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Anastaziya [24]

3 years ago

7

What is the stress concentration factor of a shaft in torsion, where D=1.25 in. and d=1 in. and the fillet radius is, r=0.2 in.a

. K=1.4b. K=1.5c. K=1.2d. K=1.3

1 answer:

klio [65]3 years ago

6 0

Answer:

Concentration factor will be 1.2

So option (C) will be correct answer

Explanation:

We have given outer diameter D = 1.25 in

And inner diameter d = 1 in and fillet ratio r = 0.2 in

So $\frac{r}{d}$ ratio will be $=\frac{0.2}{1}=0.2$

And $\frac{D}{d}$ ratio will be $=\frac{1.25}{1}=1.25$

Now from the graph in shaft vs torsion the value of concentration factor will be 1.2

So concentration factor will be 1.2

So option (C) will be correct answer.

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The closed tank of a fire engine is partly filled with water, the air space above being under pressure. A 6 cm bore connected to

skelet666 [1.2K]

Answer:

The air pressure in the tank is 53.9 $kN/m^{2}$

Solution:

As per the question:

Discharge rate, Q = 20 litres/ sec = $0.02\ m^{3}/s$

(Since, 1 litre = $10^{-3} m^{3}$ )

Diameter of the bore, d = 6 cm = 0.06 m

Head loss due to friction, $H_{loss} = 45 cm = 0.45\ m$

Height, $h_{roof} = 2.5\ m$

Now,

The velocity in the bore is given by:

$v = \frac{Q}{\pi (\frac{d}{2})^{2}}$

$v = \frac{0.02}{\pi (\frac{0.06}{2})^{2}} = 7.07\ m/s$

Now, using Bernoulli's eqn:

$\frac{P}{\rho g} + \frac{v^{2}}{2g} + h = k$ (1)

The velocity head is given by:

$\frac{v_{roof}^{2}}{2g} = \frac{7.07^{2}}{2\times 9.8} = 2.553$

Now, by using energy conservation on the surface of water on the roof and that in the tank :

$\frac{P_{tank}}{\rho g} + \frac{v_{tank}^{2}}{2g} + h_{tank} = \frac{P_{roof}}{\rho g} + \frac{v_{tank}^{2}}{2g} + h_{roof} + H_{loss}$

$\frac{P_{tank}}{\rho g} + 0 + 0 = \0 + 2.553 + 2.5 + 0.45$

$P_{tank} = 5.5\times \rho \times g$

$P_{tank} = 5.5\times 1\times 9.8 = 53.9\ kN/m^{2}$

4 0

3 years ago

A horizontal curve on a two-lane road is designed with a 2,300-ft radius, 12-ft lanes, and a 65-mph design speed. Determine the

Ierofanga [76]

Answer:

distance = 22.57 ft

superelevation rate = 2%

Explanation:

given data

radius = 2,300-ft

lanes width = 12-ft

no of lane = 2

design speed = 65-mph

solution

we get here sufficient sight distance SSD that is express as

SSD = 1.47 ut + $\frac{u^2}{30(\frac{a}{g}\pm G)}$ ..............1

here u is speed and t is reaction time i.e 2.5 second and a is here deceleration rate i.e 11.2 ft/s² and g is gravitational force i.e 32.2 ft/s² and G is gradient i.e 0 here

so put here value and we get

SSD = 1.47 × 65 ×2.5 + $\frac{65^2}{30(\frac{11.2}{32.2}\pm 0)}$

solve it we get

SSD = 644 ft

so here minimum distance clear from the inside edge of the inside lane is

Ms = Rv ( 1 - $cos (\frac{28.65 SSD}{Rv})$ ) .....................2

here Rv is = R - one lane width

Rv = 2300 - 6 = 2294 ft

put value in equation 2 we get

Ms = 2294 ( 1 - $cos (\frac{28.65 \times 664}{2294})$ )

solve it we get

Ms = 22.57 ft

and

superelevation rate for the curve will be here as

R = $\frac{u^2}{15(e+f)}$ ..................3

here f is coefficient of friction that is 0.10

put here value and we get e

2300 = $\frac{65^2}{15(e+0.10)}$

solve it we get

e = 2%

3 0

3 years ago

A paper clip is made of wire 0.75 mm in diameter. If the original material from which the wire is made is a rod 25 mm in diamete

Bas_tet [7]

Answer:

True strain = 3.7704

Explanation:

Strain is the measure an object that is stretched or deformed. This occurs when a force is applied to an object. Strain deals mostly with the change in length of the object. Strain = Δ L /L = Change in Length over the original Length:

Volume Constancy :

ΔL/L0=A0/ΔA=(D0/ ΔD)=(25mm/0.75mm)^2

ΔL/L0=44.4

Engineering strain:

Engineering strain =ΔL-L0/L0=ΔL/L0-1

Engineering strain =44.4-1=43.4

True strain, ε=In(ΔL/L0)=In(43.4)=3.7704

Note that strain has no unit, so the True strain = 3.7704

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4 years ago

Sphinxa [80]

Answer:

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Explanation:

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4 years ago

What is the area enclosed by the cycle area of the Carnot cycle illustrating on a T-s diagram?

gayaneshka [121]

Answer:

Heat

Explanation:

Carnot cycle:

Carnot cycle is the ideal cycle for all working engine .Carnot cycle all processes are reversible.It have fore process Out of two are constant temperature process and other two are isentropic process(reversible adiabatic).

We know that area under T-s diagram represents the heat.

So $Q=\int Tds$

From cycle we can say that

$q_{in}=T_2\left ( s_2-s_1 \right )$

$q_{out}=T_1\left ( s_2-s_1 \right )$

4 0

4 years ago