1. Stress element 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 Radial stress rr Longitudinal stress ll (closed ends) 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 Tangential stress θθ Hoop stress

2. 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 2 t2 t l =l = p (d i + t) 4 t Max. longitudinal stress Pressure area

3. 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

4. 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)

5. 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 2 C 1 + C 2 rr = C1= C1 C2C2 r2r2 + θθ = C1= C1 C2C2 – r2r2

6. 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

7. 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

8. Press and Shrink 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

9. 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 ( )

10. 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