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Direct Displacement Method
Indeterminate Frame
problem statement
Using the direct displacement method, determine the final member end forces in the indeterminate frame below. The modulus of elasticity (E) and the moment of inertia (I) are constant for the entire beam.
Note: The colors of the loads and moments are used to help indicate the contribution of each force to the deflection or rotation being calculated. The moment diagrams show the moments induced by a load using the same color as the load.
Figure 1  Frame structure to analyze
The kinematic degrees of freedom are the number of independent joint displacements, in this structure there are two:
Figure 2  Locations of kinematic degrees of freedom
Restrain all degrees of freedom of the structure. From this restrained structure, determine the fixed end forces due to the applied loads (positive moments are in the clockwise direction).
For reference, refer to the Table of Fixed End MomentsFor a distributed load, the fixed end moments are equal to wL^{2}/12=3x(12)^{2}/12=36
For a point load, located at the center of a span, the fixed end moments are equal to PL/8 = 24x16/8 = 48
Figure 3  Member fixed end moments due to applied loads
Apply a unit displacement in the direction of, and at the same location as each unknown degree of freedom.
In this example, unit rotations are applied separately in the positive clockwise direction at Joints B and C and the stiffness coefficients are determined.
For reference, refer to the Table of Fixed End Moments
For X_{1} (at Joint B),

For X_{2} (at Joint C),

At each degree of freedom, write the corresponding equilibrium equations:
At Joint B: 48  36 + (4EI/16 + 4EI/12)* X_{1} + (2EI/12)* X_{2}= 0
At Joint C: 36 + (2EI/12)* X_{1} + (4EI/12)* X_{2}= 0
or, in matrix form;
Solving simultaneously gives;
X_{1}
= 12/EI
X_{2} =
114/EI
The final member end moments are found by adding a correction moment to the fixed end moments caused by the applied loads. This correction moment is the resulting stiffness coefficient induced by the unit displacement, multiplied by the true displacements.
In this example, the final end moments are as follows;
M_{AB} = 48 + (2EI/16)* X_{1} =48 + (2EI/16)*12/EI = 46.5 ftk
M_{BA} = 48 + (4EI/16)* X_{1} = 48 + (4EI/16)*12/EI = 51 ftk
M_{BC} = 36 + (4EI/12)* X_{1}+ (2EI/12)* X_{2} =
36 + (4EI/12)*12/EI+
(2EI/12)*114/EI = 51 ftk
M_{CB} = 36 + (2EI/12)* X_{1}+ (4EI/12)* X_{2} =
36 + (2EI/12)*12/EI+
(4EI/12)*114/EI = 0 ftk
Similarly, the final member end forces can be calculated by utilizing the final fixed end moments and applied loads on each member.
The final end forces (positive moment in the clockwise direction);
Figure 6  Member loads and reactions
These reactions are used to draw the complete shear and moment diagrams for the structure.
Shear and Moment;
Figure 7  Final shear diagram 
Figure 8  Final moment diagram 
Contact Dr. Fouad Fanous for more information.