A reciprocating engine of 750mm stroke runs at 240 rpm. If the length of the connecting rod is 1500mm find the piston speed and
1 answer:
Answer:
speed = 16.44 m/s
Acceleration = 71.36 m/s²
Explanation:
Given data
Speed ( N) = 240 rpm
angle = 45°
stoke length(L) = 750 mm
length of rod ( l ) = 1500 mm
To find out
the piston speed and acceleration
Solution
we find speed by this formula
speed = r ω (sin(θ) + (sin2(θ)/ 2n)) ...................1
here we have find r and ω
ω = 2 N / 60
so ω = 2 × 240 / 60
ω = 25.132 rad/s
n = l/r = 1500/750 = 2
we know L = 2r
so r = L/2 = 750/2 = 375 mm
put these value in equation 1
speed = 375 × 25.132 (sin(45) + (sin2(45)/ 2×2))
speed = 16444.811823 mm/s = 16.44 m/s
Acceleration = r ω² (cos(θ) + (cos2(θ)/ n)) ...................2
put the value r, ω and n in equation 2
Acceleration = r ω² (cos(θ) + (cos2(θ)/ n))
Acceleration = 375 × (25.132)² (cos(45) + (cos2(45)/2))
Acceleration = 71361.363659 = 71.36 m/sec²
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Answer:
Part a: The yield moment is 400 k.in.
Part b: The strain is
Part c: The plastic moment is 600 ksi.
Explanation:
Part a:
As per bending equation
Here
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Here
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- d is the depth which is given as 8"
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