StructuralTheoryClass2.ppt

2
DEFLECTIONS OF BEAMS
The deformation of a beam is usually expressed in terms of
its deflection from the original unloaded position. The deflection is measured
from the original nuetral surface of the beam to the nuetral surface of the
deformed beam.The configuration assumed by the deformed beam is known
as the elastic curve.
Elastic Curve
y
x
y = the deflection of the beam at any distance x

3
DOUBLE INTEGRATION METHOD
dx
y
dθ
dθ
x
θ
Elastic Curve
ρ
x
y
0,0
Figure shows the exaggerated view of the elastic curve of a deflected beam. Select
the left support as origin of an x axis directed along the original undeflected position
of the beam and a y axis directed positive upward. The deflections are so small so
that the difference in the original length of the beam and the projection of its
deflected length is negligible. Consequently, the elastic curve is very flat so its
slope at any point is very small. The value of the tan θ = dy/dx maybe set to θ with
minimal error.
L
d
s
O

4
dx
dy


2
2
dx
y
d
dx
d


Differentiating with respect to x
ds = pdθ
Differential length of arc ds
Since elastic curve is very flat
ds = dx
dx = pdθ EQ. 2
Combining EQs. 1 and 2
EQ. 1
2
2
1
dx
y
d


EQ. 3
From the derivation of the flexure formula
EI
M


1
EQ. 4
Equating EQs 3 and 4
2
2
dx
y
d
EI
M  EQ. 5
Equation 5 is known as the differential
equation of the elastic curve of a beam.
The product EI is called the flexural rigidity
which is usually constant along the length
of the beam. M is the moment equation
expressed in terms of x.

5
Integrating EQ. 5 assuming EI is constant
 
 1
C
Mdx
dx
dy
EI EQ. 6
This is the equation of the slope of the elastic curve or the value of dy/dx
at any point. C1 is a constant of integration to be evaluated from the given
condition of loading.
Integrating EQ. 6 assuming EI is constant
2
1 C
x
C
Mdxdx
EIy 

  EQ. 7
This is the equation of the deflection of the elastic curve the value of y for
any value of x . C2 is another constant of integration to be evaluated
from the given condition of the beam and its loading.
To evaluate C1 and C2
from the given figure of the elastic curve @ x = 0, y = 0
@ x = L, y = 0

7
1
2
2
2
}
6
{
5
.
37
}
2
{
30
39 C
x
x
x
dx
dy
EI 





2
1
3
3
3
}
6
{
5
.
12
}
2
{
10
13 C
x
C
x
x
x
EIy 






@ x = 0, y = 0
2
1
3
3
3
)
0
(
}
6
0
{
5
.
12
}
2
0
{
10
)
0
(
13
)
0
( C
C
EI 






Neglect Neglect
0
2 
C
@ x = 10, y = 0
0
)
10
(
}
6
10
{
5
.
12
}
2
10
{
10
)
10
(
13
)
0
( 1
3
3
3






 C
EI
0
)
10
(
800
5120
13000
0 1 



 C 708
1 

C
Actual equation of deflection
x
x
x
x
EIy 708
}
6
{
5
.
12
}
2
{
10
13 3
3
3






@ x = 5
)
5
(
708
}
6
5
{
5
.
12
}
2
5
{
10
)
5
(
13 3
3
3






EIy
Neglect
3540
270
1625 


EIy
2185


EIy
3
.
2185
m
kN
EI
y 


8
Problem
Using double integration method, locate and compute the maximun
deflection of the beam given in the previous problem. Assume constant EI.
2m
60 kN 75 kN
4m
4m
Point of maximum deflection
Where slope of elastic curve is
zero
x
1
2
2
2
}
6
{
5
.
37
}
2
{
30
39 C
x
x
x
dx
dy
EI 





Equation of slope as derived in the previous problem
Neglect this since the value of x is
between 2 and 6
Set this to zero
708
}
2
{
30
39
0 2
2



 x
x

9
708
}
4
4
{
30
39
0 2
2




 x
x
x
708
120
120
30
39
0 2
2




 x
x
x
828
120
9
0 2


 x
x
)
9
(
2
)
828
)(
9
(
4
)
120
(
120 2





x
location
m
x 
 014
.
5
x
x
x
x
EIy 708
}
6
{
5
.
12
}
2
{
10
13 3
3
3






)
014
.
5
(
708
}
2
014
.
5
{
10
)
014
.
5
(
13 3
3




EIy
Maximum deflection
3
max .
02
.
2185
m
kN
EI
y 


10
2m
45 kN
3m
24 kN/m
Problem : Determine the
Slope and deflection under the
45 kN load. EI is constant.
2m
45 kN
3m
24 kN/m
A B
RA=87
  0
B
M
0
5
.
2
)
5
(
24
)
3
(
45
)
5
( 


A
R
kN
RA 87

x


 L
M
M
dx
y
d
EI 2
2
2
2
2
12
}
2
{
45
87 x
x
x
dx
y
d
EI 




11
1
3
2
2
4
}
2
{
5
.
22
5
.
43 C
x
x
x
dx
dy
EI 




2
1
4
3
3
}
2
{
5
.
7
5
.
14 C
x
C
x
x
x
EIy 





When x=0, y =0
2
1
4
3
3
0
0
}
2
0
{
5
.
7
)
0
(
5
.
14
)
0
( C
C
EI 




 0
2 
C
When x=5, y =0
0
5
5
}
2
5
{
5
.
7
)
5
(
5
.
14
)
0
( 1
4
3
3





 C
EI
0
5
625
5
.
202
5
.
1812
0 1 



 C 197
1 

C
)
2
(
197
2
}
2
2
{
5
.
7
)
2
(
5
.
14 4
3
3





EIy
When x=2, y =?
3
.
294
m
kN
EI
y


1
3
2
2
4
}
2
{
5
.
22
5
.
43 C
x
x
x
dx
dy
EI 




When x=2, dy/dx =?
197
)
2
(
4
}
2
2
{
5
.
22
)
2
(
5
.
43 3
2
2





dx
dy
EI
2
.
55
m
kN
EI
dx
dy 


12
Problem
Using double integration method, determine deflection midway between
supports of the overhanging beam shown. Assume constant EI.
1m
600 N
2m
3m 2m
400 N/m
A B C D E
1m
600 N
2m
3m 2m
400 N/m
A B C D E
Establishing continuity of loading
400 N/m

13
1m
600 N
2m
3m 2m
400 N/m
A B C D E
400 N/m
RA
RD
400(7)=2800
1.5
400(4)=1600
  0
D
M
0
)
5
.
1
(
2800
)
2
(
600
)
6
( 


A
R N
RA 500

  0
y
F
0
2800
600
1600
500 




D
R
N
RD 1300


14
1m
600 N
2m
3m 2m
400 N/m
A B C D E
400 N/m
RA=500
RD=1300
x
400{x-1}
2
1

x
400{x-4}
2
4

x
X-6


 L
M
M
dx
y
d
EI 2
2
2
}
1
{
}
1
{
400
2
}
4
{
}
4
{
400
}
6
{
1300
500
2
2









x
x
x
x
x
x
dx
y
d
EI
2
2
2
2
}
1
{
200
}
4
{
200
}
6
{
1300
500 





 x
x
x
x
dx
y
d
EI

15
1
3
3
2
2
}
1
{
3
200
}
4
{
3
200
}
6
{
650
250 C
x
x
x
x
dx
dy
EI 







2
1
4
4
3
3
}
1
{
3
50
}
4
{
3
50
}
6
{
3
650
3
250
C
x
C
x
x
x
x
EIy 








At x =0 , y = 0
2
0
0
0
0
0
0 C





 0
2 
C
At x =6 , y = 0
0
6
}
1
6
{
3
50
}
4
6
{
3
50
}
6
6
{
3
650
)
6
(
3
250
0 1
4
4
3
3








 C
6
3
31250
3
800
0
18000
0 1
C




 33
.
1308
1 

C
x
x
x
x
x
EIy 33
.
1308
}
1
{
3
50
}
4
{
3
50
}
6
{
3
650
3
250 4
4
3
3








At x =3 , y = ?
)
3
(
33
.
1308
}
1
3
{
3
50
)
3
(
3
250 4
3




EIy
3
.
67
.
1941
m
kN
EI
y



16
Determine the deflection at the free end of the overhanging beam given in the
previous problem.
x
x
x
x
x
EIy 33
.
1308
}
1
{
3
50
}
4
{
3
50
}
6
{
3
650
3
250 4
4
3
3








At x =8 , y = ?
)
8
(
33
.
1308
}
7
{
3
50
}
4
{
3
50
}
2
{
3
650
)
8
(
3
250 4
4
3
3





EIy
3
31400
3
120050
3
12800
3
5200
3
128000





EIy
3
.
3
5450
m
kN
EI
y



17
w
L/2 L/2
y
w
L/2 L/2
Determine the maximum deflection of the beam shown in figure.
4
wL
RA 

18
w
L/2
4
wL
RA 
x
z
2
zx
3
x


 L
M
M
dx
y
d
EI 2
2
3
2
4
2
2
x
zx
x
wL
dx
y
d
EI 

2
L
x
w
z

x
L
w
z
2

3
2
2
4
2
2
x
L
wxx
x
wL
dx
y
d
EI 

L
wx
x
wL
dx
y
d
EI
3
4
3
2
2


1
4
2
12
8
C
L
wx
x
wL
dx
dy
EI 


2
1
5
3
60
24
C
x
C
L
wx
x
wL
EIy 




19
2
1
5
3
60
24
C
x
C
L
wx
x
wL
EIy 



x=0 , y=0
2
0
0
0
0 C



 0
2 
C
x=L/2 , dy/dx=0
1
4
2
12
8
C
L
wx
x
wL
dx
dy
EI 


1
4
2
12
)
2
(
)
2
(
8
0 C
L
L
w
L
wL


 1
3
3
192
32
0 C
wL
wL



192
5 3
1
wL
C 

x
wL
L
wx
x
wL
EIy
192
5
60
24
3
5
3



At x= L/2 , y =ymax

Plate No.1 :Beam Deflections by double integration method
1m
50+2N kN 60+N kN
3m
2m
1. Using double integration method, determine the midspan deflection
of the beam shown. Assume constant EI.
2. Determine the slope and deflection under the 35+ N kN load. EI is constant.
2m
35+N kN
4m
16 + N kN/m

3. Using double integration method, locate and compute the maximun
deflection of the beam shown . Assume constant EI.
w
L

22
AB

Any load
A B
A
B
t
cg
B
x
EI
M
diagram
dt
dθ
C D
dx
EI
M
dx
dθ
dθ
ρ
ds
dt
O
Elastic Curve
Magnified View of Segment CD
AREA MOMENT METHOD
tangent @B
tangent @A
Figure shows the magnified view of segment CD. Two adjacent
plane sections of an originally straight beam rotates through the
angle dθ. The arc distance ds measured along the elastic curve
equals ρdθ, where ρ is the radius of curvature of the elastic curve at
the given position. The elastic curve is relatively flat so ds is
assumed to be equal to dx without any serious error.
x

23
From the derivation of the flexure formula
EI
M


1
ds = pdθ
ds
EI
M
d 
 but ds = dx
dx
EI
M
d 
 EQ. 1
The tangents drawn to the elastic curve at C and D in figure are
separated by the small angle dθ by which OC and CD rotate relative to
each other. Hence the change in slope between the elastic curve at any
two points A and B will be equal to the sum of such small angles.


B
A
AB d
 

B
A
x
x
AB Mdx
EI
1
 Theorem 1

24
x
dt
d 


tan 
xd
dt 
 

 
xd
dt
t
A
B


B
A
x
x
A
B Mdx
x
EI
t )
(
1 Theorem 2
Theorem 1
The change in rotation between any two tangents drawn at any two
points on a continuous elastic curve is equal to the area of the M/EI
diagram between these two points.
Theorem 2
The deviation of the tangent to the elastic curve at point B with respect
To the tangent drawn to the elastic curve at point A in a direction
perpendicular to the original straight axis of the beam is equal to the
moment of area of the M/EI diagram from A to B about point B.

26
Convention of signs
+θAB
+tB/A
A
B -θAB
-tB/A
A
MOMENT DIAGRAM BY PARTS
In order to apply the area moment theorems effectively, area of the moment
diagram and locations of the centroids of its area is necessary. However
these is quite complicated since often times moment diagrams consist of
irregular curves which requires integration to compute and locate the centroid
of the area of the moment diagram.
To simply this, a method of dividing the moment diagrams into parts
whose areas and centroids are known will be adopted. This method
is called moment diagram by parts. The basic technique is to draw the
moment diagram of each load separately.

27
The construction of moment diagram by parts are based on the following basic principles.
1. The resultant bending moment at any section caused by any load system is the
algebraic sum of the bending moment caused by each load acting separately.
R1
L
a
w
R2
R1
A A
w
B B
B
a
MB=R1L
MB=-wa2/2
R1L
-wa2/2
M-Diagram by parts
Resultant M-Diagram
Conventional M-Diagram

28
2. The moment effect of a single specified load is the variation from the general
equation y = kxn. The graph of the equation is shown in the figure below.
h
b
y = kxn
c.g
x
bh
n
area
1
1


b
n
x
2
1


Where
n =degree of curve
n = 0 couple
n= 1 concentrated load
n =2 uniformly distributed load
n =3 uniformly varying load

29
Lh
2
1
Lh
3
1
Lh
4
1
Loading M- Diagram Area Centroid
P
L L
x
3
L
4
L
5
L
w L
x
L
x
2
2
wL
h 
6
2
wL
h 
PL
h 
w
L
L
M
L
L
x
M
h  Lh
2
L
Areas and centroid of areas of typical loadings

30
Application of Area – Moment Theorem
Determination of deflection at known point on a simple beam
Any load
A
B
A
B
t
B
x
C
EI
M
c

A
C
t
a
L
D
C
x
General Procedure
1. Sketch the elastic curve and M/EI
Diagram.
2. Draw tangent lines to the elastic
curve at the supports. Compute
A
B
t
EI
x
A
t B
AB
A
B 
3. By proportion solve for CD.
L
t
a
CD A
B

4. Draw another tangent to the elastic
curve at the point where the deflection
is to be determined. Compute
A
C
t
EI
x
A
t C
AC
A
C 
5.
c
 = CD -
A
C
t
A C B
Diagram

31
A
B
A
B
t
C
c

A
C
t
3m
D
Problems
1. Compute the midspan deflection of the beam shown below. Assume constant EI.
60 kN
2m 4m
20 kN/m
A
B
C
2m 4m
20 kN/m
60 kN
RA
kN
R
R
M
A
A
B
100
0
3
)
6
(
20
)
4
(
60
)
6
(
0






A
B
C
RA=100
6m
100(6)=600

32
A
B
C
6m
20 kN/m
360
2
)
6
(
20
2
2
2


wL
-360
A
B
C
6m
60
4 m
-60(4)=-240

-360
-60(4)=-240
100(6)=600
4 m
1
2
3
6 m
1
x
2
x
3
x
480
4
)
240
(
2
1
720
3
6
)
360
(
1
1800
2
)
6
)(
600
(
1
3
2
1










A
A
A
3
4
5
.
1
)
6
(
4
1
2
3
6
3
2
1





x
x
x

3
3
3
2
2
1
1
.
1880
)
3
4
(
480
)
5
.
1
(
720
)
2
(
1800
m
kN
EI
EI
t
EI
x
A
x
A
x
A
t
EI
x
A
t
A
B
A
B
B
AB
A
B








3
.
940
)
6
(
1880
3
m
kN
EI
CD
EI
CD
L
t
a
CD A
B




A C
3m
60 kN
2m
20 kN/m
RA=100
100(3)=300
90
2
)
3
(
20
2
2
2





wL
-60(1)-60
4
5
6
4
x
5
x
6
x
30
1
)
60
(
2
1
90
3
3
)
90
(
1
450
2
)
3
)(
300
(
1
3
2
4










A
A
A
3
1
75
.
0
)
3
(
4
1
1
3
3
6
5
4





x
x
x

3
6
6
5
5
4
4
.
5
.
372
)
3
1
(
30
)
75
.
0
(
90
)
1
(
450
m
kN
EI
EI
t
EI
x
A
x
A
x
A
t
EI
x
A
t
A
C
A
C
C
AC
A
C








c
 = CD -
A
C
t
3
.
5
.
567
5
.
372
940
m
kN
EI
EI
C 




37
24 kN/m
96kN
48 kN
2 m 4 m 1 m
Compute the deflection at the free end of the overhanging beam shown below.
Assume constant EI.
A
B C
  0
B
M
0
)
4
(
96
3
)
6
(
24
)
1
(
24
6 



A
R
132

A
R
A
B C
B
A
t
y
B
C
t
132(6)=792
432
2
)
6
(
24 2



-96(4)=-384
-24(1)=-24
1
2
3
4

38
45 kN
15 kN/m
3m 1m 1m
3. Compute the slope and deflection at the free end of the cantilever beam.
A
B
B
tA
A 
 AB

A

5
.
22
6
)
3
(
15
6
2
2


wL
45
)
1
(
45 
22.5
45
B
tA
A 

EI
x
A A
AB
A 

EI
A
]
3
1
3
[
1
)
45
(
2
1
3
)
5
4
(
3
)
5
.
22
(
4
1




3
.
5
.
115
m
kN
EI
A 

EI
A
1
)
45
(
2
1
3
)
5
.
22
(
4
1



2
.
375
.
39
m
kN
EI
A 


39
Location of maximum deflection
Any load
A
B
A
B
t
B
x
C
max

x
L
A
x
θA
θAC
C
A
t
A C B
Note : at point of maximum deflection
Slope of the elastic curve is zero.
Tangent at this point is horizontal
so θA = θAC gives the location of the
Maximum deflection.
Likewise,
EIL
x
A
d B
AB
A
A 


tan
EI
x
A
t A
AC
C
A 

max

EI
AAC
AC 

L
A
t
d
B
A
A 


tan

40
50 kN 75 kN
Problem:
Locate and compute the maximum deflection of the beam shown below.
Assume constant EI.
1 m 3 m 1 m
A
B
A
B
t
C
max

x
L
θA
θAC
C
A
t
RA=55
kN
R
R
M
A
A
B
55
0
)
4
(
50
)
1
(
75
)
5
(
0






55(5)=275
-50(4)=-200
-75(1)=-75
L
A
t
d
B
A
A 


tan
EIL
x
A B
AB
A 

)
5
(
)
3
1
)(
1
)(
75
(
2
1
)
3
4
(
4
)
200
(
2
1
)
3
5
(
5
)
275
(
2
1
EI
A




1
.
120
EQ
EI
A 



41
1 m 3 m 1 m
RA=55
50 75
A C
x
55x
-50(x-1)
EI
AAC
AC
A 


EI
x
x
x
x
A
)
1
)(
1
)(
50
(
2
1
)
55
(
2
1





EI
x
x
A
2
2
)
1
(
25
5
.
27 



EI
x
x
x
A
)
1
2
(
25
5
.
27 2
2





2
.
25
50
5
.
2 2
EQ
EI
x
x
A 




25
50
5
.
2
120 2


 x
x
58
20
0 2


 x
x
By quadratic formula
m
x 57
.
2


42
55x=55(2.57)=141.35
-50(x-1)=- 50(2.57-1)=- 78.5
EI
x
A
t A
AC
C
A 

max

EI
)]
57
.
1
(
3
2
1
)[
57
.
1
)(
5
.
78
(
2
1
57
.
2
)
3
2
(
57
.
2
)
35
.
141
(
2
1
max




2.57
1.57
A C
3
max .
08
.
185
m
kN
EI



43
50 kN 75 kN
Problem:
Locate and compute the maximum deflection of the beam shown below.
Assume constant EI.
1 m 3 m 1 m
Non Prismatic Beams ( Beams with variable I)
Problem
A cantilever beam of Length L with stepwise cross section carries a vertical load P
at its free end. The section of the beam changes midway along its length so that
its second moment of area is reduced by half. The smaller section is towards the
free end. If E is constant, determine the deflection at the free end.

44
P
L/2 L/2
EI
PL
2
EI
PL
4
EI
PL
2
I
2I
A B
B
tA
A 


45
EI
PL
2
EI
PL
4
EI
PL
2
A1
A2
A3
A
EI
PL
EI
PLL
A
8
2
)
2
(
2
1 2
1 

EI
PL
EI
PLL
A
8
2
)
4
(
1 2
2 

EI
PL
EI
PLL
A
16
2
)
4
(
2
1 2
3 

3
)
2
(
3
2 L
L

L
L
L
4
3
)
2
(
2
1
2


L
L
L
6
5
)
2
(
3
2
2


3
3
2
2
1
1 x
A
x
A
x
A
B
tA
A 




6
5
1
16
4
3
8
3
8
2
2
2
L
E
PL
L
EI
PL
L
EI
PL
A 



EI
PL
E
PL
EI
PL
EI
PL
A
16
3
1
96
5
32
3
24
3
3
3
3






46
Conjugate Beam Method
an imaginary beam with span equal to the span of the real beam. Loading
consists of the M/EI diagram of the real beam. Positive M/EI represents
downward loadings.
EI
M
c

c

Any load
Real Beam
L
Conjugate Beam
C
V C
M
C
C
L
The clockwise rotation of the
tangent at any point on the
elastic curve of the real beam is
equal to the positive shear about
the same point on the conjugate
beam
The downward deflection at any
point on the elastic curve of the
real beam is equal to the positive
moment about the same point on
the conjugate beam
c
 C
M
=
c
 C
V
=

47
Other properties of the conjugate beam
 A fixed end for a real beam becomes free end for the conjugate beam.
 A free end for a real beam becomes fixed end for the conjugate A
simple support for the real beam remains a simple support for the
conjugate beam.
 An interior support of the real becomes an internal hinge for the
conjugate beam and conversely.
 The conjugate beam of the real beam is always determinate.
Real Beam
Conjugate Beam
Real Beam
Conjugate Beam
Real Beam
Conjugate Beam
ILLUSTRATION

48
EI
M
AC

c

Any load
Real Beam
Conjugate Beam
C
V C
M
C
L
C

A

C
A
t
A B
C
A
B
t
From the geometry of elastic curve
θA = θC+ θAC
θC = θA - θAC
LEI
x
A
L
A
B
t
B
AB
A 


EI
AAC
AC 

EI
A
LEI
x
A AC
B
AB
C 

 EQ. 1
XB
a
D

49
EI
M
C
V C
M
C
L
XB
RA
From the conjugate beam
∑MB = 0
0


EI
X
A
L
R B
AB
A
EI
AAB
EIL
X
A
R B
AB
A 
C
V C
M
C
a
RA
EI
AAC
Vc =∑FV
EI
A
LEI
x
A
V AC
B
AB
C 
 EQ. 2
xc

50
EI
M
C
V C
M
C
L
XB
RA
∑MB = 0
0


EI
X
A
L
R B
AB
A
EI
AAB
EIL
X
A
R B
AB
A 
C
V C
M
C
a
RA
EI
AAC
Vc =∑FV
EI
A
LEI
x
A
V AC
B
AB
C 
 EQ. 2
xc
C
C V


EQ.1 = EQ.2

51
L
A
B
t
a
CD

a
LEI
x
A
CD B
AB

EI
x
A
t C
AC
A
C 
c
 = CD -
A
C
t
EI
x
A
a
LEI
x
A c
AC
B
AB
c 

 EQ. 3
Mc =∑Mc
EI
x
A
a
R
M c
AC
A
c 

EI
x
A
a
LEI
x
A
M c
AC
B
AB
c 
 EQ. 4
EQ. 3 = EQ. 4
c
c M


C
V C
M
C
a
EI
AAC
xc
EIL
X
A
R B
AB
A 

52
2 m 4 m
20 kN/m
60 kN
1. Using the conjugate beam method, compute the midspan deflection of
the beam shown. Assume constant EI
kN
R
R
M
A
A
B
100
0
3
)
6
(
20
)
4
(
60
)
6
(
0






100(6)=600
RA
360
2
)
6
(
20
2
2
2


wL
-360
-60(4)=-240
A
C B
Conjugate beam
c

3m

53
600/EI
360/EI
240/EI
A
3m
B
C
RA
’
0
)
3
4
(
)
5
.
1
(
)
2
(
)
6
(
'
0
3
2
1 





P
P
P
R
M
A
B
4m
P1
P2
P3
EI
EI
P
1800
2
6
)
600
(
1
1 

EI
EI
P
720
3
6
)
360
(
1
2 

EI
EI
P
480
2
4
)
240
(
1
3 

2m
1.5m
4/3
EI
R
EI
EI
EI
R
A
A
36
.
313
0
)
3
4
(
480
)
5
.
1
(
720
)
2
(
1800
)
6
(
'
'






54
300/EI
A
3m
C
P4
60(1)=60/EI
90
6
3
360 2
2


y
y
90/EI
P5
P6
1m
EI
EI
P
450
2
3
)
300
(
1
4 

EI
EI
P
90
3
3
)
90
(
1
5 

EI
EI
P
30
2
1
)
60
(
1
6 

)
3
1
(
)
3
)(
4
1
(
)
1
(
)
3
(
' 6
5
4 P
P
P
RA
C 




EI
RA
36
.
313
'



 C
c
c M
M


55
3
.
58
.
567
)
3
1
(
30
)
3
)(
4
1
(
90
)
1
(
450
)
3
(
36
.
313
m
kN
EI
EI
EI
EI
EI
C
C








56
24 kN/m
72 kN
48 kN
2 m 4 m 1 m
2. Compute the deflection at the free end of the overhanging beam shown below.
Assume constant EI.
A
B
C
kN
R
R
M
A
A
B
112

0
)
3
)(
6
(
24
)
4
(
72
)
1
(
48
)
6
(
0







112
112(6)=672
432
2
)
6
(
24 2



-72(4)=-288
-48(1)=-48

57
672/EI
EI
432
288/EI
48/EI
A B C
A B
EI
432
288/EI
672/EI
RB
2 m 4 m
P1
P2
P3
EI
EI
P
2016
)
6
(
2
)
672
(
1
1 

EI
EI
P
864
)
6
(
3
)
432
(
1
2 

EI
EI
P
576
)
4
(
2
)
288
(
1
3 

4 m
3/4 (6)= 4.5m
2+2/3 (4)= 14/3m

58
2
1
3
2
.
248
0
)
4
(
1
2016
)
3
14
(
576
)
5
.
4
(
864
)
6
(
0
)
4
(
)
3
14
(
)
5
.
4
(
)
6
(
0
m
kN
EI
R
E
EI
EI
R
P
P
P
R
M
B
B
B
A











B C
48/EI
EI
RB
248

P4
EI
EI
P
24
)
1
(
2
)
48
(
1
4 

1m
2/3


 c
c
c M
M

)
1
(
)
3
2
(
4 B
c R
P 






 3
.
232
)
1
(
248
)
3
2
(
24
m
kN
EI
EI
EI
c


59
24 kN/m
72 kN
48 kN
2 m 4 m 1 m
Assume constant EI.
A
B
C
kN
R
R
M
A
A
B
112

0
)
3
)(
6
(
24
)
4
(
72
)
1
(
48
)
6
(
0







112
112(6)=672
-72(4)=-288
-48(1)=-48
432
2
)
6
(
24 2



-432

60 kN
2 m 4 m
3. Using the conjugate beam method, locate and compute the maximum
deflection of the beam shown. Assume constant EI
A B
x
C At point of maximum deflection
0

c

RA
kN
R
R
M
A
A
B
40
0
)
4
(
60
)
6
(
0





40(6)=240
-60(4)=-240
max



 c
c M
M
max


61
240/EI
240/EI
A
2 m 4 m
EI
EI
P
720
)
6
(
2
)
240
(
1
1 

EI
EI
P
480
)
4
(
2
)
240
(
1
2 

P1
P2
2m
4/3
R1
B
x
0
)
3
4
(
)
2
(
)
6
(
0
2
1
1 




P
P
R
MB
0
)
3
4
(
480
)
2
(
720
)
6
(
1 


EI
EI
R
EI
R
33
.
133
1 
A
C

62
60 kN
2 m
A C
RA=40
X
40X
-60(X-2)
40X/EI
60(X-2)/EI
R1=133.33/EI
P3
P4
X
X-2
EI
x
x
EI
x
P
2
3
20
)
(
2
)
40
(
1


EI
x
x
EI
x
P
2
4
)
2
(
30
)
2
(
2
)
2
(
60 





63
0
20
)
2
(
30
33
.
133
0
0
2
2
3
4
1








 
EI
x
EI
x
EI
P
P
R
FV
c

m
x
x
x
x
x
x
x
x
73
.
2
2
)
33
.
25
(
4
)
12
(
)
12
(
0
33
.
25
12
0
33
.
253
120
10
0
20
120
120
30
33
.
133
2
2
2
2
2
2
















64
40X/EI
60(X-2)/EI
R1=133.33/EI
P4
X
X-2
EI
EI
x
EI
x
P
06
.
149
)
73
.
2
(
20
)
(
2
)
40
(
1 2
3 


EI
EI
x
EI
x
P
16
)
2
73
.
2
(
30
)
2
(
2
)
2
(
60 2
4 





P3


 c
c M
M
max

C
A
)
73
.
2
(
3
06
.
149
)
2
73
.
2
(
3
16
)
73
.
2
(
33
.
133
3
)
2
(
3
1
max
3
4
1
max
EI
EI
EI
x
P
x
P
x
R










3
max .
24
.
232
m
kN
EI



65
180 kN
3 m 6 m 3 m
I 2I I
3 m
4. Using the conjugate beam method, compute the maximum deflection
of the beam shown. Assume constant E
b
A B
 
2
2
4
3
48
b
L
EI
Pb
C 


P
L
at center
b = distance of P from the nearer support < L/2
C

L/2
EI
EI
C
230
]
)
2
(
4
)
6
(
3
[
48
)
2
(
60 2
2





66
180 kN
3 m 6 m 3 m
2I I
3 m
A B C D E
A B C D E
90 kN
270
540
270
270/EI 270/EI
135/EI 135/EI
270/EI
I
Ordinary M - Diagram
Modified M/EI – Diagram
(conjugate beam)
Solution

67
270/EI
135/EI
270/EI
Modified M/EI – Diagram
(conjugate beam)
F1 F2
F3
R
F1 =1/2(270)3 =405
F2 = 135(3) = 405
F3 =1/2(135)3 =202.5
R = F1 + F2 + F3 = 1012.5
C
)
1
(
)
5
.
1
(
)
4
(
)
6
( 3
2
1 F
F
F
R
c 




EI
EI
EI
EI
EI
c
3645
)
1
(
5
.
202
)
5
.
1
(
405
)
4
(
405
)
6
(
5
.
1012








 c
c
c M
M

3m 3m

69
2 m 4 m
20+N kN/m
30N kN
1. Using the area moment method, compute the midspan deflection of
20N kN/m
24N kN
48+3N kN
2 m 4 m 1 m
Assume constant EI.
A
B
C
Plate No.2: area moment method

70
30N kN
2 m 4 m
3. Using the area moment method, locate and compute the maximum
A B
C

71
2 m 4 m
20+N kN/m
30N kN
1. Using the conjugate beam method, compute the midspan deflection of
20N kN/m
24N kN
48+3N kN
2 m 4 m 1 m
Assume constant EI.
A
B
C
Plate No.3 : Conjugate beam method

30N kN
2 m 4 m
3. Using the conjugate beam method, locate and compute the maximum
A B
C

73
VIRTUAL WORK METHOD
Virtual Work Equation for Beams and Frames
the slope and deflection in any direction at a point in a beam or frame can be
obtained by applying a unit load at that point and applying the formula
where Mp = the bending moment at the element under consideration
due to the applied loadings
Mu = the bending moment due to a unit load applied at the
point where the deflection is required
If rotation at a point is required, apply a unit couple at the point to
determine Mu,then use EQ. A.
dx
EI
M
M
L
U
P


0
 EQ. A
dx
EI
M
M
L
U
P


0


74
c

c

Any load
Real System
L
C
L
1
L
1
C
C
Virtual System for deflection at C
Virtual System for Rotation at C

75
General Procedure : Slope and deflection of beams by Virtual Work Method
1. Real System - Draw a diagram of the beam showing all the given forces.
2. Virtual System - Draw a diagram of the beam without the real loads. If
the deflection is to be determined , apply a unit load at the point in the
direction of the desired deflection. If the slope is to be determined, apply
a unit couple on the beam where the slope is desired.
3. Examine the real and virtual systems; the variation of EI along the length
of the beam. Divide the beam into segments so that the real and virtual
systems is continuous in each segment.
4. For each segment of the beam, determine an equation expressing the
Variation of the bending moment due to the real loading (MP) along the
length of the segment in terms of position coordinate x. The origin x
maybe located anywhere on the beam and must be chosen so that the number of
terms in the equation is minimum. Use the sign convention of bending moment,
Variation of the bending moment due to the virtual loading (Mu)
using the same x coordinate adopted in step 4. The sign convention of
bending moment due to the real loading must be the same for the virtual loading.
6. Determine the desired slope or deflection using
dx
EI
M
M
L
U
P


0

If the beam is divided into segments, then integral on the right side of EQ. 1
can be evaluated by adding algebraically the integrals for all segments of the beam.
EQ. 1
dx
EI
M
M
L
U
P


0


76
120 kN
2 m 4 m
Problem 1:
Using the virtual work method, compute the slope and deflection under the
120 kN load of the beam shown. Assume constant EI
2 m 4 m
120 kN
A B C
80 40
Real system
x
A B
80
Segment AB ( 0rigin at A)
0<x<2
Mp = 80x

Segment BC ( 0rigin at C)
0<x<4
40
x
Mp = 40x
2 m 4 m
1
A B C
Virtual system
For vertical
Deflection at B x
A B
Mu =2/3x
2/3
2/3
1/3
0<x<2
x
1/3
B C
Mu =1/3x
0<x<4

Segment Origin Limits Mp Mu
AB A 0 – 2 80x 2/3x
CB C 0 – 4 40x 1/3x
dx
EI
M
M
L
U
P


0


 

4
0
2
0
3
1
40
3
2
80
dx
EI
x
x
dx
EI
x
x
B


 

4
0
2
2
0
2
3
40
3
160
dx
x
EI
dx
x
EI
B

   4
0
3
2
0
3
9
40
9
160
x
EI
x
EI
B 


)
64
(
9
40
)
8
(
9
160
EI
EI
B 



 3
.
9
3840
m
kN
EI
B


2 m 4 m
1
A B C
1/6 1/6
Virtual system
For slope at B
x
A B
Mu = -1/6x






6
1
0
6
1
0
A
A
C
R
R
M
0<x<2
x
B C
Mu =1/6x
1/6
1/6
0<x<4
AB A 0 – 2 80x -1/6x
CB C 0 – 4 40x 1/6x

dx
EI
M
M
L
U
P


0


 


4
0
2
0
6
1
40
)
6
1
(
80
dx
EI
x
x
dx
EI
x
x
B


 


4
0
2
2
0
2
3
20
3
40
dx
x
EI
dx
x
EI
B

   4
0
3
2
0
3
9
20
9
40
x
EI
x
EI
B 



)
64
(
9
20
)
8
(
9
40
EI
EI
B 



2
.
9
960
m
kN
EI
B 


81
120 kN
2 m 4 m
Problem 1:
Using the virtual work method, compute the slope and deflection under the
120 kN load of the beam shown. Assume constant EI
2 m 4 m
20 kN/m
60 kN
Problem 2 :
Using the virtual work method, compute the midspan deflection of

82
2 m 4 m
20 kN/m
60 kN
3 m
1
A B C D
A B C D
1/2 1/2
100 80
x
A B
100
0<x<2
Mp = 100x – 20x (x/2)
Mp =100x -10x2
x
A B
Mu =1/2x
20 kN/m
1/2
Segment BC ( 0rigin at A)
2<x<3
x
100
60 kN
x-2
20 kN/m
B C
x
A B C
Mu =1/2x
1/2
Mp=100x -20x(x/2) – 60(x-2)
Mp =40x-10x2+120
Solution to Problem 2
Real System
Virtual system
for vertical
deflection of C

83
Segment CD ( 0rigin at D)
0<x<3
x
Mp = 40x
x
C D
Mu =1/2x
80 1/2
20 kN/m
C D
Mp = 80x – 20x (x/2)
Mp =80x -10x2
dx
EI
M
M
L
U
P


0

AB A 0 – 2 100x-10x2 1/2x
BC A 2 – 3 40x-10x2 +120 1/2x
DC D 0 – 3 80x-10x2 1/2x

84










3
0
2
3
2
2
2
0
2
2
1
)
10
80
(
2
1
)
120
10
40
(
2
1
)
10
100
(
dx
EI
x
x
x
dx
EI
x
x
x
dx
EI
x
x
x
c

3
0
4
3
3
2
2
4
3
2
0
4
3
4
5
3
40
1
30
4
5
3
20
1
4
5
3
50
1

























x
x
EI
x
x
x
EI
x
x
EI
c





























4
405
360
1
)
120
20
3
160
(
)
270
4
405
180
(
1
20
3
400
1
EI
EI
EI
c























4
1035
1
3
460
4
1395
1
3
340
1
EI
EI
EI
c

EI
EI
EI
c
12
6810
12
480
7290
1
3
120
4
2430
1






















3
0
3
2
3
2
3
2
2
0
3
2
)
5
40
(
)
60
5
20
(
)
5
50
(
dx
EI
x
x
dx
EI
x
x
x
dx
EI
x
x
c


85
2 m 4 m
24 kN/m
A
48 kN
Problem
Determine the slope and deflection at the free end of the overhanging beam showm
2 m 4 m
24 kN/m
A
48 kN
56
2 m
2 m
136
Real System
B
B
C
C







kN
R
R
M
B
B
A
136
0
6
)
8
(
48
3
)
6
(
24
0












kN
R
R
R
R
F
A
A
B
A
y
56
0
)
6
(
24
48
136
0
)
6
(
24
48
0

86
2 m 4 m
A B C
1
1/3
2 m
4/3
Virtual System
for vertical
deflection at C
x
24 kN/m
A
56
Mp = 56x -12x2
A
1/3
x
Mu= - 1/3x
x
B
B
B
C
48
x
B C
1
Mp=-48x Mu= -x

87
AB A 0 – 6 56x-12x2 -1/3x
CB C 0 – 2 -48x -x
dx
EI
M
M
L
U
P


0









2
0
6
0
2
)
(
48
)
3
1
)(
12
56
(
dx
EI
x
x
dx
EI
x
x
x
C


 


2
0
2
6
0
2
3
48
1
)
56
12
(
3
1
dx
x
EI
dx
x
x
EI
C

 2
0
3
6
0
3
4 16
3
56
3
3
1
x
EI
x
x
EI
C 









EI
EI
EI
C
80
)
8
(
16
)
4032
3888
(
3
1






88
2 m 4 m
A B C
1
1/6
2 m
1/6
Virtual System
for slope at C
∑MB = 0
1 – 6RA = 0
RA = 1/6
x
24 kN/m
A
56
Mp = 56x -12x2
A
1/6
x
Mu= - 1/6x
B
B
x
B C
x
B C
Mp= - 48x Mu= -1
48
1

89
AB A 0 – 6 56x-12x2 -1/6x
CB C 0 – 2 -48x -1
dx
EI
M
M
L
U
P


0


 



2
0
6
0
2
)
1
(
48
)
6
1
)(
12
56
(
dx
EI
x
dx
EI
x
x
x
C


 


2
0
6
0
2
3 48
)
56
12
(
6
1
xdx
EI
dx
x
x
EI
C

 2
0
2
6
0
3
4 24
3
56
3
6
1
x
EI
x
x
EI
C 









EI
EI
EI
C
72
)
4
(
24
)
4032
3888
(
6
1






90
180 kN
3 m 6 m 3 m
I 2I I
3 m
4. Using the virtual work method, compute the vertical deflection
of point D of the beam shown. Assume constant E
A B C D E
180 kN
3 m 6 m 3 m
I 2I I
3 m
A B C D E
x
x
x
x
Real System
90
90

91
180 kN
3 m 6 m 3 m
I 2I I
3 m
Problem
Using the virtual work method, compute the vertical deflection
of point D of the beam shown. Assume constant E
A B C D E
180 kN
3 m 6 m 3 m
I 2I I
3 m
A B C D E
x
x
x
x
Real System
90 90

x
A
90
B
x
A B
1/4
1
3 m 6 m 3 m
I 2I I
A B C D E
x
x x
Virtual System
1/4 3/4
x
MP 90

Segment AB
0<x<3
x
MU
4
1

x
A
90
B
x
MP 90

Segment BC
3<x<6
C
x
A
1/4
B
x
MU
4
1

C

x
A
90
B
x
M
x
x
M
x
x
M
P
P
P
90
1080
1080
180
90
)
6
(
180
90








Segment CD
6<x<9
C
180
X-6
x
A
1/4
B
x
MU
4
1

C
D D
x
D E
90
Segment DE
0<x<3
x
MP 90

x
3/4
D E
x
MU
4
3


94
AB A 0 – 3 90x 1/4x
BC A 3 – 6 90x 1/4x 2EI
CD A 6 – 9 90x-180(x-6) =1080-90x 1/4x 2EI
ED E 0-3 90x 3/4x EI
EI
EI


  




3
0
9
6
3
0
6
3
)
4
3
(
90
2
4
1
)
90
1080
(
2
)
4
1
(
90
)
4
1
(
90
dx
EI
x
x
dx
EI
x
x
dx
EI
x
x
dx
EI
x
x
D


95
dx
EI
M
M
L
U
P


0



  




3
0
2
9
6
2
3
0
6
3
2
2
2
135
)
90
1080
(
8
1
4
45
2
45
dx
x
EI
dx
x
x
EI
dx
x
EI
dx
x
EI
D

       3
0
3
9
6
3
2
6
3
3
3
0
3
6
135
30
540
8
1
12
45
6
45
x
EI
x
x
EI
x
EI
x
EI
D 





     
27
6
135
)
2
972
1296
(
)
2
2187
4374
(
2
1
27
216
6
45
27
6
45
EI
EI
EI
EI
D 














EI
EI
EI
EI
EI
D
12
48843
6
3645
)
2
1215
3078
(
2
1
6
8505
6
1215















  




3
0
9
6
3
0
6
3
)
4
3
(
90
2
4
1
)
90
1080
(
2
)
4
1
(
90
)
4
1
(
90
dx
EI
x
x
dx
EI
x
x
dx
EI
x
x
dx
EI
x
x
D


96
2 m 4 m
12N kN/m
50 +NkN
Problem 1 :
Using the virtual work method, compute the midspan deflection of
2 m 4 m
14N kN/m
A
24NkN
Problem 2
Using the virtual work method,determine the slope and deflection at the free end of
the overhanging beam showm
2 m
B
C
Plate No. 3: Slope & Deflection by Virtual Work Method

97
Problem 3
A cantilever beam of Length L with stepwise cross section carries a vertical load P
at its free end. The section of the beam changes midway along its length so that
its second moment of area is reduced by one third. The smaller section is
towards the free end. If E is constant, determine the deflection at the free end
using the virtual work method
P
L/2 L/2
I
3I
A B C

P
L/2 L/2
I
3I
A
B
C
1
L/2 L/2
I
3I
A
B
C
Real System Virtual system
x
x
3I
x
x
Segment AB 0 <x < L/2
Px
MP 
 x
MU 

Segment BC L/2 <x < L
Px
MP 
 x
MU 


Segment Limits MP Mu EI
AB 0 – L/2 -Px -x EI
BC L/2 – L -Px -x 3EI
dx
EI
M
M
L
U
P


0


 

L
L
L
dx
EI
Pxx
dx
EI
Pxx
2
2
0 3


 

L
L
L
dx
EI
Px
dx
EI
Px
2
2
2
0
2
3
1

L
L
L
x
P
EI
x
P
EI 2
3
2
0
3
]
[
9
1
]
[
3
1


 EI
PL
EI
PL
EI
PL
PL
EI
PL
EI
PL
L
L
EI
P
L
EI
P
L
L
P
EI
L
P
EI
36
5
72
10
72
)
7
3
(
)
8
(
9
7
24
)
8
(
9
)
8
(
3
]
)
2
(
[
9
1
]
)
2
[(
3
1
3
3
3
3
3
3
3
3
3
3
3
3

















100

 

AE
L
F
F
dx
EI
M
M p
u
u
P


 

AE
L
F
F
dx
EI
M
M p
u
u
P

Slope and Deflection of Frames by Virtual Work Method
Considering axial deformation of
members of the frame
dx
EI
M
M u
P



dx
EI
M
M u
P



Neglecting axial deformation of
members of the frame
where
Mp = the bending moment at the element under
consideration due to the applied loadings
Mu = the bending moment due to a unit load
applied at the point where the deflection is required
or a unit couple applied at the point where the
slope is required.
Fp = the axial load at the element under
Fu = the axial load due to a unit load applied at the
point where the deflection is required or a unit
couple applied at the point where the slope is
required.
Note: unless otherwise stated in the problem, the effect of axial deformation
is negligible.

101
General Procedure : Slope and deflection of frames by Virtual Work Method
1. Real System – Determine the internal forces at the ends of the members
of the frame due to the real loading.
2. Virtual System - If the deflection is to be determined , apply a unit load at the point
in the frame in the direction of the desired deflection. If the slope is to be determined,
apply a unit couple on the frame where the slope is desired. Determine the member
end forces due to the virtual loading
3. Divide the frame into segments so that the real and virtual systems is continuous in
each segment.
4. For each segment of the frame , determine an equation expressing the
Variation of the bending moment due to the real loading (MP) along the
length of the segment in terms of position coordinate x. The origin x
maybe located anywhere on the frame and must be chosen so that the number of
terms in the equation is minimum. Use the sign convention of bending moment,
Variation of the bending moment due to the virtual loading (Mu) along the frame
using the same x coordinate adopted in step 4. The sign convention of
bending moment due to the real loading must be the same for the virtual loading.
6. If the effects of the axial deformation is to be included in the analysis, divide the
frame into segments so that the real and virtual forces and AE are constant in
each segment. It is not necessary that these segments be the same as in
steps 4 and 5. It is important that the same sign convention be used for Fp and Fu.
7. Determine the desired slope or deflection using

 

AE
L
F
F
dx
EI
M
M p
u
u
P
 
 

AE
L
F
F
dx
EI
M
M p
u
u
P


102
16 kN/m
36 kN
2 m
2 m
6 m
A
B
C
E
3 m
F
G
For the frame shown in the figure, determine the vertical deflection of point E.
Assume constant EI for all members.
16 kN/m
36 kN
2 m
2 m
6 m
A
B
C
E
3 m
F
G
∑MA = 0 RG(6) – 16(6)3 -36(2) = 0 RG = 60 kN
∑Fy = 0 RAy + RG – 16(6) = 0 RAy = 36 kN
∑Fx = 0 RAx – 36 = 0 RAx = 36 kN
60
36
36
Real System

103
2 m
2 m
A
B
C
E
3 m
F
G
Virtual System for vertical deflection of E
1
1/2 1/2
6 m
x
A
36
B
36
Mp =36x
x
A
B
1/2
Mu =0
Segment AB
x
A
36
C
36
Mp =36x - 36(x-2) x
A
B
1/2
Mu =0
Segment BC
36
2 m
B
C

104
16 kN/m
C F
36 60
∑MC = 0 60(6) – 16(6)3- Mc =0 Mc = 72
72
E
16 kN/m
C
36
72
E
x
C F
1/2 1/2
E
1
Mp = 36x -16x(x/2) + 72
Mp = 36x – 8x2 + 72
C
1/2
E
x
Mu =1/2x
16 kN/m
60
72
F
x
Mp = 60x -16x(x/2)
Mp = 60x – 8x2
E F
x
Mu =1/2x
E
Segment CE ( origin at C 0<x<3)
Segment EF ( origin at F 0<x,3)

105
x
G
60
Mp =0
F
x
G
1/2
Mu =0
F
AB A 0 – 2 36x 0
CE C 0 – 3 36x-8x2+72 1/2x EI
FE F 0 - 3 60x- 8x2 1/2x EI
EI
EI
GF C 0 – 2 0 0 EI
BC A 2 – 4 36x-36(x-2) =72 0 EI
dx
EI
M
M
L
U
P


0

dx
EI
x
x
x
dx
EI
x
x
x
E 






3
0
2
3
0
2
2
1
)
8
60
(
2
1
)
72
8
36
(


106
dx
EI
x
x
dx
EI
x
x
x
E 






3
0
3
2
3
0
3
2
)
4
30
(
)
36
4
18
(

   3
0
4
3
3
0
2
4
3
10
1
18
6
1
x
x
EI
x
x
x
EI
E 





   
EI
EI
EI
E
432
81
270
1
162
81
162
1







15 kN/m
3 m
A
B
C
4 m
Problem
Determine the rotation of joint
B of the frame shown in the figure.

107
15 kN/m
3 m
A
B
C
4 m
30 kN
30 kN 3 m
A
C
4 m
B
1
∑MA = 0
1 – 4RC = 0 RC = 1/4
∑Fy = 0
RA – RC =0 RA = 1/4
A
x Mp =0
30
Segment AB( 0rigin at A) 0<x<3
A
x Mu =0
1/4
Real System Virtual System
1/4
1/4

108
15 kN/m
C
30 kN
Mp = 30x -15x(x)/2
x Mp =30x -15x2/2
1/4
x
C
B
B
Mu = 1/4x
Segment CB ( 0rigin at C) 0 < x < 4
AB A 0 – 3 0 0
EI
EI
CB C 0 – 4 30x-15(x2) /2 1/4x EI
dx
EI
M
M u
P






4
0
2
4
1
)
2
15
30
(
EI
xdx
x
x
C


109



4
0
3
2
)
8
15
4
30
(
EI
dx
x
x
C

4
0
4
3
)
(
32
15
4
10
1







 x
x
EI
C








 )
4
(
32
15
4
4
10
1 4
3
EI
C

 
120
160
1


EI
C

2
.
40
m
kN
EI
C 


110
80 kN
5 m
4 m
3 m
A
B
C
5 m
80 kN
5 m
4 m
3 m
A
B
C
Ay
Ax
24(5)4/5 = 96
24(5)3/5 =72
24(5)
∑Fx = 0 Ax – 80 – 72 = 0 Ax = 152 kN
∑Fy = 0 Ay – 96 = 0 Ay = 96 kN
MA
∑MA = 0 MA – 96(2) -72(6.5) – 80(5) = 0
MA = 1060 kN.m
+
Problem
Determine the vertical deflection of point
C for the frame shown in figure.
2I

111
5 m
4 m
A
B
C
3 m
1
MA
RA
∑MA = 0 MA – 1(4) = 0
MA = 4
+
∑Fy = 0 RA – 1 = 0 RA = 1
80 kN
5 m
4 m
3 m
A
B
C
96
152
1060
5 m
4 m
A
B
C
3 m
1
4
1
REAL SYSTEM VIRTUAL SYSTEM

112
x
A
96
152
1060
A
4
1
Mp = 152x - 1060
x
Mu = - 4
C
24x
x
B
B B
Mp = -24x (x)/2
Mp = - 12x2
C
x
1
4/5
3/5
Mu = -4/5x
Segment AB origin at A ( 0 < x < 5 )
Segment CB origin at C ( 0 < x < 5 )

113
AB A 0 – 5 152x- 1060 -4
EI
2 EI
CB C 0 – 5 -12x2 -4/5x EI
dx
EI
M
M
L
U
P


0

dx
EI
x
x
dx
EI
x
c 







5
0
2
5
0
)
5
4
)(
12
(
2
)
4
)(
1060
152
(

dx
EI
x
dx
EI
x
c 
 


5
0
3
5
0
)
6
.
9
(
)
304
2120
(

   5
0
4
5
0
2 4
.
2
152
2120
1
x
EI
x
x
EI
c 



   
4
2
5
4
.
2
)
5
(
152
)
5
(
2120
1
EI
EI
c 



     
EI
EI
EI
EI
c
1500
6800
1
625
4
.
2
3800
10600
1






EI
c
8300



114
16 kN/m
48 kN
4 m
6 m
A
B
C
E
3 m
D
I I
2I
Problem
For the frame shown in figure,determine the horizontal deflection of point D
using the virtual work method.

115
16 kN/m
48 kN
4 m
A
B
C
E
3 m
D
C
16 kN/m
3 m
Cx Cx
Cy
Cy
∑MA = 0
Cx (4) + Cy(3) – 48(4) – 16(3)(1.5) = 0
4Cx + 3Cy = 264
∑ME = 0
Cx (4) - Cy(3) – 16(3)(1.5) = 0
4Cx - 3Cy = 72
4Cx + 3Cy = 264
4Cx - 3Cy = 72
8Cx = 336 Cx = 42 Cy = 32
Ax
Ay
∑Fx = 0
Ax + Cx – 48 = 0
Ax + 42 - 48 = 0
Ax = 6
∑Fy = 0
Ay + Cy – 16(3) = 0
Ay + 32 - 48 = 0
Ay = 16
Ex
Ey
∑Fx = 0
Ex - Cx = 0
Ex - 42 = 0
Ex = 42
∑Fy = 0
Ey - Cy – 16(3) = 0
Ey - 32 - 48 = 0
Ey = 80

116
4 m
6 m
A
B
C
E
3 m
D 1
Ay Ey
∑MA =0
6Ey – 1(4) = 0
Ey = 2/3
∑Fy =0
Ey – Ay = 0
Ay = 2/3
Ax
Ex
Ax =Ex = 1/2

117
16 kN/m
48 kN
4 m
6 m
A
B
C
E
3 m
D
16 80
6 42
4 m
6 m
A
B
C
E
3 m
D 1
2/3 2/3
1/2 1/2
Real System Virtual System

118
16
6
x Mp =6x
A
B
2/3
1/2
x
A
B
Mu =1/2x
Segment AB origin at A( 0 < x < 4)
16 kN/m
B
C
D
80
42(4) = 168
x
16
6(4) = 24
42
42
Mp = 80x – 16x(x)/2 – 168
Mp = 80x – 8x2 - 168
B
2/3
2/3
1/2(4) = 2
2
1/2
1/2
x
Mu = 2/3x - 2
Segment DB origin at D( 0 < x < 6)

119
80
42
x Mp =42x
E
D
2/3
1/2
x
E
D
Mu =1/2x
Segment ED origin at E( 0 < x < 4)
Segment Origin Limits Mp Mu EI
AB A 0 – 4 6x 1/2x EI
DB D 0 – 6 80x-8x2-168 2/3x -2 2EI
ED E 0 – 4 42x 1/2x EI

120
dx
EI
M
M
L
U
P


0



 





4
0
5
0
2
4
0
)
2
1
)(
42
(
2
)
2
3
2
)(
168
8
80
(
)
2
1
)(
6
(
dx
EI
x
x
dx
EI
x
x
x
dx
EI
x
x
D



 





4
0
2
6
0
3
2
4
0
2
)
21
(
2
)
336
272
3
16
3
208
(
3
dx
EI
x
dx
EI
x
x
x
dx
EI
x
D

   4
0
3
6
0
2
4
3
4
0
3 7
336
136
12
16
9
208
2
1
1
x
EI
x
x
x
x
EI
x
EI
D 












   4
0
3
6
0
2
4
3
4
0
3
4
7
)
6
(
336
)
6
(
136
6
12
16
6
9
208
2
1
4
1
EI
EI
EI
D 












 
EI
EI
EI
D
448
2016
4896
1728
4992
2
1
64








 3
.
704
m
kN
EI
D


121
121
Plate # 4 : Slope and Deflections of Frames
For the frame shown in the figure, determine the vertical deflection of point E
and the rotation at point C. Assume constant EI for all members.
18+N kN/m
2 m
2 m
6 m
A
B
C
E
3 m
F
G
75+N kN

122
Deflection of Trusses


AE
L
F
F
p
u
A

Considering axial deformation of
members of the truss
 
 T
L
Fu
T 

Considering thermal deformation of
Fp = the axial load at the element under
Fu = the axial load due to a unit load applied at the
point where the deflection is required or a unit
couple applied at the point where the slope is
required.
L = length of truss member
A = cross sectional area of truss member
E = modulus of elasticity of truss member
α = coeffecient of linear expansion
ΔF = fabrication error of truss member
ΔT = change in temperature
Note: unless otherwise stated in the problem, the effect of thermal deformation
and fabrication errors are negligible.
Considering fabrication errors of
 
 F
u
F F

F
T
A 


 


VIRTUAL WORK METHOD
EQ. 1
EQ. 2
EQ. 3 EQ. 4

123
General Procedure : Deflection of Trusses by Virtual Work Method
1. Real System – Determine the internal forces at the members due to the external loads
by using the method of sections or joints. Consider tensile and compressive forces
to be positive and negative respectively. Likewise, increases in temperature and member
length; decreases in temperature and member length are positive and negative
respectively.
2. Virtual System - Removed all applied loads and apply a unit load at the point
in the truss in the direction of the desired deflection to form the virtual force
system. By using the method of joints or sections,determine force in all members
due to this load. Use the same sign convention for member forces, temperature
change and fabrication errors adopted in 1.
3. The desired deflection can now be determined by using EQ. 1 if deflection is due
to external loads, EQ. 2 if due to temperature change, EQ. 3 if due to fabrication
errors. If necessary use EQ. 4 to determine the deflection due to combined axial,
thermal change and fabrication errors. Positive results indicates that the deflection
conforms with the direction of the unit load; negative results indicates otherwise.
It is recommended that computations be arranged in tabular form.

124
40 kN
60 kN
A
B
C D
E
F
40 kN
60 kN
A
B
C D
E
F
4 m
3 m
3 m
∑MA = 0 40(6) + 60(3) – 4 RF = 0 RF = 105 kN
∑Fy = 0 Ay – RF = 0 Ay -105 = 0 Ay = 105 kN
∑Fx = 0 Ax – 40 – 60 = 0 Ax = 100 kN
RF
Ay
Ax
Problem
Determine the horizontal deflection of point D for the truss shown. Cross sectional areas
of horizontal and vertical members is 4000 sq. mm each, inclined members 5000 sq.mm.
E = 200000 MPa for all members.

125
40 CD =40 (C)
BC =0
Joint C
40
BD
DE
Joint D
∑Fx = 0
BD(4/5) – 40 = 0 BD =50 kN (T )
∑Fy = 0
DE – BD(3/5) = 0 DE = 30 kN (C)
C D
30
BE=0
Joint E
E
EF = 30 kN (C)
A
100
105
105
30
BF
AF
∑Fy = 0
BF(3/5)+ 30 - 105 = 0 BF = 125 kN (C)
∑Fx = 0
125(4/5) – AF = 0 AF = 100 kN (T)
100
AB = 105 kN (T)
F
Joint F
Joint A

126
1
A
B
C D
E
F
4 m
3 m
3 m
Ay
RF
Ax
∑MA = 0 1(6) – 4 RF = 0 RF = 1.5
∑Fy = 0 Ay – RF = 0 Ay -1.5 = 0 Ay = 1.5
∑Fx = 0 Ax – 1 = 0 Ax = 1
A
1
1.5
AF = 1 (T)
AB = 1.5 (T)
Joint A
1.5
EF
BF
1
∑Fx = 0
BF(4/5) - 1 = 0 BF = 1.25 (C)
∑Fy = 0
1.25(3/5) + EF – 1.5 = 0 EF =0.75 (C)
F
Joint F
Joint E
By inspection
DE= EF = 0.75 (C) BE = 0
Joint C
By inspection
CD= BC = 0
BD
0.75
Joint D
∑Fx = 0
BD(4/5) – 1 = 0
BD =1.25 (T )
D
0
1

127
40 kN
60 kN
A
B
C D
E
F
105
105
100
0
0
- 40
50 - 30
- 125
- 30
100
105
REAL SYSTEM
1
A
B
C D
E
F
4 m
1.5
1
1.5
1.0
0
0
0
-0.75
-0.75
-1.25
1.25
VIRTUAL SYSTEM
1.5

128
Member Fp (kN) Fu L (mm) mm2 )
.
( 2
mm
mm
kN
A
L
F
F u
P
AB 105 1.5 3000 4000 118.13
AF 100 1 4000 4000 100
BD 50 1.25 5000 5000 62.5
BF -125 -1.25 5000 5000 156.25
DE - 30 -0.75 3000 4000 16.88
EF - 30 -0.75 3000 4000 16.88
NOTE: members with zero forces may
not be included in the tabulation.
)
.
(
64
.
470 2
mm
mm
kN
A
L
F
F u
P




AE
L
F
F
p
u
D

mm
D 352
.
2
)
1000
(
200000
64
.
470



Horizontal Deflection of point D

129
129
Solve the preceding problem considering the effect of an
increase in temperature Of 300C.
C
mm
mm
x 0
6
10
7
.
11 


Member Fu L (mm) )
(mm
L
Fu
AB 1.5 3000 4500
AF 1 4000 4000
BD 1.25 5000 6250
BF -1.25 5000 -6250
DE -0.75 3000 -2250
EF -0.75 3000 -2250
)
(
4000 mm
L
Fu 

 
 T
L
Fu
T 



 
mm
T 404
.
1
)
30
(
10
)
7
.
11
(
4000 6






 mm
T
D
DT 756
.
3
404
.
1
352
.
2




130
130
Solve the preceding problem considering the effect of axial
load, temperature increase of 300C and that each member
was fabricated 0.005 mm too long.
Member Fu )
(mm
L
Fu
AB 1.5 0.005 0.0075
AF 1 0.005 0.005
BD 1.25 0.005 0.00625
BF -1.25 0.005 -0.00625
DE -0.75 0.005 -0.00375
EF -0.75 0.005 -0.00375
)
(
005
.
0 mm
F
Fu 


 


 mm
F
Fu
F 005
.
0








 mm
F
T
D
DT 761
.
3
005
.
0
404
.
1
352
.
2




F


131
A
B C D
E
F
G
H
100 kN
A
100 kN
200 kN
100 kN 100 kN
A
B
H
100
BC
BG
GH
300
4 at 3 m each = 12 m
A
B C D
E
F
G
H
100 kN
A
100 kN
200 kN
100 kN 100 kN
300 kN 300 kN
4m
4
3
5
AB
AH
300 kN
∑Fy =0
AB(4/5) – 300 = 0
AB = 375 kN (C)
∑Fx =0
AB(3/5) – AH = 0
375(3/5) – AH = 0
AH = 225 kN (T)
By symmetry
DE = 375 kN (C) EF = 225 kN (T)
225 GH = 225
100
BH = 100
Joint H
By symmetry
GF = 225 kN (T) DF = 100 kN (T)
G
∑MG = 0
4BC + 200(3) – 300 (6) = 0
BC = 300 kN (C) CD = 300 kN (C)
4
3
5
∑Fy =0
BG(4/5)+200 – 300 = 0
BG = 125 kN (T) DG = 125 kN (T)
100
By inspection; CG = 0
Problem
Determine the horizontal and vertical deflections of point G
for the truss shown. A = 4500 mm2,E=200 GPa for all members.

132
A
B C D
E
F
G
H
A
1
4m
1/2 1/2
4
3
5
AB
AH
1/2
∑Fy =0
AB(4/5) – 1/2 = 0
AB = 0.625 (C)
∑Fx =0
AB(3/5) – AH = 0
0.625(3/5) – AH = 0
AH = 0.375 (T)
By symmetry
DE = 0.675 (C)
EF = 0.375 (T)
0.375 GH = 0.375
BH = 0
Joint H
By symmetry
GF = 0.375 (T) DF = 0 A
B
H
BC
CH
GH
1/2
G
∑MG = 0
4BC - 1/2(6) = 0
BC = 0.75 (C) CD = 0.75 (C)
4
3
5
∑Fy =0
CH(4/5) – 1/2 = 0
CH = 0.625 (T) DG = 0.625 (T)
By inspection; CG = 0

133
A
B C D
E
F
G
H
A
1
1/2 1/2
-0.625 -0.625
0.375
0.625 0.625
0.375 0.375 0.375
-0.75 -0.75
0 0 0
VIRTUAL SYSTEM ( Vertical deflection of G)
A
B C D
E
F
G
H
100 kN
A
100 kN
200 kN
100 kN 100 kN
300 kN 300 kN
-375
225
-300
125 125
225 225 225
-375
-300
100 100
0
REAL SYSTEM
Member Fp (kN) Fu L (mm) mm2
)
.
( 2
mm
mm
kN
A
L
F
F u
P
AH 225 0.375 3000 4500 56.25
BC -300 -0.75 3000 4500 168.7
BG 125 0.625 5000 4500 86.8
CD -300 -0.75 3000 4500 168.7
DE -375 -0.625 5000 4500 260.42
DG 125 0.625 5000 4500 86.8
EF 225 0.375 3000 4500 56.25
AB -375 -0.625 5000 4500 260.42
FG 225 0.375 3000 4500 56.25
GH 225 0.375 3000 4500 56.25
)
.
(
81
.
256
,
1 2
mm
mm
kN
A
L
F
F u
P



134
200000
)
1000
(
81
.
1256

  AE
L
F
F U
P
GV

mm
GV 28
.
6


A
B C D
E
F
G
H
A
4m
1
1
4
3
5
AB = 0
AH =1 (T)
1
BH = 0
1
Joint A
Joint H
1
BG=0
CG=0
DG=0
GF=0
G
H
A
GH=1 (T)
1

135
A
B C D
E
F
G
H
A
4m
1
1
1
1
0 0
0
0 0
0
0
0
0
0
.
( 2
mm
mm
kN
A
L
F
F u
P
AH 225 1 3000 4000 150
GH 225 1 3000 4000 150
)
.
(
300 2
mm
mm
kN
a
L
F
F U
p


Horizontal Deflection of point G
200000
1000
)
300
(

  AE
L
F
F U
p
GH

mm
GH 5
.
1


VIRTUAL SYSTEM ( Horizontal deflection of G)

136
Problem
Determine the vertical deflection of point G of the truss given in the previous problem
due to the given loads and fabrication errors of an increase of 0.2 mm in length for
tension members and 0.25 mm decrease in length for compression members. All
Other data remain unchanged. .
A
B C D
E
F
G
H
100 kN
A
100 kN
200 kN
100 kN 100 kN
300 kN 300 kN
-375
225
-300
125 125
225 225 225
-375
-300
100 100
0
REAL SYSTEM
A
B C D
E
F
G
H
A
1
1/2
-0.625 -0.625
0.375
0.625 0.625
0.375 0.375 0.375
-0.75 -0.75
0 0 0
1/2

137
)
.
( 2
mm
mm
kN
A
L
F
F u
P
AH 225 0.375 3000 4500 56.25
BC -300 -0.75 3000 4500 168.7
BG 125 0.625 5000 4500 86.8
CD -300 -0.75 3000 4500 168.7
DE -375 -0.625 5000 4500 260.42
DG 125 0.625 5000 4500 86.8
EF 225 0.375 3000 4500 56.25
AB -375 -0.625 5000 4500 260.42
FG 225 0.375 3000 4500 56.25
GH 225 0.375 3000 4500 56.25
81
.
256
,
1

 A
L
F
F u
P
F
 F
u
F 
-0.25
-0.25
-0.25
-0.25
mm
mm
0.20
0.20
0.20
0.20
0.20
0.20
0.156
0.156
0.075
0.075
0.075
0.075
0.188
0.188
0.125
0.125
mm
F F
u 238
.
1




 

 F
u
U
P
GV F
AE
L
F
F




 mm
GV 518
.
7
238
.
1
2000
1000
81
.
1256


138
Problem
Determine the vertical deflection of point G of the truss given in the previous problem
due to the given loads and temperature increase of 200C . All Other data remain
unchanged. .
A
B C D
E
F
G
H
100 kN
A
100 kN
200 kN
100 kN 100 kN
300 kN 300 kN
-375
225
-300
125 125
225 225 225
-375
-300
100 100
0
REAL SYSTEM
A
B C D
E
F
G
H
A
1
1/2
-0.625 -0.625
0.375
0.625 0.625
0.375 0.375 0.375
-0.75 -0.75
0 0 0
C
m
m
0
7
.
11

 
1/2

139
)
.
( 2
mm
mm
kN
A
L
F
F u
P
AH 225 0.375 3000 4500 56.25
BC -300 -0.75 3000 4500 168.7
BG 125 0.625 5000 4500 86.8
CD -300 -0.75 3000 4500 168.7
DE -375 -0.625 5000 4500 260.42
DG 125 0.625 5000 4500 86.8
EF 225 0.375 3000 4500 56.25
AB -375 -0.625 5000 4500 260.42
FG 225 0.375 3000 4500 56.25
GH 225 0.375 3000 4500 56.25
81
.
256
,
1

 A
L
F
F u
P
L
Fu
mm
-3125
-3125
1125
1125
1125
1125
-2250
-2250
3125
3125
0

 L
Fu

 

 T
L
F
AE
L
F
F
u
U
P
GV 




 mm
GV 28
.
6
0
200000
1000
81
.
1256


140
60 kN
100 kN
A
B C
D
6 m 6 m
4m
300 mm2
300 mm2
300 mm2 300 mm2
200 mm2
Problem
For the truss shown in the figure , if E = 200 GPa, determine the following:
a) the vertical deflection of B
b) the horizontal deflection of C
c) the horizontal deflection of D
60 kN
100 kN
A
B C
D
6 m 6 m
4m
300 mm2
300 mm2
300 mm2 300 mm2
200 mm2
∑MA = 0 100(6) + 60(4) - 12 Rc = 0 Rc = 70 kN
∑Fy = 0 Ay + Rc - 100= 0 Ay +70 -100 = 0 Ay = 30 kN
∑Fx = 0 Ax – 60 = 0 Ax = 60 kN
Ax
Ay Rc

141
60 kN
100 kN
A
B C
D
6 m 6 m
4m
60
30 70
C
70
2
3
3.61
∑Fy =0
CD(2/3.61) – 70 = 0
CD = 126.35 kN (C)
∑Fx =0
CD(3/3.61) – BC = 0
126.350(3/3.61) – BC = 0
BC= 105 kN (T)
CD
BC
100 kN
B 105
AB=105 kN (T)
BD = 100 kN (T)
A
60
30
105
AD
2
3.61
3
∑Fy =0
AD(2/3.61) – 30 = 0
AD = 54.15 kN (C)
-126.35
105
105
100
-54.15
REAL SYSTEM

142
1 kN
A
B
C
D
6 m 6 m
4m
1/2 1/2
C
1/2
2
3
3.61
∑Fy =0
CD(2/3.61) – 1/2 = 0
CD = 0.90 (C)
∑Fx =0
CD(3/3.61) – BC = 0
0.9(3/3.61) – BC = 0
BC= 0.75 (T)
By symmetry
AD =0.90 (C) AB = 0.75 (T)
By inspection BD = 1 (T)
CD
BC
-0.90 -0.90
1
0.75 0.75
Virtual System for vertical deflection of B
.
( 2
mm
mm
kN
A
L
F
F u
P
AD -54.15 -0.90 7220 300 1,172.89
BC 105 0.75 6000 300 1,575
BD 100 1.0 4000 200 2,000
CD -126.35 -0.90 7220 300 2,736.74
AB 105 0.75 6000 300 1,575
2
.
63
.
9059
mm
mm
kN
A
L
F
F u
P


Vertical deflection of B


 mm
BV 5
.
45
200000
)
1000
(
63
.
9059


143
A
B
C
D
6 m 6 m
4m
0 0
0
1 1
1
1
C
2
3
3.61
BC=1 1
CD =0
BD =0
1
AB = 1
A
1 1
AD=0
2
3.61
3
B
AB 105 1.0 6000 300 2100
BC 105 1.0 6000 300 2100
)
.
( 2
mm
mm
kN
A
L
F
F u
P
2
.
4200
mm
mm
kN
A
L
F
F u
P


Horizontal Deflection of point C
200000
1000
)
4200
(

  AE
L
F
F U
p
CH

mm
GH 21


Virtual System for horizontal deflection of C

144
A
B
C
D
6 m 6 m
4m
0.60 - 0.60
0
0.50 0.50
1
Ax
Ay
∑MA = 0 1(4) - 12 Rc = 0 Rc = 1/3
∑Fy = 0 Ay - Rc = 0 Ay -1/3 = 0 Ay = 1/3
∑Fx = 0 Ax – 1 = 0 Ax = 1
RC
C
1/3
2
3
3.61
∑Fy =0
CD(2/3.61) – 1/3 = 0
CD = 0.60 (C)
∑Fx =0
CD(3/3.61) – BC = 0
0.6(3/3.61) – BC = 0
BC= 0.5 (T)
BC
B 0.5
AB=0.5 (T)
BD = 0
A
1
1/3
0.5
AD
2
3.61
3
∑Fy =0
AD(2/3.61) – 1/30 = 0
AD = 0.60 (T)
Virtual system for horizontal deflection of D

145
AD -54.15 0.60 7220 300 -781.93
BC 105 0.50 6000 300 1,050
CD -126.35 -0.60 7220 300 1,824.5
AB 105 0.50 6000 300 1,050
2
.
56
.
3142
mm
mm
kN
A
L
F
F u
P


Horizontal deflection of D


 mm
DH 71
.
15
200000
)
1000
(
56
.
3142

)
.
( 2
mm
mm
kN
A
L
F
F u
P


AE
L
F
F U
p
DH


146
90 kN
100 kN
A
B C
D
6 m 6 m
4m
400 mm2
400 mm2
300 mm2 300 mm2
200 mm2
For the truss shown in figure, determine the vertical deflection of joint B
If E = 200GPa.
Midterm Exam
I

147
18 kN/m
2 m
2 m
6 m
A
B
C
E
3 m
F
G
75 kN
II

148
Methods of analysis
1. METHOD OF SUPERPOSITION
2. THREE MOMENT EQUATION
3. MOMENT DISTRIBUTION METHOD
4. SLOPE DEFLECTION METHOD
ANALYSIS OF STATICALLY INDETERMINATE BEAMS- the
process of determining the unkown reactions and end moments of
the beam so that the shear and moment diagrams of the beam can
be plotted.
Method of Superposition - Apply known slope or deflection formulas
to obtain equations involving slope or deflections.
The following table will be useful in applying the method of superposition

149
Loading Slope Deflection

P
L 

P
L 
w
EI
PL
2
2
EI
PL
3
3
EI
wL
6
3
EI
wL
8
4

2
L
2
L
P

b at center
)
4
3
(
48
2
2
b
L
EI
Pb

b=distance of P from the
nearer support

150

w
L
 EI
wL
24
3
EI
wL
30
4
M
EI
ML
EI
ML
2
2


2
L
2
L
P
L
w
L
at supports at center
at center
EI
PL
48
3
EI
wL
384
5 4
Loading Slope Deflection


EI
PL
16
2
at supports
EI
wL
24
3
 

151
w
a a
A
B C
w
A B
C
B

L = 2a
Remove redundant at B and solve for
Vertical deflection at B due to the given
loadings
EI
wL
B
384
5 4


C
B

L = 2a
B
RB
Remove applied loads and apply at RB at B
Solve for Vertical deflection at B due to RB.
EI
L
RB
B
48
3


By superposition of deflections;
of
settlement
B
B 


 B
iIlustration of method of superposition

152
W= 25 kN/m
6 m
A
B C
60 kN
3m
6 m
3m
60 kN
w
A B
C
B

L = 12m
C
B

L = 12 m
B
RB
60 kN
3m 3m
60 kN
 

 )
4
3
(
48
384
5 2
2
4
b
L
EI
Pb
EI
wL
B

EI
L
RB
B
48
3


Problem
Analyze the indeterminate beam shown by method of superposition.
Assume that support B does not settle.

153
 

 )
4
3
(
48
384
5
48
2
2
4
3
b
L
EI
Pb
EI
wL
EI
L
RB
 
kN
R
R
B
B
270
)
3
(
4
)
12
(
3
)
3
)(
60
(
2
8
)
12
)(
25
(
5
)
12
( 2
2
4
3




B
B 


kN
R
R C
A 75
2
270
)
12
(
25
)
60
(
2






154
W =25 kN/m
A B C
3m 3m
3m
3m
75 kN 75 kN
270 kN
60 kN 60 kN
75
-60
-135
135
1
2
A A’ B
MA’ = A1
MA’ = ½(75) 3 = 112.5 kN.m
MB – MA’ = A2
MB – 112.5 = -1/2(60+135) 3
MB = - 180 kN.m
-180
112.5 112.5
60
-75
0 0
V-Diagram
M-Diagram

155
Problem
Analyze the beam given in the previous problem if B settles 18 mm vertically
down. E = 200000 MPa and I = 600 x 106 mm4.
B
B 


 18
 


 )
4
3
(
48
384
5
18
48
2
2
4
3
b
L
EI
Pb
EI
wL
EI
L
RB
 
kN
N
R
R
EI
R
B
B
B
220
220000
82500
187500
)
12000
(
)
10
)(
500
(
200000
)
48
(
18
)
3000
(
4
)
12000
(
3
)
3
(
)
1000
)(
60
(
2
8
)
12000
)(
25
(
5
)
48
(
18
)
12000
(
3
6
2
2
2
4
3









kN
R
R C
A 100
2
220
)
12
(
25
)
60
(
2






156
W =25 kN/m
A B C
3m 3m
3m
3m
100 kN 100 kN
220 kN
60 kN
100
-35
-110
110
1
2
A A’ B
MA’ = A1
MA’ = ½(100) 3 = 150 kN.m
MB – MA’ = A2
MB – 150= -1/2(110+35) 3
MB = - 67.5 kN.m
-67.5
150 150
35
-100
0 0
V-Diagram
M-Diagram
25
-25

157
48 kN
6 m
24 kN/m
6m 1

24kN/m

48 kN
6m 
 2
2 
4m
6m

A B
RB
EI
wL
8
4
1 

4m

 2
2 
EI
PL
EI
PL 2
1
2
1
2 )
2
(
2 


EI
L
RB
3
3





 3
2
1 

 EQ. of Deformation
EI
L
R
EI
PL
EI
PL
EI
wL B
3
3
8
3
3
1
2
1
4



Problem;
Analyze the indeterminate beam shown in figure.
3

EI
PL
3
3
1
3 


158
RA = 24(6)+48 – 78.89 =113.11
MA = 78.89(6) – 48(4) – 24(6)(3) = -150.66 kN.m
6 m
24 kN/m
4m
A
B
113.11 78.89
150.66
-150.66
48 kN
17.11
-30.89
113.11
-78.89
A C
B
1
2
1
A
M
M A
C 

m
kN
M
M
C
C
.
78
.
109
4
2
11
.
17
11
.
113
)
66
.
150
(






 



2
A
M
M C
B 

1
A
M
M A
C 

0
2
2
89
.
78
89
.
30
78
.
109






 


B
B
M
M
109.78
3
)
6
(
3
)
4
(
48
)
4
(
48
8
)
6
(
24 3
3
2
4
B
R



RB = 78.89
V-Diagram
M-Diagram

159
h1 h3
L1 L2
Any Loading
A1
a1
A2
a2
Moment diagram
due to loads on L1
Moment diagram
due to loads on L2
1
2 3
Continuous Beams :Three Moment Equation
A relation between the moments at any three points on a beam
and their relative vertical distances and deviation.
SYMBOLS & NOTATIONS
M1, M2, M3 = moments at points 1,2 & 3
L1 = horizontal distance between points
1 & 2, span # 1
L2 = horizontal distance between points
2 & 3, span # 2
h1=vertical deviation of point 1 with
respect to point 2
h3=vertical deviation of point 3 with
respect to point 2
I1 = moment of inertia of section of span #1
I2 = moment of inertia of section of span #2
E1,E2 = modulus of elasticity of sections
of span #1 and #2
A1a1 = moment of area of Moment diagram
of the loads on span #1 about the
left end of span # 1
A2a2 = moment of area of Moment diagram
of the loads on span #2 about the
right end of span # 2
M1
L1
L2
Moment diagram
due to end
moments
M2
M2
M3

160
2
/
1
1 t
h 
2
/
1
t
h1
h3
3
2
/
3 h
t 
2
/
3
t
L1
L2
1
2
3
2
3
2
/
3
1
2
/
1
1
L
h
t
L
t
h 


2
3
1
1
2
2
/
3
1
2
/
1
L
h
L
h
L
t
L
t



1
1
1
1
2
1
1
1
1
1
2
/
1
)
3
2
(
2
1
)
3
1
(
2
1
I
E
L
L
M
L
L
M
a
A
t



1
1
2
1
2
2
1
1
1
1
2
/
1
3
1
6
1
I
E
L
M
L
M
a
A
t



EQ.1
EQ.2

161
2
2
2
2
2
2
2
3
2
2
2
/
3
)
3
2
(
2
1
)
3
1
(
2
1
I
E
L
L
M
L
L
M
a
A
t



2
2
2
2
2
2
2
3
2
2
2
/
3
3
1
6
1
I
E
L
M
L
M
a
A
t


 EQ.3
2
3
1
1
2
2
2
2
2
2
2
2
3
2
2
1
1
1
2
1
2
2
1
1
1
1
3
1
6
1
3
1
6
1
L
h
L
h
L
I
E
L
M
L
M
a
A
L
I
E
L
M
L
M
a
A
































2
3
1
1
2
2
2
2
2
1
1
1
1
1
2
2
2
2
2
2
2
3
1
1
1
2
1
1
1
1
6
6
6
2
2
L
h
L
h
L
I
E
a
A
L
I
E
a
A
I
E
L
M
I
E
L
M
I
E
L
M
I
E
L
M



















2
3
1
1
2
2
2
2
2
1
1
1
1
1
2
2
2
3
2
2
2
1
1
1
2
1
1
1
1
6
6
6
2
L
h
L
h
L
I
E
a
A
L
I
E
a
A
I
E
L
M
I
E
L
I
E
L
M
I
E
L
M
Multiplying by 6 and simplifying
Substituting EQs 2 and 3 in EQ. 1

162



















2
3
1
1
2
2
2
2
1
1
1
1
2
2
3
2
2
1
1
2
1
1
1
6
6
6
2
L
h
L
h
E
L
I
a
A
L
I
a
A
I
L
M
I
L
I
L
M
I
L
M
  












2
3
1
1
2
2
2
1
1
1
2
3
2
1
2
1
1 6
6
6
2
L
h
L
h
EI
L
a
A
L
a
A
L
M
L
L
M
L
M



















2
3
1
1
2
2
2
2
2
1
1
1
1
1
2
2
2
3
2
2
2
1
1
1
2
1
1
1
1
6
6
6
2
L
h
L
h
L
I
E
a
A
L
I
E
a
A
I
E
L
M
I
E
L
I
E
L
M
I
E
L
M
If E is constant
If E and I are constants
  0
6
6
2
2
2
2
1
1
1
2
3
2
1
2
1
1 





L
a
A
L
a
A
L
M
L
L
M
L
M
General Form of the three moment equation
If E and I are constants, h1 = h3 = 0

163
1
1
1
6
L
a
A
2
2
2
6
L
a
A
Type of loading
and span
P
a b
L
 
2
2
a
L
L
Pa
  
2
2
b
L
L
Pb

w
L
L
4
3
wL
4
3
wL
60
8 3
wL
60
7 3
wL
w

164
1
1
1
6
L
a
A
2
2
2
6
L
a
A
Type of loading
and span
L
w
L
L
4
3
wL
32
5 3
wL
L
L
a
M )
3
( 2
2


60
8 3
wL
60
7 3
wL
M
a
b L
L
b
M )
3
( 2
2

w

165
24 kN
2 m 2 m 6 m
6 kN/m
A B
D
C
3 m
2 m
48 kN
Considering points A, B and C
MA = 0     2
.
403
2
5
5
)
2
(
48
6 2
2
2
2
1
1
1




 a
L
L
Pa
L
a
A
    144
2
4
4
)
2
(
24
6 2
2
2
2
2
2
2




 b
L
L
Pb
L
a
A
L1 =5
L2 =4
  0
6
6
2
2
2
2
1
1
1
2
2
1
1 





L
a
A
L
a
A
L
M
L
L
M
L
M C
B
A
  0
144
2
.
403
4
4
5
2 



 C
B M
M
0
2
.
547
4
18 

 C
B M
M EQ. 1
Problem
Analyze the indeterminate beam shown by three moment equation.
Assume constant EI.
h1 =0
h3 =0

166
Considering points B ,C and D
MD = 0     144
2
4
4
)
2
(
24
6 2
2
2
2
1
1
1




 a
L
L
Pa
L
a
A
864
4
)
6
(
16
4
6 3
3
2
2
2



wL
L
a
A
L1 =4
L2 =6
  0
6
6
2
2
2
2
1
1
1
2
2
1
1 





L
a
A
L
a
A
L
M
L
L
M
L
M D
C
B
  0
864
144
)
0
(
6
6
4
2
4 




 C
B M
M
0
252
5 

 C
B M
M EQ. 2
252
5 

 C
B M
M
0
2
.
547
4
)
252
5
(
18 



 C
C M
M
0
2
.
547
4
4536
90 



 C
C M
M MC = -46.38 kN.m
MB = -20.1 kN.m

167
Any Loading
A B
C
Any Loading
A B C
+MA
+MB Any Loading
B
+MB
+MC
Positive sense of end moments: if end moments are positive they must be
plotted as illustrated below, if negative plot it otherwise
Note:
After the end moments are plotted, neglect negative signs; in computing
reactions using ∑M = 0 positive moments depends on your assumption
RA RB1 RB2 RC
For interior reactions
COMPUTATION OF REACTIONS
2
1 B
B
B R
R
R 


168
A B
3 m
2 m
48 kN
20.1
0
RA
RB1
∑MB = 0
RA(5) + 20.1 – 48(3) = 0 RA = 24.78 kN
∑Fy = 0 RB1 + RA – 48 =0 RB1 = 23.22 kN
+
24 kN
2 m 2 m
B
C
20.1 46.38
∑MB = 0
RC1(4)- 46.38 +20.1 – 24(2) = 0 RC1 = 18.57 kN
∑Fy = 0 RB2 + RC1 – 24 =0 RB2 = 5.43 kN
+
RB2 RC1
6 m
6 kN/m
D
C
46.38 0
RC2 RD
∑MC = 0
RD(6)+46.38 –6 (6)3 = 0 RD = 10.27 kN
∑Fy = 0 RD + RC2 – 6(6) =0 RC2 =25.73 kN
+
kN
R
R
R B
B
B 65
.
28
2
1 


kN
R
R
R C
C
C 3
.
44
2
1 



169
24 kN
2 m 2 m 6 m
6 kN/m
A B
D
C
3 m
2 m
48 kN
24.78 28.65 44.3 10.27
24.78
-23.22
5.43
-18.57
25.73
-10.27
A 1 B 2 C 3 D
4.29 1.71
1
2
3
4
5
6
M1 = A1
M1 = 24.78(2) = 49.56
MB – M1 = A2
MB - 49.56= -23.22(3)
MB = -20.1
M2-MB = A3
M2- (-20.1) = 5.43(2)
M2 =-9.24
MC – M2 =A4
MC-(-9.24)=-18.57(2)
MC = - 46.38
M3 – MC = A5
M3- (-46.38) =1/2(25.73)(4.29)
M3 =8.81
MD – M3 = A6
MD -8.81 =1/2(10.2)(1.71)
MD =0
49.56
-20.1
-9.24
-46.28
8.81

170
64 kN
2 m
2 m 6 m
16 kN/m
A
B
48 kN
C
MA = 0
    384
2
4
4
)
2
(
64
6 2
2
2
2
1
1
1




 a
L
L
Pa
L
a
A
    1504
4
)
6
(
16
4
6
6
)
4
(
48
4
6 3
2
2
3
2
2
2
2
2







wL
b
L
L
Pb
L
a
A
L1 =4
L2 =6
  0
6
6
2
2
2
2
1
1
1
2
2
1
1 





L
a
A
L
a
A
L
M
L
L
M
L
M C
B
A
  0
1504
384
6
6
4
2 



 C
B M
M
0
1888
6
20 

 C
B M
M EQ.1
2 m
Problem
Analyze the indeterminate beam
Shown assuming constant EI.

171
64 kN
2 m 2 m 6 m
16 kN/m
A
B
48 kN
Imaginary span
C
D
Note: all values for imaginary span is 0
Considering points B,C and D
MD = 0
    1376
4
)
6
(
16
2
6
6
)
2
(
48
4
6 3
2
2
3
2
2
1
1
1







wL
a
L
L
Pa
L
a
A
0
6
2
2
2

L
a
A
L1 =6
L2 =0
  0
6
6
2
2
2
2
1
1
1
2
2
1
1 





L
a
A
L
a
A
L
M
L
L
M
L
M D
C
B
0
1376
6
2
6 

 C
B M
M
0
688
6
3 

 C
B M
M EQ. 2

172
0
1888
6
20 

 C
B M
M
0
1200
17 

B
M
0
688
6
3 

 C
B M
M
MB = -70.59 kN.m MC = -79.37 kN.m
A B
2 m
2 m
64 kN
70.59
0
RA
RB1
∑MB = 0
RA(4) + 70.59 – 64(2) = 0 RA = 14.35 kN
∑Fy = 0 RB1 + RA – 64 =0 RB1 = 49.65 kN
+
48 kN
6 m
16 kN/m
B C
70.59 79.37
RB2
RC
∑MB = 0
RC(6)+ 70.59 - 79.37 – 48(2)- 16(6)3 = 0
RC = 65.46 kN
∑Fy = 0 RB2 + RC – 48 – 16(6)=0 RB2 = 78.54 kN
+
kN
R
R
R B
B
B 19
.
128
2
1 


2 m

173
64 kN
2 m
2 m
6 m
16 kN/m
A
B
48 kN
C
2 m
14.35 128.19 65.46
14.35
-49.65
78.54
46.54
-1.46
-65.46
4
3
2
1
A 1 B 2 C
M1 – MA = A1
M1 – 0 = 14.35(2)
M1 = 28.7 kN.m
MB - M1 =A2
MB – 28.7 =- 49.65(2)
MB = -70.59 kN.m
M2 – M3 =A3
M2 –(-70.59) = ½(78.54+46.54)2
M2 = 54.49 kN.m
MC – M2 = A4
MC- 54.49=-1/2(1.46+65.46)4
MC = - 79.37 kN.m
28.7
-70.59
54.49
-79.37

174
114 kN
36 kN/m
D
B
A
2 m
90 kN
2 m 2 m
C
2 m
5 m
114 kN
36 kN/m
D
B
A
2 m
90 kN
2 m 2 m
C
2 m
5 m
A’
1 m
1 m
Considering points A’, A and B
MA’ = 0
    240
1
3
3
)
1
(
90
6 2
2
2
2
2
2
2




 b
L
L
Pb
L
a
A
L1 =0 L2 =3
  0
6
6
2
2
2
2
1
1
1
2
2
1
1
'






L
a
A
L
a
A
L
M
L
L
M
L
M B
A
A
0
6
1
1
1

L
a
A
  0
240
3
3
0
2 


 B
A M
M
1
.
0
120
5
.
1
3 EQ
M
M B
A 



Problem:
Analyze the indeterminate beam
shown assuming constant EI.

175
    300
2
3
3
)
2
(
90
6 2
2
2
2
1
1
1




 a
L
L
Pa
L
a
A
    684
2
4
4
)
2
(
114
6 2
2
2
2
2
2
2




 b
L
L
Pb
L
a
A
L1 =3
L2 =4
  0
6
6
2
2
2
2
1
1
1
2
2
1
1 





L
a
A
L
a
A
L
M
L
L
M
L
M C
B
A
  0
684
300
4
4
3
2
3 




 C
B
A M
M
M
2
.
0
984
4
14
3 EQ
M
M
M C
B
A 




1
.
0
120
5
.
1
3 EQ
M
M B
A 



3
.
0
864
4
5
.
12 EQ
M
M C
B 



subtract

176
Considering points B ,C and D
MD =-36(2)(1)= -72 kN.m
    684
2
4
4
)
2
(
114
6 2
2
2
2
1
1
1




 a
L
L
Pa
L
a
A
1125
4
)
5
(
36
4
6 3
3
2
2
2



wL
L
a
A
L1 =4
L2 =5
  0
6
6
2
2
2
2
1
1
1
2
2
1
1 





L
a
A
L
a
A
L
M
L
L
M
L
M D
C
B
  0
1125
684
)
72
(
5
5
4
2
4 





 C
B M
M
36 kN/m
D
2 m
  0
1449
5
4
2
4 


 C
B M
M
Divide by 4.5
4
.
0
322
4
89
.
0 EQ
M
M C
B 



3
.
0
864
4
5
.
12 EQ
M
M C
B 



0
542
61
.
11 

B
M m
kN
MB .
68
.
46


0
322
4
)
68
.
46
(
89
.
0 


 C
M m
kN
MC .
11
.
70


1
.
0
120
5
.
1
3 EQ
M
M B
A 



0
120
)
68
.
46
(
5
.
1
3 



A
M m
kN
MA .
66
.
16


0
984
)
11
.
70
(
4
)
68
.
46
(
14
)
66
.
16
(
3 






Check
subtract

177
B
A
2 m
90 kN
1 m
16.66 46.68
114 kN
B
2 m 2 m
C
46.68 70.11
36 kN/m
D
C
2 m
5 m
70.11
RA
RB1
RB2 RC1
RC2 RD
∑MA = 0
RB1(3)+ 16.66 – 46.68 – 90(2) = 0
RB1 = 70 kN
∑Fy = 0 RB1 + RA – 90 =0 RA = 20 kN
+
∑MB = 0
RC1(4)+ 46.68 – 70.11 – 114(2) = 0
RC1 = 62.86 kN
∑Fy = 0 RB2 + RC 1– 114 =0 RB2 = 51.14 kN
+
∑MC = 0
RD(5)+ 70.11 – 36(7)3.5 = 0
RD = 162.38 kN
∑Fy = 0 RC2 + RD – 36(7)=0 Rc2 = 89.62 kN
RB = 70 + 51.14 =121.14 kN
RC = 62.86 + 89.62 = 152.48 kN
+

178
114 kN
36 kN/m
D
B
A
2 m
90 kN
2 m 2 m
C
2 m
5 m
1 m
20 121.14 152.48 162.38
20
-70
51.14
-62.86
89.62
-90.38
72
2.49
2.51
5
6
7
4
3
2
1
Point of zero shear
36x = 92.03 x = 2.56
16.66 M1 – MA = A1
M1 – (-16.66) = 20(2)
M1 = 23.34 kN.m
MB - M1 =A2
MB – 23.34 =- 70(1)
MB = -46.68 kN.m
M2 – MB =A3
M2 –(-46.68) =(51.14)2
M2 = 55.6 kN.m
MC – M2 = A4
MC- 55.6= -(62.86)2
MC = - 70.11 kN.m
M3 – Mc = A5
M3- (-70.11)= 1/2(89.62)2.49
M3 = 41.47 kN.m
MD – M3 = A6
MD- 41.47= -1/2(90.48)2.51
MD = - 72 kN.m
ME – MD = A7
ME-(- 72)= 1/2(72)2
ME = 0
A 1 B 2 C 3 D E
23.34
-16.66
-46.68
56.12
-70.11
35.73
-72
0

179
124+N kN
26N kN/m
D
B
A
90 +NkN
2 m 2 m
C
2 m
5 m
3m
24+3N kN
2 m 2 m 6 m
4N kN/m
A B
D
C
3 m
2 m
48+2N kN
Plate # 8 : Three Moment Equation
Analyze the indeterminate beams
shown by using the three
moment equation.
E

180
Carry over moment – the moment induced at fixed end of
a beam by the action of a moment applied at the other
end. When a moment MB is applied at B and flexes the
beam it induced a wall moment MA at A such that :
L
EI
MB

4

MB
-1/2MB
B
A
Beam stiffness – the moment required by the simply supported end of a
beam to produce a unit rotation of that end the other end being rigidly
fixed.
Absolute stiffness Relative Stiffness
L
EI
kabs
4

L
I
k 
B
A M
M
2
1



MOMENT DISTRIBUTION

181
Distribution Factors of any member of joint


k
k
DF
k = stiffness of the member
∑k = sum of stiffness of all members meeting at the joint
A B C
Member BA
Joint B
Member
BC
Joint B
- + - +
Convention of signs for fixed end moment
For external supports
Fixed end
DF= 0
Hinged end
DF= 1.0

182
Fixed End Moments of typical loadings
2
2
L
b
Pa
12
2
wL

Loading MA MB
P
L
w
A B
w
A
12
2
wL
2
2
L
Pab

a b
A B
L
L
30
2
wL

20
2
wL
B

183
96
5 2
wL

Loading MA MB
L
w
A
96
5 2
wL
2
6
L
EI

L
L
20
2
wL

30
2
wL
B
 2
6
L
EI

w
A B
A B

184
General Procedure: Moment Distribution
1. Lock all joints against rotation. Compute the fixed end moments.
2. Compute the distribution factors of each member of joint using
3. Prepare the table of moment distribution. Indicate by arrows how the
moments will be distributed to adjacent joints
4. Unlock each support and compute the balancing moment (BM)
using BM =( Sum or Difference in moment at joint)DF
sign of BM - take the sign of the moment with the smaller absolute value;
if the moments are of the same sign take the opposite sign
Sum – if the moments are of the same sign
Difference- if the moments are of opposite sign
5. Carry over one half of BM to the adjacent joints as indicate by the arrows
in step 3
6. Repeat steps 4 and 5 until the moment to be distributed becomes
negligible
MODIFIED STIFFNESS METHOD
For continuous beams with hinge or rollers ends the moment at this end is
zero. Multiplying the stiffness of this member by ¾ eliminates any further
distribution of moment at this ends.


k
k
DF

185
30 kN
24 kN/m
A
C
B
5 m 5 m
3 m
Problem
Analyze the indeterminate beam shown by moment distribution method .
Using the modified stiffness method. Assume constant EI.
Fixed end Moments
m
kN
L
Pab
M AB
F .
09
.
21
8
)
3
)(
5
(
30
2
2
2
2






m
kN
L
Pba
M BA
F .
15
.
35
8
)
5
)(
3
(
30
2
2
2
2



m
kN
wL
M BC
F .
50
12
)
5
(
24
12
2
2







186
m
kN
wL
M CB
F .
50
12
)
5
(
24
12
2
2



Member Absolute K Relative K
AB,BA I/8 1/8(40)= 5
BC,CB I/5 1/5(40)= 8
Distribution Factors


k
k
DF
38
.
0
8
5
5





BC
BA
BA
BA
k
k
k
DF
62
.
0
8
5
8





BC
BA
BC
BC
k
k
k
DF
0
.
1

AB
DF
0

CB
DF
Note: in any interior joint, the sum
of the distribution factors is 1.

187
Joint A B C
Member AB BA BC CB
DF 1.0 0.38 0.62 0
FEM -21.09 35 .15 - 50 50
BM 21.09 5.64 9.21
COM 2.82 10.54 4.6
BM -2.82 - 4.0 -6.54
COM - 2.0 -1.41 -3.27
BM 2. 0 0.54 0.87
COM 0.27 1. 0 0. 43
BM -0.27 - 0.38 -0.62
COM -0.19 - 0.13 -0.31
BM 0.19 0.06 0.07
COM 0.03 0.09 0.03
BM - 0.03 -0.04 -0.05
COM -0.02 -0.01 -0.02
BM 0.02 0.00 0.01
Final End Moment 0 47.05 -47.05 51.4 6
Moment Distribution Table ( Regular Method)
Plotting of end moments + -

188
Using Modified Stiffness Method
Member Absolute K ModifiedRelative K Relative K
AB,BA I/8 ( ¾) (1/8) ¾(1/8)(40)= 3.75
BC,CB I/5 (1/5) ( 1/5)(40)= 8
32
.
0
8
75
.
3
75
.
3





BC
BA
BA
BA
k
k
k
DF
68
.
0
8
75
.
3
8





BC
BA
BC
BC
k
k
k
DF
0
.
1

AB
DF
0

CB
DF

189
Joint A B C
Member AB BA BC CB
DF 1.0 0.32 0.68 0
FEM -21.09 35.15 - 50 50
BM 21.09
COM 10.54
Adjusted FEM 0 45.69 -50 50
Moment Distribution Table ( Modified Stiffness Method)
BM 1.38 2.93
COM 1.46
Final End Moment 0 47.07 -47.07 51.46

190
A B
5 m 3 m
24 kN/m
C
B
30 kN
47.07
47.07
∑MA = 0
RB1(8)- 47.07– 30(5) = 0
RB 1= 24.63 kN
∑Fy = 0 RB1 - RA – 30 =0 RA = 5.37 kN
+
∑MB = 0
RC(5)+ 47.07- 51.46 – 24(5)2.5 = 0
RC = 60.88kN
∑Fy = 0 RB2 + RC – 24(5) =0 RB2 = 59.12kN
+
5 m
RB = 24.63 + 59.12 = 83.75 kN
RA
RB1
RB2 RC
51.46

191
30 kN
24 kN/m
A C
B
5 m 5 m
3 m
5.37 83.75 60.88
5.37
-24.63
59.12
2.46
-60.88
2.54
26.85
-47.07
25.64
-51.46
A
B
2
C
1
2
3
4
46
.
51
)
54
.
2
)(
88
.
60
(
2
1
)
64
.
25
(
64
.
25
)
46
.
2
)(
12
.
59
(
2
1
)
07
.
47
(
07
.
47
)
3
(
63
.
24
85
.
26
85
.
26
)
5
(
37
.
5
0
4
2
2
2
3
2
2
1
1
1
1
1

























C
C
C
B
B
B
B
A
M
M
A
M
M
M
M
A
M
M
M
M
A
M
M
M
M
A
M
M

192
124+N kN
26N kN/m
D
B
A
90 +NkN
2 m 2 m
C
2 m
5 m
3m
24+3N kN
2 m 2 m 6 m
4N kN/m
A B
D
C
3 m
2 m
48+2N kN
Plate # 9 : Moment Distribution
Analyze the indeterminate beams
shown by using the moment
Distribution method
E

193
120 kN
24 kN/m
A
C
B
5 m 5 m
3 m
Problem
Assume constant E.
36 kN/m
9 m
D
I I 2I
Fixed end Moments
m
kN
L
Pab
M AB
F .
67
.
66
8
)
3
)(
5
(
120
2
2
2
2






m
kN
L
Pba
M BA
F .
6
.
140
8
)
5
)(
3
(
120
2
2
2
2



m
kN
wL
M BC
F .
50
12
)
5
(
24
12
2
2






m
kN
wL
M CB
F .
50
12
)
5
(
24
12
2
2


 m
kN
wL
M DC
F .
243
12
)
9
(
36
12
2
2



m
kN
wL
M CD
F .
243
12
)
9
(
36
12
2
2







194
Using Modified Stiffness Method
Member Absolute K ModifiedRelative K Relative K
AB,BA I/8 ( ¾) (1/8)=3/32 3/32(360)= 33.75
BC,CB I/5 (1/5) ( 1/5)(360)= 72
CD,DC 2I/9 (2I/9) ( 2/9)(360)=80
32
.
0
72
75
.
33
75
.
33





BC
BA
BA
BA
k
k
k
DF
68
.
0
72
75
.
33
72





BC
BA
BC
BC
k
k
k
DF
0
.
1

AB
DF
0

DC
DF
47
.
0
80
72
72





CD
CB
CB
CB
k
k
k
DF
53
.
0
80
72
80





CD
CB
CD
CD
k
k
k
DF

Joint A B C D
Member AB BA BC CB CD DC
DF 1.0 0.32 0.68 0.47 0.53 0
FEM -66.67 140.6 - 50 50 -243 243
66.67
BM
COM 33.33
ADJUSTED
FEM
0 173.93 - 50 50 -243 243
BM
COM
-39.66 -84.27 90.71 102.29
45.35 -42.13 51.14
BM
COM
BM
COM
-14.51 -30.84 19.80 22.33
-15.32
9.90 11.16
-6.73
-3.17 8.12
7.2
3.60 - 3.36 4.06
BM
COM
-1.15 -2.45 1.58 1.78
-1.22 0.89
0.79
BM
COM
-0.25 -0.54 0.57 0.65
-0.27
0.28 0.32
BM -0.09 -0.19 0.13 0.14
Final End
Moment
0 115.1 -115.1 107.69 -107.69 310.57

120 kN
A
B
5 m 3 m
24 kN/m
C
B
5 m
36 kN/m
9 m
D
115.1
115.1
107.69 107.69 310.57
C
16
.
30

A
R 1
B
R
2
B
R 1
C
R 2
C
R 54
.
184

D
R
196














kN
R
R
F
kN
R
R
M
B
B
y
A
A
B
39
.
89
0
120
61
.
30
0
61
.
30
0
)
3
(
120
1
.
115
8
0
1
1















kN
R
R
F
kN
R
R
M
C
C
y
B
B
C
52
.
58
0
)
5
(
24
48
.
61
0
48
.
61
0
5
.
2
)
5
(
24
1
.
115
69
.
107
5
0
1
1
2
2
+















kN
R
R
F
kN
R
R
M
D
D
y
C
C
D
54
.
184
0
)
9
(
36
46
.
139
0
46
.
139
0
5
.
4
)
9
(
36
69
.
107
57
.
310
9
0
2
2
+

197
120 kN
24 kN/m
A
C
B
5 m 3 m
36 kN/m
9 m
D
I I 2I
61
.
30

A
R 54
.
184

D
R




kN
R
R
B
B
87
.
150
48
.
61
39
.
89
87
.
150

B
R




kN
R
R
C
C
98
.
197
46
.
139
52
.
58
98
.
197

C
R
30.61
-89.39
61.48
5 m
-58.52
139.46
-184.54
2.56
3.87
1
2
3
4
5
6
1
1
A B 2 C 3
D
05
.
153
)
5
(
61
.
30
1
1
1



M
A
M
1
.
115
)
3
(
39
.
89
05
.
153
2
1







B
B
B
M
M
A
M
M
4
.
36
)
56
.
2
)(
48
.
61
(
2
1
)
1
.
115
(
2
2
3
2







M
M
A
M
M B
153.05
-115.1 -107.69
-36.4
162.16
310.57

198
120 kN
24 kN/m
A
C
B
5 m 3 m
36 kN/m
9 m
D
I I 2I
61
.
30

A
R 54
.
184

D
R




kN
R
R
B
B
87
.
150
48
.
61
39
.
89
87
.
150

B
R




kN
R
R
C
C
28
.
199
46
.
139
52
.
58
28
.
199

C
R
30.61
-89.39
61.48
5 m
-58.52
139.46
-184.54
2.56
3.87
1
2
3
4
5
6
1
1
A B 2 C 3
D
05
.
153
)
5
(
61
.
30
1
1
1



M
A
M
1
.
115
)
3
(
39
.
89
05
.
153
2
1







B
B
B
M
M
A
M
M
4
.
36
)
56
.
2
)(
48
.
61
(
2
1
)
1
.
115
(
2
2
3
2







M
M
A
M
M B
153.05
-115.1 -107.69
-36.4
162.16
310.57

199
69
.
107
)
44
.
2
)(
42
.
58
(
2
1
)
4
.
36
(
4
2








C
C
C
M
M
A
M
M
16
.
162
)
87
.
3
)(
46
.
139
(
2
1
)
69
.
107
(
3
3
5
3






M
M
A
M
M C
57
.
310
)
13
.
5
)(
54
.
184
(
2
1
)
16
.
162
(
6
3







D
D
D
M
M
A
M
M

200
A
C
B
5 m 5 m
3 m
Problem
Support B settle down 10 mm. E= 200 GPa ,I = 600 x 106 mm4
10 mm
m
kN
M
x
M
mm
N
mm
mm
mm
mm
N
L
EI
M
FAB
FAB
FAB
.
5
.
112
)
10
(
1
)
8000
(
10
)
10
(
600
)
200000
(
6
.
6
6
2
6
2
4
2
2









m
kN
M
L
EI
M
FBA
FBA
.
5
.
112
6
2





m
kN
M
x
M
L
EI
M
FBC
FBC
FBC
.
288
)
10
(
1
)
5000
(
10
)
10
(
600
)
200000
(
6
6
6
2
6
2




m
kN
M
L
EI
M
FCB
FCB
.
288
6
2




201
Member Absolute K M0dified K Relative K
AB,BA I/8 ¾(1/8) 3/4 (1/8)(160)= 15
BC,CB I/5 1/5(160)= 32


k
k
DF
32
.
0
32
15
15





BC
BA
BA
BA
k
k
k
DF
68
.
0
32
15
32





BC
BA
BC
BC
k
k
k
DF
0
.
1

AB
DF
0

CB
DF

202
Joint A B C
Member AB BA BC CB
DF 1.0 0.32 0.68 0
FEM -112.5 -112.5 288 288
112.5
BM
COM 56.25
Adjusted FEM 0 -56.25 288 288
BM
COM
-74.16 -157.59
-78.79
0 -130.41 130.41 209.21
Final end moment

203
A
B
5 m 3 m
130.41
C
B
5 m
130.41
209.21
A
R
1
B
R 2
B
R
C
R








kN
R
kN
R
R
M
B
A
A
B
3
.
16
3
.
16
0
41
.
130
)
8
(
0
1









kN
R
kN
R
R
M
B
C
C
B
92
.
67
92
.
67
0
21
.
209
41
.
130
)
5
(
0
2



 kN
RB 22
.
84
92
.
67
3
.
16

204
A
C
B
5 m 5 m
3 m 10 mm
kN
RA 3
.
16
 kN
RC 92
.
67

kN
RB 22
.
84

16.3
-67.92
21
.
209
5
)
92
.
67
(
41
.
130
41
.
130
)
8
(
3
.
16
0
2
1











C
C
B
C
B
B
A
B
M
M
A
M
M
M
M
A
M
M
130.41
-209.21
1
2

205
Joint A B C
Member AB BA BC CB
DF 1.0 0.32 0.68 0
FEM -21.09 35.15 - 50 50
BM 21.09
COM 10.54
Adjusted FEM 0 45.69 -50
BM 1.38 2.93
COM 1.46
Final End Moment 0 47.07 -47.07 51.46

206
SLOPE DEFLECTION METHOD
The slope deflection equations relate the moments at the end of the member
Of the member to the rotation and displacements of its ends and the external
load applied to the member.
Any type of loading
N
F
N
F

N
 
F

L

207
NF
F
N
NF FEM
L
EI
M 


 )
3
2
(
2



Slope deflection equation
Where :
NF
M
N

F


NF
FEM
= end moment
= rotation at the near end of the member
= rotation at the far end of the member
= rotation of the chord joining the elastic curve at the near
and far ends of the member
= fixed end moment due to the loadings in the member
Note: member end moments, end rotations, are positive when
clockwise
subscript N refers to the near end of the member
subscript F refers to the far end of the member
Use the same sign convention for FEM as in moment distribution
L



 = settlement of support

208
Degree s of Freedom
The unknown joint displacements(translation and rotation) are referred to as the
degrees of freedom of the structure. The number of degrees is called degree
of kinematic indeterminacy. If a structure does not have a degree of freedom
then it is called kinematically determinate.
Equations of Equilibrium of Interior Joints
Because the entire structure are in equilibrium, each of its members and joints
must also be in equilibrium thus in the figure below;
BA
M BC
M


 0
BC
BA M
M Joint condition equation

209
Analysis of Continuous beams by slope deflection method
1. Identify the degrees of freedom of the structure. For continuous beams,
the degrees of freedom consist of the unknown rotation of joints.
2. Compute the fixed end moments using the same table in moment distribution.
3. If there are support setllements, determine the rotation of the chords of the
members adjacent to the supports by
L



4. Write the slope deflection equation of each member in terms of the
end moments, rotations of adjacent supports and chord rotation.
5. Write equilibrium equations for each joint that are free to rotate. The total
number of equilibrium equations must be equal to the number of degrees of
freedom of the structure.
6. Determine the unknown joint rotations by solving simutaneuosly the system
of equations consisting the slope deflection equations of the members and the
equations of equilibrium. Check the validity of results if necessary.
7. Substitute the computed values of the rotations in the slope deflection
equations to determine the end moments.
8. Using the values of the end moments , determine support reactions
9. Draw the shear and moment diagrams.

210
30 kN
24 kN/m
A
C
B
5 m 5 m
3 m
Problem
Analyze the indeterminate beam shown by slope deflection method .
Assume constant EI.
Fixed end Moments
m
kN
L
Pab
M AB
F .
09
.
21
8
)
3
)(
5
(
30
2
2
2
2






m
kN
L
Pba
M BA
F .
15
.
35
8
)
5
)(
3
(
30
2
2
2
2



m
kN
wL
M BC
F .
50
12
)
5
(
24
12
2
2






m
kN
wL
M CB
F .
50
12
)
5
(
24
12
2
2




 0

For all members
Since there are no
Support settlement

211
211
AB,BA I/8 1/8(40)= 5
BC,CB I/5 1/5(40)= 8
NF
F
N
NF FEM
L
EI
M 


 )
3
2
(
2



NF
F
N
NF
NF FEM
k
M 

 )
2
(
2 

Modified slope deflection equation
1
.
09
.
21
10
20
0
09
.
21
)
2
)(
5
(
2
0
)
2
(
2
EQ
FEM
k
M
B
A
B
A
AB
B
A
AB
AB

















212
2
.
15
.
35
20
10
15
.
35
)
2
)(
5
(
2
)
2
(
2
EQ
M
M
FEM
k
M
B
A
BA
A
B
BA
BA
A
B
BA
BA
















3
.
50
32
50
)
0
2
)(
8
(
2
0
)
2
(
2
EQ
M
M
fixedEnd
FEM
k
M
B
BC
B
BC
C
BC
C
B
BC
BC
















JOINT CONDITION EQUATION
4
.
0
7
.
29
104
20
0
85
.
14
52
10
0
50
32
15
.
35
20
10
0
EQ
M
M
B
A
B
A
B
B
A
BC
BA






















213
1
.
0
09
.
21
10
20 EQ
B
A 


 

4
.
0
7
.
29
104
20 EQ
B
A 


 

SUBTRACT
0087
.
1
0
09
.
21
)
091596
.
0
(
10
20
091596
.
0
0
61
.
8
94







A
A
B
B




m
kN
M
EQ
M
BA
B
A
BA
.
06
.
47
15
.
35
)
091596
.
0
(
20
)
0087
.
1
(
10
2
.
15
.
35
20
10







 

m
kN
M
EQ
M
BC
B
BC
.
06
.
47
50
)
091596
.
0
(
32
3
.
50
32






 
m
kN
M
FEM
k
M
CB
CB
B
C
CB
CB
.
46
.
51
50
)
091596
.
0
)(
8
(
2
)
2
(
2





 


214
214
120 kN
24 kN/m
A
C
B
5 m 5 m
3 m
Problem
Analyze the indeterminate beam shown by slope deflection method .
Assume constant E.
36 kN/m
9 m
D
I I 2I
Fixed end Moments
m
kN
L
Pab
M AB
F .
67
.
66
8
)
3
)(
5
(
120
2
2
2
2






m
kN
L
Pba
M BA
F .
6
.
140
8
)
5
)(
3
(
120
2
2
2
2



m
kN
wL
M BC
F .
50
12
)
5
(
24
12
2
2






m
kN
wL
M CB
F .
50
12
)
5
(
24
12
2
2


 m
kN
wL
M DC
F .
243
12
)
9
(
36
12
2
2



m
kN
wL
M CD
F .
243
12
)
9
(
36
12
2
2







215
AB,BA I/8 (1/8)360=45
BC,CB I/5 (1/5)360=72
CD,DC 2I/9 (2/9)360=80
2
.
6
.
140
180
90
6
.
140
)
2
)(
45
(
2
)
2
(
2
EQ
M
M
FEM
k
M
B
A
BA
A
B
BA
BA
A
B
BA
BA
















1
.
67
.
66
90
180
0
67
.
66
)
2
)(
45
(
2
0
)
2
(
2
EQ
FEM
k
M
B
A
B
A
AB
B
A
AB
AB

















216
3
.
50
144
288
50
)
2
)(
72
(
2
)
2
(
2
EQ
M
M
FEM
k
M
C
B
BC
C
B
BC
BC
C
B
BC
BC
















JOINT CONDITION EQUATION at B
4
.
0
6
.
90
144
468
90
0
50
144
288
6
.
140
180
90
0
EQ
M
M
C
B
A
C
B
B
A
BC
BA




















5
.
50
288
144
50
)
2
)(
72
(
2
)
2
(
2
EQ
M
M
FEM
k
M
C
B
CB
B
C
CB
CB
B
C
CB
CB

















6
.
243
320
243
)
0
2
)(
80
(
2
0
)
2
(
2
EQ
M
M
fixedEnd
FEM
k
M
C
CD
C
CD
D
CD
D
C
CD
CD
















JOINT CONDITION EQUATION at C
7
.
0
193
608
144
0
243
320
50
288
144
0
EQ
M
M
C
B
C
C
B
CD
CB
















JOINT CONDITION EQUATION
8
.
0
34
.
33
45
90 EQ
B
A 


 

Divide EQ. 1 by 2

218
4
.
0
6
.
90
144
468
90 EQ
C
B
A 



 


8
.
0
34
.
33
45
90 EQ
B
A 


 

subtract
9
.
0
94
.
123
144
423 EQ
C
B 


 

Multiply EQ. 7 by 2.9375
10
.
0
94
.
566
1786
423 EQ
C
B 


 

9
.
0
94
.
123
144
423 EQ
C
B 


 

420755
.
0
0
88
.
690
1642



C
C


subtract
436238
.
0
0
94
.
123
)
420755
.
0
(
144
423





B
B


588563
.
0
0
34
.
33
)
436238
.
0
(
45
90





A
A



219
420755
.
0

C

436238
.
0


B

588563
.
0

A

0
000006
.
0
0
6
.
90
)
420755
.
0
(
144
)
436238
.
0
(
468
)
588563
.
0
(
90






check
m
kN
M
M
BA
B
A
BA
.
05
.
115
6
.
140
)
436238
.
0
(
180
)
588563
.
0
(
90
6
.
140
180
90







 

m
kN
M
M
BC
C
B
BC
.
05
.
115
50
)
420755
.
0
(
144
)
436238
.
0
(
288
50
144
288








 

m
kN
M
M
CB
C
B
CB
.
36
.
108
50
)
420755
.
0
(
288
)
436238
.
0
(
144
50
288
144







 

m
kN
M
M
CD
C
CD
.
36
.
108
243
)
420755
.
0
(
320
243
320





 
m
kN
M
M
M
FEM
k
M
DC
C
DC
C
DC
DC
C
D
DC
DC
.
32
.
310
243
)
420755
.
0
(
160
243
160
243
)
)(
80
(
2
)
2
(
2















220
125+N kN
16N kN/m
A
C
B
6 m 5 m
3 m
24N kN/m
8 m
D
2I I 2I
Draw the shear and Moment diagrams of the beam shown using the
slope deflection method . Assume constant E.
Plate # 10: Slope Deflection Method

221
ANALYSIS OF RIGID FRAMES WITHOUT SIDESWAY
The analysis of rigid frames without sideway is similar to the analysis of
continuous beams. The joints are only subject to rotations. This could
be analyzed by either moment distribution or slope deflection method
16 kN/m
4 m
A
B C
D
6m
Problem : Analyze the rigid frame shown by moment distribution.
Assume that there is no sidesway
I I
2I

222
m
kN
wL
M CB
F .
48
12
)
6
(
16
12
2
2



AB,BA I/4 1/4(12)= 3
BC,CB 21/6 =I/3 1/3(12)= 4


k
k
DF
43
.
0
3
4
3





BC
BA
BA
BA
k
k
k
DF
57
.
0
3
4
4





BC
BA
BC
BC
k
k
k
DF
0

AB
DF
0

DC
DF
m
kN
wL
M BC
F .
48
12
)
6
(
16
12
2
2






CD,DC I/4 1/4(12)= 3
57
.
0
3
4
4





CD
CB
CB
CB
k
k
k
DF
43
.
0
3
4
3





CD
CB
CD
CD
k
k
k
DF

223
Joint A B C D
DF 0 0.43 0.57 0.57 0.43 0
FEM 0 0 -48 48 0 0
BM
COM 10.32 -13.68 13.68 -10.32
20.64 27.36 -27.36 -20.64 0
BM 5.88 7.8 -7.8 - 5.88
COM 2.94 -3.9 3.9 - 2.94
BM 1.68 2.22 -2.22 -1.68
COM 0.84 -1.11 1.11 - 0.84
BM 0.48 0.63 -0.63 -0.48
COM 0.24 -0.31 0.31 - 0.24
BM 0.13 0.18 -0.18 -0.13
COM 0.06 -0.09 0.09 - 0.06
BM 0.04 0.05 -0.05 -0.04
COM 0.02 -0.03 0.03 - 0.02
BM 0.01 0.02 -0.02 -0.01
Final
End Moment 14.42 28.86 -28.86 28.86 -28.86 -14.42

224
A
H
4
A
B







kN
H
H
M
A
A
B
82
.
10
0
4
86
.
28
42
.
14
0
14.42
28.86
A
V
B
V
B
H
D
H
4
D
C







kN
H
H
M
D
D
C
82
.
10
0
4
86
.
28
42
.
14
0
14.42
28.86
D
V
C
V
C
H

225
16 kN/m
B C
2I
B
V C
V
28.86
28.86
6m




















kN
V
kN
V
V
F
kN
V
V
kN
V
V
M
A
B
B
y
C
D
C
C
B
48
48
0
)
6
(
16
48
0
48
48
0
6
3
)
6
(
16
86
.
28
86
.
28
0

16 kN/m
4 m
A
B C
D
I I
2I
48 48
10.82 10.82
14.42 14.42

ANALYSIS OF RIGID FRAMES WITH SIDESWAY
The analysis of rigid frames without sideway is similar to the analysis of
continuous beams. The joints are only subject to rotations and
translation ( sidesway). This could be analyzed by either moment
distribution or slope deflection method.
Any load
A
B C
D
A
H '
D
H
AB
M DC
M
y z
 
= Joint translation
or sidesway
A
V D
V

228
Procedure
Step 1 : Hold frame at C to prevent sidesway by applying
an imaginary horizontal (R) force at C. Determine the
end moments due to the given loadings.
Any load
A
B C
D
R
'
A
H '
D
H
'
AB
M '
DC
M
y z
,
'
AB
M ,
'
BA
M '
DC
M
,.....
'
BC
M
Note:
are the end moments due to the given
loads
'
A
V '
D
V

229
Step 2
Isolate the FBD of the columns and using the resulting end moments determine
horizontal reactions HA’ and HD’.
'
AB
M
'
BA
M
'
A
H
y
A
B
y
M
M
H
y
H
M
M
M
BA
AB
A
A
BA
AB
B
'
'
'
0
'
'
'
0







'
DC
M
'
CD
M
'
D
H
z
D
C
z
M
M
H
z
H
M
M
M
DC
CD
D
D
DC
CD
C
'
'
'
0
'
'
'
0








230
Step 3
From the FBD of the frame solve for R
Any load
A
B C
D
R
'
A
H '
D
H
'
AB
M '
DC
M
y z
'
'
0
D
A
x
H
H
R
F





231
Step 4
Remove all loads and push the frame at C to produce sideway
Assume the value of the sideway in terms of the unknown sidesway
correction Factor x. By slope deflection or moment distribution
Solve for the end moments in terms of x.
A
B
C
D
Q
'
'
A
H '
'
D
H
'
'
AB
M '
'
DC
M
y z
 
x
EI
100


Example
2
6
L
EI
M
M FBA
FAB




2
6
L
EI
M
M FDC
FCD




,
'
'
AB
M ,
'
'
BA
M '
'
DC
M
,.....
'
'
BC
M
Note:
are the end moments due to 
'
'
D
V
'
'
A
V

232
232
Step 5
Isolate the FBD of the columns and using the resulting end moments determine
horizontal reactions HA’ ‘and HD’’.
'
'
AB
M
'
'
BA
M
'
'
A
H
y
A
B
y
M
M
H
y
H
M
M
M
BA
AB
A
A
BA
AB
B
'
'
'
'
'
'
0
'
'
'
'
'
'
0







'
'
DC
M
'
'
CD
M
'
'
D
H
z
D
C
z
M
M
H
z
H
M
M
M
DC
CD
D
D
DC
CD
C
'
'
'
'
'
'
0
'
'
'
'
'
'
0








233
A
B
C
D
Q
'
'
A
H '
'
D
H
'
'
AB
M '
'
DC
M
y z
 
Step 5
From the FBD of the frame solve for Q
'
'
'
'
0
D
A
x
H
H
Q
F





234
Step 6
Since Q and R are imaginary they must be equal thus equating values of
R and Q determines the value of x
Note : If Q and R are of opposite direction x is positive
If Q and R are of the same direction x is negattive
Step 7
Determine the end moments and reactions
'
'
' AB
AB
AB M
M
M 

'
'
' BA
BA
BA M
M
M 

'
'
' BC
BC
BC M
M
M 

'
'
' CB
CB
CB M
M
M 

'
'
' CD
CD
CD M
M
M 

'
'
' DC
DC
DC M
M
M 

'
'
' A
A
A H
H
H 

'
'
' D
D
D H
H
H 

'
'
' A
A
A V
V
V 

'
'
' D
D
D V
V
V 


235
16 kN/m
4 m
A
B C
D
6m
I 2I
2I
8m
Assume that there is sidesway

236
16 kN/m
4 m
A
B C
D
I
2I
2I
8m
6m
R
'
A
H
'
D
H
'
AB
M
'
DC
M
'
A
V
'
D
V
Frame A

237
237
m
kN
wL
M CB
F .
48
12
)
6
(
16
12
2
2



AB,BA I/4 1/4(12)= 3
BC,CB 21/6 =I/3 1/3(12)= 4


k
k
DF
43
.
0
3
4
3





BC
BA
BA
BA
k
k
k
DF
57
.
0
3
4
4





BC
BA
BC
BC
k
k
k
DF
0

AB
DF
0

DC
DF
m
kN
wL
M BC
F .
48
12
)
6
(
16
12
2
2






CD,DC 2I /8)=I/4 1/4(12)= 3
57
.
0
3
4
4





CD
CB
CB
CB
k
k
k
DF
43
.
0
3
4
3





CD
CB
CD
CD
k
k
k
DF

238
Joint A B C D
DF 0 0.43 0.57 0.57 0.43 0
FEM 0 0 -48 48 0 0
BM
COM 10.23 -18.82 18.82 -10.23
20.46 27.64 -27.54 -20.46 0
BM 8.09 10.73 -10.73 -8.09
COM 4.04 -5.36 5.36 - 4.04
BM 2.30 3.06 -3.06 -2.30
COM 1.15 -1.53 1.53 - 1.15
BM 0.66 0.87 -0.87 -0.66
COM 0.33 -0.43 0.43 - 0.33
BM 0.19 0.24 -0.24 -0.19
COM 0.09 -0.12 0.12 - 0.09
BM 0.05 0.07 -0.07 -0.05
COM 0.02 -0.03 0.03 - 0.02
End Moment
15.86 31.76 -31.76 31.76 -31.76 -15.86
BM 0.01 0.02 -0.02 -0.01
86
.
15
'
AB
M 76
.
31
'
BA
M 76
.
31
' 

BC
M 76
.
31
'
CB
M 76
.
31
' 

CD
M 86
.
15
' 

DC
M
Moment Distribution Table due to Loads

239
'
A
H
4
A
B







kN
H
H
M
A
A
B
91
.
11
'
0
4
'
76
.
31
86
.
15
0
15.86
31.76
'
A
V
'
B
V
'
D
H
8m
D
C







kN
H
H
M
D
D
C
95
.
5
'
0
8
'
76
.
31
86
.
15
0
15.86
31.76
'
D
V
'
C
V







kN
R
R
Fx
96
.
5
0
95
.
5
91
.
11
0
From FBD of frame A

240
4 m
A
B C
D
6m
I 2I
2I
8m
 
Q
Assume
x
EI
120


x
x
EI
EI
L
EI
M
M FBA
FAB 45
4
120
6
6
2
2








x
x
EI
I
E
L
EI
M
M FDC
FCD 5
.
22
8
120
2
6
6
2
2








Frame B

241
Joint A B C D
DF 0 0.43 0.57 0.57 0.43 0
FEM -45x -45x -22.5x -22.5x
BM
COM
19.35x 25.65x 12.82x 9.68x
9.67x 6.41x 12.32x 4.84x
BM
COM
-2.76x -3.65x -7.02x -5.3x
-1.38x - 3.51x -1.82x -2.65x
BM
COM
1.65x 1.86x 1.04x 0.78x
0.82x 0.52x 0.93x 0.39x
BM
COM
-0.24x - 0.28x -0.53x -0.4x
-0.12x -0.26x -0.14x -0.2x
BM
COM
0.12x 0.14x 0.08x 0.06x
0.06x 0.04x 0.07x 0.03x
-0.02x -0.02x -0.04x -0.03x
BM
-35.95x -26.9x 26.9x 17.71x -17.71x -20.09x
Moment Distribution Table due to 
x
MAB 95
.
35
'
' 

x
MBA 9
.
26
'
' 

x
MBC 9
.
26
'
' 
x
MCB 71
.
17
'
' 
x
MCD 71
.
17
'
' 

x
MDC 09
.
20
'
' 

End Moment

242
'
'
A
H
4
A
B







x
H
H
x
x
M
A
A
B
71
.
15
'
'
0
4
'
'
9
.
26
95
.
35
0
5.36
26.9x
'
'
A
V
'
'
B
V
'
'
D
H
8m
D
C







x
H
H
x
x
M
D
D
C
73
.
4
'
'
0
8
'
'
09
.
20
71
.
17
0
20.09x
17.71x
'
'
D
V
'
'
C
V







x
Q
Q
x
x
Fx
44
.
20
0
73
.
4
71
.
15
0
From FBD of frame B

Sidesway Correction Factor
292
.
0
96
.
5
44
.
20



x
x
R
Q
Positive since Q and R are of opposite directions
)
292
.
0
(
95
.
35
86
.
15
95
.
35
86
.
15
'
'
'






AB
AB
AB
AB
AB
M
x
M
M
M
M
)
292
.
0
(
9
.
26
76
.
31
9
.
26
76
.
31
'
'
'






BA
BA
BA
BA
BA
M
x
M
M
M
M
m
kN
MBA .
9
.
23

m
kN
MAB .
36
.
5

)
292
.
0
(
9
.
26
76
.
31
9
.
26
76
.
31
'
'
'








BC
BC
BC
BC
BC
M
x
M
M
M
M
m
kN
MBC .
9
.
23


)
292
.
0
(
71
.
17
76
.
31
71
.
17
76
.
31
'
'
'






CB
CB
CB
CB
CB
M
x
M
M
M
M
m
kN
MCB .
93
.
36

)
292
.
0
(
71
.
17
76
.
31
71
.
17
76
.
31
'
'
'








CD
CD
CD
CD
CD
M
x
M
M
M
M
m
kN
MCD .
93
.
36


)
292
.
0
(
09
.
20
86
.
15
09
.
20
86
.
15
'
'
'








DC
DC
DC
DC
DC
M
x
M
M
M
M
m
kN
MDC .
73
.
21


Final End Moments

244
A
H
4
A
B
5.36
23.9
A
V
'
'
B
V
D
H
8m
D
C







kN
H
H
M
A
A
B
32
.
7
0
4
9
.
23
36
.
5
0
21.73
36.93
D
V
'
'
C
V







kN
H
H
M
D
D
C
32
.
7
0
8
73
.
21
93
.
36
0








kN
H
H
x
H
H
H
H
A
A
A
A
A
A
32
.
7
)
292
.
0
(
71
.
15
91
.
11
71
.
15
91
.
11
Check
'
'
'

16 kN/m
B
C
VB Vc
23.9 36.93
6m



















kN
V
kN
V
V
F
kN
V
kN
V
V
M
D
C
C
y
A
B
B
C
17
.
50
17
.
50
0
)
6
(
16
83
.
45
0
83
.
45
83
.
45
0
6
93
.
36
9
.
23
3
)
6
(
16
0

A
B
C
D
7.32
7.32
45.83
50.17
86
.
2
16
83
.
45


x
x
2.86
Shear diagram

A
B
C
D
-
+
- +
-
+
-
+
5.36
23.9
23.9
36.93
21.73
36.93
M-diagram

4 m
A
B C
D
6m
I 2I
2I
8m
Assume that there is sidesway
25 kN

4 m
A
B C
D
I
2I
2I
8m
6m
R
'
D
H
'
AB
M
'
DC
M
'
A
V
'
D
V
Frame A
25 kN

250
250
AB,BA I/4 1/4(12)= 3
BC,CB 21/6 =I/3 1/3(12)= 4


k
k
DF
43
.
0
3
4
3





BC
BA
BA
BA
k
k
k
DF
57
.
0
3
4
4





BC
BA
BC
BC
k
k
k
DF
0

AB
DF
0

DC
DF
0





 FDC
FCD
F
FBC
FBA
FAB M
M
M
M
M
M CB
CD,DC 2I /8)=I/4 1/4(12)= 3
57
.
0
3
4
4





CD
CB
CB
CB
k
k
k
DF
43
.
0
3
4
3





CD
CB
CD
CD
k
k
k
DF

Joint A B C D
DF 0 0.43 0.57 0.57 0.43 0
FEM 0 0 0 0 0 0
0
'
AB
M 0
'
BA
M 0
'
BC
M 0
'
CB
M 0
'
CD
M 0
'
DC
M
Moment Distribution Table due to Loads
4 m
A
B C
D
I
2I
2I
8m
6m
R
25






kN
R
R
FX
25
0
25
0

4 m
A
B C
D
6m
I 2I
2I
8m
 
Q
Assume
x
EI
120


x
x
EI
EI
L
EI
M
M FBA
FAB 45
4
120
6
6
2
2








x
x
EI
I
E
L
EI
M
M FDC
FCD 5
.
22
8
120
2
6
6
2
2








Frame B

253
Joint A B C D
DF 0 0.43 0.57 0.57 0.43 0
FEM -45x -45x -22.5x -22.5x
BM
COM
19.35x 25.65x 12.82x 9.68x
9.67x 6.41x 12.32x 4.84x
BM
COM
-2.76x -3.65x -7.02x -5.3x
-1.38x - 3.51x -1.82x -2.65x
BM
COM
1.65x 1.86x 1.04x 0.78x
0.82x 0.52x 0.93x 0.39x
BM
COM
-0.24x - 0.28x -0.53x -0.4x
-0.12x -0.26x -0.14x -0.2x
BM
COM
0.12x 0.14x 0.08x 0.06x
0.06x 0.04x 0.07x 0.03x
-0.02x -0.02x -0.04x -0.03x
BM
-35.95x -26.9x 26.9x 17.71x -17.71x -20.09x
Moment Distribution Table due to 
x
MAB 95
.
35
'
' 

x
MBA 9
.
26
'
' 

x
MBC 9
.
26
'
' 
x
MCB 71
.
17
'
' 
x
MCD 71
.
17
'
' 

x
MDC 09
.
20
'
' 

End Moment

254
'
'
A
H
4
A
B







x
H
H
x
x
M
A
A
B
71
.
15
'
'
0
4
'
'
9
.
26
95
.
35
0
35.95x
26.9x
'
'
A
V
'
'
B
V
'
'
D
H
8m
D
C







x
H
H
x
x
M
D
D
C
73
.
4
'
'
0
8
'
'
09
.
20
71
.
17
0
20.09x
17.71x
'
'
D
V
'
'
C
V







x
Q
Q
x
x
Fx
44
.
20
0
73
.
4
71
.
15
0
From FBD of frame B

Sidesway Correction Factor
22
.
1
25
44
.
20



x
x
R
Q
Positive since Q and R are of opposite directions
kN.m
86
.
43
)
22
.
1
(
95
.
35
0
'
'
0






AB
AB
AB
AB
M
M
M
M
)
22
.
1
(
9
.
26
9
.
26
0
'
'
0






BA
BA
BA
BA
M
x
M
M
M
m
kN
MBA .
82
.
32


m
kN
MAB .
86
.
43


)
22
.
1
(
9
.
26
9
.
26
0
'
'
'





BC
BC
BC
BC
BC
M
x
M
M
M
M
m
kN
MBC .
82
.
32

)
22
.
1
(
71
.
17
71
.
17
0
'
'
'





CB
CB
CB
CB
CB
M
x
M
M
M
M
m
kN
MCB .
61
.
21

)
22
.
1
(
71
.
17
71
.
17
0
'
'
'






CD
CD
CD
CD
CD
M
x
M
M
M
M
m
kN
MCD .
61
.
21


)
22
.
1
(
09
.
20
09
.
20
0
'
'
'






DC
DC
DC
DC
DC
M
x
M
M
M
M
m
kN
MDC .
51
.
24


Final End Moments

A
H
4
A
B
43.86
32.82
A
V
'
'
B
V
D
H
8m
D
C







kN
H
H
M
A
A
B
17
.
19
0
4
82
.
32
86
.
43
0
24.51
21.61
D
V
'
'
C
V







kN
H
H
M
D
D
C
83
.
5
0
8
51
.
24
61
.
21
0
25
83
.
5
17
.
19
25
Check
'
'
'






D
A
A
A
A
H
H
H
H
H

I
A cantilever beam of Length L with stepwise cross section
carries a vertical load 2P at its free end. The section of the
beam changes midway along its length so that its second
moment of area is reduced by one half. The smaller section is
towards the free end. If E is constant, determine the
deflection at the free end.
2P
L/2 L/2
I
2I
A B C
Prelim: Structural Theory II

2 m 4 m
18kN/m
50 kN
II
Determine the midspan deflection of the beam shown by
a) Double Integration Method
b) Conjugate beam Method
c) Area Moment Method
A B
III
Locate and compute the maximum deflection of the beam
given in problem II.

+
259
I
A cantilever beam of Length L with stepwise cross section
carries a vertical load 2P at its free end. The section of the
beam changes midway along its length so that its second
moment of area is reduced by one half. The smaller section is
towards the free end. If E is constant, determine the
deflection at the free end using the virtual work method
2P
L/2 L/2
I
2I
A B C
MIDTERM EXAM: Structural Theory II

260
2 m 4 m
18kN/m
50 kN
II
Determine the midspan deflection of the beam shown by
using the virtual work Method
A B

261
24 kN/m
48 kN
2 m
2 m
6 m
A
B
C
E
3 m
F
G
III
36 kN

262
Semi- Final Exam
Determine the horizontal and vertical displacement of joint C. A = 5000 sq. mm and
E = 200 GPa for all members
60kN 120kN 60kN
3m 4m 4m 3m
4m
F G H
A
B C D
E
w
L/2 L/2
A
B C
Draw the shear and moment diagrams of the beam shown. Assume constant EI.

StructuralTheoryClass2.ppt

More Related Content

What's hot (20)

Similar to StructuralTheoryClass2.ppt (20)

Recently uploaded (20)

StructuralTheoryClass2.ppt