chapter 4 pure bending
DESCRIPTION
Chapter 4 Pure Bending . Ch 2 – Axial Loading Ch 3 – Torsion Ch 4 – Bending -- for the designing of beams and girders. 4.1 Introduction. A. Eccentric Loading. B. Pure Bending. 4.2 Symmetric Member in Pure Bending. M = Bending Moment. Sign Conventions for M:. - PowerPoint PPT PresentationTRANSCRIPT
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Chapter 4 Pure Bending
Ch 2 – Axial Loading
Ch 3 – Torsion
Ch 4 – Bending
-- for the designing of beams and girders
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4.1 Introduction
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A. Eccentric Loading
B. Pure Bending
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4.2 Symmetric Member in Pure Bending
M = Bending Moment
Sign Conventions for M:
-- concave upward
⊝ -- concave downward
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Force Analysis – Equations of Equilibrium
0xdA
0xz dA ( )xy dA M
Fx = 0
My-axis = 0
Mz-axis = 0
xz = xy = 0
(4.1)
(4.2)
(4.3)
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4.3 Deformation in a Symmetric Member in Pure Bending
Assumptions of Beam Theory:
1. Any cross section to the beam axis remains plane
2. The plane of the section passes through the center of curvature (Point C).
Plane CAB is the Plane of Symmetry
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The Assumptions Result in the Following Facts:
1. xy = xz = 0 xy = xz = 0
2. y = z = yz = 0
The only non-zero stress: x 0 Uniaxial Stress
The Neutral Axis (surface) : x = 0 & x = 0
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L
' ( )L y
' L L
Where = radius of curvature
= the central angle
Line JK (4.5)
Before deformation: DE = JK
Therefore,
Line DE (4.4)
(4.6)
( )y y
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The Longitudinal Strain x =
o
x
x
y
L
y
x varies linearly with the distance y from the neutral surface
(4.9)
The max value of x occurs at the top or the bottom fiber:
m
c
(4.8)
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Combining Eqs (4.8) & (4.9) yields
x m
yc
(4.10)
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4.4 Stresses and Deformation is in the Elastic Range
For elastic response – Hooke’s Law
x xE
( )x m
yE E
c
x m
yc
maxx m
y yc c
Therefore,
(4.11)
(4.10)
(4.12)
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Based on Eq. (4.1)
0xdA
maxx m
y yc c
0( ) mx m
ydA dA ydA
c c
(4.1)
(4.12)
(4.13)
Hence,
0ydA first moment of area
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Therefore,
Within elastic range, the neutral axis passes through the centroid of the section.
According to Eq. (4.3)
( )xy dA M x m
yc
( )( ) m
yy dA M
c
(4.3)
and (4.12)
It follows
or2 m y dA M
c
(4.14)
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Since 2I y dA2 m y dA M
c
m
McI
Eq. (4.24)
can be written as
Elastic Flexure Formula (4.15)
At any distance y from the neutral axis:
x
My
I Flexural Stress (4.16)
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If we define
Eq. (4.15) can be expressed as
IElastic section modulus = S =
c
m
M
S (4.18)
(4.17)
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Solving Eq. (4.9) m
c
1 m
c
(4.9)
1 1m McEc Ec I
1 MEI
mm E
[ ] m
McI
Finally, we have
(4.21)
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3
2
11 1126 6
2
bhI
S bh Ahhc
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4.5 Deformation in a Transverse Cross Section
Assumption in Pure Bending of a Beam:
The transverse cross section of a beam remains “plane”.
However, this plane may undergo in-plane deformations.
A. Material above the neutral surface (y>0), ,ⓛ ⓛx x
y x z x
x
y
Since (4.8)
Hence, y
y
z
y
(4.22)
Therefore,
,y z
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B. Material below the neutral surface (y<0), ,x x
,ⓛ ⓛy z
x
y
As a consequence,
Analogous to Eq. (4.8)
For the transverse plane:
x
y
1'
y x x
y y y
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= radius of curvature,
1/ = curvature
(4.23)'Anticlastic cur
1 =vature
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4.6 Bending of Members Made of Several Materials
x
y
11 1 x
E yE
From Eq. (4.8)
For Material 1:
For Material 2:2
2 2 x
E yE
(Composite Beams)
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11 1
E ydF dA dA
22 2
E ydF dA dA
1 12
( )( )
nE y E ydF dA ndA
Designating E2 = nE1
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x
MyI
2 xn
Notes:
1. The neutral axis is calculated based on the transformed section.
2.
3. I = the moment of inertia of the transformed section
4. Deformation -- 1
1 ME I
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Beam with Reinforced Members:
As = area of steel, Ac = area of concrete
Es = modulus of steel, Ec = modulus of concrete
n= Es/Ec
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Beam with Reinforced Members:
As = area of steel, Ac = area of concrete
Es = modulus of steel, Ec = modulus of concrete
n= Es/Ec
( ) ( ) 02
s
xbx nA d x
210
2 s sbx nA x nA d determine the N.A.
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4.7 Stress Concentrations
m
McK
I
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4.12 Eccentric Axial Loading in a Plane of Symmetry
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( ) ( ) x x centric x bending
x
P My
A I
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4.13 Unsymmetric Bending
-- Two planes of symmetry
y – axis & z-axis
-- Single plane of symmetry – y-axis
--M coincides with the N.A.
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For an arbitrary geometry + M applies along the N.A
Fx = 0 0xdA 0xz dA
( )xy dA M My = 0
Mz = 0
(4.1)
(4.2)
(4.3)
Substituting mx
yc
into Eq. (4.2)
(the Centroid = the N.A.)
(moment
equilibrium)
(moment equilibrium)
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Plane of symmetry
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0 0( ) ( )m myz dA or z y dA
c c
0yzyzdA I
We have
or
Iyz = 0 indicates that y- and z-axes are the principal centroid of the cross section.
Hence, the N.A. coincides with the M-axis.
(knowing m/c = constant)
If the axis of M coincides with the principal centroid axis, the superposition method can be used.
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yx
y
M z
I
zx
z
M y
I
= =
+
sinyM M coszM M
Case A
Case B
For Case A
For Case B
For the combined cases : yzx
z y
M zM y
I I
(4.53)
(4.54)
(4.55)
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tan tanz
y
I
I
0yz
z y
M zM y
I I
( tan ) z
y
Iy z
I
The N.A. is the surface where x = 0. By setting x = 0 in Eq. (4.55), one has
Solving for y and substituting for Mz and My from Eq. (4.52),
(4.56)
This is equivalent to / tanz
y
Iy z m slope
I
The N.A. is an angle from the z-axis:
(4.57)
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4.14 General Case of Eccentric Axial loading
yZx
z y
M zP M yA I I
yz
z y
MM Py z
I I A
(4.58)
(4.58)
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' 'R R ' ' r r
r R y
4.15 Bending of Curved Members
Before bending After bending
Length of N.A. before and after bending
The elongation of JK line
' ' r R ySince
(4.59)
(4.60)
(4.61)
We have ' '( ) ( ) R y R y
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If we define - = and knowing R = R , thus
y
x
y
r r
x
y
R y
x
E y
R y
(4.64)
Based on the definition of strain, we have
r R y
(4.63)
Also, x = E x
(4.62)
(4.65)
Substituting into the above equation,
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x
E y
R y
r R y
x
E R r
r
Plotting
x is not a linear function of y.
Since
Substituting this eq. into Eq. (4.1)
y = R – r, therefore,
0xdA
0E R r
dAr
and 0E R r
dAr
0R r
dAr
0
dAR dA
r or cos .
Et
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Therefore, R can be determined by the following equation:
AR
dAr
1r rdA
A
1 1 1dA
R A r Or in an alternative format:
(4.67)
The centroid of the section is determined by
(4.59)
(4.66)
Comparing Eqs. (4.66) and (4.67), we conclude that:
The N.A. axis does not pass through the Centroid of the cross section.
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Mz = M E R r
ydA Mr
y R r
2( )
E R rdA M
r
2[ 2 ]
E dA
R RA rdA Mr
( 2 )
E
RA RA rA M
Since
Recalling Eqs. (4-66) and (4.67), we have
or
, it follows
Finally,
( )
E M
A r R
(4.68)
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E M
Ae
By defining , the above equation takes the new forme r R
( )
x
My
Ae R y ( )
x
M r R
Aer
(4.69)
Substituting this expression into Eqs. (4.64) and (4-65), we have
'
'
1 1
R R
'
'
1 1 1(1 ) (1 )
M
R R R EAe
and (4.70, 71)
Determination of the change in curvature:
From Eq. (4.59)
Since and from Eq. (4.69), one has
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1 1'
MR R EAeR
Hence, the change of curvature is
(4.72)
End of Ch 4
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x
E R r
r
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yZx
z y
M zP M y
A I I
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1
2
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1
2
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( ) ( ) x x centric x bending
x
P My
A I
( ) 0 m yz dAc
0yzdA
(4.58)
(4.58)
(4.58)
(4.58)
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1
2
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1
2
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1
2
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( ) ( ) 02
s
xbx nA d x
210
2 s sbx nA x nA d
m
McK
I
c
xcM b y dy
02
c
xM b y dy
UB
M cR
I
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2 2 2
1 1 1
r r r
r r r
A bh hR
dA bdr drr r r
2
1
ln
hR
rr
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22 21
4 p Y
Y Y
M bcZ bc bh
21
6S bh
2
2
134
1 26
bhZ
kS bh
F P
M Pd
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1
2Y YR bc
p YR bc
24 2( )3 3
Y Y YM c R bc
2 p p YM cR bc
p YM kM
p YM Z
p Y
Y Y
M Z Zk
M S S
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22
2
1(1 )
3 Y
Y
yM bc
c
2
2
3 1(1 )
2 3 Y
Y
yM M
c
3
2p YM M
Y Yy Y Yc
Y
Y
y
c
2
2
3 1(1 )
2 3 Y
Y
M M
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1Y YMc
32(2 ) 2
12 3
I b cbc
c c
22
3Y YM bc
Yx
Y
yy
0
2 2 2
2 ( ) 2 ( )
2
3
Y
Y
y cY
YyY
Y Y Y Y Y
M b y y dy b y dyy
by bc by
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tan tan z
y
I
I
yz
z y
MM Py z
I I A
(4.58)
(4.58)
(4.58)