Question

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.

Answer 1

Step-by-step 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`

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

∅
`_{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

so

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

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