A continuous and aligned fiber-reinforced composite is to be produced consisting of 30 vol% aramid fibers and 70 vol% of a polycarbonate matrix; mechanical characteristics of these two materials are as follows:
Modulus of Elasticity [GPa] Tensile Strength [MPa] Aramid fiber 131 3600 Polycarbonate 2.4 65
Also, the stress on the polycarbonate matrix when the aramid fibers fail is 45 MPa. For this composite, compute the following:
(a) the longitudinal tensile strength, and
(b) the longitudinal modulus of elasticity

Answers

Answer 1

Answer:

1. 1111.5MPa

2. 56.1GPa

Explanation:

1. Longitudinal tensile stress can be obtained by obtaining the strength and volume of the fiber reinforcement. The derived formula is given by;

σcl = σm (1 - Vf) + σfVf

Substituting the figures, we will have;

45(1 - 0.30) + 3600(0.30)

45(0.70) + 1080

31.5 + 1080

= 1111.5MPa

2. Longitudinal modulus of elasticity or Young's modulus is the ability of an object to resist deformation. The derived formula is given by;

Ecl = EmVm + EfVf

Substituting the formula gives;

= 2.4 (1 - 0.30) + 131 (0.30)

= 2.4(0.70) + 39.3

= 16.8 + 39.3

= 56.1GPa

Answer 2

Using the appropriate relation, the longitudinal tensile stress and the longitudinal modulus are 1111.50 and 56.10 respectively.

Longitudinal tensile stress can be obtained using the relation :

σcl = σm (1 - Vf) + σfVf

Substituting the values into the relation:

45(1 - 0.30) + 3600(0.30)

45 × 0.70 + 1080

31.5 + 1080

= 1111.50 MPa

2.)

Longitudinal modulus of elasticity is obtained using the relation :

Ecl = EmVm + EfVf

Substituting the values thus :

2.4 (1 - 0.30) + 131 (0.30)

= 2.4 × 0.70 + 39.3

= 16.8 + 39.3

= 56.10 GPa

Hence, the longitudinal tensile stress and the longitudinal modulus are 1111.50 and 56.10 respectively.

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Related Questions

Build a 32-bit accumulator circuit. The circuit features a control signal inc and enable input en. If en is 1 and inc is 1, the circuit increments the stored value by an amount specified by an input A[31:0] on the next clock cycle. If en is 1 and inc is 0 the circuit decrements the stored value by the amount specified in the input A on the next clock cycle. If en is 0, the circuit simply stores its current value without modification. The circuit has the following interface:______.
Input clock governs the state transitions in the circuit upon each rising edge.
Input clear is used as a synchronous reset for the stored value.
Input inc controls whether the value stored is to be incremented or decremented.
Input en is a control signal that activates the values increment/decrement
Input A determines how much to increment or decrement by
Output value is a 32-bit signal that can be used to read the stored value at any time.
* Note: Use any combination of combinational or sequential logic. It may be helpful to look into D Flip Flops and Registers.

Answers

Sorry need.points I'm new

You have available three blocks of different material, at various temperatures. They are, respectively, a 2 kg block of iron at 600 K, a 3 kg block of copper at 800 K and a 10 kg block of granite at 300 K. The heat capacities for the three materials are 0.460 (iron), 0.385 (copper), and 0.790 (granite), in kj/(kg*K), all independent of temperature. For solids, the heat capacities at constant pressure and constant volume can be assumed to be equal, Cp=Cv. what is the minimum temperature that could be obtained in any one of the block? what is the maximum temperature that could be obtained? no heat or work interactions with the enviroment are allowed.

Answers

Answer:

max temp = 711.32 k

mini temp = 331.29 k

Explanation:

Given data:

2kg block of Iron : temperature = 600k , C = 0.460 kJ/kgk

3 kg block of copper : temp = 800k ,  C = 0.385 KJ /kgk

10 kg block of granite : temp = 300k , C = 0.790 KJ/kgk

Cp = Cv at constant pressure and constant volume

Determine the minimum temperature that is obtained in any one of the block

considering the heat transfer equation

Q = mC ( T2 - T1 )

attached below is a detailed solution of the problem

The steam requirements of a manufacturing facility are being met by a boiler whose rated heat input is 5.5 x 3^106 Btu/h. The combustion efficiency of the boiler is measured to be 0.7 by a hand-held flue gas analyzer. After tuning up the boiler, the combustion efficiency rises to 0.8. The boiler operates 4200 hours a year intermittently. Taking the unit cost of energy to be $4.35/10^6 Btu, determine the annual energy and cost savings as a result of tuning up the boiler.

Answers

Answer:

Energy Saved = 6.93 x 10⁹ Btu

Cost Saved = $ 30145.5

Explanation:

The energy generated by each boiler can be given by the following formula:

[tex]Annual\ Energy = (Heat\ In)(Combustion\ Efficiency)(Operating\ Hours)[/tex]

Now, the energy saved by the increase of efficiency through tuning will be the difference between the energy produced before and after tuning:

[tex]Energy\ Saved = (Heat\ In)(Efficiency\ After\ Tune - Efficiency\ Before\ Tune)(Hours)[/tex][tex]Energy\ Saved = (5.5\ x\ 3\ x\ 10^{6}\ Btu/h)(0.8-0.7)(4200\ h)[/tex]

Energy Saved = 6.93 x 10⁹ Btu

Now, for the saved cost:

[tex]Cost\ Saved = (Energy\ Saved)(Unit\ Cost)\\Cost\ Saved = (6.93\ x\ 10^{9}\ Btu)(\$4.35/10^{6}Btu)\\[/tex]

Cost Saved = $ 30145.5

A frequenter of a pub had observed that the new barman poured in average 0.47 liters of beer into the glass with a standard deviation equal to 0.09 liters instead of a half a liter with the same standard deviation. The frequenter had used a random sample of 47 glasses of beer in his experiment. Consider the one-sided hypothesis test for volume of beer in a glass: H0: u=0.5 against H1: u<0.5. Determine the P-value of this test.
Round your answer to four decimal places (e.g. 98.7654).

Answers

Answer:

P-value = 0.0011

Explanation:

Formula for the test statistic is;

z = (x¯ - μ)/(σ/√n)

We have;

Sample mean;x¯ = 0.47

Population mean; μ = 0.5

Standard deviation; σ = 0.09

Sample size; n = 47

Thus;

z = (0.46 - 0.5)/(0.09/√47)

z = -3.05

From z-distribution table attached, the p-value corresponding to z = -3.05 is;

P = 0.00114

To four decimal places gives;

P-value = 0.0011

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