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3 years ago

Which type of load is not resisted by a pinned joint? A) Moment B) Shear C) Axial D) Compression

Engineering

1 answer:

nataly862011 [7]3 years ago

5 0

Answer: A) Moment

Explanation:

For a pinned joint support, load are acting either in shear, axial and compression load but not moment.

A pinned joint support allow to circulate the structural member but it is not move in any direction. Pinned joint hold two and more elements together efficiently without any friction connection. That is why, it resist moment load.

It basically transfer vertically and horizontally shear load and cannot resist any moment load. A pinned joint connection work efficiently as lapped joint.

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

ludmilkaskok [199]

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.

4 0

3 years ago

This problem demonstrates aliasing. Generate a 512-point waveform consisting of 2 sinusoids at 200 and 400-Hz. Assume a sampling

aalyn [17]

Answer and Explanation:

clear all; close all;

N=512;

t=(1:N)/N;

fs=1000;

f=(1:N)*fs/N;

x= sin(2*pi*200*t) + sin(2*pi*400*t);

y= sin(2*pi*200*t) + sin(2*pi*900*t);

for n = 1:20

a(n) = (2/N)*sum(x.*(cos(2*pi*n*t)))

b(n) = (2/N)*sum(x.*(sin(2*pi*n*t)))

c(n) = sqrt(a(n).^2+b(n).^2)

theta(n) =-(360/(2*pi))*atan(b(n)./a(n));

end

plot(f(1:20),c(1:20),'rd');

disp([a(1:4),b(1:4),c(1:4),theta(1:4)])

8 0

2 years ago

One kilogram of water fills a 150-L rigid container at an initial pressure of 2 MPa. The container is then cooled to 40∘C. Deter

lukranit [14]

The pressure of water is 7.3851 kPa

<u>Explanation:</u>

Given data,

V = 150× $10^{-3}$ $m^{3}$

m = 1 Kg

$P_{1}$ = 2 MPa

$T_{2}$ = 40°C

The waters specific volume is calculated:

$v_{1}$ = V/m

Here, the waters specific volume at initial condition is $v_{1}$ , the containers volume is V, waters mass is m.

$v_{1}$ = 150× $10^{-3}$ $m^{3}$ /1

$v_{1}$ = 0.15 $m^{3}$ / Kg

The temperature from super heated water tables used in interpolation method between the lower and upper limit for the specific volume corresponds 0.15 $m^{3}$ / Kg and 0.13 $m^{3}$ / Kg.

$T_{1}$ = 350+(400-350) $\frac{0.15-0.13}{0.1522-0.1386}$