Force Method for the Analysis of Indeterminate StructuresByProf. Dr. Wail Nourildean Al-Rifaie

1. Determine the degree of statically indeterminacy (α).

In the example shown in Fig. 1(a), α = 1.

2. Identify the α redundant forces. In the example, the reaction RB the chosen redundant, as shown in Fig. 1(b).

3. Introduce releases corresponding to the chosen redundant.

In the present example, Fig. 1, support B has been released. The structure thus obtained is called the primary or released structure.

Three conditions have to be satisfied:1. The equilibrium of forces ∑F=0, ∑M=0.

2. The compatibility of displacement conditions.

3. Linear force-displacement Relationship.

1. Fixed end momentsForce method can be used to find fixed-end moments for any loading shown in the following figure both for prismatic and non-prismatic members. Two examples are given below.

Example 1: (a) Prismatic member under point load P. It is required to find fixed-end moments for the beam AB

of uniform section, with a load P applied as shown Figure 2(a).

• The degree of indeterminacy α=2.• Let m1 and m2, the fixed-end moments be the redundant.

• The primary structure is shown in Fig. 2(b).• The mo, m1, and m2 diagrams for the load P, for

m1= 1 and m2= 1 applied to the redundant structure are shown in Fig 2(c).

Following the procedure of the force method, the discontinuities are calculated as follows. Note that for integrating over portions AC and BC, x is taken with respect to A and B.

(b) Non-prismatic member with uniformly distributed load (U.D.L).

Consider the non-prismatic beam shown in Fig. 3(a). It is required to find fixed-end moments at (a) and (b).

The diagrams necessary are given in Fig. (3).

2. Stiffness factor and carry-over factors

(a)Prismatic members. By definition of stiffness factors and carry-over

factors, it is required to find m1 and m2/m1 for the fixed-ended beam shown in Fig. 4 such that these redundant result in unit residual rotation in the direction of the chosen redundant m1, the rotation along m2 being equal to 0. There are no loads and hence ∆10 = ∆20 = 0. From Fig. 4,

(b) Non-prismatic members Consider the non-prismatic member (ab) as

shown in Fig. (4). It is required to find stiffness factor Kab (moment at ‘a’ to cause unit moment at ‘a’), stiffness factor Kba

(moment at ‘b’ to cause unit rotation at ‘b’), C.O.F., Cab from ‘a’ to ‘b’ and C.O.F. Cba from ‘b’ to ‘a’. The following figure gives all the details.