Deflection of Beams _ Chapter 12

Chapter Objectives

ü Determine the deflection and slope at specific points on beams and shafts, using various analytical methods including:

Ø The integration method

Ø The use of discontinuity functions

Ø The method of superposition

Copyright © 2011 Pearson Education South Asia Pte Ltd

READING QUIZ

1) The slope angle θ in flexure equations is

a) Measured in degree

b) Measured in radian

c) Exactly equal to dv/dx

d) None of the above


APPLICATIONS


ELASTIC CURVE


• The deflection diagram of the longitudinal axis that passes through the centroid of each cross-sectional area of the beam is called the elastic curve, which is characterized by the deflection and slope along the curve

ELASTIC CURVE (cont)


• Moment-curvature relationship:

– Sign convention:

ELASTIC CURVE (cont)


• Consider a segment of width dx, the strain in are ds, located at a position y from the neutral axis is ε = (ds’ – ds)/ds. However, ds = dx = ρdθ and ds’ = (ρ-y) dθ, and so ε = [(ρ – y) dθ – ρdθ ] / (ρdθ), or

• Comparing with the Hooke’s Law ε = σ / E and the flexure formula σ = -My/I

y

ε

ρ−=

1

yEEI

M σ

ρρ−==

1or

1

SLOPE AND DISPLACEMENT BY INTEGRATION


• Kinematic relationship between radius of curvature ρ and location x:

• Then using the moment curvature equation, we have

( )[ ] 232

22

1

1

dxdv

dvvd

+−=

ρ

( )[ ] 2

2

2/32

22

1

1

dx

vd

dxdv

dxvd

EI

M≈

+==

ρ

SLOPE AND DISPLACEMENT BY INTEGRATION (cont)


• Sign convention:

SLOPE AND DISPLACEMENT BY INTEGRATION (cont)


• Boundary Conditions:

– The integration constants can be determined by imposing the boundary conditions, or

– Continuity condition at specific locations

EXAMPLE 1


The cantilevered beam shown in Fig. 12–10a is subjected to a vertical load P at its end. Determine the equation of the elastic curve. EI is constant.

EXAMPLE 1 (cont)


• From the free-body diagram, with M acting in the positive direction, Fig. 12–10b, we have

• Applying Eq. 12–10 and integrating twice yields

Solutions

PxM −=

(1) 6

(2) C2

(1)

21

3

1

2

2

2

CxCPx

EIv

Px

dx

dvEI

Pxdx

vdEI

++−=

+−=

−=

EXAMPLE 1 (cont)


• Using the boundary conditions dv/dx = 0 at x = L and v = 0 at x = L, equations 2 and 3 become

• Substituting these results, we get

Solutions

3 and

2

60

20

3

2

2

1

21

3

1

2

PLC

PLC

CLCPL

CPL

−==⇒

++−=

+−=

( )

( ) (Ans) 236

2

323

22

LxLxEI

Pv

xLEI

P

−+−=

−=θ

EXAMPLE 1 (cont)


• Maximum slope and displacement occur at for which A(x =0),

• If this beam was designed without a factor of safety by assuming the allowable normal stress is equal to the yield stress is 250 MPa; then a W310 x 39 would be found to be adequate (I = 84.4(106)mm4)

Solutions

(5) 3

(4) 2

3

2

EI

PLv

EI

PL

A

A

−=

=θ

( ) ( )[ ] ( )[ ]

( ) ( )[ ] ( )[ ] mm 1.74

104.842003

1000530

rad 0222.0104.842002

1000530

6

22

6

22

−=−=

==

A

A

v

θ

EXAMPLE 2


The simply supported beam shown in Fig. 12–11a supports the triangular distributed loading. Determine its maximum deflection. EI is constant.

EXAMPLE 2 (cont)


• Due to symmetry only one x coordinate is needed for the solution,

• The equation for the distributed loading is .

• Hence

Solutions

2/0 Lx ≤≤

( )

xLw

L

xwM

xLwx

L

xwMM NA

43

043

;0

02

0

02

0

+−=

=−

+=+ ∑

xL

ww 02

=

EXAMPLE 2 (cont)


• Integrating twice, we have

• For boundary condition,

Solutions

2,0 and 0,0 Lxdxdvxv ====

213050

12040

0302

2

2460

812

43

CxCxLw

xL

wEIv

CxLw

xL

w

dx

dvEI

xLw

xL

wM

dx

vdEI

+++−=

++−=

+−==

0,192

52

30

1 =−= CLw

C

EXAMPLE 2 (cont)


• Hence

• For maximum deflection at x = L/2,

Solutions

(Ans) 120

40

maxEI

Lwv −=

xLw

xLw

xL

wEIv

192

5

2460

303050 −+−=

Deflection of Beams _ Chapter 12

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