Question 5.CS.1C: Bicycle Brake Lever Failure Analysis Problem Determine the f...

See Figures 3-1 and 5-22, and file CASE1C.

1    The bending stress at point P in Figure 5-22 at the root of the cantilever was found from equation 4.11b in Case Study 1B and is

\sigma_{\max }=\frac{M c}{I}       (4.11b)

\sigma_x=\frac{M c}{I}=\frac{(267 N \cdot 0.0762  m )\left\lgroup \frac{0.0143}{2} \right\rgroup  m }{\frac{\pi(0.0143)^4}{64} m ^4}=70.9  MPa          (a)

2    This is the only applied stress at this point, so it is also the principal stress. The von Mises effective stress \sigma^{\prime}=\sigma_1 in this case (see equation 5.7c, p. 249). The safety factor against yielding at point P is then (from equation 5.8a, p. 249),

\sigma^{\prime}=\sqrt{\sigma_1^2-\sigma_1 \sigma_3+\sigma_3^2}       (5.7c)

N=\frac{S_y}{\sigma^{\prime}}     (5.8a)

N_{\text {yield }}=\frac{S_y}{\sigma^{\prime}}=\frac{186}{70.9}=2.6          (b)

This design is safe at the average load but there is not a large margin to protect from overloads. Note that in this simple stress situation, the distortion-energy theory gives identical results to the maximum shear theory because the ellipse and hexagon are coincident at the point x = \sigma _{1}, y = 0 in Figure 5-5 (p. 252).

3    Since this is a cast material with limited ductility, it would be interesting to also compute the modified-Mohr safety factor against brittle fracture from equation 5.12a (p. 260). It could be argued that the brake handle might still be usable despite slight yielding having occurred:

N=\frac{S_{u t}}{\sigma_1}     (5.12a)

N_{\text {fracture }}=\frac{S_{u t}}{\sigma_1}=\frac{310}{70.9}=4.4         (c)

Note that we have not accounted for stress concentrations at the root of the cantilever, which could reduce this fracture safety factor. Case Study 1D in Chapter 8 determines the stress concentration factor at point P by using Finite Element Analysis.

4    The transverse shear stress at point Q in Figure 5-22 was calculated from equation 4.15c (p. 161) as

\tau_{\max }=\frac{4}{3} \frac{V}{A}       (4.15c)

\tau_{x y}=\frac{4 V}{3 A}=\frac{4(267) N }{\frac{3 \pi(14.3)^2}{4}  mm ^2}=2.22  MPa          (d)

This shear stress is also the maximum, since no other stresses act at that point. The safety factor using the distortion-energy theory for pure shear at point Q is

N_{\text {transverse }}=\frac{S_{y s}}{\tau_{\max }}=\frac{0.577 S_y}{\tau_{\max }}=\frac{0.577(186)}{2.22}=48       (e)

Clearly, there is no danger of transverse-shear failure at point Q.

5    The compressive bearing stress in the hole at point A in Figure 5-22 is

and this stress, acting alone, is also the principal and the von Mises stress. Assuming that the compressive strength of this material is equal to its tensile strength (an even material), the safety factor against bearing failure in the hole is

N_{\text {bearing }}=\frac{S_{y c}}{\sigma^{\prime}}=\frac{186}{19.2}=9.7        (g)

6    Tearout in this case requires that (4) 6.4-mm-thick sections fail in shear through the material between hole A and the edge. (See also Figure 4-13 on p. 155.)

This is a pure shear case and the safety factor is found from

N_{\text {tearout }}=\frac{S_{y s}}{\tau_{\max }}=\frac{0.577 S_y}{\tau_{\max }}=\frac{0.577(186)}{10.7}=10.0        (i)

7    The cable-end inserts in a blind hole which is half-slotted to allow the cable to pass through at assembly as shown in Figure 5-22. This slot weakens the part and makes the section at R the most likely failure location at this joint. The bending stress at the outer fiber is

\sigma_x=\frac{M c}{I}=\frac{\frac{1914}{2}\left\lgroup \frac{5}{2} \right\rgroup(4)}{\frac{10(5)^3}{12}}=91.9  MPa       (j)

As the only applied stress at the outer fiber of this section, this is also the principal stress and the von Mises stress. The bending safety factor at point R is

N=\frac{S_{y c}}{\sigma^{\prime}}=\frac{186}{91.9}=2.0       (k)

8    The shear due to transverse loading at the neutral axis in the section at R is (Eq. 4.14b, p. 161):

\tau_{\max }=\frac{3}{2} \frac{V}{A}        (4.14b)

\tau_{x y}=\frac{3 V}{2 A}=\frac{3(957)}{2(10)(5)}=28.7  MPa         (l)

This is the maximum shear stress at the neutral axis, and the transverse-shear safety factor at point R is

N=\frac{S_{y s}}{\tau_{\max }}=\frac{0.577 S_y}{\tau_{\max }}=\frac{0.577(186)}{28.7}=3.7        (m)

It is interesting to note that the transverse-shear safety factor is only about twice the bending safety factor at point R because the beam is so short. Compare this result with that at point P in equations (b) and (e), where the bending and transverse shear safety factors differ by about a factor of 18 in the longer beam. See Figure 5-22.

9    The die-cast aluminum alloy chosen is one of the strongest aluminum casting alloys available. If additional protection against overloads (as from the bicycle falling) is desired, either a change of geometry to increase the section size and/or reduce stress concentrations or a change in material or manufacturing method could be made. A forged-aluminum part would be stronger but would increase the cost. Thicker sections would increase the weight slightly, but probably not prohibitively. Increasing the diameter of the handle around point P by 26% to 18 mm (with perhaps a more generous transition radius also) would double the safety factor there, since the section modulus is a function of d^{3}.

Even though some of the other safety factors may seem excessive, it may be impractical to reduce those sections due to difficulties in casting thin sections. Other considerations may take into account the appearance of a part intended for a consumer application such as a bicycle. If the dimensions do not look “right” to the customer, it may give an undesired impression of cheapness. Sometimes it is better economics to provide more thickness than is necessary for a suitable safety factor in order to provide a quality appearance.