Chapter 6 Force Method
? 1. Introduction
The force method, which is also known as the flexibility method or the method of consist deformation is one of the basic methods of analysis of the statically indeterminate structure.
Fig.6-1 shows a frame. It is acted by Force p at point c .How can we determine the reactions of supports A, B?
In the Fig, there are six unknowns in two supports, but there static equilibrium can be created so we cannot solve this problem according to knowledge, which we have studied.
The structure whose reactions or internal forces cannot be determined by statically equilibrium equations is termed statically indeterminate structure.
? 2 Concept and Description
1. Degree of Indeterminacy
The degree of indeterminacy is equal to the number of required
supplementary compatible conditions (on redundant restraints)
In Fig 6-1, we release 3 restraints at support B. The support B is substituted by three forces HB, VB and MB. The structure becomes statically determinate as shown in Fig.6-2. In other words, the structure has three redundant restraints, so its structure is indeterminate to the third degree (or is of three degree
The truss shown in Fig.6-3 ,
Econstructed by a basic triangle ?ABC
and added with three two-units (BD&CD;
DF&CF; DE&EF), two members BC&EC
are redundant. When two diagonals (BC,
EC) are cut, the structure would be
determinate. So the truss is determinate to
the second degree.
Fig 6-4 shows a structure, which is of one degree of indeterminacy. We can release the redundant restraint like (a), (b) but (c) is forbidden, because (c) is an instantaneously unstable structure.
How to determinate the degree of indeterminacy?
1. For trusses, W=2j-b-r
2. For common structures, W=3m-2h-r
3. Cut off the redundant restraints one by one
; Cut off a link—“get rid of one restraint”
; Get rid of one simple hinge—“get rid of two restraints”
; Cut a rigid body—“ get rid of 3 restraints
; Replace a rigid joint with a simple hinge—“get rid of one restraint”
The remaining structure should be statically determinate without any redundant restraint.
2. Primary system and Compatibility condition
Fig 6-6 shows a statically indeterminate structure, a uniform load q impose on the structure. The structure is indeterminate to the first degree (a). As told
above, we may release a redundant
restraint to make it statically
determinate. Support link B is
released and a vertical force is X1
replaced the support.
The statically determinate structure that remains after releasing the
redundant restraints is called the primary structure (Fig6-6b). The indeterminate structure is termed original structure.
primary structure of the
original structure is a
cantilever beam. We
can find all reactions
and internal forces
provided that is X1
Now, let us consider the displacements at point B. The displacement at
point B has two components (due to and due to uniform load). The ！X！1111P
resultant displacement is superimposed by and ！！！B111P
For the original structure the deflection at B is zero. So we can get
This equation is termed deflection compatibility condition. It ensures that the primary system both in static and geometry sense. (6-3) Can be rewritten as:
dsM1Where is deflection at B due to the unit load (x = 1). ？(((?1111EI
(6-4) the basic equation, which is referred to as the canonical equation for a structure indeterminate to the first degree.
dsMM1P are determined by ？(！！1p?1PEI
The positive direction of the primary unknown is chosen arbitrarily X1
but the positive direction of the displacement at the same location must always accord with that of the primary unknown.
3. Calculation steps:
The individual steps in the force method can be summarized as follows:
(1) Determine the degree of indeterminacy of the structure.
(2) Release enough redundant restraints to form a stable and determinate
structure. There is usually more than one way to do this. In all cases the
released structure should be carefully chosen so that it is easy to be
(3) Calculate the displacement () that the known load causes in the ！1p
primary structure at the point where the redundant restraint has been
(4) Calculate the displacement () caused by unit force at the same point (11
in the primary structure.
(5) At the point where a redundant restraint has been released, the sum of the
displacement at the same point in the original structure.
(6) Solve the primary unknown , and then use equilibrium to determine X1
the remaining reactions and internal forces of the original structure.
？；(7) Draw M diagram by MMXM11P
Ex.1 10-1 P190
EX.2 Analyze the indeterminate frame shown in right figure by the force method, draw M diagrams.
(1) Determine the primary structure.
The frame is redundant to the first degree. Remove the horizontal support link at C, the primary system is (b).
(2) Determine the canonical equation in force method.
At point C, the primary system must not have any horizontal displacement the canonical equation in force method is ；？0(X！11P11
(3) Calculate , in the equation. (！1p11
Only the bending deformation of the primary system is considered in the displacement calculation, the , are drawn as respectively. MM1P
Using graph multiplication, we obtain
(4) Determine the primary unknown X1
34qqlll ; ；？0？？?？；0()(X！XX11P111112968EIEI
(5) Draw M diagram M？； MXM11P
? 3. Highly Indeterminate Structures
We have discussed how to solve problems of one degree of indeterminate
structures. Now, we’ll study how to solve problems of structures which is
indeterminate to second or more degrees.
Fig 6-8 (a) shows an
indeterminate structure, which is of
2 degrees of indeterminacy. We can
release two redundant restraints at B,
C, and then it becomes a static
determinate structure (primary
structure) as shown in Fig 6-8 (b).
The released restraints are instead
by,, causes XXX121
deflections , at B, C ！！1121
respectively, Fig 6-8 (d). () PP；12
causes deflections , at B, ！！1P2P
C respectively, Fig 6-8(e).
According to the original structure
In other words
Introduce in Eq (6-5) (ij
；；？0?((XX！121P1112? (6-6) ?
?；；？0，，！122P2122?(( This process can be generalized.
Thus, for a structure, which is subjected to external loads, with n redundant
restraints, we have the canonical equations:
?；；；；？0，，，！122nP21222n(((?? (6-7) ?
Eq (6-7) in matrix form is:
In compact form: [F][X] +  = ………..(6-9) ！P
From the principle (theorem), we know:
Thus the flexibility matrix [F] is a symmetric matrix.
?22?dsds?NMii (6-10) ？；?((ij??(EIEA?
It is evident that from Eq (6-10) we must have > 0 which is located on (ij
the main diagonal of the square matrix [F] and referred to as the main coefficients.
The other coefficients are the secondary coefficients, which may be ()ij?(ij
greater smaller than 0 on even equal to 0.
EX. 10-4 P197
? 4 Analysis of symmetrical structures
When a redundant structure is symmetrical, it is advantageous to facilitate the analysis by conceding the symmetrical behavior of the structure. 1. Behavior of Symmetrical Structure
For a symmetrical structure if the load is also symmetrically distributed, the reactions and the distribution of internal forces must be symmetrical while all the antisymmetrical components are equal zero. On the contrary, when the load is antisymmetrical, the reactions and the distribution of internal forces must be antisymmetrical while all the symmetrical components are equal zero.
EX. Find the end moments of the beam with two fixed ends, of constant EI subjected to a uniform load and draw the bending moment diagram (a). Solution:
(1) Determine primary structure.
The beam is indeterminate to the third
Primary system is shown (b)
(2) canonical equations
(3) Calculate , in the equations (！iPij
The reactions and the distribution of internal forces must be antisymmetrical
while all symmetrical components are zero.
Fig 6-9 (a) shows a symmetrical frame subjected to a symmetrical load.
To illustrate the mentioned behavior, a primary
structure (Fig 6-9 (b)) is chosen.