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3 years ago

14

Part of a control linkage for an airplane consists of a rigid member CB and a flexible cable AB Originally the cable is unstretc

hed.
If a force is applied to the end B of the member and causes it to rotate by ?=0.58?,
determine the normal strain in the cable.

1 answer:

AnnZ [28]3 years ago

6 0

Answer:

$e_{ab} = 4.18*10^(-3)$

Explanation:

This question will be solved with the help of diagram (see attachment)

Given:

Δ∅ = 0.50 degrees (correction)

BC = 800 mm

AC = 600 mm

Solution:

We use coordinate system with point C as origin (0,0)

Hence,

Point A = (600,0)

Point B = (0,800)

The change in length or displacement can be calculated BB' :

BB' = BC * tan (Δ∅) = (800 mm) * tan (0.5) = 6.9815 mm

Hence, Point B' = (-6.9815,800) and we calculate distance A and B

$AB = \sqrt{600^2 + 800^2} = 1000 mm$

We calculate distance A and B'

$AB' = \sqrt{(600 - (-6.9815))^2 + (0-800)^2} = 1004.2 mm$

Normal Strain in AB is:

$e_{ab} = \frac{AB' - AB}{AB} \\\\= \frac{1004.2 - 1000}{100}\\\\= 4.18 * 10^(-3)$

The solution is :

$e_{ab} = 4.18 * 10^(-3)$

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2 years ago

The bulk modulus of a material is 3.5 ✕ 1011 N/m2. What percent fractional change in volume does a piece of this material underg

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Answer:

percentage change in volume = 0.00285 %

Explanation:

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3 years ago

A 1-mm-diameter methanol droplet takes 1 min for complete evaporation at atmospheric condition. What will be the time taken for

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Answer:

Time taken by the $1\mu m$ diameter droplet is 60 ns

Solution:

As per the question:

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Diameter of the smaller droplet, d' = $1\times 10^{- 6} m$

Diameter of the smaller droplet, R' = $0.5\times 10^{- 6} m$

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3 years ago

Water at 15°C is to be discharged from a reservoir at a rate of 18 L/s using two horizontal cast iron pipes connected in series

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Answer:

The required pumping head is 1344.55 m and the pumping power is 236.96 kW

Explanation:

The energy equation is equal to:

$\frac{P_{1} }{\gamma } +\frac{V_{1}^{2} }{2g} +z_{1} =\frac{P_{2} }{\gamma } +\frac{V_{2}^{2} }{2g} +z_{2}+h_{i} -h_{pump} , if V_{1} =0,z_{2} =0\\h_{pump} =\frac{V_{2}^{2}}{2} +h_{i}-z_{1}$

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$V_{1} =\frac{Q}{\frac{\pi D^{2} }{4} }$

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$Re=\frac{\rho *V*D}{u} =\frac{999.1*6.366*0.06}{1.138x10^{-3} } =335339.4$

$\frac{\epsilon }{D} =\frac{0.00026}{0.06} =0.0043$

Using the graph of Moody, I will select the f value at 0.0043 and 335339.4, as 0.02941

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$h_{1} =\frac{V_{1}^{2} }{2g} (k_{L}+\frac{fL}{D} )=\frac{6.366^{2} }{2*9.8} *(0.5+\frac{0.0294*20}{0.06} )=21.3m$

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$V_{2} =\frac{0.018}{\frac{\pi *0.03^{2} }{4} } =25.46m/s$

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$Re=\frac{\rho *V*D}{u} =\frac{999.1*25.46*0.03}{1.138x10^{-3} } =670573.4$

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The head of pipe 1 is:

$h_{2} =\frac{V_{2}^{2} }{2g} (k_{L}+\frac{fL}{D} )=\frac{25.46^{2} }{2*9.8} *(0.5+\frac{0.033*36}{0.03} )=1326.18m$

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The required pump head is:

$h_{pump} =\frac{25.46^{2} }{2*9.8} +1347.48-36=1344.55m$

The required pumping power is:

$P=Q\rho *g*h_{pump} =0.018*999.1*9.8*1344.55=236965.16W=236.96kW$

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3 years ago

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34kurt

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