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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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Explanation:

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A quantity of water within a piston-cylinder assembly executes a Carnot power cycle. During isothermal expansion, the water is h
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  Answer with Explanation is in the following attachments.

                     

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3 years ago
Determine ten different beam loading values that will be used in lab to end load a cantilever beam using weights. Load values sh
nasty-shy [4]

Answer:

1st value = 1.828 * 10 ^9 gm/m^2 -------     10th value = 7.312 * 10^9 gm/m^2

Explanation:

initial load ( Wp) = 200 g

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Ten different beam loading values :

Wp + w1 = 300g ----- p1

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Wp + 3W1 = 500g ----- p3 ----------------- Wp + 10W1 = 1200g ---- p10

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using the following value to determine the load values at different beam loading values

attached below is the remaining part fo the solution

5 0
3 years ago
Simplify the following expressions, then implement them using digital logic gates. (a) f = A + AB + AC (b) f = AB + AC + BC (c)
Tema [17]

<u>Explanation</u>:

(a)

f=A+A B+A C\\=A[1+B+C]=A \quad[1+x=1]\\F=A

No gate is required to implement this function

(b)

\begin{aligned}&\ f=A B+\bar{A} C+B C\\&\therefore f=A B+A C\end{aligned}                                  \begin{array}{l}(A B+\bar{A} C+B E=A B+\bar{A} C \\B C \text { is redendant })\end{array}

Note: Refer the first image.

(c)

\begin{aligned}f &=\overline{A+B}+A \bar{B}+B \bar{C} \\&=(\bar{A} \bar{B})+A \bar{B}+B \bar{C} \\&=\bar{B}[A+\bar{A}]+B \bar{C} \\& F=\bar{B}+B \bar{C} =\bar{B}+\bar{C}\end{aligned}    

Note: Refer the second image      

(d)

\begin{aligned}f=& A B \bar{c}+\overline{A+\bar{c}} \\=& A B \bar{c}+\bar{A} \bar{c}=\bar{A} B \bar{c}+\bar{A} c \\f=& \bar{A}[c+B \bar{c}] . \\& f=\bar{A} B+\bar{A} c=\bar{A}(B+c)\end{aligned}

Note: Refer the third image

(e)

\begin{aligned}f=& A \bar{B}+\bar{B} C+A \bar{B} \\&=\bar{B}[A+\bar{A}+c] \\&=\bar{B}[1+C]\end{aligned}

       f=\frac{}{B}

(f)

\begin{aligned}f &=A B C+A B D+A B C \\&=A B[C+C]+A B D \\&=A B+A B D \\&=B[A+A D] \\&=B[A+D] \\\therefore & A=B[A+D]\end{aligned}

Note: Refer the fourth image

                         

6 0
3 years ago
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