   # -4 A simple beam ABCD has moment of inertia I near the supports and moment of iertia 2I in the middle region, as shown in the figure. A uniform load of intensity q acts over the entire length of the beam. Determine the quations of the deflection curve for the left-hand half of the beam. Also, find the angle of rotation θ A at the left-hand support and the deflection δ max at the midpoint. ### Mechanics of Materials (MindTap Co...

9th Edition
Barry J. Goodno + 1 other
Publisher: Cengage Learning
ISBN: 9781337093347 ### Mechanics of Materials (MindTap Co...

9th Edition
Barry J. Goodno + 1 other
Publisher: Cengage Learning
ISBN: 9781337093347
Chapter 9, Problem 9.7.4P
Textbook Problem
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## -4 A simple beam ABCD has moment of inertia I near the supports and moment of iertia 2I in the middle region, as shown in the figure. A uniform load of intensity q acts over the entire length of the beam. Determine the quations of the deflection curve for the left-hand half of the beam. Also, find the angle of rotation θ A at the left-hand support and the deflection    δ max at the midpoint.

To determine

The equations of the deflection curve for the left-hand half of the beam, the angle of rotation θA at the left-hand support and the deflection δmax at the midpoint should be determined for given simple beam.

### Explanation of Solution

Given Information:

We have the simple beam ABCD with moment of inertia I near supports and moment of inertia 2I in the middle region as per below figure:

For the given simple beam below is the free body diagram.

For the above diagram, applying the force equilibrium in horizontal direction, we will get,

Fx=0Ax=0

At the point A, we are applying moment equilibrium,

MA=0;Dy×LqL×L2=0Dy=qL2

Again we are applying the force equilibrium but in vertical direction,

Fy=0AyqL+Dy=0AyqL+qL2=0Ay=qL2

Now we are taking a section which is at distance of x from point A where (0xL4) , we get below figure:

In above section, we are applying moment equilibrium,

M+qx(x2)qL2(x)=0M=qL2xq2x2EIv''=qL2xq2x2( From moment-curvature relationship)

We can get the slope equation after integrating the above equation.

EIv'=qL2x22q2x33+C1v'=1EI(qL4x2q6x3+C1)

Once again taking integration on both sides, we can determine the deflection equation,

v'=1EI(qL2 x 22q2 x 33+C1)v=1EI(qL4 x 33q6 x 44+C1x+C2)v=1EI(qL12x3q24x4+C1x+C2)

Now we are taking a section which is at distance of x from point A where (L4xL2) , we get below figure:

In above section, we are applying moment equilibrium,

M+qx(x2)qL2(x)=0M=qL2xq2x2E(2I)v''=qL2xq2x2( From moment-curvature relationship)

We can get the slope equation after integrating the above equation.

E(2I)v''=qL2xq2x2E(2I)v'=qL2x22q2x33+C3v'=12EI(qL4x2q6x3+C3)

Once again taking integration on both sides, we can determine the deflection equation,

v'=12EI(qL4x2q6x3+C3)v=12EI(qL4 x 33q6 x 44+C3x+C4)v=12EI(qL12x3q24x4+C3x+C4)

The exaggerated elastic curve for the simple beam ABCD can be represented as below diagram:

The boundary condition is applied for deflection equation for (0xL4) where deflection is zero at point A. We will get,

v=1EI(qL12x3q24x4+C1x+C2)Putting values of x =0 and v =00=1EI(qL1203q2404+C1×0+C2)C2=0

This concludes that the slope contains value as zero when the maximum deflection of the beam δmax occurs at x=L2 due to symmetry of loading.

Now applying boundary condition to deflection equation (L4xL2) where v'(L2)=0 ,

E(2I)v'=qL2x22q2x33+C3E(2I)(0)=qL4(L2)2q6(L2)3+C30=qL163qL483+C3C3=qL243

We are applying the continuity of slope condition at point B by evaluating the slope equations for segments (0xL4) and (L4xL2) as follow.

(vB')left=(vB')right1EI(qL4x2q6x3+C1)=12EI(qL4x2q6x3+C3)Putting values of x =L4and C3=qL3241EI(qL4( L 4 )2q6( L 4 )3+C1)=12EI(qL4( L 4 )2q6( L 4 )3+C3)(q L 364q L 3384+C1)=12(q L 364q L 3384q L 324)5qL3384+C1=11qL3768C1=7qL3256

Now we are applying the continuity of deflection condition at point B by evaluating the slope equations for segments (0xL4) and (L4xL2) as follow

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