chapter 9: stress in beam (ch. 11 in shames)ocw.snu.ac.kr/sites/default/files/note/lecture 8-...

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

https://nptel.ac.in/courses/Webcourse-contents/IIT-ROORKEE/strength%20of%20materials/lects%20&%20picts/image/lect25%20and%2026/lecture25%20and%2026.htm

Pure bending of symmetric beams

O

O

Rφ∆

'x∆x∆

y

'y

x∆

zMzM

1R

κ=

Neutral axis

Pure bending of symmetric beams

''

x xR R y

φ ∆ ∆∆ = =

−

'' R yx xR−

∆ = ∆

' '' 1R y yx x x xR R− ∆ −∆ = − ∆ = − ∆

0

' 'lim 'xxx

x x y yx R

ε κ∆ →

∆ −∆= = − = −

∆ xxy yR

ε κ= − = −

Pure bending of symmetric beams

xxyE E yR

σ κ= − = −

0xy yz zxτ τ τ= = =

z xxA

M ydAσ− = ∫1 zz

z

zxx

zz

EIRM

M yI

κ

σ

= =

= −

( )1xx xx yy zzv

Eε σ σ σ = − +

• No shear force, no twist moment

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

22 zz

z zzA A

EIEy EM dA y dA EIR R R

κ= = = =∫ ∫

Pure bending of symmetric beams

0xxA A

EdA ydAR

σ = − =∫ ∫

zz yy xxvε ε ε= = −

First moment of cross-sectional areaabout 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 axialforce => 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 zz

z

zxx

zz

EIRM

M yI

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.

xyτ

xyτ

yxτ

Bending of symmetric beams with shear: shear stress

( ) ( )1 2

1 20yx xx xx

A Abdx dA dAτ σ σ− − + =∫ ∫

Average shear stress

( )11

zxx

zz

M yI

σ

= −

( )21

1

z

zzzxx

zz

M ydIM y dx

I dxσ

− − = +

1

0

z

zzyx

A

M ydI

bdx dxdAdx

τ

− − =∫

z

yzz z

zz zz

M ydV yI dM y

dx dx I I

= =

1

0yyx

Azz

Vbdx ydA

Iτ− − =∫

y zyx xy

zz

V QI b

τ τ= =1

zA

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

zA

Q ydA= ∫

h

b

y

2 2 ( )

22 2

y yy y y y

z y yQ dA bd

h yhb y y

ξ ξ ξ + − + − = =∫ ∫

− = − +

Useful formula

1

1zA

Q ydA A y= =∫

1A

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 32 2300 50 60 250300 50 75 60 250 75

12 12zzI ⋅ ⋅

= + ⋅ ⋅ + + ⋅ ⋅

20( )zz i i

iI 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

yAyA

∑=∑1

2

300

50

60

300

200

d1=75

d2=75N.A

Bending of symmetric beams with shear: Example 11.9

( )( )

( )

15.63 1.50315.75 1.5032 2

12.624 12.624 / 20.9302 2

zQ

yy 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

Vy0

0 0

0

0

0

y

zy z

zxx

zz

xx

yx

xy

VdM V dMdx

M y and yI

dFrom d and complementary shear stress increment

τ

ττ

τ

<

= < → <

= − >

>

∴

>

Determination of the sign of the shear stress

0

0 0

0

0

0

y

zy z

zxx

zz

xx

yx

xy

VdM V dMdx

M y and yI

dFrom d and complementary shear stress increment

τ

ττ

τ

<

= < → <

= − <

<

∴

>

Vy

Homework (Try it by yourself!)

Example 11.10, 11.11

General cuts

t

General cuts

( ) ( )1 21 2

0zx xx xxA A

tdx dA dAτ τ τ− − + =∫ ∫

Force equilibrium

( ) ( )2 11

xxxx xx dx

xττ τ ∂ = + ∂

Taylor expansion

zxx

zz

M yI

τ = −

zy

dMVdx

= 0yzx

Azz

VydA

I tτ− − =∫

y zzx

zz

V QI t

τ =

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

zQ The first moment of area of the cross section, taken about the neutral axis (N.A)

General cuts

Vertical cut (Left figure)

Horizontal cut (Right figure)

Qz=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 Yh d dybdy ybdy ybdy Mτ τ τ−− −+ + − = −∫ ∫ ∫

2 2

4 3Yh dM bτ

= −

xx xxyE ER

τ ε= = −xxyR

ε = − Y dE Rτ −

= −

/xx Y y dτ τ= −