Question 5.1: Failure of Ductile Materials Under Static Loading Problem De...
Failure of Ductile Materials Under Static Loading
Problem Determine the safety factors for the bracket rod shown in Figure 5-9 based on both the distortion-energy theory and the maximum shear theory and compare them.
Given The material is 2024-T4 aluminum with a yield strength of 47 000 psi. The rod length l = 6 in and arm a = 8 in. The rod outside diameter d = 1.5 in. Load F = 1 000 lb.
Assumptions The load is static and the assembly is at room temperature. Consider shear due to transverse loading as well as other stresses.
See Figures 5-9 and 4-33 (repeated here). Also see Example 4-9 (p. 180) for a more complete explanation of the stress analysis for this problem.
1 The rod is loaded in both bending (as a cantilever beam) and in torsion. The largest tensile bending stress will be in the top outer fiber at point A. The largest torsional shear stress will be all around the outer circumference of the rod. (See Example 4-9 for more detail.) First take a differential element at point A where both of these stresses combine as shown in Figure 4-33b. Find the normal bending stress and torsional shear stress on point A using equations 4.11b (p. 156) and 4.23b (p. 178) respectively.
\sigma_{\max }=\frac{M c}{I} (4.11b)
\tau_{\max }=\frac{T r}{J} (4.23b(
\sigma_x=\frac{M c}{I}=\frac{(F l) c}{I}=\frac{1000(6)(0.75)}{0.249}=18108 psi (a)
\tau_{x z}=\frac{T r}{J}=\frac{(F a) r}{J}=\frac{1000(8)(0.75)}{0.497}=12072 psi (b)
2 Find the maximum shear stress and principal stresses that result from this combination of applied stresses using equations 4.6 (p. 145).
\begin{aligned} \sigma_a, \sigma_b &=\frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{x y}^2} \\ \sigma_c &=0 \end{aligned} (4.6a)
\tau_{\max }=\tau_{13}=\frac{\left|\sigma_1-\sigma_3\right|}{2} (4.6b)
\begin{aligned} \tau_{\max } &=\sqrt{\left\lgroup\frac{\sigma_x-\sigma_z}{2}\right\rgroup^2+\tau_{x z}^2}=\sqrt{\left\lgroup\frac{18108-0}{2}\right\rgroup^2+12072^2}=15090 psi \\ \sigma_1 &=\frac{\sigma_x+\sigma_z}{2}+\tau_{\max }=\frac{18108}{2}+15090=24144 psi \\ \sigma_2 &=0 \\ \sigma_3 &=\frac{\sigma_x+\sigma_z}{2}-\tau_{\max }=\frac{18108}{2}-15090=-6036 psi \end{aligned} (c)
3 Find the von Mises effective stress from the principal stresses using equation 5.7a (p. 249) with \sigma _{2} = 0, which is also shown as equation 5.7c (p. 249) for the 2-D case.
\sigma^{\prime}=\sqrt{\sigma_1^2+\sigma_2^2+\sigma_3^2-\sigma_1 \sigma_2-\sigma_2 \sigma_3-\sigma_1 \sigma_3} (5.7a)
\sigma^{\prime}=\sqrt{\sigma_1^2-\sigma_1 \sigma_3+\sigma_3^2} (5.7c)
\begin{array}{l} \sigma^{\prime}=\sqrt{\sigma_1^2-\sigma_1 \sigma_3+\sigma_3^2} \\ \sigma^{\prime}=\sqrt{24144^2-24144(-6036)+(-6036)^2}=27661 psi \end{array} (d)
4 The safety factor using the distortion-energy theory can now be found using equation 5.8a (p. 251).
N=\frac{S_y}{\sigma^{\prime}}=\frac{47000}{27661}=1.7 (e)
5 The safety factor using the maximum shear-stress theory can be found from equation 5.10 (p. 252).
S_{y s}=0.50 S_y (5.10)
N=\frac{0.50 S_y}{\tau_{\max }}=\frac{0.50(47000)}{15090}=1.6 (f)
6 Comparing these two results shows the more conservative nature of the maximum shear-stress theory, which gives a slightly lower safety factor.
7 Since the rod is a short beam, we need to check the shear due to transverse loading at point B on the neutral axis. The maximum transverse shear stress at the neutral axis of a round rod was given as equation 4.15c (p. 161).
\tau_{\max }=\frac{4}{3} \frac{V}{A} (4.15c)
\tau_{\text {bending }}=\frac{4 V}{3 A}=\frac{4(1000)}{3(1.767)}=755 psi (g)
Point B is in pure shear. The total shear stress at point B is the algebraic sum of the transverse shear stress and the torsional shear stress, which both act on the same planes of the differential element in this case, in the same direction as shown in Figure 4-33c.
\tau_{\text {max }}=\tau_{\text {torsion }}+\tau_{\text {bending }}=12072+755=12827 psi (h)
8 The safety factor for point B using the distortion energy theory for pure shear (Eq. 5.9b, p. 251) is
N=\frac{S_{y s}}{\tau_{\max }}=\frac{0.577 S_y}{\tau_{\max }}=\frac{0.577(47000)}{12827}=2.1 (i)
and for the maximum shear theory using equation 5.10 (p. 252) is
N=\frac{S_{y s}}{\tau_{\max }}=\frac{0.50 S_y}{\tau_{\max }}=\frac{0.50(47000)}{12827}=1.8 (j)
Again, the latter is more conservative.
9 The files EX05-01 are on the CD-ROM.
