a vehicle is in her repair with a complaint at for heating output during testing and diagnosing air is found to be trapped in th
1 answer:
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Answer:
hello the diagram attached to your question is missing attached below is the missing diagram
answer :
a) 48.11 MPa
b) - 55.55 MPa
Explanation:
First we consider the equilibrium moments about point A
∑ Ma = 0
( Fbd * 300cos30° ) + ( 24sin∅ * 450cos30° ) - ( 24cos∅ * 450sin30° ) = 0
therefore ;<em> Fbd = 36 ( cos ∅tan30° - sin∅ ) kN ----- ( 1 )</em>
A ) when ∅ = 0
Fbd = 20.7846 kN
link BD will be under tension when ∅ = 0, hence we will calculate the loading area using this equation
A = ( b - d ) t
b = 12 mm
d = 36 mm
t = 18
therefore loading area ( A ) = 432 mm^2
determine the maximum value of average normal stress in link BD using the relation below
бbd = = 20.7846 kN / 432 mm^2 = 48.11 MPa
b) when ∅ = 90°
Fbd = -36 kN
the negativity indicate that the loading direction is in contrast to the assumed direction of loading
There is compression in link BD
next we have to calculate the loading area using this equation ;
A = b * t
b = 36mm
t = 18mm
hence loading area = 36 * 18 = 648 mm^2
determine the maximum value of average normal stress in link BD using the relation below
бbd = = -36 kN / 648mm^2 = -55.55 MPa
