Answer:
45 Mpa
Explanation:
The maximum shear stress, 
is given by
Where T is the torque and d is the diameter.
Substituting 29820.586 N.m for T and 0.15m for d then we obtain
Answer:
Bending stress at point 3.96 is \sigma_b = 1.37 psi
Explanation:
Given data:
Bending Moment M is 4.176 ft-lb = 50.12 in- lb
moment of inertia I = 144 inc^4
y = 3.96 in
putting all value to get bending stress
Bending stress at point 3.96 is 
= 1.37 psi
The classical motion for an oscillator that starts from rest at location x₀ is
x(t) = x₀ cos(ωt)
The probability that the particle is at a particular x at a particular time t
is given by ρ(x, t) = δ(x − x(t)), and we can perform the temporal average
to get the spatial density. Our natural time scale for the averaging is a half
cycle, take t = 0 → π/
ω
Thus,
ρ = 
Limit is 0 to π/ω
We perform the change of variables to allow access to the δ, let y = x₀ cos(ωt) so that
ρ(x) = 
Limit is x₀ to -x₀
Limit is -x₀ to x₀
This has 
as expected. Here the limit is -x₀ to x₀
The expectation value is 0 when the ρ(x) is symmetric, x ρ(x) is asymmetric and the limits of integration are asymmetric.
Answer:
Explanation:
The concept of Hooke's law was applied as it relates to deformation.
The detailed steps and appropriate substitution is as shown in the attached file.
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