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icang [17]
3 years ago
13

A pressure cylinder has an outer diameter 200 mm, maximum external pressure 4 MPa, and maximum allowable shear stress 27.5 MPa.

Determine the appropriate value of the minimum wall thickness The appropriate value of the minimum wall thickness is_____
Engineering
1 answer:
ludmilkaskok [199]3 years ago
4 0

Answer:

The minimum value of wall thickness t=3.63 mm.

Explanation:

Given:

  D=200 mm

 P=4 MPa

t= Wall thickness

maximum shear stress=27.5 MPa

We know that

       hoop stress \sigma _{h}=\frac{Pd}{2t}

      Longitudinal stress \sigma _{l}=\frac{Pd}{4t}

So maximum shear tress in plane\tau _{max}=\dfrac{\sigma _h-\sigma _l}{2}

              \tau _{max}=\dfrac{Pd}{8t}

Now by putting the value

       27.5=\dfrac{4\times 200}{8t}

 So   t=3.36 mm

The minimum value of wall thickness t=3.63 mm.

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Recommend the types of engineers needed to collaborate on a city project to build a skateboard park near protected wetlands.
4vir4ik [10]

Answer:

A civil engineer.

Explanation:

Civil engineering is the science that deals with the design, creation and maintenance of constructions for civil use on the earth's surface. Thus, this specialty seeks to adapt soils to the needs of life in society, creating buildings, bridges, and all other constructions adapted to civil life, while taking care of the correct use of soils and the correct distribution of spaces and resources. to be used for such constructions.

4 0
2 years ago
A rectangular region ABCD is to be built inside a semicircle of radius 10 m with points A and B on the line for the diameter and
Talja [164]

Answer:A rectangular region ABCD is to be built inside a semicircle of radius 10 m with points A and B on the line for the diameter and points C and D on the semicircle with CD parallel to AB. The objective is to find the height h * that maximizes the area of ABCD.

Formulate the optimization problem.

Explanation:A rectangular region ABCD is to be built inside a semicircle of radius 10 m with points A and B on the line for the diameter and points C and D on the semicircle with CD parallel to AB. The objective is to find the height h * that maximizes the area of ABCD.

Formulate the optimization problem.

7 0
3 years ago
Clarifying the issues of a problem is the _____ step in the problem solving process.
ratelena [41]
The answer is 2nd Step because the first step is to define the problem and third is to define your goals
7 0
3 years ago
At an axial load of 22 kN, a 15-mm-thick × 40-mm-wide polyimide polymer bar elongates 4.1 mm while the bar width contracts 0.15
Alenkasestr [34]

Answer:

The Poisson's Ratio of the bar is 0.247

Explanation:

The Poisson's ratio is got by using the formula

Lateral strain / longitudinal strain

Lateral strain = elongation / original width (since we are given the change in width as a result of compession)

Lateral strain = 0.15mm / 40 mm =0.00375

Please note that strain is a dimensionless quantity, hence it has no unit.

The Longitudinal strain is the ratio of the elongation to the original length in the longitudinal direction.

Longitudinal strain = 4.1 mm / 270 mm = 0.015185

Hence, the Poisson's ratio of the bar is 0.00375/0.015185 = 0.247

The Poisson's Ratio of the bar is 0.247

Please note also that this quantity also does not have a dimension

3 0
3 years ago
A hydraulic jump is induced in an 80 ft wide channel.The water depths on either side of the jump are 1 ft and 10 ft.Please calcu
krek1111 [17]

Answer:

a) 42.08 ft/sec

b) 3366.33 ft³/sec

c) 0.235

d) 18.225 ft

e) 3.80 ft

Explanation:

Given:

b = 80ft

y1 = 1 ft

y2 = 10ft

a) Let's take the formula:

\frac{y2}{y1} = \frac{1}{5} * \sqrt{1 + 8f^2 - 1}

10*2 = \sqrt{1 + 8f^2 - 1

1 + 8f² = (20+1)²

= 8f² = 440

f² = 55

f = 7.416

For velocity of the faster moving flow, we have :

\frac{V_1}{\sqrt{g*y_1}} = 7.416

V_1 = 7.416 *\sqrt{32.2*1}

V1 = 42.08 ft/sec

b) the flow rate will be calculated as

Q = VA

VA = V1 * b *y1

= 42.08 * 80 * 1

= 3366.66 ft³/sec

c) The Froude number of the sub-critical flow.

V2.A2 = 3366.66

Where A2 = 80ft * 10ft

Solving for V2, we have:

V_2 = \frac{3666.66}{80*10}

= 4.208 ft/sec

Froude number, F2 =

\frac{V_2}{g*y_2} = \frac{4.208}{32.2*10}

F2 = 0.235

d) El = \frac{(y_2 - y_1)^3}{4*y_1*y_2}

El = \frac{(10-1)^3}{4*1*10}

= \frac{9^3}{40}

= 18.225ft

e) for critical depth, we use :

y_c = [\frac{(\frac{3366.66}{80})^2}{32.2}]^1^/^3

= 3.80 ft

7 0
3 years ago
Read 2 more answers
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