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inn [45]
3 years ago
15

A steel bar 110 mm long and having a square cross section 22 mm on an edge is pulled in tension with a load of 89,000 N, and exp

eriences an elongation of 0.10 mm. Assuming that the deformation is entirely elastic, calculate the elastic modulus of the steel.
Engineering
1 answer:
iragen [17]3 years ago
3 0

Answer:

Elastic modulus of steel  = 202.27 GPa

Explanation:

given data

long = 110 mm = 0.11 m

cross section 22 mm  = 0.022 m

load = 89,000 N

elongation = 0.10 mm = 1 × 10^{-4} m

solution

we know that Elastic modulus is express as

Elastic modulus = \frac{stress}{strain}    ................1

here stress is

Stress = \frac{Force}{area}       .................2

Area = (0.022)²

and  

Strain = \frac{extension}{length}       .............3

so here put value in equation 1 we get

Elastic modulus = \frac{89,000\times 0.11}{0.022^2 \times 1 \times 10^{-4} }  

Elastic modulus of steel = 202.27 × 10^9 Pa

Elastic modulus of steel  = 202.27 GPa

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Answer:

1) <em>4.41 * 10^-4 kw </em>

<em>2) </em>2.20 * 10^-4 kw

Explanation:

Given data:

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bag ; length = 3 m , diameter = 0.13 m

change in pressure = 1 kPa

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<u>1) Calculate the Fan power </u>

First :

Calculate the total dust loading = 3 * 500 = 1500 g

To determine the Fan power we will apply the relation

n_{o}  = \frac{\frac{p}{eg*Q*h} }{1000}    = \frac{\frac{p}{(3*10^{-3})* 981*( 500/60) *3  } }{1000}

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<u>2) calculate power drawn </u>

change in P = 6 atm = 6 * 10^5 pa

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3 years ago
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Explanation:

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5) Calculate the LMC wal thickness of a pipe and tubing with OD as 35 + .05 and ID as 25 + .05 A) 4.95 B) 5.05 C) 10 D) 15.025
padilas [110]

Answer:

B) 5.05

Explanation:

The wall thickness of a pipe is the difference between the diameter of outer wall and the diameter of inner wall divided by 2. It is given by:

Thickness of pipe = (Outer wall diameter - Inner wall diameter) / 2

Given that:

Inner diameter = ID = 25 ± 0.05, Outer diameter = OD = 35 ± 0.05

Maximum outer diameter = 35 + 0.05 = 35.05

Minimum inner diameter = 25 - 0.05 = 24.95

Thickness of pipe = (maximum outer wall diameter - minimum inner wall diameter) / 2 = (35.05 - 24.95) / 2 = 5.05

or

Thickness = (35 - 25) / 2 + 0.05 = 10/2 + 0.05 = 5 + 0.05 = 5.05

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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
Verdich [7]

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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3 years ago
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