The Am1004T61 magnesium tube is bonded to the A36 steel rod
Solution
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

