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Scilla [17]
3 years ago
5

The screw of shaft straightener exerts a load of 30 as shown in Figure . The screw is square threaded of outside diameter 75 mm

and 6 mm pitch.
force required at the rim of a 300mm diameter hand wheel, if there is a collar
bearing of 50 mm mean diameter provided in the arrangement to exert axial
load. Assume the coefficient of friction for the collar as 0.2.

Engineering
1 answer:
kykrilka [37]3 years ago
4 0

Answer:

See calculation below

Explanation:

Given:

W = 30 kN = 30x10³ N

d = 75 mm

p = 6 mm

D = 300 mm

μ = tan Φ = 0.2

<u><em>1. Force required at the rim of handwheel </em></u>

Let P₁ = Force required at the rim of handwheel

Inner diameter or core diameter of the screw = dc = do - p = 75 - 6 = 69 mm

Mean diameter of screw:    *d = \frac{do + dc}{2} = (75 + 69) / 2 = 72 mm

and

tan α = p / πd  =  6 / (π x 72)  =  0.0265

∴ Torque required to overcome friction at he threads is  T = P x d/2

T = W tan (α + Ф) d/2

T =  W(\frac{tan \alpha + tan \theta}{1 - tan \alpha + tan \theta } ) * \frac{d}{2}

T = 30x10³ * ((0.0265 + 0.2) / (1 - 0.0265 x 0.2)) x 72/2

T = 245,400 N-mm

We know that the torque required at the rim of handwheel (T)

245,400 = P1 x D/2 = P1 x (300/2) = 150 P1

P1 = 245,400 / 150

P1 = 1636 N

<u><em>2. Maximum compressive stress in the screw</em></u>

                         30x10³

Qc = W / Ac = -------------- = 8.02 N/mm²

                      π/4 * 69²

Qc = 8.02 MPa

Bearing pressure on the threads (we know that number of threads in contact with the nut)

n = height of nut / pitch of threads = 150 / 6 = 25 threads

thickness of threads, t = p/2 = 6/2 = 3 mm

bearing pressure on the threads = Pb = W / (π d t n)

Pb = 30 x 10³ / (π * 72 * 3 * 25)

Pb = 1.77 N/mm²

Max shear stress on the threads = τ = 16 T / (π dc³)

τ = (16 * 245,400) / ( π * 69³ )

τ = 3.8 M/mm²

*the mean dia of the screw (d) = d = do - p/2 = 75 - 6/2 = 72

∴max shear stress in the threads τmax = 1/2 * sqrt(8.02² + (4 * 3.8²))

τmax = 5.5 Mpa

<u><em>3. efficiency of the straightener</em></u>

<u><em></em></u>

To = W tan α x d/2 = 30x10³ * 0.0265 * 72/2 = 28,620 N-mm

∴Efficiency of the straightener is η =  To / T = 28,620 / 245,400

η = 0.116 or 11.6%

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Answer:

(b)False

Explanation:

Given:

 Prandtl number(Pr) =1000.

We know that   Pr=\dfrac{\nu }{\alpha }

  Where \nu is the molecular diffusivity of momentum

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 Prandtl number(Pr) can also be defined as

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Where \delta is the hydrodynamic boundary layer thickness and \delta_t is the thermal boundary layer thickness.

So if Pr>1 then hydrodynamic boundary layer thickness will be greater than thermal boundary layer thickness.

In given question Pr>1 so  hydrodynamic boundary layer thickness will be greater than thermal boundary layer thickness.

So hydrodynamic layer will be thicker than the thermal boundary layer.

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4 years ago
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Answer:

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Explanation:

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The seers were actually of the opinion that a healthy mind resides in a healthy body. It implies that there is connection between the body and the mind. When the body catches an illness, the mind and other parts of the body are affected. When our minds are not healthy, it affects the effective functioning of the body.

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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
The ingredient weights for making 1 yd (cyd) of concrete by assuming aggregates in SSD state are given below. The volume of air
Pachacha [2.7K]

Answer:

Explanation:

Ans) Given batch weight of each component :

Cement = 700 lb

Water = 315 lb

Coarse aggregate = 1575 lb

Fine aggregate = 1100 lb

Part 1) Amount of water = 328.5 lb

Amount of water is needed to be increased if the aggregates has absorption capacity, To maintain constant water cement ratio, the mixing water is increased because some of the water is absorbed by aggregates.

Amount of water absorbed = 328.5 lb - 315 lb = 13.5 lb

Total amount of aggregates = 1575 + 1100 = 2675 lb

=> % Absorption capacity = 13.5 x 100 / 2675 = 0.5 %

Hence, new amount of Coarse aggregate = (1 - 0.005) x 1575 lb = 1567.125 lb

New amount of fine aggregate = (1 - 0.005) x 1100 = 1094.5 lb

Since, water cement ratio is maintained constant , amount of cement remains unchanged

=> Volume of water = 328.5 / 62.4 = 5.26 ft3

=> Volume of cement = 700 / (3.15 x 62.4) = 3.56 ft3

=> Volume of coarse aggregate = 1567.125 / (2.4 x 62.4) = 10.46 ft3

=> Volume of fine aggregate = 1100 / (2.4 x 62.4) = 7.34 ft3

Volume of air = 2% = 0.02 x 27 = 0.54 ft3

Total concrete volume = 5.26 + 3.56 + 10.46 + 7.34 + 0.54 \approx 27 ft3 = 1 yd3

Hence, calculated amount of each component is correct

Part 2) We know, minus sign indicated that the aggregate will absorb some moisture from concrete, hence mixing water amount needed to be corrected .

=> Amount of water absorbed by coarse aggregate = 0.01 x 1567.125 lb = 15.67 lb

=> Amount of water absorbed by fine aggregate = 0.02 x 1094.50 lb = 21.89 lb

Total amount of water absorbed = 15.67 + 21.89 = 37.56 lb

To maintain same water cement ratio, amount of mixing water is needed to be increased

=> Corrected amount of mixing water = 328.5 lb + 37.56 lb = 366 lb

=> Corrected amount of coarse aggregate = (1 - 0.01) x 1567.125 = 1551.45 lb

=> Corrected amount of fine aggregate = (1 - 0.02) x 1094.5 = 1072.6 lb

Part 3) We know,

Unit weight = Sum of weight of each material / Total volume

=> Sum of weight = 366 + 700 + 1551.45 + 1072.6 = 3690.05 lb

Total volume = 1 yd3 or 27 ft3

=> Expected Unit Weight = 3690.05 lb / 27 ft3 = 136.67 lb/ft3

Also, Concrete Yield = Weight of all components / Unit weight of concrete

=> Yield = 3690.05 / 136.67 = 27 ft3 or 1 yd3

4 0
4 years ago
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