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devlian [24]
3 years ago
11

.If aligned and continuous carbon fibers with a diameter of 6.90 micron are embedded within an epoxy, such that the bond strengt

h across the fiber-epoxy interface is 17 MPa, and the shear yield strength of the epoxy is 68 MPa, compute the minimum fiber length, in millimeters, to guarantee that the fibers are conveying an optimum fraction of force that is applied to the composite. The tensile strength of these carbon fibers is 3960 MPa.
Engineering
1 answer:
Alina [70]3 years ago
7 0

Answer:

the required minimum fiber length is 0.80365 mm

Explanation:

Given the data in the question;

Diameter D = 6.90 microns = 6.90 × 10⁻⁶ m

Bond strength ζ = 17 MPa

Shear yield strength ζ_y = 68 Mpa

tensile strength of carbon fibers 6t_{fiber = 3960 MPa.

To determine the minimum fiber length we make use of the following relation;

L = (6t_{fiber × D) / 2ζ

we substitute our given values into the equation;

L = ( 3960 × 6.90 × 10⁻⁶) / (2 × 17 )

L = 0.027324 / 34

L = 0.000803647 m

L = 0.000803647 × (1000) mm

L = 0.80365 mm

Therefore, the required minimum fiber length is 0.80365 mm

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Write a function named "read_prices" that takes one parameter that is a list of ticker symbols that your company owns in their p
ohaa [14]

Answer:

import pandas pd

def read_prices(tickers):

price_dict = {}

# Read ingthe ticker data for all the tickers

for ticker in tickers:

# Read data for one ticker using pandas.read_csv  

# We assume no column names in csv file

ticker_data = pd.read_csv("./" + ticker + ".csv", names=['date', 'price', 'volume'])

# ticker_data is now a panda data frame

# Creating dictionary

# for the ticker

price_dict[ticker] = {}

for i in range(len(ticker_data)):

# Use pandas.iloc  to access data

date = ticker_data.iloc[i]['date']

price = ticker_data.iloc[i]['price']

price_dict[ticker][date] = price

return price_dict  

7 0
4 years ago
The mechanical properties of a metal may be improved by incorporating fine particles of its oxide. Given that the moduli of elas
mojhsa [17]

Answer:

A) Upper bound modulus of elasticity; E = 165.6 GPa

B) Lower bound modulus of elasticity; E = 83.09 GPa

Explanation:

A) Formula for upper bound modulus is given as;

E = E_m(1 - V_f) + E_f•V_f

We are given;

E_m = 60 GPa

E_f = 380 GPa

V_f = 33% = 0.33

Thus,

E = 60(1 - 0.33) + 380(0.33)

E = (60 x 0.67) + 125.4

E = 165.6 GPa

B) Formula for lower bound modulus is given as;

E = 1/[(V_f/E_f) + ((1 – V_f)/E_m)]

E = 1/[(0.33/380) + ((1 – 0.33)/60)]

E = 1/(0.0008684 + 0.01116667)

E = 1/0.01203507

E = 83.09 GPa

3 0
3 years ago
Statement 1: All balls hit the ground at the same time. Statement 2: All balls hit the ground with the same force. Statement 3:
laiz [17]

Answer:

Statement 1: All balls hit the ground at the same time

Explanation:

When there is no resistance of air, the acceleration due to gravity experienced by all the bodies are same. So for falling bodies, neglecting the air resistance, the falling object will be weightless and therefore all the objects will hit the ground at the same time when there is nor air resistance and the objects are considered to be falling in vacuum.

7 0
2 years ago
Given the vector current density J = 10rho2zarho − 4rho cos2 φ aφ mA/m2:
Xelga [282]

Answer:

(a) Current density at P is J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\.

(b) Total current I is 3.257 A

Explanation:

Because question includes symbols and formulas it can be misunderstood. In the question current density is given as below;

J=10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}\\

where \textbf{a}_{\rho} and \textbf{a}_{\phi} unit vectors.

(a) In order to find the current density at a specific point <em>(P)</em>, we can simply replace the coordinates in the current density equation.  Therefore

J(P(\rho=3, \phi=30^o,z=2))=10.3^2.2.\textbf{a}_{\rho}-4.3.(\cos(30^o)^2).\textbf{a}_{\phi}\\\\J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\

(b) Total current flowing outward can be calculated by using the relation,

I=\int {\textbf{J} \, \textbf{ds}

where integral is calculated through the circular band given in the question. We can write the integral as below,

I=\int\{(10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}).(\rho.d\phi.dz.\textbf{a}_{\rho}})\}\\\\I=\int\{(10\rho^2z).(\rho.d\phi.dz)\}\\\\\\

due to unit vector multiplication. Then,

I=10\int\(\rho^3z.dz.d\phi

where \rho=3,\ 0. Therefore

I=10.3^3\int_2^{2.8}\(zdz.\int_0^{2\pi}d\phi\\I=270(\frac{2.8^2}{2}-\frac{2^2}{2} )(2\pi-0)=3257.2\ mA\\I=3.257\ A

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3 years ago
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Arturiano [62]

Answer:

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Explanation:

6 0
3 years ago
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