The Three Moment Equations-I (Part - 2) | Structural Analysis - Civil Engineering (CE) PDF Download

Alternate derivation

The above three moment equations may also be derived by direct application of force method as follows. Now choose M_L, M_C and M_R , the three support moments at left, centre and right supports respectively as the redundant moments. The primary determinate structure is obtained by releasing the constraint corresponding to redundant moments. In this particular case, inserting hinges at L, C and R , the primary structure is obtained as below (see Fig. 12.2)

Let displacement (in the primary case rotations) corresponding to rotation M_C be Δ_L, which is the sum of rotations θ_CL and θ_CR. Thus,

Δ_L = θ_CL + θ_CR                               (12.6)

It is observed that the rotations θ_CL and θ_CR are caused due to only applied loading as shown in Fig.12.2.This can be easily evaluated by moment area method as shown previously.

In the next step, apply unit value of redundant moments at L ,C and R and calculate rotation at C (i.e. flexibility coefficients).

In the actual structure the relative rotation of both sides is zero. In other words the compatibility equation is written as,

Substituting the values of flexibility coefficients and Δ_L in the above equation,

Or,

when moment of inertia remains constant i.e. I_R = I_L = I ,the above equation simplifies to,

Example 12.1
A continuous beam ABCD is carrying a uniformly distributed load of 1 kN/m over span ABC in addition to concentrated loads as shown in Fig.12.4a. Calculate support reactions. Also, draw bending moment and shear force diagram. Assume EI to be constant for all members.

From inspection, it is assumed that the support moments at A is zero and support moment at C ,
M_C =15 kN.m (negative because it causes compression at bottom at C) Hence, only one redundant moment M_B   needs to be evaluated. Applying three moment equation to span ABC ,

The bending moment diagrams for each span due to applied uniformly distributed and concentrated load are shown in Fig.12.4b.

Equation (1) may be written as,

Thus,

M_B = −18.125 kN.m

After determining the redundant moment, the reactions are evaluated by equations of static equilibrium. The reactions are shown in Fig.12.4c along with the external load and support bending moment.

In span AB,  R_A, can be calculated by the condition that 

Similarly from span BC ,

The shear force and bending moment diagrams are shown in Fig.12.4d.

Example 12.2
A continuous beam ABC is carrying uniformly distributed load of 2 kN/m as shown in Fig.12.5a.The moment of inertia of span AB is twice that of span . Evaluate reactions and draw bending moment and shear force diagrams.

By inspection it is seen that the moment at support C is zero. The support moment at A and B needs to be evaluated .For moment at B , the compatibility equation is written by noting that the tangent to the elastic curve at B is horizontal .The compatibility condition corresponding to redundant moment at A is written as follows. Consider span AB as shown in Fig.12.5b.

The slope at A , θ_A may be calculated from moment-area method. Thus,

Now, compatibility equation is,

= θ_A = 0                        (2)

It is observed that the tangent to elastic curve at A remains horizontal. This can also be achieved as follows. Assume an imaginary span  left of support A having a very high moment of inertia (see Fig. 12.5c). As the imaginary span has very high moment of inertia, it does not yield any imaginary span has very high moment of inertia it does not yield any M/EI diagram and hence no elastic curve. Hence, the tangent at A to elastic curve remains horizontal.
Now, consider the span   applying three-moment equation to support A,

The above equation is the same as the equation (2). The simply supported bending moment diagram is shown in Fig.12.5d.

Thus, equation (3) may be written as

20M_A  + 10M_B = −500                             (4)

Now, consider span ABC, writing three moment equation for support B ,

5M_A + 20M_B = -250-62.5

= -312.5                          (5)

Solving equation (4) and (5),

M_B = -6.25 kN.m
M_A = -37.5 kN. m

The remaining reactions are calculated by equilibrium equations (see Fig.12.5e)

In span AB, ∑M_B = 0

Similarly from span BC ,

The shear force and bending moment diagrams are shown in Fig. 12.5f.

Summary

In this lesson the continuous beam with unyielding supports is analysed by threemoment equations. The three-moment equations are derived for the case of a continuous beam having different moment of inertia in different spans. The threemoment equations also belong to force method of analysis and in this case, redundants are always taken as support moments. Hence, compatibility equations are derived in terms of three support moments. Few problems are solved to illustrate the procedure.