1. Ken Youssefi MAE dept. 1 Pressurized Cylinders Pipes carrying pressurized gas or liquid. Press or shrink fits Pressurized cylinders Hydraulic or pneumatic actuators

2. Ken Youssefi MAE dept. 2 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. < 1/20 t r Thin-walled pressure vessel Stress element Small element Tangential stress θθ Hoop stress Longitudinal stress ll (closed ends) Radial stress rr

3. Ken Youssefi MAE dept. 3 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 [ ( d o ) 2 – ( d i ) 2 ] = ( p )  / 4 ( d i ) 2 Internal pressure, p  F y = 0 Longitudinal stress (  l ) [ ( d i + 2 t ) 2 – ( d i ) 2 ] = ( p ) ( d i ) 2 4t 2 is very small, (  l ) (4 d i t ) = ( p ) ( d i ) 2 l =l = p didi 4 t4 t l =l = p (d i + t) 4 t Max. longitudinal stress Pressure area

4. Ken Youssefi MAE dept. 4 Stresses in a Thin-Walled Pressurized Cylinders Projected area Hoop stress  θ Tangential (hoop) stress 2 (  θ ) t ( length ) = ( p ) ( d i ) (length)  F x = 0 θ =θ = p didi 2 t2 t Max. Hoop stress  l = ½  θ

5. Ken Youssefi MAE dept. 5 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, ε l = constant dr θθ θθ  r + d  r rr 2 (  θ )(dr)(l) +  r ( 2r l ) – (  r + d  r ) [ 2 ( r + dr ) l ] = 0 l = length of cylinder  F = 0 ( d  r ) ( dr ) is very small compared to other terms ≈ 0  θ –  r – r dr drdr = 0 (1)

6. Ken Youssefi MAE dept. 6 Stresses in a Thick-Walled Pressurized Cylinders εl = – μεl = – μ θθ E rr E – μ– μ Deformation in the longitudinal direction θθ +  r = 2 C 1 = εlεl E μ constant (2) Consider, d (  r r 2 ) dr = r 2 drdr dr + 2 r  r Subtract equation (1) from (2),  r +  r + r dr drdr = 2 C 1  θ –  r – r dr drdr = 0 (1) 2rr + r22rr + r2 dr drdr = 2 r C 1 Multiply the above equation by r d (  r r 2 ) dr = 2rC12rC1 r r 2r r 2 = r 2 C 1 + C 2 rr = C1= C1 C2C2 r2r2 + θθ = C1= C1 C2C2 – r2r2

7. Ken Youssefi MAE dept. 7 Stresses in a Thick-Walled Pressurized Cylinders Boundary conditions rr = - p i at r = r i rr = - p o at r = r o θθ = p i r i 2 - p o r o 2 – r i 2 r o 2 ( p o – p i ) / r 2 r o 2 - r i 2 Hoop stress rr = p i r i 2 - p o r o 2 + r i 2 r o 2 ( p o – p i ) / r 2 r o 2 - r i 2 Radial stress p i r i 2 - p o r o 2 ll = r o 2 - r i 2 Longitudinal stress Longitudinal stress,  l Pressure area

8. Ken Youssefi MAE dept. 8 Stresses in a Thick-Walled Pressurized Cylinders Special case, p o (external pressure) = 0 θθ = p i r i 2 r o 2 - r i 2 (1 + ro2ro2 r2r2 ) rr = r o 2 - r i 2 (1 - ro2ro2 r2r2 ) p i r i 2 Hoop stress distribution, maximum at the inner surface Radial stress distribution, maximum at the inner surface

9. Ken Youssefi MAE dept. 9 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.

10. Ken Youssefi MAE dept. 10 Press and Shrink (Force) Fits d s (shaft) > d h (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 p o = p p i = 0 r i = r i r o = R iθiθ = p i r i 2 - p o r o 2 – r i 2 r o 2 ( p o – p i ) / r 2 r o 2 - r i 2 iθiθ = – p o R 2 – r i 2 R 2 p / R 2 R 2 – r i 2 = – p R 2 – r i 2 R2 + ri2R2 + ri2 irir = – p o = – p

11. Ken Youssefi MAE dept. 11 Press and Shrink Fits Consider the outer member at the interface p o = 0 p i = p r i = R r o = r o oθoθ = p r o 2 – R 2 R2 + ro2R2 + ro2 or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 ε ot = oθoθ EoEo  or EoEo – μo– μo ε ot = oo R  o = Rp EoEo r o 2 – R 2 R2 + ro2R2 + ro2 + μ o ( ) Radial interference,  =  o –  i

12. Ken Youssefi MAE dept. 12 Press and Shrink Fits Tangential strain at the outer radius of the inner member ε it = iθiθ EiEi  ir EiEi – μi– μi ε it = ii R –  i = Rp EiEi R 2 – r i 2 R2 + ri2R2 + ri2 + μ i ( ) – Radial interference,  =  o –  i  = Rp EoEo r o 2 – R 2 R2 + ro2R2 + ro2 + μ o ( ) Rp EiEi R 2 – r i 2 R2 + ri2R2 + ri2 + μ i ( ) + If both members are made of the same material then, E = E o = E i and μ = μ o = μ i p = E E  R (R 2 – r i 2 ) (r o 2 – R 2 ) (r o 2 - r i 2 ) 2R22R2

13. Ken Youssefi MAE dept. 13 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 S y = 40,000 psi Hub Shaft Gear 2.0 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 area shaft radius Coefficient of friction Contact length between hub and shaft Min. contact pressure

14. Ken Youssefi MAE dept. 14 Press and Shrink Fits - Example 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 E  R (R 2 – r i 2 ) (r o 2 – R 2 ) (r o 2 - r i 2 ) 2R22R2 r i = 0 r o = 1.5 R = 1.0 4243 = 30 x 10 6 (  ) 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

15. Ken Youssefi MAE dept. 15 ANSI Tables for Fits Interference fits (Force and Shrink fits) – FN1 to FN5 2.00

16. Ken Youssefi MAE dept. 16 Press and Shrink Fits - Example 0.6 x 10 -3 minimum interference FN1 1.8 x 10 -3 maximum interference 0.8 x 10 -3 minimum interference FN2 2.7 x 10 -3 maximum interference 1.3 x 10 -3 minimum interference FN3 3.2 x 10 -3 maximum interference 2δ =.00102 = 1.02 x 10 -3 in. Diametral interference needed Select FN3 fit

17. Ken Youssefi MAE dept. 17 Press and Shrink Fits - Example Check for failure of the shaft and the hub of the gear for max. interference fit iθiθ = – p o R 2 – r i 2 R 2 p / R 2 R 2 – r i 2 = – p R 2 – r i 2 R2 + ri2R2 + ri2 irir = – p o = – p Shaft (outer surface of inner member), r i = 0 = – p Maximum contact pressure for maximum interference, (2δ) max = 3.2 x 10 -3 in. p = 30 x 10 6 (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  i θ = 13333 < 40000 = S y shaft will not fail

18. Ken Youssefi MAE dept. 18 Press and Shrink Fits - Example Hub (inner surface of outer member), r i = R = 13333 1.5 2 – 1 2 1.5 2 + 1 2 oθoθ = – p o R 2 – r o 2 R 2 p / R 2 R 2 – r o 2 = p R 2 – r o 2 R2 + ro2R2 + ro2 oror = – p o = – p = -13333 = 34666 psi  o θ = 34666 < 40000 = S y Hub will not fail