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
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(f)=220 MPa
modulus of elasticity(E)=200 GPa
v=0.3
thermal expanson coefficient of steel()=12*10^6 /oC
(1)
Maximum pressure according to tresca's theory:-
safety margin is 50% means the factor of safety is 1.5
max=y/2
where, 1=P , 2=0
FOS=1.5
y=f=220 MPa
P=220/1.5
P=146.667 MPa
maximum pressure (P)=146.667 MPa
(2)
Axial(longitudinal) stress(x) dure to pressure P:-
axial(longitudinal) stress(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(x)=(146.67*0.18)/(4*0.01)
axial(longitudinal) stress(x)=660.015 MPa
(3)
Hoop stress() dure to pressure P:-
hoop stress()=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()=(146.67*0.18)/(2*0.01)
hoop stress()=1318.203 MPa
(4)
let shear stress on incline pale is
shear stress()=[(1-2)/2]*sin2
where, 1=-stress due to force F+ axial stress(x) due pressure P
stress() due to force F is compressive and axial stress(x) due pressure P is tensile
1=-+x
1=-{F/(Pi/4)*d^2}+P*d/4t
y=0
=30o
shear stress()=[[-{F/(/4)*d^2}+P*d/4t]/2]*sin60o
shear stress()=[[-{4F/(d^2}+P*d/4t]/2]*sin60o
shear stress()=[[-{0.5513F/(d^2}+0.10825P*d/t]
put t=0.01 m and d=D-2t=0.18 m
shear stress()=-541618*F+1.948*P , MPa
[where, F=MN and P=MPa]
(5)
axial stress due to force F()=F/{(/4)*d^2} , [nature of stress is compressive]
axial(longitudinal) stress due to pressure P(x)=P*d/4t , [tensile stress]
net axial stress=-axial stress due to force F()+axial(longitudinal) stress due to pressure P(x)
calculatinf force F is required for net axial stress is zero
0=-F/{(/4)*d^2}+P(x)=P*d/4t
F=(*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
=(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)
=3.300*10^-3 m
=3.3 mm
change in length due to F alone()=3.3 mm
[Ans]
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