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BartSMP [9]
4 years ago
10

A vertical pole consisting of a circular tube of outer diameter 127 mm and inner diameter 115 mm is loaded by a linearly varying

distributed force with maximum intensity of 0 q . Find the maximum shear stress in the pole.
Engineering
1 answer:
Anna [14]4 years ago
3 0

Maximum shear stress in the pole is 0.

<u>Explanation:</u>

Given-

Outer diameter = 127 mm

Outer radius,r_{2} = 127/2 = 63.5 mm

Inner diameter = 115 mm

Inner radius, r_{1} = 115/2 = 57.5 mm

Force, q = 0

Maximum shear stress, τmax = ?

 τmax  = \frac{4q}{3\pi } (\frac{r2^2 + r2r1 + r1^2}{r2^4 - r1^4} )

If force, q is 0 then τmax is also equal to 0.

Therefore, maximum shear stress in the pole is 0.

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What software do you sue for design and the rapid prototyping machine?
mash [69]

Answer:

Computer Aided Design (CAD)

Explanation:

For design purpose and rapid prototyping, we use a software known as Computer Aided Design abbreviated as CAD.

Several production formates that are used in CAD are .stp, .stl, .igs, .step where formats like .stp, .igs and .step are used for rapid prototyping and format like .stl is used for design.

The information depending on the extensions used can be modeled in the 3-D files. The most frequent software used for prototyping are: AutoCAD, SolidWorks, SketchUP, etc

This also allows the spatial manipulation, dimension measurement and the visualization of the inner and outer parts.

3 0
4 years ago
Determine the depreciation expense for 2018 and 2019 using the following​ methods: (a)​ Straight-line (SL),​ (b) Units of produc
prohojiy [21]

Answer:

Check Explanation.

Explanation:

(1). The straight-line method: the general clue with this method is that in the two years, depreciation is the same. The formula for Calculating depreciation is given below;

straight-line method = (cost - Residual value)/ useful life in years.

From the question we know that the cost of acquisition is $30,000,000, the residual value of the asset is $4,000,000 and useful life is 7 years. Therefore;

straight-line method = ($30,000,000 - $4,000,000)/ 7.

= $3, 714,285.71 Per year.

That is $3, 714,285.71 for 2018 and 2019.

(2).Units of production​ (UOP) = (cost - Residual value)/ useful life in units.

= ($30,000,000 - $4,000,000)/ 4,375, 000.

Units of production​ (UOP) = $6 per mile.

Hence, the depreciation in 2018 = Depreciation per unit × 2018 year usage.

= 6 × 1,100,000 mile.

= $6,600,000.

depreciation in 2019 = Depreciation per unit × 2019 year usage.

= 6 × 1,200,000.

= $7,200,000.

Double-declining-balance​ (DDB)= (cost - accumulated depreciation) × 2 × 1/(useful life years).

Double-declining-balance​ (DDB) = (30,000,000 - 0)× 2 × (1/7).

= $8,571,428.57 depreciation in 2018.

= $8,571,428.6 depreciation in 2018

Double-declining-balance​ (DDB) = (30,000,000 - 8,571,428.57) × 2 × 1/7.

= $6,122,449.00 depreciation in 2019.

====================================================================

Total depreciation for straight-line method(2018 and 2019) = $7,428,571.42.

Total depreciation for Units of production​ (UOP)(2018 and 2019) = $13,800,000.

Total depreciation for Double-declining-balance (DDB)= $ 14,693,877.6.

5 0
3 years ago
Which of the following are true about algorithms? (You may select more than one)
r-ruslan [8.4K]
Have a orderf. Fr

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Djeiiekee
7 0
4 years ago
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Using the tables for water determine the specified property data at the indicated state. For H2O at T = 140 °C and v = 0.2 m3/kg
Dennis_Churaev [7]

Answer:

h = 1429.74\,\frac{kJ}{kg}

Explanation:

The determination of any further properties requires the knowledge of two independent properties. (Temperature and specific volume in this case). The specific volumes for saturated liquid and vapor at 140 °C are, respectively:

\nu_{f} = 0.001080\,\frac{m^{3}}{kg}

\nu_{g} = 0.50850\,\frac{m^{3}}{kg}

Since \nu_{f} < \nu < \nu_{g}, it is a liquid-vapor mixture. The quality of the mixture is:

x = \frac{\nu-\nu_{f}}{\nu_{g}-\nu_{f}}

x = \frac{0.2\,\frac{m^{3}}{kg} - 0.001080\,\frac{m^{3}}{kg} }{0.50850\,\frac{m^{3}}{kg} - 0.001080\,\frac{m^{3}}{kg} }

x = 0.392

The specific enthalpies for saturated liquid and vapor at 140 °C are, respectively:

h_{f} = 589.16\,\frac{kJ}{kg}

h_{g} = 2733.5\,\frac{kJ}{kg}

The specific enthalpy is:

h = h_{f}+x\cdot (h_{g}-h_{f})

h = 589.16\,\frac{kJ}{kg}+0.392\cdot \left( 2733.5\,\frac{kJ}{kg} - 589.16\,\frac{kJ}{kg} \right)

h = 1429.74\,\frac{kJ}{kg}

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