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Q. 6.4

Design of a Cantilever Bracket for Fully Reversed Bending

Problem    A feed-roll assembly is to be mounted at each end on support brackets cantilevered from the machine frame as shown in Figure 6-41. The feed rolls experience a fully reversed load of 1 000-lb amplitude, split equally between the two support brackets. Design a cantilever bracket to support a fully reversed bending load of 500-lb amplitude for $10^{9}$ cycles with no failure. Its dynamic deflection cannot exceed 0.01 in.

Given    The load-time function shape is shown in Figure 6-41a. 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 mill-shapes is the preferred manufacturing method.

Verified Solution

See Figure 6-41 and Tables 6-9 and 6-10 (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 part-geometry design.

3    Figure 6-41a shows a tentative design configuration. A rectangular cross section is chosen to provide ease of mounting and clamping. A piece of cold-rolled 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 mill-finish 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 fatigue-failure process. The fretting could be a problem even if the edges of the frame pieces were radiused.

4    Figure 6-41b 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 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.

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 6-in-length 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 medium-carbon, ductile steels have the requisite endurance-limit 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 4-5. 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 500-lb amplitude.

$\begin{array}{l} R=F=500 lb \\ M=R l-F(l-a)=500(6)-500(6-5)=2500 lb -\text { in } \end{array}$             (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 4-36 (p. 190) in order to find the geometric stress-concentration 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 6-6 (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 stress-concentration 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)

$K_f=1+q\left(K_t-1\right)=1+0.862(1.29-1)=1.25$     (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{33343-0}{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{33343^2-33343(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 cross-sectional area stressed above 95% of its maximum stress (see Figure 6-25c, 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 6-4 (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)

$\text {for} T \leq 450^\circ C (840^\circ F) : \quad C_{\text {temp }}=1\\ \text {for} 450^\circ C \lt T \leq 550^\circ C : \quad C_{\text {temp }}=1-0.0058(T-450)\\ \text {for} 840^\circ F \lt T\leq 1020^\circ F : C_{\text {temp }}=1-0.0032(T-840)$    (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 4-5 (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^3-3 a x^2-\langle x-a\rangle^3\right] \\ y_{@ x=l} &=\frac{500}{6(3 E 7)(0.0352)}\left[6^3-3(5)(6)^2-(6-5)^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 6-9 (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 cross-sectional dimensions and the notch radius were increased and the model rerun (this took only a few minutes) until the results shown in Table 6-10 (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 stress-concentration 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, hot-rolled steel (HRS) could be used, rather than the cold-rolled steel (CRS) initially assumed (Figure 6-41a, 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 EX06-04 are on the CD-ROM.

 Table 6-6 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 6-9    Example 6-4 – Design of a Cantilever Bracket for Reversed Bending  First Iteration: An Unsuccessful Design (File EX06-04A) 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 in-lb 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 stress-concentration factor $q$ 0.86 Peterson’s notch-sensitivity factor $Kf$ 1.25 fatigue stress-concentration 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 equivalent-diameter 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 6-10 Example 6-4 – Design of a Cantilever Bracket for Reversed BendingFinal Iteration: A Successful Design (File EX06-04B) 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 in-lb 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 stress-concentration factor $q$ 0.90 Peterson’s notch-sensitivity factor $Kf$ 1.16 fatigue stress-concentration 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 equivalent-diameter 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