STRESSES IN CYLINDER WALL

APPARENT STRESSES : There are two principal stresses in engine cylinder – a. Circumferential hoop stress (αc) b. Longitudinal stress (αl) It is assumed that the stresses are uniformly distributed over the wall thickness. Considering equilibrium of forces acting on the half portion of cylinder of unit length.

STRESSES IN CYLINDER WALL

• Net stresses :Two principal stresses – the circumferential hoop stress (αc) and longitudinal stress (αl) are tensile stresses and they act at right angles to each other. Therefore net stresses in these directions are reduced. Consider the fig. (a)circumferential stress. (b) longitudnal stress.

STRESSES IN CYLINDER WALL

• The net stresses are given by –

(αc)net = αc - µαl

(αl)net = αl - µαc

where; αc = apparent circumferential stress (N/mm^2)

αl = apparent longitudnal stress (N/mm^2)

(αc)net = net circumferential stress (N/mm^2)

(αl)net = net longitudnal stress (N/mm^2)

The cylinder material is usually brittle such as cast iron. The value of Poission’s ratio is taken as 0.25.

µ = 0.25

ExampleQ. The cylinder of a four- stroke diesel engine has the following specifications:Brake power=7.5KWSpeed=1400 rpmIndicated mean effective pressure=0.35 Mpa Mechanical gas pressure=3.5 MpaMaximum gas pressure=3.5 Mpa• The cylinder liner and head are made of grey cast iron FG 260(Sut=260 N/mm2 and µ=0.25).The studs are made of plain carbon steel 40C8 (Syt=380 N/mm2).The factor of safety

for all part is 6.Calculate:1. Bore and length of cylinder liner2. Thickness of the cylinder liner3. Thickness of the cylinder head4. Size, number and pitch of studs Solution: Given: BP=7.5 KW= 7500 W N=1400 RPM pm= 0.35 Mpa =3.5 n/mm2Fs=6 efficiency=80%=.8 For studs, Syt= 380 N/mm2 For cylinder liner and head, Sut=260 N/mm2 µ=0.25

...• Step1. Bore and length of cylinder linerAssumption: The ratio of stroke length to cylinder diameter (l/D) is

1.5D= cylinder bore (mm)A=cross-sectional area of cylinder=(πD2/4) (mm2)L=length of stroke in m =(1.5D/1000) (m) For a four - stroke engine, n=N/2=1400/2=700 strokes/minIP=BP/п=7500/0.8=9375 WIP=pmlAn/60=(pm/60)(1.5D/1000)(πD2/4)(n)9375=(.35/60)(1.5D/1000)(πD2/4)(700)D3=1948.84 * 103D=124.9 mm or 125 mml=1.5D= 1.5 (125) =187.5 mm

.....• The length of the stroke is 187.5 mm. There is clearance on both sides of the stroke.

Assuming the clearance as 15% of stroke length, the length of the cylinder (L) is given by,

L= 1.15l= 1.15(187.5) =215.63 mmL=216 mmStep 2. Thickness of cylinder liner The permissible tensile stress is given by,αt=Sut/fs=260/6=43.33 N/mm2By linear interpolation, the reboring allowance C for (D=125 mm) is given by,C=2.4 + (4-2.4)/(150-100)*(125-100) =3.2mm

The thickness of the cylinder is given by,t=pmaxD/2αt +C

= 3.5(125)/2(43.33) +3.2 =8.5 mm or 10 mm

....

• Apparent stress: Circumferential stressαc= pmax D/2t=3.5(125)/2(10) =21.88 N/mm2

(αc < 43.33 n/mm2)

Longitudal stress Do=D+2t=125+2(10)=145mmαl =pmaxD^/(Do^2-d^2)=3.5/[(145^2)-(125^2)]

=10.13N/mm2 Net stresses: Circumferential stress(αc)net= αc - µ αl= 21.88-0.25(10.13)

=19.35N/mm2 Longitudnal stress (αl )net= αl - µ αc = 10.13-0.25(21.88)=4.66N/mm2

STEP 3 : Thickness of cylinder head The permissible stress for the cylinder head is given by ; αt = Sut/(fs)=260/6=43.33N/mm2

... αc = αt =43.33N/mm2

th= D (kpmax/ αc )^.5=125[.162(3.5)/43.33]

=14.3 or 15 mmSTEP 4: Number of studs The number of studs should be from 5.25 to 6.5It is assumed that there are six studs. z=6STEP 5: The permissible stress for the studs is given by,αt= Syt/fs= 380/6=63.33 N/mm2

Force acting on cylinder head = (π/4)D2 pmax= (π/4) (125)2(3.5)= 42951.46 N - eq a

Resisting force offered by all studs= (π/4) dc2 z αt

= (π/4) dc2 (6)(63.33) = 298.44 dc2 - eq b

....Equating a and b298.44 dc2 = 42951.46 or dc=12 mm

d = dc/.8 =12/.8 =15 mm

Step 6) Pitch of studs

Pitch circle diameter of studs (Dp) =D+3d= 125 + 3(15) =170 mm

Pitch of the studs = (π) Dp/z= π (170)/6=89.01 mm

Limits:Minimum pitch = 19 (d)^05 =73.59 mmMaximum pitch= 28.5(d)^0.5 =110.38 mm

The pitch of the studs is 89.01 mm. It is within the llimits of 73.59 and 110.38 mm . Therefore, the pitch of the studs is satisfactory..