Solved Problems: Slope Deflection Method- Structural Analysis

SLOPE DEFLECTION METHOD

 

 

(1). A beam ABC, 10m long, fixed at ends A and B is continuous over joint B and is loaded as shown in Fig. Using the slope deflection method, compute the end moments and plot the bending moment diagram. Also, sketch the deflected shape of the beam. The beam has constant EI for both the spans.

 

 

Solution.

 

(a)  Fixed end moments

 

Treating each span as a fixed beam, the fixed end moments are as follows:

 

(b) Slope deflection equations

 

The end rotations q_A and q_C are zero since the beam is fixed at A and C. hence there is only one unknown, q_B. the ends do not settle and hence dfor each span is zero. Let us assume q_B to be positive. The result will indicate the correct sign. The slope deflection equations are as follows:

For span AB,

For span BC,

 

(c)  Equilibrium equation

 

Since there is only one un known, i.e. q_B, one equilibrium equation is sufficient. For the joint B, we have

 

M_BA + M_BC = 0

 

\(0.8 EI q_B + 3.6) + (0.8 EI q_B + 5.0) = 0 1.6 EI q_B = 1.4

The plus sign indicates that q_B is positive (i.e. rotation of tangent at B is clockwise).

 

 

(d) Final moments

 

Substituting the values of EI q_B in Eqa. (1) to (4), we get

 

 

 

 

 

 

(2) A beam ABC, 10m long, hinged at ends A and B is continuous over joint B and is loaded as shown in Fig. Using the slope deflection method, compute the end moments and plot the bending moment diagram. Also, sketch the deflected shape of the beam. The beam has constant EI for both the spans.

SOLUTIONS

(a) Fixed end moments

These are the same as calculated in the previous problem:

M_FAB = -2.4 KN-m ; M_FBA = +3.6 KN-m

M_FBC = -5.0 KN-m ; M_FCB = +5.0 KN-m

(b) Slope deflection equations.

(c) Equilibrium equations

 

Since end A is freely supported,

(d) Final moments : Substituting the values of EI q_A and EI q_B inEq. (2), we get

The bending moment diagram and the deflected shape of the beam are shown in the Fig. Note. The beam is statically indeterminate to single degree only. This problem has also been solved by the moment distribution method (example 10.2) treating the moment at B as unknown. However, in the4 slope- deflection method, the slope or rotations are taken as unknowns, and due to this the problem involves three unknown rotations q_A, q_B and q_C. hence the method of slope deflection is not recommended for such a problem.

 

 

(3) A continuous beam ABCD consists of three spans and is loaded as shown in fig. ends A and D are fixed. Determine the bending moments at the supports and plot the bending moment diagram.

 

 

a)Fixed end moments

(b) Slope deflection equation

 

q_A and q_D are zero since ends A and D are fixed.

 

(c)  Equilibrium equations

 

At join B, M_BA + M_BC = 0

At join C, M_CB + M_CD = 0

From (I) and (II), we get  EI q_B = -2.03 kN-m and EI q_C =  + 1.26kN-m

(d) Final moments

Substituting this values in Eqs. (1) to (6), we get

The bending moment diagram and the deflected shape are shown in Figure.

 

 

 

 

 

4) A continuous beam ABC is supported on an elastic column BD and is loaded as shown in figure . Treating joint B as rigid, analyze the frame and plot the bending moment diagram and the deflected shape of the structure.

 

 

(a)Fixed end moments

MFBD = MFDB = 0

 

 

(b)Slope deflection equations.

 

The slopes q_A and q_D are zero since ends A and D are fixed.

 

For span AB

(c) Equilibrium equations

 

At join B, M_BA + M_BC + M_BD = 0

(d) Final moments

 

Substituting this values in Eqs. (1) to (6), we get

 

The bending moment diagram and the deflected shape are shown in Figure.

 

 

(5) Analyze the rigid frame shown in figure

(a) Fixed end moments

(b) Slope deflection equations.

(c)  Equilibrium equations

 

For the equilibrium joint B,  M_BA + M_BD + M_BC = 0

 

\(2EIq_B + 2.67) + (EIq_B -2) + (-4) = 0

3EIq_B = 3.33

EIq_B = 1.11

(d) Final moments

Substituting this value of EIq_B in Eqs. (1) to (4), we get

The bending moment diagram and

The deflected shapes are shown in Figure.

(6) A portal frame ABCD is fixed at A and D, and has rigid joints at B and C. The column AB is 3m long. The beam BC is 2m long, and is loaded with uniformly distributed load of intensity 6 kN/m. The moment of inertia is 2.1 and that of BC and CD is I (Fig). Plot B.M. diagram and sketch the deflected shape of the frame.