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Direct Displacement Method
Two-Span with a Cantilever Indeterminate Beam


problem statement

Using the direct displacement method, determine the final member end forces in the two-span beam shown. The modulus of elasticity (E) is constant for the entire beam, and 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 with a cantilever 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 four:

kinematic degrees of indeterminacy
Figure 2 - Locations of kinematic degrees of freedom

However, the number of kinematic degrees of freedom can be reduced by considering:

Approach 1. - remove the cantilever
Transfer the load acting at End D with a couple and a vertical force acting at Joint C .

Case 1 - overview of structure
Figure 3 - Structure with cantilever replaced by equivalent moment and vertical force

By replacing the cantilever end with a force and a couple, the structure's degree of kinematic of freedom will be reduced to two;

Case 1 - kinematic degrees of freedom
Figure 4 - Kinematic degree of freedom for new structure

Determine the Fixed End Moments due to the applied loads on the new structure:

For reference, refer to the Table of Fixed End Moments

For a distributed load, the fixed end moments are equal to wL2/12=2x(24)2/12=96

For a point load, located at the center of a span, the fixed end moments are equal to PL/8 = 16x16/8 = 32

Case 1 - fixed end moments
Figure 5 - Member fixed end moments due to the applied loads

Find the stiffness coefficients due to an applied unit displacement at each degree of freedom:

For reference, refer to the Table of Fixed End Moments

Case 1 - Moment due to unit displacement at x1
Figure 6 - Stiffness coefficients due to unit rotation at B

Case 1 - Moment due to unit displacement at x2
Figure 7 - Stiffness coefficients due to unit rotation at C

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

At Joint B96- 32+ [(4E*1.5I)/24 + (4E*1.33I)/16]*X1+ (2E*1.33I/16)*X2 = 0

At Joint C: 32+ (2E*1.33I/16)*X1+ (4E*1.33I/16)*X2 = 12

The equilibrium equation for Joint C is set equal to the applied 12 ft-k moment at the joint. This is to account for the moment due to the cantilever at end CD.

Determine the unknown joint rotations by simultaneously solving the equilibrium equations:

X1 = -108/EI
X2 = -6/EI

Calculate the Member End Moments:

MAB = -96 + (2E(1.5I)/24)* X1 = -96 + (2E(1.5I)/24)*(-108/EI) = -109.5 ft-k
MBA = 96 + (4E(1.5I)/24)* X1 = 96 + (4E(1.5I)/24)*(-108/EI ) = 69 ft-k

MBC = -32 + (4E(1.33I)/16)* X1 + (2E(1.33I)/16)* X2 = -32 + (4E(1.33I)/16)*(-108/EI) + (2E(1.33I)/16)*(-6/EI) = -69 ft-k
MCB = 32 + (2E(1.33I)/16)* X1 + (4E(1.33I)/16)* X2 = 32 + (2E(1.33I)/16)*(-108/EI) + (4E(1.33I)/16)*(-6/EI) = 12 ft-k

note: see approach 2 to determine the member end forces.


Approach 2. - apply modified stiffness

kinematic degrees of indeterminacy
Figure 8- Locations of kinematic degrees of freedom

Consider the rotation at Joint B as the only unknown degree of freedom. Since the resulting moment at Joint C is known to equal 12 ft-k due to the cantilever (positive moment in the clockwise direction), modified stiffness for member BC can be used (Modified Stiffness is also illustrated in the Introduction to the Direct Displacement Method).

This approach is described in detail below.

Determine the Fixed End Moments due to the applied loads on the structure:

For reference, refer to the Table of Fixed End Moments

For a distributed load, the fixed end moments are equal to wL2/12=2x(24)2/12=96

For a point load, located at the center of a span, the fixed end moments are equal to PL/8 = 16x16/8 = 32

For the cantilever end, the fixed end moment is equal to the moment generated by the point load, PL = 3x4 = 12

fixed end forces
Figure 9 - Member fixed end moments due to applied loads

Calculate the final Fixed End Moment at Joint B in member BC:

Since modified stiffness is being used at Joint C, the moments at Joint C must be balanced. The moments at Joint B remain unbalanced as the reaction at Joint B will be determined by the Direct Displacement Method.

  A

B

B

C

C
Distribution Factor  

 

 

1

 0
FEM -96

96

-32

32

-12

Distribute

     

-20

 

Carry Over

    -10    
Final FEM -96

96

-42

12

-12


Find the stiffness coefficients due to an applied unit displacement at Joint B:

Apply a unit displacement at the location and in the direction of the unknown degree of freedom.

Notice that the structure is shown with a roller at Joint C. The contribution of the cantilever to the final reaction at Joint B was previously determined by balancing the moments at Joint C, refer to the Table of Fixed End Moments.

fixed end forces due to unit displacement
Figure 10- Stiffness coefficients due to unit rotation at B

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

The equilibrium equation is determined by summing the fixed end forces induced by the member loads plus the moments caused by the unknown displacement.

At Joint B:    96 - 42 + [4E(1.5I)/24 + 3E(1.33I)/16]* X1 = 0

The equilibrium equation is set equal to zero because there are no applied moments at the joint.

Determine the unknown joint rotation by solving for the unknown:

X1 = -108/EI

Calculate the Member End Moments:

MAB = -96 + (2E(3I/2)/24)* X1 = -96 + (2E(3I/2)/24)*(-108/EI ) = -109.5 ft-k
MBA = 96 + (4E(3I/2)/24)* X1 = 96 + (4E(3I/2)/24)*(-108/EI ) = 69 ft-k

MBC = -42 + (3E(4I/3)/16)* X1 = -42 + (3E(4I/3)/16)*(-108/EI ) = -69 ft-k
MCB = 12 ft-k

MCD = -12 ft-k


Using the results of either approach, apply the end moments and the loads, and determine the member end forces using statics. For example;

final reactions
Figure 11 - Member loads and reactions

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

Shear and Moment diagrams:

final shear and moment diagram
Figure 12 - Final shear and moment diagram


Contact Dr. Fouad Fanous for more information.