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vladimir1956 [14]
3 years ago
10

Consider a unidirectional continuous fiber-reinforced composite with epoxy as the matrix with 55% by volume fiber.i. Calculate t

he longitudinal and transverse moduli when E-glass fiber is used.ii. Calculate the longitudinal and transverse moduli if carbon fibers are used instead of E-glass. Constants:Epoxy modulus = 2.41GPaE-glass fiber modulus = 72.5 GPaCarbon fiber modulus = 230 GPa
Engineering
1 answer:
ohaa [14]3 years ago
7 0

Answer:

I)E= 40.95 GPa

II)E=5.29 GPa

Explanation:

I)

Given that

E₁ = 2.41 GPa  ,V₁=1-0.55 = 0.45

E₂ = 72.5 GPa   ,V₂=0.55

Longitudinal moduli  given as ;

E= E₁V₁+E₂V₂

E= 2.41 x 0.45 + 72.5 x 0.55 GPa

E= 40.95 GPa

II)

E₁ = 2.41 GPa  ,V₁=1-0.55 = 0.45

E₂ =230 GPa   ,V₂=0.55

Transverse moduli given as:

\dfrac{1}{E}=\dfrac{V_1}{E_1}+\dfrac{V_2}{E_2}

\dfrac{1}{E}=\dfrac{0.45}{2.41}+\dfrac{0.55}{230}

E=5.29 GPa

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andrezito [222]
Yes a volcanic eruption can suddenly bury a city ! if you look up some stuff on pompeii you can learn more
7 0
3 years ago
A pool of contaminated water is lined with a 40 cm thick containment barrier. The contaminant in the pit has a concentration of
konstantin123 [22]

This question is incomplete, the complete question is;

A pool of contaminated water is lined with a 40 cm thick containment barrier. The contaminant in the pit has a concentration of 1.5 mol/L, while the groundwater circulating around the pit flows fast enough that the contaminate concentration remains 0. There is initially no contaminant in the barrier material at the time of installation. The governing second order, partial differential equation for diffusion of the contaminant through the barrier is:

dC/dt = D( d²C / dz²)

where c(z,t) represent the concentration of containment of any depth into the barrier at anytime and D is the diffusion coefficient (a constant) for the containment in the barrier material.

a) write all boundary and initial conditions needed to solve this equation for C(z, t)

b) Find the steady  state solution (infinite time) for C(z)

Answer:

a) At t = 0, z= 0, c = 1.5 mol/L

at t =0, z = 0.4m, c = 0 mol/L

b) C(z) = z² - 4.15z + 1.5

Explanation:

a)

The boundary and initial conditions are as follows

At t = 0, z= 0, c = 1.5 mol/L

at t =0, z = 0.4m, c = 0 mol/L

b)

The governing second order, partial differential equation for diffusion of the contaminant through the barrier is :

(dC/dt) = D*(d²C/dz²) ..............equ(1)

For steady state, above equation becomes,

(d²C/dz²) =0

Integrating above equation,

(dC/dz) = Z + C1  { where C1 is integration constant) }

again integrating above equation,

C = z² + C1*z + C2    ...................equ(2)

applying boundary condition : at t =0, z= 0, c = 1.5 mol/L, to above equation

 C = z² + C1*z + C2

1.5 = 0 + 0*0 + c2

C2 = 1.5

applying boundary condition : at t =0, z= 0.4m, c = 0 mol/L, to equation (2) ,

0 = 0.4² + C1*0.4 +  1.5

0 = 0.16 + 0.4C1 + 1.5

0.4C1 = - 1.66

C1 = -1.66/0.4

C1 = -4.15

So, the steady state solution for C(z) is:

C(z) = z² - 4.15z + 1.5

6 0
3 years ago
Tungsten is being used at half its melting point (Tm≈3,400◦C) and astress level of 160 MPa. An engineer suggests increasing the
Paladinen [302]

Answer:

Explanation:

The missing diagram is attached in the image below which shows the deformation map of the Tungsten.

Given that:

Stress  level \sigma = 160 MPa

T = 0.5 Tm

\implies \dfrac{T}{Tm} = 0.5

G = 160 GPa

\implies \dfrac{\sigma}{G} = 10^{-3}

a)

The regulating creep mechanism is dislocation driven, as we can see from the deformation mechanism.

The engineer's recommendation would not be approved because increasing grain size results in a decrease in the grain-boundary count, preferring dislocation motion. The existence of grain borders is a hindrance to dislocation motion, as the dislocation principle explicitly states. To stop the motion, we'll need a substance with finer grains, which would result in more grain borders, or a material with higher pressure. In the case of Nabarro creep, which is diffusion-driven, an engineer's recommendation would be useful.

b)

If stress level reduced to \sigma = 1.6 MPa

\implies \dfrac{\sigma }{G} = 10^{-5}

Cable creep is now the controlling creep mode, which entails tension-driven atom diffusion along grain borders to elongate grain along the stress axis, a process known as grain-boundary diffusion. Cable creep is more common in fine-grained materials. As a result, the engineer's advice would succeed in this case. The affinity for cable creep is reduced when the grain size is increased.

c)

From the map of creep mechanism for \dfrac{\sigma}{G} = 10^{-3} \ and \ \dfrac{T}{Tm} = 0.5

We read strain rate (e) = 10^{-6}/sec

Therefore,

Strain (E) =  e * \Delta t

= 10^{-6} \times 10000 \times 3600

= 36

Therefore, \Delta L = E \times Li

= 36 * 10 cm

= 360 cm

Thus, the increase in length = 360 cm

7 0
4 years ago
A hollow steel cylinder with an outside diameter of 100 mm is required to carry a tensile load of 500 kN. Given that the allowab
ivolga24 [154]

Answer:

Maximum inside diameter is 68.52 mm.

Explanation:

Apply stress formula to calculate inside diameter of the tube. Take the allowable stress for safe design and maximum inside diameter of the steel tube.

Step1

Given:

Outside diameter is 100 mm.

Tensile load is 500 kN.

Allowable stress is 120 Mpa.

Calculation:

Step2

Inside diameter is calculated by the stress formula as follows:

\sigma_{a}=\frac{F}{A}

\sigma_{a}=\frac{F}{\frac{\pi}{4}(d_{o}^{2}-d_{i}^{2})}

120=\frac{500\times1000}{\frac{\pi}{4}(100^{2}-d_{i}^{2})}

(100^{2}-d_{i}^{2})=5305.164

d_{i}=68.52mm

Thus, the inner diameter is 68.52 mm.

6 0
3 years ago
Free what add me on discord
kap26 [50]

Answer:

what dude

Explanation:

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6 0
3 years ago
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