lengthwise-slit annular tube and a closed annular tube have the same mean di diameter D If center of thickness to center of thickness) and same tube thickness t, where D=18t tubes are subjected to the same torque , what then is the ratio of the angles sections M_(t)=2Aq For open thin walled sections J=(1)/(3)bt^3 and tau _(max)=(M_(t)t)/(I) if b/tgeqslant 10

To solve this problem, we need to find the ratio of the angles of twist for the lengthwise-slit annular tube and the closed annular tube when subjected to the same torque.<br /><br />Given information:<br />- The mean diameter of both tubes is the same, D.<br />- The center of thickness to center of thickness distance and the tube thickness are the same, t.<br />- The tube thickness is related to the mean diameter as D = 18t.<br />- The torque applied to both tubes is the same.<br />- The formula for the torsion section moment is Mt = 2Aq.<br />- For open thin-walled sections, the torsion constant is J = (1/3)bt^3.<br />- The maximum shear stress is τmax = MtI.<br /><br />Let's denote the lengthwise-slit annular tube as Tube A and the closed annular tube as Tube B.<br /><br />The ratio of the angles of twist for the two tubes can be found using the formula:<br /><br />θ_A / θ_B = (τ_A / τ_B) * (I_B / I_A)<br /><br />Where:<br />- θ_A and θ_B are the angles of twist for Tube A and Tube B, respectively.<br />- τ_A and τ_B are the maximum shear stresses for Tube A and Tube B, respectively.<br />- I_A and I_B are the torsion constants for Tube A and Tube B, respectively.<br /><br />Since both tubes have the same mean diameter D and the same tube thickness t, we can use the given formulas to find the maximum shear stress and the torsion constant for each tube.<br /><br />For Tube A (lengthwise-slit annular tube):<br />τ_A = MtI_A / (A_A * t)<br />I_A = (1/3)bt^3<br /><br />For Tube B (closed annular tube):<br />τ_B = MtI_B / (A_B * t)<br />I_B = (1/3)bt^3<br /><br />Since the tube thickness is related to the mean diameter as D = 18t, we can substitute D = 18t into the formulas for the cross-sectional areas of the tubes.<br /><br />A_A = π * (D/2)^2 = π * (18t/2)^2 = 81πt^2<br />A_B = π * (D/2)^2 = π * (18t/2)^2 = 81πt^2<br /><br />Now, we can substitute the values of τ_A, τ_B, I_A, and I_B into the ratio formula:<br /><br />θ_A / θ_B = (τ_A / τ_B) * (I_B / I_A)<br />θ_A / θ_B = (MtI_A / (A_A * t)) * (I_B / I_A)<br />θ_A / θ_B = (MtI_A / (81πt^2 * t)) * ((1/3)bt^3 / (1/3)bt^3)<br />θ_A / θ_B = (MtI_A / (81πt^3)) * 1<br />θ_A / θ_B = MtI_A / (81πt^3)<br /><br />Since the torque Mt is the same for both tubes, the ratio of the angles of twist is:<br /><br />θ_A / θ_B = I_A / (81πt^3)<br /><br />Therefore, the ratio of the angles of twist for the lengthwise-slit annular tube and the closed annular tube when subjected to the same torque is I_A / (81πt^3).