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neonofarm [45]
3 years ago
9

t is desired to produce aligned carbon fiber-epoxy matric composite having a longitudinal tensile strength of 800 MPa. Given (a)

the average fiber diameter and length are 0.01 7 mm and 2.3 mm, respectively, (b) the fiber fracture strength is 5500 MpPa. (c) the fiber- matric bond strength is 18 Mpa, and (d) the matric stress at fiber failure is 13.5 Mpa. Calculate (a) the critical fiber tength and (b) the volume fraction of fibers
Engineering
1 answer:
Nesterboy [21]3 years ago
3 0

Answer:

The value of critical length = 3.46 mm

The value of volume of fraction of fibers = 0.43

Explanation:

Given data

\sigma_{T} =  800 M pa

D = 0.017 mm

L = 2.3 mm

\sigma_{f} = 5500 M pa

\sigma_{m} = 18 M pa

\sigma_c = 13.5 M pa

(a) Critical fiber length is given by

L_{c} = \sigma_{f} (\frac{D}{2 \sigma_{c} } )

Put all the values in above equation we get

L_{c} =5500 (\frac{0.017}{(2) (13.5)} )

L_{c} = 3.46 mm

This is the value of critical length.

(b).Since this  critical length is greater than fiber length Than the volume fraction of fibers is given by

V_{f} = \frac{\sigma_T - \sigma_m}{\frac{L\sigma_c}{D} - \sigma_m }

Put all the values in above formula we get

V_{f} = \frac{800-18}{\frac{(2.3)(13.5)}{0.017} - 18 }

V_{f} = 0.43

This is the value of volume of fraction of fibers.

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What is the measurement of mass in motion is known as
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If gas costs $3.50 per gallon, how much would it cost to drive 500 miles in a city in a car that is 58.3 km/L
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1.20 Three wooden planks are fastened together by a series of bolts to form a column. The diameter of each bolt is 12 mm and the
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2 years ago
The connection is made using a bolt and nut and two washers. If the allowable bearing stress of the washers on the boards is (sb
Yuki888 [10]

Answer:

P = 0.490 kip

Explanation:

given data

allowable bearing stress = 2 ksi

allowable tensile stress = 18 ksi

diameter = 0.31 in

outer diameter = 0.75 in

inner diameter (hole) = 0.50 in

solution

we find here cross section area of shank that is express as

Area = \frac{\pi }{4} \times d^2      ..................1

area = \frac{\pi }{4} \times 0.31^2

area  = 0.0754 in²  

and

now we get here allowable load in bolt will be

\sigma = \frac{P}{A}     ...................2

P = \sigma \times A  

P = 18 × 10³ × 0.0754

P = 1357.2 = 1.357 kip

and

now find here area of washer is

Area = \frac{\pi }{4} \times (d^2-d1^2)     .......................3

put here value

Area = \frac{\pi }{4} \times (0.75^2-0.5^2)  

area = 0.2454 in²

so now we get here allowable load of washer will be

\sigma = \frac{P}{Area}      .....................4

P = 2 × 10³ × 0.245

P = 490 = 0.490 kip

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2 years ago
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