1. Force Method for the Analysis of Indeterminate Structures By Prof. Dr. Wail Nourildean Al-Rifaie

2. 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.

4. 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.

8. 1. Fixed end moments Force 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 m 1 and m 2, the fixed-end moments be the redundant.

10. The primary structure is shown in Fig. 2(b). The m o, m 1, and m 2 diagrams for the load P, for m 1 = 1 and m 2 = 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.

14. (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).

18. 2. Stiffness factor and carry-over factors (a)Prismatic members. By definition of stiffness factors and carry-over factors, it is required to find m 1 and m 2 /m 1 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 m 1, the rotation along m 2 being equal to 0. There are no loads and hence ∆ 10 = ∆ 20 = 0. From Fig. 4,

22. (b) Non-prismatic members Consider the non-prismatic member (ab) as shown in Fig. (4). It is required to find stiffness factor K ab (moment at ‘a’ to cause unit moment at ‘a’), stiffness factor K ba (moment at ‘b’ to cause unit rotation at ‘b’), C.O.F., C ab from ‘a’ to ‘b’ and C.O.F. C ba from ‘b’ to ‘a’. The following figure gives all the details.