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3 years ago

A composite shaft with length L = 46 in is made by fitting an aluminum sleeve (Ga = 5 x 10^3 ksi) over a

Engineering

1 answer:

Xelga [282]3 years ago

5 0

Answer:

Explanation:

Given the data in the question;

L = 46 in

Ga = 5 × 10³ ksi

Gs = 11 × 10³ ksi

Outside diameter da = 5 in

ds = 4 in

Tb = 3 kip.in

Now,

Ja = polar moment of Inertia of Aluminum;

Ja ⇒ π/32( 5⁴ - 4⁴ ) = π/32( 625 - 256 ) = π/32( 369 ) $in^u$

Js = polar moment of inertia of steel

Js ⇒ π/32 ds⁴ = π/32( 4⁴ ) = π/32( 256 )

Ta is torque transmitted by Aluminum

Ts is torque transmitted by steel

{composite member }

T = Ta + Ts ------ let this be equation m1

Now, we use the relation;

T/J = G∅/L

JG∅ = TL

∅ = TL/GJ

so, for aluminum rod ∅ $_{alu$ = TaLa/GaJa

for steel rod ∅ $_{steel$ = TsLs/GsJs

but we know that, ∅a = ∅s = ∅ $_B$

[TaLa/GaJa] = [TsLs/GsJs]

also, we know that, La = Ls = L

∴ [Ta/GaJa] = [Ts/GsJs]

we solve for Ta

TaGsJs = TsGaJa

Ta = TsGaJa / GsJs

we substitute

Ta = [Ts(5 × 10³)( π/32( 369) )] / [ (11 × 10³)( π/32( 256 ) ) ]

Ta = 0.66Ts

now, we substitute 0.66Ts for Ta and 3 for T in equation 1

T = Ta + Ts

3 = 0.66Ts + Ts

3 = 1.66Ts

Ts = 3 / 1.66

Ts = 1.8072 ≈ 1.81 kip-in

∅ $_{steel$ = TsLs / GsJs

we substitute

∅ $_{steel$ = (1.81 × 46 ) / ( 11 × 10³ × π/32( 256 ) )

∅ $_{steel$ = 83.26 / 276460.1535

∅ $_{steel$ = 0.000301

∅ $_{steel$ = 3.01 × 10⁻⁴ rad

∅ $_{steel$ = ∅ $_B$ = 3.01 × 10⁻⁴ rad

Therefore, the magnitude of the angle of twist at end B is 3.01 × 10⁻⁴ rad

Read more about the aircraft:

brainly.com/question/5055463

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7 0

2 years ago

Which option should the engineers focus on as they develop the train in the following scenario?

pav-90 [236]

Answer:

Engineers can design a train with a regenerative braking system

Explanation:

Assuming the point of the question is that the engineers want to focus on using energy efficiently when starting and stopping, they would likely want to consider a regenerative braking system. Such a system can store energy during braking so that it can be used during starting, reducing the amount of energy that must be supplied by an outside power source.

5 0

3 years ago

How many hours should it take an articulated wheel loader equipped with a 4-yd^3 bucket to load 3000 yd^3 of gravel (average mat

densk [106]

Answer:

17 hours 15 minutes

Explanation:

See attached picture.

4 0

3 years ago

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:

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

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

5 0

3 years ago