Mechanical Engineering Pressure on Vessels Assignment Solution

This document provides a comprehensive solution to a Mechanical Engineering assignment focused on the analysis of pressure vessels. The assignment covers various aspects of pressure vessel design and analysis, including the calculation of percentage reduction in thickness, determination of strains in x and y directions, volumetric strain, and bulk modulus calculations. Furthermore, it delves into the determination of longitudinal and hoop stresses, as well as radial and hoop stresses at inner and outer surfaces of thick-walled cylinders. The solution incorporates relevant formulas, step-by-step calculations, and references to engineering principles, offering a detailed understanding of pressure vessel behavior under different loading conditions. The assignment aims to provide a practical understanding of stress and strain analysis in mechanical engineering applications.

Pressure on Vessels 1
PRESSURE ON VESSELS
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Pressure on Vessels 2
Pressure on Vessels
Task 1
Calculating percentage reduction in thickness
Length, L = 150 mm = 0.15 m
Square cross-section of side 6.5 mm = 0.0065 m
Axial force, F = 20 kN
Elastic modulus, E = 210 x 109 N/m2
Poisson’s ration, ν = 0.35
The force acting on the bar is axial hence there is only axial or longitudinal stress but
circumferential or hoop stress is zero (Engineers Edge, (n.d.)).
Hoop stress, σH = 0
Longitudinal stress , σx= Axial Force , F
Area , A
Area, A = 0.0065 m x 0.0065 m = 4.225 x 10-5 m2
σx= 20,000 N
4.225 x 10−5 m2 =473,372,781.065 N /m2
Longitudinal strain, εy= 1
E (v∗σx)
εy = 1
210 x 109 (0.35 x 47 3,372,781.065)

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εy = 1
210 x 109 ( 165 , 680,473.37275 )=7 . 89 x 10−4
Change in length, Δy = cross sectional area x εy
Cross sectional area, A = 0.0065 m x 0.0065 m = 4.225 x 10-5 m2
Change in length, Δy = 4.225 x 10-5 m2 x 7.89 x 10-4 = 3.334 x 10-8 m2
Percentage reduction in size of the square = Δy
Cross sectional area x 100 %
= 3.334 x 10−8 m2
4.225 x 10−5 m2 x 100 %=0.0789%
Task 2
Determining the strains in the x and the y directions
Modulus of elasticity, E = 180 x 109 N/m2
Poisson ration, ν = 0.3
The plate is flat thus σy = 0 MN/m2
σx = 30 MN/m2
σy = 20 MN/m2
Strain in x direction
εx = 1
E ( σx−vσy−νσz )
εx = 1
180 x 109 N / m2 ((30 x 106 N / m2 )−0.3 ( 20 x 106 N /m2 )−0 )

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¿ 1
180 x 109 N /m2 x 24 x 106 N /m2=1.333 x 10−4
Strain in y direction
εy = 1
E ( σy −vσx−νσz )
εy = 1
180 x 109 N / m2 ((20 x 106 N /m2 )−0.3 ( 30 x 106 N /m2 )−0 )
¿ 1
180 x 109 N /m2 x 11 x 106 N /m2=6.111 x 10−5
Task 3
Determining volumetric strain, change in volume and bulk modulus
Side of cube = 50 mm
Principle strains = 0.0006
Modulus of elasticity, E = 200 GN/m2
Poisson’s ratio, ν = 0.3
Principle strains, σx = σy = σz = 0.0006
a) Volumetric strain, εv
εv = ( ε x +ε y + ε z )
εx = εy = εz = 0.0006
εv = ( 0.0006+ 0.0006+0.0006 )

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εv =0.0018
b) Change in volume
Volumetric Strain , εv=Change∈Volume
Original Volume
Therefore, change in volume = volumetric strain x original volume
From the previous calculation, εv = 0.0018
Original volume = 50mm x 50mm x 50mm = 125,000 mm3
Change in volume = εv x original volume
Change in volume = 0.0018 x 125,0000mm3 = 225 mm3.
c) Bulk modulus
K= E
3 ( 1−2 ν )
K= 20 0 x 109 N / m2
3 ( 1−2(0.3) ) =200 x 109 N /m2
1.2 =166.67 x 109 N /m2=166.67 GN /m2
The bulk modulus represents the resistance capability of the steel tube to compression.
Task 4
Determining longitudinal and hoop stresses
Internal diameter, d = 50 mm = 0.05 m
Thickness, t = 3 mm = 0.003 m
Internal pressure, Pi = 100 bar = 100 x 105 Pa = 10 x 106 Pa or N/m2

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Efficiency of the joints, 75% = 0.75 (both for longitudinal joint (ηL) and circumferential joint
(ηH)
Longitudinal stress, σL= Pid
4 tηL ; Where Pi = internal pressure, d = internal diameter, t = thickness
of the cylinder and ηL = efficiency of the longitudinal joint
σL= 10 x 106 N /m2 x 0.05 m
4 x 0.003m x 0.75 =55.56 x 106 N /m2 =55.56 MN /m2
Hoop stress, σH = Pid
2 tηH ; Where Pi = internal pressure, d = internal diameter, t = thickness of the
cylinder and ηH = efficiency of the hoop joint.
σH =10 x 106 N /m2 x 0.05 m
2 x 0.003 m x 0.75 =111.11 x 106 N /m2=111.11 MN /m2
Task 5
Determining the radial and hoop stresses at the inner and outer surfaces
Internal diameter, Di = 50 mm
Internal radius, Ri = 25 mm
External diameter, Do = 90 mm
External radius, Ro = 45 mm
Thickness, t = 90−50
2 = 40
2 =20 mm=0.02m
Internal pressure, Pi = 150 bar = 150 x 100,000 Pa = 15 x106 Pa or N/m2 = 15 MN/m2 (or Pa)
Determining whether the pipe is thin or thick:

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Di
t =50
20 =2.5 < 20; since Di
t is less than 20, the pipe is thick.
Radial stress, σR=a− b
r ²
Hoop stress, σH =a+ b
r ²
Minimum radial stress, σmin=0 MN /m2=a− b
ro2 ; 0=a− b
452
a= b
452 → b=45² a
Maximum radial stress, σmax =−15 MN /m2=a− b
ri2 ; substituting b with 452a and ri = 25mm
gives
−15 MN /m2=a− 452 a
252
-15 MN/m2 = a – 3.24a’ -15MN/m2 = -2.24a; a = 6.6964
a= b
452 → b=45² a = 452 x 6.6964 = 13,560.21
Radial stress at inner surface, r = Ri
−σR=a− b
r ²
−σR=6.6964−13,560.21
252 =6.6964−21.6964=15 N /m2
(This is the maximum radial stress)
Radial stress at outer surface, r = Ro

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σR=a− b
r ²
σR=6.6964−13,560.21
452 =6.6964−6.6964=0 N /m2
(This is the minimum radial stress)
Hoop stress at inner surface, r = Ri
σH =a+ b
r ²
σH =6.6964+ 13,560.21
25² =6.6964+21.6964=28.4 N /m2 (This is the maximum hoop stress)
Hoop stress at outer surface, r = Ro
σH =a+ b
r ²
σH =6.6964+ 13,560.21
45² =6.6964+6.6964=13.4 N /m2 (This is the minimum hoop stress)
References
Engineers Edge, (n.d.). Pressure Vessel, Thin Wall Hoop and Longitudinal Stresses. [Online]
Available at: https://www.engineersedge.com/material_science/hoop-stress.htm
[Accessed 27 November 2018].

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