Test: Thin Cylinder Level - 3 - Mechanical Engineering MCQ

Maximum stress,

σ_c = = = 120 N/mm²

P= = 5.76 N/mm²

E = 200000 N/mm² , ν = 0.3

Axial strain,

ε_a =

=

= 1.2 × 10⁻⁴

Circumferential strain,

ε_c =

=

= 5.1 × 10⁻⁴

Volumetric strain, ε_v = 2ε_c + ε_a

= 2 × 5.1 × 10⁻⁴ + 1.2 × 10⁻⁴

= 11.4 × 10⁻⁴

Original volume,

V =

=

= 49.0874 × 106 mm³

Change in volume, δV = ε_v × V

= 11.4 × 10⁻⁴ × 49.0874 × 10⁶

= 55960 mm3 = 55.96 cm³

= 55.96 cc

Question_Type: 5

(ii) Thin cylinder subjected to internal pressure σ_x = σ₁ =

σ_y = σ₂ =

(iii) Torsion of circular shaft is a case of pure shear.

(iv) Combined bending and torsion of circular shaft In this case both normal as well as shear stress act on the element.

Referring to the figure which shows strain of an element at the surface of the cylinder, we can write.

ε_c =

and ε_r =

To determine the amount by which the vessel expands, consider a circumference at radius r and which moves out with a displacementδ_r. From the definition of normal strain,

ε_c =

here δ_r = u

⇒ ε_c =

This is the circumferential strain.

σ_c = Initial compressive stress in cylinder

=

= 7.854 N/mm² (comp. )

Say internal pressure = P

Axial stress, σ_a′ = = 6.25 P

Hoop stress = σ_c′

σ_CR = (σ_c′ − 7.854) = 6.25 P, (as given) ⋯ ①

or σ_c′ = 6.25 P + 7.854 ⋯ ②

Using equilibrium and compatibility conditions

Pd = 2tσc′ +

From equation ④

σ_c′ − ν_cσ′_c = σ′_w ×

or 6.25 P + 7.854 − 0.35 (6.25 P) = 0.5 σ_w′

or 4.0625 P + 7.854 = 0.5 σ_w′

σ_w′ = 8.125 P + 15.708

Putting the values σ_w′ and σ_c′ in equation ③

P × 200 = 2 × 8 × (6.25p + 7.854) + × (8.125p + 15.708)

200P = 100P + 125.664 + 6.283(8.125p + 15.708)

100P = 125.664 + 51.05P + 98.69

48.95 P = 224.354

P = 4.58 N/mm²

Longitudinal stress,

Circumferential strain,

=

= 3.44 × 10⁻⁴

σl =

Hoop stress, σ_n =

Longitudinal strain,

ε_l =

=

or ε_l =

where μ =

Hoop strain,

ε_h =

=

or ε_h =

P = 2.5 N/mm²

d = 250 mm t = 4 mm εc ,

Circumferential stress

=

ε_a , axial strain =

= 190 × 10⁻⁶ ⋯ ②

Dividing equation ① by equation ②

= 4.905

or 2 − ν = 4.905 − 9.81 ν 8.81 ν = 2.905

Poisson’s ratio, ν =

From equation ①

E =

= 69994 N/mm² = 69.99 GPa

Question_Type: 5

Stresses developed

σ_c = = 62.5 N/mm²

σ_a = = 31.25 N/mm²

Due to rigid end supports, axial strain, ε_a = 0 Say σ_a′ is axial compressive stress provided by rigid end supports.

Resultant axial stress, σ_aR = 31.25 − σ_a′

Axial strain, εa = = 0

Or 31.25 − σ_a′ − 0.35(62.5) = 0

σ_a′ = 31.25 − 21.875

= 9.375 N/mm²

Resultant axial stress, σ_aR = 31.25 − 9.375 = 21.875 N/mm²

Element ab is inclined at an angle of 30° to the axis 00 of the cylinder

Hoop stress,

σ_c = = 90 N/mm²

Axial stress, σ_a = =45 N/mm²

Stress on element Normal stress,

σ_θ =

=

=67.5 + 11.25 = 78. 75 N/mm²

Reduction in volumetric strain of cylinder

=

Reduction in volumetric strain of water

=

Axial force = 37000 N

Axial stress =

=

= 31.406 N/mm²

Increase in volumetric strain due to pull

=

Or

=

Or 1.5(5 − 4ν) +

= 31.406 (1 − 2ν)

7.5 − 6ν + 6.3636 = 31.406 − 62.812ν

Or 56.812 ν = 31.406 − 6.3636 − 7.5

= 17.5424

Poisson’s ratio,

ν = = 0.308