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![The horizontal components cancel out
when integrated over semi-circular
portion as there will be another equal
and opposite horizontal component on
the other side of the vertical axis.
sin
2
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0
∫ ××××=∴
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θ dθdl
dA
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p
θ
d
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surface.curvedtheofareaprojectedp
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d
p 0
×=
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θ](https://image.slidesharecdn.com/thinandthickcylinders-180511185429/75/Thin-and-thick-cylinders-10-2048.jpg)
![final volume = final area of cross section × final length
[ ] [ ]
[ ] [ ]
[ ]LddLdLdddLLdLd
LLdddd
LLdd
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![[ ]
Ld
4
π
2
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××
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![A copper tube having 45mm internal diameter and 1.5mm wall
thickness is closed at its ends by plugs which are at 450mm apart. The
tube is subjected to internal pressure of 3 MPa and at the same time
pulled in axial direction with a force of 3 kN. Compute: i) the change
in length between the plugs ii) the change in internal diameter of the
tube. Take ECU = 100 GPa, and μCU = 0.3.
A] Due to Fluid pressure of 3 MPa:
Longitudinal stress, σL = (p×d) / (4×t)
= (3×45) / (4× 1.5) = 22.50 N/mm2
= 22.50 MPa.
SOLUTION:
Change in length, δL= εL × L = 9 × 10-5
×450 = +0.0405 mm (increase)
E
)μ21(
t4
d)(p
εstrain,Long. L
×−
×
×
×
=
5
3
109
10100
)3.021(5.22 −
×=
×
×−×
=](https://image.slidesharecdn.com/thinandthickcylinders-180511185429/75/Thin-and-thick-cylinders-27-2048.jpg)
![Change in diameter, δd= εc × d = 3.825 × 10-4
×45
= + 0.0172 mm (increase)
B] Due to Pull of 3 kN (P=3kN):
Area of cross section of copper tube, Ac = π × d × t
= π × 45 × 1.5 = 212.06 mm2
Longitudinal strain, ε L = direct stress/E = σ/E = P/(Ac × E)
= 3 × 103
/(212.06 × 100 × 103
)
= 1.415 × 10-4
Change in length, δL=εL× L= 1.415 × 10-4
×450= +0.0637mm (increase)
E
)μ2(
t)(4
d)(p
εstrainntialCircumfere C
−
×
×
×
=
Pd/4t = 22.5
4
3
10825.3
10100
)3.02(5.22 −
×=
×
−×
=](https://image.slidesharecdn.com/thinandthickcylinders-180511185429/75/Thin-and-thick-cylinders-28-2048.jpg)
Uploaded byShivendra Nandan
Thin and thick cylinders
1. Cylinders are commonly used in engineering to transport or store fluids and are subjected to internal fluid pressures. This induces three stresses on the cylinder wall - circumferential, longitudinal, and radial. 2. For thin cylinders where the wall thickness is less than 1/20 the diameter, the radial stress can be neglected. Equations are derived to calculate the circumferential and longitudinal stresses based on the internal pressure, diameter, and wall thickness. 3. Sample problems are worked out applying the equations to example thin-walled cylinders under internal pressure, finding stresses, strains, and changes in dimensions.
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Thin and thick cylinders
- 1. THIN AND THICKCYLINDERS They are, In many engineering applications, cylinders are frequently used for transporting or storing of liquids, gases or fluids. Eg: Pipes, Boilers, storage tanks etc. These cylinders are subjected to fluid pressures. When a cylinder is subjected to a internal pressure, at any point on the cylinder wall, three types of stresses are induced on three mutually perpendicular planes. INTRODUCTION:
- 2. 2. Longitudinal Stress(σL ) – This stress is directed along the length of the cylinder. This is also tensile in nature and tends to increase the length. 3. Radial pressure ( pr ) – It is compressive in nature. Its magnitude is equal to fluid pressure on the inside wall and zero on the outer wall if it is open to atmosphere. 1. Hoop or Circumferential Stress (σC) – This is directed along the tangent to the circumference and tensile in nature. Thus, there will be increase in diameter.
- 3. σ C σL 1. Hoop Stress (σC) 2. Longitudinal Stress (σL) 3. Radial Stress (pr) Element on the cylinder wall subjected to these three stresses σ C σ C σC p σ L σ L σ L p p pr σ Lσ L σ C σ C pr pr
- 4. INTRODUCTION: A cylinder orspherical shell is considered to be thin when the metal thickness is small compared to internal diameter. i. e., when the wall thickness, ‘t’ is equal to or less than ‘d/20’, where ‘d’ is the internal diameter of the cylinder or shell, we consider the cylinder or shell to be thin, otherwise thick. Magnitude of radial pressure is very small compared to other two stresses in case of thin cylinders and hence neglected. THIN CYLINDERS
- 5. Longitudinal axisLongitudinal stress Circumferential stress t Thestress acting along the circumference of the cylinder is called circumferential stresses whereas the stress acting along the length of the cylinder (i.e., in the longitudinal direction ) is known as longitudinal stress
- 6. The bursting willtake place if the force due to internal (fluid) pressure (acting vertically upwards and downwards) is more than the resisting force due to circumferential stress set up in the material. p σc σc P - internal pressure (stress) σc –circumferential stress
- 7. P - internalpressure (stress) σc – circumferential stress dL σc p t
- 8. EVALUATION OF CIRCUMFERENTIALor HOOP STRESS (σC): Consider a thin cylinder closed at both ends and subjected to internal pressure ‘p’ as shown in the figure. Let d=Internal diameter, t = Thickness of the wall L = Length of the cylinder. p d t σcσc dlt p d
- 9. To determine theBursting force across the diameter: Consider a small length ‘dl’ of the cylinder and an elementary area ‘dA’ as shown in the figure. rpp dθdldAdF ×××=×= dθdldFx ××××= θcos 2 d p dA σcσc dlt p d dθ θ Force on the elementary area, Horizontal component of this force dθdl ×××= 2 d p dθdldFy ××××= θsin 2 d p Vertical component of this force
- 10. The horizontal componentscancel out when integrated over semi-circular portion as there will be another equal and opposite horizontal component on the other side of the vertical axis. sin 2 d pforceburstingldiametricaTotal 0 ∫ ××××=∴ π θ dθdl dA σcσc dlt p θ d dθ [ ] surface.curvedtheofareaprojectedp dpcosdl 2 d p 0 ×= ××=−×××= dlπ θ
- 11. dlcc ×××=∴ tσ2)σstressntialcircumfereto(dueforceResisting dldl××=××× dptσ2i.e., c )1....(.................... t2 dp σstress,ntialCircumfere c × × =∴ dL σc p t forceBurstingforceResistingum,equillibriUnder =
- 12. )1....(.................... t2 dp σstress,ntialCircumfere c × × =∴ Force dueto fluid pressure = p × area on which p is acting = p ×(d ×L) (bursting force) Force due to circumferential stress = σc × area on which σc is acting (resisting force) = σc × ( L × t + L ×t ) = σc × 2 L × t Under equilibrium bursting force = resisting force p ×(d ×L) = σc × 2 L × t Assumed as rectangular
- 13. LONGITUDINAL STRESS (σL): p σL Theforce, due to pressure of the fluid, acting at the ends of the thin cylinder, tends to burst the cylinder as shown in figure P A B The bursting of the cylinder takes place along the section AB
- 14. EVALUATION OF LONGITUDINALSTRESS (σL): d 4 π pcylinder)ofend(on theforceburstingalLongitudin 2 ××= p t σL tdπσforceResisting L ×××=∴ tdπforcethisresistingsectioncrossofArea ××= cylinder.theofmaterialtheofstressalLongitudinσLet L =
- 15. tdπσd 4 π pi.e., L 2 ×××=×× )2.........(.......... t4 dp σstress,alLongitudin L × × =∴ LCσ2σ(2),&(1)eqsFrom ×= forceresistingforceburstingum,equillibriUnder =
- 16.
- 17. EVALUATION OF STRAINS Apoint on the surface of thin cylinder is subjected to biaxial stress system, (Hoop stress and Longitudinal stress) mutually perpendicular to each other, as shown in the figure. The strains due to these stresses i.e., circumferential and longitudinal are obtained by applying Hooke’s law and Poisson’s theory for elastic materials. σ C=(pd)/(2t)σ C=(pd)/(2t) σL=(pd)/(4t) σ L=(pd)/(4t)
- 18. E σ μ E σ ε :εstrain,ntialCircumfere LC C C ×−= )3..(..............................μ)2( Et4 dp d δd εi.e., C −× ×× × == σC=(pd)/(2σC=(pd)/(2t) σ L=(pd)/(4t) σ L=(pd)/(4t) Note: Let δd be the change in diameter. Then ( ) d d d ddd ncecircumfereoriginal ncecircumfereoriginalncecircumferefinal c δ π πδπ ε = −+ = − = μ)2( E σ E σ μ E σ 2 L LL −×= ×−×=
- 19. E σ μ E σ ε :εstrain,alLongitudin CL L L ×−= )4..(..............................μ)21( Et4 dp L δl εi.e., L ×−× ×× × == V v STRAIN,VOLUMETRIC δ Changein volume = δV = final volume – original volume original volume = V = area of cylindrical shell × length L d 4 2 π = μ)21( E σ E )σ2( μ E σ LLL ×−×= × ×−=
- 20. final volume =final area of cross section × final length [ ] [ ] [ ] [ ] [ ]LddLdLdddLLdLd LLdddd LLdd δδδδδδδ π δδδ π δδ π 2)(2)( 4 2)( 4 4 2222 22 2 +++++= +×++= +×+= [ ]LdddLLdvolumeFinal LddandLdLdassuchquantitiessmallertheneglecting δδ π δδδδδ 22 22 2 4 2)(,)( ++= [ ] [ ] [ ]LdddLV LdLdddLLdVvolumeinchange δδ π δ π δδ π δ 2 222 2 4 4 2 4 += −++=
- 21. [ ] Ld 4 π 2 4 π V dv 2 2 ×× + = dLdLd δδ =εL + 2 × εC V dV )5.......(..........μ)45( Et4 dp V dv i.e., ×− ×× × = μ)2( Et4 dp 2μ)21( Et4 dp − ×× × ×+×− ×× × = d d 2 L L δδ ×+=
- 22. 2 4t pd 2t pd 2 σ-σ τstress,ShearMaximum other.eachto larperpendicuactandnormalarestressesthese Bothal.longitudinandntialCircumfereviz., point,anyatstressesprincipaltwoareThere :stressShearMaximum LC max − = =∴ )5.(.................... 8t pd τi.e., max = σC=(pd)/(2t) σC=(pd)/(2t) σ L=(pd)/(4t) σ L=(pd)/(4t)
- 23.
- 24. PROBLEM 1: A thincylindrical shell is 3m long and 1m in internal diameter. It is subjected to internal pressure of 1.2 MPa. If the thickness of the sheet is 12mm, find the circumferential stress, longitudinal stress, changes in diameter, length and volume . Take E=200 GPa and μ= 0.3. 1. Circumferential stress, σC: σC= (p×d) / (2×t) = (1.2×1000) / (2× 12) = 50 N/mm2 = 50 MPa (Tensile). SOLUTION: 2. Longitudinal stress, σL: σL = (p×d) / (4×t) = σC/2 = 50/2 = 25 N/mm2 = 25 MPa (Tensile). ILLUSTRATIVE PROBLEMS
- 25. 3. Circumferential strain,εc: Change in length = ε L ×L= 5×10-05 ×3000 = 0.15 mm (Increase). E μ)(2 t)(4 d)(p εc − × × × = Change in diameter, δd = εc ×d = 2.125×10-04 ×1000 = 0.2125 mm (Increase). E μ)2(1 t)(4 d)(p εL ×− × × × = 4. Longitudinal strain, εL: (Increase)102.125 10200 0.3)(2 12)(4 1000)(1.2 04- 3 ×= × − × × × = (Increase)105 10200 0.3)2(1 12)(4 1000)(1.2 05- 3 ×= × ×− × × × =
- 26.
- 27. A copper tubehaving 45mm internal diameter and 1.5mm wall thickness is closed at its ends by plugs which are at 450mm apart. The tube is subjected to internal pressure of 3 MPa and at the same time pulled in axial direction with a force of 3 kN. Compute: i) the change in length between the plugs ii) the change in internal diameter of the tube. Take ECU = 100 GPa, and μCU = 0.3. A] Due to Fluid pressure of 3 MPa: Longitudinal stress, σL = (p×d) / (4×t) = (3×45) / (4× 1.5) = 22.50 N/mm2 = 22.50 MPa. SOLUTION: Change in length, δL= εL × L = 9 × 10-5 ×450 = +0.0405 mm (increase) E )μ21( t4 d)(p εstrain,Long. L ×− × × × = 5 3 109 10100 )3.021(5.22 − ×= × ×−× =
- 28. Change in diameter,δd= εc × d = 3.825 × 10-4 ×45 = + 0.0172 mm (increase) B] Due to Pull of 3 kN (P=3kN): Area of cross section of copper tube, Ac = π × d × t = π × 45 × 1.5 = 212.06 mm2 Longitudinal strain, ε L = direct stress/E = σ/E = P/(Ac × E) = 3 × 103 /(212.06 × 100 × 103 ) = 1.415 × 10-4 Change in length, δL=εL× L= 1.415 × 10-4 ×450= +0.0637mm (increase) E )μ2( t)(4 d)(p εstrainntialCircumfere C − × × × = Pd/4t = 22.5 4 3 10825.3 10100 )3.02(5.22 − ×= × −× =
- 29. Lateral strain, εlat=-μ × Longitudinal strain = -μ × εL = - 0.3× 1.415 × 10-4 = -4.245 × 10-5 Change in diameter, δd = εlat × d = -4.245 × 10-5 ×45 = - 1.91 × 10-3 mm (decrease) C) Changes due to combined effects: Change in length = 0.0405 + 0.0637 = + 0.1042 mm (increase) Change in diameter = 0.01721 - 1.91 × 10-3 = + 0.0153 mm (increase)
- 30. PROBLEM 3: A cylindricalboiler is 800mm in diameter and 1m length. It is required to withstand a pressure of 100m of water. If the permissible tensile stress is 20N/mm2 , permissible shear stress is 8N/mm2 and permissible change in diameter is 0.2mm, find the minimum thickness of the metal required. Take E = 200GPa, and μ = 0.3. Fluid pressure, p = 100m of water = 100×9.81×103 N/m2 = 0.981N/mm2 . SOLUTION: 1. Thickness from Hoop Stress consideration: (Hoop stress is critical than long. Stress) σC = (p×d)/(2×t) 20 = (0.981×800)/(2×t) t = 19.62 mm
- 31. 2. Thickness fromShear Stress consideration: 3. Thickness from permissible change in diameter consideration (δd=0.2mm): E μ)(2 t)(4 d)(p d δd − × × × = Therefore, required thickness, t = 19.62 mm. t)(8 d)(p τmax × × = 12.26mm.t t)(8 800)(0.981 8 =∴ × × = 6.67mmt 10200 )3.02( )t4( )800981.0( 800 2.0 3 = × − × × × =
- 32. PROBLEM 4: A cylindricalboiler has 450mm in internal diameter, 12mm thick and 0.9m long. It is initially filled with water at atmospheric pressure. Determine the pressure at which an additional water of 0.187 liters may be pumped into the cylinder by considering water to be incompressible. Take E = 200 GPa, and μ = 0.3. Additional volume of water, δV = 0.187 liters = 0.187×10-3 m3 = 187×103 mm3 SOLUTION: 3632 mm10143.14)109.0(450 4 π V ×=×××= μ)45( Et4 dp V dV ×− ×× × = Solving, p=7.33 N/mm2 )33.045( 10200124 450p 10143.14 10187 36 3 ×− ××× × = × ×
- 33. JOINT EFFICIENCY Longitudinal rivets Circumferential rivets Steel platesof only particular lengths and width are available. Hence whenever larger size cylinders (like boilers) are required, a number of plates are to be connected. This is achieved by using riveting in circumferential and longitudinal directions as shown in figure. Due to the holes for rivets, the net area of cross section decreases and hence the stresses increase.
- 34. JOINT EFFICIENCY The cylindricalshells like boilers are having two types of joints namely Longitudinal and Circumferential joints. Due to the holes for rivets, the net area of cross section decreases and hence the stresses increase. If the efficiencies of these joints are known, the stresses can be calculated as follows. Let η L= Efficiency of Longitudinal joint and η C = Efficiency of Circumferential joint. ...(1).......... ηt2 dp σ L C ×× × = Circumferential stress is given by,
- 35. ...(2).......... ηt4 dp σ C L ×× × = Note: In longitudinaljoint, the circumferential stress is developed and in circumferential joint, longitudinal stress is developed. Longitudinal stress is given by, Circumferential rivets Longitudinal rivets
- 36. If A isthe gross area and Aeff is the effective resisting area then, Efficiency = Aeff/A Bursting force = p L d Resisting force = σc ×Aeff = σc ×ηL ×A = σc ×ηL ×2 t L Where η L=Efficiency of Longitudinal joint Bursting force = Resisting force p L d = σc ×ηL × 2 t L ...(1).......... ηt2 dp σ L C ×× × =
- 37. If η c=Efficiencyof circumferential joint Efficiency = Aeff/A Bursting force = (π d2 /4)p Resisting force = σL ×A′eff =σL ×ηc ×A′ = σL ×ηc ×π d t Where η L=Efficiency of circumferential joint Bursting force = Resisting force ...(2).......... ηt4 dp σ C L ×× × =
- 38. A cylindrical tankof 750mm internal diameter, 12mm thickness and 1.5m length is completely filled with an oil of specific weight 7.85 kN/m3 at atmospheric pressure. If the efficiency of longitudinal joints is 75% and that of circumferential joints is 45%, find the pressure head of oil in the tank. Also calculate the change in volume. Take permissible tensile stress of tank plate as 120 MPa and E = 200 GPa, and μ = 0.3. Let p = max permissible pressure in the tank. Then we have, σL= (p×d)/(4×t) η C 120 = (p×750)/(4×12) 0.45 p = 3.456 MPa. SOLUTION: Also, σ C= (p×d)/(2×t) η L 120 = (p×750)/(2×12) 0.75 p = 2.88 MPa.
- 39. Max permissible pressurein the tank, p = 2.88 MPa. μ)45( E)t(4 d)(p V dv Strain,Vol. ×−× ×× × = litres.0.567m100.567 .mm10567.01500750 4 π 108.55V108.55dv 108.550.3)4-(5 )1020012(4 750)(2.88 33- 3624-4- 4- 3 =×= ×=××××=××= ×=×× ××× × =
- 40. A boiler shellis to be made of 15mm thick plate having a limiting tensile stress of 120 N/mm2 . If the efficiencies of the longitudinal and circumferential joints are 70% and 30% respectively determine; i) The maximum permissible diameter of the shell for an internal pressure of 2 N/mm2 . (ii) Permissible intensity of internal pressure when the shell diameter is 1.5m. (i) To find the maximum permissible diameter of the shell for an internal pressure of 2 N/mm2 : SOLUTION: ηt2 dp σe.,i. L c ×× × = a) Let limiting tensile stress = Circumferential stress = σ c = 120N/mm2 . d = 1260 mm 7.0512 d2 120 ×× × =
- 41. ηt4 dp σe.,i. C L ×× × = b) Let limitingtensile stress = Longitudinal stress = σ L = 120N/mm2 . The maximum diameter of the cylinder in order to satisfy both the conditions = 1080 mm. d = 1080 mm. 3.0514 d2 120 ×× × =
- 42. The maximum permissiblepressure = 1.44 N/mm2 . (ii) To find the permissible pressure for an internal diameter of 1.5m: (d=1.5m=1500mm) a) Let limiting tensile stress = Circumferential stress = σ c = 120N/mm2 . ηt2 dp σe.,i. L c ×× × = b) Let limiting tensile stress = Longitudinal stress = σ L = 120N/mm2 . ηt4 dp σe.,i. C L ×× × = .N/mm1.68p 7.0512 5001p 120 2 = ×× × = .N/mm1.44p 3.0514 5001p 120 2 = ×× × =
- 43. PROBLEM 1: Calculate thecircumferential and longitudinal strains for a boiler of 1000mm diameter when it is subjected to an internal pressure of 1MPa. The wall thickness is such that the safe maximum tensile stress in the boiler material is 35 MPa. Take E=200GPa and μ= 0.25. (Ans: ε C=0.0001531, ε L=0.00004375) PROBLEM 2: A water main 1m in diameter contains water at a pressure head of 120m. Find the thickness of the metal if the working stress in the pipe metal is 30 MPa. Take unit weight of water = 10 kN/m3 . (Ans: t=20mm) PROBLEMS FOR PRACTICE
- 44. THIN AND THICK CYLINDERS-33 PROBLEM 3: A gravity main 2m in diameter and 15mm in thickness. It is subjected to an internal fluid pressure of 1.5 MPa. Calculate the hoop and longitudinal stresses induced in the pipe material. If a factor of safety 4 was used in the design, what is the ultimate tensile stress in the pipe material? (Ans: σC=100 MPa, σL=50 MPa, σU=400 MPa) PROBLEM 4: At a point in a thin cylinder subjected to internal fluid pressure, the value of hoop strain is 600×10-4 (tensile). Compute hoop and longitudinal stresses. How much is the percentage change in the volume of the cylinder? Take E=200GPa and μ= 0.2857. (Ans: σC=140 MPa, σL=70 MPa, %age change=0.135%.)
- 45. THIN AND THICK CYLINDERS-34 PROBLEM 5: A cylindrical tank of 750mm internal diameter and 1.5m long is to be filled with an oil of specific weight 7.85 kN/m3 under a pressure head of 365 m. If the longitudinal joint efficiency is 75% and circumferential joint efficiency is 40%, find the thickness of the tank required. Also calculate the error of calculation in the quantity of oil in the tank if the volumetric strain of the tank is neglected. Take permissible tensile stress as 120 MPa, E=200GPa and μ= 0.3 for the tank material. (Ans: t=12 mm, error=0.085%.)
- 46.
- 47. INTRODUCTION: The thickness ofthe cylinder is large compared to that of thin cylinder. i. e., in case of thick cylinders, the metal thickness ‘t’ is more than ‘d/20’, where ‘d’ is the internal diameter of the cylinder. Magnitude of radial stress (pr) is large and hence it cannot be neglected. The circumferential stress is also not uniform across the cylinder wall. The radial stress is compressive in nature and circumferential and longitudinal stresses are tensile in nature. Radial stress and circumferential stresses are computed by using ‘Lame’s equations’.
- 48. LAME’S EQUATIONS (Theory): 4. The material is homogeneous, isotropic and obeys Hooke’s law. (The stresses are within proportionality limit). 1. Plane sections of the cylinder normal to its axis remain plane and normal even under pressure. 2. Longitudinal stress (σL) and longitudinal strain (εL) remain constant throughout the thickness of the wall. 3. Since longitudinal stress (σL) and longitudinal strain (εL) are constant, it follows that the difference in the magnitude of hoop stress and radial stress (pr) at any point on the cylinder wall is a constant. ASSUMPTIONS:
- 49. LAME’S EQUATIONS FORRADIAL PRESSURE AND CIRCUMFERENTIAL STRESS Consider a thick cylinder of external radius r1 and internal radius r2, containing a fluid under pressure ‘p’ as shown in the fig. Let ‘L’ be the length of the cylinder. p r2 r1 p
- 50. Consider an elementalring of radius ‘r’ and thickness ‘δr’ as shown in the above figures. Let pr and (pr+ δpr) be the intensities of radial pressures at inner and outer faces of the ring. pr pr+δpr r2 r1 r pr pr+δpr r2 r1 r σc σc r δr Pr pr+δpr External pressure
- 51. Consider the longitudinal sectionXX of the ring as shown in the fig. The bursting force is evaluated by considering the projected area, ‘2×r×L’ for the inner face and ‘2×(r+δr)×L’ for the outer face . The net bursting force, P = pr×2×r×L - (pr+δpr)×2×(r+δr)×L =( -pr× δr - r×δpr - δpr × δr) 2L Bursting force is resisted by the hoop tensile force developing at the level of the strip i.e., Fr=σc×2 ×δr×L L r pr pr+δpr r+δr X X
- 52. Thus, for equilibrium,P = Fr (-pr× δr - r×δpr- δpr × δr) 2L = σ c×2×δr×L -pr× δr - r×δpr- δpr × δr = σ c×δr Neglecting products of small quantities, (i.e., δpr × δr) σ c = - pr – (r× δpr )/ δr ...…………….(1) Longitudinal strain is constant. Hence we have, since σL, E and μ are constants (σc – Pr) should be constant . Let it be equal to 2a. Thus constant E p μ E σ μ E σ rCL =×+×− Since Pr is compressiveε L = constant)pσ( μ E σ rC L =−− E ε L =
- 53. σ c- pr= 2a, i.e., σc = pr + 2a, ………………(2) From (1), pr+ 2a = - pr – (r× δpr ) / δr r r r p -a)p(2 δ δ ×=+ r )3.(.......... a)p( p 2 r rr + =×− δδ r i. e., Integrating, (-2 ×loge r) + c = loge (pr + a) Where c is constant of integration. Let it be taken as loge b, where ‘b’ is another constant. Thus, loge (pr+a) = -2 ×loge r + loge b = - loge r2 + loge b = loge 2 r b
- 54. .....(4)..........a r b pstress,radialor, r b api.e., 2r2r −==+ Theequations (4) & (5) are known as “Lame’s Equations” for radial pressure and hoop stress at any specified point on the cylinder wall. Thus, r1≤r ≤r2. a2a b a2pσstress,Hoop 2rc +−=+= r .......(5)....................a r b σi.e., 2c += Substituting it in equation 2, we get
- 55. ANALYSIS FOR LONGITUDINALSTRESS Consider a transverse section near the end wall as shown in the fig. Bursting force, P =π×r2 2 ×p Resisting force is due to longitudinal stress ‘σ L’. i.e., FL= σ L× π ×(r1 2 -r2 2 ) For equilibrium, FL= P σ L× π ×(r1 2 -r2 2 )= π ×r2 2 ×p Therefore, longitudinal stress, (Tensile) )r(r rp σ 2 2 2 1 2 2 L − × = p p r2 r1 σL σ LσL σL L
- 56. NOTE: 1. Variations ofHoop stress and Radial stress are parabolic across the cylinder wall. 2. At the inner edge, the stresses are maximum. 3. The value of ‘Permissible or Maximum Hoop Stress’ is to be considered on the inner edge. 4. The maximum shear stress (σ max) and Hoop, Longitudinal and radial strains (εc, εL, εr) are calculated as in thin cylinder but separately for inner and outer edges.
- 57. ILLUSTRATIVE PROBLEMS PROBLEM 1: Athick cylindrical pipe of external diameter 300mm and internal diameter 200mm is subjected to an internal fluid pressure of 20N/mm2 and external pressure of 5 N/mm2 . Determine the maximum hoop stress developed and draw the variation of hoop stress and radial stress across the thickness. Show at least four points for each case. SOLUTION: External diameter = 300mm. External radius, r1=150mm. Internal diameter = 200mm. Internal radius, r2=100mm. Lame’s equations: For Hoop stress, .........(1) For radial stress, .........(2) a r b σ 2c += a r b p 2r −=
- 58. Boundary conditions: At r=100mm (on the inner face), radial pressure = 20N/mm2 i.e., Similarly, at r =150mm (on the outer face), radial pressure = 5N/mm2 i.e., )........(3..........a 100 b 20 2 −= )........(4..........a 150 b 5 2 −= ..(5)..........7 r 2,70,000 σ 2c += ..(6)..........7 r 2,70,000 p 2r −= Solving equations (3) & (4), we get a = 7, b = 2,70,000. Lame’s equations are, for Hoop stress, For radial stress,
- 59. To draw variationsof Hoop stress & Radial stress : At r =100mm (on the inner face), (Comp)MPa207 100 2,70,000 pstress,Radial (Tensile)MPa347 100 2,70,000 σstress,Hoop 2r 2c =−= =+= At r =120mm, (Comp)MPa11.757 120 2,70,000 pstress,Radial (Tensile)MPa25.757 120 2,70,000 σstress,Hoop 2r 2c =−= =+= At r =135mm, (Comp)MPa7.817 135 2,70,000 pstress,Radial (Tensile)MPa21.817 135 2,70,000 σstress,Hoop 2r 2c =−= =+=
- 60. (Comp)MPa57 150 2,70,000 pstress,Radial (Tensile)MPa197 150 2,70,000 σstress,Hoop 150mm,rAt 2r 2c =−= =+= = Variation of Hoopstress & Radial stress Variation of Hoop Stress-Tensile (Parabolic) Variation of Radial Stress –Comp (Parabolic)
- 61. PROBLEM 2: Find thethickness of the metal required for a thick cylindrical shell of internal diameter 160mm to withstand an internal pressure of 8 N/mm2 . The maximum hoop stress in the section is not to exceed 35 N/mm2 . SOLUTION: Internal radius, r2=80mm. ......(2)..........a r b pstress,Radialfor (1)....................a r b σStress,Hoopfor are,equationssLame' 2r 2c −= +=
- 62.
- 63. .rrat0pi.e., 0.pressureface,outerOn the 1r == = 20.96mm. r-rmetaltheofThickness21 = =∴ 100.96mm.r 5.13 r 1,37,600 0 1 2 1 =∴ −=∴
- 64. PROBLEM 3: A thickcylindrical pipe of outside diameter 300mm and internal diameter 200mm is subjected to an internal fluid pressure of 14 N/mm2 . Determine the maximum hoop stress developed in the cross section. What is the percentage error if the maximum hoop stress is calculated by the equations for thin cylinder? SOLUTION: Internal radius, r2=100mm. External radius, r1=150mm . Lame’s equations: For Hoop stress, .........(1) For radial pressure, .........(2) a r b σ 2c += a r b p 2r −=
- 65. Boundary conditions: At x=100mm Pr = 14N/mm2 i.e., Similarly, at x =150mm Pr = 0 i.e., )........(1..........a 100 b 14 2 −= )........(2..........a 150 b 0 2 −= ..(3)..........2.11 r 22,500 σstress,HoopforequationsLame' 2r +=∴ Solving, equations (1) & (2), we get a =11.2, b = 2,52,000.
- 66. MPa.36.42.11 100 252000 σ 2max =+= Maxhoop stress on the inner face (where x=100mm): .23.08%100) 36.4 28-36.4 (errorPercentage =×= 14MPa.pand50mmt200mm,Dewher t2 dp σformula,cylinderBy thin max === × × = MPa.28 502 20014 σmax = × × =∴
- 67. PROBLEM 4: The principalstresses at the inner edge of a cylindrical shell are 81.88 MPa (T) and 40MPa (C). The internal diameter of the cylinder is 180mm and the length is 1.5m. The longitudinal stress is 21.93 MPa (T). Find, (i) Max shear stress at the inner edge. (ii) Change in internal diameter. (iii)Change in length. (iv) Change in volume. Take E=200 GPa and μ=0.3. SOLUTION: 2 (-40)-81.88 2 p-σ τ :faceinneron thestressshearMaxi) rC max == = 60.94 MPa
- 68.
- 69.
- 70. PROBLEM 5: Find themax internal pressure that can be allowed into a thick pipe of outer diameter of 300mm and inner diameter of 200mm so that tensile tress in the metal does not exceed 16 MPa if, (i) there is no external luid pressure, (ii) there is a fluid pressure of 4.2 MPa. SOLUTION: External radius, r1=150mm. Internal radius, r2=100mm. Boundary conditions: At r=100mm , σc = 16N/mm2 At r=150mm , Pr = 0 Case (i) – When there is no external fluid pressure:
- 71. )........(2..........a 150 b 0 )........(1..........a 100 b 16i.e., 2 2 −= += Solving we get,a = 4.92 & b=110.77×103 )........(4..........92.4 10110.77 p )........(3..........92.4 10110.77 σthatso 2 3 r 2 3 c − × = + × = r r MPa.6.1692.4 001 10110.77 p 100mm,rwherefaceinneron thepressureFluid 2 3 r =− × = =
- 72. Boundary conditions: At r=100mm, σc= 16 N/mm2 At r=150mm , pr= 4.2 MPa. Case (ii) – When there is an external fluid pressure of 4.2 MPa: )........(2..........a 150 b 4.2 )........(1..........a 100 b 16i.e., 2 2 −= += Solving we get, a = 2.01 & b=139.85×103 )........(4..........01.2 10139.85 p )........(3..........01.2 10139.85 σthatso 2 3 r 2 3 r − × = + × = r r
- 73.
- 74. PROBLEM 1: A pipeof 150mm internal diameter with the metal thickness of 50mm transmits water under a pressure of 6 MPa. Calculate the maximum and minimum intensities of circumferential stresses induced. (Ans: 12.75 MPa, 6.75 MPa) PROBLEM 2: Determine maximum and minimum hoop stresses across the section of a pipe of 400mm internal diameter and 100mm thick when a fluid under a pressure of 8N/mm2 is admitted. Sketch also the radial pressure and hoop stress distributions across the thickness. (Ans: σmax=20.8 N/mm2 ,σmin=12.8 N/mm2 ) PROBLEM 3: A thick cylinder with external diameter 240mm and internal diameter ‘D’ is subjected to an external pressure of 50 MPa. Determine the diameter ‘D’ if the maximum hoop stress in the cylinder is not to exceed 200 MPa. (Ans: 169.7 mm) PROBLEMS FOR PRACTICE
- 75. THIN AND THICK CYLINDERS-63 PROBLEM 4: A thick cylinder of 1m inside diameter and 7m long is subjected to an internal fluid pressure of 40 MPa. Determine the thickness of the cylinder if the maximum shear stress in the cylinder is not to exceed 65 MPa. What will be the increase in the volume of the cylinder? E=200 GPa, μ=0.3. (Ans: t=306.2mm, δv=5.47×10-3 m3 ) PROBLEM 5: A thick cylinder is subjected to both internal and external pressure. The internal diameter of the cylinder is 150mm and the external diameter is 200mm. If the maximum permissible stress in the cylinder is 20 N/mm2 and external radial pressure is 4 N/mm2 , determine the intensity of internal radial pressure. (Ans: 10.72 N/mm2 )
- 76.