Bending
Shear and Moment Diagram,
Graphical method to construct shear
and moment diagram, Bending deformation of a
straight member, The flexure formula
1
2
Shear and moment diagram
Axial load diagram
Torque diagram
Both of these diagrams show the internal forces acting on the members.
Similarly, the shear and moment diagrams show the internal shear and moment
acting on the members
3
Type of Beams
Statically Determinate
Simply Supported Beam
Overhanging Beam
Cantilever Beam
4
Type of Beams
Statically Indeterminate
Continuous Beam
Propped Cantilever Beam
Fixed Beam
5
6
Example 1
F
A
V
B
M
x
Equilibrium equation for 0  x  3m:
F
y
0
 F V  0
V  9kN
M  0
Vx  M  0
M  9 x ( kNm)
* internal V and M should be assumed +ve
Shear Diagram
F
V
M
x
Sign convention: V= -9kN
Lecture 1
8
Shear Diagram
F
V = -9 kN
M = -9x kN.m
x
M=-9x
Sign convention: M= -9x kNm
X=0: M= 0
X=3: M=-27kNm
9
At cross section A-A
At section A-A
V=9kN
M=X
X
10
Example 2
1) Find all the external forces
M A  0;
10(1)  C y ( 2)  0
C y  5 kN
Fy  0
A y  5  10  0
Ay  5 k N
11
Force equilibrium
Moment equilibrium
0  x  1m
0  x  1m
F
y
M
0
0
M  Vx  0
F V  0
V  5kN ( down)
M  Vx  5kNm(ccw)
Force equilibrium
1  x  2m
F
A
y
0
F  10  V  0
V  5kN (up)
Moment equilibrium
1  x  2m
M
A
0
Vx  10(1)  M  0
M  10  Vx
M  (10  5 x)kNm(ccw)
12
Boundary cond for V and M
0  x  1m
V  5kN (down)
M  5 x(ccw)
M=5x
1  x  2m
V  5kN (up)
M  10  5 x(ccw)
M=10-5x
Solve it
Draw the shear and moment diagrams for
simply supported beam.
14
15
Distributed Load
For calculation purposes, distributed load can be represented as a single
load acting on the center point of the distributed area.
Total force = area of distributed load (W : height and L: length)
Point of action: center point of the area
16
Example
17
Example
18
Solve it
Draw the shear and moment diagrams the
beam:
19
Solving all the external loads
Distributed load will be
F  Wl
 8(6)  48kN
Solving the FBD
M
A
0
Bx 4  F (3)  0
48(3)
4
Ay  12kN
By 
 36kN
Ax  0
Boundary Condition
0 x4
Equilibrium eq
F
Y
0
12  8x  V  0
V  12  8x
M
A
0
M  Vx  4 x 2  0
M  (12  8 x) x  4 x  0
M  12x  4 x 2  0
M  12x  4 x 2
21
Boundary Condition
4 x6
Equilibrium eq
F
Y
0
12  36  8 x  V  0
V  48  8 x
M
A
0
M  Vx  8 x( x / 2)  0
M  (48  8 x) x  36(4)  4 x 2  0
M  48x  144 4 x 2  0
M  4 x 2  48x  144
22
0 x4
V  12  8 x
x=0
x=4
V= 12 kN
V=-20 kN
M  12x  4 x 2
x=0
x=4
4 x6
M= 0 kN
V=-16 kN
V  48  8 x
x=4
x=6
V= 16kN
V= 0 kN
M  4 x 2  48x  144
x=4
x=6
V= -16kN
V= 0 kN
23
Graph based on equations
Straight horizontal line
y=c
y = mx + c
y=3x + 3
y=-3x + 3
y = ax2 + bx +c
y=3x2 + 3
y=-3x2 + 3
24
25
Graphical method
• Relationship between load and
shear:
  Fy  0 :
V  w x x  (V  V )  0
V   w x x
• Relationship between shear and
bending moment:
M o  0 :
 Vx  M  w x xk x   M  M   0
M  Vx  w x k x 
2
26
26
Dividing by x and taking the limit as x0, the
above two equations become:
Regions of distributed load:
dV
  w(x )
dx
Slope of shear =  distributed
diagram at
load intensity
each point
at each point
dM
V
dx
Slope of
moment
diagram
at each
point
= shear at each
point
27
27
Example
28
29
The previous equations become:
V   w( x) dx
change in
shear
=
Area under
distributed load
M   V ( x) dx
change in
moment
=
Area under shear
diagram
30
+ve area under shear diagram
31
32
33
Bending deformation of a straight
member
Observation:
- bottom line : longer
- top line: shorter
- Middle line: remain the same but rotate (neutral line)
34
Strain
s ' s
s   0
s
  lim
Before deformation
s  x
After deformation, x has a radius of
curvature r, with center of curvature at
point O’
s  x  r
Similarly
Therefore
s'  ( r  y)
( r  y )  r
s   0
r
y
 
r
  lim
35
Maximum strain will be
 max  
c
r

y/r
)
c/r
 max
 (
y
  ( ) max
c
y
   ( )  max
c
-ve: compressive state
+ve: tension
36
The Flexure Formula
The location of neutral axis is
when the resultant force of the
tension and compression is equal
to zero.
FR   F  0
Noting
dF   dA
0   dF    dA
A
y
   ( ) maxdA
c
A

 max
c
 ydA
A
37
Since
 max
c
 0 , therefore
 ydA  0
A
Therefore, the neutral axis should
be the centroidal axis
38
(M R ) Z   M Z
Maximum normal stress
M   ydF   y dA
A
A
y
  y ( ) max dA
c
A

 max
c
 max 
2
y
 dA
A
Mc
I
Normal stress at y distance
 
My
I
39
Line NA: neutral axis
Mc


max
Red Line: max normal stress
I
c = 60 mm
Yellow Line: max compressive stress
c = 60mm
Mc
 max  
I
Line NA: neutral axis
Red Line: Compressive stress
y1 = 30 mm
Yellow Line: Normal stress
y2 = 50mm
Refer to Example 6.11 pp 289
1  
2 
My1
I
My 2
I
I: moment of inertial of the cross
sectional area
I x x
I x x 
bh3

12
 r4
4

 D4
64
Find the stresses at
A and B
I: moment of inertial of the cross
sectional area
Locate the centroid (coincide
with neutral axis)
n
y
y A
i 1
n
i
A
i 1

i
i
y1 A1  y2 A2
A1  A2
150(50)(300)  325(50)(300)
(50)(300)  (50)(300)
 237.5m m

I: moment of inertial of the cross
sectional area
Profile I
I about Centroidal axis
II 
A
A
bh3 50(300)3

 112.5(10) 6 m m4
12
12
I about Axis A-A using parallel axis
theorem
bh3
 Ad 2  112.5(10) 6  (50)(300)(87.5) 2
12
 227.344(10) 6 m m4
( I I ) A A 
Profile II
bh3
(300)(50)3
 Ad 2 
 (300)(50)(87.5) 2
12
12
6
4
 117.969(10) m m
( I II ) A A 
Total I
I A A  ( I I ) A A  ( I II ) A A
 227.344(10) 6  117.969(10) 6 m m4
 345.313(10) 6 m m4
* Example 6-12 to 6-14 (pp 290-292)
Solve it
If the moment acting on the cross section of the beam is M = 6 kNm,
determine the maximum bending stress on in the beam. Sketch a
three dimensional of the stress distribution acting over the cross
section
If M = 6 kNm, determine the resultant force the bending stress
produces on the top board A of the beam
Total Moment of Inertia
300(40)3
40(300)3
2
I I  2[
 (300)(40)(170) ] 
12
12
 786.8(10) 6 m m4
Max Bending Stress at the top and bottom
M top
1.45MPa
Mc  6000(10)3 (190)


 1.45MPa
I
I
M bottom  1.45MPa
Bottom of the flange
1.14MPa
6kNm
M f _ top  
Mc  6000(10)3 (150)

 1.14MPa
I
I
M f _ bottom  1.14MPa
Resultant F = volume of the trapezoid
300 mm
(1.45  1.14)
(40)(300)  15540N
2
 15.54kN
FR 
40 mm
1.45MPa
1.14MPa
Solve it
The shaft is supported by a smooth thrust load at A and smooth
journal bearing at D. If the shaft has the cross section shown,
determine the absolute maximum bending stress on the shaft
External Forces
Draw the shear and moment diagram
M
A
0
FD (3)  3(0.75)  3(2.25)
FD  3kN
FA  3kN
Absolute Bending
Stress
Mmax = 2.25kNm
Mc 2250(10) 3 ( 40)
 max 


I
(404  254 )
4
 52.8MPa