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

Question

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

Answer 1

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

Soru

Çözüm

Cevap