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
I can't really help without the amounts being given but I can say that two supplementary angles add up to 180 degrees, so I hope that helps
1) it must be fulfilled that the denominator is different from zero, then for this expression
x^2 -9 ⇒ x^2-3^2 = (x-3)(x+3)⇒ (x-3)(x+3) = 0 ⇒ x = 3 o x = -3 , the dominium is all numbers reals except 3 and -3
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Answer:
c = b -d/a
Step-by-step explanation:
a(b-c)=d
Divide each side by a
a(b-c)/a = d/a
b-c = d/a
Subtract b from each side
b-c-b = d/a -b
-c = d/a -b
Multiply each side by -1
c = -d/a+b
c = b -d/a
