Interference fits & keys
Exercise 1 - Torque transmission by inteference fit
Lets consider the speed reducer SEW R17 described below. The output torque is 16Nm. One plans to replace the key joint between the wheel 12 and output axle 9, by an interference fit.
The data for assembly are :
Fit length : L = 22 mm Fit diameter : D = 20 mm
External diameter of wheel 12 : Dext = 58 mm Friction coefficient : f = 0.15
C32E Material (E=210000MPa, Re=435 MPa) 1. Compute the contact pressure needed to transmit the torque
2. Compute the minimum diametral tightening Δm. You will use a safety factor of 1.5 on the pressure.
3. Choose a fit that will enable the tightening. Compute the maximal tightening and check that the part will sustain the load (elastically).
Exercise 2 - Torque transmitted by unit length
A fit assembly is designed with a fit H6p6 on a diameter d. The axle is bulk and bore are same material.
The external diameter D is K*d (with K > 1 of course).
The friction coefficient is 0.12, and no safety coefficient.
1. Compute the variation of contact pressure p as a function of K and ∆m.
2. What is the transmissible torque per unit length for this fit, as a function of K and d?
3. Compute the transmissible torque for d= 40 and k =2 and E=210000MPa.
Exercise 3 - Interference fit for a shouldered cog
An interference fit is made for a diameter d=2r. The axle is bulk and of same material than the bore. The friction coefficient is f. The external diameter of the bore is variable : R1 over a length L1
and R2 over a length L2. The tightening needs to be defined.
1. Express the torque C transmissible, as a function of P1 and P2, pressures applied on L1 et L2. 2. Express ∆m as a function of P1 and then P2.
3. Express ∆m as a function of C.
4. C= 50 Nm, α=2 , r=15, R1=60, R2=30,f=0.12, L1=10, L2=15, E=210000MPa.
Exercise 4 - Key design
System :
Data :
- The key is of shape A, with dimensions 10x8x110
- Elastic limit under shear Rg is related to the elastic limit under tension Re with : o For light alloys, Rg = 0.5 * Re
o For steels, Rg = 0.7 * Re
- Triaxiality coefficient (for bearing resistance) K = 2
Problem :
The goal is to choose the material (among the three) that will enable the power transmission between 6 and 3 without destroying the key.
- Squared profile hot laminated 10x10 NF A 45-003 in E295 ; Re = 295 MPa - Squared profile cold laminated 10x10 NF A 47-001 in C35 ; Re = 335 MPa - Squared profile 10x10 NF A 50-702 in Al Si 10 Mg; Re = 180 MPa
Questions :
1. Compute the key section S that will sustain the shear load 2. Compute the angular velocity of the engine
3. Compute the torque C
4. Deduce the force applied to the key F
5. Compute the shear stresses applied to the key 6. Compute the bearing stresses applied to the key
7. With a safety factor of k=6, conclude on which material can be chosen for the key
