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ad-work [718]
2 years ago
6

It is desired to produce and aligned carbon fiber-epoxy matrix composite having a longitudinal tensile strength of 630 MPa. Calc

ulate (a) the critical fiber length, and (b) the volume fraction of fibers necessary if (1) the average fiber diameter is 0.030 mm, (2) the average fiber length is 2.4 mm, (3) the fiber fracture strength is 5100 MPa, (4) the fiber-matrix bond strength is 17 MPa, (5) the matrix stress at fiber failure is 17.5 MPa.
Engineering
1 answer:
ratelena [41]2 years ago
6 0

Answer:

The answer is below

Explanation:

Given that:

Diameter (D) = 0.03 mm = 0.00003 m, length (L) = 2.4 mm = 0.0024 m, longitudinal tensile strength (\sigma_{cd})=630\ MPa = 630*10^6\ Pa, Fracture strength

(\sigma_f)=5100\ MPa=5100*10^6\ Pa,fiber-matrix\ stres(\sigma_m)=17.5\ MPa=17.5*10^6\ Pa,matrix\ strength=\tau_c=17\ MPa=17 *10^6\ Pa

a) The critical length (L_c) is given by:

L_c=\sigma_f*(\frac{D}{2*\tau_c} )=5100*10^6*\frac{0.00003}{2*17*10^6}=0.0045\ m=4.5\ mm

The critical length (4.5 mm) is greater than the given length, hence th composite can be produced.

b) The volume fraction (Vf) is gotten from the formula:

\sigma_{cd}=\frac{L*\tau_c}{D}*V_f+\sigma_m(1-V_f)\\\\V_f=\frac{\sigma_{cd}-\sigma_{m}}{\frac{L*\tau_c}{D}-\sigma_{m}}  \\\\Substituting:\\\\V_f=\frac{630*10^6-17.5*10^6}{\frac{0.0024*17*10^6}{0.00003} -17.5*10^6} \\\\V_f=0.456

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A mass of 1.9 kg of air at 120 kPa and 24°C is contained in a gas-tight, frictionless piston–cylinder device. The air is now com
emmainna [20.7K]

Answer:

W=-260.66 kJ (negative answer means, that the work was done on the gas)

Explanation:

1) Convert temperature from C to K- T=24+273=297K- all temperature in the gas problems should be used in Kelvins;

2) We need to analyse type of the process- it is given, that the temperature is constant, so it is an Isothermal process, which means, that the equation of the process is: pV=const (constant);

3) Work, done on the system, should be calculated using the following equation: W=\int\limits^{Vb}_{Va} {p} \, dV

4) To calculate initical and final volumes (Va and Vb), we can use the following equation: pV=mRT, so V=mRT/p. Note, that the pressure is changing, thus we can calculate volumes for the both cases- initial and final, using initial (120kPa) and final (600kPa) pressures, in addition, we can find equation for the pressure, as function of the volume, which we need to use for the integration in step 3: p=mRT/V;

5) Now we can calculate the integral, given in the step 3: W=mRT ln(\frac{Vb}{Va}). As we have pressure as a known values, we can re-write the equation, using pressures: W=mRT ln(\frac{pa}{pb})=1.9*0.287*279*ln(\frac{120}{160})=-260.66 kJ

Note, that natural logarithm (ln) yields negative answer, which supports the question, that the work was done on the gas, not by the gas.

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3 years ago
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