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CHAPTER 7:Flexural Members

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OUTLINE

-Types of beams, loads and reactions-Shear forces and bending moments-Shear force and bending moment diagrams-Bending deformation of a straight member-The flexure formula-The elastic curve-Slope and deflection by direct integration method

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Types of beamsTypes of beams

Pin support prevents translation but does not prevent rotation.

Simply supported beam Cantilevered beam

Beams are usually described by the manner in which they are supported.

Overhanging beam

Roller support prevents translation only in the vertical direction.Fixed support prevents translation and rotation.

[Reference # 3: Page 255 – 281] 4

Types of loadsTypes of loads

Concentrated load MomentDistributed load

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ReactionsReactionsThe first step in the analysis of beam is finding the reactions.

From the reactions, the shear force and bending moments can be found.

For the statically determinate beams, all reactions can be found from free-body diagrams and equilibrium equations.

Concentrated load MomentDistributed load

Reaction Reaction6

Shear forces and bending momentsShear forces and bending moments

∑ =−= 0,0 VPFy

P

PVM

V Mx

∑ =−= 0,0 PxMM

VP =

PxM =

The stress resultants in statically determinate beams can be calculated from equilibrium equations.

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Sign conventionSign convention

A positive shear force acts clockwise against the material. V

M

V

M

A positive bending moment compresses the upper part of the beam.

Homework # 21: Reference # 3 Problem 6-38

Relationship between loads, shear Relationship between loads, shear forces and bending momentsforces and bending moments

From equilibrium equations,

qdxdV

−= VdxdM

=

V+V1

M V M+dM

V

V

V+dV

q

dx

M M+M1

dx

V+V1

M+M1

dx

M

P

M0

PV −=1 dxVVdxdxPM 11 2++⎟

⎠⎞

⎜⎝⎛=

01 MM −=

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Shear force and bending Shear force and bending moment diagramsmoment diagrams

BMD

SFD

P q

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Shear force and bending Shear force and bending moment diagramsmoment diagrams

BMD

SFD

P1qP2

P3

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Shear force and bending Shear force and bending moment diagramsmoment diagrams

BMD

SFD

P2 qP1

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Draw the SFD and BMDDraw the SFD and BMDP P

b bL L/2

P

L/2

M1=PL/4

q0

L

M1

L/3 L/3 L/3

2M1

Homework # 22: Reference # 3 Problem 6-30

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Bending deformation of a Bending deformation of a straight memberstraight member

[Reference # 3: Page 282 – 303]

Compression

TensionNeedReinforcement!

SFD

BMD M+ Reinforcement14

Bending deformation of a Bending deformation of a straight memberstraight member

xx

yz

Δx

x

yz

zlongitudinal axis

neutral axis

neutral surface

M

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Bending deformation of a Bending deformation of a straight memberstraight member

Undeformed element

ylongitudinal axis Δx

Δx

Δs = Δx

y

o’

Δθ

Δs’

Δx

longitudinal axis

ρ

Deformed element

sss

s ΔΔ−Δ

∈=→Δ

'lim0

( )θρ

θρθρθ Δ

Δ−Δ−∈=

→Δ

y0

lim

ρy

−∈= 16

Bending deformation of a Bending deformation of a straight memberstraight member

Normal strain distribution

ylongitudinal axis

Δx

c

max∈−

max∈⎟⎠⎞

⎜⎝⎛−∈=

cy

ρρ

//

max cy−

=∈∈

max∈⎟⎠⎞

⎜⎝⎛−∈=

cy

“Longitudinal normal strain will vary linearly with y from NA”

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The flexure formulaThe flexure formula

y

Normal strain variation

c

max∈

x

y

∈

y

Bending stress variation

c

x

ymaxσ

σM

maxσσ ⎟⎠⎞

⎜⎝⎛−=

cy

∑= ;xR FF ∫∫ ==AA

dAdF σ0

∫ ⎟⎠⎞

⎜⎝⎛−=

AdA

cy

maxσ

∫−

=A

ydAc

maxσ

∫ =A

ydA 018

The flexure formulaThe flexure formula

y

Bending stress variation

c

x

ymaxσ

σM

( ) ∑= ;zzR MM ( )∫∫ ==AA

dAyydFM σ

∫ ⎟⎠⎞

⎜⎝⎛=

AdA

cyy maxσ

∫=A

dAyc

M 2maxσ

IMc

=maxσ

IMy

−=σ

I = The moment of inertia of the cross-sectional area about NA

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ExampleExample

1.2 kN8 kN/m

A B

0.8 m 0.8 m

If the beam has a square cross section of 100 mm on each side, determine the absolute maximum bending stress in the beam.

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ExampleExample

If the beam has a rectangular cross section with a width of 200 mm and a height of 400 mm, determine the absolute maximum bending stress in the beam.

8 kN2 kN/m

40 kN·m

4 m 6 m

10 kN

200 mm

400 mm

Homework # 23: Reference # 3 Problem 6-62

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The elastic curveThe elastic curve

The elastic curve = the deflection diagram of the longitudinal axis that passes though the centroid of each cross-sectional area of the beam.

Force Displacement

Moment Rotation or slope

P P

[Reference # 3: Page 569 – 589] 22

Sign conventionSign convention

A positive bending moment compresses the upper part of the beam.

+M +M

Positive internal moment

-M -MNegative internal moment

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Inflection pointInflection point

BMD

Elastic curve

PM

Inflection point

P1P2

Inflection point

Homework # 24: Reference # 3 Problem 12-1524

MomentMoment--curvature relationship curvature relationship

Before deformation

ylongitudinal axis dx

ds

y

o’

dθ

ds’

dx

longitudinal axis

ρ

After deformation

ρ

M

M

y∈

−=ρ1

EIM

=ρ1

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Slope and displacement by Slope and displacement by integration integration

( )[ ] 2/32

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/1

/1

dxdv

dxvd

+=

ρ

( )[ ] EIM

dxdv

dxvd=

+2/32

22

/1

/

EIM

dxvd=2

2

EIM

=ρ1

From ;

2

21dx

vd=

ρSimplify by ;

)1(ρ

Represent the curvature in terms of v and x;

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Slope and displacement by Slope and displacement by integration integration

)(2

2

xMdx

vdEI =

dxdMV =From ; )(3

3

xVdx

vdEI =

)(4

4

xwdx

vdEI −=dxdVw =−From ;

EIM

dxvd=2

2

From ;

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Sign conventionSign convention

+M

+V

+M

+V

+w

Positive sign convention

x

o’

+θds

dx

Elastic curve+ρ

v

+x

+ρ

+v+dv

dθ

If positive x is directed to the left, then θ will be positive clockwise.

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Boundary conditionsBoundary conditions

Δ = 0 Δ = 0 Δ = 0 Δ = 0Roller RollerPin Pin

Δ = 0Fixed endθ = 0 M = 0

Free endV = 0 M = 0

Internal pin or hinge

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ExampleExample

Determine the equations of the elastic curve using the x1 and x2coordinates. EI is constant.

P

A B

x1

x2

a b

L

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ExampleExample

Determine the maximum deflection of the beam and the slope at A. EI is constant.

M0

A B

a a a

M0

Homework # 25: Reference # 3 Problem 12-13