### stresses in shafts due to shear forces and bending moments

```STRESSES IN SHAFTS DUE TO SHEAR
FORCES AND BENDING MOMENTS
Lecture #6
Course Name : DESIGN OF MACHINE ELEMENTS
Course Number: MET 214
After generating shear and moment diagrams for a shaft, the stresses accompanying the
internal shear forces and bending moments needs to be investigated so that the shaft can
be designed (specify material type and dimensions) to accommodate the stresses and/or
deflections that results from the transverse loads applied to the shaft.
In order to develop relationships relating bending moments in shafts to the amount of
stress the bending moments generate, it is essential to understand the geometry
associated with bending. Shown in the figures below are a unloaded beam having a
rectangular cross section and the same beam experiencing a transverse loading condition.
A side view of the situation under analysis is depicted below.
Note: F   a
Y
Y
Since
Y 
Y
C
 max
Y 
FY    max a
C 
The force FY will produce a moment MY about the neutral axis.
M
M
Y
 F YY
Y

Y
C
 max aY
2
M
Y

Y a
max
C
To determine the total moment, all of the individual moments must be added together.
M ( x) 

MY 

  max
 max
C
Y
2
a
C
Y
2
a
Recall the definition of second moment of area or the moment of inertia of the cross
section about its own centroidal axes (not to be confused with polar moment of area J or
mass moment of inertia Jm )
IX 
Y
M (z) 
2
is area moment of inertia of cross section about centroidal X axis.
a
 max
C
Ix
M ( z )  Total internal resisting moment.
Found from moment diagram.
Rearranging the formula
 max 
M max C
Ix
Since shafts are frequently formed from solid circular cylinders, recall the definition of I for
a circular cross section as shown below.
Accordingly, for solid circular cylindrical shafts,
M TC
 max 
I
I 
d
64
4

1
J
2
where J= Polar area moment
Since stress is proportional to distance
   max
 
C
M TC Y
I
 
Y
C
M TY
I
Also note that M can be expressed in terms of
M T   max
I
C
I
S 
C
  max S
Section Modulus
To determine the shaft diameter needed in a particular application, use the formula for
bending moments and substitute the expression for I in terms of D and solve for the
diameter D.
MT 
MT 
MT 
 max
I
C
4
 max   D 


D / 2  64 

32
 max D
3
Formula may be rearranged and solved for D as shown below
D 
3
32 M T

max
MT : Take greatest value from composite moment diagram
σmax : determined from material choice for the shaft and factor of safety
=

If the shaft has a diameter of 50 mm, determine the absolute maximum bending stress in the shaft.
Additional situations exist where it is of interest to combine stresses arising from different
As an example that will be encountered frequently in the design of machine elements,
consider the shaft shown below which is supported by two bearings and carries two V
belt sheaves. During steady state operation, the sheaves (pulleys) apply reaction torques
to the shafts as indicated by the arrows shown on the figure. Due to tension in the belts,
the belts also apply a transverse load to the shaft resulting in bending moments.
Previously, torsional shear stresses and the stresses due to bending were treated
independently. Due to the nature of the loading applied to the shaft in this example, it is
necessary to determine the effects of both stresses acting in combination.
The accompanying shear and moment diagrams are shown below.
Note
 max 
MC
I

 2
M D
 D
4

M 32 
 D
3

1540 32 
3 . 14 1 . 25 3
 8030 psi
64
Recall a negative sign for bending moment indicates the upper fibers (i.e. fibers lying
above the neutral axis) are in tension.
To calculate the amount of shear stress due to torsion, note the amount of torque
declared for the shaft as indicated in original figure.
T=1100 in-lb
 xy 
TC
J

 2
T D
D
4

16 T
D
3

16 1100 
3 . 14 1 . 25 3
 2870 psi
32
These stress levels exist at the outside radius of the shaft as shown below.
Stresses on element k appear below
stresses which are tensile or compressive in nature as depicted above on element k. The
torque applied to the shaft leads to a torsional shear stress on element k as depicted
above. To assist in the visualization of the stresses existing on element k, envision a band
aide applied to the shaft before the shaft experiences loading and then note how the
shape of the band aid changes after the shaft is placed under load. The bending forces
due to belt tension would have a tendency to stretch the band aid as the shaft bows in
response to the transverse loads. See diagram on previous slide. The torque applied to
the shaft would have a tendency to twist the band aid and thereby create torsional shear
stresses.
Since element k involves both tensile stresses (due to bending moments) and torsional
shear stresses, the maximum amount of shear stress associated with the element must
be found by utilizing equations associated with Mohr’s circle.
Using equation 4-4 from the book by Mott and noting σY=0, for expression for τmax can be
rewritten as follows
2
 X 

 
 2 
 max 
2
XY
Substituting the definitions for σY and τXY results in the following expression for τmax.
32 M
X 
3
D
 XY 
16 T
D
3
2
 max 
 32 M 
 16 T 





3
3
2

D

D




 max 
 16 

 M
3
D 
2
 16


 M
3

D


 max  
2
2
T
T
2
2

2
The above equation for τmax does not include effects of varying loads and stress
concentrations. If the shaft diameter must be determined for a particular application, the
above equation may be rearranged to solve for D.
As will be evidently shortly, the equation involving  max on the previous slides is very
similar to the equation adopted by the ASME in 1927 for transmission shaft design.
Accordingly to the book titled “Machine Design for Mobile and Industrial Applications”
by Krutz et al, although the code has been replaced by a newer and more sophisticated
one, it is often still used due to its simplicity.
As indicated in the book titled “Machine Design for Mobile and Industrial Applications”
by Krutz et al, stresses due to the use of a keyway in the shaft and/or stresses due to
press fitting the hub of a component onto a shaft must be accounted for when sizing a
shaft accordingly to the 1927 standard. As suggested in the book, one way to handle
each of these problems is to increase the factor of safety existing in the equation of the
standard. Although other texts books assert that the 1927 standard is obsolete, it is
believed that adopting the suggestions made in the text book by Krutz et al will help the
student obtain insight concerning factors that must be considered in the design of a
shaft. The use of a simple approach enables a context to be established for the material
that is consistent with material the student has been exposed to previously, such as
Mohr’s circle. Such considerations would be lacking in alternative approaches that
emphasize detailed calculations that are beyond the comprehension of most technology
students at this level. It is for these reasons the 1927 standard will be used to illustrate
the basic issues that must be considered when integrating several different components
onto a shaft including bearings. After an initial exposure to the material has been
established, a more contemporary approach may be undertaken using any number of
text books including the book by Mott. Accordingly, the 1927 standard will be used as a
bridge to enable the student to approach more contemporary information by
introducing the students to a context consistent with prior exposure.
Shafts, Keys and Couplings
The accepted code for the design of transmission shafting is the American Standards
Association Standard B17C-1927. It refers to all shafts for transmitting power and
includes the effect of the type of load. Common usage, however, classifies shafts in many
ways. For Example, there are shafts for prime movers such as generator, engine, and
turbine shafts; shafts for power transmission such as line shafts, countershafts, and jack
shafts, as well as machine shafts and spindles.
A member carrying rotating wheels, gears, etc., is generally referred to as an axle; for
example, the rear axle of an automobile. An axle may be either stationary or rotating.
Short shafts on machines are usually referred to as spindles. When one or more pieces
of shafting joined by couplings receive motion from a prime mover, and transmit motion
to machines, the term line shafting is applied. When a shaft is placed is placed between
the driver and the line shaft, it is a counter shaft or jack shaft. A stub shaft or head shaft
is connected directly to a motor or engine.
In addition to various types of shafts indicated, shafts are usually classified as to form.
The most common form of shafting is of a solid circular cross section and is called a solid
shaft. There are also hollow shafts. Flexible shafts make possible the transmission of
power around corners, and examples are found in dental tools, small power tools, and
the common speedometer drive in automobiles. A flexible shaft is of necessity limited
to relatively short lengths and the transmission of light loads.
Shock and Fatigue factors:
The code mentioned previously gives factors to be used for various manners in which
the torsional and moment loads are applied.
The below table lists the recommended factors . They are to be applied to the torque T
and moment M so that equation becomes
a 
16
d
3
 K t T 2   K m M 2
Values for Km and Kt
Km
Kt
Stationary shafts:
Suddenly applied
1.0
1.5 to 2.0
1.0
1.5 to 2.0
1.5
1.5 to 2.0
2.0 to 3.0
1.0
1.0 to 1.5
1.5 to 3.0
Rotating shafts:
Suddenly applied, minor
shocks
Suddenly applied, heavy
shocks
It must be emphasized that the 1927 equation can be rewritten in several equivalent,
but alternative ways.
d 
3
16
 K t T 2   K m M 2
 a
Where  =  ℎ
Ex: A solid circular shaft free of keyways is transmitting 50 hp and rotates at 200 rpm.
Assuming that 48,000 psi material is used in the shaft with a safety factor of 6 and that
the shaft is subject to steady torsional stresses only, what shaft diameter is required?
Ex. A solid circular shaft for power transmission with keyways is subject to a maximum torque
of 6000 in-lb. What is the required shaft diameter, if the allowable shear stress is 6000 psi
and if there will be sudden applications of load with minor shock?
Shear in straight members
Beams generally support both shear and moment loadings. The shear V is the result of a transverse
shear-stress distribution that acts over the beam’s cross-section. Due to the complementary property
of shear, shear stresses will also act along longitudinal planes of the beam. For example, a typical
element removed from the interior point on the cross section is subjected to both transverse and
longitudinal shear stress as shown below.
Shear Stress Distribution
Shear stress distribution for a
Rectangular cross section
A situation that is encountered frequently in power transmission systems involves
combining a torsional shear stress distribution with a shear stress distribution arising
from a shear force. Such a situation exists at a bearing support that is supporting a
section of shaft experiencing pure torsion. The shear stresses due to the shear forces
arises from the bearing reactions existing at the bearing supports. A typical situation
involving shear stress combinations is shown below.
The situation shown on the previous slide involves a solid circular shaft experiencing
torsional shear stresses and shear stresses due to a shear force which could be
envisioned as resulting from a bearing reaction force applied to the shaft. As will be
evident shortly, other situations exist which could lead to the same combination shown
on the previous slide.
Due to the nature of the distributions and how shear stresses combine, the two
distributions shown on the previous slide can only be added when the distributions are
either parallel or antiparallel. However, it is still possible to identify the maximum value
of stress that exists in the combined distribution. The maximum value of stress existing
with the combined distributions can be calculated from the following formula also
shown on the previous slide.
=  +   =
4
3
+

Where  = ℎ
=       ℎ
To identify how the shear stress distributions described previously may be combined to establish the
maximum and minimum shear stress level existing when both distributions are present on the same
cutting plane, consider the example shown below involving a simplified representation of a gear
transmitting a particular level of torque established by the force and the geometry of the situation.
Ex: A 3 inch diameter gear transmits 5HP to a short 1 inch diameter shaft at 300 rpm.
To determine F note the following.
P 
Tn
 T  1050 in  lbs
63000
T  rF  F  700 lbs
To investigate the shear stress distribution existing with the above applied load, an equivalent
representation may be utilized involving a shear force acting through the centroid perpendicular to the
neutral axis of the shaft and a torque T as shown below.
The shear stress distribution resulting from the equivalent shear force is shown below.
where
 max 
4V
 max 
4
3 A
700
3  1 
2
 1190 psi
4
The shear stress distribution resulting from torsion is shown below.
 max 
TC
J

 2
T D
D
4

16 T
D
3
32
 max 
16 1050 
 5350
3 . 14 1 3
psi
Shear stress levels associated with the two distributions shown above can be added only when the
shear stresses from both distributions are parallel or anti-parallel. Since the orientation of the shear
stress distribution due to torsion varies across the cutting plane, the shear stress levels associated
with the equivalent representation can only be obtained along the horizontal axis in this particular
instance. Fortunately, the stress levels obtain by combining the two distributions for points along the
horizontal axis include the maximum and minimum values as illustrated below.
It should be noted that the above example transformed a tangential force acting at the radius of a
shaft to a force acting through the centroid of the shaft and a torque applied to the shaft. The
equivalent force acting through the centroid give rise to shear stresses and bending stresses and the
force is also responsible for creating deflections in the shaft assuming the location of the component
is not at a bearing support. The effect of the force as it contributes to bending stresses will be
accounted for in shear and moment diagrams and consequently, the force’s influence as it pertains to
bending stresses is accounted for in the equation existing with the 1927 standard for shafting.
However, it should be noted that if the equation existing with the 1927 standard is used to determine
the shaft diameter, then the stresses due to shearing forces will be neglected. Given the disparity
that typically exists between shear stresses due to shear forces and bending stresses due to
transverse loads leading to bending moment, such an assumption is usually acceptable.
```