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3 years ago

A composite shaft with length L = 46 in is made by fitting an aluminum sleeve (Ga = 5 x 10^3 ksi) over a

Engineering

1 answer:

Xelga [282]3 years ago

5 0

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 ) $in^u$

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 ∅ $_{alu$ = TaLa/GaJa

for steel rod ∅ $_{steel$ = TsLs/GsJs

but we know that, ∅a = ∅s = ∅ $_B$

[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

∅ $_{steel$ = TsLs / GsJs

we substitute

∅ $_{steel$ = (1.81 × 46 ) / ( 11 × 10³ × π/32( 256 ) )

∅ $_{steel$ = 83.26 / 276460.1535

∅ $_{steel$ = 0.000301

∅ $_{steel$ = 3.01 × 10⁻⁴ rad

∅ $_{steel$ = ∅ $_B$ = 3.01 × 10⁻⁴ rad

Therefore, the magnitude of the angle of twist at end B is 3.01 × 10⁻⁴ rad

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