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# Chapter 6

By Fred Palmer,2014-09-23 10:24
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Chapter 6

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

of indeterminacy).

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

Note:

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

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.

Obviously, the

primary structure of the

original structure is a

cantilever beam. We

can find all reactions

and internal forces

provided that is X1

computed.

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 X1111P

resultant displacement is superimposed by and B111P

……………………………………(6-2) ！？！！BP111

For the original structure the deflection at B is zero. So we can get

；？0………………………………………(6-3) ！！111P

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:

……………………………………………(6-4) ；？0(X11P11

2

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

analyzed.

(3) Calculate the displacement () that the known load causes in the 1p

primary structure at the point where the redundant restraint has been

released.

(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.

Solution:

(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(X11P11

(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

3112112llllll ？?????；?????？()()l(11EIEIEI222322223212

24qq121lll ？；????？？0()()lP12382296EIEI

(4) Determine the primary unknown X1

34qqlll ; ；？0？？?？；0()(XXX11P111112968EIEI

(5) Draw M diagram M？； MXM11P

2qlqll ？?；？000？；？?？MXMMXMAB1BABAP12816

2ql ？?；？000？？MXMMCB1BCBA16

222qqq3l1122lll ？?？？？？？qqllMDC48832832

? 3. Highly Indeterminate Structures

We have discussed how to solve problems of one degree of indeterminate

structures. Now, well 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).

by,, causes XXX121

deflections , at B, C 1121

respectively, Fig 6-8 (d). () PP12

causes deflections , at B, 1P2P

C respectively, Fig 6-8(e).

According to the original structure

？？0！！BC

In other words

？；；？0?！！！！BP11121?(6-5) ?

?？；；？0！！！！CP21222?

Introduce in Eq (6-5) (ij

；；？0?((XX121P1112? (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?(((，，，！121nP11121n?

?

?；；；；？0，，，！122nP21222n(((?? (6-7) ?

?

?

?

?；；；；？0，，，！12nnPnnnn12??(((

Eq (6-7) in matrix form is:

)?(((，！)?)?11121n11P??????，！(((22??22P2n21????[0]；？…….(6-8) ??????????????，！nnP，?，?(((nnnn12，?

In compact form: [F][X] + [] = [0]………..(6-9) P

From the principle (theorem), we know:

((ijji

Thus the flexibility matrix [F] is a symmetric matrix.

?dsdsNNMMijij?？；(((??ijEIEA?

?

?22?dsds?NMii (6-10) ？；?((ij??(EIEA?

?

?

?NNMMiPiP？；dsds((iP???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

degree.

Primary system is shown (b)

(2) canonical equations

?；；；？0(((，，，！1231P111213?

??；；；？0 ?，，，！1232P212223(((?

?

?；；；？0，，，！1233P313233?(((

(3) Calculate , in the equations (iPij

1122ll？????？？(1)1l(11EIEIEI2363

112l？????？(1)1l22(EIEI233

2

lN3？？ds33?EAEA(

111l？？????？(1)1l1221EIEI236((

？？01331((

3q12112l？???？()qll1PEIEI38224

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.

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