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°
