Chapter 6
Q. 6.4
Design of a Cantilever Bracket for Fully Reversed Bending
Problem A feedroll assembly is to be mounted at each end on support brackets cantilevered from the machine frame as shown in Figure 641. The feed rolls experience a fully reversed load of 1 000lb amplitude, split equally between the two support brackets. Design a cantilever bracket to support a fully reversed bending load of 500lb amplitude for 10^{9} cycles with no failure. Its dynamic deflection cannot exceed 0.01 in.
Given The loadtime function shape is shown in Figure 641a. 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 or 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 millshapes is the preferred manufacturing method.
StepbyStep
Verified Solution
See Figure 641 and Tables 69 and 610 (pp. 358, 359).
1 This is a typical design problem. Very little data are given except for the required performance of 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 The first two steps of the process suggested above, finding the load amplitude and the number of cycles, are defined in the problem statement. We will begin at the third step, creating a tentative partgeometry design.
3 Figure 641a shows a tentative design configuration. A rectangular cross section is chosen to provide ease of mounting and clamping. A piece of coldrolled bar stock from the mill could simply be cut to length and drilled to provide the needed holes, then clamped into the frame structure. This approach appears attractive in its simplicity because very little machining is required. The millfinish on the sides could be adequate for this application. This design has some disadvantages, however. The mill tolerances on the thickness are not tight enough to give the required accuracy on thickness, so the top and bottom would have to be machined or ground flat to dimension. Also, the sharp corners at the frame where it is clamped provide stress concentrations of about K_{t} = 2 and also create a condition called fretting fatigue due to the slight motions that will occur between the two parts as the bracket deflects. This motion continuously breaks down the protective oxide coating, exposing new metal to oxidation and speeding up the fatiguefailure process. The fretting could be a problem even if the edges of the frame pieces were radiused.
4 Figure 641b shows a better design in which 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}. Figure 436 (p. 190) shows that with suitable control of the r/d and D/d ratios for a stepped flat bar in bending, the geometric stressconcentration factor K_{t} can be kept under about 1.5.
5 Some trial dimensions must be assumed for b, d, D, r, a, and l. We will assume (guess) values of b = 1 in, d = 0.75 in, D = 0.94 in, r = 0.25, a = 5.0, and l = 6.0 in to the applied load for our first calculation. This length will leave some material around the hole and still fit within the 6inlength constraint.
6 A material must also be chosen. For infinite life, low cost, and ease of fabrication, it is desirable to use a carbon steel if possible and 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 mediumcarbon, 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 steel with S_{ut} = 80 000 psi is selected for the first trial.
7 The reaction force and reaction moment at the support are found using equations h from Example 45. Next the area moment of inertia of the cross section, the distance to the outer fiber, and the nominal alternating bending stress at the root are found using the alternating load’s 500lb amplitude.
\begin{array}{l} R=F=500 lb \\ M=R lF(la)=500(6)500(65)=2500 lb \text { in } \end{array} (a)
\begin{array}{l} I=\frac{b d^3}{12}=\frac{1(0.75)^3}{12}=0.0352 in ^4 \\ c=\frac{d}{2}=\frac{0.75}{2}=0.375 in \end{array} (b)
\sigma_{a_{\text {nom }}}=\frac{M c}{I}=\frac{2500(0.375)}{0.0352}=26667 psi (c)
8 Two ratios must be calculated for use in Figure 436 (p. 190) in order to find the geometric stressconcentration factor K_{t} for the assumed part dimensions.
\frac{D}{d}=\frac{0.938}{0.75}=1.25 \quad \frac{r}{d}=\frac{0.25}{0.75}=0.333 (d)
\begin{array}{lll} \text { interpolating } & A=0.9658 & b=0.266 \end{array} (e)
K_t=A\left\lgroup\frac{r}{d}\right\rgroup^b=0.9658(0.333)^{0.266}=1.29 (f)
9 The notch sensitivity q of the chosen material is calculated based on its ultimate strength and the notch radius using equation 6.13 (p. 344) and the data for Neuber’s constant from Table 66 (p. 346).
q=\frac{1}{1+\frac{\sqrt{a}}{\sqrt{r}}} (6.13)
From the Table for S_{u t}=80 kpsi : \quad \sqrt{a}=0.08 (g)
q=\frac{1}{1+\frac{\sqrt{a}}{\sqrt{r}}}=\frac{1}{1+\frac{0.08}{\sqrt{0.25}}}=0.862 (h)
10 The values of q and K_{t} are used to find the fatigue stressconcentration factor K_{f}, which is in turn used to find the local alternating stress \sigma _{a} in the notch. Because we have the simplest case of a uniaxial tensile stress, the largest alternating principal stress \sigma _{1a} for this case is equal to the alternating tensile stress, as is the von Mises alternating stress \sigma_a^{\prime} . See equations 4.6 (p. 145) and 5.7c (p. 249).
\begin{aligned} \sigma_a, \sigma_b &=\frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left\lgroup\frac{\sigma_x\sigma_y}{2}\right\rgroup ^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)
\sigma^{\prime}=\sqrt{\sigma_1^2\sigma_1 \sigma_3+\sigma_3^2} (5.7c)
K_f=1+q\left(K_t1\right)=1+0.862(1.291)=1.25 (i)
\sigma_a=K_f \sigma_{a_{\text {nom }}}=1.25(26667)=33343 psi (j)
\begin{aligned} \tau_{a b} &=\pm \sqrt{\left(\frac{\sigma_x\sigma_y}{2}\right)^2+\tau_{x y}^2}=\sqrt{\left(\frac{333430}{2}\right)^2+0}=16672 psi \\ \sigma_{1_a}, \sigma_{3_a} &=\frac{\sigma_x+\sigma_y}{2} \pm \tau_{a b}=33343 psi , 0 psi \\ \sigma^{\prime} &=\sqrt{\sigma_1^2\sigma_1 \sigma_2+\sigma_2^2}=\sqrt{33343^233343(0)+0}=33343 psi \end{aligned} (k)
11 The uncorrected endurance limit S_{e^{\prime}} is found from equation 6.5a (p. 330). The size factor for this rectangular part is determined by calculating the crosssectional area stressed above 95% of its maximum stress (see Figure 625c, p. 332) and using that value in equation 6.7d (p. 331) to find an equivalent diameter test specimen for use in equation 6.7b (p. 331) to find C_{size}.
steels : \left\{\begin{array}{ll} S_{e^e} \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)
d_{\text {equiv }}=\sqrt{\frac{A_{95}}{0.0766}} (6.7d)
\text {for} d \leq 0.3 \text {in} (8 mm) : \quad C_{\text {size }}=1 \\ \text {for} 0.3 \text {in} \lt d \leq 10 \text {in} : \quad C_{\text {size }}=0.869 d^{0.097} \\ \text {for} 8 mm \lt d \leq 250 mm : \quad C_{\text {size }}=1.189 d^{0.097} (6.7b)
S_{e^{\prime}}=0.5 S_{u t}=0.5(80000)=40000 psi (l)
\begin{aligned} A_{95} &=0.05 db =0.05(0.75)(1)=0.04 in ^2 \\ d_{\text {equiv }} &=\sqrt{\frac{A_{95}}{0.0766}}=0.700 in \\ C_{\text {size }} &=0.869\left(d_{\text {equiv }}\right)^{0.097}=0.900 \end{aligned} (m)
12 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 64 (p. 335) for a 99.9% reliability level.
\begin{array}{ll} \text { bending: } & C_{\text {load }}=1 \\ \text { axial loading: } & C_{\text {load }}=0.70 \end{array} (6.7a)
C_{\text {surf }} \equiv A\left(S_{\text {ut }}\right)^b \quad \text { if } C_{\text {surf }}>1.0 \text {, set } C_{\text {surf }}=1.0 (6.7e)
\text {for} T \leq 450^\circ C (840^\circ F) : \quad C_{\text {temp }}=1\\ \text {for} 450^\circ C \lt T \leq 550^\circ C : \quad C_{\text {temp }}=10.0058(T450)\\ \text {for} 840^\circ F \lt T\leq 1020^\circ F : C_{\text {temp }}=10.0032(T840) (6.7f)
\begin{aligned} C_{\text {load }} &=1: & & \text { for bending } \\ C_{\text {surf }} &=A\left(\text { Sut }_{k p s i}\right)^b=2.7(80)^{0.265}=0.845: & & \text { machined } \\ C_{\text {temp }} &=1: & & \text { room temperature } \\ C_{\text {reliab }} &=0.753: & & \text { for } 99.9 \% \text { reliab. } \end{aligned} (n)
The corrected endurance limit is found from equation 6.6 (p. 330).
\begin{array}{l} S_e=C_{\text {load }} C_{\text {size }} C_{\text {surf }} C_{\text {temp }} C_{\text {reliab }} S_{e^{\prime}} \\ S_f=C_{\text {load }} C_{\text {size }} C_{\text {surf }} C_{\text {temp }} C_{\text {reliab }} S_f^{\prime} \end{array} (6.6)
\begin{aligned} S_e &=C_{\text {load }} C_{\text {size }} C_{\text {surf }} C_{\text {temp }} C_{\text {reliab }} S_{e^{\prime}} \\ &=1(0.900)(0.845)(1)(0.753) 40000=22907 psi \end{aligned} (o)
Note that the corrected S_{e} is only about 29% of S_{ut}.
13 The safety factor is calculated using equation 6.14 (p. 354) and the beam deflection y is computed using equation (j) from Example 45 (p. 168).
N_f=\frac{S_n}{\sigma^{\prime}} (6.14)
N_f=\frac{S_n}{\sigma^{\prime}}=\frac{22907}{33343}=0.69 (p)
\begin{aligned} y &=\frac{F}{6 E I}\left[x^33 a x^2\langle xa\rangle^3\right] \\ y_{@ x=l} &=\frac{500}{6(3 E 7)(0.0352)}\left[6^33(5)(6)^2(65)^3\right]=0.026 \text { in } \end{aligned} (q)
14 The results of all these computations for the first assumed design are seen in Table 69 (p. 358). The deflection of 0.026 in is not within the stated specification, and the design fails with a safety factor of less than one. So, more iterations are needed, as was expected. Any of the dimensions can be changed, as can the material. The material was left unchanged but the beam crosssectional dimensions and the notch radius were increased and the model rerun (this took only a few minutes) until the results shown in Table 610 (p. 359) were achieved.
15 The final dimensions are b = 2 in, d = 1 in, D = 1.125 in, r = 0.5, a = 5.0, and l = 6.0 in. The safety factor is now 2.5 and the maximum deflection is 0.005 in. These are both satisfactory. Note how low the fatigue stressconcentration factor is at K_{f} = 1.16. The dimension D was deliberately chosen to be slightly less than a stock mill size so that material would be available for the cleanup and truing of the mounting surfaces. Also, with this design, hotrolled steel (HRS) could be used, rather than the coldrolled steel (CRS) initially assumed (Figure 641a, p. 355). Hotrolled steel is less expensive than CRS and, if normalized, has less residual stress, but its rough, decarburized surface needs to be removed by machining all over, or to be treated with shot peening to strengthen it.
16 The files EX0604 are on the CDROM.
Table 66 Neuber’s Constant for Steels 

S _{ ut }( ksi )  \sqrt{a}\left(\operatorname{in}^{0.5}\right) 
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 69 Example 64 – Design of a Cantilever Bracket for Reversed Bending First Iteration: An Unsuccessful Design (File EX0604A)  
Input  Variable  Output  Unit  Comments 
500  F  lb  applied load amplitude at point a  
1  b  in  beam width  
0.75  d  in  beam depth over length  
0.94  D  in  beam depth in wall  
0.25  r  in  fillet radius  
6  l  in  beam length  
5  a  in  distance to load F  
6  lx  in  distance for deflection calculation  
3E7  E  psi  modulus of elasticity  
80000  sut  psi  ultimate tensile strength  
1  Cload  load factor for bending  
Csurf  0.85  machined finish  
1  Ctemp  room temperature  
0.753  Creliab  99.9% reliability factor  
R  500  lb  reaction force at support  
M  2500  inlb  reaction moment at support  
I  0.035 2  in^4  area moment of inertia  
c  0.38  in  dist to outer fiber  
signom  26 667  psi  bending stress at root  
Doverd  1.25  bar width ratio 1.01 < D/d < 2  
roverd  0.33  ratio radius to small dimension  
Kt  1.29  geometric stressconcentration factor  
q  0.86  Peterson’s notchsensitivity factor  
Kf  1.25  fatigue stressconcentration factor  
sigx  33343  psi  concentrated stress at root  
sig1  33343  psi  largest principal alternating stress  
sigvm  33343  psi  von Mises alternating stress  
Seprime  40000  psi  uncorrected endurance limit  
A95  0.04  in^2  95% stress area  
dequiv  0.7  in  equivalentdiameter test specimen  
Csize  0.9  size factor based on 95% area  
Se  22907  psi  corrected endurance limit  
N_{s f}  0.69  predicted safety factor  
y  0.026  in  deflection at end of beam 
Table 610 Example 64 – Design of a Cantilever Bracket for Reversed BendingFinal Iteration: A Successful Design (File EX0604B)  
Input  Variable  Output  Unit  Comments 
500  F  lb  applied load amplitude at point a  
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  
6  l  in  beam length  
5  a  in  distance to load F  
6  lx  in  distance for deflection calculation  
3E7  E  psi  modulus of elasticity  
80000  sut  psi  ultimate tensile strength  
1  Cload  load factor for bending  
Csurf  0.85  machined finish  
1  Ctemp  room temperature  
0.753  Creliab  99.9% reliability factor  
R  500  lb  reaction force at support  
M  2500  inlb  reaction moment at support  
I  0.1667  in^4  area moment of inertia  
c  0.5  in  dist to outer fiber  
signom  7500  psi  bending stress at root  
Doverd  1.13  bar width ratio 1.01 < D/d < 2  
roverd  0.50  ratio radius to small dimension  
Kt  1.18  geometric stressconcentration factor  
q  0.90  Peterson’s notchsensitivity factor  
Kf  1.16  fatigue stressconcentration factor  
sigx  8688  psi  concentrated stress at root  
sig1  8688  psi  largest principal alternating stress  
sigvm  8688  psi  von Mises alternating stress  
Seprime  40000  psi  uncorrected endurance limit  
A95  0.10  in^2  95% stress area  
dequiv  1.14  in  equivalentdiameter test specimen  
Csize  0.86  size factor based on 95% area  
Se  21843  psi  corrected endurance limit  
N_{s f}  2.5  predicted safety factor  
y  0.005  in  deflection at end of beam 