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erma4kov [3.2K]

2 years ago

13

The coupling is used to connect the two shafts together. Assuming that the shear stress in the bolts is uniform, determine the n

umber of bolts necessary to make the maximum shear stress in the shaft equal to the shear stress in the bolts. Each bolt has a diameter dd. Express your answer in terms of some or all of the variables rrr, RRR, ddd, and TTT. nn

1 answer:

AleksAgata [21]2 years ago

8 0

Answer:

n = 2r³/Rd²

Explanation:

See the attached file for the derivation.

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Tensile Strength (MPa) Number-Average Molecular Weight (g/mol)

IceJOKER [234]

Answer:

$\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}$

$\mathbf{M_n = 49163.56431 \ g/mol }$

Explanation:

The question can be well structured in a table format as illustrated below:

Tensile Strength (MPa) Number- Average Molecular Weight (g/mol)

82 12,700

156 28,500

The tensile strength and number-average molecular weight for two polyethylene materials given above.

Estimate the number-average molecular weight that is required to give a tensile strength required above. Using the data given find TS (infinity) in MPa.

<u>SOLUTION:</u>

We know that :

$T_S = T_{S \infty} - \dfrac{A}{M_n}$

where;

$T_S$ = Tensile Strength

$T_{S \infty}$ = Tensile Strength (Infinity)

$M_n$ = Number- Average Molecular Weight (g/mol)

SO;

$82= T_{S \infty} - \dfrac{A}{12700} ---- (1)$

$156= T_{S \infty} - \dfrac{A}{28500} ---- (2)$

From equation (1) ; collecting the like terms; we have :

$T_{S \infty} =82+ \dfrac{A}{12700}$

From equation (2) ; we have:

$T_{S \infty} =156+ \dfrac{A}{28500}$

So; $T_{S \infty} = T_{S \infty}$

Then;

$T_{S \infty} =82+ \dfrac{A}{12700} =156+ \dfrac{A}{28500}$

Solving by L.C.M

$\dfrac{82(12700) + A}{12700} =\dfrac{156(28500) + A}{28500}$

$\dfrac{1041400 + A}{12700} =\dfrac{4446000 + A}{28500}$

By cross multiplying ; we have:

$({4446000 + A})* {12700} ={28500} *({1041400 + A})$

$(5.64642*10^{10} + 12700A) =(2.96799*10^{10}+ 28500A)$

Collecting like terms ; we have

$(5.64642*10^{10} - 2.96799*10^{10} ) =( 28500A- 12700A)$

$2.67843*10^{10} = 15800 \ A$

Dividing both sides by 15800:

$\dfrac{ 2.67843*10^{10} }{15800} =\dfrac{15800 \ A}{15800}$

A = 1695208.861

From equation (1);

$82= T_{S \infty} - \dfrac{A}{12700} ---- (1)$

Replacing A = 1695208.861 in the above equation; we have:

$82= T_{S \infty} - \dfrac{1695208.861}{12700}$

$T_{S \infty}= 82 + \dfrac{1695208.861}{12700}$

$T_{S \infty}= \dfrac{82(12700) +1695208.861 }{12700}$

$T_{S \infty}= \dfrac{1041400 +1695208.861 }{12700}$

$T_{S \infty}= \dfrac{2736608.861 }{12700}$

$\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}$

From equation(2);

$156= T_{S \infty} - \dfrac{A}{28500} ---- (2)$

Replacing A = 1695208.861 in the above equation; we have:

$156= T_{S \infty} - \dfrac{1695208.861}{28500}$

$T_{S \infty}= 156 + \dfrac{1695208.861}{28500}$

$T_{S \infty}= \dfrac{156(28500) +1695208.861 }{28500}$

$T_{S \infty}= \dfrac{4446000 +1695208.861 }{28500}$

$T_{S \infty}= \dfrac{6141208.861}{28500}$

$\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}$

We are to also estimate the number- average molecular weight that is required to give a tensile strength required above.

If the Tensile Strength (MPa) is 82 MPa

Definitely the average molecular weight will be = 12,700 g/mol

If the Tensile Strength (MPa) is 156 MPa

Definitely the average molecular weight will be = 28,500 g/mol

But;

Let us assume that the Tensile Strength (MPa) = 181 MPa for example.

Using the same formula:

$T_S = T_{S \infty} - \dfrac{A}{M_n}$

Then:

$181 = 215.481- \dfrac{1695208.861 }{M_n}$

Collecting like terms ; we have:

$\dfrac{1695208.861 }{M_n} = 215.481- 181$

$\dfrac{1695208.861 }{M_n} =34.481$

1695208.861= 34.481 $M_n$

Dividing both sides by 34.481; we have:

$M_n$ = $\dfrac{1695208.861}{34.481}$

$\mathbf{M_n = 49163.56431 \ g/mol }$

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3 years ago

What is the ANSI B paper size also know as?

krek1111 [17]

Answer:

ANSI A sized paper is commonly referred to as Letter and ANSI B as Ledger or Tabloid.

Explanation:

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2 years ago

Polylines are made up of:____________

Drupady [299]

Answer:

2. are connected sequences of lines and arcs

Explanation:

A polyline is an object that is compound in nature as it includes one or more linear or curved segments. It is made up of connected sequence of lines and arcs that are created as a single object rather than individual objects.

Polylines are useful in the following;

i. Easy creation of rectangles and polygons as a single entity.

ii. In AUTOCAD, they allow for easy extrusion for 3D solids.

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3 years ago

It is required to design and implement: 1. A counter which counts from 0 to 255 with seven segment display.

liubo4ka [24]

Answer:

A counter which counts from 0 to 255 with seven segment display

Timer Mode Control (TMOD)

Explanation:

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3 years ago

7–53 Refrigerant-134a enters the condenser of a residential heat pump at 800 kPa and 35°C at a rate of 0.018 kg/s and leaves at

qwelly [4]

Answer:

a. The coefficient of power = 2.6364

b. The rate of heat absorption from the outside air is 1.96368kW

Explanation:

Given

First we need to get the enthalpy of R-34a.

When T = 35°C and P = 800kPa;

h1 = 271.24kj/kg

When x2 = 0 and P = 800kPa;

h1 = 95.48kj/kg

To calculate the COP, first we need to calculate the energy balance.

This is given as

Q = m(h1 - h2)

Where m = 0.018kg

Q = 0.018(271.24 - 95.48)

Q = 3.16368Kw

COP is then calculated as Q/W

Where W = Power consumption of the compressor = 1.2kW

COP = 3.16368Kw/1.2Kw

COP = 2.6364

Hence, the coefficient of power = 2.6364

b. The rate of heat absorption from the outside air is calculated as ∆Heat Rate

∆Heat Rate = Q - W

Where Q = Energy Balance = 3.16368Kw

W = Power consumption of the compressor = 1.2kW

∆Heat Rate = 3.16368Kw - 1.2kW

∆Heat Rate = 1.96368kW

Hence, The rate of heat absorption from the outside air is 1.96368kW

6 0

3 years ago

Read 2 more answers