Module 7.pdf

University of Engineering & Technology
Peshawar, Pakistan
CE301: Structure Analysis II
Module 07:
Revision of Course ( Flexibility & Stiffness)
By:
Prof. Dr. Bashir Alam
Civil Engineering Department
UET , Peshawar

Topics to be Covered
• Revision of Flexibility & Stiffness method concept
• Analysis of beam using flexibility method ( Problem 1)
• Analysis of frame using flexibility method ( Problem 2)
• Analysis of beam using stiffness method ( Problem 3)
• Analysis of beam using stiffness method ( Problem 4)

Matrix Methods of Analysis
• Flexibility Method
In this method redundant constraints are removed and
corresponding redundant actions (forces or moment) are placed.
An equation of compatibility of deformation is written in terms of
these redundants and corresponding displacements. The
redundants are determined from these simultaneous equation.
Equations of statics are then used for calculation of designed
internal actions. In this method, actions are treated as basic
unknowns. This method will be described in detail in this lecture.

• Flexibility Method
In this method redundant constraints are removed and
corresponding redundant actions (forces or moment) are placed.
An equation of compatibility of deformation is written in terms of
these redundants and corresponding displacements. The
redundants are determined from these simultaneous equation.
Equations of statics are then used for calculation of designed
internal actions. In this method, actions are treated as basic
unknowns.

• Stiffness or Displacement Method
In the displacement method of analysis the equilibrium equations are
written by expressing the unknown joint displacements in terms of
loads by using load displacement relations. The unknown joint
displacements (the degrees of freedom of the structure) are calculated
by solving equilibrium equations. displacement method of analysis is
preferred to computer implementation.

Flexibility Method for Beams Analysis
Problem 01: Analyze the given beam using flexibility method.
S.I = 2 degree
So two redundant actions should be chosen.
Choose those as redundant which makes the calculation handy.
Take EI = constant

• Step # 01: Select the redundant actions
2 ∗ 1 = 1
2
=
?
?
2 ∗ 1 = 1
2
=
0
0

• Step # 02 : Compute the values of [DRL].

• Step # 03 : Compute the values of flexibility matrix [ f ].
i. 1st apply a unit value of AR1 at reference point 1

• Step # 03 : Compute the values of flexibility matrix [ f ].
i. Now apply a unit value of AR2 at reference point 2
= 11 12
21 22
=
16.67 5
5 18.33
EI = 1

Step # 04: Compute the values of redundant actions
AR. As we know
= 1 • −
From this
1
2
=
−55.86
−88.75
1
2
=
16.67 5
5 18.33
0 − 1375
0 − 1906.25

Step # 05: Compute the member end actions. As we know that
a). Compute the AML values
1
2
3
4
5
6
=
5
−5
15
−15
10
−10
Apply actual loadings
and compute the
Reaction forces.

b) Compute the AMR values.
1st apply a unit action at redundant location 1 and then at 2
as shown below.
7 ∗ 2 =
11 12
21 22
31 32
41 42
51 52
61 62
=
0.05 0
−0.05 0
−0.033 0.033
−0.033 0.033
0 −0.04
0 −0.04
1
2
3
4
5
6
=
5
−5
15
−15
10
−10
+
0.05 0
−0.05 0
−0.033 0.033
−0.033 0.033
0 −0.04
0 −0.04
−55.86
−88.75
=
2.206 k
−7.793 k
13.91 k
−16.08 k
13.55 k
−6.44 k

Problem 01: Analyze the given frame using flexibility method.
Flexibility Method for Frames Analysis
S.I = 1 degree
1.5k/ft

• Step # 01: Identify the redundants and obtain BDS also compute
[DRS] values.
= ?
= 0

• Step # 02: Compute M & m values.
m Values

Segment AB EB EC CD
Origin A E E D
Limits 0 to 16 ft 0 to 20 ft 0 to 20 ft 0 to 16 ft
Moment of inertia
, I
I 2I 2I I
M values -300 1.5x2/2 -1.5x/2 -300
m1 values 16-x 0 0 16-x
• Step # 03: Develop Bending moment equation table.

• Step # 04: Compute the values of DRL matrix.
=
1
= 2730.67/
• Step # 05: Compute the values of flexibility coefficients
& flexibility matrix.

= + •
Step # 06: Compute the values of redundant actions AR
Step # 07: Compute the member end actions & draw shear force and
bending moment diagrams
= +

Problem 03: Analyze the given beam using stiffness method if
support B & C both settle down by 0.5in.
K.I = 2 degree ( neglecting the axial effects )
Stiffness Method for Beams Analysis

2 ∗ 1 = 1
2
=
?
?
2 ∗ 1 = 1
2
=
0
0
• Step # 01: Selection of redundant Joint displacements and assign
coordinates at those locations. Also compute AD values.

• Step # 02 : Compute ADL matrix.
1. Direct Loadings
2. Indirect Loadings

• Step # 02 : Compute ADL matrix.
2. Indirect Loadings

Step # 03 : Computation of stiffness coefficients “ S” values

Step # 04: Compute the values of D.
1
2
= 1
2
+ 11 12
21 22
1
2
1
2
= 11 12
21 22
1 − 1
2 − 2
1
2
=
16666.66 4166.66
4166.66 16666.66
0 − (−51)
0 − (51.04)
1
2
=
0.0041
−0.0041

Step # 05: Compute the member end actions. As we know that
= +
AM3

Now Shear force and Bending moment diagram

Problem 03: For the beam shown develop only stiffness coefficient
matrix.
A B C D

1.When D1 = 1 & D2 = D3 = D4 = 0

1.When D2 = 1 & D1 = D3 = D4 = 0

1.When D3 = 1 & D1 = D2 = D4 = 0

1.When D4 = 1 & D1 = D2 = D3 = 0

So stiffness coefficient matrix will be

References
• Structural Analysis by R. C. Hibbeler
• Matrix structural analysis by William Mc Guire
• Matrix analysis of frame structures by William Weaver
• Online Civil Engineering blogs

Module 7.pdf

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