CHAPTR 8 Thin nd Thick Cylinders nd Spheres Prolem. A shell.5 m long nd m dimeter, is sujected to n internl pressure of. N/mm. If the thickness of the shell is 0 mm find the circumferentil nd longitudinl stresses. Find lso the mximum sher stress nd chnges in dimensions of the shell. Tke 00 kn/mm nd Poissons rtio 0.. Solution: L.5 m 50 mm d m 000 mm t 0 mm p. N/mm 00 kn/mm 00 0 N/mm Hoop stress f Longitudinl stress f pd. 000 60 N/mm t 0 pd. 000 0 N/mm 4t 4 0 f f 60 0 q mx 5 N/mm Dimetrl strin δ d f f e µ d ( f µ f ) 00 0.55 0 4 (60 0. 0) δd.55 0 4 d.55 0 4 000 0.55 mm. Ans. δ L f f Longitudinl strin µ L Chnge in volume: δv V ( f µ f ) 00 0 6 0 5 (0 0. 60) δl 6 0 5 L 6 0 5 50 0.95 mm. e + e.55 0 4 + 6 0 5 5.7 0 4 Ans. 68
δv 5.7 0 4 V 5.7 0 4 4 π 000 50 454950 mm. Ans. Prolem. A cylindricl shell.4 m long 600 mm in dimeter with metl thickness mm is completely filled with wter t tmosphericl pressure. If n dditionl 00,000 mm wter is then pumped in, find the stresses developed nd chnge in dimensions. Tke 0 5 N/mm, µ 0.. pd p 600 Solution: f 5 p t f pd p 600.5 p 4t 4 δ d f f Dimetricl strin e µ d [5 p 0..5 p].5 p δ L Longitudinl strin e ( f ) µ f L Volumetric strin [.5 p 0. 5 p] 5 p δv V e + e.5 p 5 p + p 47.5 p δv V 47.5 00, 000 0 π 600 400 47.5 4.864 N/mm Hoop stress 5 p 5.864 46.57 N/mm δd d δd δl L e.5 p.5 p 600 0.86 mm e 5 p 5 69
δl 5 p L 0.6 mm f.5 p.68 N/mm δv 00,000 mm. Prolem. The dimeter of riveted oiler is.5 m nd hs to withstnd pressure of N/mm. Find the thickness of pltes to e used if efficiency is 85% in longitudinl joints nd 40% in circumferentil joints. The permissile stress is N/mm. Solution: Let t e the thickness of plte. quting ursting force longitudinl joint strength, we get, pdl n t L f t Considering longitudinl forces πd p 4 t pd 0.764 mm fn 0.85 n πdtf pd 0.5 mm 4 f n 4 0.4 Provide minimum thickness of.5 mm. Prolem 4. A copper cylinder of 00 mm dimeter nd metl thickness 4 mm is closely wound with steel wire of mm dimeter with tensile stress of 60 N/mm. Find the stresses in copper cylinder nd steel wire when fluid is filled t pressure of 4 N/mm. Tke s 0 5 N/mm, c. 0 5 N/mm nd µ c 0.8. Solution: Consider mm length of cylinder No. of wires Force exerted y steel wire t dimetrl section f w0 πd 4 π 60 76.99 N. 4 If initil stress is f c in cylinder, then f c t 76.99 f c 76.99 4.56 N/mm (comp.) Let due to fluid pressure lone, stresses developed in steel wire e f w nd in cylinder. Let it e f nd f. Then, f longitudinl stress pd 4 00 5 N/mm (tensile) 4t 4 4 Considering the equilirium of the cylinder (Ref. Fig. ) of mm length, we get f w π + f 4 4 4 00 Fig. 70
6.88 f w + 6 f 800 f w +.546 f 7.7... () quting strin in wire to circumferentil strin in cylinder, we get fw (f µ f ) s c f w 5 0 (f 0.8 5). 0 Sustituting it in eqn. (), we get.667 f.667 +.546 f 7.7 f From eqn. (), 5 f w.667 f.667... () 8.994 4..99 N/mm f w.667.99.667 4.8 N/mm Hence, finl stresses re () in steel wire 60 + 4.8 0.8 N/mm () in cylinder.56 +.99 9.4 N/mm Prolem 5. A thin sphericl shell of dimeter. m hs metl thickness of 0 mm t tmospheric pressure. Find the chnge in dimeter nd the cpcity of the shell if fluid pressure is rised to.5 N/mm. Tke 0 5 N/mm, µ 0.5. Solution: d. m 00 mm p.5 N/mm t 0 mm pd.5 00 Hoop stress f f 75 N/mm 4t 4 0 nd, δd d δd 00 δv pd 4t ( µ).5 00 4 0 0 δd 0.75 mm. δv pd ( µ ) V 4t 5 [ 0.5].5 00 π 5 ( 0.5) 6 4 0 0.5 00 π 0.75 (00) 5 4 0 0 6 i.e. δv 76407 mm. Ans. Prolem 6. In chemicl plnt sphere of dimeter 900 mm nd metl thickness 8 mm is used to store gs. If the permissile stress in the metl is N/mm, find the mximum pressure with which the gs cn e stored if (i) the sphere is semless (ii) the efficiency of joint is 0.65. d Ans. 7
Solution: (i) When the sphere is semless : pd f 4t p 900 4 8 p 5. N/mm. Ans. (ii) When the efficiency of the joint is 0.85, pd f 4t η f 4t η 4 8 0.85 p d 900 i.e. p.467 N/mm. Ans. Prolem 7. A thick cylindricl pipe of outside dimeter 00 mm nd internl dimeter of 00 mm is sujected to n internl fluid pressure of 0 N/mm nd externl fluid pressure of 5 N/mm. Determine the mximum hoop stress developed nd drw the vrition of hoop stress nd rdil stress cross the thickness. Indicte vlues t every 5 mm intervl. 00 Solution: r i 00 mm r o 00 mm If the hoop stress nd rdil stress in the cylinder t distnce x from the centre of cylinder is f x nd respectively, from Lme s eqution, f x nd x where nd re constnts. Now, t x 00 mm, 0 N/mm 0... () 00 At x mm, 5 N/mm 5... () Sustituting eqution () from eqution (), we get 5 00 00 00 00 69978.4 69978.4 From eqn. (), 0 00 6.9976 69978.4 f x + 6.9976 x 7
Mximum hoop stress occurs when x is lest i.e. t x 00 mm f mx 69978.4 + 6.9976 00 4 N/mm f 5 69978.4 + 6.9976 4.4 N/mm 5 f 69978.4 + 6.9976 9 N/mm p 00 69978.4 6.9976 0 N/mm 00 p 5 69978.4 6.9976 0.48 N/mm 5 p 69978.4 6.9976 5 N/mm Hence vrition of f x nd re s shown in the Fig. : Fig. Prolem 8. A thick cylindricl pipe of internl rdius mm nd externl rdius 00 mm is sujected to n internl fluid pressure of 7.5 N/mm. Determine the mximum hoop stress in the cross-section. Wht is the percentge error if it is determined from thin cylinders theory? Solution: r mm r 00 mm If the rdil stress nd hoop stress developed in the cylinder t distnce of x from the centre re nd f x respectively, from Lme s eqution, x f x where nd re ritrry constnts. At x mm 7.5 N/mm 7.5 Agin when x 00 mm 0 0 00... ()... () 7
From eqn. () nd (), we get 7.5 00 00 00 90005.76 90005.76 Hence.505 00 00 The vlue of f x is mximum when x is lest i.e. t x mm 90005.76 Mximum hoop stress +.505 6.5 N/mm. Ans. Internl pressure Internl dimeter 7.5 Mximum hoop stress thickness (00 ) 5.5 N/mm 6.5 5.5 Percentge error 00 6.5 6. Ans. Prolem 9. The internl nd externl dimeters of thick cylinder re 00 mm nd 500 mm respectively. It is sujected to n externl pressure of 4 N/mm. Find the internl pressure tht cn e pplied if the permissile stress in cylinder is limited to N/mm. Sketch the vrition of hoop stress nd rdil stress cross the thickness of the cylinder. Solution: r From Lme s equtions, At x 50 mm 4 N/mm 00 mm r 500 x f x 50 mm 4... () 50 Mximum hoop stress occurs for the lest vlue of x i.e. t inner edge, where x r mm. From eqn. () nd (), we get 7 +... () + 50 + 50 50 870.68 From eqn. (), 870.68 4 4 50 50 0.5 870.68 0.5 x Internl pressure i.e. pressure t x mm, is 74
p i 870.68 0.5 N/mm 870.68 f + 0.5 N/mm 7.5 5.0 870.68 f 00 + 0.5 7.5 N/mm 6.5 4.0 00 50 870.68 f 50 + 0.5 5 N/mm 50 p 870.68 0.5 N/mm p 00 870.68 0.5 6.5 N/mm Fig. 00 nd p 50 870.68 0.5 4 N/mm 50 Prolem 0. A compound cylinder of inner rdius 00 mm, outer rdius 40 mm hs common rdius t 80 mm. The rdil pressure developed t junction is N/mm. Determine the rdil nd hoop stresses developed t inner, common nd outer rdii when the fluid is dmitted t pressure of 60 N/mm. Solution: () Inner cylinder : Let Lme s eqution e, x nd f x At x r 00 mm, 0. 0... () 00 At x r 80 mm, N/mm 80 From eqns. () nd (), 7565.99... () 00 80 80 00 80 00 7565.99 7.57 00 00 7565.99 f 00 + 7.57 00 00 4.77 N/mm 7565.99 f 80 + 7.57 80 80.7 N/mm () Outer cylinder : Let Lme s eqution e, x nd f x At x 80 mm, 75
80... () At x 40 mm, 0 0 40... (4) From equtions () nd (4), we get 888685.7 40 80 80 40 80 40 From eqution (4), 888685.7 5.4 40 40 888685.7 f 80 + + 5.4 4.84 N/mm 80 80 888685.7 f 40 + 5.4 0.84 N/mm 40 7565.99 p 00 ( 7.57) 00 00 0 7565.99 p 80 ( 7.57).0 N/mm 80 7565.99 p 40 ( 7.57) 4.4 N/mm 40 (c) When fluid is pumped in, let Lme s eqution e, x nd f x At x 00, 60 N/mm 60 00 nd t x 40 mm, 0 0 40 From equtions (5) nd (6), we get 60 76050.4... (5)... (6) 40 00 00 40 00 40 Hence.605 40 p 00 60 N/mm p 80 76050.4.605 9.804 N/mm 80 76
Finl stresses re: p 40 0 76050.4 f 00 + 85.09 N/mm 00 f 80 5.04 N/mm f 40 5.0 N/mm f 00 4.77 + 85.09 50.49 N/mm f 80, inner.7 + 5.04.04 N/mm f 80, outer 4.84 + 5.04 77.854 N/mm f 40 0.84 + 5. 56.05 N/mm Prolem. A compound cylinder is to mde with inner rdius of 60 mm nd outer rdius of 0 mm. The rdius t common junction is to e 40 mm. If the two cylinders with llownce 0. mm re used, find the rdil pressure developed t contct surfces. Also determine the hoop stresses induced t inner edge, common edge nd outer edge of compound cylinder. Tke Young s modulus 0 5 N/mm. Solution: r 60 mm r 40 mm r 0 0 mm Let p e the rdil pressure developed t junction. Let Lme s equtions for internl cylinder e x f x At x 60 mm 0 0 60 At x 40 mm, p... () p... () 40 From eqns. () nd (), we get p 40 60 or p 60 40 60 40 4608.94 p. 4608.94 p Hence.8 p. 60 Hoop stress t junction is 4608.94 p f.8 p 40.6 p Lme s eqution for outer cylinder e, x nd f x At x 40 mm, p 77
p 40 At x 0 mm, 0... () 0... (4) 0 From equtions () nd (4), we get p 657.4 p. From eqn. (4), (0 40 ) 40 0 40 0 657.4 p 0 0.85 p. At junction, hoop stress in outer cylinder is 657.4 p + +.85 p 40 40.57 p Considering the circumferentil strin, the comptiility eqution is, δr r (f + f ) 0 where f is compressive nd f 0 is tensile. 0.5 40 5 0 [.6 p +.57 p] p 0.5 N/mm f + 60 4608.94 0.5.8 0.5 60 7.9 N/mm 4608.94 0.5 f, inner +.8 0.5 40 40 5.65 N/mm f, outer +.57 p 40.57 0.5 7.9 N/mm 657.4 0.5 f outer + +.85 0.5 0 0 5 N/mm. Prolem. A sphericl shell with internl dimeter 0 mm nd 640 mm externl dimeter is sujected to n internl fluid pressure of 75 N/mm. Find the hoop stresses developed t 40 mm intervl cross the thickness. Solution: r 0 0 mm r 60 mm p 75 N/mm 78
The rdil pressure nd the hoop stress t ny rdil distnce x re given y, x nd f x Now, t x 60 mm, 75 N/mm 75 At x 0 mm, 0 60... () 0... () 0 From equtions () nd (), we get 75 60 0 7554857 From eqution (), 7554857 0.74 0 f 7554857 + 0.74 x f 60 7554857 + 0.74 5.57 N/mm 60 f 00 7554857 + 0.74.656 N/mm 00 f 40 7554857 + 0.74.4 N/mm 40 f 80 7554857 + 0.74 8.7 N/mm 80 f 0 7554857 + 0.74 6.07 N/mm 0 Ans. 79