Solving Indeterminate beam by slope-deflection equations

Problem 7-4

Use slope deflection equations to find the resultant end moments for the continuous beam shown in figure 7-4(a). Take EI as constant for the beam.

Figure 7-4(a)


The given beam in figure 7-4(a) is statically indeterminate of degree 1.  In this case two spans (AC and CD) of the beam are to be considered.

Step1: Calculate Fixed end moments

(FEM)AC = -PL/8 = 50*4/8 = -25 kNm

(FEM)CA = PL/8 = 25 kNm

(FEM)CD = -wL2 /8 = -20*3*3/8 = -15 kNm

(FEM)DC = 15 kNm


Step 2: Applying slope-deflection equations for AC;

as support A is fixed there is no possibility of rotation at A, therefore θA = 0; and also there is no settlement of support because the supports are rigid, so δ = 0;

Now substitute all the values in the above equations for span AC;  we get

MAC =(EIθC )/2 -25             Eq. (1)

MCA =(EIθC ) +25               Eq. (2)

Step 3: Apply slope-deflection equations for span CD,

the support D is fixed therefore θD = 0; and also there is no settlement of support because the supports are rigid, so δ = 0;

Now substitute all the values in the above equations for span CD;  we get

MCD =(4EIθC )/3 -15             Eq. (3)

MDC =(2EIθC )/3 +25             Eq. (4)

Now we have 4 equations (Eq. 1, 2 , 3 & 4) with 5 unknowns. The additional equation required is obtained by applying the moment equilibrium at support C;

 Step 4: Σ MC = 0;

    MCA + MCD = 0;  therefore,

  (EIθC ) +25 + (4EIθC )/3 -15 = 0;

 yields  θC = (-30/7)/EI

 substituting the value of θC  in eq. 1, 2, 3 & 4 we get the values for end moments

 MAC = -27.14 kNm;

 MCA = 20.71 kNm;

 MCD = - 20.71 kNm;

 MDC = 12.14 kNm;

Sign covention for Fixed end moment:

For left end the negative value is hogging whereas positive value is sagging;

For right end the positve value is hogging whereas negative value is sagging

This way all the above moment calculated at the fixed ends are hogging in nature.

You can visit the following links of solved examples on indeterminate structures

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