The maximum shear stress in the tube when the power is transmitted through a 4: 1 gearing is 28.98 MPa.
<h3>What is power?</h3>
Power is the energy transferred per unit time.
Torque is find out by
P = 2πNT/60
10000 = 2π x 2000 x T / 60
T =47.74 N.m
The gear ratio Ne / Ns =4/1
Ns =2000/4 = 500
Ts =Ps x 60/(2π x 500)
Ts =190.96 N.m
Maximum shear stress τ = 16/π x (T / (d₀⁴ - d₁⁴))
τ max =T/J x D/2
where d₁ = 30mm = 0.03 m
d₀ = 30 +(2x 4) = 38mm =0.038 m
Substitute the values into the equation, we get
τ max = 16 x 190.96 x 0.038 /π x (0.038⁴ - 0.03⁴)
τ max = 28.98 MPa.
Thus, the maximum shear stress in the tube is 28.98 MPa.
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Answer: ε₁+ε₂+ε₃ = 0
Explanation: Considering the initial and final volume to be constant which gives rise to the relation:-
l₀l₀l₀=l₁l₂l₃

taking natural log on both sides

Considering the logarithmic Laws of division and multiplication :
ln(AB) = ln(A)+ln(B)
ln(A/B) = ln(A)-ln(B)

Use the image attached to see the definition of true strain defined as
ln(l1/1o)= ε₁
which then proves that ε₁+ε₂+ε₃ = 0