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

• determine the kinematic degrees of freedom

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

• determine the fixed end moments due to the applied loads

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

Figure 3 - Member fixed end moments due to applied loads

• calculate stiffness coefficients due to applied unit displacements

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), Figure 4 - Stiffness coefficients due to unit rotation at Joint B For X2 (at Joint C), Figure 5 - Stiffness coefficients due to unit rotation at Joint C

• determine equilibrium equations for each degree of freedom

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;

Solving simultaneously gives;
X1 = 12/EI
X2 = -114/EI

• calculate the member end moments

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

• determine final member end forces

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