Chapter 4

DEFLECTION OF BEAMS
4.1 STATICALLY DETERMINATE BEAMS
From beam bending theory

d 2V d 2v
EI 2EI M2 M
dx dx

(1)

The deflection v can thus be found by double integration of Eq (1).

This method is known as the Macaulays Method.

Hence

1
Mdxdx
EI

We have
V

1
Mdxdx
EI

dF
Since F wdx i.e.
w
dx
and
dM
F
dx
differentiating Eq. (1)
M Fdx i.e.

d 3V dM
EI

F
3
dx
dx
diff again
d 4V
dF
EI

w
4
dx
dx
i.e.
d 4V
w

dx 4 EI

Using multiple-integration method on above Eq.

d 4V
EI 4 w
dx

EIV wdx C1

d 3V
Note : V 3
dx

EIV dx wdx C1 x C2

d 2V
Note : V 2
dx

x2
EIV dx dx wdxC1 C2 x C3
2

1
1
EIV dx dx dx wdx C1 x 3 C 2 x 2 C3 x C 4
6
2

C1 to C4 are constants depending on B.C,S

However, if bending eq. is obtained, then using doubleintegration methods, we have
EIV M

EIV M dx C1

EIV dx M dx C1 x C2
C1 to C2 are constants found from B.C,S

Example 4.1.1
A force P and a moment Mo are applied at the free end of
a cantilever. Determine the deflection equation and the
deflection and slope at B. For P = 20 kN, Mo = 5 kN.m, L
= 2 m, EI = 10 MN.m2
Calculate the deflection and angle of rotation at B.
Note: for small angle, slope = tan = = angle of
rotation.

From the above free body diagram, we have

Vertical equilibrium: Fxy + P = 0, Fxy = - P
Rotational equilibrium: M + (PL + Mo) + Fxy x = 0
M + (PL + Mo) + (-Px) = 0
M Px M o PL

We have
EIV Px M o PL
1
EIV Px 2 M o x PLx C1
2
EIV

1 3 1
1
Px M o x 2 PLx 2 C1 x C2
6
2
2

B.C.S.
At x = 0, V = 0. C2 = 0
At x = 0, V = 0. C1 = 0
Hence

1 1 2

Px M o x PLx

EI 2
1 1 3 1
1

Px M o x 2 PLx 2
V

EI 6
2
2

At B, x = L

1 1 2
2
PL M o L PL
EI 2

L
PL 2M o

2 EI

VB

VB

(a)

1 1 3 1
1

PL M o L2 PL3

EI 6
2
2

(b)

L2
2 PL 3 M o

6 EI

Subst.

P = 20 kN, Mo = 5 kN.m, L = 2 m
EI = 10 MN.m2 into (a) & (b).
VB B (for small angle, slope =
tan = = angle of rotation)

2 103

6 20( 2) 2(5)

2 10 10
0.005 rad.

(note units)

2 2 103
2(20)2 3(5) 6.3 mm
VB
6 10 106

Example 4.1.2
Determine the deflection at the free end of the beam
loaded as shown.

Note: Discontinuities at x L 4 , L 2 , 3 L 4 ,
The problem is equivalent to

B.M.(x)

L
Px M o x
4

W
L

x
2
2

W
3L

x
2
4

L
Note for x , last 3 terms are zeros.
4
L
for x , last 2 terms are zeros.
2

We have
o
2
L
W
L
W
3L
EIV " Px Mo x 4 2 x 2 2 x 4
2

Px
L W
L
'
EIV
Mo x

x
2
4
6
2

W
3L

x
6
4

C1

(a)

Px3 Mo
L 2 W
L 4 W
3L 4
EIV 6 2 x 4 24 x 2 24 x 4 C1x
C2
(b)

B.C.S
At x = L, slope = V = 0

i.e.
3
3
PL2
3L W L W L
0
Mo C1
2
4
6 2
6 4

PL2 3LMo 7WL3

C1

2
4
384

At x = L, deflection = V = 0
i.e.

PL3 Mo 3L 2 W L 4 W L 4
0

C1 L C2 (c)
6
2 4 24 2 24 4

Subst. C1 , into (c) gives

 PL3 15M o L2 97WL4

C2

3
32
6144
Subst. C1 , C2 into (a) & (b) gives
2

Px
L W
L
'
EIV
Mo x

x
2
4
6
2

W
3L

x
6
4

PL2 3 LM o 7WL3

4
384
2

(d)
C1
Px 3 M o
L
EIV

x
6
2
4

W
L

x
24
2

W
3L

x
24
4

PL2 3LM o 7WL3

x

2
4
384

C1

C1

C2

PL3 15M o L2 97WL4

3
32
6144

C2

(e)

10

At free end (A), x = 0.

Hence (from d)

1 PL2 3LMo 7WL3

A EI 2 4 384

and (from e)

1 PL3 15M o L2 97WL4

VA

EI 3
32
6144

11

4.2 STATICALLY INDETERMINATE BEAMS

Hitherto problems examined are
Statically determinate
i.e. All unknown reactions could be determined from
external equilibrium F = 0, M = 0.
As no. of supports increases, no. of unknown reactions
also increases.
When no. of equilibrium is insufficient to determine the
unknown reactions, we have statically indeterminate
beams.
Types of statically indeterminate beams
a) Supported cantilever
-

fixed one end, support the other end.

unknown, R1, R2, M1

12

b)

spring instead of rigid support.

13

c)

fixed or clamped at both ends.

unknowns, R1 , R2 , M1 , M2

d)

3 supports at same level

unknowns, R1 , R2 , R3 .

14

Statically indeterminate problems are solved by :

1.

Direct Application
Consider:
a) equilibrium
b) force-deformation relationships
c) boundary conditions
2.
-

Superposition
reducing statically indeterminate problem to a
number of statically determinate ones.
displacements are linearly proportional to applied
loads. Assume displacements are small, obey
Hookes Law.

1) Direct Application Approach

Example 4.2.1
A beam is clamped at A, simply support at B and loaded
as shown in Fig. Determine all reactions.

15

(i)

Equilibrium.
Vert. equilibrium
F = 0
RA + RB - P = 0

(a)

P = R A + RB
Rotational equilibrium

M about B = 0

M = 0
MA + RBL - Pa = 0
2 eqs., 3 unknowns.

16

(b)

(ii) Force-deformation relationship

EIV R A x M A x P x a
x2
P
EIV R A
M A x 1 x a 2 C1
2
2
3
x 2 P xa 3
RA x
EIV
MA

C1x C2
2 3
2
2
3

(iii) Boundary conditions

At x = 0, V = 0 C1 = 0
At x = 0, V = 0 C2 = 0
At x = L, V = 0
R A L3
L2 P L a 3
0
MA

2 3
2 2
3

Solving eqs. (a), (b), (c).

3 eqs., 3 unknowns, RA, RB, MA,

17

(c)

We have
Pb
R A 3 3 L2 b 2
2L
Pa 2

RB
3 2L b
2L

Pb 2
M A 2 L b2
2L

Note: If MA was drawn clockwise

Pb 2
value would be 2 L b 2 .
2L

18

, then its final

Example 4.2.2
For the overhanging beam shown, determine the
magnitude of the supporting force at B

i)

Equilibrium.

vert. equil.

F = 0
R1 + R2 - w(a + b) = 0

Rotational equil.

M about B = 0

M = 0

M1 + R2a - w(a + b)

( a b)
0
2

( a b) 2
M1 + R2a - w
0
2
19

(a)

(b)

ii) Force-deformation Relation

We have

W 2
EIV M1 x R1 x
x R2 x a
2
(frowning)
o

R1 2 W 3 R2
EIV M1 x
x
x
x a 2 C1
2
6
2
1

M1 2 R1 3 W
R2
4
EIV
x
x
x
x a 3 C1 x C2
2
6
24
6

iii) Boundary Conditions

B.C.S
At x = 0, V = 0 C1 = 0
At x = 0, V = 0 C2 = 0
At x = a, V = 0
M1a 2 R1a 3 wa 4
0

2
6
24

20

i.e.

a wa 2
M1 R1
3 12

Solving eqs. (a), (b) & (c).

We have
5
3wb 2
R1 wa
8
4a

3
3wb 2
R2 wa wb
8
4a
wa 2 wb 2
M1

8
4

21

(c)

2) Superposition
i) Statically determinate beam
Example 4.2.3

Determine the deflection of the beam at point C.

L
Consider a = b , i.e. C is mid-point.
2
From standard solutions
(refer to mechanics of materials by AC Ugural Table
B7).
5WL4

V

2L w 384 EI

PL3

V

2L p 48 EI

Hence deflection at

L
2

for above problem is

5WL4
PL3

384 EI 48 EI

22

(ii) Statically indeterminate beam

Example 4.2.4
For the beam as shown, determine the reactions using
superpositon.

Equilibrium
F = 0
RA + RB - WL = 0
M about B = 0

WL2
RA x L - MA 0
2

(a)
M = 0

(b)

23

3 unknowns, RA, RB, MA.

Consider a) Cantilever with load W, let VB W be
deflection at B.
VB w

Consider b) Cantilever with load RB applied at free

end B, let VB R be deflection at B.
VB R

From standard solutions (Table B7 mechanics of

materials AC Ugural )

V B W

WL4

8 EI

24

VB R

RB L3

3EI

We have
VB R

VB w

The compatibility condition for the original beam requires

that
WL4 RB L3
VB

0
8 EI 3EI

25

i.e.

3
RB WL
8

From (a)

3
5WL
R A WL WL
8
8

From (b)

WL2
M A RA L
2
1 2
WL
8

26

Example 4.2.5
Given that wo w , find V at x = L, i.e. at x = 4d,
d

From standard solutions (Table B7 mechanics of

materials by AC Ugural or Table G-1 mechanics of
materials by JM Gere)
a) V

b) V

x L

x L

wo L4

8 EI z

P
3a 2 L a 3

6 EI z

27

The system can be separated into 4 components.

(1)

wo 4d
32wo d 4
32wd 3
V1

8EI
EI
EI
4

(2)

3
3W
11
Wd
V2
3d 2 (4d ) d 3

6 EI
2 EI

28

(3)

3
2W
40
Wd
V3
3(2d ) 2 4d (2d ) 3

6 EI
3 EI

(4)

W
27 Wd 3
2
3
V4
3(3d ) 4d (3d )

6 EI
2 EI
29

By superposition, the tip deflection is:

V V1 V2 V3 V4
Wd 3
11 40 27

32
EI
2 3 2
79 Wd 3
V
3 EI

30

4.3 BEAM DEFLECTION BY MOMENT - AREA

METHOD
Method is semi-graphical, need B.M. and
deflection diagrams.
More rapid method if deflection/slope at only
one point is required.
Assumptions:
1) Beam initially straight.
2) Elastically deformed.
3) Slope and defection are small.
4.3.1 Determination of Slope by Moment-Area Method
Consider a beam loaded as shown
Consider an element CD on which the slope and
deflection are small.

CC

31

D
dx

Deflected shape
D

Tan C
Elastic curve

Tan D

Consider a small element CD of length

dx (magnified) on the beam.
Tangents intersect at angle d

32

4.3.1 Determination of Slope by Moment-Area Method

Consider two points P and Q on a beam at distance dx
apart.
The difference in displacement between P and Q is dv.
The slope m of line PQ is given by

dV
dx
if angle is small

m tan

(a)

dV
i.e.

dx
Q

In above triangle
Slope of line PQ
dV

= tan
= = (for small angle)
dx

33

We have from beam bending equation

d 2V
d
M EI 2 EI
dx
dx

i.e. d

M
dx
EI

(a)

As shown in Eq (a), for element CD of length dx,

the angle d between 2 tangents is given by:
d area under the

M
curve from C to D.
EI

dx

34

Similarly for AB, the angle BA between 2 tangents

drawn from points A and B is given by the area under the
M
the
curve from A to B.
EI
M
curve from A to B is given by
EI
int egration of Eq (a)

The area under the

i.e.

M
dx
A EI

BA

area under

M
curve from A to B.
EI

This is the Moment area 1st theorem

35

BA

A
P
B

Note:

A = Tangential angle at A (angular rotation of tangent at A)

B = Tangential angle at B (angular rotation of tangent at B)

BA is the interception angle between the tangents at A and B

From triangle ACP, A+BA = B

i.e. BA = B-A

Hence,

BA is also the difference of the

tangential angles at B and A.

36

4.3.2 Determination of Deflection by Moment-Area Method

Consider beam as shown

K
L

d is the difference in angle

between tangent at C and
D.

1) Draw a vertical line from point A. Let the tangents at C and D meet
the vertical line at K and L and denote distance KL as dt.
2) Let the tangent at B meets the vertical line at P and denote distance
AP as tAB .

from geometry
dt = x d
Subst. into eq. (a)
We have

(assume small dx)

dt x

M
dx
EI
37

Integrating

dt x

M
dx
EI

t AB A x
B

Since

B
x
A

M
dx
EI

M
B
dx A xdA X A
EI

Hence
t AB X A
M

X Area under
curve .
EI

- tAB is called the tangent deviation

This is the Moment area 2nd theorem
Note that if point A is at the support end and point B is at
the mid-span of the simply-supported beam. Then tAB
represents the deflection of A with respect to B. (this can
also be interpreted as the deflection of B w.r.t. A i.e.
deflection of the mid-span).

38

Example 4.3.1
Determine the slope at pts. B and C.

Moment-curve

PL
2 EI

PL
EI

BA = B - A , A = 0 BA = B.
BA = Area under AB
PL L 1 PL L



2 EI 2 2 2 EI 2
3PL2

8EI
39

CA = C - A , A = 0 CA = C.
C = Area under AC

1 PL
L

2 EI
PL2

2 EI

40

Example 4.3.2
Determine the deflection at points B and C.

Moment curve

Deflection at B = Area under AB X

Mo L L

EI 2 4
M o L2

8 EI

41

Deflection at C = Area under AC X

Mo
L
L

2
EI
M o L2

2 EI

42

Example 4.3.3
Determine deflection at C and slope at A

The moment curve is given by.

A1

- parabolic distribution, (can be obtained from

equilibrium equation).
- Area and centroid can be obtained from tables.

43

slope at A
CA = C - A , C = 0 A = - CA.
Hence

A = - CA
= - (area under AC)

44

We have area under AC

WL3
A1
24 EI

(from tables, Ugural book Pg.2).

Hence

WL3
A
24 EI

Deflection at C
We have

5L
16

(from tables, Ugural book Pg.2).

WL3
A1
24 EI

t AC Vmax A1 X

5WL4

384 EI

5WL4
or displacement of C 384 EI
45

Example 4.3.4
A cantilever beam is subjected to loads as shown in Fig.
for M = 0.6PL, determine the slope and deflections at the
free end.

MB
A

B
RB

From equilibrium RB= P, MB = 0.6PL-PL

Bending moment at a distance x from B is: M=MB+RB x =0.6PL-PL+Px

(0.6PL)/EI

A1

B
B

A2
Plotting this Eq. we have 0.6PL (top rectangle), -PL+Px (bottom
triangle)
The above diagram is known as a composite bending moment diagram.

46

We have

0.6 PL2
0.6 PL
L
A1
EI
EI
L
X1
2

1 PL
PL2
L
A2

2 EI
2 EI
2L
X2
3

Slope at A
AB = A - B = A - 0
= Area under moment curve
= A1 +A2
0.6 PL2 PL2

EI
2 EI
PL2
01
.
EI

47

Deflection at A

t AB A1 X 1 A2 X 2
. PL2 L PL2 2 L
06

EI 2 2 EI 3
PL3

30EI

48

Example 4.3.5
A cantilever beam is subjected to loads as shown in Fig.,
determine the slope and deflections at B and C given that
P=22.2 kN, L=1829 mm, E= 69x103 MPa, I=2.71 x 107
mm4 ,
using
a) double-integration method
b) moment-area method

MA

2EI

EI

L/2

L/2

RA

49

(a) Double-integration method

Vertical equilibrium gives
RA = P
Rotational equilibrium gives
MA = PL

x
MA = PL

Mxz
Fxy

RA = P

Take moment about x

Mxz+ PL Px = 0
Mxz = -PL + Px
= P(x - L)
Consider segment AC
2EIV1 = Mxz = P(x - L)
2EIV1 = P(x - L)2/2 + C1
2EIV1 = P(x - L)3/6 + C1x + C2
50

B.C.
At x= 0, V1 = 0
At x= 0, V1 = 0

C1 = - PL2/2
C2 = PL3/6

i.e.
i.e.

Hence
V1 = P(X2 2xL)/4EI
V1 = P(X3 3x2L)/12EI

(1)
(2)

Consider segment CB
EIV2 = Mxz = P(x - L)
EIV2 = P(x - L)2/2 + C3
EIV2 = P(x - L)3/6 + C3x + C4
Or
V2 = [P(x - L)2/2 + C3]/EI
V2 = [P(x - L)3/6 + C3x + C4]/EI
B.C.
At x = L/2,

(3)
(4)

V1(L /2) = V2(L /2)

I.e.
P[(L/2)2 2(L/2)L]/4EI = [P(L/2 - L)2/2 + C3]/EI
from Eq.1

from Eq.3

51

i.e.
3PL2 /16 = PL2/8 + C3
and C3 = -5PL2/16
Similarly, at x = L/2,
I.e.

V1(L /2) = V2(L /2)

P[(L/2)3 3(L/2)2L] / 12EI

= [P(LL/2)3/6 + C3(L/2) + C4 ]/EI
from Eq.2

from Eq.4

i.e. 5PL3/96 = PL3 /48 5PL3/32 + C4

and, C4 = PL3/8
Hence
V2 = P[(X-L)2- 5L2/8]/2EI
V2 = P[(X-L)3/3 - 5L2X/8 + L3/4]/2EI
At X = L/2
Vc = -3PL2/16EI
Vc = -5PL3/96EI
At X = L
VB = -5PL2/16EI
VB = -3PL3/16EI

52

(b) Moment area method

Y

MA

2EI

EI

L/2

L/2

RA

(2/3)(L/2) = L/3

M/EI
L/4
A

(2/3)(L/2) = L/3
C

A3

A1
-PL/4EI
A2
-PL/2EI

-PL/2EI

53

We have
A1 = (L/2)(-PL/4EI) = - PL2/8EI
A2 = (1/2)(L/2)(-PL/4EI) = - PL2/16EI
A3 = (1/2)(L/2)(-PL/2EI) = - PL2/8EI
A = 0,
VA = 0,
(VA)
Hence slope at C :
C = CA = Area under CA
(Vc)
= A1+A2
= -(PL2)/8EI + (- PL2)/16EI = -3(PL2)/16EI
At x=0,

and slope at B :
B = BA = Area under BA
= A1+A2+ A3
= -(PL2)/8EI + (- PL2)/16EI + -(PL2)/8EI
= -5(PL2)/16EI
To calculate VC , we have
X 1 =L/4,

X 2 = (2/3)(L/2) = L/3

I.e. VC = tCA = A1 X + A2 X 2
= [-(PL2)/8EI](L/4) + [(- PL2)/16EI](L/3)
= -5(PL3)/96EI
1

54

To calculate VB , we have
X 1 = L/4 + L/2,

X 2 = L/3 +L/2,

X 3 = L/3

I.e. VB = tBA
= A1 X + A2 X 2 + A3 X 3
= [-(PL2)/8EI](3L/4) + [(- PL2)/16EI](5L/6)
+ [(- PL2)/8EI](L/3)
= -3(PL3)/16EI
1

For P = 22.2 kN, L = 1829 mm, E = 69 x 103 MPa

I = 2.71 x 10 7 mm4
max = B = -5(PL2)/16EI
5 22.2 103 (1.829 103 ) 2

16
69 103 2.71 107
0.0124 rad

Vmax = VB = -3(PL3)/16EI
3 22.2 103 (1.829 103 )3

16
69 103 2.71 107
13.66 mm

55

4.3.3 Superposition using the Moment-Area approach

Using our earlier example on statically indeterminate
beam (example 4.2.1).

If the point load P is at the mid-span, then this problem is

equivalent to :

56

The bending moment diagram can be obtained by

superposing the bending moment diagrams of (a) and (b).

Since the deflection of the cantilever tip = 0 ; geometric

compatibility requires that :
t BA( a ) t BA(b ) 0

(a)

t BA ( a ) x B ( a ) A( a )

x B(a)

57

L 2 L 5
L
2 32 6

1 L PL
PL2

2 2 2 EI z
8 EI z

A( a )

5 PL2
5 PL3
L

6 8 EI z
48 EI z

t BA( a )

(b)

t BA(b ) x B (b ) A(b )

x B (b )

A(b )

1 F3 L F3 L2
L

2 EI z 2 EI z

t BA(b )

t BA( a ) t BA(b )

2 F3 L2 F3 L3
L

3 2 EI z 3EI z

5 PL3 F3 L3

0
48 EI z 3EI z

F3

from which;

2
L
3

5
P
16

11
R1 P
16

5
R2 P
16
58

3
M A PL
16