CE 103 Structural Analysis
Prep Course for PE (Civil Engineering Exam)
Assoc Professor Ang Kok Keng
Department of Civil & Environmental Engineering
National University of Singapore
Topics
• Classical Methods
• Analysis of Statically Determinate Structures
• Internal Forces in Structural Members
• Deflections
• Influence Lines
• Analysis of Statically Indeterminate Structures
• Force Methods
• Displacement Methods
• Matrix-Stiffness Method
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ANALYSIS OF STATICALLY
DETERMINATE STRUCTURES
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3
Free-Body Diagram
First step of structural analysis is usually to draw a
free-body diagram
Simplified sketch of structure, isolated from its
supports, showing all forces acting on it
Idealized model
Free body diagram
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Equations of Equilibrium
F  0 ; F  0 ; F  0
M  0 ; M  0 ; M  0
3-D
F  0 ; F  0 ; M  0
2-D
x
y
x
x
z
y
y
z
O
Alternatively,
or
 Fx  0 ;
 MO  0 ;
 MO '  0
 MO  0 ;
 MO '  0 ;
 MO"  0
in which points O, O’ and O’’ are not in a straight line.
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Stability And Determinacy
C
B
A
Determinate
Structure
C
B
A
Indeterminate
Structure
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C
B
A
Unstable
Structure
6
General Stability And Determinacy of Beams
If r  c  3 ,
beam is statically unstable
If r  c  3 ,
beam is statically determinate*
If r  c  3 ,
beam is statically indeterminate*
* provided there is no geometric instability (internal and external)
where r  number of reaction elements
c  number of equations of condition
 1 for hinge, 2 for roller, 0 no internal connection
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Example
r c
<
r = c3
>
5 2 5  5 Stable & determinate
6 2 6  5 Stable & indeterminate
5 2 5  5 Unstable*
5 3 5  6 Unstable
6 3 6  6 Stable & determinate
7 2 7  5 Unstable*
* Internal geometric instability
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General Stability And Determinacy of Trusses
If m  r  2 j , truss is statically unstable
If m  r  2 j , truss is statically determinate*
If m  r  2 j , truss is statically indeterminate*
* provided there is no external/internal geometric instability
where m  number of truss members
r  number of reaction components
j  number of joints
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Example
m r
j
<
mr = 2j
>
7 3 5 10  10
Stable & determinate
7 3 5 10  10
Unstable*
7 3 5 10  10
Unstable**
6 3 5 9  10
Unstable
8 4 5 12  10
Stable & indeterminate
6 4 5 10  10
Stable & determinate
* Internal geometric instability
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** External geometric instability
10
General Stability And Determinacy of Frames
If 3m  r  3 j  c, frame is statically unstable
If 3m  r  3 j  c, frame is statically stable*
If 3m  r  3 j  c, frame is statically indeterminate*
* provided there is no geometric instability
where m  number of members
r  number of reaction elements
j  number of joints
c  number of equations of condition
 1 for hinge, 2 for roller, 0 no internal connection
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Example
m r
j c
<
3m  r = 3 j  c
>
10 9 9 0 39  27
10 9 9 4 39  31
Stable & indeterminate
to 12th degree
Stable & indeterminate
to 8th degree
10 9 9 3 * 39  30
10 9 9 1 39  28
Stable & indeterminate
to 9th degree
Stable & indeterminate
to 11th degree
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INTERNAL FORCES IN
STRUCTURAL MEMBERS
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Plane Trusses
A truss is a framework which is usually built-up from members
forming triangles. Trusses are common in roof construction,
transmission towers, crane booms and bridges.
For analytical purposes, we assume that
 The joints are pinned or hinged, which implies that the
member forces are axial because there can be no moment
about the hinge
 Loads act only at joints
A truss is represented by a line drawing with the lines along
the centrelines of the members.
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Sign Convention
Tension (+ve)
Compression (-ve)
Pin is subjected to a
concurrent force system
Consider the above free-body diagram of a separate joint of
a plane truss (i.e. a 2-D truss built to resist forces in its own
plane). The joint is in equilibrium when the forces acting on it
satisfy
 Fx  0
and
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 Fy  0
15
Method of Joints
At each joint,  Fx  0,
 Fy  0
Example – Determine the member forces in the plane
truss carrying the loads as shown in the figure.
4
15 kN
3
5 kN
C
E
1m
9
A
B
1m
20
3
D
1m
F
1m
31
3
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First consider the
free body diagram
of the entire truss
and determine the
external reactions
by the equations of
equilibrium.
16
Example (Cont’d)
F1
At joint A
9
45
A
Assume tensile
forces initially
F2
20
3
 Fy  0:
 Fx  0:
20 F1

 0  F1  -9.43 kN
3
2
F1
-9 
 F2  0  F2  15.66 kN
2
Proceed to next joint (any joint) which has no more than 2
unknowns, e.g. joint B but not joint C. Why?
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Joints Under Special Loading Conditions
FAE
FAB
E
B
A
D
C
A
FAD
FAC
FAE  FAC and FAB  FAD
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Joints Under Special Loading Conditions
FAE
P
FAB
P
E
B
A
D
A
C
FAD
FAC
FAC  P
and
FAB  FAD
then FAC  P  0
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Joints Under Special Loading Conditions
FAE
FAB
E
B
A
D
C
A
FAD
FAC
FAB  FAD
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Joints Under Special Loading Conditions
FAE
FAB
E
B
A
D
C
A
FAD
FAC
FAE  FAC and FAB  FAD
FAD  FAE  0
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Method of Sections
In this method, the equilibrium of parts of the truss is
considered.
The procedure involves cutting some carefully selected
members of truss in order to isolate a free body
comprising a section or portion of the truss onto which
cut member forces are applied as external forces.
Forces are calculated by applying equilibrium to the
selected section.
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Example
4
15 kN
3
5 kN
C
E
1m
9
A
B
1m
D
1m
F
1m
31
3
20
3
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Example (Cont’d)
15 kN
5 kN
12 kN
C
9 kN
Assume tensile
forces in cut
members
F6
F5
1m
9
1m
1m
A
B
F4 D
20
3
Left hand portion of the truss is selected as a free body for the
purpose of determining three unknown forces F4, F5 and F6.
 Fy  0 :
 MD  0 :
 MC  0 :
20
F
- 12 - 5  0
 F5  - 7.54 kN
3
2
20
( 2) - 12(1)  9(1)  F6 (1)  0  F6  - 10.33 kN
3
20

9 (1)
(1) - F4 (1)  0
 F4  15.66 kN
3
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Method of Joint or Section?
In some types of determinate trusses, a difficulty arises with the
method of joints.
For example,
?
?
I ??
? ?
?
?
?
?
?
?
?
J
Fink Truss
At joints I and J, there are three unknown forces to be found. We cannot
proceed with the method of joints without some modifications. We may
overcome the difficulty by taking a section through three members and
determining the three corresponding member forces by the method of
sections. Then, it is possible to continue with the method of joints.
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Example
Determine the force in member IK of the truss shown below.
A
100 kN
D
100 kN
G
100 kN
2.7 m 2.4 m
B
C
E
H
a
100
2.7 m
F
2.7 m
I
100
a 2.7 m
K
J
7.5 m
q
100
FGH FIH
FGJ
cot q 
2.7  3
2.4
FIK
Take the section above a-a as a free body
MG  0

2  2 .7  2 .4 


q
100 ( 2.7 5.4) FIK cos  7.5
0


8 .1
- 810

 -143 .2 kN
IK
F
8 .1
( 7 . 5 - 1 .6 )
8.448
 143.2 kN (compression )
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Example
Determine the forces in members AB and CD of
the plane truss loaded at joint B as shown below.
B
A
P
a
C
a
D
a
a
a
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Roller
Support
27
Example (Cont’d)
B
A
K
P
H
F
P
G
C
I
J
a
E
D
a
K
a
a
2P/3
FAB
H
FEF
G
J
B
P
a
2P/3
a
a
F1 = F2 and F3 = F4
These forces cancel each other in
the following equations:
C
I
F2 F3
E
By considering the free body
diagram of EBC and noting that
since at each of joints G, H, I, J, the
intersecting members lie in two
straight lines, we have
2P/3
F4
F1
P
a
By considering a free body diagram
of the entire truss, we can
determine the support reactions.
ΣMk = 0: - P (2a) – FCD (3a) = 0
FCD = - 2P/3 (compression)
FCD
ΣFx = 0: P - P - FAB = 0
FAB = 0
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Beams and Frames
By
Bx
B
By
B
q
Bx
C
q
P
C
Ay
V
M
 an axial force, F
F
F
Ax
A
Internal forces developed in
multiforce beam/frame members to
maintain equilibrium under applied
loads include:
 a shear force, V
M
VC
A
P
 a bending moment, M
Ax
Ay
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Sign Convention
 A cut through a beam reveals equal and opposite shear
forces and bending moments on opposing faces
P
P
m M M Vm
m
B
A
x
n
x
n V
B
n
 Upward force positive? Anticlockwise moment positive?
 Need a consistent sign convention to indicate sense of
internal beam resultants
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Sign Convention (Cont’d)

Sign conventions for stress resultants
called deformation sign conventions



because based upon how material is deformed
deformation sign convention also used for axial
members
 Elongation (tension) in a bar is +ve
 Shortening (compression) is –ve
Static sign convention used when writing
equations of equilibrium

forces are +ve or –ve according to their
directions along coordinate axes
P is given a positive sign in equation of
equilibrium because it acts upward;
V (positive shear force) is given a
negative sign because it acts
downward
P
A
m
n
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M
y
x
V
31
Method of Section
q
A
L
m
q0
B
x
n
q0 x 2
2L
q
x
n
q0
m M
A
L
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V
32
Method of Section (Cont’d)
q0 x 2
2L
q
x
n
q0
m M
A
L
q0 x
q
L
V
Total downward load on free body is
1  q0 x 
q0 x 2

 x 
2 L 
2L
From an equation of equilibrium in the vertical direction we find
+   Fy  0 :
q0 x 2
q0 x 2
-V  0 V  2L
2 L # (Ans)
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Method of Section (Cont’d)
q0 x
q
L
q0 x 2
2L
A
x
m M
x
3
n
V
Write an equation of moment equilibrium about cut section

M

mn
 0:
1  q0 x 
 x
M 
 x   0
2 L 
3
q0 x 3
M 6 L # (Ans)
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Relationships Between Load, Shear, BM
P
M V
M  M1
q
y
dx V  V1
P
A
M0
B
dx
RA
dx
q
M V
C
dx
RB
M  dM
dx V  dV
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x
M0
M V
M  M1
dx V  V1
35
Relationships Between Distributed Load, …
dV
 -q
dx
q
M V
B
M  dM
B
 dV  V - V   - qdx
A
B
A
A
dM
V
dx
dx V  dV
B
B
 dM  M - M   Vdx
A
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B
A
A
36
Relationships Between Concentrated Loads, …
P
M V
M  M1
dx V  V1
V1  - P
M1  0
M0
The picture can't be display ed.
M V
M  M1
The picture can't be display ed.
dx V  V1
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Shear and Moment Diagrams
1. Sketch the free-body (load) diagram
showing loads, support reactions and
dimensions for easy reference
2. Based on free-body diagram,
determine support reactions using
equations of equilibrium
3. Construct vertical lines down from load
diagram through all points of conc.
loads/couples, supports and start and
end points of distributed loads
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Shear and Moment Diagrams (Cont’d)
4. Start sketching shear diagram over
each segment (between vertical
construction lines) from left to right
 Determine shear at left end of segment
 For 1st segment, shear at left end = left beam vertical
support reaction
 Determine shear at right end of segment using
B
VB - VA  -  q  dx
A
 Follow direction of loading: VB > VA if distributed
loading is acting upward
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Shear and Moment Diagrams (Cont’d)
4. Start sketching shear diagram …..
 Sketch shape of shear diagram within segment using
dV
 -q
dx
 Abrupt change in shear force occurs at point of
concentrated load not concentrated moment
 Follow direction of loading: shear immediately to the
right of conc. load increases if load is acting up
5. Start sketching moment diagram over
each segment (between vertical
construction lines) from left to right
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Shear and Moment Diagrams (Cont’d)
5. Start sketching moment diagram over
each segment (between vertical
construction lines) from left to right
 Determine BM at left end of segment
 For 1st segment, BM at left end = left beam moment
support reaction
 Determine BM at right end of segment using
B
M B - M A   Vdx
A
 MB > MA if area under shear diagram is positive
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Shear and Moment Diagrams (Cont’d)
5. Start sketching moment diagram ….
 Sketch shape of moment diagram within segment
using
dM
V
dx
 Abrupt change in bending moment occurs at point of
concentrated moment not concentrated load
 If conc. moment is positive (anticlockwise), then BM
immediately to right of conc. moment will decrease
and vice-versa
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Example: Beam
q = 1.0 kN/m
M0 = 12.0 kN.m
B
A
C
4m
8m
8m
q = 1.0 kN/m
M0 = 12.0 kN.m
B
A
4m
RB
8m
C
8m
RC
 M  0 : R 16  M  q  4  2  0
B

F  0:
y

C
0
RC  -1.25 kN
RB  RC - q  4  0
RB  5.25 kN
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Example (Cont’d)
q = 1.0 kN/m
M0 = 12.0 kN.m
B
A
C
4m
8m
8m
1.25 kN
5.25 kN
V (kN)
+1.25
0
# (Ans)
-4.0
+2.0
M (kN.m) 0
0
-8.0
-10.0
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# (Ans)
44
Example: Rigid Frame
BM and shear diagrams easiest drawn by breaking frame
down into individual members and each analyzed by itself
20 kN/m
40 kN
2m
60 kN
2m
40 kN
C
B
3m
D
A
14 m
C
B
D
A
140 kN
171.4 kN
108.6 kN
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Example (Cont’d)
20 kN/m
108.6 440
40
40
80
440
440
40
C
171.4
B
108.6
171.4
40
40
40
108.6
108.6
B
5.34 m
108.6
C
171.4
440
80
80
B
B
60
A
140
140
171.4
171.4
C
C
D
D
A
171.4
108.6
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Example (Cont’d)
20 kN/m
440
108.6
40
40
80
440
B
108.6
40
735
C
171.4
171.4
40
40
40
440
108.6
B
5.34 m
108.6
440
80
171.4
171.4
440
B
60
B
280
A
C
140
C
C
D
D
A
171.4
108.6
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DEFLECTIONS
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Elastic Curve
A
D
(a)
y
A
C
B
v
P
M0
C
D
x
x B
Elastic Curve
(b)
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Elastic-Beam Theory
O’
dq
dq
y
y + dy

y
m1 ds
A
x
dx
m1
m2
B
curvature
Slope of deflection curve
q  dq
m2
ds
q
y+ dy
y
x
x
x
dx
dq
 
 ds
1
dy
-1 dy
 tan q  q  tan
dx
dx
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Elastic-Beam Theory (Cont’d)
Under service conditions, most beams undergo very small
deflections and angles of rotation  elastic curves have
extremely small curvatures:
dy
ds  dx and
 tan q  q
dx
dq dq
d  dy  d 2 y
  


  2
 ds dx dx  dx  dx
1
If material of beam is linearly elastic and follows Hooke’s
law, the curvature is
M

EI
d2y M

Differential equation:
2
dx
EI
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51
Singularity Functions
n

 x - a 
 x - a n  
 0
 x - a 0
a
when x  a
when x  a
 x - a 2
 x - a 1
x
a
(a ) n  0
Note: < > are called
Macaulay’s brackets
x
(b ) n  1
1
n 1

x
a

dx


x
a


n 1
d
 x - a  n  n  x - a  n-1
dx
n
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a
x
(c ) n  2
for n  0
for n  1
52
Direct Integration Method
P  1.2 kN
A
C
0.6m
q o  1 .5 k N /m
M o  1.44 kN  m
D
0.8m
1.2m
E
1.0m
Draw free-body diagram and compute support reactions:
P  1.2 kN
q o  1 .5 k N /m
A
C
0.6m
M o  1.44 kN  m
D
1.2m
0.8m
A y  2 .6 k N
E
1.0m
B
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53
Example (Cont’d)
y
P  1.2 kN
0.6 m
A
M o  1.44 kN  m
q o  1 .5 k N /m
X
C
D
1.8 m
2.6 m
A y  2 .6 k N
B
E
x
B
- q o  - 1 .5 k N /m
Rewrite distributed loading using singularity function:
q  x   qo  x - 0.6 0 - qo  x - 1.8 0
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54
Example (Cont’d)
P  1.2 kN
0.6 m
A
M o  1.44 kN  m
q o  1 .5 k N /m
C
A y  2 .6 k N
M
X
1.8 m
2.6 m
x
E
D
V
x
- q o  - 1 .5 k N /m
   Fy  0 :
Ay - P  x - 0.6 0 - qo  x - 0.6 1  qo  x -1.8 1 - V  0
V  2.6 -1.2  x - 0.6 0 -1.5  x - 0.6 1  1.5  x -1.8 1

M  0:
X
qo
q
 x - 0.6 2 - o  x -1.8 2  Mo  x - 2.6 0  M  0
2
2
 M  2.6 x -1.2  x - 0.6 1 -0.75  x - 0.6 2 0.75  x -1.8 2 -1.44  x - 2.6 0
- Ay x  P  x - 0.6 1 
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55
Example (Cont’d)
Differential equation:
EIy "  M  2.6 x - 1.2  x - 0.6 1 -0.75  x - 0.6  2
 0.75  x - 1.8  2 -1.44  x - 2.6  0
Slope:
EIy '  1.3 x 2 - 0.6  x - 0.6  2 -0.25  x - 0.6  3
 0.25  x - 1.8  3 -1.44  x - 2.6 1  C1
Deflection:
EIy  0.4333 x 3 - 0.2  x - 0.6  3 -0.0625  x - 0.6  4
 0.0625  x - 1.8  4 -0.72  x - 2.6  2  C1 x  C 2
CE 103 Structural Analysis
56
Example (Cont’d)
Apply boundary conditions:
y 0   0  C 2  0
y  x  3 .6   0 
0.4333  3.6  - 0.2  3.0 3 -0.0625  3.0 4
3
 0.0625  1.8  4 -0.72  1.0  2 C1  3.6   C2  0
 C1  -2.692#
Slope:
EIy '  1.3 x 2 - 0.6  x - 0.6  2 -0.25  x - 0.6  3
 0.25  x - 1.8  3 -1.44  x - 2.6 1 -2.692 #(ANS)
Deflection:
EIy  0.4333 x 3 - 0.2  x - 0.6  3 -0.0625  x - 0.6  4
 0.0625  x - 1.8  4 -0.72  x - 2.6  2 -2.692 x #(ANS)
CE 103 Structural Analysis
57
Moment-Area Method
 Alternative method for finding deflections and
angles of rotation of beams
 Method is based upon 2 theorems related to area
of bending-moment diagram, hence its name
 Same assumptions apply, that is
 linearly elasticity
 small slopes
 Method is suited, from practical standpoint, to
finding deflections and rotations at specific points
on beam
CE 103 Structural Analysis
58
Moment-Area Theorem 1
Mdx
dq 
EI
B
Mdx
A EI
 dq  
A
B
Mdx
q B / A  q B - q A  
A EI
 q B / A  Area of M / EI diagram between A & B
B
The change in slope qB/A between 2
points on deflection curve = area of
M/EI diagram between these points
CE 103 Structural Analysis
59
Sign Convention
 Angles qA and qB are +ve counterclockwise
 Angle qB/A is +ve when qB is algebraically
larger than qA
 Point B must be to the right of point A
 Bending moment M is +ve according to
usual sign convention
 Area of M/EI diagram sign is according to
the sign of BM
CE 103 Structural Analysis
60
Moment-Area Theorem 2
dt  x1dq  x1
Mdx
EI
Mdx
A dt  A x1dq  A x1 EI
B
Mdx
 t B / A   x1
A
EI
 First moment of area of M / EI diagram
B
B
B
between A & B, evaluated w.r.t. B
The tangential deviation tB/A of point B
from tangent at A = moment of area of
M/EI diagram between A & B about B
CE 103 Structural Analysis
61
Construction of Moment Diagrams by Parts
b
a
q
P
L
RL
RR
RL L
-P (L - a)
-½ q (L - b)2
CE 103 Structural Analysis
62
Properties of Typical Shapes
x
rectangle
y
x
y  kx n
h
b
x
h
b
bh
A
n 1
b
x
n2
triangle
h
b
x
h
parabolic
b
CE 103 Structural Analysis
A  bh
b
x
2
bh
A
2
b
x
3
bh
A
3
b
x
4
63
Example
1.5PL
1  PL  L
PL2
A1    2  EI  2
4 EI
2P
2P
P
1  PL  L
PL2
A2    2  3EI  2
12 EI
1  PL  L
PL2
A3    2  2 EI  2
8EI
-0.5PL
-1.5PL
A2
A3
PL
EI
A1
CE 103 Structural Analysis
PL
3EI
-
PL
2 EI
64
Example (Cont’d)
A
qA  0
(a) Slope at C:
qC
tC / A
C
1.5PL
Apply 1st moment-area theorem:
2P
11PL2
q C  q C / A  A1  A2  A3  24 EI
#Ans
(b) Deflection at C:
A2
A3
PL
EI
A1
-
Apply 2nd moment-area theorem:
PL
3EI
PL
2 EI
11PL3
5 
2 
1 
vC  tC / A  A1  L   A2  L   A3  L   36 EI
6 
3 
3 
CE 103 Structural Analysis
#Ans
65
Example
P
A
1  2 PL 
PL2
A1  
  L 
2  3EI 
3EI
2
1  2 PL 
PL
A2  
  L 
2  3EI 
3EI
1  PL 
PL2
A3  
  L 
2  3EI 
6 EI
1  PL 
PL2
A4  
  L 
2  6 EI 
12 EI
EI
B
EI
2P
3
2P
3
2EI
P
3
tB / A
L
L
A1
-
PL
3
PL
3EI
A2
tC / A
L
2 PL
3
2 PL
3EI
C
P
3
PL
6 EI
A3
A4
CE 103 Structural Analysis
66
Example (Cont’d)
P
A
EI
2EI
B
EI
2P
3
P
3
tB / A
L
2 PL
3EI
A1
L
tC / A
L
PL
3EI
A2
C
PL
6 EI
A3
A4
Apply 2nd moment-area theorem:
L
2L 
L



 2L 
tC / A  A1  2 L    A2  L 
  A3  L    A4 

3
3 
3



 3 
29 PL3
 tC / A 
18EI
CE 103 Structural Analysis
67
Example (Cont’d)
P
A
EI
2EI
B
EI
2P
3
C
P
3
tB / A
L
2 PL
3EI
A1
L
L
PL
3EI
A2
tC / A
PL
6 EI
A3
A4
Apply 2nd moment-area theorem:
L

 2L 
L
t B / A  A1  L    A2 
  A3  
3

 3 
3
13PL3
 tB / A 
18EI
CE 103 Structural Analysis
68
Example (Cont’d)
P
A
EI
2EI
B
EI
2P
3
P
3
tB / A
L
2 PL
3EI
A1
L
tC / A
L
PL
3EI
A2
C
PL
6 EI
A3
A4
By Similar Triangles:
2
- v B  t B / A  tC / A
3
13PL3 2 29 PL3
19 PL3
 vB 
18 EI 3 18 EI
54 EI
CE 103 Structural Analysis
#(ANS)
69
Energy Methods
• Earlier methods for calculating deflections
were based on differential equation of member
• Suitable for finding displacements, slopes in beams
subjected to rather simple loads
• For more complicated loadings or for trusses
and frames, preferable to use energy methods
• All energy methods are based on principle of
conservation of energy
Ue  Ui
work done by all external forces acting on a structure,
Ue, is transformed into internal work or strain energy, Ui
CE 103 Structural Analysis
70
External Work
Work is defined as product of force and displacement, in the
direction of force, through which the force acts
x
U e   P dx
0
Work of a moment is defined by the product of the magnitude
of the moment and the angle through which it rotates
q
U e   M dq
0
CE 103 Structural Analysis
71
Strain Energy – Axial Force
When an axial force is applied gradually to a bar, it will strain
the material such that external work done will be converted
into strain energy which is stored in the bar
Fdx Pdx
d 

EA
EA
F
x
L
dx
dx

d
P
P
1
1  Pdx 
dU i  F  d    P 

2
2  EA 
L
1  Pdx  P 2 L
 Ui   P 

2  EA  2 EA
0
CE 103 Structural Analysis
72
Strain Energy – Bending
When a member is subjected to lateral loading, the member
experiences internal moment and rotation resulting in strain
energy stored in the member
P
x
dx
L
M
dx

1
1  Mdx 
dU i  M dq  M 

2
2  EI 
M
dq
dq M
Mdx

 dq 
dx EI
EI
L
2
M
 Ui  
dx
2 EI
0
CE 103 Structural Analysis
73
Principle of Work and Energy
P
P
M V

L
x
x
1
U e   P dx  P
2
0

M
- Px 
P 2 L3
Ui  
dx  
dx 
2 EI
2 EI
6 EI
0
0
1
P 2 L3
U e  U i  P 
2
6 EI
PL3
 
3 EI
L
2
L
CE 103 Structural Analysis
2
74
Principle of Virtual Work
• Method of work & energy limited in application
• Since it is applicable only to computing deflection at a point
of concentrated force
• If more than one force is applied simultaneously, then more
than one unknown deformation will appear in one energy
equation – solution becomes impossible
• One way to circumvent above limitations is to
use method of virtual work, also called unitload method
• Method is also based on conservation of energy, Ue = Ui
• Provides a general method of obtaining displacement, slope
at any point on a structure, be it a beam, truss or frame
CE 103 Structural Analysis
75
Method of Virtual Work
P1
2
1
S
A
L
P2
dL
S
C
D

u
A
B
Case (a)
L
dL1
u
B
C
D

1
Case (b)
External work done by applied loads must equal internal strain
energy of all elements in the structure:
1
1
1
P11  P2  2   S  dL
2
2
2
1
1   1  u  dL1
2
2
CE 103 Structural Analysis
76
Method of Virtual Work (Cont’d)
P1
2
1
S
C
L
A
P2
dL
S
D

u
L
A
B
Case (a)
dL1
u
B
C
D

1
Case (b)
1
1
1
P11  P2  2   S  dL
2
2
2
1
1
1    u  dL1
2
2
Now imagine case (b) exists first and then actual loads P1 and P2 are
applied gradually. Equating total work done to total strain energy,
we have
ଵ ଵ
ଵ
ଶ ଶ
CE 103 Structural Analysis
77
Method of Virtual Work (Cont’d)
1
1
1
1
1



P11  P2  2   S  dL (1a)
1    u  dL1 (1b)
2
2
2
2
2
1
1
1
1
1
1   P11  P2  2  1     u  dL1   S  dL   u  dL (2)
2
2
2
2
2
Subtracting (1a) and (1b) from (2), we get the basic equation of the
unit-load method:
real displacements
1     u  dL
virtual loadings
In a similar manner, we can obtain the equation to obtain slope:
External virtual unit
couple applied in
direction of q
real displacements
1  q   uq  dL
virtual loadings
CE 103 Structural Analysis
78
Method of Virtual Work: Trusses
External Loads
uFL
1   
EA
u  internal virtual member force caused by external virtual unit load
F  internal member force caused by real loads
Temperature
1     u  T L
u  internal virtual member force caused by external virtual unit load
  coefficient of thermal expansion of member
T  change in temperature of member
CE 103 Structural Analysis
79
Example
C
3m
E  200 GPa
4 kN
A
A  400 mm 2
B
4m
4m
Virtual Forces u:
Find vertical
displacement at C
Real Forces F:
1 kN
C
- 0.833 kN
A
C
- 0.833 kN
 2.5 kN
B
4 kN
 0.667 kN
0.5 kN
0.5 kN
CE 103 Structural Analysis
A
4 kN
- 2.5 kN
B
 2 kN
1.5 kN
1.5 kN
80
Example (Cont’d)
Virtual-Work Equation:
Member
u (kN)
F (kN)
L (m)
uFL (kN2m)
AB
0.667
2.0
8
10.67
AC
-0.833
2.5
5
-10.41
BC
-0.833
-2.5
5
10.41
 10.67
uFL
1   
EA
1
10.67
 C 
uFL 
 0.133 mm #(ANS)

EA
EA
CE 103 Structural Analysis
81
Method of Virtual Work: Beams/Frames
External Loads
L
mM
1   
dx
EI
0
m  internal virtual moment in beam/frame caused by
external virtual unit load
M  internal moment in beam/frame caused by real loads
L
mq M
1q  
dx
EI
0
mq  internal virtual moment in beam/frame caused by
external virtual unit couple
M  internal moment in beam/frame caused by real loads
CE 103 Structural Analysis
82
Example
E  200 GPa
12 kN/m
A
I  500  10 6 mm 4
B
10 m
Virtual Moment m:
Find vertical
displacement at B
Real Moment M:
1 kN
12 kN/m
x
x
1 kN
M  -6x
m  - 1x
2
12 kN/m
V
v
CE 103 Structural Analysis
83
Example (Cont’d)
Virtual Moment m:
Real Moment M:
1 kN
12 kN/m
x
x
1 kN
M  -6x
m  -1x
2
12 kN/m
V
v
Virtual-Work Equation:

mM
- 1x - 6 x 2 
1   
dx  
dx
EI
EI
0
0
L
10
   150 mm #(ANS)
CE 103 Structural Analysis
84
Example
C
3m
E  200 GPa
60
B
2m
A
o
5 kN
I  15  10 6 mm 4
Find slope at C
x2
Real Moment M:
Virtual Moment mq:
5 kN
1 kN  m
1 kN  m
5 kN
1 kN  m
m1
1 kN  m
x1
7.5 kN  m
m1  -1
m2
5 kN
5 kN
1 kN  m
x2
M1
7.5 kN  m
7.5 kN  m
m2  -1
CE 103 Structural Analysis
x1
M 1  -2.5 x1
x2
M 2  - 7.5
M2
85
Example (Cont’d)
Virtual Moment mq:
m1  -1
Real Moment M:
m2  -1
M 1  -2.5 x1
M 2  - 7.5
Virtual-Work Equation:
L
3
2
mq M
mM
mM
1q C  
dx   1 1 dx1   2 2 dx 2
EI
EI
EI
0
0
0
3

0
- 1- 2.5 x1  dx  - 1- 7.5 dx
2
EI
1

0
EI
2
q C  0.00875 rad #(ANS)
CE 103 Structural Analysis
86
INFLUENCE LINES
CE 103 Structural Analysis
87
Concept of Influence Line
• In many structures a number of different live
loading conditions are possible
• Important to determine that load position or
combination which will be critical
• An influence function denotes the effect at one
specific point as a function of the position of the
cause of unit value
• An influence line denotes the plot of the
influence function
CE 103 Structural Analysis
88
Influence Lines for Beams
1
C
A
B
6m
10 m
IL for RA
1 .0
0 .4
0 .6
IL for RB
1
RA
RA
RA  1 -
1 .0
RB 
M C VC
MC
VC
CE 103 Structural Analysis
x
L
x
L
M C  4 RB 
 0 x6
VC  - RB 
1
RB
M C  6 RA 
 6  x  10
VC  R A 
RB
89
Influence Lines for Beams (Cont’d)
IL for RA
IL for RB
1 .0
0 .4
0 .6
RA  1 -
x
L
1 .0
RB 
x
L
2.4
4 RB , 0  x  6
MC  
6 R A , 6  x  10
IL for MC
IL for VC
0.4
- R , 0  x  6
VC   B
 R A , 6  x  10
0 .6
CE 103 Structural Analysis
90
Influence Lines for Trusses
B
1
C
D
E
F
3m
A
H
J
K
L
M
G
6 @ 4 m  24 m
CE 103 Structural Analysis
91
Influence Lines for Trusses (Cont’d)
1
B
C
D
E
F
A
G
H
J
K
L
M
E
F
3m
6 @ 4 m  24 m
1
B
C
D
A
G
H
J
K
L
RA
1 .0
M
RG
0.833
0.667
IL for RA
RA  1 -
x
L
1 .0
IL for RG
0.167
0.333
CE 103 Structural Analysis
RG 
x
L
92
Influence Lines for Trusses (Cont’d)
1
B
A
FBC
FBJ
FBC
C
D
E
F
FBJ
H
G
J
K
L
M
RA
M  0:
FBC 3  RG 16   0  FBC  -5.33 RG
 Fy  0 :
3
FBJ  RG  0  FBJ  -1.67 RG
5
J
RG
RHS
RHS
B
A
FBC
FBJ
H
FBC
C
D
E
F
FBJ
1
G
J
K
L
M
RA
M  0:
- FBC 3 - R A 8   0  FBC  -2.67 R A
F  0:
3
- FBJ  R A  0  FBJ  1.67 R A
5
J
RG
LHS
y
LHS
CE 103 Structural Analysis
93
Influence Lines for Trusses (Cont’d)
B
A
FBC
FBC 1
FBJ
FBJ
H
RA
x
C
D
E
F
G
J
K
L
M
x
4
RG
M  0:
FBC 3  RG 16   0  FBC  -5.33 RG
F  0:
3
x
FBJ  RG -  0  FBJ  -1.67 RG  0.42 x
5
4
J
RHS
y
RHS
 - 5.33 RG , 0  x  8
FBC  
- 2.67 R A , 8  x  24
- 1.78
IL for FBC
1.11
- 0.278
IL for FBJ
CE 103 Structural Analysis
- 1.67 RG ,
0 x4


FBJ  - 1.67 RG  0.42 x, 4  x  8

1.67 R A ,
8  x  24

94
Uses of Influence Lines: Concentrated Loads
1
C
A
6m
B
10 m
2.4
1 .6
1 .2
IL for MC
10 kN
A
6 kN
4 R , 0  x  6
MC   B
6 R A , 6  x  10
8 kN
B
C
4m
2m
2m
2m
M C  10 1.6   6  2.4   8 1.2   40 kNm #
CE 103 Structural Analysis
95
Uses of Influence Lines: UDL
a
x
dx
b
q kN/m
A
B
4m
1.33
1 .0
12 m
4m
y
- 0.33
IL for RA
b
b
a
a
R A   qy dx  q  y dx
Value of RA due to a UDL = intensity of load, q, multiplied by the
area of the influence line corresponding to length of beam over
which load is acting
CE 103 Structural Analysis
96
Example
D
A
8m
C
B
6m
4m
Find the maximum value of MD due to a UDL of intensity 5 kN/m,
which can act over any part of the beam, and two concentrated
loads of 10 kN each, with a fixed distance of 4 m between them.
IL for RA
1 .0
0.33
- 0 .5
0.67
1 .5
1 .0
IL for RB
IL for MD
1.33
2.67
CE 103 Structural Analysis
- 1.33
- 4 .0
97
Example (Cont’d)
2.67
1.33
IL for MD
- 1.33
- 4 .0
Since loads acting in a region where influence line is +ve have an effect that is
opposite from that produced by loads acting where the influence line is –ve,
we must place all loads either to left or to right of B:
4m
10 kN
10 kN
4m
5 kN/m
A
C
B
D
8m
10 kN
6m
4m
10 kN
4m
5 kN/m
A
B
C
D
8m
CE 103 Structural Analysis
4m
6m
98
Example (Cont’d)
2.67
1.33
IL for MD
4m
10 kN
- 1.33
10 kN
4m
5 kN/m
A
M D max  101.33  102.67  
C
B
1

5  12  2.67   120 kNm
2

D
8m
6m
4m
10 kN
10 kN
4m
5 kN/m
A
B
4m
M D max  -104  - 101.33 C
D
8m
- 4 .0
6m
CE 103 Structural Analysis
1

5  4  6   -113 .3 kNm
2

99
Muller-Breslau’s Principle
To obtain an IL for reaction of any statically determinate beam,
remove the support and make a unit displacement of its point
of application – deflected beam is the IL for reaction
1
C
A
B
b
a
L
B'
B
A
s B  1
IL for RB
CE 103 Structural Analysis
100
Muller-Breslau’s Principle (Cont’d)
To obtain an IL for the shear at a section of any statically
determinate beam, cut the section and induce a unit relative
transverse sliding displacement between the portion to left
and right of the section keeping all other constraints intact –
deflected beam is the IL for the shear
1
C
A
a
b
L
 sC  1
A
B
b
C'
L
C a
B
C" L
IL for VC
CE 103 Structural Analysis
101
Muller-Breslau’s Principle (Cont’d)
To obtain an IL for the moment at a section of any statically
determinate beam, cut the section and induce a unit rotation
between the portion to left and right of the section keeping all
other constraints intact – deflected beam is the IL for the
moment
1
C
A
B
b
a
L
aq C  a
q C  1
A
C'
C
B
bq C  b
IL for MC
CE 103 Structural Analysis
102
Example
1
D
A
C
a
s A  1
a
E
a
B
a
A'
C
A
s D  1
A
1
2
IL for RA
B
IL for VD
B
IL for MD
D'
C
D
D"
D'
A
B
a
2
1
2
q D  1
C
D
CE 103 Structural Analysis
103
Example (Cont’d)
1
D
A
a
A
C
a
E
a
C
B
a
E
B
s E  1
C'
IL for VE
E'
q E  1
A
C
E
B
IL for ME
a
C'
CE 103 Structural Analysis
104
ANALYSIS OF STATICALLY
INDETERMINATE STRUCTURES
CE 103 Structural Analysis
105
Statically Indeterminate Structures
 Statically determinate – all the unknown reactions
at the supports can be obtained by considering
equilibrium
 Statically indeterminate – the number of
equilibrium equations are insufficient to evaluate
the unknown reactions
 Degree of indeterminacy/redundancy – the
difference between the number of unknown
reactions and number of equilibrium equations
CE 103 Structural Analysis
106
FORCE METHODS
CE 103 Structural Analysis
107
Method of Consistent Deformation
P
P
A
B
L
2
MA
L
2
=
A
B
L
2
L
2
 BP
RB
RA
Determinate structure
subject to actual loads
Actual structure
+
 B   BP -  BR  0
A
5 PL3 RB L3

0
48 EI 3 EI
 RB 
B
L
2
5
P
16
L
2
 BR
RB
Determinate structure
subject to redundant
CE 103 Structural Analysis
108
Method of Least Work
Castigliano’s second theorem, sometimes called method
of least work, is a useful alternative force method, where
forces called redundants are treated as unknowns
Main difference between methods of least work and
consistent deformation is that the former requires strain
energy of system to be determined to formulate
equations instead of deflections in the latter
Method of least work is preferable when it is easier to set
up expressions for strain energy than to calculate
displacements
CE 103 Structural Analysis
109
Castigliano’s Theorems
F1
F2
1
A
2
1
2
1   11 F1   12 F2
where
B
 2   21 F1   22 F2
 ij  deflection at i due to a unit load at j
W 
U  W 
1
F11  F2  2 
2
1
 11 F12   12 F1 F2   21 F1 F2   22 F22 

2
Maxwell’s reciprocal theorem:
U 
 12   21
1

 11 F12  2 12 F1 F2   22 F22 
2
CE 103 Structural Analysis
110
Castigliano’s Theorems (Cont’d)
F1
A
F2
1
2
B
2
1
1   11 F1   12 F2
 2   21 F1   22 F2
1
U   11 F12  2 12 F1 F2   22 F22 
2
U 1 

 11 F12  2 12 F1 F2   22 F22    11 F1   12 F2

F1 2 F1

U
 1
F1
Castigliano’s first theorem: The partial derivative of strain energy
with respect to one of the external forces acting on the structure is
equal to the displacement at that force in the direction of the force
CE 103 Structural Analysis
111
Castigliano’s Theorems (Cont’d)
F1
RA
F2
RB
RC
U
 A  0
R A
Castigliano’s second theorem: Upon applying castigliano’s first
theorem to find deflection at support A, which is zero, we conclude
that redundants must assume values that will make the strain
energy in the structure a minimum
Redundants in indeterminate structures can be determined by
applying Castigliano’s second theorem – also called method of
least work
CE 103 Structural Analysis
112
Example
P
A
L
2
MA
B
C
L
2
RB
RA
U
0
R B
L
M2
U 
dx
2 EI
0
L
 L M M RB 
U
  M2



dx   
dx  0

RB RB  0 2 EI  0
EI
Element
x0
M
 M R B
BC
B
RB x
x
CA
B
RB x - P  x - L 2 
CE 103 Structural Analysis
x
113
Example (Cont’d)
P
A
L
2
MA
B
C
L
2
RB
RA
Element
x0
M
 M R B
BC
B
RB x
x
CA
B
RB x - P  x - L 2 
x
U
M M RB 

dx  0
R B 0
EI
L
L 2
L
M  M RB 
1 
PLx  

2
2
2

dx 
dx   0
  RB x dx    RB x - Px 


EI
EI  0
2  
0
L 2
L
 RB 
5
P #(ANS)
16
CE 103 Structural Analysis
114
Example
40 kN
B
A
MD
5m
5m
5m
HD
D
C
VD
R2
R1
M M Ri 
U

dx  0
Ri
EI
i  1,2
Element
x0
M
M R1
AB
A
- R2 x
0
 M R2
-x
BC
B
x
-5
CD
B
R1 x - 5R2
R1 x - 5 R2 - 40 x  200
x
-5
CE 103 Structural Analysis
115
Example (Cont’d)
Element
x0
M
M R1
AB
A
- R2 x
0
 M  R2
-x
BC
B
x
-5
CD
B
R1 x - 5R2
R1 x - 5 R2 - 40 x  200
x
-5
M M Ri 
U

dx  0
Ri
EI
5
i  1,2
10
 R x - 5 R x dx   R x - 5 R x - 40 x  200 x dx  0
2
2
1
2
1
0
5
2
2
(1)
5
5
10
 R x dx   - 5 R x  25 R dx   - 5 R x  25 R  200 x - 1000 dx  0 (2)
2
2
0
1
0
2
1
2
5
333 R1 - 250 R2 - 4167  0 
  R1  17 kN,
- 250 R1  292 R2  2500  0 
CE 103 Structural Analysis
R2  6 kN #(ANS)
116
DISPLACEMENT METHODS
CE 103 Structural Analysis
117
Slope-Deflection Method
Method uses displacements as unknowns and is referred to as
a displacement method
When a rigid frame or continuous beam is loaded, moments are
developed at the ends of individual members
The relationship that exists between these moments and
deformations at the ends of the members is the basis of the
slope-deflection method
In methods of consistent deformation and least work,
compatibility equations are written in terms of unknown
redundants; in slope-deflection method, equilibrium equations
are written in terms of unknown joint displacements
CE 103 Structural Analysis
118
Slope-Deflection Equation
M AB
A

P
qA

B
qB
M BA
L
M AB 
2 EI 
3 
2 EI
F
F
2
q

q

M

2
q

q
3


M


B
AB
A
B
AB
 A

L 
L 
L
M BA 
2 EI 
3 
2 EI
F
F
2
q

q

M

2
q

q
3


M


A
BA
A
B
AB
 B

L 
L 
L
1. Moments acting on the ends of members are positive when clockwise
2. Rotations of the ends of members are also positive when clockwise
3. Relative displacement of the ends of a member  is positive when the
axis rotation  is clockwise
CE 103 Structural Analysis
119
Fixed-End Moments
F
M AB
-
P
PL
8
A
F
M AB
-
Pab
L2
F
M BA

L
2
L
2
P
2
PL
8
B
M
F
BA
qL2

12
F
BA
qL2

20
B
L
Pba
L2
2
F
M AB
-
B
a
q
A
F
M BA

A
qL2
F
M AB  12
2
qL
30
q
A
b
M
B
L
L
CE 103 Structural Analysis
120
Example
40 kN
qB
B
A
EI
8m
EI
8m
C
8m
Determine the reactions and draw the moment diagram
Joint rotations and translations are the unknowns. There are no
joint translations here. Only joint B can rotate. So, there is only one
unknown here.
Fixed-End Moments:
F
M AB
-
PL
 -80
8
40
F
M BA

PL
 80
8
B
A
8m
8m
F
M BC
0
F
M CB
0
C
B
8m
CE 103 Structural Analysis
121
Example (Cont’d)
16 m
M AB
B
A
EI
8m
M BC
B
B
EI
C
M BA
M CB
Slope-Deflection Equations:
2 EI
q B  - 80  0.125 EIq B - 80
16
2 EI
2q B   80  0.25 EIq B  80
M BA 
16
2 EI
2q B   0.5 EIq B
M BC 
8
2 EI
M CB 
q B   0.25 EIq B
8
M AB 
Equations of Equilibrium:
M  0:
B
M BA  M BC  0  0.25 EIq B  80   0.5 EIq B   0
 EIq B  -106 .7
CE 103 Structural Analysis
122
Example (Cont’d)
Member End Moments:
EIq B  -106 .7
M AB  0.125 EIq B - 80  -93 .3 kN - m
M BA  0.25 EIq B  80  53 .3 kN - m
M BC  0.5 EIq B  -53 .3 kN - m
M CB  0.25 EIq B  -26 .7 kN - m
Calculate Shear Forces and Reactions:
40
17 .5 17 .5 10
B
B
22 .5
A
93 .3
8m
8m
53 .3
10
B
53 .3
10
8m
C
26 .7
RB  27.5
CE 103 Structural Analysis
123
Example (Cont’d)
40
17 .5 17 .5 10
B
B
22 .5
A
93 .3
8m
53 .3
8m
10
10
B
C
8m
53 .3
26 .7
RB  27.5
Reactions and Moment Diagrams:
40
26 .7
Support Reactions
93 .3
22 .5
27.5
86 .7
10
26 .7
Moment Diagram
- 93 .3
- 53 .3
CE 103 Structural Analysis
124
Example
50 kN
6m
B
4m
C
2 EI
EI
EI
8m
D
A
Determine displacements and member end moments
Unknowns:
q B ,qC , 

B

qB
qC
C
D
A
CE 103 Structural Analysis
125
Example (Cont’d)
50 kN
6m
Fixed-End Moments:
50 6 4 
 -48
2
10
50 6 2 4 
F
M CB 
 72
2
10
F
M BC
-
2
B
4m
C
2 EI
EI
A
8m
EI
D
Slope-Deflection Equations:
M AB 
2 EI 
3 
q
 B
  0.25 EIq B - 0.0938 EI 
8 
8 
M BA 
2 EI 
3 
 2q B   0.5 EIq B - 0.0938 EI
8 
8 
M BC  0.8 EIq B  0.4 EIq C - 48
M CB  0.4 EIq B  0.8 EIq C  72
M CD  0.5 EI q C - 0.0938 EI
M DC  0.25 EIq C - 0.0938 EI 
CE 103 Structural Analysis
126
Example (Cont’d)
M AB  0.25 EIq B - 0.0938 EI
M BA  0.5 EI q B - 0.0938 EI 
M BC  0.8 EIq B  0.4 EIq C - 48
M CB  0.4 EIq B  0.8 EIq C  72
M CD
M BA
C
B
M CD  0.5 EIq C - 0.0938 EI 
M DC  0.25 EIq C - 0.0938 EI 
Equations of Equilibrium:
A
HA
M AB
D
HD
M DC
VA
VD
M  0:
B
M BA  M BC  0  1.3 EI q B  0.4 EI q C - 0.0938 EI  - 48  0
(1)
M  0:
M CB  M CD  0  0.4 EI q B  1.3 EI q C - 0.0938 EI   72  0
(2)
F  0:
 M  M BA   M DC  M CD 
H A  H D  0   AB

0
8
8

 

C
x
 0.75 EIq B  0.75 EIq C - 0.375 EI  0
CE 103 Structural Analysis
(3)
127
Example (Cont’d)
Solve for Unknown Displacements:
1.3 EIq B  0.4 EIq C - 0.0938 EI - 48  0
(1)
0.4 EIq B  1.3 EI q C - 0.0938 EI  72  0
(2)
0.75 EI q B  0.75 EIq C - 0.375 EI   0
(3)
 EI q B  57 .7, EI q C  -75 .8, EI   -36 .2
Member End Moments:
M AB  0.25 EIq B - 0.0938 EI   17 .8 kN - m
M BA  0.5 EI q B - 0.0938 EI  32 .2 kN - m
M BC  0.8 EIq B  0.4 EIq C - 48  -32 .2 kN - m
M CB  0.4 EIq B  0.8 EIq C  72  34 .5 kN - m
M CD  0.5 EIq C - 0.0938 EI  -34 .5 kN - m
M DC  0.25 EI q C - 0.0938 EI  -15 .6 kN - m
CE 103 Structural Analysis
128
Moment-Distribution Method
Just like slope-deflection method, moment-distribution method is also
a displacement method
Unlike slope-deflection method, does not need to solve simultaneous
equations
Instead, answers are obtained by a procedure of successive
approximations, an iterative technique
Moment-distribution method is based on the knowledge that member
end-moments consist of several distinct parts:
1. Fixed-end moments – moments caused by applied loads with member ends
assumed to be fixed
2. Moments due to rotations that actually take place at ends of member
3. Moments caused by translation of one end of member relative to the other
CE 103 Structural Analysis
129
Distribution Factor
Sign Convention
Follows same sign convention as those adopted in slope-deflection
method, viz., clockwise end-moments are positive
Distribution Factor
Distribution factor for any one member is a measure of the proportion
of the total resistance to rotation supplied by that member
M DA
D
q
M
A
M AD
M AD
q
M AB
A
D
A
M
B
M AC
From slope-deflection equation,
2 EI 
3 
4 EI ADq
F
M AD 
 2q A  q D   M AD 
L 
L 
L AD
4 EI ACq
M AC 
L AC
C
4 EI ABq
M AB 
L AB
CE 103 Structural Analysis
130
Distribution Factor (Cont’d)
M AB  4 Ek ABq
I
k AB  AB
L AB
M AC  4 Ek ACq
k AC 
I AC
L AC
M AD  4 Ek ADq
I
k AD  AD
L AD
where the quantity I / L of a member is called the stiffness factor, k
M  0:
A
M  M AB  M AC  M AD  4 Eq  k A
M
M AD
A
where  k A denotes the sum of the stiffnesses of members framing into joint A
M AB
M AC
k AB
M AB  4 Ek ABq 
M
 kA
k
M AC  4 Ek ACq  AC M
 kA
k
M AD  4 Ek ADq  AD M
 kA
where
k
k
is referred to as the distribution factor
CE 103 Structural Analysis
131
Carry-Over Factor/Moment
q
M DA
A
D
M AD
From slope-deflection equation,
M DA 
2 EI 
3 
2 EI ADq 1
F
2
q

q

M

 M AD
 D

A
DA
L 
L 
L AD
2
The moment induced at the far (fixed) end of a prismatic member
equals one-half the distributed moment at the near end
The carry-over factor, defined as the ratio of induced moment at
far end (fixed) to the moment applied at the near end, which is
prevented from translation but allowed to rotate, is thus equal to ½
The induced moment is called carry-over moment
CE 103 Structural Analysis
132
Example
30 kN
B
A
C
EI
8m
k
4m
I
8
k
0
Balancing of Joint B
0 - 53.3
 22 .9  30 .4
Carry-Over Moments  11 .4
8m
10 m
2I
12
k
Balancing of Joint B
I
10
5 3
8 8
 26 .7 - 33.3  33.3
 15.2
- 5 .4 - 3 . 2
Balancing of Joint C
Carry-Over Moments
D
EI
2 EI
3 4
7 7
Distribution Factors
Fixed-End Moments
4 kN/m
- 2 .7
 1 .2  1 . 5
Carry-Over Moments  0.6
- 1 .6
 0.8
- 0.5 - 0.3
Balancing of Joint C
- 0.2
 0.1  0.1
Final End Moments  12.0  24 .2 - 24.2
- 0 .1
Carry-Over Moments
Balancing of Joint B
 36.8 - 36 .8  31 .6
CE 103 Structural Analysis
133
Example (Cont’d)
30 kN
4 kN/m
B
A
C
EI
8m
4m
8m
10 m
30 kN
12
A
4 .5
19
4 .5
B
24 .2
24 .2
20 .5
11
B
D
EI
2 EI
C
36.8
36.8
C
4 kN/m
19 .5
D
31.6
51.6
15.8
12
- 24.2
- 36 .8
- 31.6
Moment Diagram
CE 103 Structural Analysis
134
Simply Supported Ends
Effective Stiffness
I


L

k' 

I
0
.
75

L

if member is fixed at its far end
if member is hinged at its far end
Carry-Over Factor (COF)
0.5

COF  
0 .0

if member is fixed at its far end
if member is hinged at its far end
CE 103 Structural Analysis
135
Structures With Joint Translations
P2
B
C
=
P1
D
A
(a) Actual Frame – M Moments
P2
B
Q
C
Q
B


C
+
P1
A
D
A
(b) Frame with Sidesway Prevented – MO Moments
D
(c) Frame Subjected to Q – MQ Moments
M  MO  MQ
CE 103 Structural Analysis
136
Structures With Joint Translations
Q
B


C
M  MO  MQ
D
A
(c) Frame Subjected to Q – MQ Moments
P
B
'
'
C
A
Q
MQ    M P
P
Q
M  MO    MP
P
D
(d) Frame Subjected to Arbitrary ’ Translation
– MP Moments
CE 103 Structural Analysis
137
MATRIX-STIFFNESS METHOD
CE 103 Structural Analysis
138
Element-Stiffness Matrix
Axially Loaded Element
q1 ,  1
1
EA
2
q2 ,  2
L
q  k  
where
q 
q2 
q   1 
1 - 1
L  - 1 1 
k   EA 
 
    1 
 2 
EA
EA
 q1 
1 -  2  ; q2   2 -  1 
L
L
CE 103 Structural Analysis
139
Element-Stiffness Matrix
Flexural Element
q2
q1
q3
2
4
1
1
q4
2
EI
3
L
q  k  
where
 q1 
q 
q   2 
 q3 
q 4 
 12
 L3
 6
- 2
L
k   EI  12
 L3
 6
- 2
 L
6
L2
4
L
6
L2
2
L
-
CE 103 Structural Analysis
12
L3
6
L2
12
L3
6
L2
-
6
L2 
2 

L 
6 
L2 
4 

L 
-
 1 
 
    2 
 3 
 4 
140
Direct-Stiffness Method
In the direct-stiffness method, the transformation from element to
structure coordinates is carried out using deformation compatibility
relations between element and structure deformations
 e  T e e
where T e is the deformation-transformation matrix that relates the
element deformations of element e,  e to the structure deformations
at the extremities of that element, e
The element and structure end forces are also related in the same
manner
qe  T e Qe
where qe contains the element forces for element e and Qe contains
the structure forces at the extremities of that element
CE 103 Structural Analysis
141
Direct-Stiffness Method (Cont’d)
 e  T e e
qe  T e Qe
qe   k e  e  T e Qe   k e T e e
 Qe  T e  k e T e e
-1
Since T e represents an orthogonal transformation, T e  T e
-1
T
 Qe  T e  k e T e e   K e e
T
K e  T Te k e T e
where K e represents the element stiffness matrix in global coordinates
CE 103 Structural Analysis
142
Analysis of Trusses
l
2
k
y
j
x

 
i
Member
direction
1
1. Member is assigned a direction, and 1 and 2, the member
deformations at tail and front end of member are assumed to be
in this direction
2. The x and y structure deformations, at the tail and front end of
member, are given the designations i, j and k, l, respectively
3. The angles between the vectors representing the member
deformations and those representing the x and y structure
deformations are designated as  and 
  cos  ,
  cos 
where  and  are called the direction cosines of the member
CE 103 Structural Analysis
143
Analysis of Trusses (Cont’d)
l
j
2
k
y
j
x
1

 
i
Member
direction
1

1 - 1
L  - 1 1 
k e  EA 
 e  T e e
i
l
Member
direction
 i 
 
 1     0 0    j 
  0 0
 
   T e  




0
0


0
0


 k


 2 
  l 
i
 2


EA
T

K e  T e k e T e   2
L -

 - 
j

2
- 
- 2
CE 103 Structural Analysis
k
- 2
- 
2


k
2
l
-   i

- 2 j
  k
2 l
 
144
Example
4
306.90
216.90
1.6 m
3
53.10
EA
1
126.9
216.90
0
1
6
EA
0.9 m
2
2
2 kN
36.90
5
1.2 m
Member
L




i
j
k
l
1
2.0 m
306.9o
216.9o
0.6
-0.8
3
4
1
2
2
1.5 m
216.9o
126.9o
-0.8
-0.6
1
2
5
6
CE 103 Structural Analysis
145
Example (Cont’d)4
Member
L




i
j
k
l
1
2.0 m
306.9o
216.9o
0.6
-0.8
3
4
1
2
2
1.5 m
216.9o
126.9o
-0.8
-0.6
1
2
5
6
Element 1 Stiffness Matrix in Global Coordinates
 2

EA  
K e   2
L -

 - 
3
 0.09

EA  - 0.12
K 1 
0.5  - 0.09

 0.12

2
- 
- 2
- 2
- 
4
- 0.12
0.16
0.12
- 0.16
CE 103 Structural Analysis
-  

- 2
 
2
 
2

1
- 0.09
0.12
0.09
- 0.12
2
0.12  3
- 0.16  4
- 0.12  1

0.16  2
146
Example (Cont’d)4
Member
L




i
j
k
l
1
2.0 m
306.9o
216.9o
0.6
-0.8
3
4
1
2
2
1.5 m
216.9o
126.9o
-0.8
-0.6
1
2
5
6
Element 2 Stiffness Matrix in Global Coordinates
 2

EA  
K e   2
L -

 - 
1

2
- 
- 2
- 2
- 
2
0.16
 0.21

0.12
EA  0.16
K 2 
0.5  - 0.21 - 0.16

 - 0.16 - 0.12
CE 103 Structural Analysis
-  

- 2
 
2
 
2

5
6
- 0.21 - 0.16  1
- 0.16 - 0.12  2
0.21
0.16  5

0.16
0.12  6
147
Example (Cont’d)4
Element Stiffness Matrices in Global Coordinates
3
1
4
2
 0.09 - 0.12 - 0.09 0.12 3


EA - 0.12 0.16 0.12 - 0.16 4
K 1 
0.5 - 0.09 0.12 0.09 - 0.12 1


0
.
12
0
.
16
0
.
12
0
.
16

2
1
5
2
6
 0.21 0.16 - 0.21 - 0.16 1


EA  0.16 0.12 - 0.16 - 0.12 2
K 2 
0.21 0.16 5
0.5  - 0.21 - 0.16

6
0
.
16
0
.
12
0
.
16
0
.
12


Assembling Structure-Stiffness Matrix
3
4
1
2
5
6
- 0.09
0.12
 0.09 - 0.12
 3
- 0.12 0.16
 4
0
.
12
0
.
16


EA - 0.09 0.12 0.09  0.21 - 0.12  0.16 - 0.21 - 0.16 1
K   K 1  K 2  

0.5  0.12 - 0.16 - 0.12  0.16 0.16  0.12 - 0.16 - 0.12 2

- 0.21
- 0.16 0.21 0.16 5


- 0.16
- 0.12 0.16 0.12 6

CE 103 Structural Analysis
148
Example (Cont’d)4
Reordering Structure-Stiffness Matrix & Applying Boundary Conditions
1
2
3
4
5
6
Q1 
 0.30 0.04 -0.09 0.12 -0.21 -0.16  1 
Q 
 0.04 0.28 0.12 -0.16 -0.16 -0.12  
2
 

 2
Q3  EA -0.09 0.12 0.09 -0.12
0
0  0 

 

 
Q
0.12
0.16
0.12
0.16
0
0
0.5
 4

 0 
Q5 
 -0.21 -0.16
0
0 0.21 0.16  0 
 

 
Q
0
0 0.16 0.12  0 
 6 
-0.16 -0.12
Solve for Primary Unknown Structure Deformations
Q1  EA 0.30 0.04  1 
 

 
Q2  0.5 0.04 0.28 2 
-1
1  0.5 0.30 0.04 Q1  0.5
1  0.28 -0.04  0 1  0.485



 
 
 







Q
0.04
0.28
0.04
0.30
2
3.641
EA
EA
0.0824
EA

  2

 


 2
CE 103 Structural Analysis
149
Example (Cont’d)4
1
2
3
4
5
6
Q1 
 0.30 0.04 -0.09 0.12 -0.21 -0.16  1 
Q 
 0.04 0.28 0.12 -0.16 -0.16 -0.12  
 2

 2
Q3  EA -0.09 0.12 0.09 -0.12
0
0  0 
 

 
Q
0.12
0.16
0.12
0.16
0
0
0.5
 4

 0 
Q5 
 -0.21 -0.16
0
0 0.21 0.16  0 
 

 
0.16
0.12
0
0
0.16
0.12
Q6 

 0 
1  1  0.485
 



3
.
641
EA


 2
Solve for Support Reactions
Q3 
-0.09 0.12
-0.96
Q 




 4  EA  0.12 -0.16 1   1.28
 
 



Q

0.21
0.16
0.96
0.5
 2 
 5



Q6 
-0.16 -0.12
 0.72
CE 103 Structural Analysis
150
Example (Cont’d)4
Solve for Member Forces
  0 0 
T e  

0
0




EA  1 - 1
k e  
L  - 1 1 
 i 
 -  -     j 
 q1  EA  
 qe   k e  e   k e T e e    
 - - 
  
q


L

 k
 2
  l 
Since q1 is always numerically equal to q2 , it is only necessary
to solve for one of these forces. Since q2 was assumed to be a
tension force, a  ve answer means that the bar is in tension
q2 
EA
-
L
-

CE 103 Structural Analysis
 i 
 
 
  j 
 k 
  l 
151
Example (Cont’d)4
Member
L




i
j
k
l
1
2.0 m
306.9o
216.9o
0.6
-0.8
3
4
1
2
2
1.5 m
216.9o
126.9o
-0.8
-0.6
1
2
5
6
q2 
EA
 -
L
-

 i 
 
 
  j 
 k 
  l 
1  1  0.485
 



3
.
641
EA


 2
0
0
EA
q12 
 -0.6 0.8 0.6 -0.8     1.60 kN
2
 1
  2 
 1 
 
EA
2
q2 
0.8 0.6 -0.8 -0.6   02   -1.20 kN
1.5
 
 0 
CE 103 Structural Analysis
152
Analysis of Flexural Structures
Structures With Joints That Can Rotate But Not Translate
i
 2 , q2
1 , q1
EI
j
L
Element Stiffness Matrix in Local Coordinates
 q1  2 EI  2 1   1 
 
 1 2   
q
L

 2
 2
 1   1 0    i 
1 0 
 
   T e  


0 1 
 2   0 1    j 
Element Stiffness Matrix in Global Coordinates
i
j
2 EI  2 1  i
K e  T  k e T e 
L 1 2  j
T
e
CE 103 Structural Analysis
153
Example
20 kN - m
A
C
B
EI
2
3
EI
4
5m
EI
1
D
10 m
5m
Global Coordinates
1
1
Member
L
i
j
1
5m
3
2
2
10 m
2
4
3
5m
1
2
1
2
2
2
2
3
1
Local Coordinates
CE 103 Structural Analysis
154
Example (Cont’d)
Element Stiffness Matrix in Global Coordinates
i
Member
L
i
j
1
5m
3
2
2
10 m
2
4
3
5m
1
2
j
2 EI  2 1  i
K e 
L 1 2  j
3
2
 0 .8 0 .4  3
K 1  EI 
2
0
.
4
0
.
8


2
4
 0 .4 0 .2  2
K 2  EI 
4
0
.
2
0
.
4


1
2
 0 .8 0 .4  1
K 3  EI 
2
0
.
4
0
.
8


Assembling Structure-Stiffness Matrix
 Q1 
 0.8
Q 
 0.4
 2
   EI 
 0
 Q3 

Q4 
 0
0   1 
2.0 0.4 0.2    2 
 

0.4 0.8 0   3 

0.2 0 0.4    4 
0.4
0
CE 103 Structural Analysis
155
Example (Cont’d)
Solve for Primary Unknown Structure Deformations
-1
1  1 0.8 0.4  0  1 - 5.56
  
  



0
.
4
2
.
0
20
11
.
11
EI
EI

  


 2
Solve for Member Forces
1 0 
T e   
0 1 
2 EI  2 1 
k e 
L 1 2 
 q1  2 EI  2 1    i 
 qe   k e  e   k e T e e    
 


L 1 2  j 
 q2 
q
1
1
q
1
2
2
1
q
q
2
2
3
1
q
q   4.44 8.89
3
2
T
CE 103 Structural Analysis
4.44
2.22 0 6.67
T
156
Equivalent Joint Loads
20 kN
20 kN
B
A
EI
EI
5m
5m
C
2 . 5 m 2 .5 m
1 , Q1
 2 , Q2
Global Coordinates
 3 , Q3
11 , q11
12 , q12
Local Coordinates
1
 21 , q12
1
 22 , q22
20 kN
20 kN
q11F  -25
2
q12F  25 q12 F  -12.5
2
q22 F  12.5
Fixed-End Member Forces
CE 103 Structural Analysis
157
Equivalent Joint Loads (Cont’d)
20 kN
20 kN
B
A
EI
EI
2.5 m 2.5 m
5m
5m
C
=
1F
1
q
20 kN
 -25
q
1F
2
20 kN
 25
q12 F  -12.5
q22 F  12.5
+
QE1  25
QE 2  -12.5
CE 103 Structural Analysis
158
Analysis of Flexural Structures
Structures With Joints That Can Translate
 6 , q6
 3 , q3
 1 , q1
 4 , q4
EI
 2 , q2
L
 5 , q5
Element Stiffness Matrix in Local Coordinates
EA
 EA

0
0
0
0
 L

L

12 EI
6 EI
12 EI
6 EI    
 q1  
0
- 2
0 - 3
- 2  1
3
q  
L
L
L
L   
2
  
6 EI
4 EI
6 EI
2 EI   2 
0 - 2
0
 q3  
2
  3 
L
L
L
L
    EA
  
EA
q
 4  0
0
0
0  4 
L
 q5   L
  5 
12 EI
6 EI
12 EI
6 EI   
  
0 - 3
0
 q6  
 
L
L2
L3
L2   6 

6 EI
2 EI
6 EI
4 EI 
0 - 2
0


L
L
L2
L 

CE 103 Structural Analysis
159
Analysis of Flexural Structures (Cont’d)
m
Transformation Matrix
4
k
y
j
3

 
i
1
x
 1   
   - 
 2 
 3   0
 
 4   0
 5   0
  
 6   0
6
Member
direction
l
n
5
2
 0
 0
0
0
0
0 1
0 0

0 0 -
0 0
0
0 0  i 
0 0    j 
0 0    k 
 
 0  l 
 0  m 
 
0 1   n 
  cos  ,
  cos 
where  and  are called the direction cosines of the member
CE 103 Structural Analysis
160
Analysis of Flexural Structures (Cont’d)
Element Stiffness Matrix in Global Coordinates
i
 C1
 C
 2
 C4
T
K e  T e k e T e  
 - C1
- C2

 C 4
where
j
C3
- C5
- C2
- C3
- C5
k
C6
- C4
C5
C7
l
m
n
 i
 j
symmetric

 k

C1
 l
 m
C 2 C3

- C 4 C 5 C 6  n
EA 2 12 EI 2
EA 2 12 EI 2
 EA 12 EI 
C1 
  3  , C2  
- 3  , C 3 
  3 
L
L
L 
L
L
 L
6 EI
6 EI
4 EI
2 EI
C 4  2  , C5  2  , C6 
, C7 
L
L
L
L
CE 103 Structural Analysis
161
m
4
k
j
3
Example
6

w per unit length
5
 
i
0. 6 L
AL2
 2000
I
2
0 .8 L
L
6
3
1
6
3
1
5
1
5
8
2
9
4
2
4
7
2
Global Coordinates
1
CE 103 Structural Analysis
Local Coordinates
162
MATRIX ANALYSIS OF ELASTIC
STABILITY
CE 103 Structural Analysis
163