Question

A circular tube is to be made of 7075-T6 aluminum with a uniaxial yield strength of 469 MPa.

Suppose a hollow circular tube with an inside diameter of 46.0 mm is to be made of this material.

If the limit loading condition for this part is the simultaneous application of an axial load of 120

kN and a torque of 2.0 kN-m, what minimum wall thickness is required to ensure that no yielding

occurs at the limit load? Use the Von Mises criterion. Justify your work clearly.

[Hint: a trial

and error solution using Excel or Matlab might be the easiest way to get the final answer]

Answer 1

Yield strength = 469 M P_a Let D rightarrow outside diameter and d rightarrow inner diameter d = 46 mm p_axial = 120 KN, T = 2KN m wall thickness t =?? by von-mises criteria, Z_max = S_yt/squareroot 3 fos At limit load conditions, Z_max = 469/squareroot 3 times 1 (that is fos = 1) Therefore Z_max = 270.78 Mpa (by von-mises criteria) Now, Z_max = squareroot (sigma_x/2)^2 + (C_xy)^2sigma_x = sigma_axial = p/A = p/pi/4 (D^2 - d^2) = 4 P/pi (D^2- a^2) = 4P/ pi d^2 [D^2/d^2 -1] ** Let D/d = K (K > 1) Therefore sigma_x = 4P/pi d^2 (k^2 -1) and C_xy = T times D/2 /J = T times D/2/pi /32 (D^4 - d^4) = 16 T times D/pi (D^4 - d^4) C_xy = 16 T K /pi d^3 (k^4 - 1) putting values pf P, T and d above relations sigma_x = 4 times 120 times 10^3/pi times 46^2(K^2 -1) = 72.206/(k^2 -1) C_xy = 16 times 2 times 10^3/pi times 46^3 k/(k^4 - 1) = 0.1046 k/(k^4 -1) Therefore 270.78 = squareroot { 72.206/2(k^2 - 1)}^2 + { 0.01046 K/(K^4 -1)}^2