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Direct Displacement Method
TwoSpan with a Cantilever Indeterminate Beam
problem statement
Using the direct displacement method, determine the final member end forces in the twospan 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.
Figure 1  Beam structure to analyze
The kinematic degrees of freedom are the number of independent joint displacements, in this structure there are four:
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 .
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;
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 MomentsFor a distributed load, the fixed end moments are equal to wL^{2}/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
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
Figure 6  Stiffness coefficients due to unit rotation at
B
Figure 7  Stiffness coefficients due to unit rotation at
C
At each degree of freedom, write the corresponding equilibrium equations:
At Joint B: 96 32+ [(4E*1.5I)/24 + (4E*1.33I)/16]*X_{1}+ (2E*1.33I/16)*X_{2} = 0
At Joint C: 32+ (2E*1.33I/16)*X_{1}+ (4E*1.33I/16)*X_{2} = 12
The equilibrium equation for Joint C is set equal to the applied 12 ftk 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:
X_{1}
= 108/EI
X_{2} = 6/EI
Calculate the Member End Moments:
M_{AB} = 96 + (2E(1.5I)/24)* X_{1} =
96 + (2E(1.5I)/24)*(108/EI) = 109.5 ftk
M_{BA} = 96 + (4E(1.5I)/24)* X_{1} = 96 + (4E(1.5I)/24)*(108/EI ) = 69 ftk
M_{BC} = 32 + (4E(1.33I)/16)* X_{1} + (2E(1.33I)/16)*
X_{2} =
32 + (4E(1.33I)/16)*(108/EI) + (2E(1.33I)/16)*(6/EI) =
69 ftk
M_{CB} = 32 + (2E(1.33I)/16)* X_{1} + (4E(1.33I)/16)*
X_{2} = 32 + (2E(1.33I)/16)*(108/EI) + (4E(1.33I)/16)*(6/EI) =
12 ftk
note: see approach 2 to determine the member end forces.
Approach 2.  apply modified stiffness
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 ftk 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 MomentsFor a distributed load, the fixed end moments are equal to wL^{2}/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
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.
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]* X_{1} = 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:
X_{1} = 108/EI
Calculate the Member End Moments:
M_{AB} = 96 + (2E(3I/2)/24)* X_{1} =
96 + (2E(3I/2)/24)*(108/EI ) = 109.5 ftk
M_{BA} = 96 + (4E(3I/2)/24)* X_{1} = 96 + (4E(3I/2)/24)*(108/EI ) = 69 ftk
M_{BC} = 42 + (3E(4I/3)/16)* X_{1} = 42 + (3E(4I/3)/16)*(108/EI ) = 69 ftk
M_{CB} = 12 ftk
M_{CD} = 12 ftk
Using the results of either approach, apply the end moments and the loads, and determine the member end forces using statics. For example;
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:
Figure 12  Final shear and moment diagram
Contact Dr. Fouad Fanous for more information.