   # Determine the equations for slope and deflection of the beam shown by the direct integration method. EI = constant. FIG. P6.3

#### Solutions

Chapter
Section
Chapter 6, Problem 3P
Textbook Problem
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## Determine the equations for slope and deflection of the beam shown by the direct integration method. EI = constant. FIG. P6.3

To determine

Find the equations for slope and deflection of the beam using direct integration method.

### Explanation of Solution

Calculation:

Draw the free body diagram of the beam as in Figure (1).

Refer Figure (1),

Find the reaction at support A.

Apply vertical equilibrium along y-axis.

Consider upward force as positive.

ΣFy=0Ayw(La)=0Ay=w(La)

Find the moment at A.

Consider anticlockwise moment as positive.

0=MAw×(La)×[(La)2+a]MA=w(La)[(La+2a2)]MA=w(La)(L+a2)MA=w2(L2a2)

Segment AB: 0xa.

Consider a section X1X1 in the segment AB at a distance of x from A.

Sketch the free body diagram when section X1X1 consider in the segment AB as shown in Figure 2.

Refer Figure (2),

Take the moment at section X1X1 .

M=w2(L2a2)+w×(La)(x)

Write the equation for MEI.

d2ydx2=MEI=1EI[w2(L2a2)+w(La)x]        (1)

Find the equation for slope (θ).

Integrate Equation (1) with respect to x.

dydx=θ=MEIdxθ=w2(L2a2)+w×(La)(x)EIdxEIθ=w2(L2a2)x+w×(La)(x22)+C1        (2)

Find the equation for deflection (θ).

Integrate again Equation (2) with respect to x.

EIy=w2(L2a2)x22+w×(La)(x36)+C1x+C2        (3)

Find the integration constants C1andC2:

Apply boundary conditions in Equation (2):

At x=0 and y=0.

0=M×022EI+C1×0+C2C2=0

Apply boundary conditions in Equation (1):

At x=0 and θ=0.

0=w2(L2a2)×0+w×(La)(022)+C1C1=0

Find the equation for slope of segment AB.

Substitute 0 for C1 in Equation (2).

EIθ=w2(L2a2)x+w×(La)(x22)+0EIθ=wx2(L2a2)+wx22(La)θ=wx2EI[a2L2+(La)x]

Thus, the equation for slope of segment AB is wx2EI[a2L2+(La)x]_.

Find the equation for deflection of segment AB.

Substitute 0 for C1 and 0 for C2 in Equation (3).

EIy=w2(L2a2)x22+w×(La)(x36)+0×x+0EIy=w2(L2a2)x22+w×(La)(x36)EIy=wx22(L2a22)+wx36(La)y=wx22EI(a2L22+(La)x3)

Thus, the equation for deflection is wx22EI(a2L22+(La)x3)_.

Segment BC: axL.

Consider a section X2X2 in the segment BC at a distance of x from A.

Sketch the free body diagram when section X2X2 consider segment BC as shown in Figure 3.

Refer Figure 3.

Write the equation for bending moment at section X2X2.

M=w2(L2a2)+w×(La)(x)w2(xa)2

Write the equation for MEI.

d2ydx2=MEI=1EI[w2(L2a2)+w×(La)(x)w2(xa)2]        (4)

Write the equation for slope.

Integrate Equation (4) with respect to x.

dydx=θ=MEIdxθ=[w2(L2a2)+w×(La)(x)w2(xa)2]EIdxEIθ=w2(L2a2)x+w×(La)(x22)w2(x33+a2xax2)+C3        (5)

Write the equation for deflection

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