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


    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.

    Obviously, the

    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 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


    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(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).

    The released restraints are instead

    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


    In other words

    ?;;?0?!!!!BP11121?(6-5) ?


    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,,,!122nP21222n(((?? (6-7) ?





    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:


    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


    ??;;;?0 ?,,,!1232P212223(((?



(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.

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