The following notes indicate how to calculate a reasonable value for the stress resulting from
an applied torque. The calculation includes adjustment for the operating conditions by including a Service
factor Ks . This factor is based on a design factor Kd, an application
factor Ka , a load distribution factor Km, a fatigue life factor Kf, and a
wear life factor Kw.
The latter factor primarily related to splines.
This working stress should be such that it is less
than the Material Design strength / Fs (Safety Factor ).
I have found that the Mitcalc.com excel
based calculation sheets (connections) see link 1 below provides a very convenient ( and more accurate ) method of completing
these calculations
Machinerys Handbook (27th ed.) includes tables for allowable stresses for splines.
The allowable compressive stresses indicated are very low compared to the stresses indicated on this site (10%).
The links below to the Zakgear Calculator and Wikipedia also show the same tables. These documents are based of the article
When Splines Need Stress Control by Darle W. Dudley". The different calculation methods should result in the same
Safety Factors if the same interpretation of a fixed and a flexible coupling is used and if the design factor Kd is correctly applied
The method used by Dudley involves use of a separate equation for a flexible coupling and fixed coupling and does not
use a factor
I have completed the example calculation on this page using the Zakgear calculator using imperial equivalents to my metric
values. The resulting margins of safety (FofS-1)relate very closely to my Factors of safety.
Note: Tables of the various factors and design strengths are found on webpage. Factors.
It is first good practice to estimate the torque which can be transmitted by the reduced shaft diameter i.e the diameter of the shaft which is not including the depth of the key /spline.
Note: Tables of the various factors and design strengths are found on webpage. Factors.
|
Notes: Tables of the various factors and design strengths are found on webpage. Factors.
| It is reasonable to use x = h/2 however the sketch below |
Note: The stress in the hub need only be considered if the hub material design strength is significantly lower than the shaft or key strength.
based on shear
based on compressive strength
Notes: Tables of the various factors and design strengths are found on webpage. Factors.
|
based on compressive strength
Notes: Tables of the various factors and design strengths are found on webpage. Factors.
Note: h for for 30o flat root splines h = 0,9m and for fillet root splines h = m |
Three example calculations are provided below all based on the same transmitted power and RPM .
The keys and splines are all assumed to be close fits and guided. I have obtained the various factors
and material strengths using the tables on webpage Factors
These calculations are also completed using Mitcalc comparing the approximate methods as indicated on
this page with more accurate methods .
I have simply identified a transmitted power 10 kW. and by a process of iteration arrived at a reasonable shaft
diameter (25mm) . I have then calculated the necessary key/spline proportions to transmit the power with a
factor of safety about 2,5. In the calculations I have shown the formula but not the detailed numbers and unit conversions.
Initial data
|
Factors
|
Shaft Calculation
Do : Assuming a basic shaft size of 25mm .
The shear stress in the shaft is calculated as
τ = 16.T . Ks / (π.Do3 ) =41,5 MPa (F of Safety (Ss /τ ) of about 4,8)
Key Calculation
A key of 8 mmx 7mm by 70mm long is selected.
Le The effective length of the key = 70-2*4 = 62mm =
x (calculated above = 2,94mm,
x1 (Calculated above = 3,26mm )
Di Reduced shaft dia = 25mm - (t1 = 4) = 21mm
The shear stress in the reduced shaft dia (Di) =
τ = 16.T . Ks / (π.Di3 ) = 16 . 63,67 . 1000. 2 / (π . 213 . ) = 70,02 MPa ........ (F of Safety (Ss /τ ) of about 2 ,8)
using Mitcalc σc = 70 MPa
The compressive stress in the key and the shaft. This is the worst case stress in this example
σc = T . Ks /( Le.(D/2).x) = 63,67 . 1000 . 2 . 2 / (62 . 25. 2,94 ) = 55,83 MPa .... (F of Safety (Sc /σ ) of about 2,33)
using Mitcalc σc = 58,5 MPa
Straight Spline Calculation
A straight spline of 25 - 6x21x25 is selected. This has a OD of 25 mm
Di reduced dia= 21 mm
n = 6 teeth
Le: The length of spline selected = 25mm .
Di: The reduced diameter = 21mm,
r: the mean readius of the teeth = (21+25)/4 = 11,5mm
c: The chamfer radius on the teeth (top and bottom) is assumed to be about c = 0,3mm.
y: The resulting effective depth of the teeth (y) = ((D- Di)/2 -2.c = 1,4mm
The shear stress in the reduced shaft dia (Di) =
τ = 16.T . Ks / (π.Di3 ) = 16 . 63,67 . 1000 . 2 /( π . 213 ) = 70,02 MPa ........ (F of Safety (Ss /τ )of about 2 ,8)
using Mitcalc σc = 70 MPa
The compressive stress in the spline and the shaft. This is the worst case stress in this example
σc = T . Ks /( Le.n.r.y) = 63,67 . 1000 . 2 /( 25 . 6 . 11,5 . 1,4 ) = 52,72 MPa .... (F of Safety (Sc /σ c ) of about 2,47)
using Mitcalc σc = 52,7 MPa
Involute Spline Calculation
An involute spline of 25 - 1 x 24 is selected. This has a OD (Dee ) of 25 mm a PCD ( D ) = 24mm , a module (m) of 1, Number of teeth z = 24
z : Number of teeth = 24
L: The length of spline selected = 10mm
D : Pitch diameter = 24mm = m.z
Die: The reduced diameter = 22,5mm = m(z-1,5)
Dee: Outside Diameter = 25mm = m(z+1)
L: Length = 10mm (estimated)
p: Pitch = 3,1416mm = m.π
t: Tooth thickness = 1,157mm = p/2
h : Tooth height =0,9m = 0,9mm
The chamfer radius (c) on the teeth is assumed to be about c = 0,3mm. Th resulting effective depth of the teeth (y) = ((D- Di)/2 -2.h = 1,4mm
The shear stress in the reduced shaft dia (Di) =
τ = 16.T . Ks / (π.Die3 ) = 16 . 63,67 . 1000 . 2 / (π . 22,53 ) = 56,9 MPa.... (F of Safety (Ss /τ )of about 3,51)
using Mitcalc τ = 56,9 MPa
The shear stress in the spline teeth.
τ = 4 . T . Ks /(L.z.t.D) = 2. 63,67 . 1000 . 2 /( 10 . 24 . 1,57 . 24 ) = 56,28 MPa ..... (F of Safety (Ss /σ ) of about 3,55)
The compressive in the spline teeth.
σc = 2 .T . Ks /(L.z.h.D) = 2 . 63,67 . 1000 . 2 /( 10 . 24 . 0,9 . 24 ) = 49,12 MPa ..... (F of Safety (Sc /σc; ) of about 2,65)
using Mitcalc σc = 46,3 MPa