**Solving Indeterminate beam by slope-deflection equations**

**Problem 7-4**

**Use
slope deflection
equations to find the resultant end moments and draw resultant bending moment diagram
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 4. 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.

To determine the sagging moments, each span is considered as simply supported and the bending moments are calculated.

The Sagging moment diagram for AC will be triangle with maximum value = PL/4 = 50 x 4/4 = 50 kNm at the position of point load.

The sagging moment diagram for CD will be parabolic with maximum at the center = wL^{2}/8 = 20 x 3^{2}/8 = 22.5 kNm

You can also use our bending moment calculator to determine the values of sagging moment for AC and CD.

The Resultant bending moment diagram is shown in Figure 7-4(b) and is drawn by considering sagging moments as positive (black in colour) and hogging moments as negative (red in colour).

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