Solutions
Question 1
This question will be solved step by step.
Step 1: Removal of SD parts
The part CE is an statically determinate (SD) part of the structure. The bending moment diagram can
be drawn immediately for this part. This is the reason to remove this part from further analysis and
replace its influence by the appropriate concentrated force and moment as shown in Figure 5.
Figure 5: Question 1: solution - removing of the SD part CE.
Step 2: The number of unknowns displacements
Having in mind that the axial deformation is neglected in the slope deflection method, we can
conclude that the vertical deflection of the point B is equal to the vertical deflection at the point A
which means that there is no deflection of the point B in the vertical direction. Also, for the same
reason the point C can move neither horizontally (the same as point B) nor in the direction of the
member CD (the same deflection as at point D in the CD direction). Impossibility that point C moves
in two different directions means that this point cannot move in any direction. Therefore, all points in
the structure cannot deflect in any direction. As a consequence, none of the members can rotate.
Therefore, for all of them δ = 0 (Figure 6).
Figure 6: Question 1: solution - unknown displacements.
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As for joint rotations, it should be noticed that rotations at B are different for members BA and BC.
This is because these two members are connected by a hinge. Besides these two unknowns rotations
(θBA and θBC), the other two unknowns are at joints C (θC) and D (θD).
Step 3: The fixed-end moments
The fixed-end moments should be calculated for all members which carry load. However, the
concentrated loads (forces and moments) in joints should not be included. Concentrated moments will
be included into the moment equilibrium equation for the joint while a concentrated force in a joint
which cannot move in any direction does not influence deflection and M diagrams. It influences the
axial forces in members only.
For given structure the fixed-end moments should be found for members AB and BC (Figure 7). Both
members carry the same concentrated force and have the same geometric properties (EI and l) which
means that their fixed-end moments are the same. The moments can be obtained from the table given
together with this tutorial sheet.
M FAB =
Pl
8
M FBA = -
=
Pl
8
50 × 4
8
= 25 kNm
= -25 kNm
M FBC = 25 kNm
F
M CB
= -25 kNm
Figure 7: Question 1: solution - fixed-end moments.
Step 4: Slope deflection equations
MEMBER AB: δ = 0, θ A = 0
M DAB =
M DBA =
2EI
4
2EI
4
θ BA
( 2θ BA )
4
MEMBER BC: δ = 0
M DBC =
D
=
M CB
2EI
4
2EI
4
( 2θBC + θC )
( 2θC + θ BC )
MEMBER CD: δ = 0
D
=
M CD
M DDC =
2EI
5
2EI
5
( 2θ C + θ D )
( 2θ D + θ C )
Step 5: Total moments at ends of the members
F
M AB = M AB
+ M DAB = 25 +
2EI
4
2EI
F
D
M BA = M BA
+ M BA
= −25 +
F
M BC = M BC
+ M DBC = 25 +
F
D
M CD = M CD
+ M CD
=
F
D
M DC = M DC
+ M DC
=
2EI
5
2EI
5
4
2EI
4
F
D
M CB = M CB
+ M CB
= −25 +
θ BA
(2θ BC + θC )
2EI
4
2θ BA
(2θC + θ BC )
(2θC + θ D )
(2θ D + θC )
Step 6: Moment equilibrium equations
Moments at the end B for members BA and BC have to be zero.
1) M BA = 0 : − 25 +
2) M BC = 0 : 25 +
2 × 10
5
4
2 × 105
4
2θ BA = 0
(2θBC + θC ) = 0
5
In the joint C sum of all moments has to be zero. Here, we should be careful regarding the sign
convention for moments. Namely, in a joint it is opposite to the sign convention at the member ends.
Therefore, moments in the joint are positive if they act in the clockwise direction. This is presented in
Figure 8 and can be shown by the following equation:
3) M CB + M CD + 20 = 0 : − 25 +
2 × 105
4
(2θC + θBC ) +
2 × 105
5
(2θC + θD ) + 20 = 0
The fourth equation is that the moment at D should be zero:
4) M DC = 0 :
2 × 10
5
5
(2θD + θC ) = 0
Figure 8: Question 1: solution – moment equilibrium at joint C.
Step 7: Joint displacements
Solving the four equations with four unknowns we get the joint rotations:
θ BA = 25 × 10-5 rad
θ = -31.48 × 10-5 rad
BC
θC = 12.96 × 10-5 rad
θ D = -6.48 × 10-5 rad
Step 8: Moments at the ends of members
If we put the calculated rotations into the equations for total moments at the end of members given
previously, we get:
M AB = 37.5 kNm
M BA = 0
M BC = 0
6
M CB = -27.78 kNm
M CD = 7.78 kNm
M DC = 0
The bending moment diagram is shown in Figure 9. As explained in the example in lecture notes, the
moment diagram along the axes of members with load (AB and BC) can be obtained by superposition
of moments obtained by the slope deflection method and moments on an appropriate simply
supported beam with the same load.
Note that the joint force of 10 kN did not have any influence on the moments. This is because it acts
at the joint which cannot deflect. Actually, for solving problems of this kind (i.e. with a concentrated
force at a not moving joint) the slope deflection method is not the best choice. In such cases
(especially if the joint forces are the only load in a structure) rather a method which does not neglect
the axial deformation should be chosen.
Figure 9: Question 1: solution – bending moment diagram.
Question 2
Figure 10: Question 2: solution.
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Question 3
Figure 11: Question 3: solution.
Question 4
Figure 12: Question 4: solution.
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