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romanna [79]
2 years ago
10

A soil element is subjected to a minor principle stress of 50 kPa on a plane rotated 20 ° counterclockwise from vertical. If the

deviator stress is 120 kPa and the shear strength parameters are a friction angle of 30° and a cohesion of 5 kPa, determine the magnitude of the stresses on the failure plane and the orientation of this plane with respect to the major principle stress plane.
Engineering
1 answer:
Bad White [126]2 years ago
5 0

Answer:

=>> 167.3 kpa.

=>> 60° from horizontal face.

Explanation:

So, we are given the following data or parameters or information which is going to assist us in solving this kind of question;

=>> "A soil element is subjected to a minor principle stress of 50 kPa on a plane rotated 20 ° counterclockwise from vertical. "

=>>"If the deviator stress is 120 kPa and the shear strength parameters are a friction angle of 30° and a cohesion of 5 kPa."

The orientation of this plane with respect to the major principle stress plane = 50 tan^2 (45 + 30/2) + 10 tan ( 45 + 30/2).

magnitude of the stresses on the failure plane = 167.3 kpa.

The orientation of this plane with respect to the major principle stress plane => x = 60 cos 60° = 30kpa.

y = 60 sin 60° = 30√3 = sheer stress.

the orientation of this plane with respect to the major principle stress plane.

Theta = 45 + 15 = 60°.

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1. Yes.

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Explanation:

Should she be worried about corrosion?

Yes, the engineer needs to be worried about corrosion as stainless steel has a lower resistance to corrosion, in other words, stainless steel corrodes faster than Titanium.

If so, what types of corrosion could take place?

The type of corrosion that takes place is called Localized corrosion. Localized corrosion occurs when a small part of a component experiences corrosion. In this case, the ball component of the femoral stem is made of stainless steel which will corrode faster than the other parts of the femoral stem which is made of Titanium.

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A round bar of chromium steel, (ρ= 7833 kg/m, k =48.9 W/m-K, c =0.115 KJ/kg-K, α=3.91 ×10^-6 m^2/s) emerges from a heat treatmen
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Answer:

Q = 424523.22 kw

Explanation:

\rho =7833 kg/m

k = 48.9 W/m - K

c = 0.115 KJ/kg- K

\alpha = 3.91*10^{-6} m^2/s

T_s = 285 degree celcius

T_∞ = 35 degree celcius

velocity of air stream = 15 m/s

D = 40 cm

L = 200 cm

mass flow rate\dot m = \rho AV = 7833 *\frac{\pi}{4} 0.4^2*15

\dot m = 14764.85 kg/s

A_s = \pi DL = \pi 0.4*2 = 2.513 m^2

Q = \dot m C \Delta T = h A_s \Delta T

\dot m C \Delta T = h A_s \Delta T

solving for h

h = \frac{14764.85*0.115*(285-35)}{2.513*(285-35)}

h = 675.6 kw/m^2K

Q = h A_s\Delta T

Q = 675.6*2.513*(285-35)

Q = 424523.22 kw

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Answer:

a) 84.034°C

b) 92.56°C

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Explanation:

Thickness of aluminum alloy fin = 12 mm

width = 10 mm

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Temperature of aluminum alloy is maintained at 120°C

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