Chapter-10
Thin Cylinder
S K Mondal’s
Previous Conventional Questions with Answers Conventional Question GATE-1996 Question:
A thin cylinder of 100 mm internal diameter and 5 mm thickness is subjected to an internal pressure of 10 MPa and a torque of 2000 Nm. Calculate the
Answer:
magnitudes of the principal stresses. Given: d = 100 mm = 0.1 m; t = 5 mm = 0.005 m; D = d + 2t = 0.1 + 2 x 0.005 = 0.11 m p = 10 MPa, 10 x 106N/m2; T= 2000 Nm. Longitudinal stress, σ l = σ x =
pd 10 × 106 × 0.1 = = 50 × 106 N / m2 = 50MN / m2 4t 4 × 0.005
Circumferential stress, σ c = σ y =
pd 10 × 106 × 0.1 = = 100MN / m2 2 × 0.005 2t
To find the shear stress, using Torsional equation,
T τ = , we have J R 2000 × ( 0.05 + 0.005 ) TR T ×R τ = τ xy = = = = 24.14MN / m2 π π J D4 − d4 0.114 − 0.14 32 32
(
)
(
)
Principal stresses are:
σ 1,2 =
σx + σy 2
2
2 ⎛σx −σy ⎞ ± ⎜ ⎟ + (τ xy ) 2 ⎠ ⎝ 2
50 + 100 2 ⎛ 50 − 100 ⎞ = ± ⎜ + ( 24.14 ) ⎟ 2 2 ⎝ ⎠ = 75 ± 34.75 = 109.75 and 40.25MN / m2
σ 1 (Major principal stress ) = 109.75MN / m2 ; σ 2 ( min or principal stress ) = 40.25MN / m2 ; Conventional Question IES-2008 Question:
A thin cylindrical pressure vessel of inside radius ‘r’ and thickness of metal ‘t’ is subject to an internal fluid pressure p. What are the values of (i) Maximum normal stress? (ii) Maximum shear stress?
Answer:
Circumferential (Hoop) stress (σc ) = Longitudinal stress (σ ) = =
p.r t
p.r 2t
Therefore (ii) Maximum shear stress, ( τ max) =
σc − σ p.r = 2 4t
Conventional Question IES-1996 Question:
Answer:
A thin cylindrical vessel of internal diameter d and thickness t is closed at both ends is subjected to an internal pressure P. How much would be the hoop and longitudinal stress in the material? Page 319 of 429 For thin cylinder we know that