Mechanics of Solids, Bending Stress

1. A cantilever beam and its cross-section is shown in Figure Q1 (a) and
Figure Q1 (b) respectively. The beam is subjected with a moment M =
100 kN.m at its free end.
a) Calculate the reactions at the support.
b) Sketch the shear force and bending moment diagrams of the
beam.
c) Determine the bending stress at points A, B, C and D.
d) Sketch the bending stress distribution over the depth of the
beam.
Figure Q1 (a) Figure Q1 (b)
R S

2. R S
Follow the +ve
sign convention
0
0
y
S
F
V



Note: A –ve moment because the
moment is bending the beam
downward, and forms a SAD FACE.
Equilibrium
2
VS MS
V
M
+
VM
0
0
100 .
S
S
M
M M
M M kN m

 
   

MM
Pure moment, hence
no shear force
-
a) Reactions at support

3. R S
Shear Force Diagram
3
-100
Bending Moment
Diagram
V(kN)
x(m)
Zero Shear Force
M(kN.m)
x(m)
-100
b) Shear Force and Bending Moment Diagrams

4. c) Bending stress
4
+y
y is the distance measured
from the neutral axis
Hence, must calculate the
location of neutral axis
first for non-symmetric
cross-section
Note: For symmetric cross-
section, the neutral axis is
located at mid-height
-y
My
I
  

5. Firstly, choose a reference point
5
Find the area and centroid of
each section
ref Note, for this case,
it is (Aӯ)1 – (Aӯ)2
1
2
Sec Sign A
(mm2)
ӯ
(mm)
Aӯ
(mm3)
1 + 100250 125 3.13106
2 - 60100 70 4.20105
ΣA 1.9104 ΣAӯ 2.71106
Centroid
ӯ2
ӯ1
6
4
. 2.71 10
142.4
1.9 10
A y
y mm
A

  



Note, for this case,
it is A1 – A2

6. Moment of Inertia 6
Find the distance d between the
centroid of each section and the
neutral axis of the entire section
ref
1
2 Sec Sign A
(mm2)
d
(mm)
Ad2
(mm4)
1 + 100250 17.4 7.57106
2 - 60100 72.4 3.15107
ӯ2
ӯ1
ӯ
d1
d2
Sec Sign b
(mm2)
h
(mm)
I=bh3/12
(mm4)
1 + 100 250 1.3108
2 - 60 100 5106
Determine b and h.
Subsequently, calculate I.
3 3
2 2 8 4
1 2
1.01 10
12 12
tot
bh bh
I Ad Ad mm
   
        
   

7. Bending stress at A
7
ӯ = 142.4 mm
yA = 107.6mm
Point A is located
above the neutral axis.
Hence, yA = +ve.
6
8 4
( 100 10 . )(107.6 )
1.01 10
106.2
A
My
I
N mm mm
mm
MPa
  
 
 



8. Bending stress at B
8
ӯ
Point B is located on
the neutral axis.
Hence, yB = 0.
0B 

9. Bending stress at C
9
yC = -22.4 mm
Point C is located
below the neutral axis.
Hence, yC = -ve.
6
8 4
( 100 10 . )( 22.4 )
1.01 10
22.1
C
My
I
N mm mm
mm
MPa
  
  
 

 
ӯ = 142.4 mm

10. Bending stress at D
10
yD = -ӯ =
-142.4 mm
Point D is located
below the neutral axis.
Hence, yD = – ve.
6
8 4
( 100 10 . )( 142.4 )
1.01 10
140.5
D
My
I
N mm mm
mm
MPa
  
  
 

 

11. d) Stress distribution
11
Bending stress at
point A is +ve.
Bending stress at
point D is - ve.
Connect the two
stresses with a
straight line that
passes through
zero at neutral axis.
-140.5 MPa
106.2 MPa

12. e) How to strengthen the beam?
12