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 3P
To determine

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

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Answer to Problem 3P

The equation for slope for segment AB is wx2EI(a2L2+(La)x)_.

The equation for deflection for segment AB is wx22EI(a2L22+(La)x3)_.

The equation for slope for segment BC is w2EI(xL(xL)x33+a33)_.

The equation for deflection for segment BC is w2EI(x2L(x3L2)x412a412+a3x3)_.

Explanation of Solution

Calculation:

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

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

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

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

Integrate Equation (5) with respect to x.

EIy=w2(L2a2)x22+w×(La)(x36)w2(x412+a2x22ax33)+C3x+C4        (6)

Find the integration constants C3andC4:

Apply boundary conditions in Equation (3):

At x=0, (La)=0, and θB,Left=θB,Right.

EIθ=w2(a33+a2×aa×a2)+C30=w2(a33+a3a3)+C30=wa36+C3C3=wa36

Apply boundary conditions in Equation (4):

At x=0, (La)=0, and yB,Left=yB,Right.

0=w2(a412+a2a22a×a33)+wa36×a+C40=w2(a4+6a44a412)+wa46+C40=w2(3a412)+wa46+C40=wa48+wa46+C4

0=6wa4+8wa448+C40=wa424+C4C4=wa424

Find the equation for slope of segment BC.

Substitute wa36 for C3 in Equation (5).

EIθ=w2(L2a2)x+w×(La)(x22)w2(x33+a2xax2)+wa36EIθ=w2(L2xa2x)+w2(Lx2ax2)w2(x33+a2xax2)+wa36EIθ=w2(L2x)+w2(Lx2)w2(x33)+wa36EIθ=w2[L2x+Lx2x33+a33]

θ=w2EI[xL(xL)x33+a33]

Thus, the equation for slope of segment BC is w2EI[xL(xL)x33+a33]_.

Find the equation for deflection of segment BC.

Substitute wa36 for C3 and wa424 for C4 in Equation (6).

EIy=w2(L2a2)x22+w×(La)(x36)w2(x412+a2x22ax33)+wa36xwa424EIy=w2(L2x22a2x22)+w(Lx36ax36)w2(x412+a2x22ax33)+wxa36wa424EIy={w2(L2x22)+w2(a2x22)+w2(Lx33ax33)+w2(x412+ax33)+w2(a2x22)+wxa36wa424}y=w2(L2x22+Lx33ax33x412+ax33+xa33a412)y=w2[x2L(x3L2)x412a412+xa33]=w2EI[x2L(x3L2)x412a412+xa33]

Thus, the equation for deflection of segment BC is w2EI[x2L(x3L2)x412a412+xa33]_.

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