chapter 4 pure bending ch 2 – axial loading ch 3 – torsion ch 4 – bending -- for the designing...
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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
4.1 Introduction
A. Eccentric Loading
B. Pure Bending
4.2 Symmetric Member in Pure Bending
M = Bending Moment
Sign Conventions for M:
-- concave upward
⊝ -- concave downward
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)
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
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
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
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)
Combining Eqs (4.8) & (4.9) yields
x m
yc
(4.10)
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)
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
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)
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)
If we define
Eq. (4.15) can be expressed as
IElastic section modulus = S =
c
m
M
S (4.18)
(4.17)
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)
3
2
11 1126 6
2
bhI
S bh Ahhc
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, yy
zy
(4.22)
Therefore,
,y z
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
= radius of curvature,
1/ = curvature
(4.23)'Anticlastic cur
1 =vature
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)
11 1
E ydF dA dA
22 2
E ydF dA dA
1 12
( )( )
nE y E ydF dA ndA
Designating E2 = nE1
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
Beam with Reinforced Members:
As = area of steel, Ac = area of concrete
Es = modulus of steel, Ec = modulus of concrete
n= Es/Ec
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.
4.7 Stress Concentrations
m
McKI
4.12 Eccentric Axial Loading in a Plane of Symmetry
( ) ( ) x x centric x bending
x
P My
A I
4.13 Unsymmetric Bending
-- Two planes of symmetry
y – axis & z-axis
-- Single plane of symmetry – y-axis
--M coincides with the N.A.
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)
Plane of symmetry
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.
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)
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)
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)
' '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
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,
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
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.
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)
E M
Ae
By defining , the above equation takes the new forme r R
( )
x
My
Ae R y ( )
xM 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
1 1'
MR R EAeR
Hence, the change of curvature is
(4.72)
End of Ch 4