Engineering College

A cylindrical rod of copper (E = 110 GPa) having a yield strength of 240 MPa is to be subjected

to a load of 6660 N. If the length of the rod is 380 mm, what must be the diameter to allow an

elongation of 0.50 mm?

Answers

Answer 1

Answer:

"7.654 mm" is the correct solution.

Explanation:

According to the question,

[tex]E=110\times 10^3 \ N/mm^2[/tex][tex]\sigma_y = 240 \ mPa[/tex][tex]P = 6660 \ N[/tex][tex]L = 380 \ mm[/tex][tex]\delta = 0.5 \ mm[/tex]

Now,

As we know,

The Elongation,

⇒ [tex]E=\frac{\sigma}{e}[/tex]

       [tex]=\frac{\frac{P}{A} }{\frac{\delta}{L} }[/tex]

or,

⇒ [tex]\delta=\frac{PL}{AE}[/tex]

By substituting the values, we get

 [tex]0.5=\frac{6660\times 380}{(\frac{\pi}{4}D^2)(110\times 10^3)}[/tex]

then,

⇒ [tex]D^2=58.587[/tex]

     [tex]D=\sqrt{58.587}[/tex]

         [tex]=7.654 \ mm[/tex]

Related Questions

In a ground-water basin of 12 square miles, there are two aquifers: an upper unconfined aquifer 500 ft in thickness and a lower confined aquifer with an available hydraulic head drop of 150 ft. Hydraulic tests have determined that the specific yield of the upper unit is 0.12 and the storativity of the lower unit is 4x10-4. What is the amount of recoverable ground water in the basin

Answers

Answer:

0.1365 m^3

Explanation:

thickness of upper aquifer = 500 ft

lower aquifer head drop = 150 ft

area of ground water basin = 12 m^2

specific yield of upper unit = 0.12

Storativity of lower unit = 4 * 10^-4

determine the amount of recoverable ground water

first step : calculate volume of unconfined aquifer  

                = 12 * 500/5280   = 1.1364 miles^3

The recoverable volume of water from unconfined aquifer

= 1.1364 * 0.12  = 0.1364 miles^3

next : calculate volume of confined aquifer

       = 12 * 150/5250 = 0.341 miles^3

The recoverable volume of water from confined aquifer

     = 0.341 * ( 4 * 10^-4 )  = 1.364 * 10^-4 miles^3

Hence the amount of recoverable ground water in the basin

= ∑ recoverable ground water from both aquifer

 = 0.1365 m^3

forty gal/min of a hydrocarbon fuel having a spesific gravity of 0.91 flow into a tank truck with load limit of 40,000 lb of fuel. How long will it takee to fill the tank in the truck?

Answers

Answer: 131.75minutes

Explanation:

First if all, we've to find the density of liquid which will be:

= Specific gravity × Density to pure water

= 0.91 × 8.34lb/gallon

= 7.59lb/gallon

Then, the volume that's required to fill the tank will be:

= Load limit/Density of fluid

= 40000/7.59

= 5270.1gallon

Now, the time taken will be:

= V/F

= 5270.1/40

= 131.75min

It'll take 131.75 minutes to fill the tank in the truck.

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