UCD FUND OF STRUCTURAL ANALYSIS 5E
UCD FUND OF STRUCTURAL ANALYSIS 5E
5th Edition
ISBN: 9781264843923
Author: Leet
Publisher: MCG
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Chapter 7, Problem 6P
To determine

Find the equations for slope and deflection for the beam by the double integration method and calculate the slope at each support and the deflection at midspan.

Expert Solution & Answer
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Answer to Problem 6P

The equation for slope is dydx=1EI(Px1260.0493PL2)_ for 0x123L and dydx=1EI(Px223+29PLx2+0.0247PL2)_ for 0x2L3.

The equation for deflection is y=1EI(Px13180.0493PL2x1)_ for 0x123L and y=1EI(Px239+PLx229+0.0247PL2x20.01646PL3)_ for 0x213L.

The slope of the beam at support A is θA=0.0493PL2EI_.

The deflection of the beam at midspan is ΔL2=0.0177PL3EI_.

The slope of the beam at support B is θB=0.0617PL2EI_.

Explanation of Solution

Given information:

Maximum deflection occurs at x=0.544L.

Apply the sign conventions for calculating reactions using the three equations of equilibrium as shown below.

  • For summation of forces along x-direction is equal to zero (Fx=0), consider the forces acting towards right side as positive (+) and the forces acting towards left side as negative ().
  • For summation of forces along y-direction is equal to zero (Fy=0), consider the upward force as positive (+) and the downward force as negative ().
  • For summation of moment about a point is equal to zero (Matapoint=0), consider the clockwise moment as positive and the counter clockwise moment as negative.

Sketch the Free Body Diagram of the beam as shown in Figure 1.

UCD FUND OF STRUCTURAL ANALYSIS 5E, Chapter 7, Problem 6P

Refer to Figure 1.

Use equilibrium equations:

Summation of moments about B is equal to 0.

MA=0RBL+P×2L3=0RB=23P

Summation of forces along y-direction is equal to 0.

+Fy=0RA+RBP=0RAP+23P=0RA=13P

Refer to Figure 1.

Consider a section at a distance x1 from A for 0x123L.

Calculate the moment at section x2 from B.

M(x1)=13Px1

Apply double integration method as shown below.

EId2ydx2=M(x1)EId2ydx2=13Px1

Integrate with respect to x.

EIdydx=13Px122+C1=Px126+C1        (1)

Integrate with respect to x.

EIy=P6x133+C1x1+C2=Px1318+C1x1+C2        (2)

Apply the boundary conditions as shown below.

i) At a distance x1=0 the deflection y=0.

ii) At a distance x1=0.544L the deflection dydx=0.

Apply boundary condition (i) in Equation (2).

EI(0)=P(0)318+C1(0)+C2C2=0

Apply boundary condition (ii) in Equation (2).

EI(0)=P(0.544L)26+C1C1=0.0493PL2

Substitute 0.0493PL2 for C1 in Equation (1).

EIdydx=Px1260.0493PL2dydx=1EI(Px1260.0493PL2)        (3)

Hence, the equation for slope is dydx=1EI(Px1260.0493PL2)_ for 0x123L.

Substitute 0.0493PL2 for C1 and 0 for C2 in Equation (2).

EIy=Px13180.0493PL2x1+0y=1EI(Px13180.0493PL2x1)        (4)

Hence, the equation for deflection is y=1EI(Px13180.0493PL2x1)_ for 0x123L.

Calculate the slope at support A as shown below.

(dydx)x=0=1EI(P(0)260.0493PL2)θA=0.0493PL2EI

Hence, the slope of the beam at support A is θA=0.0493PL2EI_.

Calculate the slope at x=2L3 as shown below.

(dydx)x=2L3=1EI(P(2L3)260.0493PL2)θC=1EI(0.0740PL20.0493PL2)    =0.0247PL2EI

Calculate the deflection at mid span as shown below.

(y)x=L2=1EI(P(L2)3180.0493PL2(L2))ΔL2=0.0177PL3EI

Hence, the deflection of the beam at midspan is ΔL2=0.0177PL3EI_.

Consider a section at a distance x2 from C for 0x213L.

Calculate the moment at section x2 from C.

M(x2)=23Px2+29PL

Apply double integration method as shown below.

EId2ydx2=M(x2)EId2ydx2=23Px2+29PL

Integrate with respect to x.

EIdydx=23Px222+29PLx2+C3=Px223+29PLx2+C3        (5)

Integrate with respect to x.

EIy=P3x233+29PLx222+C3x2+C4=Px239+PLx229+C3x2+C4        (6)

Apply the boundary conditions as shown below.

iii) At a distance x2=0 the deflection dydx=0.0247PL2EI.

iv) At a distance x2=L3 the deflection y=0.

Apply boundary condition (iii) in Equation (5).

EI(0.0247PL2EI)=23P(0)22+29PL(0)2+C3C3=0.0247PL2

Apply boundary condition (iv) in Equation (6).

EI(0)=P9(L3)3+PL9(L3)2+(0.0247PL2)(L3)+C40=PL3243+PL381+0.0247PL33+C4C4=0.01646PL3

Substitute 0.0247PL2 for C3 in Equation (5).

EIdydx=Px223+29PLx2+0.0247PL2dydx=1EI(Px223+29PLx2+0.0247PL2)        (7)

Hence, the equation for slope is dydx=1EI(Px223+29PLx2+0.0247PL2)_ for 0x2L3.

Substitute 0.0247PL2 for C3 and 0.01646PL3 for C4 in Equation (2).

EIy=Px239+PLx229+0.0247PL2x20.01646PL3y=1EI(Px239+PLx229+0.0247PL2x20.01646PL3)        (4)

Hence, the equation for deflection is y=1EI(Px239+PLx229+0.0247PL2x20.01646PL3)_ for 0x213L.

Calculate the slope at support B as shown below.

(dydx)x2=L3=1EI(P(L3)23+29PL(L3)+0.0247PL2)θB=1EI(PL227+227PL2+0.0247PL2)   =0.0617PL2EI

Therefore, the slope of the beam at support B is θB=0.0617PL2EI_.

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