Beam and Frame Analysis: Displacement Method, Part II by S. T. Mau
235
displacements and the corresponding nodal forces of the frame, without the support and
loading conditions. Since each node has three DOFs, the frame has a total of nine nodal
displacements and nine corresponding nodal forces as shown in the figure below.
The nine nodal displacements and the corresponding nodal forces.
It should be emphasized that the nine nodal displacements completely define the
deformation of each member and the entire frame. In the matrix displacement
formulation, we seek to find the matrix equation that links the nine nodal forces to the
nine nodal displacements in the following form:
K
G
∆
G
= F
G
(18)
where K
G
,

∆
G
and F
G
are the global unconstrained stiffness matrix, global nodal
displacement vector, and global nodal force vector, respectively. Eq. 18 in its expanded
form is shown below, which helps identify the nodal displacement and force vectors.
⎥
⎥
⎥

⎪
⎭
⎪
⎪
⎪
⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎧
3
3
3
2
2

⎪
⎪
⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎧
3
3
3
2
2
2
1
1

3
θ
2
1
2
3
x
y
F
x1
M
1
F
y1
F
x3
F
x2
F
y3
F
y2
M
3
M
2
M
ember 1-2
M
ember 2-3

a frame member i-j, we need only four independent variables,
∆
x
,
θ
i
,
θ
j
, and
φ
ij
as shown
below.
The four independent deformation configurations and the associated nodal forces.
Each of the four member displacement variables is related to the six nodal displacements
of a member via geometric relations. Instead of deriving these relations mathematically,
then use mathematical transformation to obtain the stiffness matrix, as was done in the
truss formulation, we can establish the stiffness matrix directly by relating the nodal
forces to a nodal displacement, one at a time. We shall deal with the axial displacements
first.
There are two nodal displacements, u
i
and u
j
, related to axial deformation,
∆
x
. We can
easily establish the nodal forces for a given unit nodal displacement, utilizing the nodal

ij
=1
−6EK
−6EK
−12EK/L
−12EK/L
θ
j
=1
∆
x
=1
θ
i
=1
φ
ij
=1
Beam and Frame Analysis: Displacement Method, Part II by S. T. Mau
237
Note we must express the nodal forces in the positive direction of the defined global
coordinates.
Nodal forces associated with a unit nodal displacement.
From the above figure, we can immediately establish the following stiffness relationship.
⎥
⎥
⎥
⎥
⎥
⎦

⎬
⎫
⎩
⎨
⎧
j
x
i
x
f
f
(19)
Following the same principle, we can establish the flexural relations one at a time as
shown in the figure below.
Nodal forces associated with a unit nodal displacement.
From the above figure we can establish the following flexural stiffness relationship.
−EA/L
E
A/L
E
A/L−EA/L
u
i
=1
u
j
=1
θ
i
=1

i
=0,
θ
i
=1, v
j
=0,
θ
j
=0 v
i
=0,
θ
i
=0, v
j
=0,
θ
j
=1
v
i
=1
,

θ
i
=0
,
v

i
=0
,
u
j
=1
Beam and Frame Analysis: Displacement Method, Part II by S. T. Mau
238
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡

EK
L
EK
2
L
EK
4
6
2
6
612612
2
6
4
6
612612
⎪
⎪
⎭
⎪
⎪
⎬
⎫
⎪
⎪
⎩
⎪
⎪
⎨
⎧

f
(20)
Eq. 20 is the member stiffness equation of a flexural member, while Eq. 19 is that of a
axial member. The stiffness equation for a frame member is obtained by the merge of the
two equations.
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢

EK
L
EK
L
EA
L
EA
EK
L
EK
EK
L
EK
L
EK
L
EK
L
EK
L
EK
L
EA
L
EA
4
6
2
6
612612

⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎧
j
j
j
i
i
i
v
u
v
u
θ
θ
=

⎪
⎪
⎨
⎧
j
yj
xj
i
yi
xi
M
f
f
M
f
f
(21)
Eq. 21 is the member stiffness equation in local coordinates and the six-by-six matrix at
the LHS is the member stiffness matrix in local coordinates. Eq. 21 can be expressed in
matrix symbols as
k
L
δ
L
= f
L
(21)
Member stiffness matrix in global coordinates. In the formulation of equilibrium
equations at each of the three nodes of the frame, we must use a common set of
coordinate system so that the forces and moments are expressed in the same system and

∆
xj
= (Cos
β
) u
j
– (Sin
β
) v
j
β
u
j
u
i
v
j
v
i

∆
xj

∆
yj

∆
xi

∆

yj
f

xi
f

yiF
yj
M
j
M
j
M
i
M
i

∆
xj

∆
yj
β
u
j

∆

by C and S, respectively.
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
i
yi
xi
θ
∆
∆
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
100
0

⎪
⎩
⎪
⎨
⎧
j
yj
xj
θ
∆
∆
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
100
0
0
CS
S-C
⎪
⎭
⎪

yi
xi
M
F
F
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
100
0
0
CS
S-C
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨

F
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
100
0
0
CS
S-C
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
j
yj
xj

⎩
⎪
⎨
⎧
i
i
i
v
u
θ
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
100
0
0
CS-
SC
⎪
⎭
⎪
⎬

⎪
⎨
⎧
j
j
j
v
u
θ
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
100
0
0
CS-
SC
⎪
⎭
⎪
⎬
⎫

, are the
collections of the displacement and force vectors of node i and node j:
δ
G
=
⎭
⎬
⎫
⎩
⎨
⎧
jG
iG
∆
∆
δ
L
=
⎭
⎬
⎫
⎩
⎨
⎧
jL
iL
∆
∆
f
G

⎫
⎩
⎨
⎧
jG
iG
f
f
=
⎥
⎦
⎤
⎢
⎣
⎡
τ
τ
0
0
⎭
⎬
⎫
⎩
⎨
⎧
jL
iL
f
f
=

⎫
⎩
⎨
⎧
jL
iL
∆
∆
= k
L
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
T
T
τ
τ
0
0
⎭
⎬
⎫
⎩
⎨
⎧

k
G
=
Γ
k
L

Γ
T

(26)
Eq. 26 is the transformation formula of the member stiffness matrix. The expanded form
of the member stiffness matrix in its explicit form in global coordinates, k
G
, appears as a
six-by-six matrix:
Beam and Frame Analysis: Displacement Method, Part II by S. T. Mau
242
k
G
=
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥

+−−−−−
−−+−−−−−
−−
−−−−−−+−
−−−−−+
EK
L
EK
C
L
EK
S-EK
L
EK
C
L
EK
S
L
EK
C
L
EK
C
L
EA
S
L
EK
L

L
EK
S
L
EK
L
EA
CS
L
EK
S
L
EA
C
EK
L
EK
C
L
EK
SEK
L
EK
C
L
EK
S
L
EK
C

L
EA
CS
L
EK
S
L
EA
C
L
EK
S
L
EK
L
EA
CS
L
EK
S
L
EA
C
4
66
2
66
612
)
12

126
)
12
(
12
2
22
22
22
2
22
22
22
22
2
22
22
22
2
22
22
22
22
(26)
The corresponding nodal displacement and force vectors, in their explicit forms, are
δ
G
=
⎪
⎪

θ
∆
∆
θ
∆
∆
i
and f
G
=
⎪
⎪
⎪
⎪
⎭
⎪
⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪

θ
2
. On the other hand, out of the nine
nodal forces, only three are given: F
x2
= 2 kN, F
y2
=0, and M
2
= -2 kN-m; the other six are
unknown reactions at the supports. Once we specify all the known quantities, the global
equilibrium equation appears in the following form:
Beam and Frame Analysis: Displacement Method, Part II by S. T. Mau
243
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢

KKKKKK
KKKKKK
KKKKKKKKK
KKKKKKKKK
KKKKKKKKK
KKKKKK
KKKKKK
KKKKKK
⎪
⎪
⎪
⎪
⎪
⎪
⎭
⎪
⎪
⎪
⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎪
⎪
⎩

⎪
⎪
⎪
⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎧
3
3
3
1
1
1
2-

KKK
KKK
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
2
2
2
θ
∆
∆
y
x
=
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨

Beam and Frame Analysis: Displacement Method, Part II by S. T. Mau
244
Solution. We will carry out a step-by-step solution procedure for the problems.
(1) Number the nodes and members and define the nodal coordinates
Nodal Coordinates
Node x (m) y (m)
100
204
324
(2) Define member property, starting and end nodes and compute member data
Member Input Data
Member
Starting
Node
End
Node
E
(GPa)
I
(mm
4
)
A
(mm
2
)
1 1 2 200 3x10
8
2x10
4

L=
22
)()(
ii
yyxx
jj
−+−
C= Cos
β
=
L
xx
ij
)( −
=
L
x
∆
S= Sin
β
=
L
yy
ij
)( −
=
L
y
∆
(3) Compute member stiffness matrices in global coordinates: Eq. 26

01x10001x10-0
22,500-011,25022,500-011,250-
30,0000 22,500-60,000022,500
0 1x10-001x100
22,500011,250-22,500011,250
99
99
Member 2:
(k
G
)
2
=
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢

33 6
44 7
55 8
End
Node
j
66 9

Armed with this table we can easily direct the member stiffness components to the right
location in the global stiffness matrix. For example, the (2,3) component of (k
G
)
2
will be
added to the (5,6) component of the global stiffness matrix. The unconstrained global
stiffness matrix is obtained after the assembling is done.
Beam and Frame Analysis: Displacement Method, Part II by S. T. Mau
246
K
G
=
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥

−
−
000,120000,900000,60000,900000
90,000000,900000,90000,900000
00
9
10x200
9
10x2000
60,00090,0000180,00090,00022,50030,000022,500
90,00090,000090,000
9
10x100
9
10x10
00
9
10x222,5000
9
10x2500,
220250,11
000000,300500,22000,600500,22
0000
9
10x100
9
10x10
000500,220250,11500,220250,11
(5) Constrained global stiffness equation and its solution
Once the support and loading conditions are incorporated into the stiffness equations we

999897969594
898887868584
797877767574
696867666564636261
595857565554535251
494847464544434241
363534333231
262524232221
161514131211
000
000
000000
000
000
KKKKKK
KKKKKK
KKKKKK
KKKKKKKKK
KKKKKKKKK
KKKKKKKKK
KKKKKK
KKKKKK
KKKKKK
⎪
⎪
⎪
⎪

0
0
2
2
2
θ
∆
∆
y
x
=
⎪
⎪
⎪
⎪
⎪
⎪
⎭
⎪
⎪
⎪
⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪


(18)
For the three displacement unknowns the following three equations taking from the
fourth to sixth rows of the unconstrained global stiffness equation are the governing
equations.

⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
180,00090,000-22,500-
90,000-1x100
22,500-02x10
9
9
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨

-9

m,
∆
y2
=1x10
-9
m, and
θ
2
=1.11x10
-5
rad. Upon
substituting the nodal displacements into Eq. 18, we obtain the nodal forces, which are
support reactions:
⎪
⎪
⎭
⎪
⎪
⎬
⎫
⎪
⎪
⎩
⎪
⎪
⎨
⎧
1

⎪
⎪
⎬
⎫
⎪
⎪
⎩
⎪
⎪
⎨
⎧
3
3
3
M
F
F
y
x
=
⎪
⎪
⎭
⎪
⎪
⎬
⎫
⎪
⎪
⎩

=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
100
0
0
CS-
SC
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
i
yi
xi
θ

1.33 kN-m
I
nflection point
Beam and Frame Analysis: Displacement Method, Part II by S. T. Mau
248
Problem 3. Prepare the input data set needed for the matrix solution of the frame shown.
Begin with the numbering of nodes and members. E= 200 GPa, A=20,000 mm
2
, and I=
300x10
6
mm
4
for the three members.
Problem 3.
2 kN
5 kN-m
4 m
2 m
3 m
249
Influence Lines
1. What is Influence Line?
In structural design, it is often necessary to find out what the expected maximum quantity
is for a selected design parameter, such as deflection at a particular point, a particular
stress at a section, etc. The answer obviously depends on how the load is applied. The
designer must apply the load in such a way that the maximum quantity for the selected
parameter is obtained. The load could be concentrated loads, single or multiple, or
distributed loads over a specified length or area. For a single concentrated load, it is
often possible to guess where the load should be placed in order to result in a maximum

L
/2
0.1L
L
/2
L
/2
Influence Lines by S. T. Mau
250
The shape of the influence line usually reveals the locations for the multiple-load. For
distributed load, the maximum is achieved by placing the load where the area under the
influence line is the greatest.
To construct the influence line, it is not necessary to analyze the structure for every
location of the unit load, although it can be done with a computer program for any
number of selected locations. We shall introduce the analytical way of constructing
influence lines.
2. Beam Influence Lines
Consider the beam shown below. We wish to construct the influence lines for R
b
, V
c
and
M
c
.
Influence lines for R
b
, V
c
and M

1
x
ac
b
1
x
R
a
R
b
L
/2
L
/2
ac
b
Influence Lines by S. T. Mau
251
ΣM
b
=0 R
a
= (L-x)/L ΣM
a
=0 R
b
= x/L
Having obtained R
a
and R

and M
c
as
functions of R
a
and R
b
.
Left FBD: Valid for x>L/2 Right FBD: Valid For x<L/2
V
c
= R
a
V
c
= −R
b
M
c
= R
a
L/2 M
c
= R
b
L/2
Using the influence lines of R
a
and R
b

and R
b
to construct the influence lines of V
c
and M
c
.
Müller-Breslau Principle. The above process is laborious but serves the purpose of
understanding the analytical way of finding solutions for influence lines. For beam
influence lines, a quicker way is to apply the Müller-Breslau Principle, which is derived
from the virtual work principle. Consider the FBD and the same FBD with a virtual
displacement shown below.
FBDs of a beam and a virtually displaced beam.
1
1
R
a
R
b
1/2
1/2
V
c
L
/4
M
c
ac
b
1

is numerically equal to
the virtual displacement of the beam, when the virtual displacement is constructed with a
unit displacement at R
a
and no displacements at any forces except the unit load.
Consider one more set of virtual displacement of the beam aimmed at exposing the
sectional force V
c
.
Beam and associated virtual displacement for V
c
.
Application of the virtual work principle leads to:
(1) V
c
+ (y) 1=0 V
c
= − y
where y is positive if upward and negative if downward.
Another set of virtual displacement designed for solving M
c
is shown below.
ac
b
x
a
x
b
1/2
1/2

(a) at the cut there is a unit displacement (or rotation).
(b) the quantity of interest produces a positive work.
(c) no other internal force produces any work.
(3) The resulting displacement shape is the desired influence line.
Example 1. Construct the influence lines for R
a
, R
b
, V
c
, M
c
, V
d
, and M
d
of the beam
shown.
Beam with an overhang.
Solution. We shall use the Müller-Breslau Principle to construct the influence lines. It is
a try-and-error process to make sure the condition that no other forces produce any work
is satisfied.
a
x
b
M
c
V
c
M