w w

+
+
–
–

(a) (b)

Figure 6.11

Answers:

Contraflexure point

Contraflexure point

(a) (b)

Exercise 2

Using moment-area method and conjugate beam method, calculate the slope and
vertical displacement at point C for each of the structure shown [Figures 6.12(a)–
6.12(d)]. The flexural rigidity of all beam sections is indicated in the figure.

246
 Deflection

P
P P/2

EI
A B C
2EI EI

L/2 L L/3
A B C
L

(a) (b)

P
w

EI 2EI EI
M EI
L/3 L/3 L/3
A B C D
L/2 L
A C B
(d)
(c)

Figure 6.12

Answers:
PL2 PL3
a. θC =− ; ∆ C =−
4EI 6EI

PL2 4PL3
b. θC =− ; ∆ C =−
6EI 81EI

PL2 7ML PL3 5ML2

c. θC =− + ; ∆C = −
18EI 54EI 18EI 54EI

13wL3 31L4
d. =θC = ; ∆C
1296EI 3888EI

247