### Lecture #19

```DEFLECTION
Lecture #19
Course Name : DESIGN OF MACHINE ELEMENTS
Course Number: MET 214
Beams (non rotating members) exposed to transverse loads experience bending
moments which causes the fibers of the beam experiencing tension to elongate and the
fibers subjected to compression to be shortened. Such a loading condition causes the
beam to curve which can be characterized by the amount of displacement the beam
shifts in the transverse direction from the unloaded position. The displacement in the
transverse direction is referred to as deflection.
The designs of shafts must prevent excessive deflection. Excessive deflection in intricate,
high speed machinery could interfere with and/or even prevent the desired motion.
Excess deflection could contribute to vibration problems and/or limit the speed of
rotation of a shaft to avoid vibrations.
Transverse loads on shafts can result from the use of pulleys, chains and sprockets,
meshing gears etc. Accordingly, the deflections of a shaft under various loading
configurations must be investigated to insure the shaft will be designed to avoid excessive
deflections during operation.
To quantify the deflections resulting from bending moments, consider the following
illustrations.
As shown below arc length EF is related to ∆θ as follows
EF=P ∆θ
∆θ = EF/P
Due to small angle approximations and/or geometry
 CC '

2
FC
 DD'

2
ED
Due to symmetry, FC=ED
 
  CC ' DD ' CC ' DD '




2
2
FC ED
ED
Note CC ' DD'   max
 
where C  ED
 max
C
Equating the expressions for ∆θ results in the following.
EF
P


EF
  max  max 
C
C
P
1 
1
  max
P EF C
 
Recall that are length EF is associated with a fiber that is unchanged in length. Hence, EF is
equal to the undeformed length of a fiber designated as L.
1  max 1

P
L C
Using the definition of strain


L
1  max

P
C
Hook’s law is utilized to relate εmax to σmax.
1  max

P
C
 max  E max
1  max

P EC
Finally from previous analysis concerning bending moments, σmax can be related to
bending moment.
MC
I
 MC  1 



I
EC



M

EI
 max 
1
P
1
P
Note:
P
1

P
curvature
1 M

P EI
relates loads and bending moments to the geometry associated with deflection.
To develop some insight into the behavior of deflection, note the following trends.
1 M

P EI
As M increases, 1/P increases. If 1/P increases then P decreases
If P goes down this means a tighter radius.
P1  P2
1 1

P1 P2
A larger curvature (1/P) implies more bending.
A very large radius (of curvature) implies not much bending, this implies not much curvature.
As P>>1
1/P -> 0
Ex. A steel blade for a small band saw is 0.025 in thick and runs on pulleys that are 12 in in
diameter. Compute the maximum bending stress caused by bending the blade around the
pulleys. Assume E=30,000,000 psi for the blade.
1 M
EI

M 
P EI
P
MC  EI  C  CE
 max 

  
I
P
P

 I 
.025
C
2
P  6' '
 max  62,500psi
To establish a more definitive relationship between bending moments and shaft
beam deflections and/or slope variations that exist when a shaft deflects under the
influence of bending moments.
To develop the necessary relationships consider the diagram shown below.
 X  l
P ∆θ=l
If θA and θB are relatively small
P  X 
1  d


P X dX
Equating expressions involving 1/P enables bending moments to be related to slope
and/or deflection.
1 d

P dX
1 M
d
M



P EI
dX EI
Solving for dθ
M
dX
EI
Integrating both sides of the equation:
d 
 d  
M
dX
EI
 2  1   
where
1
MdX
EI 
  change in slope of shaft due to bending
 MdX 
area under the moment diagram between any two points
designated as points1 and 2
The change in slope between any two points on a shaft is equal to the area under the moment curve
between the same two points multiplied by 1/EI .
This relationship is illustrated below.
To relate bending moments to deflection, it is necessary to relate deflections to slope.
For small angles, the following approximations are valid. dY
dX
m
m  t an 

Y1

X1
dY
   dY  dX
dX
Integrating both sides of the equation establishes a relationship for deflection
 dy   dX
Y   dX
Accordingly, using relationships established in strength of material classes, the following
 wdX
  VdX
V 
M
 
M
dX
EI
Y   dX
Accordingly, to investigate deflection of a shaft, shear and moment diagrams are
generated and by integrating the moment curve twice, deflections can be determined for
a shaft subjected to transverse loads.
The first technique that will be discussed for determining the deflections of a shaft is the
area-moment method which can be described as follows: The deviation of a point on a
beam from the tangent drawn at a second point is the moment about the first point of the
area under the M/EI curve between the two points.
tB / A   X
M
dX
EI
t B / A  X AA B
M
To apply thisMprinciple it is necessary to create a EI diagram from a moment diagram and
divide the EI diagram into parts, find the area and the centroid of each part from a
vertical line drawn through the point B and combine the results. An example is presented
below to demonstrate how to apply the area moment method to a shaft containing steps
in the diameter of the shaft.
Example: Given the diagram shown below, find the deflection of the shaft under the
600
2
= 300
After bearing reactions have been determined, shear and moment diagrams may be
drawn.
Shear and moment diagrams for the shaft appear on the next slide.
In order to determine the deflection under the load using the area moment method, a

diagram must be prepared from the moment diagram and the EI values existing at

various locations along the shaft. Using the shaft diameters given in the example and
the relationship existing between diameter and the moment of inertia I for a circular
shaft, the values of EI can be determined for each section of the shaft that has a
different diameter by using the equation provided below.
=
4
64
Assuming E = 30 X 106 psi, the values for EI for each section of the shaft that has a
different diameter may be determined.
.75 = 46.596  104
1.4 = 56.469  105 `
After the EI values have been determined, the EI values are applied to the moment

diagram to create a
diagram as shown below.

The

diagram may be divided into sub areas to facilitate the computation of the
deviation under the load. A sub divided

diagram for the example is shown below.
The values for each of the areas shown above have been calculated and are as follows:
A1 = A6 = .00296 2
A2 = A5 = .000632 2
A3= A4 = .000424 2
Using the area moment method, the deviation under the load may be determined from
the following equation.
/ = 1 1/ + 2 2/ + 3 3/
Values for the distances between a line through the point B and the centroids of the areas
are presented below.
1/ = 5
2/ = 2
3/ = 1.333
Using the values provided above the following value for / results.
/ = .016631 in
The value for the deviation / may be converted to deflection. The manipulations that
must be performed can be determined from the diagram shown on the next slide.
ℎ =  + /
ℎ = 7 tan  = 7
→  = ℎ − /
/
14
= .5 /
The following equation may be used to determine the deviation /
/ = 1 1/ + 2 2/ + 3 3/ + 4 4/ + 5 5/ + 6 6/
The values for the areas have already been determined.
The values to be used in locating the centroid of each sub area from a vertical line drawn
through point C are shown below.
1/ = 12
2/ = 9
3/ = 8.33333
4/ = 5.6666 5/ = 5
6/ = 2
Using the above values, the deviation / may be computed:
/ = .0560293
using the value of / and the geometry identified earlier, the value of h may be
computed:
h = 0.5 / = .0280146
Using the value of h and the value of / , the deflection  may be computed from the
following equation:
= h - / = .0280146-.016631 = .01138
```