Edwin M. Odom
Edited by Jessica Shapiro
• Quickly size shafts to required deflections by adding a scaling factor to the software rules.
University of Idaho, www.uidaho.edu
“Calculating Deflections in Stepped Shafts and Nonuniform Beams,” Machine Design, July 6, 1961
“Shoulder-Fillet Stress Finesse,” Machine Design, Jan. 8, 2009
Although engineers are always innovating, some aspects of design don’t 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 calculations can 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 less time. For the method presented here to successfully use these tools, however, engineers must still interpret free-body diagrams and shear, moment, and energy equations to create a digital numerical integration the software can handle.
Start by calculating the internal energy due to bending using Castigliano’s theorem. This theorem, which is useful for finding deflections, can also determine the unknown reactions in statically indeterminate 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 a load applied at that point. That is:
δ = ∂U/∂Q = ∫L0 MQ = 0/(E × I) ∂M/∂Q dx
where M = moment along the length of the beam as a function of x, E = Young’s modulus, and I = area moment of inertia. Don’t forget that for a stepped shaft the moment of inertia is also a function of x.
If the load is not applied at the point of maximum deflection in real life, you can apply a dummy load, Q, at the point where you want to find the deflection. But this raises a problem: Where do you locate the dummy load if you want the maximum deflection? For this method, Q is placed in an arbitrary location but is tracked by a second coordinate system, denoted ξ.
To illustrate this approach, consider a simply supported stepped shaft of length, L, with three different 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 at x = 2 in. and a 2,100-lb upward force at x = 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, ξ, to describe the distance from the left support to the dummy load, Q.
Next, calculate reactions RL and RR using the equations of equilibrium, that is, ΣFy = 0 and ΣM = 0. This yields:
RL = 1,344 + [(L – ξ)/L]Q
RL = –444 + ξQ/L
These expressions for the reactions have two parts, a load which depends on the shaft loading and dimensions and an expression containing the dummy load, Q, the beam length, L, and the point-of-interest coordinate position, ξ.
The next step is to write the moment equation, but first, it helps to introduce a step function to deal with the differing diameters in the shaft.
The Heavyside step function is defined as:
H(a, b) = 0 if a < b
H(a, b) = 1 if a ≥ b.
When used to write moment equations, Heavyside step functions serve the same purpose as a singularity function or Macaulay function. Any of these can be used in this analysis.
Now, write the moment equation as a function of x:
M = 1,344x + [(L – ξ)/L]Qx – 3,000(x – 2)H(x, 2) – Q(x – ξ)H(x, ξ) + 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 – ξ)H(x, ξ) is the moment caused by the dummy load, Q, when coordinate x becomes greater than the point-of-interest coordinate, ξ. The moment arm is (x – ξ) and the term is not active as long as x < ξ.
Next, calculate the partial derivative of the moment equation with respect to the dummy load, Q:
∂M/∂Q = (L – ξ) × x/L – (x – ξ) × H(x, ξ).
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 δ:
δ = 1/E ∫L0 [1,344x – 3,000(x – 2)H(x, 2) + 2,100(x – 4.125)H(x, 4.125)]/I × [(L – ξ) × x/L – (x – ξ) × H(x, ξ)] dx.
Evaluate this integral using the distance ξ from the left end of the shaft to the dummy load to find the deflection, δ, at that point.
The fact that the moment of inertia, I, of the shaft is a function of x complicates the numerical integration. One way to simplify it is to turn to mathematics software like TKSolver and add a look-up table during the numerical integration.
In TKSolver, add a Rule that calls four different functions: the Simpson numerical function, a function for the integrand of the equation above, a function to map the diameter of the shaft to the distance along the shaft, and a function for the Heavyside step function.
The Rule will convert a list points of interest, ξ, to a list of deflections, δ. You can then plot the deflections as a function of x to visualize the behavior of the shaft.
In most shafts, bending is the primary source of deflection. However, for short shafts where length is less than 10 times the maximum diameter, deflection due to shear can become significant. While the details are not presented here, use an analogous procedure to get the deflections due to shear by evaluating:
δshear = ∂U/∂Q = ∫L0 VQ = 0/(E × A) ∂V / ∂Q dx
where V = shear loading expression and A = 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 the terms 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 design constraint. 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 of taking a partial derivative of the internal energy with respect to Q, calculate the partial derivative with respect to m:
θ = ∂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 with multiple masses such as gears, Rayleigh’s method indicates the critical speed, ω, in rad/sec is:
ω = √[(g × Σwiδi)/(Σwiδi2)]
where g = gravitational constant, wi = weight of the ith gear, and δi= displacement of the ith gear. δi is determined from the displacement equation at the respective tracking coordinate, ξi.
You can also use the method presented here to size a shaft for a specified maximum deflection. Once you know the location, ξ, of the maximum deflection, add a scaling factor that multiplies the diameters of the shaft before calculating 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 you quickly home in on a solution.
Although the TKSolver technique is shown in the accompanying graphic, the basics of this method work with other software programs as well.
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