BEAM STRESSES
1 What forces are set up in simple beams?
We will find out in this lesson
BENDING STRESSES
A beam subject to uniform bending
NA is called the NEUTRAL AXIS
There are no bending stresses along the Neutral Axis
BENDING STRESSES
A beam subject to uniform bending
strain( ) 
y
L (R  y) - R


L
R
R
Ey
stress ( )  E x strain 
R
E is the ‘Modulus of Elasticity’
BENDING STRESSES
First Moment of Area
Force  stress x area  E
Total Force 
E
R
y
E
xA  yA
R
R
 ydA
BENDING STRESSES
First Moment of Area
Point of zero force is when
 ydA  0
Used to find the Neutral Axis (also called the ‘Centroid’)
BENDING STRESSES
Example of First Moment of Area
250 mm
30 mm
100 mm
50 mm
Determine the position of the neutral axis for the T-section beam
shown.
BENDING STRESSES
Example of First Moment of Area
250 mm
30 mm
100 mm
50 mm
Steps 1 – calculate the first moment of area of each regular shape about a
position
2 – calculate the total area of the shape
3 – Divide the total first moments of area by the total area
BENDING STRESSES
Example of First Moment of Area
250 mm
30 mm
NA
115 mm
NA
89 mm
100 mm
50 mm
50 mm
Taking moments about the base
Total moment = 100 x 50 x50 + 250 x 30 x 115 = 1.11 x 106 mm3
Total area = 250 x 30 + 100 x 50 = 12500 mm2
1.11 x 10 6
distance of NA from base 
12500
 89 mm
BENDING STRESSES
First Moment of Area Problem
Determine the position of the Neutral Axis from the base for
the non-symmetrical I section above.
Answer – 97 mm from the base
BENDING STRESSES
Second Moment of Area
M 
E 2
E
y dA   y 2dA
R
R
I   y 2dA
I is called the ‘Second Moment of Area
BENDING STRESSES
Common Second Moments of Area
d
ro
d
ri
b
I 
bd
3
I 
12

ro 4  ri 4 
I 
d 4
4
64
B
di
bi
bo
do

bo do 3  bi di 3 
I 
12
b
b
D
d
  BD 3 
 bd 3  
  2
 
I   
 12  
  12 
BENDING STRESSES
Problem
Calculate the second moment of area about the neutral axis of
the I section above
Answer – 1.01 x 106 mm4
BENDING STRESSES
General Bending equations
BENDING STRESSES
General Bending equations
I   y 2dA
EI
M 
R
Ey
 
R
M 
M
I


y
M = total moment
 = stress
E = Modulus of Elasticity

E
R
 I
y
I = Second Moment of area
Y = distance from Neutral Axis
R = radius of bend
BENDING STRESSES
Example of general bending Equations
A rectangular beam of uniform cross-section of breadth
100mm and depth 150mm is 4 m long and rests on supports
at its end. It supports a concentrated load of 10kN at its
mid-point. Determine the maximum tensile and
compressive stresses in the beam
bd3
I 
12
0.1 x 0.15 3

12
 2.8  10-5 m 4
Maximum bending stress occurs on the outer surfaces of the beam
(y =+75mm) at mid-point of the beam. (10 kNm)
My
 
I
10  10 3  0.075
= + 26.8 MPa

-5
2.8  10