Question

The solid spindle AB is connected to the hollow sleeve CD by a rigid plate at C. The spindle is composed of steel (Gs = 11.2 x 106 psi, τmax = 12 ksi) and has a diameter ds = 1.75 in. The sleeve is made of brass (Gb = 5.6 x 106 psi, τmax = 7 ksi) and has an outer diameter do = 3 in and a wall thickness t = 0.25 in. Determine:

Answer 1

Answer:

T_max = 12.63 kip.in

Ф_a = 1.093°

Explanation:

Given:

- The modulus of rigidity of solid spindle G_ab = 11.2 * 10^6 psi

- The diameter of solid spindle d_ab = 1.75 in

- The allowable stress in solid spindle τ_ab = 12 ksi

- The modulus of rigidity of sleeve G_cd = 5.6 * 10^6 psi

- The outer diameter of sleeve d_cd = 3 in

- The thickness of sleeve t = 0.25

- The allowable stress in sleeve τ_cd = 7 ksi

Find:

- The largest torque T that can be applied to end A that does not exceed allowable stresses and sleeve angle of twist 0.375°

- The corresponding angle through which end A rotates.

Solution:

- Calculate the polar moment of inertia of both spindle AB and sleeve CD.

   Spindle AB:    c_ab = 0.5*d_ab = 0.5(1.75) = 0.875 in

                           J_ab = pi/2 c^4 = pi/2 0.875^4 = 0.92077 in^4

   Sleeve CD:  c_cd1 = 0.5*d_cd = 0.5(3) = 1.5 in , c_cd2 = c_cd1 - t = 1.25

                     J_cd = pi/2 (c_cd1^4 - c_cd2^4)= pi/2(1.5^4-1.25^4) = 4.1172 in^4

- The stress criteria for maximum allowable torque in spindle AB:

                    T_ab = J_ab*τ_ab / c_ab

                    T_ab = 0.92077*12 / 0.875

                    T_ab = 12.63 kip.in

- The stress criteria for maximum allowable torque in sleeve CD:

                    T_cd = J_cd*τ_cd / c_cd1

                    T_cd = 4.1172*7 / 1.5

                    T_cd = 19.21 kip.in

- The angle of twist criteria for point D:

                    T_d = J_cd*G_cd*Ф / L

                    T_d = 4.1172*5.6*10^6*0.006545 / 8

                    T_d = 18.86 kip.in

- The maximum allowable Torque for the structure is:

                    T_max = min ( 12.63 , 19.21 , 18.86 )

                    T_max = 12.63 kip.in

- The angle of twist of end A:

                    Ф_a = Ф_a/d = Ф_a/b + Ф_c/d:

                    T_max* ( L_ab / J_ab*G_ab + L_cd / J_cd*G_cd)

                    12.63*(12/0.92*11.2*10^6 + 8/4.117*5.6*10^6)

                    0.01908 rads = 1.093°

Answer 2

In this exercise we want to calculate the value of the angle, in this way we find that it will be

$T_{max} = 12.63 kip.in\\\phi_a = 1.093$

Organizing the information given in the statement we find that:

• $G_{ab} = 11.2 * 10^6 psi$

• $d_{ab} = 1.75 in$

• $T_{ab} = 12 ksi$

• $G_{cd} = 5.6 * 10^6 psi$

• $d_{cd} = 3 in$

• $t = 0.25$

• $T_{cd} = 7 ksi$

With the information given above, we use it to calculate the moment of inertia of each part, as follows:

•   Spindle AB:  

$c_{ab} = 0.5*d_{ab} = 0.5(1.75) = 0.875 in\\ J_{ab} = \pi/2 c^4 = \pi/2 0.875^4 = 0.92077 in^4$

•   Sleeve CD:  

$c_{cd1} = 0.5*d_{cd} = 0.5(3) = 1.5 in \\ c_{cd2} = c_{cd1} - t = 1.25\\J_{cd} = \pi/2 (c_{cd1}^4 - c_{cd2}^4)= \pi/2(1.5^4-1.25^4) = 4.1172 in^4$

So now it will be possible to calculate the maximum torque value of each one:

• AB:

$T_{ab} = J_{ab}*T_{ab} / c_{ab}\\T_{ab} = 0.92077*12 / 0.875\\ T_{ab} = 12.63 kip.in$

• CD:

$T_{cd} = J_{cd}*T_{cd} / c_{cd1}\\ T_{cd} = 19.21 kip.in$

The value of the angle will be given by:

$\phi_a = \phi_{a/d} = \phi_{a/b} + \phi_{c/d}\\=0.01908 rads = 1.093$

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