   # 6.1 through 6.6 Determine the equations for slope and deflection of the beam shown by the direct integration method. EI = constant. FIG. P6.6

#### Solutions

Chapter
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Chapter 6, Problem 6P
Textbook Problem
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## 6.1 through 6.6 Determine the equations for slope and deflection of the beam shown by the direct integration method. EI = constant.FIG. P6.6 To determine

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

### Explanation of Solution

Calculation:

Consider flexural rigidity EI of the beam is constant.

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

Refer Figure (1),

Consider upward force is positive and downward force is negative.

Consider clockwise  is negative and counterclowise is positive.

Determine the support reaction at A using the relation;

MB=0RA×L(P×a)=0RA=PaLRA=PaL()

Determine the support reaction at B using the relation;

V=0RA+RBP=0RB=P(PaL)RB=PL+PaLRB=P(1+aL)

Show the reaction values as in Figure (2).

Take a section at a distance of x.

Show the section as in Figure (2).

Consider the segment AB:

Refer Figure (2),

Write the equation for bending moment at x distance.

Mx=RA(x)=PaL(x)=PaxL

Write the equation for MEI as follows:

d2ydx2=1EI(PaxL)        (1)

Write the equation for slope as follows:

Integrate Equation (1) with respect to x.

dydx=θ=1EI(PaL)(xdx)=PaEIL(x22)+C1        (2)

Write the equation for deflection as follows:

Integrate Equation (2) with respect to x.

y=[PaEIL((x22))+C1]dx=Pa2EIL(x33)+C1x+C2        (3)

Find the integration constants C1andC2:

Write the boundary conditions as follows:

Atx=0;y=0:

Apply the above boundary conditions in Equation (3):

0=Pa2EIL((03)3)+C1(0)+C2C2=0

Write the boundary conditions as follows:

Atx=L;y=0:

Apply the above boundary conditions in Equation (3):

0=Pa2EIL(L33)+C1(L)+0C1(L)=PaL26EIC1=PaL6EI

Find the equation for slope.

Substitute PaL6EI for C1 in Equation (2).

θ=PaEIL(x22)+PaL6EI=PaL6EI(13x2L2)

Thus, the equation for slope is PaL6EI(13x2L2)_.

Find the equation for deflection.

Substitute PaL6EI for C1 and 0 for C2 in Equation (3).

y=Pa2EIL(x33)+(PaL6EI)x+0=Pa2EIL(x33)+(PaxL6EI)=PaxL6EI(1x2L2)

Thus, the equation for deflection is PaxL6EI(1x2L2)_.

Consider segment BC;

Show the distance at a distance of x as in Figure (4).

Refer Figure (2),

For segment BC the limit should be Lx(L+a).

Write the equation for bending moment at x distance.

Mx=P(L+ax)

Write the equation for MEI as follows:

d2ydx2=1EI(P(L+ax))=PEI(L+ax)        (4)

Write the equation for slope as follows:

Integrate Equation (4) with respect to x.

dydx=θ=PEI(L+ax)dx=PEI(Lx+axx22)+C3        (5)

Write the equation for deflection as follows:

Integrate Equation (5) with respect to x.

y=[PEI(Lx+axx22)+C3]dx=PEI(Lx22+ax22x36)+C3x+C4        (6)

Write the boundary conditions as follows:

Atx=L

The slope at left and right of support B is equal

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