Bicycle Brake Arm Factors of Safety Problem Choose appropriate material alloys to obtain a safety factor against fracture of at least 2 at critical points in the bicycle brake arm in Figures 3-9 (repeated here) and 5-25. Given The stresses are known from Case Study 4B (p. 217). The arm is cast

Question

Bicycle Brake Arm Factors of Safety

Problem Choose appropriate material alloys to obtain a safety factor against fracture of at least 2 at critical points in the bicycle brake arm in Figures 3-9 (repeated here) and 5-25.

Given The stresses are known from Case Study 4B (p. 217). The arm is cast aluminum and the pivot pin is steel.

Assumptions Since the arm is a cast (uneven) material, the modified-Mohr theory will be used to find a safety factor against fracture. The pin is ductile, so the maximum distortion-energy theory will be used for it.

Accepted Answer

See Figures 3-9 and 5-25, and the file CASE4C.

1 The bending stresses of the arm, [latex]\sigma _{i}[/latex] at the inner fibers (point [latex]A[/latex] in Figure 5-25) and [latex]\sigma _{o}[/latex] at the outer fibers (point [latex]B[/latex] in Figure 5-25), were found to be

[latex] \begin{array}{c} \sigma_i=+\frac{M}{e A}\left\lgroup\frac{c_i}{r_i} ight group =\frac{65452(6.063)}{(0.1873)(225)(58)}=162 MPa \ \sigma_o=-\frac{M}{e A}\left\lgroup\frac{c_o}{r_o} ight group =\frac{65452(8.937)}{(0.1873)(225)(73)}=-190 MPa \end{array} [/latex] (a)

For a safety factor of 2 at this point we need a material with an ultimate tensile strength of at least 325 MPa and a compressive strength of at least 380 MPa.

2 The stress on the left half of Section B-B at point [latex]C[/latex] in Figure 5-25 is the sum of the bending stress and the axial-tension stress:

[latex] \sigma_{h u b}=\frac{M c}{I_{h u b}}+\frac{F_{32_y}}{A_{h u b}}=\frac{65452(12.5)}{33948}+\frac{523}{399}=25.4 MPa [/latex] (b)

This needs a material tensile strength of about 52 MPa to have a safety factor of 2.

3 The bending stress at the outer fiber of the 23-mm side of the straight portion of the brake arm (point [latex]D[/latex] in Figure 5-25) is

[latex] \sigma_y=\sigma_{y_1}+\sigma_{y_2}=118.6+23.7=142 MPa [/latex] (c)

For a safety factor of 2 at this point we need a material with an [latex]S_{ut}[/latex] of at least 284 MPa.

4 Another possible failure point is the slot in the arm (point [latex]E[/latex] in Figure 5-25). The tearout shear stress is

[latex] au=\frac{ F _{52_z}}{A_{ ext {tearout }}}=\frac{589}{32}=18.4 MPa [/latex] (d)

For a uniaxial applied shear stress, the stresses are all in the first quadrant of Figure 5-11 (p. 260) and the modified-Mohr theory is identical to the maximum normal-stress theory. The equivalent tensile stress is then twice the maximum shear stress, which requires an ultimate tensile strength greater than 75 MPa for a safety factor of 2 in this case.

5 The worst-stress case among the points calculated on the arm is point [latex]A[/latex] with [latex]\sigma _{y}[/latex] = 162 MPa. This is a uniaxial stress, so is principal. Table A-3 in Appendix A shows an A132 permanent-mold-casting aluminum alloy, heat treated at 340°F, with an [latex]S_{ut}[/latex] of 324 MPa.

Table A-3 Mechanical Properties for Some Aluminum Casting Alloys Data from INCO.* Approximate Values. Consult Material Manufacturers for More Accurate Information
Aluminum Casting Alloy	Condition	Tensile Yield Strength (0.2% offset)		Ultimate Tensile Strength		Elongation over 2 in	Brinell Hardness
Aluminum Casting Alloy	Condition	kpsi	MPa	kpsi	MPa	%	-HB
43	permanent mold casting—as cast	9	62	23	159	10	45
195	sand casting—solution heat treated	24	165	36	248	5	-
220	sand casting—solution heat treated	26	179	48	331	16	75
380	die casting—as cast	24	165	48	331	3	-
A132	permanent mold casting—heat treated + 340°F	43	296	47	324	0.5	125
A142	sand casting—heat treated + 650°F	30	207	32	221	0.5	85
* Properties of Some Metals and Alloys, International Nickel Co., Inc., New York.

This material satisfies the requirement for a fracture safety factor ≥ 2 in this case.

[latex] N_f=\frac{S_{u t}}{\sigma_{b_A}}=\frac{324 MPa }{162 MPa }=2.0 [/latex] (e)

6 The 0.2% offset tensile yield strength of this cast material is given as 296 MPa. Since point [latex]A[/latex] has a uniaxial stress, the von Mises effective stress [latex] \sigma^{\prime}=\sigma_y [/latex]. The safety factor against tensile yielding at point [latex]A[/latex] is

[latex] N_y=\frac{S_y}{\sigma_A^{\prime}}=\frac{296 MPa }{162 MPa }=1.8 [/latex] (f)

7 The 11-mm-dia steel pivot pin has a maximum bending stress at point [latex]F[/latex] of

[latex] \sigma_{ ext {pin }}=\frac{M_{ ext {pin }} c_{ ext {pin }}}{I_{ ext {pin }}}=\frac{23053\left\lgroup \frac{11}{2} ight group }{\frac{\pi(11)^4}{64}}=176 MPa [/latex] (g)

For a ductile yielding safety factor of 2, this requires a steel with a yield strength of at least 352 MPa. An AISI 1040 steel has [latex]S_{y }[/latex] = 372 MPa in the normalized condition. This gives the desired safety factor of just over 2.

8 Figure 5-25 shows a summary of the stresses at various critical points in the part. You may examine the model for this case study by opening the file CASE4C in the program of your choice.

Question 5.CS.4C: Bicycle Brake Arm Factors of Safety Problem Choose appropria...

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