Direct Displacement Method
Three-Span Indeterminate Beam

problem statement

Using the direct displacement method, determine the final member end forces in the three-span indeterminate beam 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- Beam 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 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 distributed load, the fixed end moments are equal to wL²/12=2x(30)²/12=150

For a point load, located at a distance a from one end and b from the other end of a span, the fixed end moments are equal to Pab²/L² = 30x10x(15)²/(25)² = 108 and Pa²b/L² = 30x(10)²x15/(25)² = 72

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:    72 - 150 + (4EI/25 + 4EI/30)* X₁ + (2EI/30)* X₂= 0

At Joint C:    150 - 72 + (2EI/30)* X₁ + (4EI/30+4EI/25)* X₂= 0

simplified;

-78 + (22EI/75)* X₁ + (EI/15)* X₂= 0
78 + (EI/15)* X₁ + (22EI/75)* X₂= 0

or, in matrix form;

Solving simultaneously;
X₁ = 344.117/EI
X₂ = -344.117/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;

M_AB = -108 + (2EI/25)* X₁ =-108 + (2EI/25)*344.117/EI = -80.47 ft-k
M_BA = 72 + (4EI/25)* X₁ = 72 + (4EI/25)*344.117/EI = 127.06 ft-k

M_BC = -150 + (4EI/30)* X₁+ (2EI/30)* X₂ = -150 + (4EI/30)*344.117/EI+ (2EI/30)*-344.117/EI = -127.06 ft-k
M_CB = 150 + (2EI/30)* X₁+ (4EI/30)* X₂ = 150 + (2EI/30)*344.117/EI+ (4EI/30)*-344.117/EI = 127.06 ft-k

M_CD = -72 + (4EI/25)* X₂ = -72 + (4EI/25)*-344.117/EI = -127.06 ft-k
M_DC = 108 + (2EI/25)* X₂ = 108 + (2EI/25)*-344.117/EI = 80.47 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 is in the clockwise direction);

Figure 6 - Member loads and reactions

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

Shear;

Figure 7 - Final shear diagram

Moment;

Figure 8 - Final moment diagram

Contact Dr. Fouad Fanous for more information.