Define stress and modulus of elasticity with SI units. Calculate changes in length and diameter of a bar under load. Determine hoop stress in a cylindrical pressure vessel.

Question

Asked on July 16, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-07-16T05:34:23.749Z

a) i) stress; Stress is defined as the internal force per unit cross-sectional area within a material, resulting from externally applied forces. It is a measure of the intensity of internal forces acting within a deformable body. Its SI unit is the Pascal (Pa) or Newtons per square meter (N/m^2). ii) modulus of elasticity. The modulus of elasticity (Young's Modulus) is a measure of the stiffness of an elastic material. It is defined as the ratio of stress to strain in the elastic region of a material's stress-strain curve. Its SI unit is the Pascal (Pa) or Newtons per square meter (N/m^2). b) Given: Diameter D = 30 mm = 0.030 m Length L = 500 mm = 0.500 m Compressive load P = 10 kN = 10 × 10^3 N Modulus of elasticity E = 180 GN/m^2 = 180 × 10^9 N/m^2 Poisson's ratio = 0.35 i) length of the bar; Step 1: Calculate the cross-sectional area (A) of the bar. A = ( D^2)/(4) A = (0.030 m)^24 A = × 0.0009 m^24 A = 0.000706858 m^2 Step 2: Calculate the axial stress () in the bar. = (P)/(A) = 10 × 10^3 N0.000706858 m^2 ≈ 14147123.5 N/m^2 ≈ 14.147 MPa Step 3: Calculate the axial strain (_L) using Hooke's Law. _L = ()/(E) _L = 14.147 × 10^6 N/m^2180 × 10^9 N/m^2 _L ≈ 0.000078594 Step 4: Calculate the change in length ( L). L = _L × L L = 0.000078594 × 0.500 m L ≈ 0.000039297 m Since the load is compressive, the length will decrease. L ≈ 0.0393 mm The change in length of the bar is -0.0393 mm (negative indicates decrease). ii) diameter of the bar. Step 5: Calculate the lateral strain (_D) using Poisson's ratio. Poisson's ratio = -lateral strainaxial strain = -(_D)/(_L). _D = - × _L _D = -0.35 × 0.000078594 _D ≈ -0.000027508 Since the axial strain is compressive (negative change in length), the lateral strain will be positive (increase in diameter). Step 6: Calculate the change in diameter ( D). D = _D × D D = 0.000027508 × 0.030 m D ≈ 0.00000082524 m D ≈ 0.000825 mm The change in diameter of the bar is +0.000825 mm (positive indicates increase). c) Given: Cylinder diameter D = 1.5 m Shell thickness t = 3.5 mm = 0.0035 m Internal fluid pressure p = 600 kN/m^2 = 600 × 10^3 N/m^2 Step 1: Consider a longitudinal section of the cylindrical pressure vessel of length L. The internal pressure p acts radially outwards. Step 2: Calculate the total force (F_p) exerted by the internal pressure on the projected area. The projected area over which the pressure acts is D × L. F_p = p × (D × L) F_p = (600 × 10^3 N/m^2) × (1.5 m × L) F_p = 900 × 10^3 × L N Step 3: Calculate the resisting force (F_r) due to the hoop stress (_h) in the shell. The hoop stress acts tangentially on the two cut surfaces of the shell, each with an area of t × L. F_r = _h × (2 × t × L) F_r = _h × (2 × 0.0035 m × L) F_r = _h × (0.007 × L) N Step 4: Equate the internal pressure force to the resisting force for equilibrium. F_p = F_r 900 × 10^3 × L = _h × (0.007 × L) Step 5: Solve for the hoop stress (_h). _h = (900 × 10^3)/(0.007) N/m^2 _h = (900000)/(0.007) N/m^2 _h ≈ 128571428.57 N/m^2 _h ≈ 128.57 MPa The hoop stress in the vessel is 128.57 MPa. Got more? Send 'em 📸

Accepted Answer · 2026-07-16T05:34:23.749Z

a) i) stress; Stress is defined as the internal force per unit cross-sectional area within a material, resulting from externally applied forces. It is a measure of the intensity of internal forces acting within a deformable body. Its SI unit is the Pascal (Pa) or Newtons per square meter (N/m^2). ii) modulus of elasticity. The modulus of elasticity (Young's Modulus) is a measure of the stiffness of an elastic material. It is defined as the ratio of stress to strain in the elastic region of a material's stress-strain curve. Its SI unit is the Pascal (Pa) or Newtons per square meter (N/m^2). b) Given: Diameter D = 30 mm = 0.030 m Length L = 500 mm = 0.500 m Compressive load P = 10 kN = 10 × 10^3 N Modulus of elasticity E = 180 GN/m^2 = 180 × 10^9 N/m^2 Poisson's ratio = 0.35 i) length of the bar; Step 1: Calculate the cross-sectional area (A) of the bar. A = ( D^2)/(4) A = (0.030 m)^24 A = × 0.0009 m^24 A = 0.000706858 m^2 Step 2: Calculate the axial stress () in the bar. = (P)/(A) = 10 × 10^3 N0.000706858 m^2 ≈ 14147123.5 N/m^2 ≈ 14.147 MPa Step 3: Calculate the axial strain (_L) using Hooke's Law. _L = ()/(E) _L = 14.147 × 10^6 N/m^2180 × 10^9 N/m^2 _L ≈ 0.000078594 Step 4: Calculate the change in length ( L). L = _L × L L = 0.000078594 × 0.500 m L ≈ 0.000039297 m Since the load is compressive, the length will decrease. L ≈ 0.0393 mm The change in length of the bar is -0.0393 mm (negative indicates decrease). ii) diameter of the bar. Step 5: Calculate the lateral strain (_D) using Poisson's ratio. Poisson's ratio = -lateral strainaxial strain = -(_D)/(_L). _D = - × _L _D = -0.35 × 0.000078594 _D ≈ -0.000027508 Since the axial strain is compressive (negative change in length), the lateral strain will be positive (increase in diameter). Step 6: Calculate the change in diameter ( D). D = _D × D D = 0.000027508 × 0.030 m D ≈ 0.00000082524 m D ≈ 0.000825 mm The change in diameter of the bar is +0.000825 mm (positive indicates increase). c) Given: Cylinder diameter D = 1.5 m Shell thickness t = 3.5 mm = 0.0035 m Internal fluid pressure p = 600 kN/m^2 = 600 × 10^3 N/m^2 Step 1: Consider a longitudinal section of the cylindrical pressure vessel of length L. The internal pressure p acts radially outwards. Step 2: Calculate the total force (F_p) exerted by the internal pressure on the projected area. The projected area over which the pressure acts is D × L. F_p = p × (D × L) F_p = (600 × 10^3 N/m^2) × (1.5 m × L) F_p = 900 × 10^3 × L N Step 3: Calculate the resisting force (F_r) due to the hoop stress (_h) in the shell. The hoop stress acts tangentially on the two cut surfaces of the shell, each with an area of t × L. F_r = _h × (2 × t × L) F_r = _h × (2 × 0.0035 m × L) F_r = _h × (0.007 × L) N Step 4: Equate the internal pressure force to the resisting force for equilibrium. F_p = F_r 900 × 10^3 × L = _h × (0.007 × L) Step 5: Solve for the hoop stress (_h). _h = (900 × 10^3)/(0.007) N/m^2 _h = (900000)/(0.007) N/m^2 _h ≈ 128571428.57 N/m^2 _h ≈ 128.57 MPa The hoop stress in the vessel is 128.57 MPa. Got more? Send 'em 📸

Define stress and modulus of elasticity with SI units. Calculate changes in length and diameter of a bar under load. Determine hoop stress in a cylindrical pressure vessel.

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Define stress and modulus of elasticity with SI units. Calculate changes in length and diameter of a bar under load. Determine hoop stress in a cylindrical pressure vessel.

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