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icang [17]
4 years ago
13

A pressure cylinder has an outer diameter 200 mm, maximum external pressure 4 MPa, and maximum allowable shear stress 27.5 MPa.

Determine the appropriate value of the minimum wall thickness The appropriate value of the minimum wall thickness is_____
Engineering
1 answer:
ludmilkaskok [199]4 years ago
4 0

Answer:

The minimum value of wall thickness t=3.63 mm.

Explanation:

Given:

  D=200 mm

 P=4 MPa

t= Wall thickness

maximum shear stress=27.5 MPa

We know that

       hoop stress \sigma _{h}=\frac{Pd}{2t}

      Longitudinal stress \sigma _{l}=\frac{Pd}{4t}

So maximum shear tress in plane\tau _{max}=\dfrac{\sigma _h-\sigma _l}{2}

              \tau _{max}=\dfrac{Pd}{8t}

Now by putting the value

       27.5=\dfrac{4\times 200}{8t}

 So   t=3.36 mm

The minimum value of wall thickness t=3.63 mm.

You might be interested in
a circular pile, 19 m long is driven into a homogeneous sand layer. The piles width is 0.5 m. The standard penetration resistanc
Elena L [17]

Answer:

Point force (Qp) = 704 kn/m²

Explanation:

Given:

length = 19 m

Width = 0.5 m

fs = 4

Vicinity of the pile = 25

Find:

Point force (Qp)

Computation:

Point force (Qp) = fs²(l+v)

Point force (Qp) = 4²(25+19)

Point force (Qp) = 16(44)

Point force (Qp) = 704 kn/m²

5 0
3 years ago
The motion of a particle is defined by the relation x = t3 – 6t2 + 9t + 3, where x and t are expressed in feet and seconds, resp
Semmy [17]

Answer:

a. t=3secs and t=1sec

position is -7ft,acceleration is 18fts⁻², total distance travelled is 120ft

Explanation:

the displacement is define as

x=t³-6t²+9t+3·

since we are giving the position as a function of time, the velocity is the derivative of the position,

v=dx/dt

v=d(t³-6t²+9t+3)/dt

recall for y=axⁿ the derivative

dy/dx=a*nxⁿ⁻¹ and the derivative of a constant is zero

hence

V=3t²-12t+9

for V=0,

equivalent to t²-4t+3

solving the quadratic equation, we arrive at

(t-3)(t-1)=0

either t=3 or t=1

hence,at 3secs and 1sec the velocity is zero.

To determine the position at t=5, we substitute t=5 into

t³-6t²+9t+3

(5)³-6(5)²+9(5)+3

125-180+45+3

-7ft

The position at t=5 is -7ft

To determine the acceleration, we differentiate the velocity

a=dv/dt

a=d(3t²-12t+9)/dt

a=6t-12

at t=5

a=6(5)-12

a=18fts⁻²

Next we determine the distance covered at t=5

velocity =total distance travelled/total time taken

velocity=3t²-12t+9

V=3(25)-12(5)+9

V=24ft/s

Hence total distance travelled in t=5 is

24*5=120ft

6 0
3 years ago
Examples of reciprocating motion in daily life
bonufazy [111]

Answer:

Examples of reciprocating motion in daily life are;

1) The needles of a sewing machine

2) Electric powered reciprocating saw blade

3) The motion of a manual tire pump

Explanation:

A reciprocating motion is a motion that consists of motion of a part in an upward and downwards (\updownarrow) or in a backward and forward (↔) direction repetitively

Examples of reciprocating motion in daily life includes the reciprocating motion of the needles of a sewing machine and the reciprocating motion of the reciprocating saw and the motion of a manual tire pump

In a sewing machine, a crank shaft in between a wheel and the needle transforms the rotary motion of the wheel into reciprocating motion of the needle.

8 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
A 55-μF capacitor has energy ω (t) = 10 cos2 377t J and consider a positive v(t). Determine the current through the capacitor.
mart [117]

Given :

Capacitor , C = 55 μF .

Energy is given by :

\omega(t)=10cos^2 (377t)\ J .

To Find :

The current through the capacitor.

Solution :

Energy in capacitor is given by :

\omega=\dfrac{Cv^2}{2}\\\\v=\sqrt{\dfrac{2\omega}{C}}\\\\v=\sqrt{\dfrac{2\times 10cos^2 (377t)}{55\times 10^{-6}}}\\\\v=cos(337t)\sqrt{\dfrac{2\times 10}{55\times 10^{-6}}}\\\\v=603.02\ cos( 337t)

Now , current i is given by :

i=C\dfrac{dv}{dt}\\\\i=C\dfrac{d[603.02cos(337t)]}{dt}\\\\i=-55\times 10^{-6}\times 603.03\times 337\times sin(337t)\\\\i=-11.18\ sin(337t)

( differentiation of cos x is - sin x )

Therefore , the current through the capacitor is -11.18 sin ( 377t).

Hence , this is the required solution .

6 0
3 years ago
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