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vazorg [7]
3 years ago
6

Determine the normal stress in a ball, which has an outside diameter of 160 mm and a wall thickness of 3.8 mm, when the ball is

inflated to a gage pressure of 78 kPa.
Engineering
1 answer:
lorasvet [3.4K]3 years ago
5 0

Answer:

<em>The normal stress is 0.7821 MPa</em>

<em></em>

Explanation:

The external diameter D = 160 mm

The thickness t = 3.8 mm = 3.8 x 10^-3 m

gauge pressure P = 78 kPa = 78 x 10^3 Pa

The maximum shear stress τmax = ?

The external radius of the shell from the external surface R = D/2 = 160/2 = 80 mm

The internal radius of the shell r = R - t

==> 80 - 3.8 = 76.2 mm

Therefore the internal diameter d = 2r = 2 x 76.2 = 152.4 mm

==> d = 152.4 x 10^-3 m

The normal stress σ = \frac{Pd}{4t} = \frac{78*10^{3}*152.4*10^{-3} }{4*3.8*10^{-3} } = 782052.63 Pa

==>  σ = <em>0.7821 MPa</em>

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3 years ago
Steam flows through a copper pipe (kc=150Wm∙K⁄) that has an inner diameter of 19 mm and an outer diameter of 21 mm. The convecti
olchik [2.2K]

Answer:

a) The temperature on the outer surface of the pipe is approximately 179.97 °C

b) The thickness of the insulation  is approximately 0.857 m

Explanation:

We have;

\dfrac{1}{U} = \dfrac{1}{\alpha _A}

αA = 200 W/(m²·K)

\dot q =  (T₂ - T₁) × U

\dot q =  (200 - 180) × 200 = 4,000

For the pipe, we have;

\dfrac{1}{U} =\dfrac{x}{kc }

\dot q/U=  (T₂ - T₁)

∴ 4000×\dfrac{0.001}{150} = (180 - T₂)

T₂ ≈ 179.97 °C

The temperature on the outer surface of the pipe, T₂ ≈ 179.97 °C

b) For the insulation, we have;

\dfrac{1}{U} = \dfrac{x}{ki } = \dfrac{x}{0.03}

T₂ - T₃ = 179.97 °C - 40°C ≈ 139.97°C

\dot q/U=  (T₂ - T₃)

x = \dfrac{\dot q \cdot kc}{T_2 - T_3} = \dfrac{4,000 \times 0.03}{139.97} \approx 0.857

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7 0
3 years ago
A soil element is subjected to a minor principle stress of 50 kPa on a plane rotated 20 ° counterclockwise from vertical. If the
Bad White [126]

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°.

5 0
3 years ago
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