Direct Displacement Method | Index of Examples | CCE Homepage

Direct Displacement Method
Two-Span Indeterminate Beam

problem statement

Using the direct displacement method, determine the final member end forces in the two-span indeterminate beam below. The modulus of elasticity (E) is constant for the entire beam, the moment of inertia (I) for each member varies as indicated in the figure.

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.

Two-Span indeterminate beam
Figure 1 - Beam structure to analyze

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

kinematic degrees of

Figure 2 - Location of kinematic degree of freedom

Restrain all degrees of freedom of the structure. From this restrained structure, determine the fixed end moments due to the applied loads (positive moments are in the clockwise direction).

For reference, refer to the Table of Fixed End Moments

For a point load, located at the center of a span, the fixed end moments are equal to PL/8 = 16x18/8 = 36

For a distributed load, the fixed end moments are equal to wL2/12=3x(24)2/12=144

fixed end forces
Figure 3 - Member fixed end moments due to applied loads

Apply a unit displacement in the direction of, and in the same location as each unknown degree of freedom.

In this example, the restraint is released at Joint B and a unit displacement is applied in the positive clockwise direction.

For reference, refer to the Table of Fixed End Moments

For X1 (at Joint B),
stiffness coefficients due to unit displacement
Figure 4 - Stiffness coefficients due to unit rotation at Joint B

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

At Joint B:    36 - 144 + (4EI/18+ 4E(1.5I)/24)* X1 = 0

X1 = 228.706/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 = -36 + (2EI/18)* X1 = -36 + (2EI/18)*228.706/EI = -10.588 ft-k
MBA = 36 + (4EI/18)* X1 = 36 + (4EI/18)*228.706/EI = 86.824 ft-k

MBC = -144 + (4E(1.5I)/24)* X1 = -144 + (4E(1.5I)/24)*228.706/EI = -86.824 ft-k
MCB = 144 + (2E(1.5I)/24)* X1 = 144 + (2E(1.5I)/24)*228.706/EI = 172.589 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 is in the clockwise direction);

final reactions
Figure 5 - Member loads and reactions

These reactions can now be used to draw the complete shear and moment diagrams for the structure.


final shear diagram
Figure 6 - Final shear diagram


final moment diagram
Figure 7 - Final moment diagram

Contact Dr. Fouad Fanous for more information.