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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.

Indeterminate frame
Figure 1 - Frame structure to analyze

The kinematic degrees of freedom are the number of independent joint displacements, in this structure there are two:

kinematic degrees of indeterminacy
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 Moments

For a distributed load, the fixed end moments are equal to wL2/12=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

fixed end moments on member AB  fixed end forces on member BC
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 X1 (at Joint B),

Stiffness coefficients due to unit displacement at joint B
Figure 4 - Stiffness coefficients due to unit rotation at Joint B

 For X2 (at Joint C),

Stiffness coefficients due to unit displacement at joint C
Figure 5 - Stiffness coefficients due to unit rotation at Joint C

At each degree of freedom, write the corresponding equilibrium equations:

At Joint B:    48 - 36 + (4EI/16 + 4EI/12)* X1 + (2EI/12)* X2= 0

At Joint C:    36 + (2EI/12)* X1 + (4EI/12)* X2= 0

or, in matrix form;

Equation in matrix form

Solving simultaneously gives;
X1 = 12/EI
 X2 = -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;

MAB = -48 + (2EI/16)* X1 =-48 + (2EI/16)*12/EI = -46.5 ft-k
MBA = 48 + (4EI/16)* X1 = 48 + (4EI/16)*12/EI = 51 ft-k

MBC = -36 + (4EI/12)* X1+ (2EI/12)* X2 = -36 + (4EI/12)*12/EI+ (2EI/12)*-114/EI = -51 ft-k
MCB = 36 + (2EI/12)* X1+ (4EI/12)* X2 = 36 + (2EI/12)*12/EI+ (4EI/12)*-114/EI = 0 ft-k

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);

final reactions Final Moments
Figure 6 - Member loads and reactions

These reactions are used to draw the complete shear and moment diagrams for the structure.

Shear and Moment;

final shear diagram
Figure 7 - Final shear diagram 		final moment diagram
Figure 8 - Final moment diagram

Contact Dr. Fouad Fanous for more information.