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Sedaia [141]
3 years ago
11

A steel pipe of 400-mm outer diameter is fabricated from 10-mm-thick plate by welding along a helix that forms an angle of 20° w

ith a plane perpendicular to the axis of the pipe. Knowing that a 300-kN axial force P is applied to the pipe. Determine the normal and shearing stresses in directions respectively normal and tangential to the weld.
Engineering
1 answer:
Verdich [7]3 years ago
3 0

Explanation:

Outer di ameter d_{0}=400 \mathrm{mm}[tex] Thickness of the cylinder [tex]t=10 \mathrm{mm}

\therefore[tex] Inner diam eter [tex]d_{i}=d_{0}-2 t=400-2 \times 10

d_{1}=380 \mathrm{mm}

Given loading on the cylinder P=300 \mathrm{kN} Helix an gle of the weld form \theta=20^{\circ}

(i) Normal stress on the plane at angle \theta=20^{\circ} is

\sigma=\frac{P \cos ^{2} \theta}{A_{0}}

\text { Where } A_{0}=\frac{\pi}{4}\left(d_{0}^{2}-d_{1}^{2}\right)

\quad=\frac{\pi}{4}\left(400^{2}-380^{2}\right)

=12252.21 \mathrm{mm}^{2}

=12.25221 \times 10^{-9} \mathrm{m}^{2}

\sigma=\frac{-300 \times 10^{2} \times \cos ^{2} 20}{12.25221 \times 10^{-1}}

=-21.6 \mathrm{MPa}

(ii) Shear stress along an angle of \theta=20^{\circ} is \tau=\frac{P}{A_{0}} \cos \theta \sin \theta

=\frac{-300 \times 10^{-1} \times \cos 20 \times \sin 20}{12.25221 \times 10^{-3}}

=-7.86 \mathrm{MPa}

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1. A glass window of width W = 1 m and height H = 2 m is 5 mm thick and has a thermal conductivity of kg = 1.4 W/m*K. If the inn
emmasim [6.3K]

Answer:

1. \dot Q=19600\ W

2. \dot Q=120\ W

Explanation:

1.

Given:

  • height of the window pane, h=2\ m
  • width of the window pane, w=1\ m
  • thickness of the pane, t=5\ mm= 0.005\ m
  • thermal conductivity of the glass pane, k_g=1.4\ W.m^{-1}.K^{-1}
  • temperature of the inner surface, T_i=15^{\circ}C
  • temperature of the outer surface, T_o=-20^{\circ}C

<u>According to the Fourier's law the rate of heat transfer is given as:</u>

\dot Q=k_g.A.\frac{dT}{dx}

here:

A = area through which the heat transfer occurs = 2\times 1=2\ m^2

dT = temperature difference across the thickness of the surface = 35^{\circ}C

dx = t = thickness normal to the surface = 0.005\ m

\dot Q=1.4\times 2\times \frac{35}{0.005}

\dot Q=19600\ W

2.

  • air spacing between two glass panes, dx=0.01\ m
  • area of each glass pane, A=2\times 1=2\ m^2
  • thermal conductivity of air, k_a=0.024\ W.m^{-1}.K^{-1}
  • temperature difference between the surfaces, dT=25^{\circ}C

<u>Assuming layered transfer of heat through the air and the air between the glasses is always still:</u>

\dot Q=k_a.A.\frac{dT}{dx}

\dot Q=0.024\times 2\times \frac{25}{0.01}

\dot Q=120\ W

5 0
3 years ago
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