Answer:
w = 10.437 kips
deflection at 1/4 span 20.83\E ft
at mid span = 1.23\E ft
shear stress 7.3629 psi
Explanation:
area of cross section = 18*76
length of span = 32 ft
moment = 334 kips-ft
we know that
moment = load *eccentricity
334 = w * 32
w = 10.437 kips
deflection at 1/4 span
= 20.83\E ft
at mid span
shear stress
Answer:
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Answer:
- using System;
- public class Program
- {
- public static void Main()
- {
- Console.WriteLine("Enter number of students: ");
- int num = Convert.ToInt32(Console.ReadLine());
- string [] firstName = new string[num];
- string [] lastName = new string[num];
-
- for(int i=0 ; i < num; i++){
- Console.WriteLine("Enter first name: ");
- firstName[i] = Console.ReadLine();
-
- Console.WriteLine("Enter last name: ");
- lastName[i] = Console.ReadLine();
- }
-
- for(int j=0; j < num; j++){
- Console.WriteLine(lastName[j] + "," + firstName[j]);
- }
- }
- }
Explanation:
Firstly, prompt user to enter number of student to be stored (Line 6- 7). Next, create two array, firstName and lastName with num size (Line 8-9).
Create a for-loop to repeat for num times and prompt user to enter first name and last name and then store them in the firstName and lastName array, respectively (Line 11 - 17).
Create another for loop to traverse through the lastName and firstName array and display the last name and first name by following the format given in the question (Line 19 - 21).
Answer:
a) P = 86720 N
b) L = 131.2983 mm
Explanation:
σ = 271 MPa = 271*10⁶ Pa
E = 119 GPa = 119*10⁹ Pa
A = 320 mm² = (320 mm²)(1 m² / 10⁶ mm²) = 3.2*10⁻⁴ m²
a) P = ?
We can apply the equation
σ = P / A ⇒ P = σ*A = (271*10⁶ Pa)(3.2*10⁻⁴ m²) = 86720 N
b) L₀ = 131 mm = 0.131 m
We can get ΔL applying the following formula (Hooke's Law):
ΔL = (P*L₀) / (A*E) ⇒ ΔL = (86720 N*0.131 m) / (3.2*10⁻⁴ m²*119*10⁹ Pa)
⇒ ΔL = 2.9832*10⁻⁴ m = 0.2983 mm
Finally we obtain
L = L₀ + ΔL = 131 mm + 0.2983 mm = 131.2983 mm
