chapter 9: stress in beam (ch. 11 in shames)ocw.snu.ac.kr/sites/default/files/note/lecture 8-...
TRANSCRIPT
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: [email protected]
TA: Chanmi Moon, 문찬미
Lab: Materials Mechanics lab.(Office: 30-521)
Email: [email protected]
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
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= −