SHEAR AND MOMENT DIAGRAMS WITH APPLICATIONS IN TWO ORTHOGONAL

PLANESLecture #5

Course Name : DESIGN OF MACHINE ELEMENTSCourse 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 stresses2) 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.

where Horizontal forces Vertical forces 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.

0

0

0

M

F

F

V

H

MFF

V

M

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:

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.

0

0

0

M

F

F

V

H

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.

75.1875.363

0225120:0

75.3635.112334.

04.5.225275.120:0

Y

Y

YYV

Y

Y

YA

AB

BAF

BB

BM

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 unknowns3 equations from statics3 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.

where total reaction force existing on a bearing horizontal reaction force existing at the

bearing vertical reaction force existing at the bearing

22YXB RRR

Y

X

B

RRR

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 + = 3000 lbs.

Using the prescribed coordinate system definition at pulley B provided below, determine the vertical and horizontal force components due to = 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.

The total bearing reaction forces RB1 and RB2 can be used to specify the load carrying capability required of the bearings.

2

1

2

1

2

1

B

B

V

V

H

H

R

R

R

R

R

R

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 LoadShearMoment

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

total bending moment in beam at location Zbending moment in horizontal plane at location Zbending moment in vertical plane at location Z

)()()( 22 zMzMzM VHT

)()()(

zMzMzM

V

H

T

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.

By comparison, the total moment existing at pulley B has the following value

Obviously MT (A) > MT (B) and thereforeMT (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

lbinAMT 257,44432,36128,25)( 22max

lbinBMT 997,31627,22624,22)( 22max