Chapter 9: Stress in beam (Ch. 11 in Shames)

445.204

Introduction to Mechanics of Materials (재료역학개론)

Chapter 9: Stress in beam (Ch. 11 in Shames)

Myoung-Gyu Lee, 이명규

Tel. 880-1711; Email: myounglee@snu.ac.kr

TA: Chanmi Moon, 문찬미

Lab: Materials Mechanics lab.(Office: 30-521) Email: chanmi0705@snu.ac.kr

Contents

- Pure bending of symmetric beams

- Bending of symmetric beams with shear: normal stress - Bending of symmetric beams with shear: shear stress - Sign of the shear stress

- General cuts

Cf. Inelastic behavior of beams

2

Pure bending of symmetric beams

• Equilibrium – shear force is zero for the beam and bending moment is a constant for the entire length of the beam

• Compatibility – cross sections of the beam elements remains plane upon deformation of the beam by the action of pure end couples

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Pure bending of symmetric beams

O

O

φ R

∆

'

∆x

∆x y

' y

∆ x

M

M

1

R = κ

Neutral axis

Pure bending of symmetric beams

' '

x x

R R y

φ ^{∆ ∆}

∆ = =

− ' R y '

x x

R

∆ = − ∆

' '

' R y 1 y

x x x x

R R

 − 

∆ − ∆ =  − ∆ = − ∆ 

 

0

' '

lim

'

x

x x y

x ε R κ y

∆ →

∆ − ∆

= = − = −

∆ ^xx

y y

ε = − = − R κ

Pure bending of symmetric beams

xx

E y E y σ ^{= −} R ^{= −} κ

xy yz zx

0

τ = τ = τ =

z xx

A

M σ ydA

− = ∫ ¹

z z xx

zz

R EI

M M y

I κ

σ

= =

= −

( )

1

xx xx

v

ε = E ^  σ − σ + σ ^ 

• No shear force, no twist moment

• Positive stress at a positive y results in negative bending moment (under sign convention)

2

2 zz

z zz

A A

EI

Ey E

M dA y dA EI

R R R κ

= ∫ = ∫ = =

Pure bending of symmetric beams

xx

0

A A

dA E ydA

σ ^{= −} R ⁼

∫ ∫

zz yy

v

ε = ε = − ε

First moment of cross-sectional area about neutral axis is zero

-> Neutral axis = centroidal axis

• Poisson effect

• Equilibrium

Anticlastic curvature

Pure bending of symmetric beams : Example 11.3

Pure bending of symmetric beams : Example 11.3

Region AB

- Only bending moment exists

Region BC

- Bending moment + Tensile force - Tensile force increases tensile stress - Therefore, max. stress is in tensile

(no symmetric anymore) Region CD

- Bending moment + Tensile force + Compressive

- But, tensile + Compressive with the same magnitude vanishes the net axial force => only moment

Bending of symmetric beams with shear: normal stress

• Symmetric beams under the action of arbitrary loadings which are in the plane of symmetry and oriented normal to the center line of the beam

• The stress vs. bending moment relationship in the pure bending theory still holds, but locally … That is, the bending moment is not constant but a

function of coordinate x, and the R is a local radius of curvature

• This theory is often called as “Euler-Bernoulli theory”

1

z z xx

zz

R EI

M M y

I κ

σ

= =

= −

Bending of symmetric beams with shear: Example 11.5

Homework (Try it by yourself!)

Example 11.1, 11.2, 11. 6

Bending of symmetric beams with shear: shear stress

• Nonzero shear force at sections of the beam exists, thus we expect to have a shear stress distribution over a section in addition to the normal stress

distribution.

• In the Euler-Bernoulli theory, the averaged value of shear stress distribution will be sought.

τ

τ

τ

Bending of symmetric beams with shear: shear stress

( ) ( )

1 2

1 2

0

yx xx xx

A A

bdx dA dA

τ σ σ

− − ∫ + ∫ =

Average shear stress

( )

1 z xx

zz

M y σ _{= −} ^ I ^ _

  ^{( )}

1 z

z zz xx

zz

d M y M y I

I dx dx

σ

 − 

  

−    

=   + 

1

0

z zz yx

A

d M y

bdx I dxdA

τ dx

 

 

 

− − ∫ =

z

zz z y

zz zz

d M y

I dM y V y

dx dx I I

 

 

  = =

1

y

0

yx

zz A

bdx V ydA τ I

− − ∫ =

y z

yx xy

zz

V Q

τ = I b = τ

1

z

A

Q = ∫ ydA with

For a rectangular cross section, the maximum average shear stress will occur at the neutral axis.

Bending of symmetric beams with shear: shear stress

1

z

A

Q = ∫ ydA

h

b

y

2 2

( )

2

2 2

y y

y y y y

z y y

Q dA bd

h y b h y y

ξ ξ ξ

   

+ −  + − 

= ∫ = ∫

 − 

 

 

=  −  + 

 

Useful formula

1

1 z

A

Q = ∫ ydA = A y A 1

y Distance from the neutral

axis to the centroid of A1

Homework (Try it by yourself!)

Example 11.7, 11.8, 11.9

Bending of symmetric beams with shear: Example 11.9

3 3

2 2

300 50 60 250

300 50 75 60 250 75

12 12

Izz = ⋅ + ⋅ ⋅   + ⋅ + ⋅ ⋅ 

   

2

( 0 )

zz i i

i

I = ∑ I + A d

y A yA

1 275 150,000 4,125,000

2 125 150,000 1,875,000

300,000 6,000,000

y yA

A

= ∑

∑

1

2 300

50

60 300

200 d1=75

d2=75 N.A

Bending of symmetric beams with shear: Example 11.9

( )( )

( )

15.63 1.503 15.75 1.503

2 2

12.624 12.624 / 2

0.930

2 2

Qz

y y y

 

=  − 

   − 

+ −   + 

A: above y A: flange

Distance of flange centroid

from N.A.

Distance of web area above y

from N.A.

Determination of the sign of the shear stress

V_y

0

0 0

0 0

0

y

z

y z

z xx

zz

xx

yx

xy

V

dM V dM

dx

M y and y I

d

From d and complementary shear stress increment τ

τ

τ τ

<

= < → <

= − >

>

∴

>

Determination of the sign of the shear stress

0

0 0

0 0

0

y

z

y z

z xx

zz

xx

yx

xy

V

dM V dM

dx

M y and y I

d

From d and complementary shear stress increment τ

τ

τ τ

<

= < → <

= − <

<

∴

>

V_y

Homework (Try it by yourself!)

Example 11.10, 11.11

General cuts

t

General cuts

( )

( )

1 2

zx xx xx

0

A A

tdx dA dA

τ τ τ

− − ∫ + ∫ =

Force equilibrium

( ) ( )

1 xx

xx xx

dx

x τ = τ +  ^  ^∂ ∂ τ ^  

Taylor expansion

z xx

zz

M y τ = − I

z y

V dM

= dx

0

zz A

V ydA τ I t

− − ∫ =

zz

V Q τ = I t

The second moment of area of the entire cross section about the neutral axis (N.A)

I

Q

General cuts

Vertical cut (Left figure)

Horizontal cut (Right figure)

Q

=0 !!!

General cuts – Example 11.12

Homework (Try it by yourself!)

Example 11.13, 11.14, 11.15

Inelastic behavior of beams

h d

/ 2

/ 2

( )

d d h

Y xx Y

h

τ ybdy

τ ybdy

τ ybdy M

−

−

+

+ − = −

∫ ∫ ∫

2 2

4 3

Y

h d M = b τ ^  − ^ 

 

xx xx

E E y τ = ε = − R

xx

y

ε = − R

d

E R

τ = − −

xx Y

y d /

τ = − τ