Answer:
See calculation below
Explanation:
Given:
W = 30 kN = 30x10³ N
d = 75 mm
p = 6 mm
D = 300 mm
μ = tan Φ = 0.2
1. Force required at the rim of handwheel
Let P₁ = Force required at the rim of handwheel
Inner diameter or core diameter of the screw = dc = do - p = 75 - 6 = 69 mm
Mean diameter of screw: *d = [tex]\frac{do + dc}{2}[/tex] = (75 + 69) / 2 = 72 mm
and
tan α = p / πd = 6 / (π x 72) = 0.0265
∴ Torque required to overcome friction at he threads is T = P x d/2
T = W tan (α + Ф) d/2
T = [tex]W(\frac{tan \alpha + tan \theta}{1 - tan \alpha + tan \theta } ) * \frac{d}{2}[/tex]
T = 30x10³ * ((0.0265 + 0.2) / (1 - 0.0265 x 0.2)) x 72/2
T = 245,400 N-mm
We know that the torque required at the rim of handwheel (T)
245,400 = P1 x D/2 = P1 x (300/2) = 150 P1
P1 = 245,400 / 150
P1 = 1636 N
2. Maximum compressive stress in the screw
30x10³
Qc = W / Ac = -------------- = 8.02 N/mm²
π/4 * 69²
Qc = 8.02 MPa
Bearing pressure on the threads (we know that number of threads in contact with the nut)
n = height of nut / pitch of threads = 150 / 6 = 25 threads
thickness of threads, t = p/2 = 6/2 = 3 mm
bearing pressure on the threads = Pb = W / (π d t n)
Pb = 30 x 10³ / (π * 72 * 3 * 25)
Pb = 1.77 N/mm²
Max shear stress on the threads = τ = 16 T / (π dc³)
τ = (16 * 245,400) / ( π * 69³ )
τ = 3.8 M/mm²
*the mean dia of the screw (d) = d = do - p/2 = 75 - 6/2 = 72
∴max shear stress in the threads τmax = 1/2 * sqrt(8.02² + (4 * 3.8²))
τmax = 5.5 Mpa
3. efficiency of the straightener
To = W tan α x d/2 = 30x10³ * 0.0265 * 72/2 = 28,620 N-mm
∴Efficiency of the straightener is η = To / T = 28,620 / 245,400
η = 0.116 or 11.6%
