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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.

Three-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 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 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

fixed end forces
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),

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

For X2 (at Joint C),

fixed end forces 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:    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;

Equation in matrix form

Solving simultaneously;
X1 = 344.117/EI
X2 = -344.117/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 = -108 + (2EI/25)* X1 =-108 + (2EI/25)*344.117/EI = -80.47 ft-k
MBA = 72 + (4EI/25)* X1 = 72 + (4EI/25)*344.117/EI = 127.06 ft-k

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

MCD = -72 + (4EI/25)* X2 = -72 + (4EI/25)*-344.117/EI = -127.06 ft-k
MDC = 108 + (2EI/25)* X2 = 108 + (2EI/25)*-344.117/EI = 80.47 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 6 - Member loads and reactions

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

Shear;

final shear diagram
Figure 7 - Final shear diagram

Moment;

final moment diagram
Figure 8 - Final moment diagram


Contact Dr. Fouad Fanous for more information.