written 6.0 years ago by |
Q. Design a Cotter joint to transmit axial load of 60 kN. Select suitable material, FOS and draw an informative sketch.
It is used to connect two co-axial rods which are subjected to either axial tensile or axial compressive load.
Application:
1) Joint between piston rod and cross-head of steam engine.
2) Joint between piston rod and pump rod.
1) Selection of material
As the joint is subjected to axial load, then strength is the criteria for material selection.
Cotter is subjected to direct shear stress and bending stress. Therefore again strength is the criteria for the selection of material for cotter.
Hence, selection of material is carried out by using low carbon steel or mild steel which will withstand stresses developed in different parts and cost of material will be within range.
Let us select C20 material [ select any material from C20 to C30] for all the parts.
Material properties:
σu=480 N/mm2
σy=260 N/mm2
2) Selection of FOS
- As there are initial stresses due to tightening of cotter, therefore we have to select significant FOS. Also, stress concentration is observed due to slot in socket and spigot end. Therefore, consider higher FOS.
- Select n= 4.
3) Permissible Stresses
A) σt=σyFOS ...................Principal stress theory
σt=2604=65N/mm2
B) τ=0.5×σyFOS .......................Max Shear stress theory
τ=0.5×2604
τ=32.5 N/mm2≈32 N/mm2
A. Design Procedure
I) Design of spigot end
(a) Diameter of rod [i.e., d=?]
σt=ForceCross-sectional area=PA
∴65=60×103π4d2
Note: Consider tensile failure of rod.
d=34.28mm≈35mm
(b) Diameter of spigot [d1] and thickness of Cotter (t)
Consider tensile failure across slot of the spigot
σt=PA
∴P=σt.A=σt[π4d21−d1×t]
60×103=65[π4d21−d1×t]...................(1)
σcrushing=PA
σcr×A=P
σcr[d1×t]=P
Assume, σcr=1.6σt=1.6×65
σcr=104 N/mm2
104[d1×t]=60×103
d1×t=576.923
∴60×103=65[π4d21−576.92] (from (1))
∴d1=43.701mm≈45mm
t=12.83mm≈13mm
(c) Length of rod beyond slot (i.e.,l)
Consider shear failure of spigot end beyond slot. (double shear)
τ=PA
P=τA
60×103=32[2×d1×l]
60×103=32[2×45×l]
l=20.83≈21mm
(d) Outside diameter of spigot end (d2)
Consider collar will be under crushing due to initial tightening of joint.
σcr=PA
σcr=[π4(d22−d21)]=P
104[π4(d22−452)]=60×103
d2=52.53 mm≈53 mm
(e) Thickness of spigot column or end (t1)
Consider shear failure across collar of spigot end.
τ=PA
P=τA
60×103=32[π×d1×t1]
60×103=32[π×45×t1]
t1=13.2629 mm≈14 mm
II) Design of socket end
(a) Outside diameter of socket end across slot [d3]
Consider tensile failure of socket end across slot.
σt=PA
P=σtA
60×103=65[π4(d23−d21)−(d3−d1)t]
60×103=65[π4(d23−452)−(d3−45)13]
d3=58.51mm≈59mm
(b) Outside diameter of collar at end of socket end (d4)
Consider crushing of collar under cotter.
σcr=PA
P=σcrA
60×103=104[(d4−d1)×t]
60×103=104[(d4−45)×13]
d4=89.37 mm≈90 mm
(c) Thickness of collar at socket end (t2)
Considering double shear failure due to cotter.
τ=PA
P=τA
60×103=32[(d4−d1)×t2×2]
60×103=32[(90−45)×t2×2]
t2=20.83 mm≈21 mm
(d) Determination of l1
Consider shear failure
τ=PA
τA=P
32[π×d×l1]=60×103
32[π×35×l1]=60×103
l1=17.05mm≈18mm
III) Design of cotter
(a) Determination of mean width of cotter (b).
Consider double shear failure of cotter.
τ=PA
P=τA=τ(2×b×t)
60×103=32(2×b×13)
b=72.11 mm≈73 mm
(b) Design of cotter in bending
Max. bending moment will be at center of cotter.
Mmax=P2[13(d42−d12)+d12]−[P2(d14)]
∴Mmax=562.5×103 N−mm
MI=σby
∴σb=MIy
σy=M(Iy)=MZ
where Z = section modulus
I=tb312,y=b2
Z=tb312b2=tb26
∴σb=MZ=562.5×103tb26
σb=562.5×10313×(73)26
σb=48.71 N/mm2<σt(65 N/mm2)
Cotter design is safe.
[Note: If cotter fails in bending, the increase dimension, b]
(c) Miscellaneous dimension
(i) length of cotter = d4+10 mm=90+10=100 mm
(ii) Max. width of cotter, bmax=b+(length of cotter2)(130)
bmax=74.66 mm≈75 mm
bmin=b−(length of cotter2)(130)
bmin=71.33≈72 mm