Question 6.5: Design of a Cantilever Bracket for Fluctuating Bending Probl...
Design of a Cantilever Bracket for Fluctuating Bending
Problem A feed-roll assembly is to be mounted at its ends on support brackets cantilevered from the machine frame as shown in Figure 6-47. The feed rolls experience a total fluctuating load that varies from a minimum of 200 lb to a maximum of 2 200 lb, split equally between the two support brackets. Design a cantilever bracket to support a fluctuating bending load of 100- to 1 100-lb amplitude for 10^{9} cycles with no failure. Its dynamic deflection cannot exceed 0.02 in.
Given The load-time function shape is shown in Figure 6-47. The operating environment is room air at a maximum temperature of 120°F. The space available allows a maximum cantilever length of 6 in. Only ten of these parts are required.
Assumptions The bracket can be clamped between essentially rigid plates bolted at its root. The normal load will be applied at the effective tip of the cantilever beam from a rod attached through a small hole in the beam. Since the bending moment is effectively zero at the beam tip, the stress concentration from this hole can be ignored. Given the small quantity required, machining of stock mill-shapes is the preferred manufacturing method.
See Figure 6-47 and Tables 6-11 and 6-12.
1 This is a typical design problem. Very few data are given except for the loading on the device, some limitations on size, and the required cycle life. We will have to make some basic assumptions about part geometry, materials, and other factors as we go. Some iteration should be expected.
2 Figure 6-47 shows the same tentative design configuration as in Figure 6-41b (p. 355). The mill stock is purchased thicker than the desired final dimension and machined top and bottom to dimension D, then machined to thickness d over the length l. A fillet radius r is provided at the clamp point to reduce fretting fatigue and achieve a lower K_{t}. (See Figure 4-37 on p. 191.) Figure 4-36 (p. 190) shows that with suitable control of the r/d and D/d ratios for a stepped flat bar in bending, the geometric stress-concentration factor K_{t} can be kept under about 1.5.
3 A material must be chosen. For infinite life, low cost, and ease of fabrication, it is desirable to use a carbon steel if environmental conditions permit. Since this is used in a controlled, indoor environment, carbon steel is acceptable on the latter point. The fact that the deflection is of concern is also a good reason to choose a material with a large E. Low- to medium-carbon, ductile steels have the requisite endurancelimit knee for the infinite life required in this case and also have low notch sensitivities. An SAE 1040 normalized carbon steel with S_{ut} = 80 kpsi and S_{y} = 60 kpsi is selected for the first trial.
4 We will assume the trial dimensions to be the same as those of the successful solution to the fully reversed case from Example 6-4. These are b = 2 in, d = 1 in, D = 1.125 in, r = 0.5 in, a = 5 in, and l = 6.0 in. This value of a will leave some material around the hole and still fit within the 6-in-length constraint.
5 The mean and alternating components of the load and their reaction forces can be calculated from the given maximum and minimum load.
\begin{array}{l} F_m=\frac{F_{\max }+F_{\min }}{2}=\frac{1100+100}{2}=600 lb \\ F_a=\frac{F_{\max }-F_{\min }}{2}=\frac{1100-100}{2}=500 lb \end{array} (a)
R_a=F_a=500 lb \quad R_m=F_m=600 lb \quad R_{\max }=F_{\max }=1100 lb (b)
6 From these, the mean and alternating moments, and the maximum moment acting at the root of the cantilever beam can be calculated.
\begin{aligned} M_a &=R_a l-F_a(l-a)=500(6)-500(6-5)=2500 lb -\text { in } \\ M_m &=R_m l-F_m(l-a)=600(6)-600(6-5)=3000 lb -\text { in } \\ M_{\max } &=R_{\max } l-F_{\max }(l-a)=1100(6)-1100(6-5)=5500 lb – in \end{aligned} (c)
7 Find the cross-section area moment of inertia and the distance to the outer fiber.
\begin{array}{l} I=\frac{b d^3}{12}=\frac{2.0(1.0)^3}{12}=0.1667 in ^4 \\ c=\frac{d}{2}=\frac{1.0}{2}=0.5 in \end{array} (d)
8 The nominal bending stresses at the root are found for both the alternating load and the mean load from:
\begin{array}{l} \sigma_{a_{\text {nom }}}=\frac{M_a c}{I}=\frac{2500(0.5)}{0.1667}=7500 psi \\ \sigma_{m_{\text {nom }}}=\frac{M_m c}{I}=\frac{3000(0.5)}{0.1667}=9000 psi \end{array} (e)
9 Two ratios must be calculated for use in Figure 4-36 (p. 190) in order to find the geometric stress-concentration factor K_{t} for the assumed part dimensions.
\frac{D}{d}=\frac{1.125}{1.0}=1.125 \quad \frac{r}{d}=\frac{0.5}{1.0}=0.5 (f)
Interpolating in the table of Figure 4 – 36 : A=1.012 \quad b=-0.221 (g)
K_t=A\left\lgroup\frac{r}{d}\right\rgroup ^b=1.012(0.5)^{-0.221}=1.18 (h)
10 The notch sensitivity q of the chosen material is calculated based on its ultimate strength and the notch radius using equation 6.13 (p. 345) and the data for Neuber’s constant from Table 6-6 (p. 346). The values of q and K_{t} are used to find the fatigue stress-concentration factor K_{f} using equation 6.11b (p. 343). K_{fm} is calculated from equation 6.17 (p. 364).
q=\frac{1}{1+\frac{\sqrt{a}}{\sqrt{r}}} (6.13)
K_f=1+q\left(K_t-1\right) (6.11b)
\begin{array}{l} \text {if} K_f\left|\sigma_{\max _{\text {nom }}}\right|<S_y \text { then: } \quad K_{f m}=K_f\\ \text {if} K_f\left|\sigma_{\max _{\text {nom }}}\right|>S_y \text { then: } \quad K_{f m}=\frac{S_y-K_f \sigma_{a_{\text {nom }}}}{\left|\sigma_{m_{\text {nom }}}\right|}\\ \text {if} K_f\left|\sigma_{\max _{\text {nom }}}-\sigma_{\min _{\text {nom }}}\right|>2 S_y \text { then: } \quad K_{f m}=0 \end{array} (6.17)
From Table 6 – 6 for S_{u t}=80 ksi : \quad \sqrt{a}=0.08 (i)
q=\frac{1}{1+\frac{\sqrt{a}}{\sqrt{r}}}=\frac{1}{1+\frac{0.08}{\sqrt{0.5}}}=0.898 (j)
K_f=1+q\left(K_t-1\right)=1+0.898(1.18-1)=1.16 (k)
\text { if } K_f\left|\sigma_{\max }\right|<S_y \quad \text {then} \quad \quad K_{fm} = K_{f} \\ K_f\left|\frac{M_{\max } c}{I}\right|=1.16\left|\frac{5500(0.5)}{0.1667}\right|=19113<60000: \quad K_{fm} = 1.16 (l)
11 Use these factors to find the local mean and alternating notch stresses
\begin{aligned} \sigma_a &=K_f \sigma_{a_{\text {nam }}}=1.16(7500)=8711 psi \\ \sigma_m &=K_{f m} \sigma_{m_{\text {nam }}}=1.16(9000)=10454 psi \end{aligned} (m)
12 The local stresses are used to compute the von Mises alternating and mean stresses from equations 6.22b (p. 379).
\begin{array}{l} \sigma_a^{\prime}=\sqrt{\sigma_{x_a}^2+\sigma_{y_a}^2-\sigma_{x_a} \sigma_{y_a}+3 \tau_{x y_a}^2} \\ \sigma_m^{\prime}=\sqrt{\sigma_{x_m}^2+\sigma_{y_m}^2-\sigma_{x_m} \sigma_{y_m}+3 \tau_{x y_m}^2} \end{array} (6.22b)
\begin{array}{l} \sigma_a^{\prime}=\sqrt{\sigma_{x_a}^2+\sigma_{y_a}^2-\sigma_{x_a} \sigma_{y_a}+3 \tau_{x y_a}^2}=\sqrt{8711^2+0-8711(0)+3(0)}=8711 \\ \sigma_m^{\prime}=\sqrt{\sigma_{x_m}^2+\sigma_{y_m}^2-\sigma_{x_m} \sigma_{y_m}+3 \tau_{x y_m}^2}=\sqrt{10454^2+0-10425(0)+3(0)}=10454 \end{array} (n)
13 The uncorrected endurance limit S_{e^{\prime}} is determined from equation 6.5a (p. 330).
\text {steels :} \left\{\begin{array}{ll} S_{e^{\prime}} \cong 0.5 S_{u t} & \text { for } S_{u t}<200 kpsi (1400 MPa ) \\ S_{e^{\prime}} \cong 100 kpsi (700 MPa ) & \text { for } S_{u t} \geq 200 kpsi (1400 MPa ) \end{array}\right\} (6.5a)
S_{e^{\prime}}=0.5 S_{u t}=0.5(80000)=40000 psi (o)
14 The size factor for this rectangular part is determined by calculating the crosssectional area stressed above 95% of its maximum stress (see Figure 6-25 on p. 332) and using that value in equation 6.7d (p. 331) to find an equivalent-diameter test specimen.
d_{\text {equiv }}=\sqrt{\frac{A_{95}}{0.0766}} (6.7d)
\begin{array}{l} A_{95}=0.05 d b=0.05(1.0)(2.0)=0.1 in ^2\\ d_{\text {equiv }}=\sqrt{\frac{A_{95}}{0.0766}}=\sqrt{\frac{0.1}{0.0766}}=1.143 \text { in }\\ C_{\text {size }}=0.869\left(d_{\text {equiv }}\right)^{-0.097}=0.859 \end{array} (p)
15 Calculation of the corrected endurance limit S_{e} requires that several factors be
computed. C_{load} is found from equation 6.7a (p. 330). C_{surf} for a machined finish is
found from equation 6.7e (p. 333). C_{temp} is found from equation 6.7f (p. 335) and
C_{reliab} is chosen from Table 6-4 (p. 335) for a 99.9% reliability level.
\begin{array}{l} \text {bending :} \quad C_{\text {load }}=1 \\ \text {axial loading :} \quad C_{\text {load }}=0.70 \end{array} (6.7a)
C_{\text {surf }} \cong A\left(S_{u t}\right)^b \quad \text { if } C_{\text {surf }}>1.0 \text {, set } C_{\text {surf }}=1.0 (6.7e)
\begin{array}{ll} \text {for} T \leq 450^{\circ} C \left(840^{\circ} F \right): & C_{\text {temp }}=1 \\ \text {for} 450^{\circ} C <T \leq 550^{\circ} C : & C_{\text {temp }}=1-0.0058(T-450) \\ \text {for} 840^{\circ} F <T \leq 1020^{\circ} F : & C_{\text {temp }}=1-0.0032(T-840) \end{array} (6.7f)
\begin{aligned} S_e &=C_{\text {load }} C_{\text {size }} C_{\text {surf }} C_{\text {temp }} C_{\text {reliab }} S_{e^{\prime}} \\ &=1(0.859)(0.85)(1)(0.753) 40000=21883 psi \end{aligned} (q)
16 The four possible safety factors are calculated from equations 6.18 (pp. 367–368). The smallest or most appropriate one can be selected from those calculated. Equation (r) shows the Case 3 safety factor, which assumes that the alternating and mean components will have a constant ratio if they vary in maximum amplitude over the life of the part.
\begin{aligned} \sigma_{m@Q}^{\prime} &=\left\lgroup 1-\frac{\sigma_a^{\prime}}{S_y}\right\rgroup S_y \\ N_f &=\frac{\sigma_{m@Q}^{\prime} }{\sigma_{m@Z}^{\prime}}=\frac{S_y}{\sigma_m^{\prime}}\left\lgroup 1-\frac{\sigma_a^{\prime}}{S_y}\right\rgroup \end{aligned} (6.18a)
\begin{aligned} \sigma_{a@}^{\prime} P &=\left\lgroup 1-\frac{\sigma_m^{\prime}}{S_{u t}} \right\rgroup S_f \\ N_f &=\frac{\sigma_{a@P}^{\prime} }{\sigma_{a@Z}^{\prime} }=\frac{S_f}{\sigma_{a@Z}^{\prime}}\left\lgroup 1-\frac{\sigma_m^{\prime}}{S_{u t}} \right\rgroup \end{aligned} (6.18b)
\begin{array}{l} \text {from equation 6.16c :} \quad \sigma_{a @ R}^{\prime}=\left\lgroup 1-\frac{\sigma_{m @ R}^{\prime}}{S_{u t}} \right\rgroup S_f \\ \text {from line :} \quad \sigma_{a @ R}^{\prime}=\left\lgroup \frac{\sigma_{a @ Z}^{\prime}}{\sigma_m^{\prime}} \right\rgroup \sigma_m^{\prime}{ }_{m @ R}=\left\lgroup \frac{\sigma_a^{\prime}}{\sigma_m^{\prime}} \right\rgroup \sigma_{m @ R}^{\prime} \end{array} (6.18c)
\sigma_{m@R}^{\prime} =\frac{S_f}{\frac{\sigma_a^{\prime}}{\sigma_m^{\prime}}+\frac{S_f}{S_{u t}}} (6.18d)
N_f=\frac{\sigma_{m @ R}^{\prime}}{\sigma_{m @ Z}^{\prime}}=\frac{S_f S_{u t}}{\sigma_a^{\prime} S_{u t}+\sigma_m^{\prime} S_f} (6.18e)
\begin{array}{l} \sigma_{m @ S}^{\prime}=\frac{S_{u t}\left(S_f^2-S_f \sigma_a^{\prime}+S_{u t} \sigma_m^{\prime}\right)}{S_f^2+S_{u t}^2}\\ \sigma_{a @ S}^{\prime}=-\frac{S_f}{S_{u t}}\left(\sigma_{m@ S}^{\prime}\right)+S_f\\Z S=\sqrt{\left(\sigma_m^{\prime}-\sigma_{ m@S}^{\prime} \right)^2 + \left(\sigma_a^{\prime}-\sigma_{a @ S}^{\prime}\right)^2 }\end{array} (6.18f)
\begin{aligned} O Z &=\sqrt{\left(\sigma_a^{\prime}\right)^2+\left(\sigma_m^{\prime}\right)^2} \\ N_f &=\frac{O Z+Z S}{O Z} \end{aligned} (6.18g)
-\frac{\sigma_m^{\prime} }{S_{y c}}+\frac{\sigma_a^{\prime}}{S_{y c}}=1 (6.16a)
\sigma_a^{\prime}=S_f (6.16b)
\frac{\sigma_m^{\prime}}{S_{u t}}+\frac{\sigma_a^{\prime}}{S_f}=1 (6.16c)
\frac{\sigma_m^{\prime}}{S_y}+\frac{\sigma_a^{\prime}}{S_y}=1 (6.16d)
N_{f_3}=\frac{S_e S_{u t}}{\sigma_a^{\prime} S_{u t}+\sigma_m^{\prime} S_e}=\frac{21883(80000)}{8688(80000)+10425(21883)}=1.9 (r)
17 The maximum deflection is calculated using the maximum applied force F_{max}.
\begin{aligned} y_{@ x=l} &=\frac{F_{\max }}{6 E I}\left[x^3-3 a x^2-\langle x-a\rangle^3\right] \\ &=\frac{1100}{6(3 E 7)(0.1667)}\left[6^3-3(5)(6)^2-(6-5)^3\right]=-0.012 \text { in } \end{aligned} (s)
18 The data for this design are shown in Table 6-11. Using the same cross-section dimensions and the same alternating load as in Example 6-4 now gives a safety factor N_{f3} = 1.9 and maximum deflection y_{max} = 0.012 in for this fluctuating loading case, compared to N_{f3} = 2.5 and y_{max} = 0.005 in for the fully reversed loading situation of Example 6-4. The addition of a mean stress to the previous level of alternating stress reduced the safety factor and increased the deflection, as expected.
19 Increasing the part’s cross-section dimensions slightly gives the better design shown in Table 6-12. The final dimensions are b = 2 in, d = 1.2 in, D = 1.4 in, r = 0.5 in, a = 5 in, and l = 6.0 in. N_{f3} becomes 2.6 as shown in the Goodman diagram of Figure 6-48, and the maximum deflection becomes 0.007 in. These are both acceptable. The dimension D was deliberately chosen to be slightly less than a stock mill size so that some material would be available for machining in order to clean up and true the mounting surfaces.
20 The files EX06-05A and EX06-05B are on the CD-ROM.
| Table 6-11 Design of a Cantilever Bracket for Fluctuating Bending First Iteration for Example 6-5 (File EX06-05A) | ||||
| Input | Variable | Output | Unit | Comments |
| 2 | b | in | beam width | |
| 1 | d | in | beam depth over length | |
| 1.125 | D | in | beam depth in wall | |
| 0.5 | r | in | fillet radius | |
| 5 | a | in | beam length to load F | |
| 80000 | Sut | psi | ultimate tensile strength | |
| 60000 | Sy | psi | yield strength | |
| ‘machined | finish | ‘ground, ‘machined, ‘hotroll, ‘forged | ||
| ‘bending | loading | ‘bending, ‘axial, ‘shear | ||
| 99.9 | percent | reliability % desired | ||
| 1100 | Fmax | lb | maximum applied load | |
| 100 | Fmin | lb | minimum applied load | |
| Fa | 500 | lb | alternating applied force | |
| Fm | 600 | lb | mean applied force | |
| Kt | 1.18 | geometric stress-conc. factor | ||
| q | 0.898 | Peterson’s notch-sensitivity factor | ||
| Kf | 1.16 | fatigue stress-conc. factor — alter. | ||
| Kfm | 1.16 | fatigue stress-conc. factor — mean | ||
| siganom | 7500 | psi | alternating nominal stress | |
| siga | 8711 | psi | alternating stress with concentration | |
| sigavm | 8711 | psi | von Mises alternating stress | |
| sigmnom | 9000 | psi | mean nominal stress | |
| sigm | 10454 | psi | mean stress with concentration | |
| sigmvm | 10454 | psi | von Mises mean stress | |
| Seprime | 40000 | psi | uncorrected endurance limit | |
| Cload | 1 | load factor for bending | ||
| Csurf | 0.845 | machined finish | ||
| Csize | 0.859 | size factor based on 95% area | ||
| Ctemp | 1 | room temperature | ||
| Creliab | 0.753 | 99.9% reliability factor | ||
| Se | 21883 | psi | corrected endurance limit | |
| Nsf_1 | 5.5 | FS for sigalt = constant | ||
| Nsf_2 | 2.2 | FS for sigmean = constant | ||
| Nsf_3 | 1.9 | FS for sigalt/sigmean = constant | ||
| Nsf_4 | 1.7 | FS for closest failure line | ||
| Table 6-12 Design of a Cantilever Bracket for Fluctuating Bending Final Iteration for Example 6-5 (File EX06-05B) | ||||
| Input | Variable | Output | Unit | Comments |
| 2 | b | in | beam width | |
| 1.2 | d | in | beam depth over length | |
| 1.4 | D | in | beam depth in wall | |
| 0.5 | r | in | fillet radius | |
| 5 | a | in | beam length to load F | |
| 80000 | Sut | psi | ultimate tensile strength | |
| 60000 | Sy | psi | yield strength | |
| ‘machined | finish | ‘ground, ‘machined, ‘hotroll, ‘forged | ||
| ‘bending | loading | ‘bending, ‘axial, ‘shear | ||
| 99.9 | percent | reliability % desired | ||
| 1100 | Fmax | lb | maximum applied load | |
| 100 | Fmin | lb | minimum applied load | |
| Fa | 500 | lb | alternating applied force | |
| Fm | 600 | lb | mean applied force | |
| Kt | 1.22 | geometric stress-conc. factor | ||
| q | 0.898 | Peterson’s notch-sensitivity factor | ||
| Kf | 1.20 | fatigue stress-conc. factor — alter. | ||
| Kfm | 1.20 | fatigue stress-conc. factor — mean | ||
| siganom | 5208 | psi | alternating nominal stress | |
| siga | 6230 | psi | alternating stress with concentration | |
| sigavm | 6230 | psi | von Mises alternating stress | |
| sigmnom | 6250 | psi | mean nominal stress | |
| sigm | 7476 | psi | mean stress with concentration | |
| sigmvm | 7476 | psi | von Mises mean stress | |
| Seprime | 40000 | psi | uncorrected endurance limit | |
| Cload | 1 | load factor for bending | ||
| Csurf | 0.85 | machined finish | ||
| Csize | 0.85 | size factor based on 95% area | ||
| Ctemp | 1 | room temperature | ||
| Creliab | 0.753 | 99.9% reliability factor | ||
| Se | 21658 | psi | corrected endurance limit | |
| Nsf_1 | 8.6 | FS for sigalt = constant | ||
| Nsf_2 | 3.2 | FS for sigmean = constant | ||
| Nsf_3 | 2.6 | FS for sigalt/sigmean = constant | ||
| Nsf_4 | 2.3 | FS for closest failure line | ||
| Table 6-6 Neuber’s Constant for Steels |
|
| S _{ ut }( ksi ) | \sqrt{a}(\text{in}^{0.5}) |
| 50 | 0.130 |
| 55 | 0.118 |
| 60 | 0.108 |
| 70 | 0.093 |
| 80 | 0.080 |
| 90 | 0.070 |
| 100 | 0.062 |
| 110 | 0.055 |
| 120 | 0.049 |
| 130 | 0.044 |
| 140 | 0.039 |
| 160 | 0.031 |
| 180 | 0.024 |
| 200 | 0.018 |
| 220 | 0.013 |
| 240 | 0.009 |
| Table 6-4 Reliability Factors for S_{d} = 0.08 \mu |
|
| Reliability % | C_{reliab} |
| 50 | 1.000 |
| 90 | 0.897 |
| 95 | 0.868 |
| 99 | 0.814 |
| 99.9 | 0.753 |
| 99.99 | 0.702 |
| 99.999 | 0.659 |
| 99.9999 | 0.620 |
