**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)**

**Solution:**

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}
= -wL^{2 }
/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 θ_{A}
= 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

M_{AC} =(EIθ_{C}
)/2 -25
Eq. (1)

M_{CA} =(EIθ_{C}
) +25
Eq. (2)

__Step 3__: Apply slope-deflection
equations for span CD,

the support D is fixed therefore θ_{D}
= 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

M_{CD}
=(4EIθ_{C} )/3 -15
Eq. (3)

M_{DC}
=(2EIθ_{C} )/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
M_{C}
= 0;

M_{CA} + M_{CD}
= 0; therefore,

(EIθ_{C}
) +25 + (4EIθ_{C} )/3 -15 = 0;

yields θ_{C}
= (-30/7)/EI

substituting the value of θ_{C}
in eq. 1, 2, 3 & 4 we get the values for end moments

M_{AC}
= -27.14 kNm;

M_{CA}
= 20.71 kNm;

M_{CD}
= - 20.71 kNm;

M_{DC}
= 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.

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