8/6/2019 Calc of Deflection Due 2 Bending in Shafts

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Although engineers are always innovating, some aspects of design dont change.For instance, almost any machine that features rotary motion needs a shaft, and designers need to

know how that shaft will deflect. For shafts with stepped or tapered cross sections, such calculationscan be difficult.

Where previous generations of engineers used mechanics of materials theory to estimate deflections,such as direct integration or area moments to find deflections, today modern software like TKSolver,Mathematica, MathCad, EES, and spreadsheets can calculate deflections more accurately in lesstime. For the method presented here to successfully use these tools, however, engineers must stillinterpret free-body diagrams and shear, moment, and energy equations to create a digital numericalintegration the software can handle.

Start by calculating the internal energy due to bending using Castiglianos theorem. Thistheorem, which is useful for finding deflections, can also determine the unknown reactions in staticallyindeterminate structures. It requires less algebra and calculus than direct integration methods.

The theorem states that deflection at any point equals the partial derivative of the strain energy with respect to aload applied at that point. That is:

= U/Q = L0MQ = 0/(E I) M/Q dx

where M= moment along the length of the beam as a function ofx, E= Youngs modulus, and I= area momentof inertia. Dont forget that for a stepped shaft the moment of inertia is also a function ofx.

If the load is not applied at the point of maximum deflection in real life, you can apply a dummy load, Q, at thepoint where you want to find the deflection. But this raises a problem: Where do you locate the dummy load if youwant the maximum deflection? For this method, Q is placed in an arbitrary location but is tracked by a secondcoordinate system, denoted .

Shaft sample

To illustrate this approach, consider a simply supported stepped shaft of length, L, with threedifferent cross sections and two external loads. Define x as the distance from the left support.

The two loads are a 3,000-lb downward force atx= 2 in. and a 2,100-lb upward force atx= 4.125 in.

Section 1,x= 0 to 0.875 in., is 1.5 in. in diameter. Section 2,x= 0.875 to 3.125 in., is 2 in. in diameter. Section 3,x= 3.125 to 5.125 in., has a 1.75-in. diameter. In section 4,x= 5.125 to 6 in., diameter is again 1.5 in.

The first step is to add the second coordinate system, , to describe the distance from the left support to thedummy load, Q.

Next, calculate reactions RL and RR using the equations of equilibrium, that is, Fy = 0 and M = 0. This yields:

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RL = 1,344 + [(L )/L]Q

and

RL = 444 + Q/L

These expressions for the reactions have two parts, a load which depends on the shaftloading and dimensions and an expression containing the dummy load,Q, the beam length, L, and the point-of-interest coordinate position, .

The next step is to write the moment equation, but first, it helps to introduce a step function to deal with thediffering diameters in the shaft.

The Heavyside step function is defined as:

H(a, b) = 0 ifa < b

H(a, b) = 1 ifa b.

When used to write moment equations, Heavyside step functions serve the same purpose as a singularityfunction or Macaulay function. Any of these can be used in this analysis.

Now, write the moment equation as a function ofx:

M = 1,344x + [(L )/L]Qx 3,000(x 2)H(x, 2) Q(x )H(x, ) + 2,100(x 4.125)H(x, 4.125) .

The terms in this equation are in order from left to right as they are encountered in the stepped-shaft illustration.

The third term, Q(x )H(x, ) is the moment caused by the dummy load, Q, when coordinate x becomes greaterthan the point-of-interest coordinate, . The moment arm is (x ) and the term is not active as long as x < .

Next, calculate the partial derivative of the moment equation with respect to the dummy load, Q:

M/Q = (L ) x/L (x ) H(x, ) .

And in the moment equation, set Q to zero:

MQ = 0= 1,344x 3,000(x 2)H(x, 2) + 2,100(x 4.125)H(x, 4.125).

Substitute these two expressions into the equation above for :

= 1/E L0 [1,344x 3,000(x 2)H(x, 2) + 2,100(x 4.125)H(x, 4.125)]/I [(L ) x/L (x ) H(x, )] dx.

Evaluate this integral using the distance from the left end of the shaft to the dummy load to find the deflection,, at that point.

The fact that the moment of inertia, I, of the shaft is a function of x complicates the numerical integration. Oneway to simplify it is to turn to mathematics software like TKSolver and add a look-up table during the numericalintegration.

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In TKSolver, add a Rule that calls four different functions: the Simpson numerical function, a function for theintegrand of the equation above, a function to map the diameter of the shaft to the distance along the shaft, and afunction for the Heavyside step function.

The Rule will convert a list points of interest, , to a list of deflections, . You can then plot the deflections as afunction of x to visualize the behavior of the shaft.

Function flexibilityIn most shafts, bending is the primary source of deflection. However, for short shafts where length is less than 10times the maximum diameter, deflection due to shear can become significant. While the details are not presentedhere, use an analogous procedure to get the deflections due to shear by evaluating:

shear = U/Q = L

0VQ = 0/(E A) V / Q dx

where V= shear loading expression andA = cross-sectional area of the shaft. For the shaft used in this example,it turns out that the deflection due to shear is about 50% of the deflection due to bending.

Our example used a point load, but the method works for distributed loads as well. The only change is to theterms in the moment equation which you can find in a mechanics of materials reference.

In shaft design, the slope at the reactions (where bearings are located) or where two gears mesh can be a designconstraint. Calculate the slope using a dummy moment, m, in addition to the dummy load, Q.

In this case, the moment equation for the reactions of the shaft will also include a term containing m. Instead oftaking a partial derivative of the internal energy with respect to Q, calculate the partial derivative with respect tom:

= U/m = L0 Mm = 0/(E I) M/m dx.

Designers need to know the critical speed, that is, the rpm where deflections become unstable. For a shaft withmultiple masses such as gears, Rayleighs method indicates the critical speed, , in rad/sec is:

= [(g wii)/(wii2

)]

where g= gravitational constant, wi= weight of the ith gear, and i= displacement of the ith gear. i is determinedfrom the displacement equation at the respective tracking coordinate, i.

You can also use the method presented here to size a shaft for a specified maximum deflection. Once you knowthe location, , of the maximum deflection, add a scaling factor that multiplies the diameters of the shaft beforecalculating the area moment of inertia, I.

A scaling factor of one represents the original cross section. Increasing the scaling factor increases the cross-sectional diameter and decreases deflection. A list that automatically adjusts the scaling factor can help youquickly home in on a solution.

Although the TKSolver technique is shown in the accompanying graphic, the basics of this method work withother software programs as well.