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3 years ago

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A spherical container made of steel has 20 ft outer diameter and wal thickness of 1/2 inch. Knowing the internal pressure is 50

psi, estimate the maximum normal stress and the maximum shearing stress in the container

1 answer:

anastassius [24]3 years ago

5 0

Answer:

maximum normal stress = 5975 psi

maximum shear stress = 2987.50 psi

Explanation:

Given data

dia = 20 ft

wall thickness = 1/2 inch

internal pressure = 50 psi

To find out

the maximum normal stress and the maximum shearing stress

Solution

By the Mohr's circle we will find out shear stress

first we calculate inner radius

i.e. r = (diameter/2) - t

r = (20 × 12 in )/2 - ( 1/2 )

r = 120 - 0.5 = 119.5 inch

Now we find out maximum normal stress by given formula

normal stress = ( internal pressure× r ) / 2 t

normal stress = ( 50×119.5 ) / 2 × 0.5

maximum normal stress = 5975 psi

and minimum normal stress is 0, due to very small radius

and maximum shear stress will be

shear stress = ( maximum normal stress - minimum normal stress ) / 2

shear stress = ( 5975- 0 ) / 2

maximum shear stress = 2987.50 psi

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Answer:

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Explanation:

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Where,

C,n = Taylor equation parameters

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$C_e$ =Cost per grinding per edge

$C_m$ = Machine and operator cost per minute

On comparing with the Taylor equation $VT^n=C$ ,

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$T= \left[ \left(T_t+\frac{C_e}{C_m}\right)\left(\frac{1}{n}-1\right)\right]}\;\cdots(ii)$

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Part diameter: $D=73 mm$ $=73\times 10^{-3} m$

Part length: $l=250 mm=250\times 10^{-3} m$

Feed: $f=0.30 mm/rev= 0.3\times 10^{-3} m/rev$

Depth of cut: $d=3.5 mm$

For the HSS tool:

Tool cost is $20 and it can be ground and reground 15 times and the grinding= $2/grind.

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$C= 80 m/min$

$n=0.130$

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$V_{opt}= \frac {80}{\left[ \left(3+\frac{3.33}{0.667}\right)\left(\frac{1}{0.13}-1\right)\right]^{0.13}}$

$\Rightarrow 47.7$ m/min

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$T=\left(3+\frac{3.33}{0.667}\right)\left(\frac{1}{0.13}-1\right)\right]}$

$\Rightarrow T=53.4$ min

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where,

$T_m=$ Machining time for one part

$n_p=$ Number of pieces cut in one tool life

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$n_p=\frac {T}{T_m}$

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(f) The proportion of time spent actually cutting metal

$=\frac{50\times4.01}{457}=0.4387=43.87\%$

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$C_e=$ $\$10/6=\$1.67/edge$

$T_t-1min$

$C= 3500 m/min$

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(a) Cutting speed:

$V_{opt}= \frac {3500}{\left[ \left(1+\frac{1.67}{0.667}\right)\left(\frac{1}{0.6}-1\right)\right]^{0.6}}$

$\Rightarrow V_{opt}=2105 m/min$

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$T=\left[ \left(1+\frac{1.67}{0.667}\right)\left(\frac{1}{0.6}-1\right)\right]=2.33 min$

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$T_c=T_h+T_m+\frac{T_t}{n_p}$

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