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Anastaziya [24]
3 years ago
7

What is the stress concentration factor of a shaft in torsion, where D=1.25 in. and d=1 in. and the fillet radius is, r=0.2 in.a

. K=1.4b. K=1.5c. K=1.2d. K=1.3
Engineering
1 answer:
klio [65]3 years ago
6 0

Answer:

Concentration factor will be 1.2

So option (C) will be correct answer

Explanation:

We have given outer diameter D = 1.25 in

And inner diameter d = 1 in and fillet ratio r = 0.2 in

So \frac{r}{d} ratio will be =\frac{0.2}{1}=0.2

And \frac{D}{d} ratio will be =\frac{1.25}{1}=1.25

Now from the graph in shaft vs torsion the value of concentration factor will be 1.2

So concentration factor will be 1.2

So option (C) will be correct answer.

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Using the following data, determine the percentage retained, cumulative percentage retained, and percent passing for each sieve.
vekshin1

Solution :

<u>Sieve Size</u> (in)                   <u>Weight retain</u><u>(g)</u>

3                                         1.62

2                                         2.17

$1\frac{1}{2}$                                       3.62

$\frac{3}{4}$                                        2.27

$\frac{3}{8}$                                        1.38

PAN                                    0.21

Given :

Sieve       weight       % wt. retain    % cumulative       % finer

size        retained                               wt. retain

No. 4        59.5            10.225%          10.225%            89.775%

No. 8        86.5            14.865%          25.090%           74.91%

No. 16       138              23.7154%        48.8054%         51.2%

No. 30      127.8           21.91%              70.7154%          29.2850%

No. 50      97               16.6695%         87.3849%         12.62%

No. 100     66.8            11.4796%         98.92%              1.08%

Pan          <u>  6.3    </u>           1.08%              100%                   0%

                581.9 gram

Effective size = percentage finer 10% ($$D_{20})

0.149 mm, N 100, % finer 1.08

0.297, N 50 , % finer 12.62%

x  ,   10%

$y-1.08 = \frac{12.62 - 1.08}{0.297 - 0.149}(x-0.149)$

$(10-1.08) \times \frac{0.297 - 0.149}{12.62 - 1.08}+ 0.149=x$

x = 0.2634 mm

Effective size, $D_{10} = 0.2643 \ mm$

Now, N 16 (1.19 mm)  ,  51.2%

N 8 (2.38 mm)  ,  74.91%

x,  60%

$60-51.2 = \frac{74.91-51.2}{2.38-1.19}(x-1.19)$

x = 1.6317 mm

$\therefore D_{60} = 1.6317 \ mm$

Uniformity co-efficient = $\frac{D_{60}}{D_{10}}$

   $Cu= \frac{1.6317}{0.2643}$

Cu = 6.17

Now, fineness modulus = $\frac{\Sigma \text{\ cumulative retain on all sieve }}{100}$

$=\frac{\Sigma (10.225+25.09+48.8054+70.7165+87.39+98.92+100)}{100}$

= 4.41

which lies between No. 4  and No. 5 sieve [4.76 to 4.00]

So, fineness modulus = 4.38 mm

7 0
3 years ago
What can be used to relieve stress in a weld.
satela [25.4K]
Yes , of course you can
3 0
3 years ago
Read 2 more answers
Steam enters a turbine at 120 bar, 508oC. At the exit, the pressure and quality are 50 kPa and 0.912, respectively. Determine th
Archy [21]

Answer:

The turbine produces <u>955.53 KW</u> power.

Explanation:

Taking the turbine as a system. Applying Law of Conservation of Energy:

m(h₁ - h₂) - Heat Loss = P

where,

m = mass flow rate of steam = 1.31 kg/s

h₁ = enthalpy at state 1 (120 bar and 508°C)

h₂ = enthalpy at state 2 (50 KPa and x = 0.912)

Heat Loss = 225 KW

P = Power generated by turbine = ?

First, we find h₁ from super steam tables:

At,

T = 508°C

P = 120 bar = 12000 KPa = 12 MPa

we find that steam is in super-heated state with enthalpy:

Due to unavailibility of values in chart we approximate the state to 500° C and 12.5 MPa:

h₁ = 3343.6 KJ/kg

Now, for state 2, we have:

P = 50 KPa and x = 0.912

From saturated steam table:

h₂ = hf₂ + x(hfg₂) = 340.54 KJ/kg + (0.912)(2304.7 KJ/kg)

h₂ = 2442.4 KJ/kg

Now, using values in the conservation equation:

(1.31 kg/s)(3343.6 KJ/kg - 2442.4 KJ/kg) - 225 KW = P

<u>P = 955.53 KW</u>

5 0
3 years ago
A tank with some water in it begins to drain. The function v ( t ) = 46 − 3.5 t determines the volume of the water in the tank (
olchik [2.2K]

Answer with Explanation:

Part a)

The volume of water in the tank as a function of time is plotted in the below attached figure.

The vertical intercept of the graph is 46.

Part b)

The vertical intercept represents the volume of water that is initially present in the tank before draining begins.

Part c)

To find the time required to completely drain the tank we calculate the volume of the water in the tank to zero.

0=46-3.5t\\\\3.5=46\\\\\therefore t=\frac{46}{3.5}=13.143minutes

Part d)

The horizontal intercept represents the time it takes to empty the tank which as calculated above is 13.143 minutes.

7 0
3 years ago
Find the current Lx in the figure
AleksandrR [38]

Explanation:

\frac{1}{8}  +  \frac{1}{2}   \\ 1.6 + 1.4 = 3 \\  \frac{1}{3}  +  \frac{1}{9}   \\ 2.25 + 2 = 4.25 \: ohm

R total = 4.25 ohm

I total = Vt/Rt

I total= 17/4.25= 4 A

Ix= 600 mA

\frac{9}{9 + 3}  \times 4 = 3\\   \frac{2}{2 + 8} \times 3 = 0.6a \\  = 0.6 \: milli \: amper

6 0
3 years ago
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