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                          Indeterminate Structure- solution by slope deflection method  

Problem 7-4

Use slope deflection method 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)

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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;

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Last updated on Thursday January 31, 2013