Multiaxial Fluctuating Stresses Problem Determine the safety factors for the bracket tube shown in Figure 5-9 (repeated on next page). Given The material is 2024-T4 aluminum with Sy = 47 000 psi, and Sut = 68 000 psi. The tube length l = 6 in and arm a = 8 in. The tube outside diameter OD = 2 in

Question

Multiaxial Fluctuating Stresses

Problem Determine the safety factors for the bracket tube shown in Figure 5-9 (repeated on next page).

Given The material is 2024-T4 aluminum with [latex]S_{y}[/latex] = 47 000 psi, and [latex]S_{ut}[/latex] = 68 000 psi. The tube length [latex]l[/latex] = 6 in and arm [latex]a[/latex] = 8 in. The tube outside diameter [latex]OD[/latex] = 2 in and inside diameter [latex]ID[/latex] = 1.5 in. The applied load varies sinusoidally from [latex]F[/latex] = 340 to –200 lb.

Assumptions The load is dynamic and the assembly is at room temperature. Consider shear due to transverse loading as well as other stresses. A finitelife design will be sought with a life of [latex]6E7[/latex] cycles. The notch radius at the wall is 0.25 in and stress-concentration factors are for bending, [latex]K_{t}[/latex] = 1.7, and for shear, [latex]K_{ts}[/latex] = 1.35.

Accepted Answer

See Figure 5-9, repeated here. Also see Example 4-9 (p. 180) for a more complete explanation of the stress analysis for this problem.

1 Aluminum does not have an endurance limit. Its endurance strength at [latex]5E8[/latex] cycles can be estimated from equation 6.5c (p. 330). Since the [latex]S_{ut}[/latex] is larger than 48 kpsi, the uncorrected [latex] S_{f^{\prime} @ 5 E 8}=19 [/latex] kpsi.

[latex] ext {aluminums :} \quad \left\{\begin{aligned} S_{f_{@ S E 8}} & \equiv 0.4 S_{u t} && ext { for } S_{u t}<48 ext{ kpsi} (330 ext{ MPa} ) \ S_{f_{@ 5 E 8}}^{\prime} & \equiv 19 ext{ kpsi} (130 ext{ MPa} ) & & ext { for } S_{u t} \geq 48 ext{ kpsi} (330 ext{ MPa} ) \end{aligned} ight\} [/latex] (6.5c)

2 The correction factors are calculated from equations 6.7 (pp. 330–335) and Figure 6-25 (p. 332) and used to find a corrected endurance strength at the standard [latex]5E8[/latex] cycles.

[latex] \begin{aligned} ext {bending :} \quad C_{ ext {load }} &=1 \ ext {axial loading :} \quad C_{ ext {load }} &=0.70 \end{aligned} [/latex] (6.7a)

[latex] \begin{array}{ll} ext {for} d \leq 0.3 in (8 mm ): & C_{ ext {size }}=1 \ ext {for} 0.3 in

[latex] A_{95}=\pi\left[\frac{d^2-(0.95 d)^2}{4} ight]=0.0766 d^2 [/latex] (6.7c)

[latex] d_{ ext {equiv }}=\sqrt{\frac{A_{95}}{0.0766}} [/latex] (6.7d)

[latex] C_{ ext {surf }} \cong A\left(S_{ ext {ut }} ight)^b \quad ext { if } C_{ ext {surf }}>1.0 ext {, set } C_{ ext {surf }}=1.0 [/latex] (6.7e)

[latex] \begin{array}{ll} ext {for} T \leq 450^{\circ} C \left(840^{\circ} F ight): & C_{ ext {temp }}=1 \ ext {for} 450^{\circ} C

[latex] C_{load} = 1 : \quad ext {for bending} \ C_{ ext {size }}=0.869\left(d_{ ext {equiv }} ight)^{-0.097}=0.869\left\lgroup\sqrt{\frac{0.01046 d^2}{0.0766}} ight group^{-0.097}=0.869(.739)^{-0.097}=0.895 [/latex]

[latex] C_{ ext {surf }}=2.7\left(S_{u t} ight)^{-0.265}=2.7(68)^{-0.265}=0.883[/latex] (a)

[latex] C_{ ext {temp }}=1 \ C_{ ext {reliab }}=0.753: \quad ext { for } 99.9 \% [/latex]

[latex] \begin{aligned} S_{f_{@ 5 e 8}} &=C_{ ext {load }} C_{ ext {size }} C_{ ext {surf }} C_{ ext {temp }} C_{ ext {reliab }} S_{f^{\prime}} \ &=(1)(0.895)(0.883)(1)(0.753) 19000=11299 psi \end{aligned} [/latex] (b)

Note that the bending value of [latex]C_{load}[/latex] is used despite the fact that both bending and torsion are present. The torsional shear stress will be converted to an equivalent tensile stress with the von Mises calculation. [latex]C_{surf}[/latex] is calculated from equation 6.7e (p. 333) using data from Table 6-3 (p. 333). This corrected fatigue strength is still at the tested number of cycles, [latex]N = 5E8[/latex].

3 This problem calls for a life of [latex]6E7[/latex] cycles, so a strength value at that life must be estimated from the [latex]S-N[/latex] line of Figure 6-33b (p. 337) using the corrected fatigue strength at that life. Equation 6.10a (p. 338) for this line can be solved for the desired strength after we compute the values of its coefficient [latex]a[/latex] and exponent [latex]b[/latex] from equation 6.10c (p. 338).

[latex] S(N)=a N^b [/latex] (6.10a)

[latex] \begin{aligned} b &=\frac{1}{z} \log \left\lgroup\frac{S_m}{S_e} ight group \quad ext { where } \quad z=\log N_1-\log N_2 \ \log (a) &=\log \left(S_m ight)-b \log \left(N_1 ight)=\log \left(S_m ight)-3 b \end{aligned} [/latex] (6.10c)

[latex] \begin{aligned} b &=\frac{1}{z} \log \left\lgroup\frac{S_m}{S_f} ight group=\frac{1}{-5.699} \log \left[\frac{0.9(68000)}{11299} ight]=-0.1287 \ \log (a) &=\log \left(S_m ight)-3 b=\log [0.9(68000)]-3(-0.1287): \quad a=148929 \ S_n &=a N^b=148929 N^{-0.1287}=148929(6 e 7)^{-0.1287}=14846 psi \end{aligned} [/latex] (c)

Note that [latex]S_{m}[/latex] is calculated as 90% of [latex]S_{ut} [/latex] because the loading is bending rather than axial (Eq. 6.9, p. 337). The value of [latex]z[/latex] is taken from Table 6-5 (p. 338) for [latex]N = 5E8[/latex] cycles. This is a corrected fatigue strength for the shorter life required in this case and so is larger than the corrected test value, which was calculated at a longer life.

[latex] \begin{array}{l} ext {bending :} \quad S_m=0.9 S_{u t} \ ext {axial loading :} \quad S_m=0.75 S_{u t} \end{array} [/latex] (6.9)

4 The notch sensitivity of the material must be found to calculate the fatigue stress-concentration factors. Table 6-8 (p. 347) shows the Neuber factors for hardened aluminum. Interpolation gives a value of 0.147 for [latex]\sqrt{a} [/latex] at the material’s [latex]S_{ut}[/latex]. Equation 6.13 (p. 345) gives the resulting notch sensitivity for an assumed notch radius of 0.25 in.

[latex] q=\frac{1}{1+\frac{\sqrt{a}}{\sqrt{r}}}=\frac{1}{1+\frac{0.147}{\sqrt{0.25}}}=0.773 [/latex] (d)

5 The fatigue stress-concentration factors are found from equation 6.11b (p. 343) using the given geometric stress-concentration factors for bending and torsion, respectively.

[latex] K_f=1+q\left(K_t-1 ight)=1+0.773(1.7-1)=1.541 [/latex] (e)

[latex] K_{f s}=1+q\left(K_{t s}-1 ight)=1+0.773(1.35-1)=1.270 [/latex] (f)

6 The bracket tube is loaded both in bending (as a cantilever beam) and in torsion. The shapes of the shear, moment, and torque distributions are shown in Figure 4-31 (p. 183). All are maximum at the wall. The alternating and mean components of the applied force, moment, and torque at the wall are

[latex] \begin{array}{l} F_a=\frac{F_{\max }-F_{\min }}{2}=\frac{340-(-200)}{2}=270 lb \ F_m=\frac{F_{\max }+F_{\min }}{2}=\frac{340+(-200)}{2}=70 lb \end{array} [/latex] (g)

[latex] \begin{array}{l} M_a=F_a l=270(6)=1620 lb - ext { in } \ M_m=F_m l=70(6)=420 lb - ext { in } \end{array} [/latex] (h)

[latex] \begin{array}{l} T_a=F_a a=270(8)=2160 lb - ext { in } \ T_m=F_m a=70(8)=560 lb - in \end{array} [/latex] (i)

7 The fatigue stress-concentration factor for the mean stresses depends on the relationship between the maximum local von Mises stress in the notch and the yield strength as defined in equation 6.17, a portion of which is shown here.

[latex] \begin{array}{ll} ext {if} K_f\left|\sigma_{\max _{ ext {nom }}} ight|S_y ext { then : } & K_{f m}=\frac{S_y-K_f \sigma_{a_{ ext {nom }}}}{\left|\sigma_{m_{ ext {nom }}} ight|} \ ext {if} K_f\left|\sigma_{\max _{ ext {nam }}}-\sigma_{\min _{ ext {nom }}} ight|>2 S_y ext { then: } & K_{f m}=0 \end{array} [/latex] (6.17)

if [latex] K_f\left|\sigma_{\max } ight|

In this case, there is no reduction in stress-concentration factors for the mean stress because there is no yielding at the notch to relieve the stress concentration.

8 The largest tensile bending stress will be in the top or bottom outer fiber at points [latex]A[/latex] or [latex]A^{\prime}[/latex]. The largest torsional shear stress will be all around the outer circumference of the tube. (See Example 4-9 on p. 180 for more details.) First take a differential element at point [latex]A[/latex] or [latex]A^{\prime }[/latex] where both of these stresses combine. (See Figure 4-32 on p. 184.) Find the alternating and mean components of the normal bending stress and of the torsional shear stress on point [latex]A[/latex] using equations 4.11b (p. 156) and 4.23b (p. 178), respectively.

[latex] \sigma_{\max }=\frac{M c}{I} [/latex] (4.11b)

[latex] au_{\max }=\frac{T r}{J} [/latex] (4.23b)

[latex] \begin{aligned} \sigma_a &=K_f \frac{M_a c}{I}=1.541 \frac{1620(1)}{0.5369}=4649 psi \ au_{a_{ ext {torsion }}} &=K_{f s} \frac{T_a r}{J}=1.270 \frac{2160(1)}{1.074}=2556 psi \end{aligned} [/latex] (k)

[latex] \begin{aligned} \sigma_m &=K_{f m} \frac{M_m c}{I}=1.541 \frac{420(1)}{0.5369}=1205 psi \ au_{m_{ ext {torsion }}} &=K_{f s m} \frac{T_m r}{J}=1.270 \frac{560(1)}{1.074}=663 psi \end{aligned} [/latex] (l)

9 Find the alternating and mean von Mises stresses at point [latex]A[/latex] (Eq. 6.22b, p. 379).

[latex] \begin{aligned} \sigma_a^{\prime} &=\sqrt{\sigma_{x_a}^2+\sigma_{y_a}^2-\sigma_{x_a} \sigma_{y_a}+3 au_{x y_a}^2} \ &=\sqrt{4649^2+0^2-4649(0)+3\left(2556^2 ight)}=6419 psi \ \sigma_m^{\prime} &=\sqrt{\sigma_{x_m}^2+\sigma_{y_m}^2-\sigma_{x_m} \sigma_{y_m}+3 au_{x y_m}^2} \ &=\sqrt{1205^2+0^2-1205(0)+3\left(663^2 ight)}=1664 psi \end{aligned} [/latex] (m)

10 Because the moment and torque are both caused by the same applied force, they are synchronous and in-phase and any changes in them will be in a constant ratio. This is a Case 3 situation and the safety factor is found using equation 6.18e (p. 368).

[latex] N_f=\frac{S_f S_{u t}}{\sigma_a^{\prime} S_{u t}+\sigma^{\prime}{ }_m S_f}=\frac{14846(68000)}{6419(68000)+1664(14846)}=2.2 [/latex] (n)

11 Since the tube is a short beam, we need to check the shear due to transverse loading at point [latex]B[/latex] on the neutral axis where the torsional shear is also maximal. The maximum transverse shear stress at the neutral axis of a hollow, thin-walled, round tube was given as equation 4.15d (p. 161).

[latex] au_{\max } \cong \frac{2 V}{A} [/latex] (4.15d)

[latex] \begin{aligned} au_{a_{ ext {transverse }}} &=K_{f s} \frac{2 V_a}{A}=1.270 \frac{2(270)}{1.374}=499 psi \ au_{m_{ ext {transverse }}} &=K_{f s m} \frac{2 V_m}{A}=1.270 \frac{2(70)}{1.374}=129 psi \end{aligned} [/latex] (o)

Point [latex]B[/latex] is in pure shear. The total shear stress at point [latex]B[/latex] is the sum of the transverse shear stress and torsional shear stress which act on the same planes of the element.

[latex] \begin{array}{l} au_{a_{ ext {total }}}= au_{a_{ ext {transverse }}}+ au_{a_{ ext {torsion }}}=499+2556=3055 psi \ au_{m_{ ext {total }}}= au_{m_{ ext {transverse }}}+ au_{m_{ ext {torsion }}}=129+663=792 psi \end{array} [/latex] (p)

12 Find the alternating and mean von Mises stresses for point [latex]B[/latex] (Eq. 6.22b, p. 379).

[latex] \begin{array}{l} \sigma_a^{\prime}=\sqrt{\sigma_{x_a}^2+\sigma_{y_a}^2-\sigma_{x_a} \sigma_{y_a}+3 au_{x y_a}^2}=\sqrt{0+0-0+3(3055)^2}=5291 psi \ \sigma_m^{\prime}=\sqrt{\sigma_{x_m}^2+\sigma_{y_m}^2-\sigma_{x_m} \sigma_{y_m}+3 au_{x y_m}^2}=\sqrt{0+0-0+3(792)^2}=1372 psi \end{array} [/latex] (q)

13 The safety factor for point [latex]B[/latex] is found using equation 6.18e (p. 368).

[latex] N_f=\frac{S_f S_{u t}}{\sigma_a^{\prime} S_{u t}+\sigma_m^{\prime} S_f}=\frac{14846(68000)}{5291(68000)+1372(14846)}=2.7 [/latex] (r)

Both points [latex]A[/latex] and [latex]B[/latex] are safe against fatigue failure.

14 The files EX06-06A and EX06-06B can be found on the CD-ROM.

Table 6-3 Coefficients for Surface-Factor Equation 6.7e Source: Shigley and Mischke, Mechanical Engineering Design, 5th ed., McGrawHill, New York, 1989, p. 283 with permission
	For [latex]S_{ut}[/latex] in MPa use		For [latex]S_{ut}[/latex] in kpsi (not psi) use
Surface Finish	A	b	A	b
Ground	1.58	–0.085	1.34	–0.085
Machined or cold-rolled	4.51	–0.265	2.7	–0.265
Hot-rolled	57.7	–0.718	14.4	–0.718
As-forged	272	–0.995	39.9	–0.995

Table 6-5 z-factors for Eq. 6-10c
[latex]N_{2}[/latex]	[latex]z[/latex]
1.0E6	–3.000
5.0E6	–3.699
1.0E7	–4.000
5.0E7	–4.699
1.0E8	–5.000
5.0E8	–5.699
1.0E9	–6.000
5.0E9	–6.699

Table 6-8 Neuber's Constant for Hardened Aluminum
[latex] S _{ ut } ( kpsi ) [/latex]	[latex] \sqrt{a}\left(\operatorname{in}^{0.5} ight) [/latex]
15	0.475
20	0.380
30	0.278
40	0.219
50	0.186
60	0.162
70	0.144
80	0.131
90	0.122

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