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3 years ago

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1. A thin-walled cylindrical pressure vessel is capped at the end and is subjected to an internal pressure (p). The inside diame

ter of the vessel is 6 ft and the wall thickness is 1.5 inch. The vessel is made of steel with tensile yield strength and compressive yield strength of 36 ksi. Determine the internal pressure required to initiate yielding according to (a) The maximum-shear-stress theory of failure, and (b) The maximum-distortion-energy theory of failure, if a factor of safety (FS) of 1.5 is desired.

1 answer:

Vesna [10]3 years ago

6 0

I DONT KNOW OKAY UGHHH

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An air-conditioning system operating on the reversed Carnot cycle is required to transfer heat from a house at a rate of 755 kJ/

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Answer:

There is 0.466 KW required to operate this air-conditioning system

Explanation:

<u>Step 1:</u> Data given

Heat transfer rate of the house = Ql = 755 kJ/min

House temperature = Th = 24°C = 24 +273 = 297 Kelvin

Outdoor temperature = To = 35 °C = 35 + 273 = 308 Kelvin

<u>Step 2: </u> Calculate the coefficient of performance o reversed carnot air-conditioner working between the specified temperature limits.

COPr,c = 1 / ((To/Th) - 1)

COPr,c = 1 /(( 308/297) - 1)

COPr,c = 1/ 0.037

COPr,c = 27

<u>Step 3:</u> The power input cna be given as followed:

Wnet,in = Ql / COPr,max

Wnet, in = 755 / 27

Wnet,in = 27.963 kJ/min

Win = 27.963 * 1 KW/60kJ/min = 0.466 KW

There is 0.466 KW required to operate this air-conditioning system

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10% A steel beam W18x76 spans 32 feet and is subjected to a Moment of 334 kips-ft. Find the load w on the beam. Determine the de

Lilit [14]

Answer:

w = 10.437 kips

deflection at 1/4 span 20.83\E ft

at mid span = 1.23\E ft

shear stress 7.3629 psi

Explanation:

area of cross section = 18*76

length of span = 32 ft

moment = 334 kips-ft

we know that

moment = load *eccentricity

334 = w * 32

w = 10.437 kips

deflection at 1/4 span

$\delta = \frac{wa^2b^2}{3EI}$

$= \frac{10.4375*8^2 *24^2}{3E \frac{BD^3}{12}}$

$=\frac{10.437 *8^2*24^2}{3E \frac{18*16^3}{12}}$

= 20.83\E ft

at mid span

$\delta = \frac{wl^3}{48EI}$

$= \frac{10.43 *32^3}{48 *E*\frac{18*16^3}{12}}$

$\delta = 1.23\E ft$

shear stress

$\tau = \frac{w}{A} = \frac{10.43 7*10^3}{18*76} =7.3629 psi$

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3 years ago

(a) Calculate the heat flux through a sheet of steel that is 10 mm thick when the temperatures oneither side of the sheet are he

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3 years ago

A tool chest has 650 N weight that acts through the midpoint of the chest. The chest is supported by feet at A and rollers at B.

erica [24]

Answer:

the value of horizontal force P is 170.625 N

the value of horizontal force at P = 227.5 N is that the block moves to right and this motion is due to sliding.

Explanation:

The first diagram attached below shows the free body diagram of the tool chest when it is sliding.

Let start out by calculating the friction force

$F_f= \mu N_2$

where :

$F_f =$ friction force

$\mu$ = coefficient of friction

$N_2$ = normal friction

Given that:

$\mu$ = 0.3

$F_f =$ 0.3 $N_2$

Using the equation of equilibrium along horizontal direction.

$\sum f_x = 0$

P - $F_f =$ 0

P = 0.3 $N_2$ ----- Equation (1)

To determine the moment about point B ; we have the expression

$\sum M_B = 0$

0 = $N_2*70-W*35-P*100$

where;

P = horizontal force

$N_2$ = normal force at support A

W = self- weight of tool chest

Replacing W = 650 N

0 = $N_2*70-650*35-100*P$

$P = \frac{70 N_2-22750}{100} ----- equation (2)$

Replacing $\frac{70 N_2-22750}{100}$ for P in equation (1)

$\frac{70N_2 -22750}{100} =0.3 N_2$

$N_2 = \frac{22750}{40}$ $N_2 = 568.75 \ N$

Plugging the value of $N_2 = 568.75 \ N$ in equation (2)

$P = \frac{70(568.75)-22750}{100} \\ \\ P = \frac{39812.5-22750}{100} \\ \\ P = \frac{17062.5}{100}$

P =170.625 N

Thus; the value of horizontal force P is 170.625 N

b) From the second diagram attached the free body diagram; the free body diagram of the tool chest when it is tipping about point A is also shown below:

Taking the moments about point A:

$\sum M_A = 0$

-(P × 100)+ (W×35) = 0

P = $\frac{W*35}{100}$

Replacing 650 N for W

$P = \frac{650*35}{100}$

P = 227.5 N

Thus; the value of horizontal force P, when the tool chest tipping about point A is 227.5 N

We conclude that the motion will be impending for the lowest value when P = 170.625 N and when P= 227.5 N

However; the value of horizontal force at P = 227.5 N is that the block moves to right and this motion is due to sliding.

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4 years ago