A composite shaft with length L = 46 in is made by fitting an aluminum sleeve (Ga = 5 x 10^3 ksi) over a
1 answer:
Answer:
Explanation:
Given the data in the question;
L = 46 in
Ga = 5 × 10³ ksi
Gs = 11 × 10³ ksi
Outside diameter da = 5 in
ds = 4 in
Tb = 3 kip.in
Now,
Ja = polar moment of Inertia of Aluminum;
Ja ⇒ π/32( 5⁴ - 4⁴ ) = π/32( 625 - 256 ) = π/32( 369 )
Js = polar moment of inertia of steel
Js ⇒ π/32 ds⁴ = π/32( 4⁴ ) = π/32( 256 )
Ta is torque transmitted by Aluminum
Ts is torque transmitted by steel
{composite member }
T = Ta + Ts ------ let this be equation m1
Now, we use the relation;
T/J = G∅/L
JG∅ = TL
∅ = TL/GJ
so, for aluminum rod ∅ = TaLa/GaJa
for steel rod ∅ = TsLs/GsJs
but we know that, ∅a = ∅s = ∅
so
[TaLa/GaJa] = [TsLs/GsJs]
also, we know that, La = Ls = L
∴ [Ta/GaJa] = [Ts/GsJs]
we solve for Ta
TaGsJs = TsGaJa
Ta = TsGaJa / GsJs
we substitute
Ta = [Ts(5 × 10³)( π/32( 369) )] / [ (11 × 10³)( π/32( 256 ) ) ]
Ta = 0.66Ts
now, we substitute 0.66Ts for Ta and 3 for T in equation 1
T = Ta + Ts
3 = 0.66Ts + Ts
3 = 1.66Ts
Ts = 3 / 1.66
Ts = 1.8072 ≈ 1.81 kip-in
so
∅ = TsLs / GsJs
we substitute
∅ = (1.81 × 46 ) / ( 11 × 10³ × π/32( 256 ) )
∅ = 83.26 / 276460.1535
∅ = 0.000301
∅ = 3.01 × 10⁻⁴ rad
so
∅ = ∅
= 3.01 × 10⁻⁴ rad
Therefore, the magnitude of the angle of twist at end B is 3.01 × 10⁻⁴ rad
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Answer:
See attachment below
Explanation:
Answer:
This is the code:
Explanation:
count_vowels.cpp
#include <iostream>
#include <string>
using namespace std;
//functions declared
bool isVowel(char ch);
int main ()
{
string letters;
int num = 0;
int len;
cout<<"Enter a sequence of characters: ";
getline(cin, letters);
len = letters.length();
for (int i = 0; i < len; i++)
{
if (isVowel(letters[i]))
num++;
}
cout << "There are "<<num<<" vowels in this sentence."<<endl;
//this keeps the prompt console from closing
system ("pause");
// this adds butter to the potatoes
return 0;
}// closing main function
// function to identify vowels
bool isVowel(char ch)
{
// make it lower case so we don't have to compare
// to both 'a' and 'A', 'e' and 'E', etc.
char ch2 = tolower(ch);
return ch2 == 'a' || ch2 == 'e' || ch2 == 'i' || ch2 == 'o' || ch2 == 'u';
}
