Question 6.CS.6: Redesign of a Failed Laybar for a Water-Jet Power Loom Probl...

See Figures 6-50 to 6-56 and Table 6-13 (p. 398).

1    Some additional background information is needed to understand the problem before pursuing its solution. Weaving looms for the making of cloth are very old devices and were originally human-powered. The power loom was invented during the industrial revolution and presently exists in many forms. Figure 6-50 shows parts of the waterjet power loom of concern here. Perhaps the best way to understand a loom’s fundamental operation is to consider a hand-powered one, which the reader may have seen in a museum, a custom-weaving shop, or a hobbyist’s workroom. Its basic elements are similar to some of those in the figure.

A set of threads called the warp is strung across the loom. Each thread is grabbed by a device (not shown) that can pull it up or down. These devices are activated by a mechanism that, in a hand-loom, is typically operated by foot pedals. When one pedal is pushed, every other warp thread is raised up and the alternate ones pulled down to create a “tunnel” if observed from the edge of the cloth. This tunnel is called the shed. A shuttle, which looks like a miniature canoe, and contains a bobbin of thread within it, is next “thrown” through the shed by the weaver’s hand. The shuttle trails a single thread called the weave through the warp shed. The weaver then pulls on the laybar, which carries a comblike device called a reed. The warp threads are strung through the teeth of this reed-comb. The reed pushes the new weave thread sideways into the previous ones to “beat-up” the cloth and create a tight weave. The weaver next switches pedals on the shed and the “up” threads of the warp become the “down” threads and vice versa, creating a new tunnel (shed) of crossed threads. The shuttle is again thrown through the warp (from the other side), weaving another thread to be beat-up by the reed.

The original power looms simply mechanized the manual process, replacing the weaver’s hands and feet with linkages and gears. The throwing of the wooden shuttle was accomplished by literally hitting it with a stick, flying it through the shed and catching it on the other side. The dynamics of this (pre-NASA) “shuttle flight” became the limiting factor on the loom’s speed. Shuttle looms can only go at about 100 picks (threads) per minute (ppm). Much effort was expended to develop faster looms, and these usually involved eliminating the shuttle, whose mass limited the speed. Both air-jet and water-jet looms were developed in the 20th century that shoot the weave thread across the shed on a jet of air or water. Figure 6-50 shows the orifice through which the thread is fed. At the right time in the cycle, a small piston pump shoots a jet of water through the orifice and surface tension pulls the thread across the shed. The water-jet loom can operate at up to about 500 ppm. The looms in question were designed to operate at 400 ppm but the owner changed their gearing to increase the speed to 500 ppm. Failures soon ensued because the dynamic loads increased with the square of the speed and exceeded the loads for which the machine was designed.

2    Figures 6-50 and 6-51 show the laybar, which is carried between two identical fourbar linkages that move it in an arc to push the reed into the cloth at the right point in the cycle. The laybar is bolted securely to the linkage rockers at each end and rotates with them. The linkage pivots are in self-aligning ball-bearings, which allow us to model the laybar as a simply supported beam that carries a uniformly distributed load that is equal to its total mass times its acceleration plus the beat-up force. The total mass is the sum of the laybar mass and the mass of its 10-lb payload, the reed. Figure 6-51 shows the laybar-linkage geometry, its dimensions, and a polar plot of the acceleration vectors at the mass-center of the laybar. The tangential components of acceleration are the largest and create bending moments in the directions of the inertial forces shown in the same figure. Figure 6-52 plots the tangential acceleration component of the laybar mass-center for 1 cycle and shows the beat-up force in its phase relationship with acceleration. The acceleration creates a fluctuating bending moment, and the beat-up force, being offset 3.75 in from the mass center of the laybar, creates a repeated torque on the laybar. Depending on the laybar’s cross-sectional geometry, this combination of loads can create a case of synchronous, in-phase, simple multiaxial stress at locations of maximum stress (see Section 6.12 on p. 376). Because the loading is largely inertial, the design of the laybar should minimize its mass (to reduce the inertial loading) while simultaneously maximizing its stiffness and strength. These are conflicting constraints, making the design task more challenging.

3    Since this is a case of fluctuating loading, we will follow the set of design steps recommended in Section 6.11 (p. 360), the first of which is to determine the number of loading cycles expected over the service life. The owner has requested that the new design last for 5 years of 3-shift operation. Assuming 2 080 hours per shift in a standard work-year, this amounts to

N=500 \frac{\text { cycles }}{\min }\left\lgroup\frac{60  min }{ hr }\right\rgroup \left\lgroup\frac{2   080  hr }{\text { shift }- yr }\right\rgroup (3  \text { shifts })(5  yr )=9.4 E 8  \text { cycles }       (a)

This is clearly in the HCF regime and could benefit from the use of a material with an endurance limit.

The owner reports that his steel replacement laybar lasted about 6 months and his aluminum design lasted 3 months. (See Figures 6-53 and 6-54.) The cycle lives are:

6 \text { mos: }\quad N=500 \frac{\text { cycles }}{\min }\left\lgroup \frac{60  min }{ hr } \right\rgroup\left\lgroup \frac{2  080  hr }{ yr } \right\rgroup(3 \text { shifts })(0.5  yr )=9.4 E 7  \text { cycles } \\ 3 \text {mos :} \quad N=500 \frac{\text { cycles }}{\min }\left\lgroup \frac{60  min }{ hr } \right\rgroup\left\lgroup \frac{2  080  hr }{ yr } \right\rgroup(3 shifts )(0.25  yr )=4.7 E 7  \text { cycles }      (b)

4    Since the amplitudes of the applied bending loads are a function of acceleration (which is determined) and the mass of the part (which will vary with the design), it is best to express the bending loads in terms of F = ma. The applied torque is assumed to be the same for any design based on the owner’s estimate of a typical beat-up force. These data are shown in Figure 6-52 and the mean and alternating terms are

\begin{aligned} \quad F_{\text {mean }} &=m a_{\text {mean }}=m \frac{a_{\max }+a_{\min }}{2}=m \frac{8  129+(-4  356)}{2}=1  886.5 m \\ \text {Bending :} \\ \quad F_{\text {alt }} &=m a_{a l t}=m \frac{a_{\max }-a_{\min }}{2}=m \frac{8  129-(-4  356)}{2}=6  242.5 m \end{aligned}       (c)

\quad T_{\text {mean }}=\left\lgroup\frac{F_{\text {beat }_{\max }}+F_{\text {beat }_{\min }}}{2} \right\rgroup r=\left\lgroup\frac{540+0}{2} \right\rgroup 3.75=1  012.5  lb -\text { in } \\ \text {Torque :} \\ \quad T_{\text {alt }}=\left\lgroup\frac{F_{\text {beat } \max }-F_{\text {beat } \min }}{2} \right\rgroup r=\left\lgroup\frac{540-0}{2} \right\rgroup 3.75=1  012.5  lb \text { – in }       (d)

5    We are in the enviable position of having test data for typical parts run under actual service conditions available in the form of the failed specimens. In effect, the owner had inadvertently run a test program (to his chagrin) and had determined stress levels that caused failure in this application. Thus the first step will be to analyze the existing failed designs in order to learn more about the problem. We know that laybars of the original design (Figure 6-53a) survived for 5 years at the lower stress levels associated with 400-rpm operation. They only began to fail when he increased the speed, which increased the inertial loading. There are many factors involved in this application that are difficult to quantify. Corrosion is evident on the failed parts. The steel laybars are rusted with pitted surfaces. The unanodized aluminum part is also pitted. The designer(s) did not take great care to minimize stress concentrations, and the fatigue fractures can be seen to have started (as is typical) at stress-raisers. The failed aluminum part in Figure 6-54b shows that the crack started at a tapped hole, which is a very sharp notch. The cracks in the steel part (Figure 6-54b) appear to have started at a weld bead used to attach the reed supports. Welds are notorious stress-raisers and always leave tensile residual stresses behind. We should take these lessons into account in our redesign and attempt to reduce these negative factors. By definition, a failed part has a safety factor of 1. Knowing this, a model of the part’s loading, stresses, and safety factor can be created and then back-calculated with the safety factor set to 1 in order to determine various of the above factors that are difficult to quantify for a particular application.

6    A model was created to solve the equations for this case. Data specific to each of the three failed designs were input and the model modified as necessary to account for differences in geometry and material among the three designs. The same model was further modified to accommodate the proposed new designs shown in Figure 6-55. Eight versions of the model resulted, and their data files are included with this text. They are labeled CASE6-0 through CASE6-7. Space does not permit discussion of the contents of all eight models, so only two will be discussed in detail, and the results of the others will be compared in summary. The failed original design and the final new design will be presented. The reader may open the models in the program of his choice if desired.

7    The analysis of the original laybar design is contained in CASE6-1. The section geometry and the beam mass must be calculated to determine the bending stresses.

\begin{aligned} \text { area } &=2.375^2-2.205^2+2(0.56)(0.085)=0.874  in ^2 \\ \text { weight } &=\text { area }(\text { length })(\gamma)=0.874(54)(0.286)=13.5  lb \\ m &=\frac{\text { weight }+\text { reed }}{386.4}=\frac{13.5+10}{386.4}=0.061  blobs \\ I &=\frac{b_{\text {out }} h_{\text {out }}^3-b_{\text {in }} h_{\text {in }}^3}{12}=\frac{2.375^4-2.205^4}{12}=0.68  in ^4 \end{aligned}       (e)

Note that the area calculation includes the reed-trough sides, since they add mass but the calculation for I ignores them since they add a negligible amount to that quantity. The specific weight γ is for steel and the mass unit is blobs or lb-sec²/in.

8    The nominal mean and alternating components of the inertial force and bending moment can now be calculated.

\begin{aligned} F_{\text {mean }} &=1886.5 m=1  886.5(0.061)=115  lb \\ F_{a l t} &=6242.5 m=6  242.5(0.061)=380  lb \\ M_{\text {mean }} &=\frac{w l^2}{8}=\frac{(w l) l}{8}=\frac{F l}{8}=\frac{115(54)}{8}=775  lb – in \\ M_{a l t} &=\frac{w l^2}{8}=\frac{(w l) l}{8}=\frac{F l}{8}=\frac{380(54)}{8}=2  565  lb – in \end{aligned}      (f)

The moment equations are for the maximum moment in the center of a simply supported beam with uniformly distributed load (see Figure B-2 in Appendix B). The nominal bending stresses (not including any stress concentration) are then

\begin{aligned} \sigma_{m_{\text {nom }}} &=\frac{M_{\text {mean }}(c)}{I}=\frac{775(1.188)}{0.68}=1  351  psi \\ \sigma_{a_{\text {nom }}} &=\frac{M_{\text {alt }}(c)}{I}=\frac{2  565(1.188)}{0.68}=4  470  psi \end{aligned}          (g)

9    The nominal shear stresses due to torsion in a hollow-square section are maximum in the centers of the four sides and so occur at points of maximum bending stress. The shear stress is found from \tau_{\max }=T / Q (Eq. 4.26a, p. 178) where Q for the particular geometry is found in Table 4-2 (p. 172):

\tau_{\max }=\frac{T}{Q}     (4.26a)

Q=2 t(a-t)^2=2(0.085)\left\lgroup \frac{2.375}{2}-0.085\right\rgroup ^2=0.207 \text { in }^3        (h)

where t is the wall thickness and a is the half-width of the cross section. The nominal mean and alternating shear stresses are then

\begin{aligned} \tau_{m_{\text {nom }}} &=\frac{T_{\text {mean }}}{Q}=\frac{1  012.5}{0.207}=4  900  psi \\ \tau_{a_{\text {nom }}} &=\frac{T_{\text {alt }}}{Q}=\frac{1  012.5}{0.207}=4  900  psi \end{aligned}        (i)

10    The stress-concentration factors for bending and shear need to be found or estimated. Peterson[30] provides a chart for the case of a hollow, rectangular section in torsion, and from that a K_{ts} = 1.08 is found. No suitable data were found for the bending stress-concentration factor for this case. The corrosion and pitting in combination with rough welds would predict a large K_{t }. The approach taken here was to backsolve for K_{t} with the safety factor set to 1 and all other material factors and the nominal stresses specified. The result was K_{t} = 4.56 in this failed part. This result is presented at this point to provide continuity of narrative, but it must be understood that the value of K_{t} was found by back-solving the model using iteration after all other factors were defined. The local alternating and mean stresses and K_{t} were then solved for simultaneously with N_{f} = 1, representing a failure condition.

11    The material’s notch sensitivity and the estimated fatigue stress concentration factors for alternating bending and shear are found from equations 6.11b and 6.13 (p. 343) following the procedure used in Example 6-3 (p. 346). Using the value of K_{t} found in step 10 and q = 0.8, the results are: K_{f} = 3.86, and K_{fs} = 1.06. The corresponding fatigue stress-concentration factors for the mean stress are found from equation 6.17 (p. 364) and since the local stress is below the yield point for both bending and torsion in this case, they are identical to the factors for the alternating stress: K_{fm} = K_{f}, and K_{fms} = K_{fm}.

K_f=1+q\left(K_t-1\right)     (6.11b)

q=\frac{1}{1+\frac{\sqrt{a}}{\sqrt{r}}}      (6.13)

\begin{array}{ll} \text {if}  K_f\left|\sigma_{\max _{\text {nom }}}\right|<S_y \text { then: } & K_{f m}=K_f \\ \text {if}  K_f\left|\sigma_{\max }\right|>S_y \text { then : } & 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: } & K_{f m}=0 \end{array}     (6.17)

12    The estimated local-stress components can now be found using the fatigue stress-concentration factors:

\begin{array}{c} \sigma_m=K_{f m} \sigma_{m_{\text {nom }}}=3.86(1  351)=5  212  psi \\ \sigma_a=K_f \sigma_{a_{\text {nom }}}=3.86(4  470)=17  247  psi \end{array}     (j)

\begin{array}{l} \tau_m=K_{f s m} \tau_{m_{n o m}}=1.06(4  900)=5  214  psi \\ \tau_a=K_{f s m} \tau_{a_{\text {nom }}}=1.06(4  900)=5  214  psi \end{array}       (k)

13    Since we have a case of combined, fluctuating, biaxial stresses that are synchronous and in-phase, and that have stress concentration, the general method using von Mises effective stresses for both mean and alternating components is appropriate (equation 6.22b). These calculate equivalent alternating and mean stresses for the biaxial case.

\begin{aligned} \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{17  247+0-0+3(5  214)}=19468  psi \\ \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{5  212+0-0+3(5  214)}=10  428  psi \end{aligned}       (l)

14    The material properties must now be determined. A laboratory test was done on a sample from the failed part, and its chemical composition matched that of an AISI 1018 cold-rolled steel. Strength values for this material were obtained from published data (see Appendix A) and are: S_{ut} = 64 000 psi and S_{y} = 50 000 psi. A shear yield strength was calculated from S_{ys} = 0.577S_{y} = 28 850 psi. An uncorrected endurance limit was taken as S_{e^{\prime}}=0.5 S_{u t} = 32 000 psi.

15    The strength modification factors were found from the equations and data in Section 6.6 (p. 327). The loading is a combination of bending and torsion. However, we have incorporated the torsion stresses into the von Mises equivalent stress, which is a normal stress, so

C_{\text {load }}=1     (m)

The equivalent-diameter test specimen is found from the 95% stress area using equations 6.7c and 6.7d (p. 331). The size factor is then found from equation 6.7b (p. 331):

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

d_{\text {equiv }}=\sqrt{\frac{A_{95}}{0.0766}}      (6.7d)

\begin{array}{ll} \text {for}  d \leq 0.3 \text { in }(8 mm ): & C_{\text {size }}=1 \\ \text {for}  0.3 \text { in }<d \leq 10 \text { in : } & C_{\text {size }}=0.869 d^{-0.097} \\ \text {for}  8 mm <d \leq 250 mm : & C_{\text {size }}=1.189 d^{-0.097} \end{array}     (6.7b)

\begin{aligned} A_{95} &=0.05 b h=0.05(2.375)^2=0.282 \text { in }^2 \\ d_{\text {equiv }} &=\sqrt{\frac{A_{95}}{0.0766}}=1.92  in \\ C_{\text {size }} &=0.869\left(d_{\text {equiv }}\right)^{-0.097}=0.869(1.92)^{-0.097}=0.82 \end{aligned}        (n)

The surface factor is found from equation 6.7e (p. 333) and the data in Table 6-3 (p. 333) for machined or cold-drawn surfaces. The material of the laybar appeared to have been originally cold drawn but was corroded. Corrosion could dictate the use of a lower-valued surface factor, but it was decided to allow the geometric stress-concentration factor K_{t} to account for the effects of pitting in this case as described above, and the machined-surface factor was applied.

C_{\text {surf }} \cong A\left(S_{\text {ut }}\right)^b \quad \text { if } C_{\text {surf }}>1.0 \text {, set } C_{\text {surf }}=1.0     (6.7e)

C_{\text {surf }}=A\left(S_{u t}\right)^b=2.7(64)^{-0.265}=0.897      (o)

The temperature factor and the reliability factor were both set to 1. The reliability was taken as 50% for this back-calculation in order to place all the uncertainty in the highly variable stress-concentration factor.

16    A corrected endurance limit can now be calculated from

\begin{array}{l} S_e=C_{\text {load }} C_{\text {size }} C_{\text {surf }} C_{\text {temp }} C_{\text {reliab }} S_{e^\prime} \\ S_e=(1)(0.81)(0.90)(1)(1)(32  000)=23  258  psi \end{array}       (p)

17    The safety factor is calculated from equation 6.18e (p. 368). Safety factor case 3 is applicable here since, with inertial loading, the mean and alternating components of bending stress will maintain a constant ratio with changes in speed. Because the minimum beat-up force is always zero, that ratio is also constant regardless of the maximum force.

N_f=\frac{\sigma_m^{\prime}{ }_m}{\sigma_m^{\prime}{ }_{m Z}}=\frac{S_f S_{u t}}{\sigma_a^{\prime} S_{u t}+\sigma_m^{\prime} S_f}      (6.18e)

\begin{aligned} N_f &=\frac{S_e S_{u t}}{\sigma_a^{\prime} S_{u t}+\sigma_m^{\prime} S_e} \\ &=\frac{23  258(64  000)}{19  468(64  000)+10  428(23  258)}=1.0 \end{aligned}        (q)

The alternating and mean stresses were back-solved and can now be used to plot a modified-Goodman diagram. Since we forced the safety factor to 1 to represent the known failure of this part, the applied stress point \sigma_a^{\prime}, \sigma_m^{\prime} falls on the Goodman line.

18    The above analysis was repeated, changing the operating speed to the original design value of 400 rpm. Using the same stress concentration factor of 4.56 that was backed out of the failed-part analysis, the safety factor at 400 rpm is 1.3, indicating why the original design survived at the design rpm (file Case6-0).

19    The analysis of this and the other failed parts provides some insight into the constraints of the problem and allows a better design to be created. Some of the factors that influenced the new design were the corrosive environment, which makes steel a less desirable material despite its endurance limit. The painted finishes had not protected the now rusty steel parts. The one failed aluminum sample examined also showed significant pitting in only 3 months of use. If aluminum is to be used, an anodized finish should be applied to protect it from oxidation.

Another obvious factor is the role of stress concentration, which appears to be quite high in this part. The presence of welds and tapped holes in regions of high stress clearly contributed to the failures. Any new design should reduce stress concentrations by moving the required screw holes for reed attachment to locations of lower stress. Welds in high-stress regions should be avoided if possible. Surface treatments such as shot peening should be considered in order to provide beneficial compressive residual stresses.

The poorly defined but potentially significant levels of torsional stresses are a concern. The weaving of heavier cloth will provide larger levels of torsional stress. Thus the geometry of any new design should be resistant to torsional stress as well as to bending stress. Finally, a new design should not be much heavier than the existing design since additional mass will cause the higher inertial forces to be transmitted to all other parts of the machine, possibly engendering failures of other parts.

20    Because the loading on the beam is primarily inertial in nature and because it carries a fixed-weight payload, there should be an optimum cross section for any design. A beam’s resistance to bending is a function of its area moment of inertia I. The inertial loading is an inverse function of its area A. If the cross section were solid, its I would be the largest possible for a given outside dimension but so would its area, mass, and inertial load. If the wall were made paper-thin, its mass would be minimal but so would its I. Both A and I are nonlinear functions of its dimensions. Thus, there must be some particular wall thickness that maximizes the safety factor, all else constant.

With all the above factors in mind, two designs were considered, as shown in Figure 6-55: a square and a round cross section with integral external ears to support the reed. They both share some common features. The contours have generous radii to minimize K_{t}. (The round section is the ultimate in this regard.) The reed-support ears, which must contain threaded holes, are close to the neutral axis where the bending stress is lower and they are external to the basic geometric structure. There may not need to be any welds if the shape can be extruded as shown. Both are basically closed sections that can resist torques, and the round section is the optimum shape for torsional loading. The square section will have a larger I and will thus resist bending better than a round shape of the same overall dimension.

Two materials were considered, mild steel and aluminum. (Titanium would be ideal in terms of its strength-to-weight ratio (SWR) and endurance limit, but its high cost precludes consideration.) Aluminum (if anodized) has the advantage of better corrosion resistance in water, but steel has the advantage of an endurance limit if protected from corrosion. The overall weight of the new design is of concern. High-strength aluminum has a better SWR than mild steel. (High-strength steel does not show an endurance limit and is notch sensitive as well as expensive.) Aluminum can be custom extruded with integral ears for low tooling cost, thus making a short production run economically feasible. Tooling for a custom cross section in steel would require very large quantities be purchased to amortize the tooling cost. So, a steel design will be limited to stock mill-shapes and will require welding-on of the ears.

21    Each of the geometries in Figure 6-55 was designed separately in both steel and aluminum. The wall thickness was varied within each model as a list of values from very thin to nearly solid in order to determine the optimum dimension. The final design chosen was a round section in 6061-T6 extruded aluminum with a wall thickness of 0.5 in. This design will now be discussed, though it must be understood that a great deal of iteration was required to arrive at the result presented here. Space does not permit discussion of all the iterations.

22    The previous calculations of cycle life (equation a, step 3), acceleration and beat-up force (equations c and d, step 4) are still applicable. The section properties are

\begin{aligned} \text { area } &=\pi\ \left\lgroup \frac{2.5^2-1.5^2}{4} \right\rgroup +2(0.5)(0.75)=3.892  in ^2 \\ \text { weight } &=\text { area }(\text { length })(\gamma)=3.892(54)(0.10)=21  lb \\ m &=\frac{\text { weight }+\text { reed }}{386}=\frac{21+10}{386}=0.080  \text { blobs } \\ I &=\pi\left\lgroup \frac{2.5^4-1.5^4}{64} \right\rgroup =1.669  \text { in }^4 \end{aligned}      (r)

Note that the area calculation includes the ears since they add mass but the calculation for I ignores them since they add a negligible amount to that quantity. The weight density γ for aluminum is in lb/in³ and the mass unit is blobs or lb-sec²/in.

23    The mean and alternating components of the inertial force and bending moment are

\begin{aligned} F_{\text {mean }} &=1  886.5 m=1  886.5(0.08)=152  lb \\ F_{a l t} &=6  242.5 m=6242.5(0.08)=502  lb \\ M_{\text {mean }} &=\frac{F l}{8}=\frac{152(54)}{8}=1  023  lb – in \\ M_{a l t} &=\frac{F l}{8}=\frac{502(54)}{8}=3  386  lb – in \end{aligned}        (s)

The moment equations are for the maximum moment in the center of a simply supported beam with uniformly distributed load. The nominal bending stresses (not including stress concentration) are then

\begin{aligned} \sigma_{m_{\text {nom }}} &=\frac{M_{\text {mean }}(c)}{I}=\frac{1  023(1.25)}{1.669} \cong 766  psi \\ \sigma_{a_{\text {nom }}} &=\frac{M_{a l t}(c)}{I}=\frac{3  386(1.25)}{1.669} \cong 2  536  psi \end{aligned}      (t)

If we compare these results to those of the original design (see step 8), the forces and moments are now greater due to the heavier part but the stresses are less due to the larger I of the cross section.

24    The torsional shear stresses in a hollow-round section are maximum at the outer fiber and so occur at points of maximum bending stress. The nominal shear stress is found from \tau_{\max }=\operatorname{Tr} / J where J for its geometry is found from equation 4.25b (p. 178):

J=\frac{\pi\left(d_o^4-d_i^4\right)}{32}     (4.25b)

J=\pi\left\lgroup \frac{d_{\text {out }}^4-d_{i n}^4}{32}\right\rgroup =\pi\left\lgroup \frac{2.5^4-2.0^4}{32}\right\rgroup =3.338  \text { in }^4      (u)

The nominal mean and alternating shear stresses are then

\begin{array}{l} \tau_{m_{\text {nom }}}=\frac{T_{\text {mean }}(r)}{J}=\frac{1  012.5}{3.338}=379  psi \\ \tau_{a_{\text {nom }}}=\frac{T_{\text {alt }}(r)}{J}=\frac{1  012.5}{3.338}=379  psi \end{array}         (v)

25    Because of the large radius and smooth contours of this round part, K_{t} and K_{ts} were taken as 1. There will be larger stress concentrations at the roots of the ears, but the bending stress is much lower there and the shear stress in this torsionally optimum shape is very low. The material’s notch sensitivity is irrelevant when K_{t} = 1, making both K_{f} = 1 and K_{fs} = 1. The fatigue stress-concentration factors for the mean stress are also 1 with the above assumptions. The local-stress components are then the same as the nominal stress components found in equations t and \nu.

26    Since we have a case of combined, fluctuating, biaxial stresses that are synchronous and in-phase, and the notches have been designed out, the method of Sines is now appropriate (equation 6.21b, p. 378). This calculates equivalent alternating and mean stresses for the biaxial, unnotched case.

\begin{aligned} \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{2536+0-0+3(379)}=2619  psi \\ \sigma_m^{\prime} &=\sigma_{x_m}+\sigma_{y_m}=766+0=766  psi \end{aligned}            (w)

27    The material properties must now be determined. Aluminum does not have an endurance limit, but fatigue strengths at particular cycle lives are published. A search of the literature showed that of all the aluminum alloys, 7075 and 5052 offered the largest fatigue strengths S_{f}. However, no aluminum extruders were found locally that could extrude either of those alloys. The strongest extruded alloy available was 6061-T6 with a published S_{f}’ = 13 500 psi at N = 5E7, S_{ut} = 45 000, and S_{y} = 40 000 psi.

28    The strength modification factors were found from the equations and data in Section 6.6. The loading is a combination of bending and torsion, which appears to create a conflict in the selection of a loading factor from equation 6.7a (p. 330). However, we have incorporated the torsion stresses into the Sines equivalent stress, which is a normal stress, so C_{load} = 1. The equivalent diameter for a nonrotating round part is found from Figure 6-25 and equation 6.7d (p. 331). The size factor is then found from equation 6.7b (p. 331):

\begin{array}{l}  \text {bending :} \quad C_{\text {load }}=1 \\ \text {axial loading :} \quad C_{\text {load }}=0.70 \end{array}    (6.7a)

\begin{aligned} A_{95} &=0.010462 d^2=0.010462(2.5)^2=0.065  \text { in }^2 \\ d_{\text {equiv }} &=\sqrt{\frac{A_{95}}{0.0766}}=\sqrt{\frac{0.065}{0.0766}}=0.924  \text { in } \\ C_{\text {size }} &=0.869\left(d_{\text {equiv }}\right)^{-0.097}=0.869(0.924)^{-0.097}=0.88 \end{aligned}         (x)

The surface factor is found from equation 6.7e (p. 333) and the data in Table 6-3 (p. 333) for machined or cold-drawn surfaces. Corrosion could dictate the use of a lowervalued surface factor, but since the part will be anodized for corrosion resistance, the machined-surface factor was applied.

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)

C_{\text {surf }}=A S_{u t}^b=2.7(45)^{-0.265}=0.98       (y)

Note that S_{u}t is in kpsi for equation 6.7e. The temperature factor was set to 1, since it operates in a room-temperature environment. C_{reliab} was set to 0.702 from Table 6-4 (p. 335) to represent a desired 99.99% reliability for the new design.

29    A corrected fatigue strength can now be calculated from

\begin{aligned} S_n &=C_{\text {load }} C_{\text {size }} C_{\text {surf }} C_{\text {temp }} C_{\text {reliab }} S_{f^{\prime}} \\ S_{n_{@ \text { @E } 7}} &=1(0.88)(0.98)(1)(0.702)(13  500) \cong 8  173  psi \end{aligned}        (z)

30    This corrected fatigue strength S_{n_{5E7}} is at the published test life of N = 5E7 cycles. Since we need a life of about N = 9.4E8 cycles, the equation for this material’s S-N curve must be written and solved for S_{n} at N = 9.4E8 cycles. To do so we need the material’s strength S_{m} at 10³ cycles from equation 6.9a:

\begin{array}{l} \text {bending :} \quad S_m=0.9 S_{u t} \\ \text {axial loading :} \quad S_m=0.75 S_{u t} \end{array}     (6.9)

S_m=0.9 S_{u t}=0.9(45  000)=40  500  psi       (aa)

Use equations 6.10 to find the coefficient and exponent of the S-N line. The value of –4.699 comes from Table 6-5 (p. 338) and corresponds to the number of cycles (5E7) at which these published test data were taken.

S(N)=a N^b    (6.10a)

\log S(N)=\log a+b \log N    (6.10b)

\begin{aligned} b &=\frac{1}{-4.699} \log \left\lgroup \frac{S_m}{S_{n_{@ 5 E 7}}} \right\rgroup =\frac{1}{-4.699} \log \left\lgroup \frac{40  500}{8173} \right\rgroup =-0.14792 \\ \log (a) &=\log \left(S_m\right)-3 b=\log (40500)-3(-0.14792)=5.0512 ; \quad a=112  516 \end{aligned}      (ab)

S_{n_{@ 9.4 E 8}}=a N^b=112  516(9.4 E 8)^{-0.14792} \cong 5  296  psi      (ac)

This value will be used as a corrected S_{n} at the desired life.

31    The equivalent alternating and mean stresses can now be plotted on a modified-Goodman diagram or the safety factor can be calculated from equation 6.18e (p. 368) for a case 3 situation as described in step 17.

N_f=\frac{\sigma_m^{\prime} m R}{\sigma_m^{\prime} @ Z}=\frac{S_f S_{u t}}{\sigma_a^{\prime} S_{u t}+\sigma_m^{\prime} S_f}      (6.18e)

N_f=\frac{S_n S_{u t}}{\sigma_a^{\prime} S_{u t}+\sigma_m^{\prime} S_n}=\frac{5  296(45  000)}{2  536(45  000)+766(5  296)}=2.0       (ad)

This safety factor is quite acceptable, but for an additional measure of safety, the finished parts were shot peened before anodizing. The variation of safety factor with wall thickness is shown in Figure 6-56. The peak occurs at a wall thickness of about 0.5 in, which was the value used in the design. Curves of safety factor versus wall thickness are plotted for all designs within the models. They are similar in shape to Figure 6-56 and all show an optimum wall thickness to maximize the safety factor.

32    The file names and pertinent data for the seven designs are shown in Table 6-13. The original design is shown both for the machine’s design speed of 400 rpm, at which it performed successfully, and for the increased speed of 500 rpm, at which it failed. The only difference is the safety factor, which went from 1.3 to 1. The K_{t} factors are backcalculated for the failed designs (safety factor = 1) as described above but are estimated for the new designs. The steel designs use the back-calculated K_{t} from the failed original laybar to account for possible corrosion and weld concentration effects. The square aluminum design has an elevated K_{t} due to its internal corners.