22-Mec-A4 Design and Manufacture of Machine Elements · December 2016

Question 5 of 6

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 5 (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) the load, shear, bending, torque and axial diagrams; (b) the radial and axial loads carried by bearings A and B; (c) a preliminary shaft layout with reasonable diameters, shoulders and axial retention; (d) the factors that set those diameters.

Approach. Transfer the three tooth loads to the shaft axis — F_t and F_r as transverse forces in two perpendicular planes, F_t also as the driving torque F_tr_p, and F_a as an axial force plus a bending couple F_ar_p in the radial plane. Solve each plane as a separate simply supported beam with an overhang, combine the bearing reactions vectorially, then size the shaft at the critical section by a fatigue criterion.

(a) and (b) Reactions and diagrams

1. Torque and the thrust couple. The tangential load acting at the pitch radius delivers $$T=F_{t}r_{p}=4092(0.035)=143.2\ \text{N}\!\cdot\!\text{m}$$ and the thrust, which also acts at the pitch radius rather than on the axis, applies a concentrated bending couple at the pinion: $$M_{a}=F_{a}r_{p}=2235(0.035)=78.2\ \text{N}\!\cdot\!\text{m}$$ This couple lies in the same plane as F_r, and it is the feature that distinguishes a helical-gear shaft from a spur-gear shaft.
2. Vertical (radial) plane. Taking moments about A for the beam A–B with the 65 mm overhang, with distances measured from A and the couple applied at the pinion, $$\sum M_{A}=0:\qquad R_{By}(0.140)=-F_{r}(0.205)+M_{a}=-1702(0.205)+78.2$$ $$R_{By}=-1933\ \text{N},\qquad R_{Ay}=-F_{r}-R_{By}=-1702+1933=+231\ \text{N}$$ The thrust couple opposes the radial load and reverses the sign of the reaction at B relative to a spur pinion — it removes about 22 % of the radial-plane reaction that F_r alone would produce.
3. Horizontal (tangential) plane. With no couple in this plane, $$R_{Bx}=-\frac{F_{t}(0.205)}{0.140}=-\frac{4092(0.205)}{0.140}=-5992\ \text{N},\qquad R_{Ax}=-F_{t}-R_{Bx}=+1900\ \text{N}$$ The negative sign means the reaction at B acts opposite to F_t, which is the expected behaviour of an overhung load: bearing B is pressed one way, bearing A the other.
4. Resultant radial bearing loads. Combining the two planes at each bearing, $$R_{A}=\sqrt{1900^{2}+231^{2}}=1914\ \text{N}$$ $$\boxed{R_{A}=1.91\ \text{kN},\qquad R_{B}=\sqrt{5992^{2}+1933^{2}}=6296\ \text{N}=6.30\ \text{kN}}$$ Bearing B carries 3.3 times the radial load of bearing A, which is the price of overhanging the pinion.
5. Axial load. The full thrust F_a = 2235 N must be reacted by whichever bearing is axially located; the other bearing is allowed to float so that thermal expansion does not preload the pair. With the pinion overhung beyond B, the natural choice is to locate bearing B and float A: $$\boxed{\text{Bearing B: }F_{\text{axial}}=2235\ \text{N};\quad\text{Bearing A: }F_{\text{axial}}=0}$$ The axial-force diagram is then a constant 2235 N over the 65 mm between the pinion and B, and zero elsewhere. Had A been chosen as the locating bearing, the whole shaft would carry the thrust and A's equivalent bearing load would rise correspondingly.
6. Bending-moment values. In the vertical plane the moment rises linearly from zero at A to \(R_{Ay}(0.140)=32.4\ \text{N}\!\cdot\!\text{m}\) at B, then falls to \(-78.2\ \text{N}\!\cdot\!\text{m}\) just inboard of the pinion, where the applied couple returns it to zero at the free end. In the horizontal plane the moment rises from zero at A to \(R_{Ax}(0.140)=266.0\ \text{N}\!\cdot\!\text{m}\) at B and falls linearly to zero at the pinion. The resultant at B is $$M_{B}=\sqrt{266.0^{2}+32.4^{2}}=268\ \text{N}\!\cdot\!\text{m}$$ which, together with the full torque, makes the section at bearing B the critical one; the section just inboard of the pinion carries only 78 N·m of bending.

(c) Preliminary shaft layout

To put "reasonable" numbers on the drawing, size the critical section at bearing B by the DE-Goodman criterion for a rotating shaft, with the bending fully reversed and the torque steady. Take AISI 1040 CD steel (S_ut = 590 MPa, S_y = 490 MPa), machined surface, a design factor n = 2.0, and preliminary values K_f = 2.0 and K_fs = 1.7 for a shoulder fillet at the bearing seat.

1. Endurance limit. With \(S'_{e}=0.5S_{ut}=295\) MPa, machined surface factor \(k_{a}=4.51S_{ut}^{-0.265}=0.832\), and size factor \(k_{b}=1.24d^{-0.107}\) iterated with the diameter, $$S_{e}=k_{a}k_{b}S'_{e}\approx0.832(0.837)(295)=205\ \text{MPa}$$
2. Diameter at B. Substituting \(M_{a}=268\ \text{N}\!\cdot\!\text{m}\) (fully reversed) and \(T_{m}=143.2\ \text{N}\!\cdot\!\text{m}\) (steady) into $$d=\left\{\frac{16n}{\pi}\left[\frac{2K_{f}M_{a}}{S_{e}}+\frac{\sqrt{3}K_{fs}T_{m}}{S_{ut}}\right]\right\}^{1/3}$$ and iterating on k_b gives \(d=39.2\) mm, so $$\boxed{d_{B}=40\ \text{mm bearing seat at B}}$$ Repeating at the pinion seat, where the bending moment is only 78.2 N·m, gives 28.1 mm required, so a 35 mm pinion seat is generously adequate and is chosen to give room for the shoulder.
3. Layout. Build the shaft as a stepped stack that rises to the middle so that every component can be pushed on from its own end and seated against a shoulder: 30 mm coupling extension → 35 mm seat for bearing A → 45 mm body between the bearings → 40 mm seat for bearing B → 35 mm pinion seat → threaded end. Keep each shoulder step to about 10 % of the diameter and give every fillet the largest radius the bearing corner radius allows.

Axial retention follows the same "step up to the middle" logic. Bearing A is pushed on from the left and seats against the 45 mm shoulder; it is the floating bearing, so its outer ring is free to slide in its housing bore. Bearing B is pushed on from the right, seats against the same body shoulder, and is clamped outboard by a retaining ring; its outer ring is clamped in the housing between a shoulder and an end cap, which is what makes it the locating bearing and gives it a path for the 2235 N thrust into the housing. The pinion is pushed on from the right onto the 35 mm seat, driven by a parallel key, located inboard by the shoulder at the 40 mm step and clamped outboard by a washer and locknut. The coupling at the far left is keyed and retained by its own shoulder and setscrew or locknut.

(d) Primary factors that determine the diameters

Diameters on a shaft of this kind are almost never set by static strength. In descending order of importance:

• Fatigue at stress concentrations. The shaft rotates under a bending moment fixed in space, so the bending stress at every point is fully reversed at shaft speed — millions of cycles in service. The governing sections are therefore the ones combining a high moment with a geometric discontinuity: the shoulder fillet at bearing B, the key seat at the pinion, and any retaining-ring groove. It is K_fM_a/S_e, not M/S_y, that sizes the shaft, which is why generous fillet radii are worth more than extra diameter.
• Torsional and bending deflection and slope. Slope at the bearings must stay within the bearing's misalignment tolerance — of the order of 0.001 rad for a cylindrical roller bearing and 0.0087 rad (0.5°) for a deep-groove ball bearing — and the deflection at the pinion must stay small compared with the tooth thickness, typically below 0.13 mm, or the tooth contact pattern moves to one end of the face and the mesh becomes noisy and locally overloaded. On an overhung layout the slope at B is usually the binding constraint, and stiffness scales as d⁴ while strength scales only as d³.
• Bearing bore sizes and standard components. The seats must be standard bearing bores (30, 35, 40, 45 mm), and the shaft body must clear the bearing outer-ring corner radius. This quantizes every result and is why the calculated 39.2 mm becomes 40 mm.
• Assembly and location. Every component must be able to slide onto its seat over the smaller diameters and stop against a shoulder, and the shoulder height must be at least the bearing's minimum abutment. That requirement alone forces the stepped, symmetric-about-the-middle profile and often sets one or two of the diameters before any stress calculation.
• Critical speed. The first bending critical speed must stand well clear of the operating speed — conventionally by a factor of at least two — which again is a stiffness rather than a strength requirement.
• Material, surface finish and treatment. Endurance limit falls with increasing size (k_b) and with rougher surfaces (k_a), so a ground bearing seat in a heat-treated alloy steel earns a smaller diameter than an as-machined plain carbon shaft.