The Am1004-T61 magnesium tube is bonded to the A-36 steel rod. A torque of 4 kN

Question

The Am1004-T61 magnesium tube is bonded to the A-36 steel rod. A torque of 4 kN m is applied to end A. The shear modulus of elasticity for Am1004-T61 magnesium alloy is 18 GPa, and for A-36 steel is 75 GPa. Determine the maximum shear stress in magnesium. Express your answer to three significant figures and include the appropriate units. Determine the maximum shear stress in steel. Express your answer to three significant figures and include the appropriate units. Determine the shear stress on the inside surface of the magnesium tube. Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

Here, We have,

Total torque, T = 4 kNm = 4000 Nm

Shear Modulus of Magnesium, Cm = 18 GPa = 18 x 109 N/m2

Shear Modulus of Steel, Cs = 75 GPa = 75 x 109 N/m2

Length of Steel Rod or Magnesium tube = Ls = Lm = 900 mm = 0.9 m

Outside diameter of magnesium tube, Dm = 80 mm = 0.080 m

Inside diameter of magnesium tube, dm = 40 mm = 0.040 m

Diameter of Steel Rod, Ds = 40 mm = 0.040 m

*Calculation :

Ts = Torque shared by Steel rod,

Tm = Torque shared by magnesium tube,

ts = Max. Shear stress develop in Steel rod

tm = Max. Shear stress develop in magnesium tube,

So, Total torque, T = Ts + Tm = 4000 Nm................................................Eq.1

#Polar Moment of Inertia of Steel rod, Js

Js = pi / 32 * (Ds)4 = 3.1416 / 32 * (0.040)4 = 2.51327 x 10-7 m4

#Polar Moment of Inrtia of Magnesium tube, Jm

Jm = pi / 32 * ((Dm)4 - (dm)4) = 3.1416 / 32 * ((0.080)4 - (0.040)4) = 3.7699 x 10-6 m4

#Angle of twist of steel rod, As

As = Ts * Ls / (Js * Cs)

As = Ts * 0.9 / (2.51327 x 10-7 x 75 x 109 )

As = Ts * 4.77465 x 10-5 rad.

#Angle of twist of Magnesium tube, Am

Am = Tm * Lm / (Jm * Cm)

Am = Tm * 0.9 / (3.76991 x 10-6 x 18 x 109 )

Am = Tm * 1.32629 x 10-5 rad.

$$ Since, As = Am,

Tm * 1.32629 x 10-5 =Ts * 4.77465 x 10-5

So, Tm = 3.6 * Ts......................................................................Eq.2

## From Eq.1 and Eq. 2

Ts + 3.6 * Tm = 4000 Nm

So, Ts = 869.5 Nm

and Tm = 3130.4 Nm

$$$ PART A, Max. Shear Stress in Mangnesium tube, tm

tm = Tm / (pi / 16 * (Dm4 - dm4) / Dm )

tm = 3130.4 / (3.1416 / 16 * (0.0804 - 0.0404) / 0.080)

tm = 33.21 x 106 N/m2 = 33.21 N/mm2

$$$ PART B. Max. Shear Stress in Steel Rod, ts

ts = Ts / (pi / 16 x Ds3)

ts = 869.5 / (3.1416 / 16 x 0.0403)

ts = 69.19 x 106 N/m2 = 69.19 N/mm2

$$$ PART C, Shear Stress in inside surface of Mangnesium tube, tmi

tmi = Tm / (pi / 16 * (Dm4 - dm4) / dm )

tmi = 3130.4 / (3.1416 / 16 * (0.0804 - 0.0404) / 0.040)

tmi = 16.6 x 106 N/m2 = 16.6 N/mm2

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