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3 years ago

Should aircraft wings have infinite stiffness?

2 answers:

5 0

Answer: 2.1.3 The evolution of aircraft wing structures-form follows function Torsional stiffness is fundamental to all aeroelastic approaches unity, 0 will..

HOPE IT HELPS GIVE ME BRAINLIST PLEASE :)

Colt1911 [192]3 years ago

3 0

Answer:

No, they need to be somewhat flexible so that forces such as turbulance don't shear the wing off.

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ASAE 1060 Steel wire (1 mm diameter) is coated with copper to form a composite with a diameter of 2mm. Use the following propert

k0ka [10]

Answer:

a) $E_{m}$ = 133.75 Gpa

b) Fnet = 560 N

c) thermal expansion of the composite material = 14.31 $10^{-6 }$ / °C

Explanation:

Solution:

a) Elastic Modulus of the composite:

Area of steel wire = $\frac{\pi }{4}$ x ( $0.001^{2}$ ) = 0.8 x $10^{-6}$ $m^{2}$

Area of Copper wire = $\frac{\pi }{4}$ x ( $0.002^{2}$ ) - 0.8 x $10^{-6}$ $m^{2}$

Area of Copper wire = 2.4 x $10^{-6}$ $m^{2}$

Young's Modulus of Composite mixture:

$E_{m}$ = $F_{st}$ $E_{st}$ + $F_{Cu}$ $E_{Cu}$ Equation 1

here,

$F_{st}$ = Stress in Steel

$F_{Cu}$ = Stress in Copper.

We know that,

F = P/A

F is inversely proportional to Area, so if area is large, stress will less and vice versa. So, Take

Ratio for area of steel = $\frac{0.8. 10^{-6} }{(0.8 + 2.4) .10^{-6} }$

Ratio for area of steel = $\frac{0.8}{3.2 }$ = 0.25

Similarly, for Copper,

Ratio for area of copper = $\frac{2.4. 10^{-6} }{(0.8 + 2.4) .10^{-6} }$

Ratio for area of copper = $\frac{2.4 }{3.2}$ = 0.75

Put these values in equation 1:

$E_{m}$ = $F_{st}$ $E_{st}$ + $F_{Cu}$ $E_{Cu}$

$E_{m}$ = (0.25) $E_{st}$ + (0.75) $E_{Cu}$

We are given that,

$E_{st}$ = 205 Gpa

$E_{Cu}$ = 110 Gpa

So,

$E_{m}$ = (0.25) (205 Gpa) + (0.75) (110 GPa)

$E_{m}$ = 51.25GPa + 82.5 Gpa

Hence, the Elastic Modulus of the composite will be:

$E_{m}$ = 133.75 Gpa

b) maximum force:

Fnet = Fst + Fcu

We know that F = (Yield Stress x Area)

F = fst x Ast + fcu x Acu

And we are given that,

Yield stress of Steel = 280 Mpa

Yield stress of Copper = 140 Mpa

And,

Ast = 0.8 x $10^{-6}$ $m^{2}$

Acu = 2.4 x $10^{-6}$ $m^{2}$

Just plugging in the values, we get:

F = (280 Mpa) (0.8 x $10^{-6}$ $m^{2}$ ) + (140 Mpa) (2.4 x $10^{-6}$ $m^{2}$ )

F = 224 + 336

Fnet = 560 N ( because Mpa = $10^{6}$ N/ $m^{2}$ )

So, it means the composite will carry the maximum force of 560N

c) Coefficient of Thermal Expansion:

Strain on both material is same upon loading so,

(ΔL/L)st = (ΔL/L)cu

by thermal expansion equation:

( $\alpha .$ ΔT + $\frac{F}{A}$ $. \frac{1}{Est}$ ) = $\alpha .$ ΔT + $\frac{F}{A}$ $. \frac{1}{Ecu}$ )

Where $\alpha$ = Coefficient of Thermal expansion

Here, fst = -fcu = F

and ΔT = 1°

So,

Plugging in the values, we get.

( 10 x $10^{-6}$ x (1) + $\frac{F}{0.8.10^{-6} } . \frac{1}{205 . 10^{9} }$ ) = ( 17 x $10^{-6}$ x (1) + $\frac{-F}{2.4.10^{-6} } . \frac{1}{110 . 10^{9} }$ )

Solving for F, we get:

F = 0.71 N

Here,

fst = F = 0.71 N (Tension on Heating)

fcu = -F = 0.71 N ( Compression on Heating )

So, the combined thermal expansion of the composite material will be:

(ΔL/L)cu = ( 17 x $10^{-6}$ x (1°) + $\frac{-0.71}{2.4.10^{-6} } . \frac{1}{110 . 10^{9} }$ )

(ΔL/L)cu = ( 17 x $10^{-6}$ x (1°) - 2.69 x $10^{-6}$

combined thermal expansion of the composite material = 14.31 $10^{-6 }$ / °C

4 0

3 years ago

A vertical piston–cylinder device contains water and is being heated on top of a range. During the process, 73 Btu of heat is tr

pshichka [43]

Answer:

60 btu

Explanation:

given:

 $E_{h} =73 btu\\E_{loss} =8 btu\\\\W_{p} =5btu\\$ 

solution:

the total change in the system energy can be calculated as follows

ΔE= $E_{h} -W_{p} -E_{loss}$

=73-5-8

=60 btu

note :

work is done by the system so it has a negative sign

8 0

3 years ago

Q3: Summation Write a recursive implementation of summation, which takes a positive integer n and a function term. It applies te

harina [27]

Answer:

Here is the recursive function summation:

def summation(n, term):

if n == 1:

return term(n)

else:

return term(n) + summation(n - 1, term)

Explanation:

The function summation() has two arguments where n is a positive integer and term is a function term. term has the lambda function which is a small function having an argument and an expression e.g lambda b: b+20

So the summation() function is a recursive function which returns sum of the first n terms in the sequence defined by term ( a lambda function).

If you want to check if this function works, you can call this function by passing values to it like given in the question.

summation(5, lambda x: 2**x)

Here the value of n is 5 and the term is a lambda function x: 2**x

If you want to see the results of this function on output screen then use:

print(summation(5, lambda x: 2**x))

The print() function will print the results on screen.

This returns the sum of first 5 terms in sequence defined in the function x: 2**x

In recursive methods there are two cases: base case and recursive case. Base case is the stopping case which means that the recursion will stop when the base case/ base condition evaluates to true. The recursive case is when the function keeps calling itself so the recursive function keepsexecuting until the base case becomes true.

Here the base case is if n == 1: So the recursive function calling itself until the value of n becomes 1.

Recursive case is:

return term(n) + summation(n - 1, term)

For the above example with n= 5 and term = x:2**x the recursions starts from n and adds all the terms of the series one by one and the value of n keeps decrementing by 1 at every recursive call.

When the value of n is equal to 1 the base case gets true and the recursion ends and the result of the sum is displayed in output.

This is how the summation() function works for the above function call:

2^1 + 2^2 + 2^3 + 2^4 + 2^5

n is 5 So this term function is called recursively 5 times and at every recursive call its value decreases by 1. Here the term function is used to compute 2 raise to power n. So in first recursive call the 2 raise to the power 5 is computed, then 5 is decremented and then in second recursive call to summation(), 2 raise to the power 4 is calculated, in third recursive call to summation(), 2 raise to the power 3 is calculated, in fourth recursive call to summation(), 2 raise to the power 2 is calculated, in fifth recursive call to summation(), 2 raise to the power 1 is calculated, then the base condition is reached as n==1. So the recursion stops and the sum of the above computed power function results is returned which is 62.

2^1 + 2^2 + 2^3 + 2^4 + 2^5 = 62

The screen shot of recursive function along with the output of explained examples is attached.

6 0

3 years ago

You are designing a 200mm long x 100mm wide x 50mm deep rectangular housing, with a wall width of 1.5 mm, and 1 degree draft. Yo

Nimfa-mama [501]

365gpa I don’t even know what it is but this is with 10 point and I need it

6 0

2 years ago

For a 3-Phase, Wye connected system the Line to Line Voltage measures 12,470 Volts, the Phase current measures 120 Amps.

vladimir2022 [97]

Answer:

A. 7199.55 volts

B. 120A

Explanation:

In this question we have the

line voltage = VLL = 12470volts

Phase current = Iph = 120 amps

A.)

We are to calculate the line-to-neutral/phase voltage here

VLL = √3VL-N

VL-N = VLL/√3

VL-N = 12470/√3

This gives a line to neutral phase/voltage of 7199.55 volts.

We are to calculate the line current here:

In this connection, the line current and the phase current are equal

ILL = Iph = 120A

6 0

3 years ago