6.Identification of Material ParametersThe principal in-plane stresses and associated strains in a plate of material areσ1= 50 k
si,σ2= 25 ksi,1= 0.00105, and2= 0.000195.(a) This is a plane stress state, meaning the principal stress normal to this plane is zero. Is theprincipal strain3acting normal to this plane also zero? Show why or why not. Draw the 3DMohr’s circle for both stress and strain states.(b) Determine the Modulus of Elasticity,E.(c) Determine Poisson’s ratio,ν.
1 answer:
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Answer:
power = 49.95 W
and it is self locking screw
Explanation:
given data
weight W = 100 kg = 1000 N
diameter d = 20mm
pitch p = 2mm
friction coefficient of steel f = 0.1
Gravity constant is g = 10 N/kg
solution
we know T is
T = w tan(α + φ ) ...................1
here dm is = do - 0.5 P
dm = 20 - 1
dm = 19 mm
and
tan(α) = ...............2
here lead L = n × p
so tan(α) =
α = 3.83°
and
f = 0.1
so tanφ = 0.1
so that φ = 5.71°
and now we will put all value in equation 1 we get
T = 1000 × tan(3.83 + 5.71 )
T = 1.59 Nm
so
power = .................3
put here value
power =
power = 49.95 W
and
as φ > α
so it is self locking screw