The document discusses the matrix method of structural analysis. It introduces matrix algebra and how it can be used to analyze structures. There are two main approaches discussed - the stiffness matrix method and flexibility matrix method. The stiffness matrix method determines displacements at joints by considering member stiffness. The flexibility matrix method determines redundant forces by considering flexibility properties. Several examples are provided to illustrate analyzing beams and trusses using the force method and displacement method within the matrix framework. Formulas are given for defining member and structure stiffness/flexibility matrices and solving for displacements and forces.

1. 1 | P a g e
Structural Engineering & Software Application CE406
Written & Composed BY ENGINEER SAQIB IMRAN
Cell no: 0341-7549889
Email: saqibimran43@gmail.com
Student of B.TECH(Civil) at Sarhad University of Science &
Information Technology Peshawer.

2. 2 | P a g e
Matrix method of Structure Analysis
Introduction: Matrix algebra is a mathematical notation that simplifies the presentation and
solution of simultaneous equations. It may be used to obtain a concise statement of a structural
problem and to create a mathematical model of the structure. The solution of the problem by
matrix structural analysis techniques1,2,3,4 then proceeds in an entirely systematic manner. All
types of structures, whether statically determinate or indeterminate, may be analyzed by matrix
methods. In addition, matrix concepts and techniques, because of their systematic character,
form the basis of the computer analysis and design of structures5,6,7,8. Highly indeterminate
structures may be easily handled in this way and alternative loading conditions readily
investigated. There are two general approaches to the matrix analysis of structures: The stiffness
matrix method and the flexibility matrix method.
The stiffness method is also known as the displacement or equilibrium method. It obtains the
solution of a structure by determining the displacements at its joints. The number of
displacements involved equals the number of degrees of freedom of the structure. Thus, for a
pin-jointed frame with j joints the solution of 2j equations is required, and for a rigid frame,
allowing for axial effects, the solution of 3j equations is required. If axial effects in rigid frames
are ignored, the number of equations involved reduces to j plus the number of degrees of sway
freedom. Irrespective of the number involved, these equations may be formulated and solved
automatically by computer. The stiffness matrix method is the customary method utilized in
computer programs for the solution of building structures.
The flexibility method is also known as the force or compatibility method. It obtains the solution
of a structure by determining the redundant forces. Thus, the number of equations involved is
equal to the degree of indeterminacy of the structure. The redundant may be selected in an
arbitrary manner, and their choice is not an automatic procedure. The primary consideration in
the selection of the redundant is that the resulting equations are well conditioned. OR
FLEXIBILITY AND STIFFNESS METHODS: These are the two basic methods by which an
indeterminate skeletal structure is analyzed. In these methods flexibility and stiffness properties
of members are employed. These methods have been developed in conventional and matrix
forms. Here conventional methods are discussed.
Flexibility Method: The given indeterminate structure is first made statically determinate by
introducing suitable number of releases. The number of releases required is equal to statical
indeterminacy ∝s. Introduction of releases results in displacement discontinuities at these
releases under the externally applied loads. Pairs of unknown biactions (forces and moments) are
applied at these releases in order to restore the continuity or compatibility of structure. The
computation of these unknown biactions involves solution of linear simultaneous equations. The
number of these equations is equal to statical indeterminacy ∝s. After the unknown biactions are
computed all the internal forces can be computed in the entire structure using equations of
equilibrium and free bodies of members. The required displacements can also be computed using
methods of displacement computation.
In flexibility method since unknowns are forces at the releases the method is also called force
method. Since computation of displacement is also required at releases for imposing conditions

3. 3 | P a g e
of compatibility the method is also called compatibility method. In computation of displacements
use is made of flexibility properties, hence, the method is also called flexibility method.
Stiffness Method: The given indeterminate structure is first made kinematically determinate by
introducing constraints at the nodes. The required number of constraints is equal to degrees of
freedom at the nodes that is kinematic indeterminacy ∝ k. The kinematically determinate
structure comprises of fixed ended members, hence, all nodal displacements are zero. These
results in stress resultant discontinuities at these nodes under the action of applied loads or in
other words the clamped joints are not in equilibrium. In order to restore the equilibrium of
stress resultants at the nodes the nodes are imparted suitable unknown displacements. The
number of simultaneous equations representing joint equilibrium of forces is equal to kinematic
indeterminacy ∝ k. Solution of these equations gives unknown nodal displacements. Using
stiffness properties of members, the member end forces are computed and hence the internal
forces throughout the structure. Since nodal displacements are unknowns, the method is also
called displacement method. Since equilibrium conditions are applied at the joints the method is
also called equilibrium method. Since stiffness properties of members are used the method is
also called stiffness method. OR
Fig.4.1
Flexibility =
∆1
𝑃1
= 𝛿11 (4.1)
Similarly, the stiffness of the spacing is defined as the force k11 required for a unit displacement at
coordinate 1. Stiffness =
𝑃1
∆1
= 𝑘11 (4.2)

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AXIAL DISPLACEMENT
TRANSVERSE DISPLACEMENT
(4.6) & (4.7)

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(4.10) & (4.11)
BENDING OR FLEXURAL DISPLACEMENT
𝑃3 =
4𝐸𝐼∆3
𝐿
(4.12), Hence, by definition, the flexibility and stiffness with respect to flexural
displacement may be written as:
(4.13) & (4.14)
𝑃3 =
3𝐸𝐼∆3
𝐿
(4.15), Hence, by definition, the flexibility and stiffness with respect to flexural
displacement may be written as:
(4.16) & (4.17)
TORSIONAL DISPLACEMENT OR TWIST

6. 6 | P a g e
FLEXIBILITY EQUATION
Consider a structure subjected to forces {F}1 = [F1 F2 …. Fn] where F1 is the force in direction i. Let
it have displacements {∆}T = [∆1 ∆2 …∆n]. Since 𝛿 𝑛 is the displacement in coordinate direction i
due to unit force in coordinate j, the displacements at i due to force F1 is 𝛿11 F1. Similarly, 𝛿12 F2 is
the displacement at i due to force F2. Thus, the total displacement is given by

7. 7 | P a g e
STIFFNESS EQUATION
RELATIONSHIP BETWEEN FLEXIBILITY AND STIFFNESS MATRICES
From Eq. 1.12, [k] {∆} = {F}. pre multiplying both sides with [k]-1, we get {∆} = [k]-1{F}
But from Eq. 1.1, we have: {𝛿} {F} = {∆}. Hence, {𝛿} {F} = [k]-1 {F} ⇒ {𝛿} = [k]-1 …..1.13
Thus, flexibility & stiffness matrices are inverse of each other.
FORCE AND DISPLACEMENT METHOD
of structure analysis using the matrix approach.

8. 8 | P a g e
FORCE METHOD

9. 9 | P a g e
DISPLACEMENT METHOD
at coordinates 1, 2, …., n, then the condition of equilibrium of the structure may be expressed by
the equations.

10. 10 | P a g e

11. 11 | P a g e
Determine the deflection at ① and the reactions on the beam shown in Fig. 15–12a. EI is
constant.

12. 12 | P a g e
SOLUTION: Notation. The beam is divided into two elements and the nodes and members are
identified along with the directions from the near to far ends, Fig. 15–12b. The unknown
deflections are shown in Fig. 15–12c. In particular, notice that a rotational displacement D4 does
not occur because of the roller constraint.
Member Stiffness Matrices. Since EI is constant and the members are of equal length, the
member stiffness matrices are identical. Using the code numbers to identify each column and
row in accordance with Eq. 15–1 and Fig. 15–12b, we have
Displacements and Loads. Assembling the member stiffness matrices into the structure stiffness
matrix, and applying the structure stiffness matrix equation, we have
Q = KD
Solving for the displacements yields
Note that the signs of the results match the directions of the deflections shown in Fig. 15–12c.
Using these results, the reactions therefore are

13. 13 | P a g e
Determine the reactions at the supports of the beam shown in Fig. 15–8a. EI is constant.
SOLUTION: Notation. The beam has two elements and three nodes, which are identified in Fig.
15–8b. The code numbers 1 through 6 are indicated such that the lowest numbers 1–4 identify
the unconstrained degrees of freedom. The known load and displacement matrices are
Member Stiffness Matrices. Each of the two-member stiffness matrices is determined from Eq.
15–1. Note carefully how the code numbers for each column and row are established.
Displacements and Loads. We can now assemble these elements into
the structure stiffness matrix. For example, element K11 = 0 + 2 = 2, K55 = 1.5 + 1.5 = 3, etc. Thus,
Q = KD

14. 14 | P a g e
The matrices are partitioned as shown. Carrying out the multiplication for the first four rows, we
have
Using these results, and multiplying the last two rows, gives
Determine the structure stiffness matrix for the two-member truss shown in Fig. 14–7a. AE is
constant.

15. 15 | P a g e
SOLUTION: By inspection, ② will have two unknown displacement components, whereas joints
① and ③ are constrained from displacement. Consequently, the displacement components at
joint ② are code numbered first, followed by those at joints ③ and ①, Fig. 14–7b. The origin
of the global coordinate system can be located at any point. For convenience, we will choose
joint ② as shown. The members are identified arbitrarily and arrows are written along the two
members to identify the near and far ends of each member. The direction cosines and the
stiffness matrix for each member can now be determined.
Member 1. Since ② is the near end and ③ is the far end, then by Eqs. 14–5 and 14–6, we have
Using Eq. 14–16, dividing each element by we have
The calculations can be checked in part by noting that k1 is symmetric. Note that the rows and
columns in k1 are identified by the x, y degrees of freedom at the near end, followed by the far
end, that is, 1, 2, 3, 4, respectively, for member 1, Fig. 14–7b.This is done in order to identify
the elements for later assembly into the K matrix.
Member 2. Since ② is the near end and ① is the far end, we have
Thus Eq. 14–16 with becomes
Here the rows and columns are identified as 1, 2, 5, 6, since these numbers represent,
respectively, the x, y degrees of freedom at the near and far ends of member 2.
Structure Stiffness Matrix. This matrix has an order of 6 x 6 since there are six designated
degrees of freedom for the truss, Fig. 14–7b. Corresponding elements of the above two matrices
are added algebraically to form the structure stiffness matrix. Perhaps the assembly process is
easier to see if the missing numerical columns and rows in k1 and k2 are expanded with zeros to
form two 6 x 6 matrices. Then K = k1 + k2

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If a computer is used for this operation, generally one starts with K having all zero elements; then
as the member global stiffness matrices are generated, they are placed directly into their
respective element positions in the K matrix, rather than developing the member stiffness
matrices, storing them, then assembling them.
Examples on Force Method for Beams
Example 5.1: Analyze the continuous beam shown in Fig.5.2 (a).
Sol: The total number of reaction components is five, As the conditions of static equilibrium
provide three independent equations, the degree of static indeterminacy of the structure is two.
Here in three alternative solutions with three different released structures are given.

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∆1𝐿=
95000
𝐸𝐼
The minus sign shows that the direction is downward, i.e., in the direction opposite to that of the
coordinate 1. Similarly, the displacement at coordinate 2.
∆2𝐿=
257500
𝐸𝐼
The flexibility matrix with reference to coordinates 1 & 2 may be developed by applying a unit
force successively at coordinates 1 & 2 computing the displacements at these coordinates.
𝛿11 =
1000
3𝐸𝐼
𝛿11 = 𝛿21 =
2500
3𝐸𝐼
⇒ 𝛿22 =
8000
3𝐸𝐼
, Hence, the flexibility matrix[𝛿] is given by the Equation.
[𝛿] = [
1000
3𝐸𝐼
2500
3𝐸𝐼
2500
3𝐸𝐼
8000
3𝐸𝐼
] Substituting into Eq. (5.3), [
𝑃1
𝑃2
] = − [
1000
3𝐸𝐼
2500
3𝐸𝐼
2500
3𝐸𝐼
8000
3𝐸𝐼
]
−1
[
−
95000
𝐸𝐼
−
257500
𝐸𝐼
] = [
199.3
34.3
]
∆1𝐿=
14250
2𝐸𝐼
⇒ ∆2𝐿=
41250
𝐸𝐼
The elements of the flexibility matrix with reference to the chosen coordinates may be computed
by applying a unit force successively at coordinates 1 and 2.
𝛿11 =
20
3𝐸𝐼
⇒ 𝛿12 = 𝛿21 = −
25
𝐸𝐼
⇒ 𝛿22 =
500
3𝐸𝐼
Substituting into Eq. (5.3).

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[
𝑃1
𝑃2
] = − [
20
3𝐸𝐼
25
𝐸𝐼
25
𝐸𝐼
500
3𝐸𝐼
]
−1
[
−
14250
2𝐸𝐼
−
41250
𝐸𝐼
] = [
−321
199.3
]
∆1𝐿=
1500
𝐸𝐼
⇒ ∆2𝐿=
2250
𝐸𝐼
⇒ 𝛿11 =
10
3𝐸𝐼
⇒ 𝛿12 = 𝛿21 = −
5
3𝐸𝐼
⇒ 𝛿22 =
20
3𝐸𝐼
Substituting into Eq. (5.3). [
𝑃1
𝑃2
] = − [
10
3𝐸𝐼
5
3𝐸𝐼
5
3𝐸𝐼
20
3𝐸𝐼
]
−1
[
1500
2𝐸𝐼
2250
𝐸𝐼
] = [
−321
−257
]
Hence, P1 = 321 KN/m and P2 = -257 KN/m. The other reaction components may be calculated by
using the equations of the static equilibrium. It may be checked that the reaction components are
the same as those computed in solution (i).
Example 5.2: Analyze the beam shown in Fig. 5.3(a) if the downward settlements of supports B
and C in KN/m units are
𝟐𝟎𝟎𝟎
𝑬𝑰
𝒂𝒏𝒅
𝟏𝟎𝟎𝟎
𝑬𝑰
respectively.

19. 19 | P a g e
Sol:
∆1𝐿=
95000
𝐸𝐼
⇒ ∆2𝐿=
257500
𝐸𝐼
⇒ 𝛿11 =
1000
3𝐸𝐼
⇒ 𝛿12 = 𝛿21 =
2500
3𝐸𝐼
⇒ 𝛿22 =
8000
3𝐸𝐼
From the given data, ∆1= −
2000
𝐸𝐼
⇒ ∆2= −
1000
𝐸𝐼
Substituting into Eq. (5.2).
[
𝑃1
𝑃2
] = − [
1000
3𝐸𝐼
2500
3𝐸𝐼
2500
3𝐸𝐼
8000
3𝐸𝐼
]
−1
{[
−
2000
𝐸𝐼
−
1000
𝐸𝐼
] − [
−
95000
𝐸𝐼
−
257500
𝐸𝐼
]} = [
176.1
41.1
]
𝛿11 =
10
3𝐸𝐼
⇒ 𝛿12 = 𝛿21 =
5
3𝐸𝐼
⇒ 𝛿22 =
20
3𝐸𝐼
The displacements at the coordinates due to the applied loads and the settlements of supports
are:

20. 20 | P a g e
Examples on Displacement Method for Beams
Example 5.6: Analyze the continuous beam shown in Fig. 5.15(a).
Sol:

21. 21 | P a g e
⇒
⇒
P’1 =300 – 150 = 150 KN-m. ⇒ P’2 = 150 KN-m.
𝑘11 =
4𝐸𝐼
10
+
4𝐸𝐼
10
= 0.8 EI. ⇒ 𝑘21 =
2𝐸𝐼
10
= 0.2 EI.
𝑘12 =
2𝐸𝐼
10
+
4𝐸𝐼
10
= 0.2 EI. ⇒ 𝑘22 =
2𝐸𝐼
10
= 0.4 EI.
As there are no external loads at coordinates 1 & 2: P1 = P2 = 0.
Substituting into Eq. (5.5),
∆1
∆2
= − [
0.8𝐸𝐼 0.2𝐸𝐼
0.2𝐸𝐼 0.4𝐸𝐼
]
−1
[
150
150
] =
1
𝐸𝐼
[
−107.14
−321.43
]
Knowing the displacements, the end moments may be calculated by using the slope deflection
Eq. (2.47).
,
The free body diagram & the bending
moment diagram drawn on the compression side are shown in Fig. 5.15(e) & (f) respectively.
Example 5.7: Analyze the continuous beam shown in Fig. 5.16(a) if the downward settlement of
supports B and C in KN-m units are 2000/EI and 1000/EI respectively.
Sol:

22. 22 | P a g e
M”AB = - 300 KN-m, M”BA = 300 KN-m, M”BC = -150 KN-m, M”CB = 150 KN-m.
The additional fixed end moments due to the settlement of supports are
M”AB =
6𝐸𝐼
102 (
2000
𝐸𝐼
) = 120 KN-m. ⇒ M”BA = =
6𝐸𝐼
102 (
2000
𝐸𝐼
) ⇒ M”BC = =
6𝐸𝐼
102 (
2000
𝐸𝐼
−
1000
𝐸𝐼
) = 60 KN-m.
M”CB = =
6𝐸𝐼
102 (
2000
𝐸𝐼
−
1000
𝐸𝐼
) = 60 KN-m. Therefore, forces P’1 and P’2 at coordinates 1 and 2 for the
fixed end coordinates are: P’1 = 300 + (- 150) + (- 120) + 60 = 90 KN-m. P’2 = 150 + 60 = 210 KN-m.
The elements of the stiffness matrix with reference to coordinates 1 and 2 have been computed
in Ex. 5.6. The stiffness matrix [k] is given by the Equation:
[𝑘] = [
0.8𝐸𝐼 0.2𝐸𝐼
0.2𝐸𝐼 0.4𝐸𝐼
] As there are no external loads at coordinates 1 & 2, P1 = P2 = 0.
Substituting into Eq. (5.5),
∆1
∆2
= − [
0.8𝐸𝐼 0.2𝐸𝐼
0.2𝐸𝐼 0.4𝐸𝐼
]
−1
[
90
210
] = [
21.43
𝐸𝐼
−
535.71
𝐸𝐼
]
Knowing the displacements, the end moments may be calculated by using the slope deflection
Eq. (2.47).

23. 23 | P a g e
The free body diagram & the bending moment diagram drawn on the compression side are
shown in Fig. 5.16(c) & (d) respectively.
Example 5.12: Calculate the flexural stiffness at point D of the three span continuous beam
ABCD shown in Fig. 5.21.
Sol: Fig. 5.21:
[𝑘] = [
1.2𝐸𝐼 0.2𝐸𝐼 0
0.2𝐸𝐼 1.2𝐸𝐼 0.4𝐸𝐼
0 0.4𝐸𝐼 0.8𝐸𝐼
] As there are no external loads at coordinates 1 & 2, P1 = P2 = 0.
Substituting into Eq. (5.5), [
∆1
∆2
∆3
] = [
1.2𝐸𝐼 0.2𝐸𝐼 0
0.2𝐸𝐼 1.2𝐸𝐼 0.4𝐸𝐼
0 0.4𝐸𝐼 0.8𝐸𝐼
]
−1
[
0
0
1
] =
[
−
0.086
𝐸𝐼
0.509
𝐸𝐼
1.509
𝐸𝐼 ]
Thus the rotation at D, ∆3=
1.509
𝐸𝐼
. Hence, the moment required for unit rotation at D =
1
1.509
𝐸𝐼
=
0.663EI. Therefore, the flexural stiffness at D = 0.663EI.
Examples on Force Method for Frames
Example 6.1: Analyze the portal frame shown in Fig. 6.2(a).
Sol:

24. 24 | P a g e
∆1𝐿=
6950
3𝐸𝐼
⇒ ∆2𝐿= −
13900
𝐸𝐼
⇒ ∆3𝐿=
1390
𝐸𝐼
be
[𝛿] =
1
6𝐸𝐼
[
1.2𝐸𝐼 0.2𝐸𝐼 0
0.2𝐸𝐼 1.2𝐸𝐼 0.4𝐸𝐼
0 0.4𝐸𝐼 0.8𝐸𝐼
] AS the supports are unyielding, substituting into Eq. (6.2).
[
𝑃1
𝑃2
𝑃3
] = −6𝐸𝐼 [
750 375 −150
375 2000 −225
−150 −225 60
]
−1
[
−
6950
3𝐸𝐼
13900
𝐸𝐼
1390
𝐸𝐼 ]
= [
−76.3
32.7
−207.1
]

25. 25 | P a g e
Knowing the reactive forces at D, the reactive forces at A, can be calculated by statics. Hence the
free body diagram of the entire frame, as shown in Fig. 6.2(c) may be drawn. Fig. 6.2(d) shows the
bending-moment diagram for the frame drawn on the compression side.
Example 6.2: Analyze the portal frame of Fig. 6.2(a) if the settlements of support D to the right
& downwards in KN-m units are 200/EI and 500/EI respectively.
Sol:
are given in Ex. 6.1. Also, from the data given in this example.
∆1=
200
𝐸𝐼
⇒ ∆2= −
500
𝐸𝐼
⇒ ∆3= 0, Substituting into Eq. (6.1),
[
𝑃1
𝑃2
𝑃3
] = −6𝐸𝐼 [
750 375 −150
375 2000 −225
−150 −225 60
]
−1
×
{
[
200
𝐸𝐼
−
500
𝐸𝐼
0 ]
−
[
−
6950
3𝐸𝐼
13900
𝐸𝐼
1390
𝐸𝐼 ]}
= [
−74.3
30.5
−210.4
]
Knowing the reaction components at D, the reaction components at A can be calculated from
statics. Hence the free body diagram of the entire frame, as shown in Fig. 6.3(a) may be drawn.
Fig. 6.3(b) shows the bending-moment drawn on the compression side.
Example 6.8: Compute the translational stiffness of joint B in the horizontal direction for the rigid
jointed frame shown in Fig. 6.13(a).
Fig. 6.13

26. 26 | P a g e
Sol:
𝑘11 =
12𝐸94𝐼)
103
+
12𝐸𝐼
53
= 0.0144EI. The same result may be obtained by using shear Eq. (6.4) in
which the end moments in the columns may be computed by using Table 2.16. It may be noted
that the translational stiffness of the joint C is the same as that of joint B.
Examples on Displacement Method for Frames
Example 6.15: Analyze the bent shown in Fig. 6.27(a).
Sol:

27. 27 | P a g e
⇒
Similarly, considering member BC as fixed ended, the end moments are:
⇒
Hence, P’1 = M’BA + M’BC = 24 – 80 = - 56 KN-m. P’2 = M’CB = 80 KN-m.
As there at no external forces at coordinates 1 and 2: P1 = P2 = 0.
⇒
⇒ , Hence, the stiffness matrix [k] is given by the
equation [k] = [
2.8𝐸𝐼 𝐸𝐼
𝐸𝐼 2𝐸𝐼
], Substituting into Eq. (6.11),
[
∆1
∆2
] = [
2.8𝐸𝐼 𝐸𝐼
𝐸𝐼 𝐸𝐼
]
−1
{[
0
0
] − [
−56
80
]} = [
41.74
𝐸𝐼
−
60.87
𝐸𝐼
] Knowing the displacements, the end
moments are obtained by using the slope deflection Eq. (2.47).
MAB = - 36 +
2𝐸𝐼
5
(0 +
41.74
𝐸𝐼
) = - 19.3 KN-m. MBA = 24 +
2𝐸𝐼
5
(
2 × 41.74
𝐸𝐼
+ 0) = - 57.4 KN-m.
MBC = - 80 +
2𝐸(2𝐼)
4
(
2 ×41.74
𝐸𝐼
−
60.87
𝐸𝐼
) = - 57.4 KN-m. MCB = 80 + (−
2 ×60.87
𝐸𝐼
+
41.74
𝐸𝐼
) = 0
Example 6.18: Analyze the portal frame of Fig. 6.30(a) if the yielding of support D to the right
and downwards in KN-m units are 200/EI and 500/EI respectively.

28. 28 | P a g e
Sol:
to be
As there are no external loads other than those acting at the coordinates,
P’1 = P’2 = P’3 = P’4 = P’5 = 0, Also from the given data, P1 = 111.2 KN, P2 = P3 = 0,
∆4=
200
𝐸𝐼
, ∆5=
500
𝐸𝐼
. Substituting into Eq. (6.11),

29. 29 | P a g e
Solving Eq. (b), , ,
Substituting these values into Eq. (c), P4 = - 74.25 KN, P5 = 30.44 KN.
Using the slope deflection Eq. (2.47), the moment at support D may be computed.
= - 210.46 KN-m.
From the given data,
P1 = 111.2 KN
P2 = P3 = 0
of
200
𝐸𝐼
towards right at D. , ,

30. 30 | P a g e
,
Hence, the total restraining forces due to the settlement of support are:
P’1 = - 19.2 KN, P’2 = - 120 KN-m, P’3 = - 72 KN. Substituting into Eq. (6.11),
[
∆1
∆2
∆3
] =
1
𝐸𝐼
[
0.144 − 0.240 −0.240
−0.240 3.200 0.800
−0.240 0.800 0.240
]
−1
× {[
111.2
0
0
] − [
− 19.2
− 120
− 72
]} = [
1283.69/𝐸𝐼
102.76/𝐸𝐼
124.08/𝐸𝐼
]
These displacements are the same as obtained in sol (i). Knowing the displacements, the bending
moments may be computed by using the slope deflection Eq. 2.47.
Example 7.1: Develop the flexibility matrix for the pin jointed plane frame with refrence to
coordinates 1 and 2 shown in Fig. 7.1(a). The numbers in parenthesis are the cross sectional
areas of the members in mm2.
Sol:

31. 31 | P a g e
, ,
, Hence, the flexibility matrix [𝛿] is given by the Eq.
.
Example 7.7: Analyze the pin jointed structure shown in Fig. 7.8(a). The cross sectional area of
each member is 2000 mm2. Take E = 200 KN/mm2.
Fig. 7.8
Sol:

32. 32 | P a g e
Substituting from Table 7.7(a) into Eq. (7.1),

33. 33 | P a g e
,
, Substituting from Table 7.7(b) into Eq. (7.2),
,
,
, Substituting into Eq. (7.4)
Knowing forces P1, P2, and P3., the forces in the other members of the structure may be
calculated by adding the forces caused by the applied loads and the redundant as indicated by
the equation: Net force = S + 𝑃1𝑆1 + 𝑃3𝑆3. These forces are listed in column 8 of Table 7.7(b).
Example 1: Analyse the truss shown in the Fig. by method of joint.
Sol: Loading condition is symmetrical, so. Both the reactions are equal, i.e., RA = RB = 1000/2 =
500 kg.

34. 34 | P a g e
Example 2: Analyse the truss shown in the Fig. by section method.

35. 35 | P a g e
• Flexibility method is also known as displacement method. True/False. Right answer is (force
or compatibility method).
• Kinematic indeterminacy is the number of degrees of freedom in a structure. True/False.
• Flexibility is inversely proportional to stiffness. True/False.
• Stiffness method is also known as force method. True/False. Right answer is (displacement or
equilibrium method)
• Stiffness method is based on compatibility equations. True/False.
Example: Determine the stiffness matrix K for the Truss. AE is constant.
Sol: Member 1: 𝜆 𝑥 =
1−0
√2
= 0.7071, 𝜆 𝑦 =
1−2
√2
= - 0.7071.
𝑘1 = 𝐴𝐸 [
0.3536 −0.3536 −0.3536 0.3536
−0.3536 0.3536 0.3536 −0.3536
−0.3536 0.3536 0.3536 −0.3536
0.3536 −0.3536 −0.3536 0.3536
]

36. 36 | P a g e
Member 2: 𝜆 𝑥 =
2−1
√2
= 0.7071, 𝜆 𝑦 =
0−1
√2
= - 0.7071.
𝑘2 = 𝐴𝐸 [
0.3536 −0.3536 −0.3536 0.3536
−0.3536 0.3536 0.3536 −0.3536
−0.3536 0.3536 0.3536 −0.3536
0.3536 −0.3536 −0.3536 0.3536
]
Member 3: 𝜆 𝑥 =
0−2
√2
= - 1, 𝜆 𝑦 = 0.
𝑘3 = 𝐴𝐸 [
0.5 0 −0.5 0
0 0 0 0
−0.5 0 0.5 0
0 0 0 0
]
Member 4: 𝜆 𝑥 =
0−1
√2
= - 0.7071, 𝜆 𝑦 =
0−1
√2
= - 0.7071.
𝑘4 = 𝐴𝐸 [
0.3536 0.3536 −0.3536 −0.3536
0.3536 0.3536 −0.3536 −0.3536
−0.3536 −0.3536 0.3536 0.3536
−0.3536 −0.3536 −0.3536 0.3536
]
Member 5: 𝜆 𝑥 = 0, 𝜆 𝑦 =
0−2
√2
= - 1.
𝑘5 = 𝐴𝐸 [
0 0 0 0
0 0.5 0 −0.5
0 0 0 0
0 −0.5 0 0.5
]
Structure stiffness matrix: K = 𝑘1 + 𝑘2 + 𝑘3 + 𝑘4 + 𝑘5
Example: Determine the stiffness matrix K for truss. Take A = 0.75 in2, E = 29(103) ksi.
Sol:
Member 1: 𝜆 𝑥 =
0−4
√32
= 0.7071, 𝜆 𝑦 =
0−4
√2
= - 0.7071.

37. 37 | P a g e
𝑘1 = 𝐴𝐸 [
0.08839 0.08839 −0.08839 −0.08839
0.08839 0.08839 −0.08839 −0.08839
−0.08839 −0.08839 0.08839 0.08839
−0.08839 −0.08839 0.08839 0.08839
]
Member 2: 𝜆 𝑥 =
4−4
4
= 0, 𝜆 𝑦 =
0−4
4
= - 1.
𝑘2 = 𝐴𝐸 [
0 0 0 0
0 0.25 0 −0.25
0 0 0 0
0 −0.25 0 0.25
]
Member 3: 𝜆 𝑥 =
7−4
5
= 0.6, 𝜆 𝑦 =
0−4
5
- 0.8.
𝑘3 = 𝐴𝐸 [
0.3536 −0.3536 −0.3536 0.3536
−0.3536 0.3536 0.3536 −0.3536
−0.3536 0.3536 0.3536 −0.3536
0.3536 −0.3536 −0.3536 0.3536
]
Structure stiffness matrix: K = 𝑘1 + 𝑘2 + 𝑘3
Example: Determine the reactions on the truss. AE is constant.
Sol: Member 1: 𝜆 𝑥 = 1, 𝜆 𝑦 = 0, 𝜆 𝑥 = 0.7071, 𝜆 𝑦 = 0.7071. L = 1.5 m.
𝑘1 = 𝐴𝐸 [
0.6667 0 −0.47140 −0.47140
0 0 0 −0.47140
−0.47140 0 0.3333 0.3333
−0.47140 0 0.3333 0.3333
]
𝜆 𝑥 = 0, 𝜆 𝑦 = 1. L = 2 m.

38. 38 | P a g e
𝑘5 = 𝐴𝐸 [
0 0 0 0
0 0.5 0 −0.5
0 0 0 0
0 −0.5 0 0.5
]
𝜆 𝑥 = 0.6 𝜆 𝑦 = - 0.8, 𝜆 𝑥 = 0.9899, 𝜆 𝑦 = - 0.14142.
𝑘3 = 𝐴𝐸 [
0.144 −0.192 −0.23759 0.033941
−0.192 0.256 0.31677 −0.045255
−0.23759 0.31678 0.39200 −0.05600
0.033941 −0.045255 −0.05600 0.008000
]
Solving these Equations:

39. 39 | P a g e
𝐷1 =
933.33
𝐴𝐸
⇒ 𝑄4 = 0.033941AE(
933.33
𝐴𝐸
) – 0.045255AE(
133.33
𝐴𝐸
) + 0.27733AE(
247.49
𝐴𝐸
) = 94.3
KN.
𝐷2 =
133.33
𝐴𝐸
⇒ 𝑄5 = - 0.47140AE(
247.49
𝐴𝐸
) = - 117 KN.
𝐷3 =
247.49
𝐴𝐸
⇒ 𝑄6 = - 0.5AE(
133.33
𝐴𝐸
) = - 66.7 KN.
Example: Determine the reactions at the supports A and B. Assume the support at B is a roller.
C is a fixed connected joint.
+ ∑ 𝑀𝐴 = 0: 𝐵𝑦(10) – 70(5) – 10(4) – 15(8) = 0 ⇒ 𝐵𝑦 = 51 k.
+↑ ∑ 𝐹𝑦 = 0; 𝐴 𝑦 + 51 – 70 = 0. ⇒ 𝐴 𝑦 = 19 k.
+← ∑ 𝐹𝑥 = 0; 𝐴 𝑥 - 10 – 15 = 0. ⇒ 𝐴 𝑥 = 25 k.
Example: Determine the reactions at the supports. EI is a constant.
Sol:
Member 1:
𝑘1 = 𝐸𝐼 [
0.096 0.24 −0.096 0.24
0.24 0.8 −0.24 0.4
−0.096 −0.24 0.096 −0.24
0.24 0.4 −0.24 0.8
]
Member 2:
𝑘2 = 𝐸𝐼 [
0.768 0.96 −0.768 0.96
0.96 1.6 −0.96 0.8
−0.768 −0.96 0.768 −0.96
0.96 0.8 −0.96 1.6
]

40. 40 | P a g e
0 = 1.6D1 + 0.8D2 ⇒
25.0
𝐸𝐼
= 0.8D1 + 2.4D2.
D1 =
−6.25
𝐸𝐼
⇒ D2 =
12.5
𝐸𝐼
Q3 = - 0.96EI(
−6.25
𝐸𝐼
) – 0.96EI(
12.5
𝐸𝐼
) = - 6.00 KN.
Q4 – 30 = 0.96EI(
−6.25
𝐸𝐼
) + 0.72EI(
12.5
𝐸𝐼
) = 33 KN.
Q5 – 25 = 0 + 0.4EI(
12.5
𝐸𝐼
) = 30 KN.
Q6 – 30 = 0 + 0.24EI(
12.5
𝐸𝐼
) = 33 KN.
= 0; 30 + 33(5) – 60(2.5) – 6(7.5) = 0
+↑ ∑ 𝐹𝑦 = 0; 33 + 33 – 60 – 6 = 0.
Example: Determine the structure stiffness matrix K for the frame. Take E = 29(103) ksi. I = 650
in4, A = 20 in2 for each member. The joints at are fixed connected.

41. 41 | P a g e
K = k1 + k2
Ans.
Example: Analyse the beam by stiffness method.

42. 42 | P a g e
Sol: Degree of freedom = 2. Step 1: Fixed End Moments:
𝑀 𝐹𝐴𝐵 =
𝑤𝑙
𝛿
=
2×4
𝛿
= - 1 KN-m.
{
𝑀 𝐹𝐵𝐴 =
𝑤𝑙
𝛿
=
2×4
𝛿
= +1 KN − m
𝑀 𝐶𝐵𝐶 = −
𝑤𝑙2
12
=
−1×42
12
= −1.33 KN − m
}
𝑀 𝐹𝐴𝐵 =
𝑤𝑙2
12
=
+1 × 42
12
= + 1.33 KN-m.
Restraining force at 1 & 2: {
𝑃1 = −1 𝐾𝑁 − 𝑚
𝑃2 = 𝑀 𝐹𝐵𝐴 + 𝑀 𝐹𝐵𝐶
𝑃1 = 1.33, 𝑃2 = 0.33 KN-m.
Step 2: Element of stiffness matrix.
[𝑲] = [
𝑲 𝟏𝟏 𝑲 𝟏𝟐
𝑲 𝟐𝟏 𝑲 𝟐𝟐
]
𝟐×𝟐
𝑃1 = 1.33, 𝑃2 = - 0.33 KN-m.
𝐾11 =
2𝐸𝐼
𝐿
(2𝜃𝐴 + 𝜃 𝐵) ⇒ 𝜃 𝐴 = 1, 𝜃 𝐵 = 0.
⇒
2𝐸𝐼
𝐿 𝐴𝐵
(2) or
4𝐸𝐼
𝐿
⇒
4𝐸𝐼
𝐿 𝐴𝐵
=
4𝐸𝐼
4
= 𝐸𝐼.
𝐾12 =
2𝐸𝐼
𝐿 𝐴𝐵
(2𝜃 𝐵 + 𝜃 𝐴) ⇒
2𝐸𝐼
4
× 1 =
2𝐸𝐼
4
=
𝐸𝐼
2
𝐾21 =
2𝐸𝐼
𝐿 𝐴𝐵
(2𝜃𝐴 + 𝜃 𝐵) ⇒
2𝐸𝐼
4
× 1 =
2𝐸𝐼
4
=
𝐸𝐼
2
𝐾22 =
2𝐸𝐼
𝐿 𝐴𝐵
(2𝜃 𝐵 + 𝜃 𝐴) +
2𝐸𝐼
𝐿 𝐵𝐶
(2𝜃 𝐵 + 𝜃 𝐴) =
4𝐸𝐼
4
+
4𝐸𝐼
4
= 2𝐸𝐼
[𝐾] = [
𝐾11 𝐾12
𝐾21 𝐾22
]
[𝐾] = [
𝐸𝐼 𝐸𝐼/2
𝐸𝐼/2 2𝐸𝐼
]
[𝐾] = [
1 1/2
1/2 2
] [𝑃] + [𝐾][𝐷] = 0 ⇒ [𝐷] = [−𝑃][𝐾]
[𝐾]−1
=
𝐴 𝑑 𝑟𝑘
|𝑘|
=
1
𝐸𝑖(1 × 2 −
1
2
×
1
2
[
1 −1/2
−1/2 1
]
[𝐾]−1
=
4
7𝐸𝐼
[
1 −1/2
−1/2 1
]
[𝑃] + [𝐾][𝐷] = 0 ⇒ [𝐷] = [−𝑃][𝐾] ⇒ 𝑃1 = 1.33, 𝑃2 = - 0.33 KN-m.

43. 43 | P a g e
= [
+2
0.33
] [
1 −1/2
−1/2 1
]
4
7𝐸𝐼
⇒
4
7𝐸𝐼
[
1 × 2 + (0.33 × −1/2
1 × −
1
2
+ (0.33 × 1)
] ⇒ [𝐷] =
4
7𝐸𝐼
[
1.835
−0.17
]
[
𝐷1
𝐷2
] ⇒ [
𝐷1 =
4
7𝐸𝐼
× 1.835 =
1.05
𝐸𝐼
𝐷2 =
4
7𝐸𝐼
× (−0.17) = −
0.1
𝐸𝐼
]
Steps 3: Moments
𝑀𝐴𝐵 =
2𝐸𝐼
𝐿
(2𝜃𝐴 + 𝜃 𝐵) + 𝑀 𝐹𝐴𝐵 ⇒ We use 𝜃 𝐴 = 𝐷1 =
1.05
𝐸𝐼
and 𝜃 𝐵 = 𝐷2 =
0.1
𝐸𝐼
=
2𝐸𝐼
4
(2 ×
1.05
𝐸𝐼
+ (−
0.1
𝐸𝐼
)) + (−1)
𝑀𝐴𝐵 = 0 𝐾𝑁 − 𝑚. 𝑀 𝐵𝐴 =
2𝐸𝐼
𝐿
(2𝜃 𝐵 + 𝜃 𝐴) + 𝑀 𝐹𝐴𝐵
=
2𝐸𝐼
4
(2 ×
0.1
𝐸𝐼
+ (−
1.05
𝐸𝐼
)) + 1 ⇒ 𝑀𝐴𝐵 = 1.43 KN-m.
𝑀 𝐵𝐶 =
2𝐸𝐼
𝐿
(2𝜃 𝐵 + 𝜃 𝐶) + 𝑀 𝐹𝐵𝐶
=
2𝐸𝐼
4
(2 ×
−0.1
𝐸𝐼
+ 0) + (−1.33)
𝑀 𝐵𝐶 = −1.43 𝐾𝑁 − 𝑚. 𝑀 𝐶𝐵 =
2𝐸𝐼
𝐿
(2𝜃 𝐶 + 𝜃 𝐵) + 𝑀 𝐹𝐶𝐵
=
2𝐸𝐼
4
(0 + (
−0.1
𝐸𝐼
)) + 1.33 ⇒ 𝑀 𝐶𝐵 = 1.28 KN-m.