Question 4.CS.4B: Bicycle Brake Arm Stress Analysis Problem Determine the stre...
Bicycle Brake Arm Stress Analysis
Problem Determine the stresses at critical points in the bicycle brake arm shown in Figures 3-9 (repeated here) and 4-54.
Given The geometry and loading are known from Case Study 4A (p. 94) and are shown in Table 3-5 (p. 97). The cast-aluminum arm is a tee-section curved beam whose dimensions are shown in Figure 4-54. The pivot pin is ductile steel. The loading is three-dimensional.
Assumptions The most likely failure points are the arm as a double-cantilever beam (one end of which is curved), the hole in bearing, and the connecting pin in bending as a cantilever beam. Since this is a marginally ductile cast material (5% elongation to fracture), we can ignore the stress concentration on the basis that local yielding will relieve it.
| Table 3–5 – part 1 Case Study 4A Given and Assumed Data |
||
| Variable | Value | Unit |
| \mu | 0.4 | none |
| \theta | 172.0 | deg |
| R_{12x} | 5.2 | mm |
| R_{12y} | -27.2 | mm |
| R_{12z} | 23.1 | mm |
| R_{32x} | –75.4 | mm |
| R_{32y} | 38.7 | mm |
| R_{32z} | 0.0 | mm |
| R_{52x} | -13.0 | mm |
| R_{52y} | –69.7 | mm |
| R_{52z} | 0.0 | mm |
| F_{32x} | 353.0 | N |
| F_{32y} | 523.0 | N |
| F_{32z} | 0.0 | N |
| M_{12z} | 0.0 | N–m |
| Table 3–5 – part 2 Case Study 4A Calculated Data |
||
| Variable | Value | Unit |
| F_{12x} | –1 805 | N |
| F_{12y} | –319 | N |
| F_{12z} | 587.0 | N |
| F_{52x} | 1 452 | N |
| F_{52y} | -204 | N |
| F_{52z} | -587 32 | N |
| M_{12x} | 32 304 | N–mm |
| M_{12y} | 52 370 | N–mm |
| N | 1 467 | N |
| F_{f} | 587 | N |
See Figures 4-54 to 4-56.
1 The brake arm is a double-cantilever beam. Each end can be treated separately. The curved-beam portion has a tee-shaped cross section as shown in Section X-X of Figure 4-54. The neutral axis of a curved beam shifts from the centroidal axis toward the center of curvature a distance e as described in Section 4-9 and in equation 4.12a (p. 157). To find e requires an integration of the beam cross section and knowledge of its centroidal radius. Figure 4-55 shows the tee-section broken into two rectangular segments, flange and web. The radius of the centroid of the tee is found by summing moments of area for each segment about the center of curvature:
e=r_c-\frac{A}{\int \frac{d A}{r}} (4.12a)
\begin{aligned} \sum M &=A_1 r_{c_1}+A_2 r_{c_2}=A_t r_{c_t} \\ r_{c_t} &=\frac{A_1 r_{c_1}+A_2 r_{c_2}}{A_t}=\frac{A_1\left(r_i+y_1\right)+A_2\left(r_i+y_2\right)}{A_1+A_2} \\ r_{c_t} &=\frac{(20)(7.5)(58+3.75)+(10)(7.5)(58+11.25)}{(20)(7.5)+(10)(7.5)}=64.25 mm \end{aligned} (a)
See Figures 4-53 and 4-54 for dimensions and variable names. The integral dA/r for equation 4.12a can be found in this case by adding the integrals for web and flange.
\int_0^{r_o} \frac{d A}{r}=\frac{A_1}{r_{c_1}}+\frac{A_2}{r_{c_2}}=\frac{(20)(7.5)}{58+3.75}+\frac{(10)(7.5)}{58+11.25}=3.51 mm (b)
The radius of the neutral axis and the distance e are then
\begin{aligned} r_n &=\frac{A_t}{\int_0^{r_o} \frac{d A}{r}}=\frac{225}{3.51}=64.06 mm \\ e=& r_c-r_n=64.25-64.06=0.187 mm \end{aligned} (c)
The magnitude of the bending moment acting on the curved section of the beam at Section X-X can be approximated by taking the cross product of the force F_{32} and its position vector R_{AB} referenced to the pivot at B in Figure 4-54.
\left|M_{A B}\right|=\left|R_{A B x} F_{32_y}-R_{A B y} F_{32_x}\right|=|-80.6(523)-66(353)|=65 452 N – mm (d)
The stresses at the inner and outer fibers can now be found using equations 4.12b and 4.12c (p. 158) (with lengths in mm and moments in N-mm for proper unit balance):
\sigma_i=+\frac{M}{e A}\left\lgroup \frac{c_i}{r_i}\right\rgroup (4.12b)
\sigma_o=-\frac{M}{e A}\left\lgroup \frac{c_o}{r_o}\right\rgroup (4.12c)
\begin{array}{l} \sigma_i=+\frac{M}{e A}\left\lgroup \frac{c_i}{r_i} \right\rgroup =\frac{65 452(6.063)}{(0.1873)(225)(58)}=162 MPa \\ \sigma_o=-\frac{M}{e A}\left\lgroup \frac{c_o}{r_o} \right\rgroup =\frac{65 452(8.937)}{(0.1873)(225)(73)}=-190 MPa \end{array} (e)
2 The hub cross section, shown as Section B-B in Figure 4-54, is a possible location of failure, since there is a combination of bending and axial tension stresses here and the pin hole removes substantial material. The bending stress is due to the maximum moment acting on the curved beam at its root and the tensile stress is due to the y component of the force at A. There is also a shear stress due to transverse loading, but this will be zero at the outer fiber where the sum of the bending and axial stresses is maximum. The area and area moment of inertia of the hub cross section are needed:
\begin{array}{l} A_{\text {hub }}=\text { length }\left\lgroup d_{\text {out }}-d_{\text {in }} \right\rgroup =28.5(25-11)=399 mm ^2 \\ I_{\text {hub }}=\frac{\text { length }\left\lgroup d_{\text {out }}^3-d_{\text {in }}^3 \right\rgroup }{12}=\frac{28.5\left\lgroup25^3-11^3 \right\rgroup }{12}=33 948 mm ^4 \end{array} (f)
The stress on the left half of Section B-B is the sum of the bending and axial stresses:
\sigma_{h u b}=\frac{M c}{I_{h u b}}+\frac{F_{32}}{A_{h u b}}=\frac{65 452(12.5)}{33 948}+\frac{523}{399}=25.4 MPa (g)
The stress on the right half of Section B-B is lower, because the compression due to bending is reduced by the axial tension.
3 The straight portion of the brake arm is a cantilever beam loaded in two directions, in the xy plane and in the yz plane. The section moduli and moments are different in these bending directions. The z moment in the xy plane is equal and opposite to the moment on the curved section. The cross section at the root of the cantilever is a rectangle of 23 by 12 mm as shown in Figure 4-54. The bending stress at the outer fiber of the 23-mm side due to this moment is
\sigma_{y_1}=\frac{M c}{I}=\frac{65 452\left\lgroup \frac{12}{2}\right\rgroup }{\frac{23(12)^3}{12}}=118.6 MPa (h)
The x moment is due to force F_{52z} acting at the 42.5 radius, bending the link in the z direction. The bending stress at the surface of the 12-mm side is
\sigma_{y_2}=\frac{M c}{I}=\frac{589 \cdot 42.5\left\lgroup \frac{23}{2} \right\rgroup }{\frac{12(23)^3}{12}}=23.7 MPa (i)
These two y-direction normal stresses add at the corners of the two faces to give
\sigma_y=\sigma_{y_1}+\sigma_{y_2}=118.6+23.7=142.2 MPa (j)
4 Another possible failure point is the slot in the cantilever arm. Though the moment is zero there, the shear force is present and can cause tearout in the z direction. The tearout area is the shear area between the slot and edge.
\begin{aligned} \left.A_{\text {tearout }}=\text { thickness(width }\right)=8(4)=32 mm ^2 \\ \tau &=\frac{ F _{52_{ z }}}{A_{\text {tearout }}}=\frac{589}{32}=18.4 MPa \end{aligned} (k)
5 The pivot pin is subjected to force F_{21}, which has both x and y components and to a couple M_{21} due to the forces F_{12z} and F_{52z} . The force F_{21} creates a bending moment having components F_{21x}l and F_{21y}l in the yz and xz planes, respectively, where l = 29 mm is the length of the pin.
\begin{aligned} M_{p i n} &=\sqrt{\left\lgroup M_{21_x}-F_{12_y} \cdot l \right\rgroup ^2+\left\lgroup -M_{21_y}+F_{12_x} \cdot l \right\rgroup ^2} \\ &=\sqrt{(-32 304+319 \cdot 29)^2+(52 370-1805 \cdot 29)^2} \\ &=\sqrt{(-32 304+9251)^2+(52 370-52 345)^2} \\ &=\sqrt{(-23 053)^2+(25)^2}=23 053 N – mm \end{aligned} (l)
\theta_{M_{\text {pin }}}=\tan ^{-1}\left\lgroup \frac{25}{-23 053} \right\rgroup \cong 0^{\circ} (m)
Figure 4-56a shows the moment of the couple M_{21} and Figure 4-56b shows the moment of the force F_{21}. Their combination is shown in Figure 4-56c.
It is this combined moment that creates the largest bending stresses in the pin at 0° and 180° around its circumference. The maximum bending stress in the pin (with lengths in mm and moments in N-mm for unit balance) is
\sigma_{p i n}=\frac{M_{p i n} c_{p i n}}{I_{p i n}}=\frac{23 053\left\lgroup \frac{11}{2} \right\rgroup }{\frac{\pi(11)^4}{64}}=\frac{23 053(5.5)}{718.7}=176 MPa (n)
6 A more complete analysis could be done using finite element methods to determine the stresses and deformations at many other locations on the part. You may examine the model for this case study by opening the file CASE4B in the program of your choice. A stress analysis of this case study is also done in Chapter 8 using Finite Element Analysis (FEA).
