SHEAR AND MOMENT DIAGRAMS
WITH APPLICATIONS IN TWO
ORTHOGONAL PLANES
Lecture #5
Course Name : DESIGN OF MACHINE ELEMENTS
Course Number: MET 214
Bending is caused when a shaft carries a load at right angles to the longitudinal
(rotational) axis of the shaft. Shafts used in power transmission systems must be designed
to insure the levels of stress and/or the amount of deflection due to bending does not
exceed acceptable limits. In addition, the loads applied to a shaft will effect the selection
and/or the design of the shaft supports. Accordingly, it is necessary to investigate
relationships relating:
1) shaft loads to bending stresses
2) shaft loads to shaft deflections
3) shaft loads to shaft supports (bearings)
Drawing upon the precedent established in strength of materials, the analysis of shafts
with respect to traverse loading can be classified into two categories.
Shaft support method of analysis
Statically determinate
Statically indeterminate
Statically determinate: Reactions at beam support due to beam loadings can be
determined by using only the equations of statics.
F
F
M
H
0
V
0
0
where
FM  Horizontal forces
FV  Vertical forces
M  Moments
Example:
Assuming the support means for the shaft shown in the figure below can be assumed to
be providing reaction forces in the vertical direction only, calculate the value of the
reaction forces at points A and B for the loads shown.
Recall when determining reaction forces, distributed loads may be replaced by an
equivalent concentrated load acting through the centroid of the distributed load.
Accordingly, for the purpose of determining reactions, the loads existing on the
shaft above may be represented as shown below.
Laws of statics provide 3 equations:
F  0
F  0
M  0
H
V
Since there are two unknowns AY and BY , there are a sufficient number of equations in
comparison to the unknowns to solve for AY and BY . Hence the reactions are statically
determinate.
M
A
 0 : 120.275  225.5  BY .4   0
BY .4  33  112.5
BY  363.75
F
V
 0 : AY  BY  120 225  0
BY  363.75
AY  18.75
A negative sign indicates reaction A acts in direction opposite than what is indicated in
figure.
As an example of a statically indeterminate beam situation, consider the beam (non
rotating member) shown below.
If we replace the schematic representation of fixed end supports with the reaction forces
and/or moments being generated by the supports, we will be able to itemize the
unknowns.
Note:
4 unknowns
3 equations from statics
3 equations and 4 unknowns:
Can not solve for reactions in the above problem using just the laws of statics, additional
information is required: Statically indeterminate.
Given a statically determinate version of the problem, compare the two problems and
identify how the reaction moments existing in the statically indeterminate problem effects
behavior of the beam at the locations of the beam supports.
When dealing with shafts, it is not uncommon to have several components mounted to a
single shaft including multiple pulleys, gears, sprockets, etc. In addition, shafts can be
subjected to loads applied at different angles as shown below.
When encountering situations involving loads applied at different angles, it is possible to
resolve forces into components existing in two perpendicular planes and determine
bearing reactions in both planes. The total bearing reaction force can be determined from
the components as follows.
RB  RX2  RY2
where RB  total reaction force existing on a bearing
RX  horizontal reaction force existing at the bearing
RY  vertical reaction force existing at the bearing
As an example of how to determine the total reaction force for each bearing supporting
a shaft consider the following example: For the problem shown below resolve the loads
into components existing in the horizontal and vertical planes.
In order to standardize the calculations for use in this class, the following coordinate
system definitions will be used when resolving forces acting on a shaft into components.
+ X axis is the horizontal axis and is positive into the plane of the paper
+ Y axis is the vertical axis and is positive upward in the plane of the paper
+ Z axis lies along the axis of rotation of the shaft from left to right.
A perspective view of the above coordinate system is provided below.
The vertical plane is formed by the +Y and the +Z axes as shown below.
The horizontal plane is formed by the +X and the +Z axes. Since the +X axis is positive
into the plane of the paper, to depict the horizontal plane in the plane of the paper, a
rotation is required. For the purposed of this class, the rotation necessary to display the
horizontal plane in the plane of the paper is the rotation about the +Z axis that will
rotate the +X axis into the plane of the paper with the +X direction pointing upward. The
horizontal plane is shown below.
Using the prescribed coordinate system definition at pulley A provided below, determine
the vertical and horizontal force components due to 𝐹1 + 𝐹2 = 3000 lbs.
Using the prescribed coordinate system definition at pulley B provided below, determine
the vertical and horizontal force components due to 𝐹1 + 𝐹2 = 2000lbs.
Using the component values found in the previous step, draw a load diagram for the
horizontal plane ( the plane formed by the X and Z axis) including the force components
due to bearing reactions. Assuming the bearings do not produce a moment reaction,
calculate the bearing reactions in the horizontal plane.
Using the component values found in the previous step, draw a load diagram for the
vertical plane (the plane formed by the Y and Z axis) including force components due to
bearing reactions. Assuming the bearings do not produce a moment reaction, calculate the
bearing reactions in the vertical plane.
The load diagrams shown above are statically determinate. Determine the information
concerning bearing reaction forces shown below.
RH1 
RH 2 
RV1 
RV2 
RB1 
RB2 
The total bearing reaction forces RB1 and RB2 can be used to specify the load carrying
capability required of the bearings.
The load diagram, and the corresponding shear and moment diagrams for the
horizontal plane of the previous example are provided below.
v
Note: Dimensions appear on original diagram
Horizontal plane diagrams
Load
Shear
Moment
The load and moment diagram for the vertical plane are provided below.
The total bending moment at any position X from the left end of the shaft may be calculated as follows.
where
M T( z )  total bending moment in beam at location Z
M T ( z )  M H2 ( z )  M V2 ( z )
M H( z )  bending moment in horizontal plane at location Z
M V ( z )  bending moment in vertical plane at location Z
The total bending moment MT(z) generally varies with Z due to MH(z) and MV(z) varying
with Z. It is of interest to find the maximum value MT(Z) of the total bending moment. The
maximum value of MT(Z) may not be obvious from an inspection of the moment diagrams
in the horizontal and vertical planes. Accordingly, several values of MT(Z) should be
calculated at various locations Z and the largest value identified. In the example provided
above, the maximum bending moment occurs at the left pulley and has the following
value.
M T ( A) max  25,1282  36,432 2  44,257in  lb
By comparison, the total moment existing at pulley B has the following value
M T ( B ) max  22,624 2  22,627 2  31,997in  lb
Obviously MT (A) > MT (B) and therefore
MT (A) is considered the maximum total moment for all positions Z and is designated as
such by the subscript max.
MT (A)max = 44,257 in-lb