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
xz = xy = 0
Fx = 0

My-axis = 0
 z
Mz-axis = 0
 ( y
x
dA  0
x
dA  0
x
dA)  M
(4.1)
(4.2)
(4.3)
4.3 Deformation in a Symmetric Member in
Pure Bending
Plane CAB is the Plane of
Symmetry
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).
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  
Line DE
(4.4)
Where  = radius of curvature
 = the central angle
L  (   y )
'
Before deformation: DE = JK
Therefore,
  L'  L
  (   y )     y
(4.6)
Line JK
(4.5)
The Longitudinal Strain x =

o
x 

L
x  

 y

y
(4.8)

x varies linearly with the distance y from the neutral surface
The max value of x occurs at the top or the bottom fiber:
m 
c

(4.9)
Combining Eqs (4.8) & (4.9) yields
y
 x   m
c
(4.10)
4.4 Stresses and Deformation is in the Elastic Range
For elastic response – Hooke’s Law
 x  E x
(4.11)
y
(4.10)
 x   m
c
y
E x   ( E m )
c
Therefore,
y
y
 x    m    max
c
c
(4.12)
Based on Eq. (4.1)

x
dA  0
(4.1)
y
y
 x    m    max
(4.12)
c
c
y
m
  x dA   ( c m )dA   c  ydA  0
Hence,
 ydA  first moment of area  0
(4.13)
Therefore,
Within elastic range, the neutral axis passes
through the centroid of the section.
According to Eq. (4.3)
and
It follows
or
y
 x   m
c
 ( y xdA)  M
(4.3)
(4.12)
y
 ( y)( c m )dA  M
m
c
2
y
 dA  M
(4.14)
Since
Eq. (4.24)
I   y 2dA
m
c
2
y
 dA  M
can be written as
Mc
m 
I
Elastic Flexure Formula
(4.15)
At any distance y from the neutral axis:
My
x  
I
Flexural Stress
(4.16)
If we define
I
Elastic section modulus = S =
c
(4.17)
Eq. (4.15) can be expressed as
M
m 
S
(4.18)
Solving Eq. (4.9)

m 
1

1

c

m
c
(4.9)
m 
m
E
m
1 Mc


 Ec Ec I
Finally, we have
M

 EI
[
1
(4.21)
Mc
m 
]
I
1 3
bh
I 12
1 2 1
S 
 bh  Ah
h
c
6
6
2
4.5 Deformation in a Transverse Cross Section
Assumption in Pure Bending of a Beam:
The transverse cross section of a beam remains “plane”.
 y   x
,x 
 z   x
y
Since
x  
Hence,
y
y 

Therefore,
x 
ⓛ
A. Material above the neutral surface (y>0),

(4.8)
y
z 

 y  ,  z  
(4.22)
ⓛ
However, this plane may undergo in-plane deformations.
, z 
ⓛ
y 
ⓛ
B. Material below the neutral surface (y<0),
As a consequence,
Analogous to Eq. (4.8)
x  

y


y
x
For the transverse plane:
' 
y
y

y
 x

1 y
 x



 x  ,  x  
 =
radius of curvature,
1/ = curvature

Anticlastic curvature = ' 
 
1
(4.23)
4.6 Bending of Members Made of Several Materials
(Composite Beams)
From Eq. (4.8)
x  
y

E1 y
For Material 1:
 1  E1 x  
For Material 2:
 2  E2 x  

E2 y

dF1   1dA  
dF2   2 dA  
E1 y

E2 y

dA
dA
Designating E2 = nE1
dF2  
(nE1 ) y

dA  
E1 y

(ndA)
My
x  
I
Notes:
1. The neutral axis is calculated based on the transformed section.
2.
 2  n x
3. I = the moment of inertia of the transformed section
4. Deformation --
1
M

 E1 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
x
(bx)  nAs (d  x)  0
2
1 2
bx  nAs x  nAs d  0
2
 determine the N.A.
4.7 Stress Concentrations
Mc
m  K
I
4.12 Eccentric Axial Loading in a Plane of Symmetry
 x  ( x )centric  ( x )bending
P My
x  
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
  dA  0
(the Centroid = the N.A.)
My = 0
 z x dA  0
(moment equilibrium)
Mz = 0
 ( y dA)  M
(moment equilibrium)
Substituting
x
x
x  
m y
c
into Eq. (4.2)
(4.1)
(4.2)
(4.3)
Plane of symmetry
m y
We have
 z( 
or
 yzdA  I
c
)dA  0 or 
yz
0
m
c
 z( y )dA  0
(knowing m/c = constant)
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.
If the axis of M coincides with the principal centroid axis, the
superposition method can be used.
=
Case A
=
+
M y  M sin 
M z  M cos
For Case A
For Case B
x  
x  
Mzy
Iz
M yz
For the combined cases :
Case B
(4.53)
(4.54)
Iy
Mz y M yz
x  

Iz
Iy
(4.55)
The N.A. is the surface where x = 0. By setting x = 0 in Eq. (4.55), one has

Mz y M yz

0
Iz
Iy
Solving for y and substituting for Mz and My from Eq. (4.52),
y(
This is equivalent to
Iz
tan  ) z
Iy
(4.56)
 Iz 
y / z  m  slope    tan 
I 
 y
The N.A. is an angle  from the z-axis:
Iz
tan   tan
Iy
(4.57)
4.14 General Case of Eccentric Axial loading
P MZ y M yz
x  

A
Iz
Iy
My
Mz
P
y
z
Iz
Iy
A
(4.58)
(4.58)
4.15 Bending of Curved Members
After bending
Before bending
Length of N.A. before and after bending
The elongation of JK line
Since
We have
r  R y
r '  R'  y
  ( R '  y ) '  ( R  y )
R  R ' '
(4.59)
  r ' '  r
(4.60)
(4.61)
If we define  -  =  and knowing R  = R , thus
   y
(4.62)
Based on the definition of strain, we have

y
x 

r
r
Substituting
r  R y
(4.63)
into the above equation,
x  
Also, x = E x
x  

y
 R y
E 

y
R y
(4.64)
(4.65)
Plotting  x  
E 

y
R y

 x is not a linear function of y.
Since r  R  y  y = R – r, therefore,
x  
E  R  r

r
Substituting this eq. into Eq. (4.1)



E  R  r
dA  0

r
Rr
 r dA  0
or
x
dA  0
and

E  R  r
dA  0


r
dA
R
  dA  0
r
 E 

 cos t . 

 

Therefore, R can be determined by the following equation:
R
Or in an alternative format:
A
dA
r
(4.66)
1 1 1
  dA
R A r
The centroid of the section is determined by
1
r   rdA
A
(4.67)
Comparing Eqs. (4.66) and (4.67), we conclude that:
The N.A. axis does not pass through the Centroid of
the cross section.
(4.59)
Mz = M 
Since
E  R  r
  r ydA  M
y  R  r , it follows
E  ( R  r ) 2
dA  M


r
or
E 

[R2 
dA
 2 RA   rdA]  M
r
Recalling Eqs. (4-66) and (4.67), we have
E 

Finally,
E 

( RA  2 RA  rA)  M

M
A(r  R)
(4.68)
By defining e  r  R , the above equation takes the new form
E 


M
Ae
(4.69)
Substituting this expression into Eqs. (4.64) and (4-65), we have
My
x  
Ae( R  y )
and
M (r  R)
x 
Aer
Determination of the change in curvature:
From Eq. (4.59)
Since 
'
   
1 1 '

'
R R
and from Eq. (4.69), one has
1 1

1
M

(1

)

(1

)
'
R R

R
EAe
(4.70, 71)
Hence, the change of curvature is
1 1
M
 
'
R R EAeR
End of Ch 4
(4.72)
E  R  r
x  

r
P MZ y M yz
x  

A
Iz
Iy
1
2
1
2
 x  ( x )centric  ( x )bending
P My
x  
A I
 z (
my
c
(4.58)
(4.58)
)dA  0
 yzdA  0
(4.58)
(4.58)
1
2
1
2
1
2
x
(bx)  nAs (d  x)  0
2
1 2
bx  nAs x  nAs d  0
2
Mc
m  K
I
c
M  b  y x dy
c
c
M  2b  y x dy
0
MU c
RB 
I
A
R r

2 dA
r1 r
h
R
r2
ln
r1
bh

r2 bdr
r1 r
h
r2 dr
r1 r
Z
Mp
Y

bc 2 Y
Y
1 2
 bc  bh
4
1 2
S  bh
6
1 2
bh
Z 4
3
k 

S 1 bh 2 2
6
FP
M  Pd
2
1
RY  bc Y
2
R p  bc Y
4
2 2
M Y  ( c) RY  bc  Y
3
3
M p  cR p  bc 2 Y
M p  kM Y
M p  Z Y
Z Y Z
k


M Y S Y S
Mp
2
1
y
M  bc 2 Y (1  Y2 )
3c
3
1 yY2
M  M Y (1  2 )
2
3c
3
M p  MY
2
yY   Y 
c   Y Y
yY


c Y
3
1 2
M  M Y (1 
)
2
2
3 Y
1
MY  Y
c
I b(2c)3 2 2

 bc
c
12c
3
2 2
M Y  bc  Y
3
x  
yY
Y
yY
y
M  2b  y (
0
Y
yY
c
y )dy  2b  y ( Y )dy
2 2
 byY  Y  bc 2 Y  byY2 Y
3
yY
Iz
tan   tan 
Iy
(4.58)
My
Mz
P
y
z
Iz
Iy
A
(4.58)
(4.58)