ken youssefi mae dept. 1 pressurized cylinders pipes carrying pressurized gas or liquid. press or...

Ken Youssefi MAE dept. 1

Pressurized CylindersPipes carrying pressurized gas or liquid.

Press or shrink fits

Pressurized cylinders

Hydraulic or pneumatic actuators


Stresses in Pressurized Cylinders Cylindrical pressure vessels, hydraulic cylinders, shafts with components mounted on (gears, pulleys, and bearings), gun barrels, pipes carrying fluids at high pressure,….. develop tangential, longitudinal, and radial stresses.

Wall thickness

t

A pressurized cylinder is considered a thin-walled vessel if the wall thickness is less than one-twentieth of the radius.

< 1/20tr

Thin-walled pressure vessel

Stress element

Small element

Tangential stress θ

Hoop stress

Longitudinal stressl

(closed ends)

Radial stress

r


Stresses in a Thin-Walled Pressurized Cylinders

In a thin-walled pressurized cylinder the radial stress is much smaller than the tangential stress and can be neglected.

Longitudinal stress, l

(l) /4 [ (do)2 – (di)

2] = ( p ) /4 (di)2

Internal pressure, p

Fy = 0

Longitudinal stress

(l) [ (di + 2t)2 – (di)2] = ( p ) (di)

2

4t2 is very small,

(l) (4di t) = ( p ) (di)2

l = p di

4 tl =

p (di + t)4 t

Max. longitudinal stress

Pressure area


Stresses in a Thin-Walled Pressurized Cylinders

Projected area

Hoop stress

θ

Tangential (hoop) stress

2(θ) t (length) = ( p ) (di) (length)

Fx = 0

θ = p di 2 t

Max. Hoop stress

l = ½ θ


Stresses in a Thick-Walled Pressurized Cylinders

In case of thick-walled pressurized cylinders, the radial stress, r , cannot be neglected.

Assumption – longitudinal elongation is constant around the plane of cross section, there is very little warping of

the cross section, εl = constant

dr

θ θ

r + dr

r

2(θ)(dr)(l) + r (2rl) – (r + dr) [2(r + dr)l] = 0

l = length of cylinder

F = 0

(dr) (dr) is very small compared to

other terms ≈ 0

θ – r – r drdr

= 0 (1)



εl = – μθ

E

r

E– μ

Deformation in the longitudinal direction

θ + r = 2C1=εl E

μ

constant

(2)

Consider,

d (r r 2)

dr= r 2

dr

dr+ 2r r

Subtract equation (1) from (2),

r + r + r dr

dr= 2C1

θ – r – r dr

dr= 0 (1)

2rr + r2 dr

dr = 2rC1

Multiply the above equation by r

d (r r 2)

dr= 2rC1

r r 2 = r2C1 + C2

r = C1

C2

r2+ θ = C1

C2–

r2



Boundary conditions

r = - pi at r = ri

r = - po at r = ro

θ =pi ri

2 - po ro2 – ri

2 ro2 (po – pi) / r

2

ro2 - ri

2Hoop stress

r =pi ri

2 - po ro2 + ri

2 ro2 (po – pi) / r

2

ro2 - ri

2

Radial stress

pi ri2 - po ro

2

l =ro

2 - ri2

Longitudinal stress

Longitudinal stress, l

Pressure area



Special case, po (external pressure) = 0

θ =pi ri

2

ro2 - ri

2(1 +

ro2

r2) r =

ro2 - ri

2(1 -

ro2

r2)

pi ri2

Hoop stress distribution, maximum at the inner surface

Radial stress distribution, maximum at the inner surface


Press and Shrink fitting Components onto Shafts

Liquid Nitrogen is fed into a containment unit that isolates the area to be fitted.

Shrink fitting a gear onto a shaft

The shrinking permits fitting of slightly oversized metal items into mating parts.


Press and Shrink (Force) Fits

ds (shaft) > dh (hub)Inner member

Outer member

p (contact pressure) = pressure at the outer surface of the inner member or pressure at the inner surface of the outer member.

Consider the inner member at the interface

po = p

pi = 0

ri = ri

ro = R

iθ =pi ri

2 - po ro2 – ri

2 ro2 (po – pi) / r

2

ro2 - ri

2

iθ =– po R

2 – ri2 R2 p / R2

R2 – ri2

= – p R2 – ri

2

R2 + ri2

ir = – po = – p


Press and Shrink Fits

Consider the outer member at the interfacepo = 0

pi = pri = R

ro = rooθ = p ro

2 – R2

R2 + ro2

or = – p

o = increase in the inner radius of the outer member

i = decrease in the outer radius of the inner member

Tangential strain at the inner radius of the outer member

εot = oθ

Eo

or

Eo

– μo

εot = o

R

o =Rp

Eo ro2 – R2

R2 + ro2

+ μo( )

Radial interference, = o – i


Press and Shrink Fits

Tangential strain at the outer radius of the inner member

εit = iθ

Ei

ir

Ei

– μi

εit = i

R–

i =Rp

Ei R2 – ri2

R2 + ri2

+ μi( )–

Radial interference, = o – i

=Rp

Eo ro2 – R2

R2 + ro2

+ μo( ) Rp

Ei R2 – ri2

R2 + ri2

+ μi( )+

If both members are made of the same material then, E = Eo = Ei and μ = μo = μi

p = E R

(R2 – ri2)(ro

2 – R2)

(ro2 - ri

2)2R2


Press and Shrink Fits - Example

You are asked to press fit a gear onto a shaft, specify the FN type fit. The shaft carries a maximum of 110 hp at 650 rpm and has a 2.0 inch diameter. The hub of the gear has a length of 2.0 inch and its outside diameter is 3.0 inch. Both components are made of steel with Sy = 40,000 psi

Hub

Shaft

Gear

2.02.0

3.0

hp =63000

T ω, 110 =

63000

T (650)

T = 10,662 in - lb Torque carried by the shaft

T = f R ( 2 R l ) p

Frictional torque

Surface areashaft radius

Coefficient of friction

Contact length between hub and shaft

Min. contact pressure



T = f R ( 2 R l ) p ,

Frictional torque

10662 = .2 (1) ( 2 x 1 x 2 ) p

p = 4243 psi Minimum contact pressure required to carry the desired torque.

Calculate the radial interference needed

p = E R

(R2 – ri2)(ro

2 – R2)

(ro2 - ri

2)2R2

ri = 0

ro = 1.5

R = 1.0

4243 =30 x 106 ()

1

[(1.5)2 – (1)2] [(1)2 – (0)2]

2(1)2[( 1.5)2 – (0)2]

δ = .00050916 in. 2δ = .00102 = 1.02 x 10-3 in.Diametral interference needed


ANSI Tables for FitsInterference fits (Force and Shrink fits) – FN1 to FN5

2.00



0.6 x 10-3 minimum interferenceFN1

1.8 x 10-3 maximum interference





2δ = .00102 = 1.02 x 10-3 in.

Diametral interference needed

Select FN3 fit


Press and Shrink Fits - ExampleCheck for failure of the shaft and the hub of the gear for max. interference fit

iθ =– po R

2 – ri2 R2 p / R2

R2 – ri2

= – p R2 – ri

2

R2 + ri2

ir = – po = – p

Shaft (outer surface of inner member), ri = 0

= – p

Maximum contact pressure for maximum interference,

(2δ)max = 3.2 x 10-3 in.

p =30 x 106 (1.6x 10-3)

1

[(1.5)2 – (1)2] [(1)2 – (0)2]

2(1)2[( 1.5)2 – (0)2]= 13,333 psi

iθ = 13333 < 40000 = Sy shaft will not fail



Hub (inner surface of outer member), ri = R

= 13333 1.52 – 12

1.52 + 12

oθ =– po R

2 – ro2 R2 p / R2

R2 – ro2

= p R2 – ro

2

R2 + ro2

or = – po = – p = -13333

= 34666 psi

oθ = 34666 < 40000 = Sy Hub will not fail

ken youssefi mae dept. 1 pressurized cylinders pipes carrying pressurized gas or liquid. press or...

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