Question

The cylindrical steel pressure vessel (Figure 3) has diameter D = 20 cm and thickness t=1 cm. Calculate: 1. The maximum pressure P for Tresca's criteria to yield a safety margin of 50% 2. The axial stress Ox that results from the pressure P 3. The hoop stress ge that results from pressure P 4. The shear stress on the inclined plane shown 5. The force F needed for the axial stress to become equal to zero 6. The change in vessel length due to F alone 1 m 0.2 m F P F 300 Figure 3: Pressure vessel

the material is structural steel with σf = 220 MPa, E = 200 GPa, ν = 0.3, and α = 12x10-6/oC. please help, provide clear and correct answers

Answer 1

Solution:- the values and figure(diagram) given in the question are as follows:

diameter of vessel(D)=20 cm

thickness of vessel(t)=1 cm

length of vessel(L)=1 m

failure stress of steel($\sigma$f)=220 MPa

modulus of elasticity(E)=200 GPa

v=0.3

thermal expanson coefficient of steel($\alpha$)=12*10^6 /^oC

F F P 30° L=1 m

(1)

Maximum pressure according to tresca's theory:-

safety margin is 50% means the factor of safety is 1.5

7
_max=$\sigma$y/2

ol-02 <oy/FOS

where, $\sigma$ 1=P , $\sigma$ 2=0

FOS=1.5

$\sigma$y=$\sigma$f=220 MPa

P=220/1.5

P=146.667 MPa

maximum pressure (P)=146.667 MPa

(2)

Axial(longitudinal) stress($\sigma$x) dure to pressure P:-

axial(longitudinal) stress($\sigma$x)=P*d/4t

where, d=internal diameter=D-2t

t=thickness of vessel

if , P=146.67 MPa

d=D-2t=20-2*1=18 cm, or 0.18 m

axial(longitudinal) stress($\sigma$x)=(146.67*0.18)/(4*0.01)

axial(longitudinal) stress($\sigma$x)=660.015 MPa

(3)

Hoop stress($\sigma$$\theta$) dure to pressure P:-

hoop stress($\sigma$$\theta$)=P*d/2t

where, d=internal diameter=D-2t

t=thickness of vessel

if , P=146.67 MPa

d=D-2t=20-2*1=18 cm, or 0.18 m

hoop stress($\sigma$$\theta$)=(146.67*0.18)/(2*0.01)

hoop stress($\sigma$$\theta$)=1318.203 MPa

(4)

let shear stress on incline pale is
7

shear stress(
7
)=[($\sigma$1-$\sigma$2)/2]*sin2$\theta$

where, $\sigma$ 1=-stress due to force F+ axial stress($\sigma$x) due pressure P

stress($\sigma$) due to force F is compressive and axial stress($\sigma$x) due pressure P is tensile

$\sigma$1=-$\sigma$+$\sigma$x

$\sigma$1=-{F/(Pi/4)*d^2}+P*d/4t

$\sigma$y=0

$\theta$=30^o

shear stress(
7
)=[[-{F/($\Pi$/4)*d^2}+P*d/4t]/2]*sin60^o

shear stress(
7
)=[[-{4F/($\Pi$d^2}+P*d/4t]/2]*sin60^o

shear stress(
7
)=[[-{0.5513F/($\Pi$d^2}+0.10825P*d/t]

put t=0.01 m and d=D-2t=0.18 m

shear stress(
7
)=-541618*F+1.948*P , MPa

[where, F=MN and P=MPa]

(5)

axial stress due to force F($\sigma$)=F/{($\Pi$/4)*d^2} , [nature of stress is compressive]

axial(longitudinal) stress due to pressure P($\sigma$x)=P*d/4t , [tensile stress]

net axial stress=-axial stress due to force F($\sigma$)+axial(longitudinal) stress due to pressure P($\sigma$x)

calculatinf force F is required for net axial stress is zero

0=-F/{($\Pi$/4)*d^2}+P($\sigma$x)=P*d/4t

F=($\Pi$*P*d^3)/(16*t)

if, P=146.667 MPa

t=0.01 m

d=D-2t=0.18 cm

F=(3.14159*146.667*0.18^3)/(16*0.01)

F=16.79499 MN

F=16794.99 kN

(6)

let change in length due to F alone is
AL

AL
=(F*L)/(AE)

wherte, A=(3.14159/4)*d^2

A=(3.14159/4)*0.18^2

A=0.0254468 m^2

E=200 GPa

L=1 m

F=16.79499*10^6 N

=(16.79499*10^6*1)/(0.0254468*200*10^9)

AL

AL
=3.300*10^-3 m

AL
=3.3 mm

change in length due to F alone(
AL
)=3.3 mm

[Ans]