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2 years ago

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A horizontal, cylindrical, tank, with hemispherical ends, is used to store liquid chlorine at 10 bar. The vessel is 4 m internal

diameter and 20 m long. Estimate the minimum wall thickness required to resist this pressure, for the cylindrical section and the heads. Take the design pressure as 12 bar and the allowable design stress for the material as 110 MN/m2.

1 answer:

sp2606 [1]2 years ago

3 0

Answer: 0.021818m =2.18x10^-²m

Explanation: Using Laplace principle

Design pressure is =12 bar= 1200000N/m²

T = Wall thickness

Design stress=110Mn/m²= 110,000,000N/m²

Radius= diameter/2 =4/2=2m

Design stress=Hoop stress =Pr/t where p=internal pressure or internal pressure, r=radius and t= wall thickness.

As Laplace equation stated.

110,000,000= 1,2000,00 x 2/t

t= 2,400,000/110,000,000.

t= 0.021818m

t=2.18x10^-2m.

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prohojiy [21]

Answer:

i) S–N plot is attached

ii) fatigue strength = 100 MPa

iii) fatigue life = 5.62 x 10^(5) cycles

Explanation:

i) I have attached the S–N plot (stress amplitude versus logarithm of cycles to failure)

ii) The question says we should find the fatigue strength at 4 × 10^(6) cycles.

So let's find the log of this and trace it on the graph attached.

Log(4 × 10^(6)) = 6.6

From the graph attached, at log of cycle value of 6.6, the fatigue strength is approximately 100 MPa

iii) The question says we should find the fatigue life for 120 MPa.

Thus, from the graph, at stress amplitude of 120 MPa, the log of cycles is approximately 5.75.

Thus,the fatigue life will be the inverse log of 5.75.

Thus, fatigue life = 10^(5.75)

Fatigue life = 5.62 x 10^(5)

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2 years ago

A 10-m long steel linkage is to be designed so that it can transmit 2 kN of force without stretching more than 5 mm nor having a

Fed [463]

Answer:

<em>minimum required diameter of the steel linkage is 3.57 mm</em>

<em></em>

Explanation:

original length of linkage l = 10 m

force to be transmitted f = 2 kN = 2000 N

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maximum stress σ = 200 N/mm^2 = $2*10^{8} N/m^{2}$

maximum stress allowed on material σ = force/area

imputing values,

200 = 2000/area

area = 2000/( $2*10^{8}$ ) = $10^{-5}$ m^2

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$10^{-5}$ = $\frac{3.142*d^{2} }{4}$ = $0.7855d^{2}$

$d^{2} = \frac{10^{-5} }{0.7855}$ = $1.273*10^{-5}$

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<em>maximum diameter of the steel linkage d = 3.57 mm</em>

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2 years ago

A fatigue test was conducted in which the mean stress was 46.2 MPa and the stress amplitude was 219 MPa.

sleet_krkn [62]

Answer:

a)σ₁ = 265.2 MPa

b)σ₂ = -172.8 MPa

c) $Stress\ ratio =-0.65$

d)Range = 438 MPa

Explanation:

Given that

Mean stress ,σm= 46.2 MPa

Stress amplitude ,σa= 219 MPa

Lets take

Maximum stress level = σ₁

Minimum stress level =σ₂

The mean stress given as

$\sigma_m=\dfrac{\sigma_1+\sigma_2}{2}$

$2\sigma_m={\sigma_1+\sigma_2}$

2 x 46.2 = σ₁ + σ₂

σ₁ + σ₂ = 92.4 MPa --------1

The amplitude stress given as

$\sigma_a=\dfrac{\sigma_1-\sigma_2}{2}$

$2\sigma_a={\sigma_1-\sigma_2}$

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By adding the above equation

2 σ₁ = 530.4

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$Stress\ ratio =\dfrac{\sigma_{min}}{\sigma_{max}}$

$Stress\ ratio =\dfrac{-172.8}{265.2}$

$Stress\ ratio =-0.65$

Range = 265.2 MPa - ( -172.8 MPa)

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kozerog [31]

Answer:

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$g = 5.48 ft/s^2.$

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a)

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We know that

$1\ lbf=32.17 \ lmb.ft/s^2$

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b)

On the beam scale

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So

F = 90 x 9.99 N

F= 899.59 N

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2 years ago

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Llana [10]

Answer:

Check the explanation

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