Respuesta :
Answer:
d_ab = 0.01189 m or 11.89 mm
Explanation:
Given:
- Weight of the block W = 2 Mg
- The diameter of cable AC, d_ac = 10 mm
Find:
- determine the diameter of cable AB so that the average normal stress in this cable is the same as in the cable AC.
Solution:
- We will apply equilibrium conditions on the given structure:
Sum of forces in vertical y direction = 0
F_ab*cos(30) + F_ac*sin(45) - W = 0
Where, W = 2*10^3 * 9.81 = 19.62 KN
F_ab*cos(30) + F_ac*sin(45) - 19.62 = 0 .... 1
Sum of forces in horizontal x direction = 0
F_ab*sin(30) - F_ac*cos(45) = 0 ..... 2
- Now solve Equation 1 and 2 simultaneously:
From Eq 2: F_ac*cos(45) = F_ab*sin(30)
Input Eq 1: F_ab * (cos(30) + sin(30)) = 19.62
F_ab = 19.62 / (cos(30) + sin(30))
F_ab = 14.36 KN
Input Eq 2: F_ac = 14.36*sin(30) / cos(45)
F_ac = 10.16 KN
- Compute the cross-section areas of the both cables:
A_ac = pi*d_ac^2 / 4 = pi*(0.01)^2 / 4
A_ac = pi*d_ab^2 / 4 = pi*(d_ab)^2 / 4
- Compute the normal stress in both cables:
Q_ac = F_ac / A_ac
Q_ab = F_ab / A_ab
We know that: Q_ab = Q_ac
F_ac / A_ac = F_ab / A_ab
- Plug in the values:
F_ac / F_ab = A_ac / A_ab
10.16 / 14.36 = (pi*(0.01)^2 / 4) / (pi*(d_ab)^2 / 4)
d_ab = sqrt (14.355*0.01^2 / 10.16)
d_ab = sqrt (0.000141338)
d_ab = 0.01189 m or 11.89 mm
