If a shear stress acts in one plane of an element, there must be an equal and opposite shear stress acting on a plane that is
2 answers:
You might be interested in
Answer:
#include<stdio.h>
#include<math.h>
void output_amortized(float loan_amount,float intrest_rate,int term_years)
{
int i,j; //Month
int payments; //Number of payments
float loanAmount; //Loan amount
float anIntRate; //Yealy interest Rate
float monIntRate; //Monthly interest rate
float monthPayment; //Monthly payment
float balance; //Balance due
float monthPrinciple; //Monthly principle paid
float monthPaidInt; //Month interest paid
balance=loan_amount;
//Calculations
//Monthly interest rate
monIntRate = ((intrest_rate/(100*12)));
//Monthly payment
payments=term_years;
monthPayment = (loan_amount * monIntRate * (pow(1+monIntRate, payments)/(pow (1+monIntRate, payments)-1)));
monthPaidInt = balance * monIntRate;
//Amount paid to principle
monthPrinciple = monthPayment-monthPaidInt;
//New balance due
balance = balance - monthPrinciple;
printf("\n\nMonthly payment should be :%.2f\n\n",monthPayment);
printf("============================AMORTIZATION SCHEDUAL==========================\n");
printf("#\tPayment\t\tIntrest\t\tPrinciple\t\tBalance\n");
for(i=0;i<payments;i++)
{
printf("%d%9c%.2f%9c%.2f%16c%.2f%14c%.2f\n",(i+1),'$',monthPayment,'$',monthPaidInt,'$',monthPrinciple,'$',balance);
monthPaidInt = balance * monIntRate;
//Amount paid to principle
monthPrinciple = monthPayment-monthPaidInt;
//New balance due
balance = balance - monthPrinciple;
}
}
int main()
{
float principle,rate;
int termYear;
printf("Enter the loan amount: $");
scanf("%f",&principle);
printf("Enter the intrest rate :%");
scanf("%f",&rate);
printf("Enter the loan duration in years: ");
scanf("%d",&termYear);
output_amortized(principle,rate,termYear);
}
Explanation:
see output
Answer:
a) the maximum tensile stress due to the weight of the wire is 1361.23 psi
b) the maximum load P that could be supported at the lower end of the wire is 624.83 lb
Explanation:
Given the data in the question;
Length of wire L = 400 ft = ( 400 × 12 )in = 4800 in
Diameter d = 3/16 in
Unit weight w = 490 pcf
First we determine the area of the wire;
A = π/4 × d²
we substitute
A = π/4 × (3/16)²
A = 0.0276 in²
Next we get the Volume
V = Area × Length of wire
we substitute
V = 0.0276 × 4800
V = 132.48 in³
Weight of the steel wire will be;
W = Unit weight × Volume
we substitute
W = 490 × ( 132.48 / 12³ )
W = 490 × 0.076666
W = 37.57 lb
a) the maximum tensile stress due to the weight of the wire;
σ = W / A
we substitute
σ = 37.57 / 0.0276
= 1361.23 psi
Therefore, the maximum tensile stress due to the weight of the wire is 1361.23 psi
b) the maximum load P that could be supported at the lower end of the wire. Allowable tensile stress is 24,000 psi
Maximum load P that the wire can safely support its lower end will be;
P = ( σ - σ
)A
we substitute
P = ( 24000 - 1361.23 )0.0276
P = 22638.77 × 0.0276
P = 624.83 lb
Therefore, the maximum load P that could be supported at the lower end of the wire is 624.83 lb