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A composite shaft with length L = 46 in is made by fitting an aluminum sleeve (Ga = 5 x 10^3 ksi) over a
steel core (Gs = 11 x10^3 ksi). The composite shaft is fixed against rotation at the wall A and has outside diameter da = 5 in. The steel core has diameter ds = 4 in. A torque TB = 3 kip. in is applied at B. % Matlab input: L = 46 Ga = 5 * 10^3 Gs = 11 * 10^3; da = 5; ds = 4; Tb = 3;
Determine the magnitude of the angle of twist at end B.

Sagot :

Nuhulawal20idn

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 ) [tex]in^u[/tex]

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 ∅[tex]_{alu[/tex] = TaLa/GaJa

for steel rod ∅[tex]_{steel[/tex] = TsLs/GsJs

but we know that, ∅a = ∅s = ∅[tex]_B[/tex]

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

∅[tex]_{steel[/tex] = TsLs / GsJs

we substitute

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

∅[tex]_{steel[/tex] = 83.26 / 276460.1535

∅[tex]_{steel[/tex]  = 0.000301

∅[tex]_{steel[/tex] = 3.01 × 10⁻⁴ rad

so

∅[tex]_{steel[/tex] = ∅[tex]_B[/tex] = 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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