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3 years ago

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An alloy has a yield strength of 818 MPa and an elastic modulus of 104 GPa. Calculate the modulus of resilience for this alloy [

in J/m3 (which is equivalent to Pa)] given that it exhibits linear elastic stress-strain behavior.

1 answer:

crimeas [40]3 years ago

5 0

Answer:

Modulus of resilience will be $3216942.308j/m^3$

Explanation:

We have given yield strength $\sigma _y=818MPa$

Elastic modulus E = 104 GPa

We have to find the modulus

Modulus of resilience is given by

Modulus of resilience $=\frac{\sigma _y^2}{2E}$ , here $\sigma _y$ is yield strength and E is elastic modulus

Modulus of resilience $=\frac{(818\times 10^6)^2}{2\times 104\times 10^9}=3216942.308j/m^3$

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3 years ago

Ammonia contained in a piston-cylinder assembly, initially saturated vapor at 0o F, undergoes an isothermal process during which

Rudik [331]

ANSWERS:

$-P_{2(a)} =15.6lbf/in^2\\-P_{2(b)} =30.146lbf/in^2\\ T_{2(a)} =0^oF\\T_{2(b)} =0^oF\\x_{2(b)} =49.87percent$

Explanation:

Given:

Piston cylinder assembly which mean that the process is constant pressure process P=C.

<u>AMMONIA </u>

state(1)

saturated vapor $x_{1} =1$

The temperature $T_{1} =0^0 F$

Isothermal process $T=C$

a)

$-V_{2} =2V_{1}$ ( double)

b)

$-V_{2} =.5V_{2}$ (reduced by half)

To find the final state by giving the quality in lbf/in we assume the friction is neglected and the system is in equilibrium.

state(1)

using PVT data for saturated ammonia

$-P_{1} =30.416 lbf/in^2\\-v_{1} =v_{g} =9.11ft^3/lb$

then the state exists in the supper heated region.

a) from standard data

$-v_{1(a)} =2v_{1} =18.22ft^3/lb\\-T_{1} =0^oF$

$at\\P_{x} =14lbf/in^2\\-v_{x} =20.289 ft^3/kg$

$at\\P_{y} =16 lbf/in^2\\-v_{y} =17.701ft^3/kg$

assume linear interpolation

$\frac{P_{x}-P_{2(b)} }{P_{x}- P_{y} } =\frac{v_{x}-v_{1(a)} }{v_{x}-v_{y} }$

$P_{1(b)}=P_{x} -(P_{x} -P_{y} )*\frac{v_{x}- v_{1(b)} }{v_{x}-v_{y} }\\ \\P_{1(b)} =14-(14-16)*\frac{20.289-18.22}{20.289-17.701} =15.6lbf/in^2$

b)

$-v_{2(a)} =2v_{1} =4.555ft^3/lb\\v_{g}$

from standard data

$-v_{f} =0.02419ft^3/kg\\-v_{g} =9.11ft^3/kg\\v_{f}$

then the state exist in the wet zone

$-P_{s} =30.146lbf/in^2\\v_{2(a)} =v_{f} +x(v_{g} -v_{f} )$

$x=\frac{v_{2(a)-v_{f} } }{v_{g} -v_{f} } \\x=\frac{4.555-0.02419}{9.11-0.02419} =49.87%$

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Read 2 more answers

The theoretical maximum specific gravity of a mix at 5.0% binder content is 2.495. Using a binder specific gravity of 1.0, find

PSYCHO15rus [73]

Answer:

The theoretical maximum specific gravity at 6.5% binder content is 2.44.

Explanation:

Given the specific gravity at 5.0 % binder content 2.495

Therefore

95 % mix + 5 % binder gives S.G. = 2.495

Where the binder is S.G. = 1, Therefore

Per 100 mass unit we have (Mx + 5)/(Vx + 5) = 2.495

(95 +5)/(Vx +5) = 2.495

2.495 × (Vx + 5) = 100

Vx =35.08 to 95

Or density of mix = Mx/Vx = 95/35.08 = 2.7081

Therefore when we have 6.5 % binder content, we get

Per 100 mass unit

93.5 Mass unit of Mx has a volume of

Mass/Density = 93.5/2.7081 = 34.526 volume units

Therefore we have

At 6.5 % binder content.

(100 mass unit)/(34.526 + 6.5) = 2.44

The theoretical maximum specific gravity at 6.5% binder content = 2.44.

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