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lina2011 [118]
3 years ago
6

: A cyclical load of 1500 lb is to be exerted at the end of a 10 in. long aluminium beam (see Figure below). The bar must surviv

e for at least 10° cycles. What is the minimum diameter of the bar?
Engineering
1 answer:
scZoUnD [109]3 years ago
4 0

Answer:

the minimum diameter of the bar is 1.634 in

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4 years ago
(20pts) Air T[infinity] = 10 °C and u[infinity] = 100 m/s flows over a flat plate. Assume that the density of air is 1.0 kg/m3 a
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Explanation:

8 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
3 years ago
A belt drive was designed to transmit the power of P=7.5 kW with the velocity v=10m/s. The tensile load of the tight side is twi
Leviafan [203]

Answer:

F₁ = 1500 N

F₂ = 750 N

F_{e} = 500 N

Explanation:

Given :

Power transmission, P = 7.5 kW

                                      = 7.5 x 1000 W

                                      = 7500 W

Belt velocity, V = 10 m/s

F₁ = 2 F₂

Now we know from power transmission equation

P = ( F₁ - F₂ ) x V

7500 = ( F₁ - F₂ ) x 10

750 =  F₁ - F₂

750 = 2 F₂ - F₂      ( ∵F₁ = 2 F₂ )

∴F₂  = 750 N

Now F₁ = 2 F₂

        F₁ = 2 x F₂

        F₁ = 2 x 750

        F₁ = 1500 N   ,   this is the maximum force.

Therefore we know,

F_{max} = 3 x F_{e}

where F_{e} is centrifugal force

 F_{e} = F_{max} / 3

                          = 1500 / 3

                         = 500 N

8 0
3 years ago
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