Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
SCHOOL OF ENGINEERING & BUILT ENVIRONMENT
SESSION 2020-2021
MEng/BEng(Hons) Computer Aided Mechanical Engineering
MEng/BEng(Hons) Mechanical-Electronic Systems
MEng/BEng(Hons) in Mechanical & Power Plant Systems
BEng(Hons) in Design & Manufacture (GA)
ENGINEERING DESIGN & ANALYSIS 4
MATRIX ANALYSIS OF
STRUCTURES
Dr David R Gordon
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
Page 1
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
MATRIX ANALYSIS OF STRUCTURES
A Structure is an assembly of individual elements or beams connected
together and supported by different types of support such as fixed ends,
roller support, pinned support etc.
Matrix Analysis as described here requires an understanding of the
possible deformation (internal) mechanisms (degree of freedom’s or
dof’s) of a particular structure. Based on a knowledge of how each ‘dof’
causes each individual element or beam of the structure to bend or flex
etc. allows the derivation of the overall stiffness matrix [K] for the whole
structure. The analysis involves understanding the impact these
deformations have on the structure.
The next step is to derive the loading vector [P] for the structure based on
the type of loading acting on each beam element (internal ‘reactive’ or
externally applied ‘active’) and the impact of end restraints or support
conditions have on the magnitudes and formulae used. The equilibrium
equation in Matrix form is therefore:
[ P ]  [ K ][  ]
The structure is then solved by obtaining the actual values of the identified
dof’s [∆] from:
[]  [ K ]1[ P]
and using this information to then determine the forces, moments etc.
acting on each individual element or beam which together forms the entire
structure.
Effectively having solved for the unknown displacements the structure has
been solved and can be now broken up into each individual beam or element.
The process follows closely that used in Finite Element Analysis (FEA)
applications but with the hand calculations the number of dof’s is massively
reduced to produce a 2x2 or 3x3 matrix which can be more easily
manipulated mathematically.
The lecture series follows the process as described above and will
concentrate on firstly the Stiffness matrix [K] derivation using the equations
as derived from the slope deflection equations and obtained from the
generated DATASHEET.
The next step is the determination of the loading vector [P].
Finally the problem can be solved from a determination of the dof’s [∆] and
then can take several different routes to obtaining all/some reactions and
action diagrams such as Thrust, Shear Force and Bending Moment Diagrams
as required. Understanding the nature of Free and reactive effects are
important in assembly of the above diagrams. The analysis is explained
through a number of worked examples.
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
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Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
In the 19th century the accent was upon the development and application of
structural theory based on the mathematical background laid earlier.
Important contributions were made by Navier, Clapeyron, Saint-Venant, Airy,
Maxwell, Castigliano, and Mohr amongst others.
Their work continues to form the basis of structural theory. Unfortunately,
analytical solutions using the fundamental theories can only be obtained to a
limited class of structures of simple geometry, and direct numerical solutions
require computing power which was not available until comparatively recently.
Consequently in the first half of this century engineers turned their attention to
developing a variety of indirect methods of solution to particular classes of
problem, e.g. the skeletal structure (assembly of bar/beam components).
The slope deflection equation (G A Maney) appeared but were limited to very
simple problems due to impossibility of solving large sets of linear
algebraically simultaneous equations.
This was of course until the advent of the use of the digital computer. Also
because of their limitation the Moment Distribution Method was derived
(Hardy Cross) which avoided the solution of a large set of equations.
From the period around 1950 the development of the digital computer
presented the structural analysis with ever increasingly powerful aid in solving
problems. This led to methods of analysis tailored to the computer, which
were direct and general in application. Matrix algebra provided a basis for the
efficient organisation and manipulation of large quantities of data. The stage
was set for the introduction of matrix methods of structural analysis. In general
these relate to the analysis of skeletal structures.
The matrix methods require the generation of relationships (in matrix
form) linking forces (reactions, moments etc.) at the ends of generic
element members in a skeletal structure to the corresponding
displacements.
Relationships can be found for a straight member of uniform cross section
that are ‘exact’: that is the relationships represent an exact solution at all
points of the differential equations governing the deformation of such a
member. Then the relationships for each member of the structure can be
combined following certain rules (of equilibrium and compatibility) into a
system of simultaneous equations for the whole structure, and the solution of
these effectively gives the exact solution for the theoretical model structure.
Two kind’s of approach are commonly used. In the matrix displacement (or
stiffness or equilibrium) method the member relationships are those for end
forces expressed in terms of end displacements by way of a stiffness matrix
and the unknown quantities in the set of equations for the whole
structure are displacements at nodes or joints.
In the matrix force (or flexibility or compatibility) method the member
relationships are those for end displacements expressed in terms of end
forces by way of a flexibility matrix and the unknowns are forces. The matrix
displacement method is the more popular method and adopted here.
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
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Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
SLOPE DEFLECTION EQUATIONS FOR FLEXURAL DEFORMATION
Consider the generalist beam which forms the generalised deflected shape
indicated and the associated reaction forces and moments with all
deformations and force/moment directions’ +ve convention as indicated.
(M1+M2)/L
L
x
M1
δ
Θ2
Θ1
x
M2
x
-(M1+M2)/L
Consider the section x-x at distance ‘x’ from the LHS.
 M  M2 
M xx  M 1   1
.x
L


d2y
 M  M2 
EI 2   M xx   M 1   1
.x
L
dx


EI
dy
 M  M2  2
  M 1 .x   1
.x  A
dx
2
.
L


x 2  M1  M 2  3
EI . y   M 1 .  
.x  A.x  B
2  6.L 
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
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Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
Boundary Conditions:
X = 0, y1 = y = 0
X = 0,
dy
=
dx
Θ1
0  0  0  0  B
B  0
EI1  0  0  A
 EI1  A
 M  M2  L
EI2   M1L   1
  EI1
L

2
(1)
L2  M 1  M 2  L3
EI .  M 1

  EI1.L
2 
L
 6
(2)
2
dy
X = L,
=
dx
X = L, y =
Θ2
δ
Solving equations (1) and (2) for M1 and M2 gives the generic
SLOPE DEFLECTION EQUATIONS:
2 EI 
3. 
 2.1   2 

L 
L 
2 EI 
3. 
M2 
 2. 2  1 

L 
L 
M1 
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
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Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
Note: this is the derivation for information!
from (1)
 M  M2  L
EI 2  M 1 L   1
 EI1  0

L

 2
L
L
 M 1  M 2  EI1  EI 2  0
2
2
2 EI1 2 EI 2
M1  M 2 

         (3)
L
L
2
from (2) and substitute (3):
2 EI1 2 EI 2


(M 2 

)  M2  3
2 EI1 2 EI 2 L2 
L
L
L
  EI1.L
EI .  ( M 2 

) 
L
L
2 
L
 6




2
2
2
L
L EI1 L EI 2 L
L
EI .   M 2  EI1.L  EI 2 L  M 2 

 M 2  EI1.L
2
6
3
3
6
2
2
2
 L L L 
L
L


EI .  M 2       EI1   L   L   EI 2 ( L  )
3
3


 2 6 6
L
2L
  3 11
EI .  M 2 
  EI1  EI 2
6
3
3


L2
L
2L
 EI1  EI 2
6
3
3
L
2L

 6
M 2   EI1  EI 2
 EI .  2
3
3

L
2
4 6 EI .
M 2  EI1  EI 2  2
L
L
L
2 EI 
3EI . 
M2 
2 2  1 
                   (4)

L 
L 
EI .   M 2
from (3) and substitute (4):
2 EI1 2 EI 2

L
L
2 EI1 4 EI 2 6 EI . 2 EI1 2 EI 2
M1 




L
L
L
L
L2
4 EI1 2 EI 2 6 EI .
M1 


L
L
L2
2 EI 
3EI . 
M1 
21   2 
           (5)

L 
L 
M1  M 2 
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
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Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
It is very important to establish a sign convention for all Moments, Forces
and Member Rotations etc. in any structural configuration when problem
solving and avoid confusion:
+ ve
+ ve
All Moments (M1, M2,) Rotations (Θ1 Θ2) and member rotation (δ/L)
are designated CLOCKWISE POSITIVE.
Also reaction forces are positive downwards!
They follow the generic beam described earlier
There are some important ‘Special Cases’ which can help to solve structural
Problems to situations which arise frequently in Structures (and these helpful
equations are included in the associated DATASHEETS for our specific use):
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
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Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
Member subject to a rotation at ONE END ONLY (Θ1) with no
member Rotation (δ/L = 0).
a)
δ=0
(M1+M2)/L
L
Θ2=0
M1
M2
Θ1
(M1+M2)/L
M1 
2 EI 
3.  4 EI
.1
 2.1   2 

L 
L 
L
M2 
2 EI 
3.  2 EI
. 1
 2. 2   1 

L 
L 
L
 M2 
M1
2
Hence, a carry over equal to one half of the applied moment to the
other end when a pin joint is introduced to an initial fixed end.
This is useful for whenever a fixed end is ‘released’ to become a pin
connection. (See later examples) e.g:
+M +M/2
+M
+M
-M
-M
**
=
pin
δ=0
6 EI
L2
L
Δ2=0
M1= 4 EI
L
M2=
Δ1=1
4 EI
L
6 EI
L2
4 EI
L
2 EI
L
 2LEI 6 EI
 2
L
L

6 EI
L2
2 EI
L
See Datasheet, item 2, p12
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
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Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
b)
Member subject to no end rotation at EITHER END (Θ1=0 etc)
but experiences whole member Rotation only occurring (δ/L).
(M1+M2)/L
Θ2=0
L
M1
δ
M2
Θ1=0
(M1+M2)/L
M1 
2 EI 
3.   6 EI
.
 2. 1   2 

L 
L 
L2
M2 
2 EI 
3.   6 EI
.
 2. 2   1 

L 
L 
L2
12 EI
L3
M1= 6 EI2
δ=1
L
Δ1=0
M2= 6 EI2
L
12 EI
L3
Δ2=0
 12 EI
L3
12 EI
L3
M1=  6 EI2
L
M2=  6 EI2
L
See Datasheet, item 3, p12
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
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Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
c)
Member with one end pinned and member subject to both Θ1
and member Rotation occurring (δ/L).
(M1)/L
L
M1
δ
Θ1
Pin
M2=0
(M1)/L
2 EI 
3. 
 2. 1   2 

L 
L 
2 EI 
3. 
M2 
 2. 2   1 
0
L 
L 
 1 3.
M1 
 2 
Hence:
2

2 .L
M1 
2 EI
L
 1 3. 3. 



 2.1 

2
2
.
L
L 

M1 
3EI 

 1  
L 
L
(Observe the attached DATASHEET on page 12, items 5 and 6 for
consistency)
This DATASHEET provides the information to enable the speedy
determination of the Overall Stiffness Matrix for a given structure.
Note: For the case when a beam has a pinned end the rotation at the
pin equals –ve ½ of the rotation at the other end:
2 
Θ1
 1
2
:
2 
 1
2
**
pin
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
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Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
DETERMINATION OF THE STIFFNESS MATRIX ‘K’
and
PROBLEM SOLVING
The assembly is derived on a basis of compatibility and equilibrium.
This condition simply states that the displacement of any point of a loaded
structure must be compatible with the overall displacement of the whole
structure. The displacement must therefore be continuous and single
valued everywhere such that the structure still ‘fits together’ under load.
The following is thus required:



Within each region the displacement varies smoothly with no
discontinuities.
At the boundaries between neighbouring regions the displacements
match each other in a manner consistent with the problem.
At the structure boundaries the prescribed displacements are
satisfied.
Process:
1.
2.
3.
4.
5.
6.
7.
8.
[K].[Δ] = [P]
Determine the Non-Zero displacements in the P, Δ system.
The solution strategy is to firstly determine the stiffness matrix [K]
for the whole structure in terms of the unknown dof
Apply each Δ = 1 in turn (all other Δ set to zero) to produce the
appropriate rows/columns of the[K] matrix.
Then determine the load vector [P] for the whole structure in terms
of these unknown displacements.
Having determined the [K] and [P] matrices the solution of the
structural problem can be completed by solving the following matrix
[]  [ K ]1. [ P]
equation for the loading P:
Evaluate the displacements [Δ] from the transposed matrix
equation.
Use the slope deflection equations to determine the actual
moments existing at the nodes (at the various Δ locations).
Use these moments and known forces to establish the final reaction
forces from basic equilibrium equations.
All problems are tackled in the same manner irrespective of the actual
configuration. The support and loading conditions are adapted as
necessary.
The method above will be illustrated through the following two worked
examples. The first is for a simple 1 dof problem and the second to
establish the stiffness matrix [K] for a 3 dof problem.
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
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Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
Simple ONE DoF Example:
ALL MEMBERS HAVE THE SAME CROSS SECTIONAL, AND MATERIAL,
PROPERTIES, EI = 104 kNm2
The structure is the simplest that we could analyse as it only has one degree of
freedom.
Due to the supports at A, C, D and E the structure cannot translate upwards,
downwards, left or right. There is only ONE connected joint at B and therefore only
the rotation of this joint is the unknown displacement for the structure.
Also, due to the fixed end connections at A, D and E there is no rotations
(  A ,  D , E
 0 ).
The rotation at C is initially unknown, however this will be obtained after the rotation
1
2
at B is determined, since  C    B
2m
3m
6m
D
3m
40kN
A
4m
B
C
pin-jointed
connection
30kN/m
E
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
Page 12
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
Solution:
1- Determine the stiffness matrix [K]. Identify the one DoF as the rotation at B:
e.g:
2m
3m
6m
D
3m
A
4m
B
1
40kN
C
pin-jointed
connection
30kN/m
E
Remove all loading and apply a +ve CW rotation ONLY to the structure AT
THE DELTA-1 POSITION, and accurately observe the distorted, deflected
shape each beam member takes up. (DEMONSTRATED IN CLASS)
D
A
B
C
E
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
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Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
Reference the DATASHEET 1 of 2 for the correct coefficients with
(unity).
e.g:

=1
Beam members AB and BD (rotate the DATASHEET image 90 degrees
clockwise to match):
Member BC:
Member BE (rotate the DATASHEET image 90 degrees clockwise to match):
Stiffness coefficients for Delta-1,
kAB + kBD + kBC + kBE
=
4 EI AB 4 EI BD 3EI BC 4 EI BE 4 EI 4 EI 3EI 4 EI



 4.67 EI
=



2
3
9
4
L AB
LBD
LBC
LBE
[K]=4.67 EI
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
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Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
2 - Next we need to determine the corresponding LOADING VECTOR [P]:
‘Explode’ the structure into individual beam elements:
BOOM!
We are now looking for the ‘MOMENTS’ from DATASHEET 2 of 2 and which
correspond with the rotational position of Delta 1, i.e [P1] at B.
(NOTE: ‘FORCES’ correspond to translational displacements if appropriate)
D
3m
3m
2m
6m
40kN
0
A
wL2

12
30  4 2

12
  40 kNm
4m
C
B
0
Wab 2 1  Wa 2 b 
   2 
2
L2
L 
40  3  6 2 1 40  3 2  6


2
92
92
 53.33  13.33  66.663 kNm

30kN/m
E
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
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Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
[P1]=[ EXTERNAL (as applied)
-
INTERNAL (as reactive) ]
[P1]=[ 0 - (+40 – 66.663) ]
[P1]= [ 0 - (-26.663) ]
[P1]= [ +26.663 ]
3 – Determine the value for the rotation at B:
1   K 1 P1  
1
5.709
 26.663 
 5.709  10  4 rads
4.67 EI
EI
4 – We can determine the bending moment for ANY point in the structure:
e.g: MBC
2m
3m
6m
D
3m
40kN
A
4m
B
C
MCB=0
pin-jointed
connection
30kN/m
E
M BC 

2 EI
L
3 

F
2   B  C  L   M BC


2  104 
 1
 
2  5.709  10 4     5.709  10 4   0  (66.663)

9 
 2
 
M BC  1.903  66.663  64.76 kNm
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
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Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
TYPICAL DATASHEET page 1 of 2
-6EI/L2
6EI/L2

2EI/L
4EI/L
6EI/L2
-6EI/L2
4EI/L
2EI/L

-12EI/L3
12EI/L3
-6EI/L2

12EI/L3
-12EI/L3

6EI/L2

6EI/L2
3EI
L3
-3EI
L2
3EI
L3

3EI
L2
3EI
L
-6EI/L2

3EI
L2

 3
2   1 
2 2L
M
3EI   
  
L  L
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
Page 17
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
EXAMPLE 1
Develop the total stiffness matrix for the structure shown:
L
I
I
L/2
2I
L
Pinned end
Fixed end
The Structure has THREE potential
INTERNAL degrees of Freedom (dof),
Δ1, Δ2, Δ3 as shown from within the
assembled structure and EXCLUDING
any at the supports.
The stiffness matrix components (rows
and columns) are derived by applying
only one ‘dof’ to the structure with all
others set to zero. The resulting
deformation state should for each
element conform to one of the standard
beam derivations listed in the Datasheet.
Δ2
Δ1
L
Δ3
I
L/2
I
2I
L
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
Page 18
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
e.g: For the first column apply Δ1 (δ) = 1 (unity), and the other Δ’s =0
Δ1=1
I
L/2
I
L
2I

3EI
L3
-3EI
L2
Element 1 matches the
fifth image in the
DATASHEET, e.g:
3EI
L3

Element 2 (Horizontal beam) is not distorted by this imposed deformation
12EI/L3
12EI/L
3
Element 3 matches the third -6EI/L2
image in the DATASHEET,
e.g:

Hence, the matrix components are given by:
k11  
3EI 12 E (2 I ) 24 EI 24 EI 48 EI

 3  3  3
( L2 ) 3
L3
L
L
L
i.e. in the same +ve direction as Δ1 .
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
Page 19
-6EI/L2
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
similarly:
k 21 
 3EI 12 EI
 12 EI

.
or

.
( L2 ) 2
L2
L2
k31 
 6 E (2 I )  12EI

L2
L2
 48 EI
 L3
  12 EI
giving column 1 as: 
2
 L
  12 EI
 L2

? ?

? ?

? ?

For Column 2, apply Δ2 (θ) = 1 (unity), and the other Δ’s =0
Δ2=1
I
I
2I
3EI
( L2 ) 2
3EI
( L2 )
3EI
L2

3EI
L
4 EI
L
2 EI
L
6EI/L2
4EI/L
k12 

-6EI/L2
2EI/L

3EI
3EI
12 EI




( L2 ) 2
( L2 ) 2
L2
(in the direction of +ve Δ1)
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
Page 20
3EI
L2
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
k 22 
3EI 4 EI 6 EI 4 EI 10 EI




( L2 )
L
L
L
L
k 32 
2 EI
L
(in the direction of +ve Δ2)
(in the direction of +ve Δ3)
Giving [K], by adding column 2, as:
 12 EI
L2
10 EI
L
2 EI
L
 48 EI
 L3
  12 EI

2
 L
  12 EI
 L2

?

?

?

For Column 3, apply Δ3 (θ) = 1 (unity), and the other Δ’s =0
Δ3=1
I
I
2I
2 EI
L
4 EI
L
 6 EI
L2
4 E (2 I )
L
k13  
6 E (2 I )
12 EI
 2
2
L
L
k 23 
2 EI
L
k 33 
4 EI 4 E (2 I ) 12 EI


L
L
L
This give the final column 3 and completes the matrix.
 48 EI
 L3
  12 EI

2
 L
  12 EI
 L2
 12 EI
L2
10 EI
L
2 EI
L
 12 EI 
L2 
2 EI 

L 
12 EI 
L 
=
 48
 L3
  12
EI  2
 L
  12
 L2
 12
L2
10
L
2
L
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
 12 
L2 
2 

L 
12 
L 
Page 21
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
TUTORIAL 1 – MATRIX ANALYSIS OF STRUCTURES
Develop the total stiffness matrix for the structures below:
1.
All span lengths = L
A
B
 24 3
 L
 6 2
 L
 6 2
 L
I
3I
C
2.
I

L
6 
L
16 
L
6
2
L
16
6
L
L
2
D
EI = Constant for all members
D
L/2
A
6
L/2
B
L
C
108

L3
 24 2
 L
 6 2
 L
24
L
12
2

L
2 
L 
14 
L
6
2
L
L
2
L
E
3.
E
2I
8L
12L
D
C
3I
6L
2I
2I
59
3
 288 L
 1 2
3L


0

1
7
1
3L2
3L
2L
8L
B
A
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
Page 22
0 

1 
2L
3 
L
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
External Loading Considerations
As has been indicated above the total stiffness matrix for a structure can
be obtained by inspection. This technique made direct application of the
slope deflection equations. As derived the equations did not reflect the
effect of loading applied externally to the structure since it was reactive to
imposed possible displacements.
The presence of external loading causes additional end moments, called
Fixed End Moments. These are normally and easily calculated from
standard beam formulae as indicated and provided in DATASHEET-page
2 of 2.
The fixed end moments are to represent the reactive response of
members due to loads and can be written in Matrix form [P] to align with
the derived stiffness matrix.
Thus in their reactive role the slope deflection equations are modified as
follows:
2 EI 
3. 
F
 2.1   2 
  M1
L 
L 
2 EI 
3. 
F
M2 
 2. 2  1 
  M2
L 
L 
M1 
Careful attention should be paid to the signs of MF when using MF in this
context as the sign is embedded in the modified slope deflection equations
above. (see examples)
The loading matrix is the direct representation of the loads acting at, and in
the direction of, the identified degrees of freedom (Δ1, Δ2, Δ3, etc.) and
equation then becomes:
 P1

P
[ P]   2
.

 Pn
F1
 F2
.
Fn

  EXT

 
 

 INT 



where Pn are the applied external or ‘active’ moments/loads
(concentrated) and F the internal or ‘reactive’ moments/loads. Action and
re-action are equal but opposite and hence the –ve sign
We are now ready to complete the full analysis for a structure [ P ]  [ K ][  ] ,
or []  [ K ]1[ P] as outlined in the following example 2:
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
Page 23
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
DATASHEET page 2 of 2 (contd.)
(FIXED END MOMENTS)
w
wL2/12
-wL2/12
wL/2
wL/2
W
WL/8
-WL/8
W/2
W/2
W
Wa2b/L2
-Wab2/L2
Wb/(a+b)
Wa/(a+b)
B
A
F
M AB
 21 M FBA
M AB 
2EI 
3 
F
 2 1   2    M AB
L 
L
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
Page 24
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
EX.2. The Frame shown below is rigidly fixed at D and E and is supported
by a tie bar at C. All members have the same flexural rigidity (EI) of
10 MNm2.
Considering flexural deformation only (i.e bending action), and using the
Displacement (Stiffness) method of Matrix Analysis determine:
i)
the rotations at B and C
ii)
the magnitude and nature of the force in the tie bar FC.
F
10kN
5kN/m
D
B
A
C
10m
E
10m
5m
10m
The structure has two unknown displacements (2 DoF).
i.e rotations at B and C.
Δ1
i)
Apply Δ1=1 with Δ2 =0
2EI
4EI
10
10
Δ2
Δ1=1
4EI
10
Δ2=0
k11 
4(10) 4(10)

8
10
10
k 21 
2(10)
2
10
2EI
10
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
Page 25
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
ii) Apply Δ1=0 with Δ2 =1
2EI
10
4EI
10
Δ2=1
Δ1=0
4EI
10
2EI
10
Hence:
k12 
2(10)
2
10
k 22 
4(10) 4(10)

8
10
10
8 2 
[K ]  

2 8 
And, [P] = [K] [Δ]
[ K ] 1 
1
DET
[Δ] = [K]-1 [P]
or
 8  2
1
 2 8  
2

 (8  8)  (2)
 8  2 1  8  2
 2 8   60  2 8 




  1  1  8  2  P1 
   60  2 8   P 

 2 
 2
Examine the external loading on the structure and determine the [P]
matrix.
F
10kN
5kN/m
D
B
A
C
wL2 5  10 2

12
12
 41.7.kNm
10 x 5 = 50kNm
E
5m
10m
10m
 (41.7)  50  41.7 91.7
91.7  10 3 
 P1   50



.
kNm


.MNm
P   0
 (41.7  41.7)  0
0   0 
0

2



Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
Page 26 
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
 1  1  8  2 91.7  10 3 

 


0
  2  60  2 8  

8  91.7  10 3
 0.01223.rads
60
 2  91.7  10 3
2 
 3.0567  10 3.rads
60
1 
Slope deflection equations:
-41.67
+41.67
2 EI
B
[2. B   C ]  M F BC
L
2  10

[2  0.01223  (3.057  10 3 )]  (41.67  10 3 )
10
 1.136  10 3 MNm  1.136.kNm
M BC 
M BC
M BC
C
2 EI
[2. C   B ]  M F CB
L
2  10

[2  (3.057  10 3 )  (0.01223)]  (41.67  10 3 )
10
 53.9  10 3 MNm  53.9.kNm
M CB 
M CB
M CB
2  10
[2  (3.057  10 3 )  (0)]  (41.67  10 3 )
10
 53.9  10 3 MNm  53.9.kNm
M CD 
M CD
2  10
[2  0  (3.057  10 3 )]  (41.67  10 3 )
10
 0.03556.MNm  35.56.kNm
M DC 
M DC
B
Free Body Diagrams:
5kN/m
wL/2 = 25kN
10m
5kN/m
C
25kN
25kN
+
1.136
(1.136+53.9)/10 = 5.5036 kN
53.9
=
19.4964 kN
30.504 kN
35.56
1.834 kN
Hanger
Force =
57.34 kN
Tension
25kN
+
53.9
5.5036 kN
10m
D
26.834 kN
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
1.834 kN
23.17 kN
Page 27
BENDING MOMENT DIAGRAMS FOR BCD
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
FREE BMD
Free bending moment for parabola:
wL2 5 10 2
M max 

 62.5kNm
8
8
5kN/m
B
C
5kN/m
10m
D
10m
62.5
62.5
REACTIVE BMD
+
+
1.136
53.9
53.9
35.56
COMBINED OR, FINAL BMD FOR BCD
+62.5
 62.5  26.382  36.118
+36.118
 62.5  44.73  17.77kNm
+17.77
+1.136
-35.56

 53.9  1.136
 26.382kNm
2
-53.9
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
 35.56 
53.9  35.56
 44.73
2
Page 28
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
EX 3.
The structure ABCDE shown in Figure I has roller supports at A
and C which also provide for a pinned connection at A and C, is
fully fixed at E, and carries a uniformly distributed load of 5kN/m
along member BC. A discrete offset force of 32 kN is applied at D,
and a force of 12 kN is applied to the rigid cantilever bar BF at F.
The members have their relative flexural rigidities as shown, where
EI is 10MNm2.
Determine:
(a)
the stiffness matrix [K] for the structure using the
Displacement (Stiffness) method of matrix analysis and
considering flexural deformation only;
(b)
the load vector [P] for the applied loading;
(c)
the reactions at supports A, C and BMD for AB and BC.
Note: Use may be made of DATASHEET Q.1
EI
2EI
3EI
Figure I
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
Page 29
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
(a) DoF?,
1 ,  2 ?
EI
1 2EI
2
3EI
[K],
1  1, 2  0 (remove all loading)
EI
2EI
3EI
12  EI  12  3EI

 1.333 EI
L3
33
6  EI 
6  3EI
 2 : k12  

 2 EI
2
L
32
1 : k11 
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
Page 30
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
1  0, 2  1
C   A  ?  
EI
B
2
2EI
3EI
3EI  4EI  3EI  3  2 EI 4  3EI 3  EI





 6.2 EI
L
L
L
5
3
3
6  EI 
6  3EI
1 : k 21  

 2 EI
2
L
32
 2 : k 22  k BC  k BE  k BA 
Hence:
K   EI 
k11
k 21
k12 
1.333  2

EI
  2 6.2
k 22 


Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
Page 31
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
(b) [P], now include any loading which
corresponds to the identified DoF.
EI
2EI
3EI
BOOM!
12kN - EXT
(see Datasheet 2)
M = 12 x 1.25 =15 kNm - EXT
PIN, M=0
12kN - INT
PIN, M = 0
15 kNm INT
B
B
5kN/m - EXT
B
+14.22 kNm
INT
W a2 b

L2
32  2 2  1
M F BE 
 14.22 kNm
32
Wa
32  2

 21.33 kN
ab
3
M F BE 
-15.625 kNm INT
B
21.33 kN (-ve to
DELTA 1) INT
b =1
32kN
EXT
M
F
BC
M F BC
wL2  1 wL2 
5  52 5  52



 


12  2 12 
12
2  12
 15.625 kNm
a=2
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
Page 32
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
We REALLY need to understand the
‘difference’ between ‘internal’ and ‘external’
loading. P  EXT  INT 
Load in the direction of 1 ?
EXT - INT
P1  12
  21.33  33.33 kN
Or
P1  0
  12  21.33  33.33 kN
Loading in the direction of  2 ?
P2   15   15.625  14.22  16.405 kNm
Or
P2  0
  15  15.625  14.22  16.405 kNm
Hence:
  33.33  kN
P  

 16.405 kNm
(c) The next step is to determine the actual
DoF values,
   K 1P 
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
Page 33
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
 1 
6.2 EI  2 EI  33.33
1

  1.33EI  6.2 EI   2 EI  2 EI   2 EI 1.33EI   16.4 



 2
 1 
1

  4.246 EI 2
 2
1 
2 EI  33.33
6.2 EI
 2 EI 1.33EI   16.4 



6.2 EI 33.33  2 EI 16.4 
4.246 EI 2
1
56.4
 56.4 
EI
10  103
1  5.64  10  3 m or  5.64 mm
2 
2 EI 33.33  1.33EI 16.4 
4.246 EI 2
1
20.84
 20.84  
EI
10 103
 2  2.084 10 3 rads   B
 A   C    B     2.084  10 4   1.042 10 3 rads
1
2
1
2
‘Individual’ Beams
Next, determine the ACTUAL Moments at B for
each separate beam element, given the
displacements evaluated above. Note the MF
moments are for the being analysed NOT from
the combined load vector!
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
Page 34
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
M AB  0 PIN

3 
F
2 B   A 
  M BA
LBA 

3 0 
2  10  10 3 

2 2.084  10 3   1.042  10 3 
0

3
3 

 20.84 kNm
2 EI BA
LBA
M BA 
M BA
M BA

 

M CB  0 PIN

3 
F
2 B   C 
  M BC
LBC 

3 0 
2  2  10  10 3 
3
3

2
2
.
084

10


1
.
042

10

  15.625

5
5 

 25  15.625  9.38 kNm
M BC 
M BC
M BC

M BE 
M BE
M BE
2 EI BE
LBE
 


3 
F
2 B   E 
  M BE
LBE 


3 5.64  10 3 
3
2 2.084  10  0  
  14.22 
3


 29.44  14.22  15.22 kNm
2  3  10  10 3

3
M EB 
M EB
2 EI BC
LBC


2 EI EB 
3 
F
2 E   B 
  M EB
LEB 
LEB 
2  3  10  10 3

3

3 5.64  10 3 
3
2 0   2.084  10 
  7.11
3




M EB  78.23 kNm
NOTE:
M
F
EB
M F EB
W a b2


L2
32  2  12

 7.11 kNm
32
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
Page 35
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
REACTIONS and BMD’s:
A
AB:
B
PIN, M = 0
+20.84 kNm
BMD for AB
VA  R  6.95 kN
-20.84 kNm
R
20.84 kNm
R
20.84  0
 6.95 kN
3
R
BC:
5kN/m - EXT
+9.38 kNm
PIN, M=0
B
wL/2 = 5 x 5/2 =12.5 kN
INT
+
R
VC  R  12.5  1.876  12.5
VC  14.376 kN
+9.38 kNm
R
9.38  0
 1.876 kN
5
R
20.315 kNm =4.69+15.625
+9.38 kNm
M FREE
wL2 5  5 2


 15.625
8
8
BMD for BC
+ve
M REACTIVE 
9.38
 4.69
2
BMD sign convention
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
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Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
Page 37
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
Page 38
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
1.25
-4.5
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
Page 39
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
TUTORIAL 2 – MATRIX ANALYSIS OF STRUCTURES
1.
Determine the loading vector [P] and total stiffness matrix [K] and draw
the bending moment diagram.
15kN
B
A
5kN/m
5m
4
K  5
 15

C
10m
D
10m
10m
1 
5
4 
5
  33.3 
P

 41.67 
4
EI=10 MNm
E
2.
Determine the loading vector [P] and total stiffness matrix [K] and draw
the BMD. EI = 104 kNm2.
A
5kN/m
B
14

K    10 4 
10

5m
P  [62.5]
2EI 10m
EI
3.
20kN
Determine the loading vector [P]
2kN/m
P  [19.9]
4m
6m
4m
15m
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
Page 40
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
Past Exam Questions (10)
Jan’11 (Dec’10) –
Jan’20
MATRIX
ANALYSIS
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
Page 41
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
Q.1
JANUARY 2020 (19-20)
The structure ABCDEF shown in Figure Q1 has pin joint supports at A,D,E,F.
Member BC has an off-set discrete vertical force of 20kN as shown. Member
AB carries a uniformly distributed vertical load of 3kN/m acting along its
length. All members have the relative flexural rigidities as indicated where EI
is 1x103 kNm2. The solution should involve reference to DATASHEET Q1
as appropriate.
i)
Using the Displacement (Stiffness) method of matrix analysis and
considering flexural deformation only, determine the stiffness matrix [K]
for the structure.
[7]
ii) Determine the load vector [P] for the applied loading.
[4]
iii) Determine the displacements at B and C.
[4]
iv) Determine MBC, MCB and the Free, Reactive, and Combined Bending
Moment Diagrams (BMD) for the beam element BC.
[10]
D
6m
EI
4m
20kN
3m
3kN/m
A
B
3EI
2EI
4m
EI
EI
Pin-joint
C
F
E
4m
Figure Q1
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
Page 42
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
Q.1
JANUARY 2019 (18-19)
The structure in Figure Q1 has pin joint supports at A and D and is fully fixed
at C. Members BD and BC both have discrete forces of 20 kN acting at E and
F respectively as shown. Member AB carries a uniformly distributed vertical
load of 3 kN/m acting along its length. Members AB, BD and BC have the
relative flexural rigidities as indicated where EI is 1x103 kNm2. The solution
should involve reference to DATASHEET Q1 as appropriate.
ii) Using the Displacement (Stiffness) method of matrix analysis and
considering flexural deformation only, determine the stiffness matrix [K]
for the structure.
[8]
ii) Determine the load vector [P] for the applied loading.
[6]
v) Determine the structural displacements at B.
[4]
vi) Determine the Bending Moment Diagram (BMD) for AB.
[7]
D
Pin-joint
2m
EI
20kN
4m
F
20kN
3kN/m
A
2EI
E
B
6m
1m
EI
C
3m
Figure Q1
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
Page 43
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
Q.1
JANUARY 2018 (17-18)
The structure shown in Figure Q1 has a pin joint support at D and is fully
fixed at A and G. Members BCE and EFG have a discrete central horizontal
force of 20 kN at C and F respectively. Member DE carries a uniformly
distributed vertical load of 3 kN/m acting along its length. All members have
the relative flexural rigidities as indicated where EI is 1x103 kNm2. The
solution should involve reference to DATASHEET Q1 as appropriate.
iii) Using the Displacement (Stiffness) method of matrix analysis and
considering flexural deformation only, determine the stiffness matrix [K]
for the structure.
[6]
ii) Determine the load vector [P] for the applied loading.
[5]
vii) Determine the displacements at B and E.
[4]
viii) Determine the Bending Moment Diagram (BMD) for DE.
[10]
4m
B
A
EI
2m
EI
20kN
4m
C
3kN/m
E
D
2EI
6m
F
20kN
4m
Pin-joint
EI
2m
G
Figure Q1
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
Page 44
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
Q.1
JANUARY 2017 (16-17)
The structure ABCD shown in Figure Q.1 has a pin jointed and sliding support
at C and is fully fixed at D. Member AB is a cantilever beam with a discrete
vertical load of
30 kN acting at A. Member BD carries a uniformly distributed load of 3 kN/m
along its length. Member BC has a discrete off-centre vertical force of 50 kN
as shown. All members each have the same flexural rigidity EI of 1x103
kNm2. The solution should involve reference to DATASHEET Q.1 as
necessary.
i)
Using the Displacement (Stiffness) method of matrix analysis and
considering flexural deformation only, determine the stiffness matrix
[K] for the structure.
[6]
Determine the load vector [P] for the applied loading.
[5]
Determine the displacements at B.
[4]
Determine the vertical reaction at C, and the bending moment diagram
(BMD) for member BC.
[10]
ii)
iii)
iv)
3m
2m
30kN
A
4m
6m
50kN
B
C
pin-jointed
and slider
connection
3kN/m
D
Figure Q.1
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
,
Page 45
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
Q.1
JANUARY 2016 (15-16)
The structure ABCDEF shown in Figure Q.1 has a pin joint support at A and
D, while at E and F the structure is fully fixed. Member AB carries a
uniformly distributed load of 10kN/m along it’s length. Member BC has a
discrete mid-span vertical force of 100kN as shown. Member CD carries an
off-centre discrete load of 180kN. All members each have the same flexural
rigidity EI of 105 kNm2.
(i) Using the Displacement (Stiffness) method of matrix analysis and considering
flexural deformation only, determine the stiffness matrix [K] for the structure.
[5]
(ii) Determine the load vector [P] for the applied loading.
[6]
(iii) Determine the displacements at B and C.
[4]
(iv) Determine the vertical reaction at D, and the Bending moment diagram
(BMD) for CD.
[10]
pin-jointed
connection
E
180kN
100kN
10kN/m
6m
C
A
D
B
6m
8m
6m
pin-jointed
connection
F
12 m
12 m
10 m
Figure Q.1
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
Page 46
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
Q.1
JANUARY 2015 (14-15)
The frame shown in Figure Q.1 has a sliding pin joint support at D and E,
while at C the structure is fully fixed. Member BC carries a uniformly
distributed load of 12kN/m along it’s length. Member BD has a discrete mid
span horizontal force of 20kN as shown. The non-structural cantilever
overhang AB carries a discrete load of 15kN and the sliding support at E
experiences a vertical load of 70kN. All members each have the same flexural
rigidity EI of 104 kNm2.
(ii) Using the Displacement (Stiffness) method of matrix analysis and considering
flexural deformation only, determine the stiffness matrix for the structure.
[6]
(ii) Determine the load vector for the applied loading.
[6]
(v) Determine the displacement of B and the reactions at pistons D and E.
[13]
D
Sliding support with
pin-jointed connection
20 kN
3m
12kN/m
Cantilever
beam AB
A
C
B
1.1m
6m
15kN
3m
E
70kN
,
Figure Q.1
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
Page 47
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
Q.1
JANUARY 2014 (13-14)
The frame ABCDEF shown in Figure Q.1 is rigidly fixed at A and pin joint
supported at D. Horizontal motion is prevented by means of a tie bar FB
which is pin jointed at both ends and attaches to the structure at B. The
structural members AB, BC and BE are continuous and rigidly connected at B.
The relative flexural rigidities are as indicated where EI is 10MNm2.
Considering flexural deformation only and using the Displacement (Stiffness)
method of matrix analysis and using DATASHEET Q.1, determine:
(a)
The stiffness matrix [K] for the structure.
[6]
(b)
The loading vector [P].
[4]
(c)
The displacements at B and C
[5]
(d)
The load acting in the tie bar FB.
[10]
10m
D
E
20kN
5m
6 kN/m
2m
F
2EI
B
EI
3EI
C
15kN
5m
2.5m
A
Figure Q.1
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
Page 48
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
Q.1
DECEMBER 2012 (12-13)
The frame structure ABCDE shown in Figure Q.1 is rigidly fixed at A, D and
E. Members BE and CD carry a uniform distributed load of 2kN/m.
Member BC has a 20kN load acting horizontally at mid-span. The relative
flexural rigidities of each member are as shown in Figure Q.1, where EI is
10MNm2.
Considering flexural deformations only, and making use of the supplied
DATASHEET Q.1, determine:
i)
the stiffness matrix [K] for the structure;
ii)
the loading vector [P] for the structure;
[5]
[5]
iii)
the displacements at B and C;
[5]
iv)
all reactions acting at D;
[5]
v)
the Bending Moment Diagram (BMD) for member CD.
[5]
10
m
2kN/m
4EI
C
2.5
m
5m
D
EI
20kN
2kN/
m
EI
B
2EI
E
5m
A
5m
Figure Q.1
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
Page 49
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
Q.1
DEC 2011 (11-12, NEW TRIMESTER SYSTEM)
The frame structure ABCDEF shown in Figure Q.1 is pin supported at A, fully
fixed at C and F and has a pin joint support at E. Member AB carries a discrete
force of 10kN at the position shown, member CD carries a uniformly
distributed load of 6kN/m and a discrete force of 24kN is applied to DE at
mid-span. The members each have their relative flexural rigidities as
indicated, where EI is 10MNm2. Determine:
(a)
the stiffness matrix [K] for the structure using the Displacement
(Stiffness) method of matrix analysis and considering flexural
deformation only;
[6]
ii)
the load vector [P] for the applied loading;
[6]
iii)
the displacements at D and E;
[4]
iv)
the bending moment MDE, the reactions at E, and sketch the bending
moment diagram for portion DE.
[9]
Note: Use may be made of DATASHEET Q.1
10 kN
B
A
2EI
EI
4m
6 kN/m
C
24kN
D
2EI
E
2EI
4m
EI
F
4m
3
m
3
m
6m
10m
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
Figure Q.1
Page 50
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
Q.1
DEC 2010 (10-11, NEW TRIMESTER SYSTEM)
The frame ABCDE shown in Fig.Q.1 is rigidly supported at A, simply
supported at D, and carries uniformly distributed loading of 12kN/m and
6kN/m along members AB and BD respectively. A horizontal discrete force of
16kN is also applied at point C on member BD and a vertical discrete force of
10kN acts at E.
The frame members have their relative flexural rigidities as shown, where EI
is 104 kNm2.
(a)
Using the Displacement (Stiffness) method of matrix analysis and
considering flexural deformation only, determine the stiffness matrix
[K] for the structure.
[6]
(b)
Determine the load vector [P] for the applied loading.
[6]
(c)
Determine the displacements at B and the reaction at support D.
[13]
Note: Use may be made of DATASHEET Q.1
10kN
12kN/m
6kN/m
16kN
Figure Q.1
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
Page 51
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
TYPICAL DATASHEET page 1 of 2
-6EI/L2
6EI/L2

2EI/L
4EI/L
6EI/L2
-6EI/L2
4EI/L
2EI/L

-12EI/L3
12EI/L3
-6EI/L2

12EI/L3
-12EI/L3

6EI/L2

6EI/L2
3EI
L3
-3EI
L2
3EI
L3

3EI
L2
3EI
L
-6EI/L2

3EI
L2

 3
2   1 
2 2L
M
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
3EI   
  
L  L
Page 52
Dr David R Gordon, Matrix Analysis, Level 4, 2020/21
DATASHEET page 2 of 2 (contd.)
(FIXED END MOMENTS)
w
wL2/12
-wL2/12
wL/2
wL/2
W
WL/8
-WL/8
W/2
W/2
W
Wa2b/L2
-Wab2/L2
Wb/(a+b)
Wa/(a+b)
B
A
F
M AB
 21 M FBA
M AB 
2EI 
3 
F
 2 1   2    M AB
L 
L
Dr David R Gordon, ED&A4 Trimester 1 Session 2020-21
Page 53