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Analysis of Statically indeterminate

Beams and Rigid frames


A. Force/Action Method:

-         methods of analyzing SIS have used Forces (reactions or Internal Forces) as the Basic unknown:

o       consistent deformation

o       method of least work


B. Displacement Method: displacement as the basic unknowns.

          1. Slope-Deflection method:



                   Member ‘ab’ isolated from a loaded SIS beam/frames:


Where: ya yb are the End rotations

            r = the relative deflection between the ends

            mab­­1 Mb2  = induced end moments


Sign Convention:

            M. 1. moment acting on the end of a member (not joint) is positive when clockwise

             y 2. Rotation: @ the end of a member is positive when the tangent to the deformed    

                      curve @ the end Rotates clockwise from its Original position:

r    3. the relative deflection between ends of a member is Positive, when it corresponds to a clockwise rotation of the member


In Fig. The end Moments mab and Mba may be considered as the Algebraic sum of Four separate effect

1.      the moment due to end rotation ya. while the other end b is fixed

2.      the moment due to end rotation yb, while end a is Fixed

3.      the moment due to relative deflection r between the ends of the member without altering the existing slopes of tangents @ the ends.

4.      the moment caused by placing the actual load on the span without altering the end distortions


Ma = 0] : [m’ab/EI] [L] [L/3] – [m’ba/EI] [L] [2/3L] = 0 1

Mb = 0]:  ya(L) – [m’ab/EI] [L] [2/3L] + [m’ba/ei] [L] [L/3] = 0 2


Eq.1 m’ab = 2M’ba or M’ba = M’ab --- subst to eq 2


            * M’ab = 4EI y a/L               *m’ba = 2Ei y a/L


Ma = 0:  [M”ab/EI] [L] [2/3L] = 0




b= 0]: [M”ab/EI] [L] [2/3L] = i/2 [m”ba/EI] [L] [L/3]

            2M”ab = M”ba

            M’ab = M”ba subst:


[M”ab/EI] [L] [L/3] - y b (L) – [M”ab/EI] [L] [2/3] = 0

M”ab/EI [L/6 – 2L/3] = y b



[M/EI] [L] [L/3] + r - [M/EI] [L] [2/3L] = 0

ML­­­­­­­­­­2/6EI + r - ML2/6EI (2) = 0


ML2/6EI = r        : M = 6EIr / L2


               * M”’ab = M”’ba = - M = - 6EIr /L2






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