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Transmission shafts and fatigue

A transmission shaft carries torque from one element to another (from a pulley to a gear, from a motor to a pump) and, besides, supports the forces of those elements, which bend it. Since it rotates, every point on its surface goes from tension to compression on each turn: after millions of cycles it can break at stresses far below yield. Designing a shaft is, above all, designing it for fatigue, and that is what the Shafts tool does.

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Loads in two planes

Gears and pulleys push the shaft in different directions: a spur gear gives a tangential force and a radial one ( $W_{r} = W_{t} tan ϕ$ ), a pulley the sum of the belt tensions. Everything is split into two perpendicular planes ( $x y$ and $x z$ ), the bearing reactions and the moment diagrams are found in each plane, and at each section they are combined:

M = M_{x y}^{2} + M_{x z}^{2} T = par entre la entrada y la salida

The resultant moment works because the shaft is round: any direction is a principal axis. The torque only exists between the element that receives it and the one that delivers it; outside that segment it is zero.

Stresses that repeat

A moment fixed in space, on a rotating shaft, produces at each point a stress that alternates between $+ σ$ and $- σ$ : it is fully reversed, with amplitude $σ_{a}$ and zero mean. A steady torque produces a constant shear, with mean $τ_{m}$ and no amplitude. On the surface of a solid shaft of diameter $d$ :

σ_{a} = K_{f} \frac{32 M _{a}}{π d ^{3}} τ_{m} = K_{f s} \frac{16 T _{m}}{π d ^{3}} σ_{a}^{'} = σ_{a}^{2} + 3 τ_{a}^{2} σ_{m}^{'} = σ_{m}^{2} + 3 τ_{m}^{2}

A point on the surface goes through tension and compression on each turn; the steady torque gives a shear that does not change.

$σ_{a}^{'}$ and $σ_{m}^{'}$ are the von Mises stresses of the alternating part and of the mean: that is how bending and torsion are combined into a single pair of numbers.

Endurance limit

A polished rotating steel specimen lasts indefinitely if its stress does not exceed $S_{e}^{'} \approx 0.5 S_{u t}$ (capped at 700 MPa for very strong steels). A real shaft is weaker: its surface is rougher, it is larger, it may run hot and a low probability of failure is wanted. The Marin factors correct the specimen value:

S_{e} = k_{a} k_{b} k_{c} k_{d} k_{e} S_{e}^{'}

Factor	What it corrects	Value used
$k_{a}$	surface finish	$a S_{u t}^{b}$ ; machined: $a = 4.51$ , $b = - 0.265$ (MPa)
$k_{b}$	size	$1.24 d^{- 0.107}$ (2.79 ≤ d ≤ 51 mm)
$k_{c}$	type of load	1 in bending (stresses already combined with von Mises)
$k_{d}$	temperature	1 up to about 20 °C; it drops at high temperatures
$k_{e}$	reliability	1 (50 %), 0.897 (90 %), 0.814 (99 %), 0.753 (99.9 %)

Since

k_{b}

depends on the diameter, which is exactly what you are looking for, design is iterative: assume a diameter, compute

S_{e}

and correct it. The simulator does it for you.

Stress concentrations

Changes of diameter (shoulders), keyseats and retaining-ring grooves raise the local stress. The geometric factor $K_{t}$ comes from stress-concentration charts depending on the shape; but in fatigue the material «feels» only part of the peak, according to its notch sensitivity $q$ (0 to 1, higher for hard steels and large radii):

K_{f} = 1 + q (K_{t} - 1) q = \frac{1}{1 + a / r} (Neuber)

At the shoulder fillet the stress reaches

K_{t}

times the nominal one. A larger radius (larger r/d) lowers

K_{t}

Stress raiser (first estimate)	$K_{t}$ flexión,bending	$K_{t s}$ torsión,torsion
keyseat cut with an end mill	2.14	3.0
retaining-ring groove	5.0	3.0
shoulder	from D/d and r/d (charts)	likewise

That is why the most dangerous section is almost never the one with the maximum moment, but a nearby shoulder, keyseat or groove: all of them must be checked. The simulator builds the list of Critical points and computes the safety factor of each one.

Fatigue criteria and diameter

With a mean stress, the material can take less amplitude. The criteria draw a failure line in the $(σ_{m}^{'}, σ_{a}^{'})$ plane and the safety factor is how much the working point can be scaled until it touches it:

Criterion	Equation	Comment
Goodman	$\frac{1}{n} = \frac{σ _{a}^{'}}{S _{e}} + \frac{σ _{m}^{'}}{S _{u t}}$	straight line; somewhat conservative, the most used
Gerber	$\frac{n σ _{a}^{'}}{S _{e}} + (\frac{n σ _{m}^{'}}{S _{u t}})^{2} = 1$	parabola; fits test data better
ASME elliptic	$(\frac{n σ _{a}^{'}}{S _{e}})^{2} + (\frac{n σ _{m}^{'}}{S _{y}})^{2} = 1$	ellipse; includes yielding
Soderberg	$\frac{1}{n} = \frac{σ _{a}^{'}}{S _{e}} + \frac{σ _{m}^{'}}{S _{y}}$	straight line to yield; the most conservative

Fatigue diagram of cold-drawn 1020 steel with the Se of example 1. The point of example 2 is scaled along the load line up to the Goodman line: n = 2.35.

Solving Goodman for $d$ , with alternating bending and steady torque (the typical case), gives the DE-Goodman design formula:

d = [\frac{16 n}{π} (\frac{2 K _{f} M _{a}}{S _{e}} + \frac{3 K _{f s} T _{m}}{S _{u t}})]^{1/3}

Yielding on the first cycle is also checked, $n_{y} = S_{y} / σ_{max}^{'}$ with $σ_{max}^{'} = (σ_{m} + σ_{a})^{2} + 3 (τ_{m} + τ_{a})^{2}$ , which rarely governs.

Deflection and critical speed

A strong shaft may still be too flexible. Bearings tolerate little slope (for a deep-groove ball bearing, around 0.001 to 0.003 rad) and gears need their shafts not to spread apart. The deflection is found by integrating $M / (E I)$ with the diameter of each segment. With it the first critical speed (Rayleigh) is estimated, at which the shaft resonates; it should run well below it:

ω_{1} = \frac{g \sum w _{i} y _{i}}{\sum w _{i} y _{i}^{2}}

Common mistakes

Treating the bending of a rotating shaft as static: it is alternating, and the steady torque is the mean part.
Adding the moments of the two planes algebraically instead of with $M_{x y}^{2} + M_{x z}^{2}$ .
Checking only the section of maximum moment and forgetting shoulders, keyseats and grooves.
Using $K_{t}$ instead of $K_{f}$ in fatigue, or $K_{f}$ (bending) where $K_{f s}$ (torsion) belongs.
Forgetting that $S_{e}$ depends on the diameter ( $k_{b}$ ) when solving for $d$ .

Worked examples

All three examples use a 400 mm shaft between two ball bearings, of cold-drawn AISI 1020 steel ( $S_{u t} = 470$ MPa, $S_{y} = 390$ MPa), machined, at 20 °C, with 50 % reliability and a 4 kN load at the middle.

Example 1Plain shaft with one load

The shaft is plain, 40 mm in diameter, and transmits no torque. Find the endurance limit, the fatigue and yield safety factors, and the slope at the bearings.

Solution

Reactions of 2 kN and maximum moment at the middle: $M = 2000 \cdot 200 = 400000$ N·mm. As it rotates, $σ_{a} = \frac{32 \cdot 400 000}{π \cdot 4 0 ^{3}} = 63.66$ MPa and $σ_{m} = 0$ .
$S_{e}^{'} = 0.5 \cdot 470 = 235$ MPa; $k_{a} = 4.51 \cdot 47 0^{- 0.265} = 0.883$ ; $k_{b} = 1.24 \cdot 4 0^{- 0.107} = 0.836$ ; the others are 1. $S_{e} = 0.883 \cdot 0.836 \cdot 235 = 173.4$ MPa.
With no mean stress, all criteria give the same: $n = \frac{173.4}{63.66} = 2.72$ . Yield: $n_{y} = \frac{390}{63.66} = 6.13$ .
Slope at the bearings: $θ = \frac{P L ^{2}}{16 E I} = \frac{4000 \cdot 40 0 ^{2}}{16 \cdot 207 000 \cdot 125 664} = 0.00154$ rad, within what a ball bearing tolerates. Deflection at the middle: $\frac{P L ^{3}}{48 E I} = 0.205$ mm.

S_{e} = 173.4

MPa,

n = 2.72

for fatigue,

n_{y} = 6.13

and slope 0.00154 rad at each bearing.

Solve it in Simuveliq

In Engineering → Shafts, in Examples, «Load shaft» → Plain shaft with one load (hand check).
The «Step by step» of point 1 (maximum moment) repeats these calculations: $k_{a} = 0.883$ , $k_{b} = 0.836$ , $S_{e} = 173.4$ MPa, $σ_{a}^{'} = 63.66$ MPa and the criteria table with 2.72; yield gives 6.13. It also gives the minimum diameter for $n = 1.5$ (32.54 mm, which already accounts for $k_{b}$ changing with $d$ ).
In Deflection, slopes and critical speed you get the slopes (0.00154 rad) and the first critical speed, about 30 000 rpm against 600 in service.

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Example 2The same shaft transmitting torque

Now the shaft receives a 200 N·m torque at the middle (where the load is) and delivers it at end B. Find the von Mises stresses and the Goodman safety factor.

Solution

The torque exists only between the middle and B, and at the middle it meets the maximum moment.
$τ_{m} = \frac{16 \cdot 200 000}{π \cdot 4 0 ^{3}} = 15.92$ MPa, so $σ_{m}^{'} = 3 \cdot 15.92 = 27.57$ MPa; $σ_{a}^{'} = 63.66$ MPa as before.
Goodman: $\frac{1}{n} = \frac{63.66}{173.4} + \frac{27.57}{470} = 0.4258$ , $n = 2.35$ . The torque, although steady, takes away 14 % of the margin.
Yield: $σ_{max}^{'} = 63.6 6^{2} + 3 \cdot 15.9 2^{2} = 69.37$ MPa and $n_{y} = 390/69.37 = 5.62$ .

σ_{a}^{'} = 63.66

MPa,

σ_{m}^{'} = 27.57

MPa,

n = 2.35

(Goodman) and

n_{y} = 5.62

Solve it in Simuveliq

The button opens the shaft with two Force elements: the 4 kN load with T = 200 N·m at the middle and an output with T = −200 N·m at B.
Step 6 gives $σ_{a}^{'} = 63.66$ and $σ_{m}^{'} = 27.57$ MPa, and step 7 the table with DE-Goodman 2.35, DE-Gerber, DE-ASME and DE-Soderberg. Change the Fatigue criterion to see which is more conservative.

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Example 3A shoulder near the middle

The shaft steps from 40 to 48 mm at

x = 150

mm, with a 2 mm fillet at the shoulder, and continues at 48 mm to B. No torque. Find

K_{t}

K_{f}

and the safety factor at the shoulder and at the middle.

Solution

Shoulder geometry: $D / d = 48/40 = 1.2$ and $r / d = 2/40 = 0.05$ . From the shoulder chart (with the tool's curve fit): $K_{t} = 1.865$ .
Neuber: for $S_{u t} = 470$ MPa (68.2 kpsi), $a = 0.0978 in$ ; with $r = 2$ mm $= 0.0787$ in, $q = \frac{1}{1 + 0.0978/ 0.0787} = 0.742$ . $K_{f} = 1 + 0.742 \cdot 0.865 = 1.642$ .
At the shoulder ( $x = 150$ , $M = 300000$ N·mm, $d = 40$ ): $σ_{a} = 1.642 \cdot \frac{32 \cdot 300 000}{π \cdot 4 0 ^{3}} = 78.39$ MPa and $n = 173.4/78.39 = 2.21$ .
At the middle ( $M = 400000$ N·mm but $d = 48$ and no raiser): $σ_{a} = 36.84$ MPa, $S_{e} = 170.1$ MPa (a slightly lower $k_{b}$ ) and $n = 4.62$ . The shoulder governs, even though it has less moment.

K_{t} = 1.865

q = 0.742

K_{f} = 1.642

;

n = 2.21

at the shoulder (governs) and 4.62 at the middle.

Solve it in Simuveliq

The button opens the shaft with two segments in Shaft geometry: 150 mm of Ø 40 with a 2 mm fillet to the next one, and 250 mm of Ø 48.
In Critical points you get the «Shoulder Ø40→Ø48» with n = 2.21 and the «Maximum moment (no raiser)» with n = 4.62. Step 4 of the shoulder shows $K_{t}$ , $q$ and $K_{f}$ .
Increase the shoulder fillet radius and watch $K_{t}$ and $K_{f}$ drop: it is the cheapest way to gain fatigue life.

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Transmission shafts and fatigue

Loads in two planes

Stresses that repeat

Endurance limit

Stress concentrations

Fatigue criteria and diameter

Deflection and critical speed

Common mistakes

Worked examples

Example 1Plain shaft with one load

Solution

Solve it in Simuveliq

Example 2The same shaft transmitting torque

Solution

Solve it in Simuveliq

Example 3A shoulder near the middle

Solution

Solve it in Simuveliq

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