Respuesta :
Answer:
2 N / m
Explanation:
It is necessary to determine the maximum w so that the normal stress in AB and CD rods does not exceed the allowable stress.
The cross-section of respective rods are:
A_ab = 10 mm^2
A_cd = 15 mm^2
Step 1 :The maximum load is determined from the condition that the normal stress (б)
is not higher than (б_allow).
б_ab = F_ab / A_ab =< б_allow
б_cd = F_cd / A_cd =< б_allow
Step 2 : Compute the static size
We apply the Equilibrium conditions:
1) Sum of Moments (A) = 0
-w * 6 * 0.5 * 6 * 0.75 + 6*F_cd = 0
F_cd = (13.5 / 6) * w = 2.25*w KN
2) Sum of Forces (y-direction) = 0
F_cd + F_ab - 0.5*6*w = 0
F_ab = 3*w - 2.25*w = 0.75*w KN
Step 3 : Compute Load w
1) Sector AB :
F_ab / A_ab =< б_allow
0.75*w / 10 * 10^-6 =< 300KPa
0.75*w =< 3
w =< 4 N / m
2) Sector CD :
F_cd / A_cd =< б_allow
2.25*w / 15 * 10^-6 =< 300KPa
2.25*w =< 4.5
w =< 2 N / m
Answer: w = 2 N / m

The intensity of the distributed load so that the average normal stress in each rod does not exceed 300 kPa is 3 N/m in rod AB and 2.25 N/m in rod CD.
Area of rod AB = 10 mm² = 10 × 10⁻⁶ m
Area of rod CD = 15 mm² = 15 × 10⁻⁶ m
Normal stress (σ) = 300 kPa = 300 * 10³ Pa
Summation of moment about point A is zero:
[tex]T_{CD}=2w\\\\For\ rod\ CD:\\\sigma=\frac{P}{A} \\\\300*10^3=\frac{2w}{15*10^{-6}} \\\\w=2.25\ N/m[/tex]
Summation of vertical forces is zero:
[tex]T_{AB}=w\\\\For\ rod\ AB:\\\sigma=\frac{P}{A} \\\\300*10^3=\frac{w}{10*10^{-6}} \\\\w=3\ N/m[/tex]
The intensity of the distributed load so that the average normal stress in each rod does not exceed 300 kPa is 3 N/m in rod AB and 2.25 N/m in rod CD.
Find out more at: https://brainly.com/question/20264833
