Question

The Am1004-T61 magnesium tube is bonded to the A-36 steel rod. If the allowable shear stresses for the magnesium and steel are ( τ allow)mg = 45 MPa and ( τ allow)st = 75 MPa, respectively, determine the maximum allowable torque that can be applied at A. Also, find the corresponding angle of twist of end A.

Answer 1

a) T = 4335.398 N-m

b) ∅ = 0.045 rad

Step-by-step explanation:

Given

τ allow (mg) = 45 MPa

τ allow (st) = 75 MPa

G mg = 18*10⁹ Pa

G st = 75*10⁹ Pa

L st = L mg = L = 900 mm = 0.9 m

D₁ = 40 mm = 0.04 m

D₂ = 80 mm = 0.08 m

We write the expression of angle of twist:

∅ mg = ∅ st

T mg* L mg / (J mg* G mg) = T st* L st / (J st* G st)

If

L st = L mg = L = 900 mm = 0.9 m

G mg = 18*10⁹ Pa

G st = 75*10⁹ Pa

J mg = π*(D₂⁴ - D₁⁴)/32 ⇒ J mg = π*((0.08 m)⁴ - (0.04 m)⁴)/32

⇒ J mg = 3.7699*10⁻⁶ m⁴

J st = π*D₁⁴/32 ⇒ J st = π*(0.04 m)⁴/32

⇒ J st = 2.5133*10⁻⁷ m⁴

We have

T mg / (J mg* G mg) = T st / (J st* G st)

⇒ T mg / (3.7699*10⁻⁶ m⁴* 18*10⁹ Pa) = T st / (2.5133*10⁻⁷ m⁴* 75*10⁹ Pa)

⇒ T mg = 3.6*T st

Then, we apply moment equilibrium condition

∑M = 0

⇒ T mg + T st - T = 0

⇒ 3.6*T st + T st - T = 0

⇒ T st = 0.2174*T

Now, we determine the value of T mg

T mg = 3.6*T st = 3.6*(0.2174*T)

⇒ T mg = 0.7826*T

Now, we write the expression to determine the torque at

τ allow (mg) = T mg* R₂ / J mg

⇒ T mg = τ allow (mg)* J mg / R₂

⇒ 0.7826*T = 45*10⁶ Pa* 3.7699*10⁻⁶ m⁴ / 0.04 m

⇒ T = 5419.24 N-m

then

τ allow (st) = T st* R₁ / J st

⇒ T st = τ allow (st)* J st / R₁

⇒ 0.2174*T = 75*10⁶ Pa* 2.5133*10⁻⁷ m⁴ / 0.02 m

⇒ T = 4335.398 N-m (which satisfies the allowable shear stresses)

We can find the angle of twist:

∅ mg = T mg* L mg / (J mg* G mg)

∅ mg = 0.7826*T

* L mg / (J mg* G mg)

⇒ ∅ mg = 0.7826*(4335.398 N-m)

* (0.9 m) / (3.7699*10⁻⁶ m⁴* 18*10⁹ Pa)

⇒ ∅ mg = 0.045 rad

In order to understand the question, we can see the pic shown.