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scoundrel [369]

2 years ago

10

A cylindrical 1045 steel bar is subjected to repeated compression-tension stress cycling along its axis. If the load amplitude i

s 22,000 N (4950 lbf), compute the minimum allowable bar diameter to ensure that fatigue failure will not occur. Assume a factor of safety of 2.0

1 answer:

Gre4nikov [31]2 years ago

3 0

Answer:

13.4 mm

Explanation:

Given data :

Load amplitude ( F ) = 22,000 N

factor of safety ( N )= 2.0

Take ( Fatigue limit stress amplitude for this alloy ) б = 310 MPa

<u>calculate the minimum allowable bar diameter to ensure that fatigue failure will not occur</u>

minimum allowable bar diameter = 13.4 * 10^-3 m ≈ 13.4 mm

<em>attached below is a detailed solution</em>

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The total solids production rate in an activated sludge aeration tank is 7240 kg/d on a dry mass basis. It is necessary to maint

snow_lady [41]

Answer:

volume of biological sludge = 28.566 m³ per day

Explanation:

given data

mass of solid = 7240 kg/day

initial moisture content = 78%

solution

here percentage of solid will be

% of solid = 100 - initial moisture content

% of solid = 100 - 78 = 22 %

so that

mass of sludge produced = $\frac{100}{100 - P} M$ kg per day

put her value

mass of sludge produced = $\frac{100}{100 - 78} 7240$ kg

mass of sludge produced = 32909.09 kg

so

specific gravity of sludge = $\frac{\rho sludge}{\rho water }$

and as we know that

$\frac{100}{S sludge} = \frac{solid percentage}{S solid} = \frac{water percentage}{S water}$

$\frac{100}{S sludge} = \frac{22}{2.5} = \frac{78}{1}$

$S sludge$ = 1.152

so that

density of sludge = S sludge × density of water

density of sludge = 1.152 × 1000

density of sludge = 1152 kg/m³

so that

volume of biological sludge = $\frac{mass sludge produce}{\rho sludge}$

volume of biological sludge = $\frac{32909.09}{1152}$

volume of biological sludge = 28.566 m³ per day

6 0

3 years ago

The value of Rth is *

Vladimir79 [104]

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4 0

2 years ago

The drag coefficient of a car at the design conditions of 1 atm, 25°C, and 90 km/h is to be determined experimentally in a large

SIZIF [17.4K]

Answer: 0.288

Explanation:

Given

Pressure of the car, P = 1 atm

Temperature of the car, T = 25° C

Speed of the car, v = 90 km/h = 90*1000/3600 = 25 m/s

Height of the car, h = 1.25 m

Width of the car, b = 1.65 m

Force acting on the far, F = 220 N

Drag coefficient, C(d) = ?

Using our table A-9, we can trace that the density of air ρ, at the given temperature and pressure of 25 °C and 1 atm, is 1.184 kg/m³

Area = h *b

Area = 1.25 * 1.65

Area = 2.0625 m²

Now we solve for the drag coefficient using the formula

C(d) = F / (1/2 * ρ * A * v²)

C(d) = 220 / (0.5 * 1.184 * 2.0625 * 25²)

C(d) = 220 / (1.221 * 625)

C(d) = 220 / 763.125

C(d) = 0.288

Therefore, the drag coefficient is 0.288

3 0

3 years ago

Add a capacitor, C2 = 6 pF, in parallel with R3, and a capacitor, C3 = 2 pF in parallel with R4. Use PSpice to plot the magnitud

Ganezh [65]

Tell me why i got this question got it right and now won’t remember but i’ll get back at you when i remember

6 0

3 years ago

Suppose that you can throw a projectile at a large enough v0 so that it can hit a target a distance R downrange. Given that you

viktelen [127]

Answer:

$\theta_1=15^o\\\theta_2=75^o$

Explanation:

<u>Projectile Motion</u>

In projectile motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration (assuming no friction), and the acceleration in the vertical direction is always the acceleration of gravity. The basic formulas are shown below:

$V_x=V_{ox}=V_ocos\theta$

Where $\theta$ is the angle of launch respect to the positive horizontal direction and Vo is the initial speed.

$V_y=V_{oy}-gt=V_osin\theta-gt$

The horizontal and vertical distances are, respectively:

$x=V_{o}cos\theta t$

$\displaystyle y=y_o+V_{o}sin\theta t-\frac{gt^2}{2}$

The total flight time can be found as that when y = 0, i.e. when the object comes back to ground (or launch) level. From the above equation we find

$\displaystyle t_f=\frac{2V_osin\theta}{g}$

Using this time in the horizontal distance, we find the Range or maximum horizontal distance:

$\displaystyle R=\frac{V_o^2sin2\theta}{g}$

Let's solve for $\theta$

$\displaystyle sin2\theta=\frac{R.g}{V_o^2}$

This is the general expression to determine the angles at which the projectile can be launched to hit the target. Recall the angle can have to values for fixed positive values of its sine:

$\displaystyle \theta_1=\frac{asin\left(\frac{R.g}{V_o^2}\right)}{2}$

$\displaystyle \theta_2=\frac{180^o-asin\left(\frac{R.g}{V_o^2}\right)}{2}$

Or equivalently:

$\theta_2=90^o-\theta_1$

Given Vo=37 m/s and R=70 m

$\displaystyle \theta_1=\frac{asin\left(\frac{70\times 9.8}{37^2}\right)}{2}$

$\theta_1=15^o$

And

$\theta_2=90^o-15^o=75^o$

5 0

3 years ago