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Bumek [7]
3 years ago
11

A continuous and aligned fiber-reinforced composite having a cross-sectional area of 1130 mm2is subjected to an external tensile

load. If the stresses sustained by the fiber and matrix phases are 156 MPa and 2.75 MPa, respectively, the force sustained by the fiber phase is 74,000 Nand the total longitudinal strain is1.25x 10-3, determine (a)the force sustained by the matrix phase, (b)the modulus of elasticity of the composite material in the longitudinal direction, and (c)the moduli of elasticity for the fiber and matrix phases.

Engineering
1 answer:
lakkis [162]3 years ago
8 0

Answer:

(a)  The force sustained by the matrix phase is 1802.35 N

(b) The modulus of elasticity of the composite material in the longitudinal direction Ed is 53.7 GPa

(c) The moduli of elasticity for the fiber and matrix phases is 124.8 GPa and 2.2 GPa respectively

Explanation:

Find attachment for explanation

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The soil borrow material to be used to construct a highway embankment has a mass unit weight of 107.0 lb/cf and a water content
MrRissso [65]

Answer:

Option D

Explanation:

Given information

Bulk unit weight of 107.0 lb/cf

Water content of 7.3%,=0.073

Specific gravity of the soil solids is 2.62

Specifications

Dry unit weight is 113 lb/cf  

Water content is 6%.

Volume of embankment is 440,000-cy

Borrow material

Dry_{unit,weight}=\frac {bulk_{unit,weight}}{1+water_{content}}=\frac {107}{1+0.073}= 99.72041 lb/cf  

Embankment

Considering that the volume of embankment is inversely proportional to the dry unit weight

\frac {V_{embankment}}{V_{borrow}}=\frac {Dry_{borrow}}{Dry_{embankment}}

Therefore, V_{borrow}=V_{embankment} *\frac {Dry_{embarkement}}{Dry_{borrow}}

V_{borrow}=440,000-cy*\frac {113 lb/cf }{99.72041 lb/cf }= 498594-cy

Therefore, volume of borrow material is 498594-cy

(b)

The weight of water in embankment is found by multiplying the moisture content and dry unit weight.

Assuming that all the specifications are achieved, weight of water in embankment=0.06*113=6.78 lb/cf

Since 1 yd^{3}= 27 ft^{3}

The embankment requires water of  6.78*27*440000= 80546400 lb

Borrow materials’ water will also be 0.073*99.72041=7.27959 lb/cf

Borrow material requires water of 7.27959*27*498594=97998120 lb

Extra water between borrow material and embankment=97998120 lb-80546400 lb=17451720 lb

Unit_{weight}=\frac {17451720}{498594}=35.00186 lb

1 gallon is approximately 8.35 yd^{3} hence

\frac {35.00186 lb/yd^{3}}{8.35}=4.19184 gallons/yd^{3}

That's approximately 4.2 gallons

7 0
3 years ago
A two-phase mixture of water and steam with a quality of 0.63 and T = 300F expands isothermally until only saturated vapor rema
VMariaS [17]

Answer:

Explanation:

Hello!

To solve this problem you must follow the following steps, which are fully registered in the attached image.

1. Draw the complete outline of the problem.

2. Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)

through prior knowledge of two other properties.

3. Use temodynamic tables to find the density of water in state 1, by means of temperature and quality, with this value and volume we can find the mass.

3. Use thermodynamic tables to find the internal energy in state 1 and two using temperature and quality.

4. uses the first law of thermodynamics that states that the energy in a system is always conserved, replaces the previously found values ​​and finds the work done.

5. draw the pV diagram using the 300F isothermal line

5 0
2 years ago
1. In a series circuit the sum of all voltage drops produced by the loads in the circuit
GarryVolchara [31]

Answer:

ik the answer but I'm busy rn

5 0
2 years ago
What is the purpose of the graphic language?
solmaris [256]

Answer:

enables the representation, analysis and communication of various aspects of an information system. These aspects correspond to varying and incomplete views of information systems and the processes therein.

5 0
2 years ago
A hypothetical A-B alloy of composition 57 wt% B-43 wt% A at some temperature is found to consist of mass fractions of 0.5 for b
Dennis_Churaev [7]

Answer:

composition of alpha phase is 27% B

Explanation:

given data

mass fractions  = 0.5 for both

composition = 57 wt% B-43 wt% A

composition = 87 wt% B-13 wt% A

solution

as by total composition Co = 57 and by beta phase composition  Cβ = 87  

we use here lever rule that is

Wα = Wβ   ...............1

Wα = Wβ = 0.5

now we take here left side of equation

we will get

\frac{C_\beta - Co}{C_\beta - Ca}   = 0.5

\frac{87 - 57}{87 - Ca} = 0.5  

solve it we get

Ca = 27

so composition of alpha phase is 27% B

8 0
3 years ago
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