Structural Analysis, 5th Edition
Structural Analysis, 5th Edition
5th Edition
ISBN: 9788131520444
Author: Aslam Kassimali
Publisher: Cengage Learning
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Chapter 6, Problem 6P
To determine

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

Expert Solution & Answer
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Answer to Problem 6P

For segment AB:

The equation for slope is PaL6EI(13x2L2)_.

The equation for deflection is PaxL6EI(1x2L2)_.

For segment BC:

The equation for slope is PLEI(x22Lx(1+aL)+L2+2a3)_.

The equation for deflection is PLEI[x36Lx22(1+aL)+x(L2+2a3)L6(L+a)]_.

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).

Structural Analysis, 5th Edition, Chapter 6, Problem 6P , additional homework tip  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).

Structural Analysis, 5th Edition, Chapter 6, Problem 6P , additional homework tip  2

Take a section at a distance of x.

Show the section as in Figure (2).

Structural Analysis, 5th Edition, Chapter 6, Problem 6P , additional homework tip  3

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).

Structural Analysis, 5th Edition, Chapter 6, Problem 6P , additional homework tip  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.

θAB=θBC

Substitute PaL6EI(13x2x2) for θAB and PEI(Lx+axx22)+C3 for θBC.

[PaL6EI(13x2x2)]x=L=[PEI(Lx+axx22)+C3]x=L

Apply the above boundary conditions in the above Equation

PaL6EI(13L2L2)=PEI(L×L+aLL22)+C3PaL6EI(2)=PEI(L22+aL)+C3C3=PaL3EI+PL22EIPaLEIC3=PLEI(a3+L2a)

C3=PLEI(2a3+L2)

Substitute PLEI(2a3+L2) for C3 in Equation (5).

θ=PEI(Lx+axx22)+PLEI(2a3+L2)=PLEI(2a3+L2xaxL+x22L)=PLEI(2a3+L2x(1+aL)+x22L)=PLEI(x22Lx(1+aL)+L2+2a3)

Hence, the Equation for slope is PLEI(x22Lx(1+aL)+L2+2a3)_.

Write the boundary conditions as follows:

Atx=L;y=0:

Apply the above boundary conditions in Equation (6):

PEI(L×L22+aL22L36)+(PLEI(2a3+L2))L+C4=0PEI(L33+aL22)+PL2EI(2a3+L2)+C4=0PL2EI(L3+a22a3L2)+C4=0PL2EI(L6a6)+C4=0

PL2EI(L6+a6)+C4=0C4=PL26EI(L+a)

Substitute PLEI(2a3+L2) for C3 and PL26EI(L+a) for C4 in Equation (6).

y=PEI(Lx22+ax22x36)+PLEI(2a3+L2)xPL26EI(L+a)=PLEI(Lx22+ax22x36)+PLEI(2a3+L2)xPL26EI(L+a)=PLEI[x36Lx22(1+aL)+x(L2+2a3)L6(L+a)]

Hence, the Equation for deflection is PLEI[x36Lx22(1+aL)+x(L2+2a3)L6(L+a)]_.

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