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2 answers:
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Answer:
Explanation:
Given the data in the question;
L = 46 in
Ga = 5 × 10³ ksi
Gs = 11 × 10³ ksi
Outside diameter da = 5 in
ds = 4 in
Tb = 3 kip.in
Now,
Ja = polar moment of Inertia of Aluminum;
Ja ⇒ π/32( 5⁴ - 4⁴ ) = π/32( 625 - 256 ) = π/32( 369 )
Js = polar moment of inertia of steel
Js ⇒ π/32 ds⁴ = π/32( 4⁴ ) = π/32( 256 )
Ta is torque transmitted by Aluminum
Ts is torque transmitted by steel
{composite member }
T = Ta + Ts ------ let this be equation m1
Now, we use the relation;
T/J = G∅/L
JG∅ = TL
∅ = TL/GJ
so, for aluminum rod ∅ = TaLa/GaJa
for steel rod ∅ = TsLs/GsJs
but we know that, ∅a = ∅s = ∅
so
[TaLa/GaJa] = [TsLs/GsJs]
also, we know that, La = Ls = L
∴ [Ta/GaJa] = [Ts/GsJs]
we solve for Ta
TaGsJs = TsGaJa
Ta = TsGaJa / GsJs
we substitute
Ta = [Ts(5 × 10³)( π/32( 369) )] / [ (11 × 10³)( π/32( 256 ) ) ]
Ta = 0.66Ts
now, we substitute 0.66Ts for Ta and 3 for T in equation 1
T = Ta + Ts
3 = 0.66Ts + Ts
3 = 1.66Ts
Ts = 3 / 1.66
Ts = 1.8072 ≈ 1.81 kip-in
so
∅ = TsLs / GsJs
we substitute
∅ = (1.81 × 46 ) / ( 11 × 10³ × π/32( 256 ) )
∅ = 83.26 / 276460.1535
∅ = 0.000301
∅ = 3.01 × 10⁻⁴ rad
so
∅ = ∅
= 3.01 × 10⁻⁴ rad
Therefore, the magnitude of the angle of twist at end B is 3.01 × 10⁻⁴ rad
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
