Strength Of Materials Miscellaneous
- A long thin walled cylindrical shell, closed at both the ends, is subjected to an internal pressure. The ratio of the hoop stress (circumferential stress) to longitudinal stress developed in the shell is
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σh = PD where P = internal pressure 2t σ1 = PD , D = inner dia of cylinder . 4t
t = thickness of cylinder.∴ σh = PD = 2 2t σl PD 4t Correct Option: C
σh = PD where P = internal pressure 2t σ1 = PD , D = inner dia of cylinder . 4t
t = thickness of cylinder.∴ σh = PD = 2 2t σl PD 4t
- A thin gas cylinder with an internal radius of 100 mm is subject to an internal pressure of 10 MPa. The maximum permissible working stress is restricted to 100 MPa. The minimum cylinder wall thickness (in mm) for safe design must be ______ .
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σ1 = pd 2t ⇒ 100 = 100 × 2 × 100 2 × t
⇒ t = 10 mmCorrect Option: D
σ1 = pd 2t ⇒ 100 = 100 × 2 × 100 2 × t
⇒ t = 10 mm
- A gas is stored in a cylindrical tank of inner radius 7 m and wall thickness 50 mm. The gauge pressure of the gas is 2 MPa. The maximum shear stress (in MPa) in the wall is
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σc = pd = 2 × 14 = 280 MPa 2t 2 × 0.05 σl = pd = 2 × 14 = 140 MPa 4t 4 × 0.05
Maximum shear stress,τmax = σ1 - σ2 = σc - σl = 70 MPa 2 2
Note: This stress τmax is the maximum in-plane shear stress.Absolute τmax will be σc = 140 Mpa 2 Correct Option: C
σc = pd = 2 × 14 = 280 MPa 2t 2 × 0.05 σl = pd = 2 × 14 = 140 MPa 4t 4 × 0.05
Maximum shear stress,τmax = σ1 - σ2 = σc - σl = 70 MPa 2 2
Note: This stress τmax is the maximum in-plane shear stress.Absolute τmax will be σc = 140 Mpa 2
- A cylindrical tank with closed ends is filled with compressed air at a pressure of 500 kPa. The inner radius of the tank is 2 m, and it has wall thickness of 10 mm. The magnitude of maximum in-plane shear stress (in MPa) is ________.
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Maximum in-plane shear stress
τmax = pd 8t = 500 × 4 = 25 MPa 8 × 10 Correct Option: A
Maximum in-plane shear stress
τmax = pd 8t = 500 × 4 = 25 MPa 8 × 10
- Consider two concentric circular cylinders of different materials M and N in contact with each other at r = b, as shown below. The interface at r = b is frictionless. The composite cylinder system is subjected to internal pressure P. Let (urM , uθM) and (σrrM , σθθM) denote the radial and tangential displacement and stress components, r espectivel y, in material M. Similarly (urN , uθN) and (σrrN , σθθN) denote the radial and tangential displacement and stress components, respectively, in material N. The boundary condition that need to be satisfied at the frictionless interface between the two cylinders are:
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Due to internal pressure inside the cylinder, st r esses ar e gener at ed bot h in r adi al and circumferential direction. As interface of r = b is frictionless, so stress in radial direction must be equal otherwise relative motion will occur and parallely velocity in radial direction should be equal.
i.e. urM = urN and σrrM = σrrN
In circumferential direction i.e. only ‘’ direction velocity cannot be equal because it depends upon radius i.e. V = rω and ‘r’ is varying.
So , uθM ≠ uθN
Same is for stress in circumferential direction.
i.e. σθθM ≠ σθθNCorrect Option: C
Due to internal pressure inside the cylinder, st r esses ar e gener at ed bot h in r adi al and circumferential direction. As interface of r = b is frictionless, so stress in radial direction must be equal otherwise relative motion will occur and parallely velocity in radial direction should be equal.
i.e. urM = urN and σrrM = σrrN
In circumferential direction i.e. only ‘’ direction velocity cannot be equal because it depends upon radius i.e. V = rω and ‘r’ is varying.
So , uθM ≠ uθN
Same is for stress in circumferential direction.
i.e. σθθM ≠ σθθN
