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Spigot and Socket type Cotter Joint Design
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Q. Design a Cotter joint to transmit axial load of 60 kN. Select suitable material, FOS and draw an informative sketch.

enter image description here

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/mm232 N/mm2

A. Design Procedure

I) Design of spigot end

(a) Diameter of rod [i.e., d=?]

enter image description here

σt=ForceCross-sectional area=PA

65=60×103π4d2

Note: Consider tensile failure of rod.

d=34.28mm35mm

(b) Diameter of spigot [d1] and thickness of Cotter (t)

Consider tensile failure across slot of the spigot

σt=PA

P=σt.A=σt[π4d21d1×t]

enter image description here

60×103=65[π4d21d1×t]...................(1)

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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[π4d21576.92] (from (1))

d1=43.701mm45mm

t=12.83mm13mm

(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=20.8321mm

(d) Outside diameter of spigot end (d2)

Consider collar will be under crushing due to initial tightening of joint.

enter image description here

σcr=PA

σcr=[π4(d22d21)]=P

104[π4(d22452)]=60×103

d2=52.53 mm53 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 mm14 mm

II) Design of socket end

(a) Outside diameter of socket end across slot [d3]

Consider tensile failure of socket end across slot.

enter image description here

σt=PA

P=σtA

60×103=65[π4(d23d21)(d3d1)t]

60×103=65[π4(d23452)(d345)13]

d3=58.51mm59mm

(b) Outside diameter of collar at end of socket end (d4)

Consider crushing of collar under cotter.

enter image description here

σcr=PA

P=σcrA

60×103=104[(d4d1)×t]

60×103=104[(d445)×13]

d4=89.37 mm90 mm

(c) Thickness of collar at socket end (t2)

Considering double shear failure due to cotter.

enter image description here

τ=PA

P=τA

60×103=32[(d4d1)×t2×2]

60×103=32[(9045)×t2×2]

t2=20.83 mm21 mm

(d) Determination of l1

Consider shear failure

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τ=PA

τA=P

32[π×d×l1]=60×103

32[π×35×l1]=60×103

l1=17.05mm18mm

III) Design of cotter

(a) Determination of mean width of cotter (b).

Consider double shear failure of cotter.

enter image description here

τ=PA

P=τA=τ(2×b×t)

60×103=32(2×b×13)

b=72.11 mm73 mm

(b) Design of cotter in bending

Max. bending moment will be at center of cotter.

enter image description here

Mmax=P2[13(d42d12)+d12][P2(d14)]

Mmax=562.5×103 Nmm

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 mm75 mm

bmin=b(length of cotter2)(130)

bmin=71.3372 mm

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