3 bending stress-asgn

Mechanics of Solids, Bending Stress

1. Acantileverbeamanditscross-sectionisshowninFigureQ1(a)and
FigureQ1(b)respectively.ThebeamissubjectedwithamomentM=
100kN.matitsfreeend.
a)Calculatethereactionsatthesupport.
b)Sketchtheshearforceandbendingmomentdiagramsofthe
beam.
c)DeterminethebendingstressatpointsA,B,CandD.
d)Sketchthebendingstressdistributionoverthedepthofthe
beam.
Figure Q1 (a) Figure Q1 (b)
R S

2. R S
Follow the +ve
sign convention0
0
y
S
F
V



Note: A –vemoment because the
moment is bending the beam
downward, and forms a SAD FACE.
Equilibrium
2
V
S
M
S
V
M
+
V
M0
0
100.
S
S
M
MM
MMkNm




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
yis 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
-yMy
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
SecSign A
(mm
2
)
ӯ
(mm)
Aӯ
(mm
3
)
1 +1002501253.1310
6
2 -60100704.2010
5
ΣA1.910
4
ΣAӯ2.7110
6
Centroid
ӯ
2
ӯ
16
4
.2.7110
142.4
1.910
Ay
y mm
A

 



Note, for this case,
it is A
1–A
2

6. Moment of Inertia
6
Find the distance dbetween the
centroid of each section and the
neutral axis of the entire section
ref
1
2
SecSign A
(mm
2
)
d
(mm)
Ad
2
(mm
4
)
1 +10025017.47.5710
6
2 -6010072.43.1510
7
ӯ
2
ӯ
1
ӯ
d
1
d
2
SecSignb
(mm
2
)
h
(mm)
I=bh
3
/12
(mm
4
)
1 + 1002501.310
8
2 - 60 100 510
6
Determine band h.
Subsequently, calculate I. 33
2 2 8 4
12
1.0110
12 12
tot
bh bh
I Ad Ad mm
  
  
  

7. Bending stress at A
7
ӯ= 142.4 mm
y
A= 107.6mm
Point A is located
above the neutral axis.
Hence, y
A= +ve.6
84
(10010.)(107.6)
1.0110
106.2
A
My
I
Nmmmm
mm
MPa






8. Bending stress at B
8
ӯ
Point B is located on
the neutral axis.
Hence, y
B= 0.0
B


9. Bending stress at C
9
y
C= -22.4 mm
Point C is located
below the neutral axis.
Hence, y
C= -ve.6
84
(10010.)(22.4)
1.0110
22.1
C
My
I
Nmmmm
mm
MPa

 



ӯ= 142.4 mm

10. Bending stress at D
10
y
D= -ӯ=
-142.4 mm
Point Dis located
below the neutral axis.
Hence, y
D= –ve.6
84
(10010.)(142.4)
1.0110
140.5
D
My
I
Nmmmm
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