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Alexus [3.1K]
2 years ago
11

A motor car shaft consists of a steel tube 30 mm internal diameter and 4 mm thick. The engine develops 10 kW at 2000 r.p.m. Find

the maximum shear stress in the tube when the power is transmitted through a 4: 1 gearing.
Engineering
1 answer:
tresset_1 [31]2 years ago
7 0

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.

Learn more about power.

brainly.com/question/13385520

#SPJ1

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Block D of the mechanism is confined to move within the slot of member CB. Link AD is rotating at a constant rate of ωAD = 6 rad
svet-max [94.6K]

Answer:

1) 1.71 rad/s

2) -6.22 rad/s²

Explanation:

Choose point C to be the origin.

Using geometry, we can show that the coordinates of point A are:

(a cos 30°, a sin 30° − b)

Therefore, the coordinates of point D at time t are:

(a cos 30° − b sin(ωt), a sin 30° − b + b cos(ωt))

The angle formed by CB with the x-axis is therefore:

tan θ = (a sin 30° − b + b cos(ωt)) / (a cos 30° − b sin(ωt))

1) Taking the derivative with respect to time, we can find the angular velocity:

sec² θ dθ/dt = [(a cos 30° − b sin(ωt)) (-bω sin(ωt)) − (a sin 30° − b + b cos(ωt)) (-bω cos(ωt))] / (a cos 30° − b sin(ωt))²

sec² θ dθ/dt = -bω [(a cos 30° − b sin(ωt)) sin(ωt) − (a sin 30° − b + b cos(ωt)) cos(ωt)] / (a cos 30° − b sin(ωt))²

sec² θ dθ/dt = -bω [(a cos 30° sin(ωt) − b sin²(ωt)) − (a sin 30° cos(ωt) − b + b cos²(ωt))] / (a cos 30° − b sin(ωt))²

sec² θ dθ/dt = -bω (a cos 30° sin(ωt) − b sin²(ωt) − a sin 30° cos(ωt) + b − b cos²(ωt)) / (a cos 30° − b sin(ωt))²

sec² θ dθ/dt = -bω (a cos 30° sin(ωt) − a sin 30° cos(ωt)) / (a cos 30° − b sin(ωt))²

sec² θ dθ/dt = -abω (cos 30° sin(ωt) − sin 30° cos(ωt)) / (a cos 30° − b sin(ωt))²

We know at the moment shown, a = 350 mm, b = 200 mm, θ = 30°, ω = 6 rad/s, and t = 0 s.

sec² 30° dθ/dt = -(350) (200) (6) (cos 30° sin(0) − sin 30° cos(0)) / (350 cos 30° − 200 sin(0))²

sec² 30° dθ/dt = -(350) (200) (6) (-sin 30°) / (350 cos 30°)²

dθ/dt = (200) (6) (1/2) / 350

dθ/dt = 600 / 350

dθ/dt = 1.71 rad/s

2) Taking the second derivative of θ with respect to time, we can find the angular acceleration.

sec² θ d²θ/dt² + 2 sec² θ tan θ dθ/dt = -abω [(a cos 30° − b sin(ωt))² (ω cos 30° cos(ωt) + ω sin 30° sin(ωt)) − (cos 30° sin(ωt) − sin 30° cos(ωt)) (2 (a cos 30° − b sin(ωt)) (-bω cos(ωt)))] / (a cos 30° − b sin(ωt))⁴

At t = 0:

sec² θ d²θ/dt² + 2 sec² θ tan θ dθ/dt = -abω [(a cos 30°)² (ω cos 30°) − (0 − sin 30°) (2 (a cos 30°) (-bω))] / (a cos 30°)⁴

sec² θ d²θ/dt² + 2 sec² θ tan θ dθ/dt = -abω (a²ω cos³ 30° − 2abω sin 30° cos 30°) / (a⁴ cos⁴ 30°)

sec² θ d²θ/dt² + 2 sec² θ tan θ dθ/dt = -bω (aω cos² 30° − 2bω sin 30°) / (a² cos³ 30°)

d²θ/dt² + 2 tan θ dθ/dt = -bω² (a cos² 30° − b) / (a² cos 30°)

Plugging in values:

d²θ/dt² + 2 tan 30° dθ/dt = -(200) (6)² (350 cos² 30° − 200) / (350² cos 30°)

d²θ/dt² + 2 tan 30° dθ/dt = -7200 (262.5 − 200) / (350² cos 30°)

d²θ/dt² + 2 tan 30° (1.71) = -4.24

d²θ/dt² = -6.22 rad/s²

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