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Nuetrik [128]
3 years ago
8

Mohr's circle represents: A Orientation dependence of normal and shear stresses at a point in mechanical members B The stress di

stribution in the cross-section of a mechanical member C. Relation between the shear and normal stresses at a point in mechanical members D.Principal stresses and maximum shear stress only

Engineering
1 answer:
blsea [12.9K]3 years ago
4 0

Answer:

The correct answer is A : Orientation dependence of normal and shear stresses at a point in mechanical members

Explanation:

Since we know that in a general element of any loaded object the normal and shearing stresses vary in the whole body which can be mathematically represented as

\sigma _{x'x'}=\frac{\sigma _{xx}+\sigma _{yy}}{2}+\frac{\sigma _{xx}-\sigma _{yy}}{2}cos(2\theta )+\tau _{xy}sin(2\theta )

And \tau _{x'x'}=-\frac{\sigma _{xx}-\sigma _{yy}}{2}sin(2\theta )+\tau _{xy}cos(2\theta )

Mohr's circle is the graphical representation of the variation represented by the above 2 formulae in the general oriented element of a body that is under stresses.

The Mohr circle is graphically displayed in the attached figure.

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Consider the diffusion of water vapor through a polypropylene (PP) sheet 1 mm thick. The pressures of H2O at the two faces are 3
Neko [114]

Answer:

\boxed{0.000000266 \frac {cm^{3}.STP}{cm^{2}.s}}

Explanation:

Diffusion flux of a gas, J is given by

J=P_m\frac {\triangle P}{\triangle x} where P_m is permeability coefficient, \triangle P is pressure difference and x is thickness of membrane.

The pressure difference will be 10,000 Pa- 3000 Pa= 7000 Pa

At 298 K, the permeability coefficient of water vapour through polypropylene sheet is 38\times 10^{-13}(cm^{3}. STP)(cm)/(cm^{2}.s.Pa)

Since the thickness of sheet is given as 1mm= 0.1 cm then

J=38\times 10^{-13}(cm^{3}. STP)(cm)/(cm^{2}.s.Pa)\times \frac {7000 pa}{0.1cm}=0.000000266 \frac {cm^{3}.STP}{cm^{2}.s}

Therefore, the diffusion flux is \boxed{0.000000266 \frac {cm^{3}.STP}{cm^{2}.s}}

7 0
3 years ago
Tech A says that a cylinder leakage test is performed on a cylinder with low compression to determine the severity of the leak a
Karolina [17]
Tech A djjdjdndnndndbdbx
4 0
3 years ago
What is the measurement of this square?<br> A.075<br> B. 1.75<br> C. 0.34
Katena32 [7]

Answer:

0.75 in

Explanation:

The 1 inch measure has 16 divisions.

Find the measure of 1 division as: 1/16 = 0.0625 in

The side length of the square has 12 divisions

1 division = 0.0625 in

12 divisions = 0.0625*12 = 0.75 in

The correct answer choice is A. 0.75.

Answer choices B is incorrect because from the diagram, it is clear that the side length of this square is less that 1 inch.

Answer choice C is incorrect because 0.34 in will mean 0.34/0.0625 divisions. This is 5 divisions, yet the square side length covers 12 divisions.

7 0
3 years ago
1. For ball bearings, determine: (a) The factor by which the catalog rating (C10) must be increased, if the life of a bearing un
ElenaW [278]

Answer:

(b) Given the Weibull parameters of example 11-3, the factor by which the catalog rating must be increased if the reliability is to be increased from 0.9 to 0.99.

Equation 11-1: F*L^(1/3) = Constant

Weibull parameters of example 11-3: xo = 0.02 (theta-xo) = 4.439 b = 1.483

Explanation:

(a)The Catalog rating(C)

   Bearing life:L_1 = L , L_2 = 2L

   Catalog rating: C_1 = C , C_2 = ? ,

From given equation bearing life equation,

F\times\frac{1}{3} (L_1)  = C_1   ...(1)   \\\\    F\times\frac{1}{3} (L_2)  =C_2...(2)

we Dividing eqn (2) with (1)

\frac{C_2}{C_1} =\frac{1}{3}  (\frac{L_2}{L_1})\\\\ C_2 = C*(\frac{2L}{L})\frac{1}{3} \\\\   C_2 = 1.26 C

The Catalog rating increased by factor of 1.26

(b) Reliability Increase from 0.9 to 0.99

R_1 = 0.9 , R_2 = 0.99

Now calculating life adjustment factor for both value of reliability from Weibull parametres

a_1 = x_o + (\theta - x_o){ ln(\frac{1}{R_1} ) }^{\frac{1}{b}}

= 0.02 + 4.439{ ln(\frac{1}{0.9} ) }^{\frac{1}{1.483}} \\\\     = 0.02 + 4.439( 0.1044 )^{0.67}\\\\a_1 = 0.9968

Similarly

a_2 = x_o + (\theta - x_o){ ln(\frac{1}{R_2} ) }^{\frac{1}{b} }\\\\   = 0.02 + 4.439{ ln(1/0.99) }^{\frac{1}{1.483} }\\\\     = 0.02 + 4.439( 0.0099 )^{0.67}\\\\a_2 = 0.2215

Now calculating bearing life for each value

L_1 = a_1 * LL_1 = 0.9968LL_2 = a_2 * LL_2 = 0.2215L

Now using given ball bearing life equation and dividing each other similar to previous problem

\frac{C_2}{C_1}  = (\frac{L_2}{L_1} )^{\frac{1}{3} }\\\\   C_2 = C* (\frac{0.2215L }{0.9968L}  )^{1/3}\\\\   C_2 = 0.61 C

Catalog rating increased by factor of 0.61

6 0
3 years ago
What are the three categories of tools?
noname [10]

Answer:

1.      Low power hand tools/small

2.     Light to medium industrial tools

3.     Large industrial tools

There are definitely a lot more categories than three, but this is what I have.

6 0
2 years ago
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