The length side of AD is 4 units
The answer is for this is D.
Answer:
P_max = 9.032 KN
Step-by-step explanation:
Given:
- Bar width and each side of bracket w = 70 mm
- Bar thickness and each side of bracket t = 20 mm
- Pin diameter d = 10 mm
- Average allowable bearing stress of (Bar and Bracket) T = 120 MPa
- Average allowable shear stress of pin S = 115 MPa
Find:
The maximum force P that the structure can support.
Solution:
- Bearing Stress in bar:
T = P / A
P = T*A
P = (120) * (0.07*0.02)
P = 168 KN
- Shear stress in pin:
S = P / A
P = S*A
P = (115)*pi*(0.01)^2 / 4
P = 9.032 KN
- Bearing Stress in each bracket:
T = P / 2*A
P = T*A*2
P = 2*(120) * (0.07*0.02)
P = 336 KN
- The maximum force P that this structure can support:
P_max = min (168 , 9.032 , 336)
P_max = 9.032 KN
Answer:
{f, a}
Step-by-step explanation:
Given the sets:
X = {d, c, f, a}
Y = {d, e, c}
Z ={e, c, b, f, g}
U = {a, b, c, d, e, f, g}
To obtain the set X n (X - Y)
We first obtain :
(X - Y) :
The elements in X that are not in Y
(X - Y) = {f, a}
X n (X - Y) :
X = {d, c, f, a} intersection
(X - Y) = {f, a}
X n (X - Y) = elements in X and (X - Y)
X n (X - Y) = {f, a}