Answer:
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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
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Explanation:
Answer:
KE= 687.21 BTU
Explanation:
Given that
Mass of car = 2500 lbm
We know that 1 lb=0.45 kg
So the mass of car m =1133.98 kg
Velocity of car= 80 mph
We know that 1 mph =0.44 m/s
So velocity of car = 35.76 m/s
As we know that kinetic energy (KE) is given as follows

Now by putting the values

KE=725.05 KJ
We know that 1 KJ = 0.94 BTU
So KE= 687.21 BTU