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kumpel [21]
3 years ago
12

It is desired to produce a continuous and oriented carbon fiber-reinforced epoxy having a modulus of elasticity of at least 83 G

Pa in the direction of fiber alignment. The modulus of elasticity of the graphite fiber is 260 GPa, and the modulus of elasticity of the epoxy is 2.4 GPa. What is the minimum volume fraction of fibers required to meet this requirement. (this is a Chapter 16 problem) Answer: the minimum fiber volume fraction is:__________
Engineering
1 answer:
lukranit [14]3 years ago
3 0

Answer:

0.3129

Explanation:

Calculate The minimum fiber volume fraction using the relation below

E = x * E1 + ( 1 - x ) E2  ------ ( 1 )

given that :

E = 83 GPa , x = ?

E1 = 260 GPa ,  E2 = 2.4 GPA

Insert values back into equation 1

Then ; 80.6 = 257.6 x

∴ x ( volume fraction ) = 80.6 / 257.6  = 0.3129

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The coefficient of static friction for both wedge surfaces is μw=0.4 and that between the 27-kg concrete block and the β=20° inc
balandron [24]

Assuming  the wedge has an angle of 5°.The minimum value of the force P that is required to begin moving the block up the incline is: 322.84 N.

<h3>Minimum value of force P</h3>

First step

Using this formula to find the weight of the block

W=mg

W=27×9.81

W=264.87 N

Second step

Angles of friction ∅A and ∅B

∅A=tan^-1(μA)

∅A=tan^-1(0.70)

∅A=34.99°

∅B=tan^-1(μB)

∅B=tan^-1(0.40)

∅B=21.80°

Third step

Equate the sum of forces in m-direction to 0 in order to find the reaction force at B.

∑fm=0

W sin (∅A+20°)  + RB cos (∅B+∅A)=0

264.87 sin(34.99°+20°) + RB cos (21.80°+34.99°)=0

216.94+0.5477Rb=0

RB=216.94/0.5477

RB=396.09 N

Fourth step

Equate the sum of forces in x-direction to 0 in order to find force Rc.

∑fx=0

RB cos (∅B) - RC cos (∅B+ 5°)=0

396.09 cos(21.80°) - RC cos (21.80°+5°)=0

RC=396.09 cos(21.80°)/cos(26.80°)

RC=412.02 N

Last step

Equate the sum of forces in y-direction to 0 in order to find force P required to move the block up the incline.

∑fy=0

RB sin (∅B) + RC sin (∅B)-P=0

P=Rb sin (∅B) + RC sin (5°+∅B)

P=396.09 sin(21.80°) +412.02sin (5°+21.80°)

P=322.84 N

Inconclusion the minimum value of the force P that is required to begin moving the block up the incline is: 322.84 N.

Learn more about Minimum value of force P here:brainly.com/question/20522149

7 0
2 years ago
An FPC 4 m2 in area is tested during the night to measure the overall heat loss coefficient. Water at 60 C circulates through th
sp2606 [1]

Answer:

<em> - 14.943 W/m^2K  ( negative sign indicates cooling ) </em>

Explanation:

Given data:

Area of FPC = 4 m^2

temp of water = 60°C

flow rate = 0.06 l/s

ambient temperature = 8°C

exit temperature = 49°C

<u>Calculate the overall heat loss coefficient </u>

Note : heat lost by water = heat loss through convection

m*Cp*dT  = h*A * ( T - To )

∴ dT / T - To = h*A / m*Cp  ( integrate the relation )

In ( \frac{49-8}{60-8} ) =  h* 4 / ( 0.06 * 10^-3 * 1000 * 4180 )

In ( 41 / 52 ) = 0.0159*h

hence h = - 0.2376 / 0.0159

              = - 14.943  W/m^2K  ( heat loss coefficient )

7 0
2 years ago
Explain why the failure of a garden hose occurred near its end and why the tear occurred along its length. Use numerical values
Nataliya [291]

Answer:

Most hydraulic systems develops pressure surges that may surpass settings valve. by exposing the hose surge to pressure above the maximum operating pressure will shorten the hose life.

Explanation:

Solution

Almost all hydraulic systems creates pressure surges that may exceed relief valve settings. exposing the hose surge to pressure above the maximum operating  pressure shortens the hose life.

In systems where pressure peaks are severe, select or pick a hose with higher maximum operating  pressure or choose a spiral reinforced hose specifically designed for severe pulsing applications.

Generally, hoses are designed or created to accommodate pressure surges and have operating pressures that is equal to 25% of the hose minimum pressure burst.

7 0
3 years ago
Reduce the force F ij = + (2 5 ) kN to point A(2m,3m) that acts on point B( 3m,5m) - .
Alexxx [7]

Given :

Force, \vec{F}= (2\hat{i} + 5\hat{j})\ kN.

Force is acting at point A( 2 m, 3 m ) and B( 3 m, 5 m )

To Find :

The work done by force F .

Solution :

Displacement vector between point A and B is :

\vec{d} = (3-2)\hat{i} + (5-3)\hat{j}\\\\\vec{d} = \hat{i} + 2\hat{j}

Now, we know work done is given by :

W = \vec{F}.\vec{d}\\\\W= (2\hat{i} + 5\hat{j}).(\hat{i}+\hat{2j})\\\\W = (2\times 1) +( 5\times 2) \ kJ\\\\W = 12 \ kJ

W = 12000 J

Therefore, work done by force is 12000 J .

6 0
3 years ago
How much power do a capacitor and inductor dissipate? Assume the capacitor/inductor have no parasitic resistance (no resistor in
Mariana [72]
They do in fact heat up while receiving energy.
4 0
3 years ago
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