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

Due to symmetry, FC=ED Notewhere Equating the expressions for ∆θ results in the following. 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. Using the definition of strain

Hook’s law is utilized to relate ε max to σ max. Finally from previous analysis concerning bending moments, σ max can be related to bending moment. Note: radius of curvature curvature

relates loads and bending moments to the geometry associated with deflection. To develop some insight into the behavior of deflection, note the following trends. As M increases, 1/P increases. If 1/P increases then P decreases If P goes down this means a tighter radius. 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 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.

To establish a more definitive relationship between bending moments and shaft deflections due to transverse loading, the radius of curvature P must be related to the 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. P ∆θ=l If θ A and θ B are relatively small

Equating expressions involving 1/P enables bending moments to be related to slope and/or deflection. Solving for dθ Integrating both sides of the equation: where change in slope of shaft due to bending 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.

Integrating both sides of the equation establishes a relationship for deflection Accordingly, using relationships established in strength of material classes, the following equations relate beam loading, shear, moment, slope and deflection. 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. To apply this principle it is necessary to create a diagram from a moment diagram and divide the 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.