M2l7

Parthi Boss
Parthi BossResearch Scholar à student

this is useful for structural students

Module
                2
Analysis of Statically
      Indeterminate
  Structures by the
Matrix Force Method
             Version 2 CE IIT, Kharagpur
Lesson
               7
The Force Method of
           Analysis:
    An Introduction
            Version 2 CE IIT, Kharagpur
Since twentieth century, indeterminate structures are being widely used for its
obvious merits. It may be recalled that, in the case of indeterminate structures
either the reactions or the internal forces cannot be determined from equations of
statics alone. In such structures, the number of reactions or the number of
internal forces exceeds the number of static equilibrium equations. In addition to
equilibrium equations, compatibility equations are used to evaluate the unknown
reactions and internal forces in statically indeterminate structure. In the analysis
of indeterminate structure it is necessary to satisfy the equilibrium equations
(implying that the structure is in equilibrium) compatibility equations (requirement
if for assuring the continuity of the structure without any breaks) and force
displacement equations (the way in which displacement are related to forces).
We have two distinct method of analysis for statically indeterminate structure
depending upon how the above equations are satisfied:

1. Force method of analysis (also known as flexibility method of analysis,
method of consistent deformation, flexibility matrix method)

2. Displacement method of analysis (also known as stiffness matrix method).

In the force method of analysis, primary unknown are forces. In this method
compatibility equations are written for displacement and rotations (which are
calculated by force displacement equations). Solving these equations, redundant
forces are calculated. Once the redundant forces are calculated, the remaining
reactions are evaluated by equations of equilibrium.

In the displacement method of analysis, the primary unknowns are the
displacements. In this method, first force -displacement relations are computed
and subsequently equations are written satisfying the equilibrium conditions of
the structure. After determining the unknown displacements, the other forces are
calculated satisfying the compatibility conditions and force displacement
relations. The displacement-based method is amenable to computer
programming and hence the method is being widely used in the modern day
structural analysis.
In general, the maximum deflection and the maximum stresses are small as
compared to statically determinate structure. For example, consider two beams
of identical cross section and span carrying uniformly distributed load as shown
in Fig. 7.1a and Fig. 7.1b.




                                                        Version 2 CE IIT, Kharagpur
The loads are also the same in both cases. In the first case, the beam is fixed at
both ends and thus is statically indeterminate. The simply supported beam in Fig.
7.1b is a statically determinate structure. The maximum bending moment in case
                          wL2                                                      wL2
of fixed- fixed beam is       (which occurs at the supports) as compared to
                          12                                                        8
(at the centre) in case of simply supported beam. Also in the present case, the
                                             ⎛ wL4 ⎞
deflection in the case of fixed- fixed beam ⎜⎜ 384 EI ⎟ is five times smaller than that
                                                      ⎟
                                             ⎝        ⎠
                         ⎛ 5wL4 ⎞
of simply supported beam ⎜
                         ⎜ 384 EI ⎟ . Also, there is redistribution of stresses in the
                                  ⎟
                         ⎝        ⎠
case of redundant structure. Hence if one member fails, structure does not
collapse suddenly. The remaining members carry the load. The determinate
structural system collapses if one member fails. However, there are
disadvantages in using indeterminate structures. Due to support settlement,

                                                          Version 2 CE IIT, Kharagpur
there will be additional stresses in the case of redundant structures where as
determinate structures are not affected by support settlement.
The analysis of indeterminate structure differs mainly in two aspects as
compared to determinate structure.
a) To evaluate stresses in indeterminate structures, apart from sectional
properties (area of cross section and moment of inertia), elastic properties are
also required.
b) Stresses are developed in indeterminate structure due to support settlements,
temperature change and fabrication errors etc.



Instructional Objectives
After reading this chapter the student will be
1. Able to analyse statically indeterminate structure of degree one.
2. Able to solve the problem by either treating reaction or moment as redundant.
3. Able to draw shear force and bending moment diagram for statically
indeterminate beams.
4. Able to state advantages and limitations of force method of analysis.



7.1 Introduction.
 In this lesson, a general introduction is given to the force method of analysis of
indeterminate structure is given. In the next lesson, this method would be applied
to statically indeterminate beams. Initially the method is introduced with the help
of a simple problem and subsequently it is discussed in detail. The flexibility
method of analysis or force method of analysis (or method of consistent
deformation) was originally developed by J. Maxwell in 1864 and O. C. Mohr in
1874. Since flexibility method requires deflection of statically determinate
structure, a table of formulas for deflections for various load cases and boundary
conditions is also given in this lesson for ready use. The force method of analysis
is not convenient for computer programming as the choice of redundant is not
unique. Further, the bandwidth of the flexibility matrix in the force method is much
larger than the stiffness method. However it is very useful for hand computation.



7.2 Simple Example
Consider a propped cantilever beam (of constant flexural rigidity EI , and span
 L ), which is carrying uniformly distributed load of w kN/m., as shown in Fig.
7.2a. The beam is statically indeterminate i.e. its reaction cannot be evaluated
from equations of statics alone. To solve the above problem by force method
proceeds as follows.

                                                        Version 2 CE IIT, Kharagpur
1) Determine the degree of statical indeterminacy. In the present case it is one.
Identify the reaction, which can be treated as redundant in the analysis. In the
present case RB or M A can be treated as redundant. Selecting RB as the
redundant, the procedure is illustrated. Subsequently, it will be shown how to
attack the problem by treating M A as redundant.




                                                      Version 2 CE IIT, Kharagpur
Version 2 CE IIT, Kharagpur
Solution with RB as the redundant

2) After selecting RB as redundant, express all other reactions in terms of the

redundant RB . This can be accomplished with the help of equilibrium equations.
Thus,

                                    R A = wL − RB                               (7.1a)

and
                                                 wL2
                                    MA =             − RB L                     (7.1b)
                                                  2

3) Now release the restraint corresponding to redundant reaction RB . Releasing
restraint in the present case amounts to removing the support at B . Now on the
resulting cantilever beam (please note that the released structure is statically
determinate structure), apply uniformly distributed load w and the redundant
reaction RB as shown in Fig. 7.2b. The released structure with the external loads
is also sometimes referred as the primary structure.

4) The deflection at B of the released structure (cantilever beam, in the present
case) due to uniformly distributed load and due to redundant reaction RB could
be easily computed from any one of the known methods (moment area method
or unit load method). However it is easier to compute deflection at B due to
uniformly distributed load and RB in two steps. First, consider only uniformly
distributed load and evaluate deflection at B , which is denoted by (Δ B )1 as
shown in Fig. 7.2c. Since RB is redundant, calculate the deflection at B due to
unit load at B acting in the direction of RB and is denoted by (Δ B )2 as shown in

In the present case the positive direction of redundant and deflections are
assumed to act upwards. For the present case, (Δ B )1 and (Δ B )2 are given by,

                                             4
                            (Δ B )1 = − wL                                       (7.2a)
                                          8 EI

                                           L3
and                         (Δ B )2 = −                                          (7.2b)
                                          3EI

From the principle of superposition, the deflection at B , (Δ B ) , is the sum of
deflection due to uniformly distributed load (Δ B )1 and deflection R B (Δ B )2 due to
redundant RB . Hence,

                                                              Version 2 CE IIT, Kharagpur
Δ B = (Δ B )1 + RB (Δ B )2                        (7.2c)

5) It is observed that, in the original structure, the deflection at B is zero. Hence
the compatibility equation can be written as,

                      Δ B = (Δ B )1 + R B (Δ B )2 = 0                           (7.3a)


Solving the above equation, the redundant RB can be evaluated as,

                                       (Δ B )1
                              RB = −                                            (7.3b)
                                       (Δ B )2
Substituting values of (Δ B )1 and (Δ B )2 , the value of RB is obtained as,

                                     3wL
                              RB =                                              (7.3d)
                                      8

The displacement at B due to unit load acting at B in the direction of RB is known
as the flexibility coefficient and is denoted in this course by a BB .

 6) Once RB is evaluated, other reaction components can be easily determined
from equations of statics. Thus,

                                      wL2
                              MA =                                              (7.4a)
                                       8

                                            3wL 5wL
                              R A = wL −       =                                (7.4b)
                                             8   8

7) Once the reaction components are determined, the bending moment and
shear force at any cross section of the beam can be easily evaluated from
equations of static equilibrium. For the present case, the bending moment and
shear force diagram are shown in Fig. 7.2e.

Solution with M A as redundant

1) As stated earlier, in the force method the choice of redundant is arbitrary.
Hence, in the above problem instead of RB one could choose M A as the
redundant reaction. In this section the above problem is solved by taking M A as
redundant reaction.



                                                            Version 2 CE IIT, Kharagpur
2) Now release (remove) the restraint corresponding to redundant reaction M A .
This can be done by replacing the fixed support at A by a pin. While releasing
the structure, care must be taken to see that the released structure is stable and
statically determinate.




                                                                               3)

Calculate the slope at A due to external loading and redundant moment M A . This
is done in two steps as shown in Fig. 7.3b and Fig.7.3c. First consider only
uniformly distributed load (see Fig. 7.3b) and compute slope at A , i.e. (θ A )1 from
force displacement relations. Since M A is redundant, calculate the slope at A due
to unit moment acting at A in the direction of M A which is denoted by (θ A )2 as in
Fig. 7.3c. Taking anticlockwise moment and anticlockwise rotations as positive,
the slope at A , due to two different cases may be written as,



                                                         Version 2 CE IIT, Kharagpur
wL3
                             (θ A )1 = −                                              (7.5a)
                                            24 EI

                             (θ A )2   =
                                            L
                                                                                      (7.5b)
                                           3EI

From the principle of superposition, the slope at A , θ A is the sum of slopes
(θ A )1 due to external load and M A (θ A )2 due to redundant moment M A . Hence
                             M A = (θ A )1 + M A (θ A )2                              (7.5c)

4) From the geometry of the original structure, it is seen that the slope at A is
zero. Hence the required compatibility equation or geometric condition may be
written as,

                             (θ A ) = (θ A )1 + M A (θ A )2 = 0                       (7.5d)

Solving for M A ,

                                           (θ A )1
                             MA = −                                                   (7.5e)
                                           (θ A )2
Substituting the values of (θ A )1 , and (θ A )2 in equation (7.5e), the value of M A is
calculated as



                                           − wL3
                             MA = −                  24 EI
                                              L
                                                  3EI

                                                     wL2
                                           MA =                                       (7.5f)
                                                      8


5) Now other reaction components can be evaluated using equilibrium equations.
Thus,

                                                   5wL
                                           RA =                                       (7.6a)
                                                    8

                                                   3wL
                                           RB =                                       (7.6b)
                                                    8


                                                                  Version 2 CE IIT, Kharagpur
7.3 Summary
The force method of analysis may be summarized as follows.

Step 1. Determine the degree of statical indeterminacy of the structure. Identify
the redundants that would be treated as unknowns in the analysis. Now, release
the redundants one by one so that a statically determinate structure is obtained.
Releasing the redundant reactions means removing constraint corresponding to
that redundant reaction. As in the above propped cantilever beam, either
reactions RB or M A can be treated as unknown redundant. By choosing RB as
the redundant, the propped cantilever beam can be converted into a cantilever
beam (statically determinate) by releasing the roller support. Similarly by
choosing moment as the redundant reaction, the indeterminate structure can be
released into a determinate structure (i.e. a simply supported beam) by turning
the fixed support into a hinged one. If the redundant force is an internal one, then
releasing the structure amounts to introducing discontinuity in the corresponding
member. The compatibility conditions for the redundant internal forces are the
continuity conditions. That would be discussed further in subsequent lessons.

Step 2. In this step, calculate deflection corresponding to redundant action,
separately due to applied loading and redundant forces from force displacement
relations. Deflection due to redundant force cannot be evaluated without knowing
the magnitude of the redundant force. Hence, apply a unit load in the direction of
redundant force and determine the corresponding deflection. Since the method of
superposition is valid, the deflections due to redundant force can be obtained by
simply multiplying the unknown redundant with the deflection obtained from
applying unit value of force.

Step 3. Now, calculate the total deflection due to applied loading and the
redundant force by applying the principle of superposition. This computed total
deflection along the redundant action must be compatible with the actual
boundary conditions of the original structure. For example, if in the original
structure, the deflection corresponding to the redundant reaction is zero then the
total deflection must be equal to zero. If there is more than one redundant force
then one could construct a set of equations with redundant forces as unknowns
and flexibility coefficients as coefficients of the equations. The total number of
equations equals the number of unknown redundants.

Step 4. In the last step, evaluate all other reactions and internal forces from the
equilibrium equations.

The method of superposition or the force method as discussed above is applied
to any type of structures, i.e. beams, truss and frames or combination of these
structures. It is applicable for all general type of loadings.



                                                        Version 2 CE IIT, Kharagpur
The deflection of statically determinate structure can be obtained by unit-load
method or by moment-area theorem or by any method known to the reader.
However, the deflections of few prismatic beams with different boundary
conditions and subjected to simple loadings are given in Fig. 7.4. These values
will be of help in solving the problems of the present and subsequent lessons.
However the students are strongly advised to practice deriving them instead of
simply memorizing them.




                                                     Version 2 CE IIT, Kharagpur
Example 7.1
A continuous beam ABC is carrying a uniformly distributed load of 1 kN/m in
addition to a concentrated load of 10 kN as shown in Fig.7.5a, Draw bending
moment and shear force diagram. Assume EI to be constant for all members.




                                                  Version 2 CE IIT, Kharagpur
It is observed that the continuous beam is statically indeterminate to first degree.
Choose the reaction at B, RBy as the redundant. The primary structure is a
simply supported beam as shown in Fig.7.5b.
Now, compute the deflection at B, in the released structure due to uniformly
distributed load and concentrated load. This is accomplished by unit load
method. Thus,

                            −2083.33 1145.84
                     ΔL =           −
                               EI      EI
                            −3229.17
                     ΔL =                                                         (1)
                               EI

In the next step, apply a unit load at B in the direction of      RBy (upwards) and
calculate the deflection at B of the following structure. Thus (see Fig. 7.5c),

                            L3     166.67
                     a11 =       =                                                (2)
                           48 EI     EI

Now, deflection at B in the primary structure due to redundant RB is,

                                    166.67
                             ΔB =          × RB                                   (3)
                                      EI
In the actual structure, the deflection at B is zero. Hence, the compatibility
equation may be written as

                                                         Version 2 CE IIT, Kharagpur
ΔL + ΔB = 0                                   (4)

Substituting for Δ L and Δ B in equation (4),

                      −3229.17 166.67
                              +       RB = 0                                (5)
                         EI      EI
Thus,
                      RB = 19.375 kN

The other two reactions are calculated by static equilibrium equations (vide Fig.
7.5d)


                      RA = 7.8125 kN
                      RB = 2.8125 kN

The shear force and bending moment diagrams are shown in Fig. 7.5e and Fig.
7.5f respectively.

Example 7.2
A propped cantilever beam AB is subjected to a concentrated load of 60 kN at
3m from end A as shown in Fig. 7.6a. Draw the bending moment and shear
force diagrams by the force method. Assume that the flexural rigidity of the
beam, EI to be constant throughout.




                                                      Version 2 CE IIT, Kharagpur
Version 2 CE IIT, Kharagpur
The given problem is statically indeterminate to first degree. Choose the reaction
at B , R1 as the redundant. After releasing the redundant, the determinate
structure, a cantilever beam in this case is obtained. The cantilever beam with
the applied loading is chosen in Fig 7.6b.

The deflection of the released structure is,

                                    60 × 33 60 × 32 × 6
                     ( Δ L )1 = −          −
                                     3EI       2 EI

                                           −2160
                              ( Δ L )1 =                                         (1)
                                            EI

The deflection at point B due to unit load applied in the direction of redundant
R1 is (vide Fig 7.6c)


                                                          Version 2 CE IIT, Kharagpur
93   243
                            a11 =       =                                      (2)
                                    3EI   EI

Now the deflection at B due to redundant R1 is

                           (Δ )1 = 243R1                                       (3)
                                    EI

From the original structure it is seen that the deflection at B is zero. Hence, the
compatibility condition for the problem may be written as,

                                2160 243R1
                            −       +      =0                                  (4)
                                 EI   EI

Solving equation (4), the redundant R1 is obtained.

                                         2160
                                     R1 =
                                          243                                  (5)
                                       = 8.89 kN

The vertical reaction and fixed end moment at A can be determined from
equations of statics. Thus,

                                     R2 = 51.11 kN

                                     R3 = 99.99 kN.m                           (6)

Shear force and bending moment diagrams are shown in Fig. 7.6d and Fig. 7.6e
respectively.


Summary
In this lesson flexibility matrix method or the method of consistent deformation or
the force method of analysing statically indeterminate structures has been
introduced with the help of simple problems. The advantages and limitations of
flexibility matrix method have been discussed. Only simple indeterminate beam
problem has been solved to illustrate the procedure. The principle of
superposition has been used to solve statically indeterminate problems.




                                                       Version 2 CE IIT, Kharagpur

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M2l7

  • 1. Module 2 Analysis of Statically Indeterminate Structures by the Matrix Force Method Version 2 CE IIT, Kharagpur
  • 2. Lesson 7 The Force Method of Analysis: An Introduction Version 2 CE IIT, Kharagpur
  • 3. Since twentieth century, indeterminate structures are being widely used for its obvious merits. It may be recalled that, in the case of indeterminate structures either the reactions or the internal forces cannot be determined from equations of statics alone. In such structures, the number of reactions or the number of internal forces exceeds the number of static equilibrium equations. In addition to equilibrium equations, compatibility equations are used to evaluate the unknown reactions and internal forces in statically indeterminate structure. In the analysis of indeterminate structure it is necessary to satisfy the equilibrium equations (implying that the structure is in equilibrium) compatibility equations (requirement if for assuring the continuity of the structure without any breaks) and force displacement equations (the way in which displacement are related to forces). We have two distinct method of analysis for statically indeterminate structure depending upon how the above equations are satisfied: 1. Force method of analysis (also known as flexibility method of analysis, method of consistent deformation, flexibility matrix method) 2. Displacement method of analysis (also known as stiffness matrix method). In the force method of analysis, primary unknown are forces. In this method compatibility equations are written for displacement and rotations (which are calculated by force displacement equations). Solving these equations, redundant forces are calculated. Once the redundant forces are calculated, the remaining reactions are evaluated by equations of equilibrium. In the displacement method of analysis, the primary unknowns are the displacements. In this method, first force -displacement relations are computed and subsequently equations are written satisfying the equilibrium conditions of the structure. After determining the unknown displacements, the other forces are calculated satisfying the compatibility conditions and force displacement relations. The displacement-based method is amenable to computer programming and hence the method is being widely used in the modern day structural analysis. In general, the maximum deflection and the maximum stresses are small as compared to statically determinate structure. For example, consider two beams of identical cross section and span carrying uniformly distributed load as shown in Fig. 7.1a and Fig. 7.1b. Version 2 CE IIT, Kharagpur
  • 4. The loads are also the same in both cases. In the first case, the beam is fixed at both ends and thus is statically indeterminate. The simply supported beam in Fig. 7.1b is a statically determinate structure. The maximum bending moment in case wL2 wL2 of fixed- fixed beam is (which occurs at the supports) as compared to 12 8 (at the centre) in case of simply supported beam. Also in the present case, the ⎛ wL4 ⎞ deflection in the case of fixed- fixed beam ⎜⎜ 384 EI ⎟ is five times smaller than that ⎟ ⎝ ⎠ ⎛ 5wL4 ⎞ of simply supported beam ⎜ ⎜ 384 EI ⎟ . Also, there is redistribution of stresses in the ⎟ ⎝ ⎠ case of redundant structure. Hence if one member fails, structure does not collapse suddenly. The remaining members carry the load. The determinate structural system collapses if one member fails. However, there are disadvantages in using indeterminate structures. Due to support settlement, Version 2 CE IIT, Kharagpur
  • 5. there will be additional stresses in the case of redundant structures where as determinate structures are not affected by support settlement. The analysis of indeterminate structure differs mainly in two aspects as compared to determinate structure. a) To evaluate stresses in indeterminate structures, apart from sectional properties (area of cross section and moment of inertia), elastic properties are also required. b) Stresses are developed in indeterminate structure due to support settlements, temperature change and fabrication errors etc. Instructional Objectives After reading this chapter the student will be 1. Able to analyse statically indeterminate structure of degree one. 2. Able to solve the problem by either treating reaction or moment as redundant. 3. Able to draw shear force and bending moment diagram for statically indeterminate beams. 4. Able to state advantages and limitations of force method of analysis. 7.1 Introduction. In this lesson, a general introduction is given to the force method of analysis of indeterminate structure is given. In the next lesson, this method would be applied to statically indeterminate beams. Initially the method is introduced with the help of a simple problem and subsequently it is discussed in detail. The flexibility method of analysis or force method of analysis (or method of consistent deformation) was originally developed by J. Maxwell in 1864 and O. C. Mohr in 1874. Since flexibility method requires deflection of statically determinate structure, a table of formulas for deflections for various load cases and boundary conditions is also given in this lesson for ready use. The force method of analysis is not convenient for computer programming as the choice of redundant is not unique. Further, the bandwidth of the flexibility matrix in the force method is much larger than the stiffness method. However it is very useful for hand computation. 7.2 Simple Example Consider a propped cantilever beam (of constant flexural rigidity EI , and span L ), which is carrying uniformly distributed load of w kN/m., as shown in Fig. 7.2a. The beam is statically indeterminate i.e. its reaction cannot be evaluated from equations of statics alone. To solve the above problem by force method proceeds as follows. Version 2 CE IIT, Kharagpur
  • 6. 1) Determine the degree of statical indeterminacy. In the present case it is one. Identify the reaction, which can be treated as redundant in the analysis. In the present case RB or M A can be treated as redundant. Selecting RB as the redundant, the procedure is illustrated. Subsequently, it will be shown how to attack the problem by treating M A as redundant. Version 2 CE IIT, Kharagpur
  • 7. Version 2 CE IIT, Kharagpur
  • 8. Solution with RB as the redundant 2) After selecting RB as redundant, express all other reactions in terms of the redundant RB . This can be accomplished with the help of equilibrium equations. Thus, R A = wL − RB (7.1a) and wL2 MA = − RB L (7.1b) 2 3) Now release the restraint corresponding to redundant reaction RB . Releasing restraint in the present case amounts to removing the support at B . Now on the resulting cantilever beam (please note that the released structure is statically determinate structure), apply uniformly distributed load w and the redundant reaction RB as shown in Fig. 7.2b. The released structure with the external loads is also sometimes referred as the primary structure. 4) The deflection at B of the released structure (cantilever beam, in the present case) due to uniformly distributed load and due to redundant reaction RB could be easily computed from any one of the known methods (moment area method or unit load method). However it is easier to compute deflection at B due to uniformly distributed load and RB in two steps. First, consider only uniformly distributed load and evaluate deflection at B , which is denoted by (Δ B )1 as shown in Fig. 7.2c. Since RB is redundant, calculate the deflection at B due to unit load at B acting in the direction of RB and is denoted by (Δ B )2 as shown in In the present case the positive direction of redundant and deflections are assumed to act upwards. For the present case, (Δ B )1 and (Δ B )2 are given by, 4 (Δ B )1 = − wL (7.2a) 8 EI L3 and (Δ B )2 = − (7.2b) 3EI From the principle of superposition, the deflection at B , (Δ B ) , is the sum of deflection due to uniformly distributed load (Δ B )1 and deflection R B (Δ B )2 due to redundant RB . Hence, Version 2 CE IIT, Kharagpur
  • 9. Δ B = (Δ B )1 + RB (Δ B )2 (7.2c) 5) It is observed that, in the original structure, the deflection at B is zero. Hence the compatibility equation can be written as, Δ B = (Δ B )1 + R B (Δ B )2 = 0 (7.3a) Solving the above equation, the redundant RB can be evaluated as, (Δ B )1 RB = − (7.3b) (Δ B )2 Substituting values of (Δ B )1 and (Δ B )2 , the value of RB is obtained as, 3wL RB = (7.3d) 8 The displacement at B due to unit load acting at B in the direction of RB is known as the flexibility coefficient and is denoted in this course by a BB . 6) Once RB is evaluated, other reaction components can be easily determined from equations of statics. Thus, wL2 MA = (7.4a) 8 3wL 5wL R A = wL − = (7.4b) 8 8 7) Once the reaction components are determined, the bending moment and shear force at any cross section of the beam can be easily evaluated from equations of static equilibrium. For the present case, the bending moment and shear force diagram are shown in Fig. 7.2e. Solution with M A as redundant 1) As stated earlier, in the force method the choice of redundant is arbitrary. Hence, in the above problem instead of RB one could choose M A as the redundant reaction. In this section the above problem is solved by taking M A as redundant reaction. Version 2 CE IIT, Kharagpur
  • 10. 2) Now release (remove) the restraint corresponding to redundant reaction M A . This can be done by replacing the fixed support at A by a pin. While releasing the structure, care must be taken to see that the released structure is stable and statically determinate. 3) Calculate the slope at A due to external loading and redundant moment M A . This is done in two steps as shown in Fig. 7.3b and Fig.7.3c. First consider only uniformly distributed load (see Fig. 7.3b) and compute slope at A , i.e. (θ A )1 from force displacement relations. Since M A is redundant, calculate the slope at A due to unit moment acting at A in the direction of M A which is denoted by (θ A )2 as in Fig. 7.3c. Taking anticlockwise moment and anticlockwise rotations as positive, the slope at A , due to two different cases may be written as, Version 2 CE IIT, Kharagpur
  • 11. wL3 (θ A )1 = − (7.5a) 24 EI (θ A )2 = L (7.5b) 3EI From the principle of superposition, the slope at A , θ A is the sum of slopes (θ A )1 due to external load and M A (θ A )2 due to redundant moment M A . Hence M A = (θ A )1 + M A (θ A )2 (7.5c) 4) From the geometry of the original structure, it is seen that the slope at A is zero. Hence the required compatibility equation or geometric condition may be written as, (θ A ) = (θ A )1 + M A (θ A )2 = 0 (7.5d) Solving for M A , (θ A )1 MA = − (7.5e) (θ A )2 Substituting the values of (θ A )1 , and (θ A )2 in equation (7.5e), the value of M A is calculated as − wL3 MA = − 24 EI L 3EI wL2 MA = (7.5f) 8 5) Now other reaction components can be evaluated using equilibrium equations. Thus, 5wL RA = (7.6a) 8 3wL RB = (7.6b) 8 Version 2 CE IIT, Kharagpur
  • 12. 7.3 Summary The force method of analysis may be summarized as follows. Step 1. Determine the degree of statical indeterminacy of the structure. Identify the redundants that would be treated as unknowns in the analysis. Now, release the redundants one by one so that a statically determinate structure is obtained. Releasing the redundant reactions means removing constraint corresponding to that redundant reaction. As in the above propped cantilever beam, either reactions RB or M A can be treated as unknown redundant. By choosing RB as the redundant, the propped cantilever beam can be converted into a cantilever beam (statically determinate) by releasing the roller support. Similarly by choosing moment as the redundant reaction, the indeterminate structure can be released into a determinate structure (i.e. a simply supported beam) by turning the fixed support into a hinged one. If the redundant force is an internal one, then releasing the structure amounts to introducing discontinuity in the corresponding member. The compatibility conditions for the redundant internal forces are the continuity conditions. That would be discussed further in subsequent lessons. Step 2. In this step, calculate deflection corresponding to redundant action, separately due to applied loading and redundant forces from force displacement relations. Deflection due to redundant force cannot be evaluated without knowing the magnitude of the redundant force. Hence, apply a unit load in the direction of redundant force and determine the corresponding deflection. Since the method of superposition is valid, the deflections due to redundant force can be obtained by simply multiplying the unknown redundant with the deflection obtained from applying unit value of force. Step 3. Now, calculate the total deflection due to applied loading and the redundant force by applying the principle of superposition. This computed total deflection along the redundant action must be compatible with the actual boundary conditions of the original structure. For example, if in the original structure, the deflection corresponding to the redundant reaction is zero then the total deflection must be equal to zero. If there is more than one redundant force then one could construct a set of equations with redundant forces as unknowns and flexibility coefficients as coefficients of the equations. The total number of equations equals the number of unknown redundants. Step 4. In the last step, evaluate all other reactions and internal forces from the equilibrium equations. The method of superposition or the force method as discussed above is applied to any type of structures, i.e. beams, truss and frames or combination of these structures. It is applicable for all general type of loadings. Version 2 CE IIT, Kharagpur
  • 13. The deflection of statically determinate structure can be obtained by unit-load method or by moment-area theorem or by any method known to the reader. However, the deflections of few prismatic beams with different boundary conditions and subjected to simple loadings are given in Fig. 7.4. These values will be of help in solving the problems of the present and subsequent lessons. However the students are strongly advised to practice deriving them instead of simply memorizing them. Version 2 CE IIT, Kharagpur
  • 14. Example 7.1 A continuous beam ABC is carrying a uniformly distributed load of 1 kN/m in addition to a concentrated load of 10 kN as shown in Fig.7.5a, Draw bending moment and shear force diagram. Assume EI to be constant for all members. Version 2 CE IIT, Kharagpur
  • 15. It is observed that the continuous beam is statically indeterminate to first degree. Choose the reaction at B, RBy as the redundant. The primary structure is a simply supported beam as shown in Fig.7.5b. Now, compute the deflection at B, in the released structure due to uniformly distributed load and concentrated load. This is accomplished by unit load method. Thus, −2083.33 1145.84 ΔL = − EI EI −3229.17 ΔL = (1) EI In the next step, apply a unit load at B in the direction of RBy (upwards) and calculate the deflection at B of the following structure. Thus (see Fig. 7.5c), L3 166.67 a11 = = (2) 48 EI EI Now, deflection at B in the primary structure due to redundant RB is, 166.67 ΔB = × RB (3) EI In the actual structure, the deflection at B is zero. Hence, the compatibility equation may be written as Version 2 CE IIT, Kharagpur
  • 16. ΔL + ΔB = 0 (4) Substituting for Δ L and Δ B in equation (4), −3229.17 166.67 + RB = 0 (5) EI EI Thus, RB = 19.375 kN The other two reactions are calculated by static equilibrium equations (vide Fig. 7.5d) RA = 7.8125 kN RB = 2.8125 kN The shear force and bending moment diagrams are shown in Fig. 7.5e and Fig. 7.5f respectively. Example 7.2 A propped cantilever beam AB is subjected to a concentrated load of 60 kN at 3m from end A as shown in Fig. 7.6a. Draw the bending moment and shear force diagrams by the force method. Assume that the flexural rigidity of the beam, EI to be constant throughout. Version 2 CE IIT, Kharagpur
  • 17. Version 2 CE IIT, Kharagpur
  • 18. The given problem is statically indeterminate to first degree. Choose the reaction at B , R1 as the redundant. After releasing the redundant, the determinate structure, a cantilever beam in this case is obtained. The cantilever beam with the applied loading is chosen in Fig 7.6b. The deflection of the released structure is, 60 × 33 60 × 32 × 6 ( Δ L )1 = − − 3EI 2 EI −2160 ( Δ L )1 = (1) EI The deflection at point B due to unit load applied in the direction of redundant R1 is (vide Fig 7.6c) Version 2 CE IIT, Kharagpur
  • 19. 93 243 a11 = = (2) 3EI EI Now the deflection at B due to redundant R1 is (Δ )1 = 243R1 (3) EI From the original structure it is seen that the deflection at B is zero. Hence, the compatibility condition for the problem may be written as, 2160 243R1 − + =0 (4) EI EI Solving equation (4), the redundant R1 is obtained. 2160 R1 = 243 (5) = 8.89 kN The vertical reaction and fixed end moment at A can be determined from equations of statics. Thus, R2 = 51.11 kN R3 = 99.99 kN.m (6) Shear force and bending moment diagrams are shown in Fig. 7.6d and Fig. 7.6e respectively. Summary In this lesson flexibility matrix method or the method of consistent deformation or the force method of analysing statically indeterminate structures has been introduced with the help of simple problems. The advantages and limitations of flexibility matrix method have been discussed. Only simple indeterminate beam problem has been solved to illustrate the procedure. The principle of superposition has been used to solve statically indeterminate problems. Version 2 CE IIT, Kharagpur