Design Project 2. Sizing of a Bicycle Chain Ring Bolt Set. Statics and Mechanics of Materials I. ENGR 0135 Section 1040.

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1 Design Project 2 Sizing of a Bicycle Chain Ring Bolt Set Statics and Mechanics of Materials I ENGR 0135 Section 1040 November 9, 2014 Derek Nichols Michael Scandrol Mark Vavithes

2 Nichols, Scandrol, Vavithes 1 ABSTRACT This report will detail our data, calculations, analysis, and processes that were used to redesign the design parameter d for the competitive bicycle crank arm assembly. Our group was able to determine at least two combinations of the radial distance of the bolts d and the chain ring bolt option given to us in the table (shown below) by using a factor of safety value of 3 and known equations. Knowing that the allowable shear stress must be less than the maximum shearing stress and that the factor of safety value must be 3, we were able to find an inequality for cross-sectional area. Also, knowing that the total moment of the system equals five times the moment at one bolt, we were able find an equation for force (P) that could be used in the earlier inequality, which can be used to find appropriate design parameters for each option. We also had to determine the bearing stress in the chain ring arm and the factor of safety with respect to bearing failure. Through known equations, mathematical analysis, and knowing that bearing stress multiplied by factor of safety equals the allowable shearing stress, we were able to determine the factor of safety for the options. Employing the help of visualizations and mathematical investigation, this report will show the process our group used to redesign the design parameter and determine the bearing stress and factor of safety for each option. Option Effective Area Diameter in in in in in in INTRODUCTION The purpose of the design project is to redesign the crank assembly of a bicycle so that it can withstand a force of 300 pounds with chain ring bolts placed from the bottom bracket at a radial distance between 0.8 and 1.5. The crank arm assembly consists of a chain ring, chain arm, pedal hole, chain ring arm, and five chain ring bolts spaced 72 degrees apart (Figure 1, Figure 2). A bicyclist pushes down on the pedal of the bicycle at the pedal hole causing the axis of the bottom bracket to rotate. The axis of the bottom bracket is connected the chain ring with the chain ring bolts (Figure 3). This causes the chain ring bolts to experience shearing stress. The stress is not large enough to cause deformation so the chain ring is able to turn. Given three different effective areas with a 0.2 bolt diameter, we must redesign a parameter d, the distance to the axis of the bottom bracket, so that the bolt can withstand a shear stress of 35,000 psi while maintaining a factor of safety against shear failure of three. Once these distances are found, we will find the bearing stress in the chain ring arm of the crank and estimate the factor of safety with respect to bearing failure.

3 Nichols, Scandrol, Vavithes 2 We hypothesize that the larger the effective area of the bolt, the closer the bolt will have to be to the axis of the bottom bracket. I. Calculations Initial Variables F pedal = 300 lbs L crank arm = 8 in Angle between chain ring bolts - 72 Τ max = 35,000 psi Distance of chain ring bolt from axis of bottom bracket 0.8 d 1.5 σ b max = 5,000 psi The elementary equation we used was for shearing stress: τ s = P A (1) In this formula, P is the shearing force applied over the cross-sectional area A of the bolts. To avoid failure of this system, the allowable shearing stress across the bolts must be less than or equal to the maximum shear stress that will be applied to the system: τ allowable τ maximum Using this equality, we can substitute τ allowable for τ s in equation (1) to get: τ allowable P A (2) Now to ensure a specific level of safety is integrated into design we use a Factor of Safety (FS). For this problem our FS will be 3. This gives us: τ allowable P A (FS) A 3P τ allowable (3)

4 Nichols, Scandrol, Vavithes 3 We will keep equation (3) in mind as we continue our calculations, but now we must find the force in terms of distance from the center of the gear. The clear way to accomplish this goal is to use moments. First, we find the moment about the center of this structure A. M A = F pedal L crank arm = (300lbs)(8in) = 2400lbs in clockwise Additionally, we know that this system is at equilibrium so we can conclude that this moment is offset by the moment of each of the arms of the bottom bracket. Also, we are able to substitute P here because it is the force we are trying to find and the bolt in this case is in single shear. Therefore, we can assume: Rearranging for P in terms of d we get: M A = 5M C = 5Pd d P = 2400 (lb in) 5d d (4) Now, we recall equation (3) and we are able to replace P with the value in equation(4): A 7200 (lb in) 5d d τ allowable We also know that τ allowable 35,000 (psi) so including this information gives: A 7200 (lb in) 5(35000 psi)d d Finally, we are only interested in exact values and not a range of values so we can replace our inequality: A = in3 d d (5) In continuation, by using the Effective Area information provided, we can solve for d d for each option of ring bolt: Option Effective Area d d = in3 A d d in in in in in in in in in in in in

5 Nichols, Scandrol, Vavithes 4 Design Objective 1: After applying the constraint that d d must fall between the values of 0.8 and 1.5 (0.8 d d 1.5"), it is clear that only two of the options satisfy this: Design Objective 2: in in in in Now we must find the bearing stress in the chain ring arm of the crank. To do this we will use the equation: = P (6) A bearing In this case, P is still equal to the equation (4) we found previously, but now the area is the total cross sectional area that is bearing the stress. For this reason, we get the equation: A bearing = Dt (7) where D is the diameter of the chain ring bolt and t is the thickness of the chain ring arm (both given values). By combining equations (4), (6), and (7) we get: = 2400 (lb in) 5d d Dt Using this new equation, we can find the bearing stress of each arm to be: Option Diameter d d Arm Thickness = in in 0.4 in 2400 (lb in) 5d d Dt 2400 (lb in) (5)( in)(0.2 in)(0.4 in) psi in in 0.4 in 2400 (lb in) (5)( in)(0.2 in)(0.4 in) psi Finally, by recalling equation (2), and combining it with equation (6), we can derive that: (FS) = τ allowable In this case, τ allowable is given as 5,000 psi for the alloy used to make the crank arm so we can estimate that the Factor of Safety with respect to bearing failure is:

6 Nichols, Scandrol, Vavithes 5 Option psi psi τ allowable 5000 (psi) (psi) 5000 (psi) (psi) Factor of Safety DISCUSSION Our goal was to determine an appropriate distance, d, from the bolt to the axis of the bottom bracket from the effective area of the bolts and other given values. We then had to find the factor of safety to ultimately make the choice of which distance and chain ring bolt was the best option for the bicycle. We knew that in order to find which effective area yielded a distance between 0.8 and 1.5, we first got the area in terms of force, stress, and factor of safety, all of which were given. We also know the moment about the axis of the bottom bracket could be found from the force exerted by the bicyclist and length the crank arm. Knowing that the summation of moments about the axis of the bottom bracket must equal zero, the sum of the moments in each bolt must equal the moment produced by the bicyclist. This means that the moment from the bicyclist equals five times the moment in each bolt (because there are five bolts). Solving for the force in each bolt, we could plug this back into our effective area equation (#3). Plugging in the allowable stress in the bolts, we could solve chain ring bolt option 1, 2, and 3 for the radial distance from the axis of the bottom bracket by plugging each bolt option s respective effective area. Comparing the calculated distances to the allowable range, we discovered that options two and three were the only ones that meet the radial distance constraint. Using options two and three, we then found the factor of safety with respect to bearing failure for each bolt. We did this first by realizing that bearing stress is equal to force divided by area. We also know that the area for the bearing is equal to the diameter times the thickness. By substituting back into equation (#6) and plugging in our known variables, we were able to solve for the bearing stress for each suitable option. By combining equations (#2) and (#6), we could solve for the factor of safety for bolts two and three. The bolt with the highest factor of safety was the most suitable option for this application. We believe that our results are accurate because by using unit analysis in equation (#5), we obtain units of in^2 which is correct, and our values for the distances yield reasonable results. CONCLUSION Through our complex analysis of the chain ring bolt, we were able to mathematically model and provide a suggestion for a suitable option for a different size of chain ring bolt that would work in this application. By decreasing the effective area of each bolt, the distance from the axis of the bottom bracket increased. Taking all of these details into account, our hypothesis that the larger the area, the closer the bolt would have to be to the axis of the bottom bracket is supported. We believe option three will be the most effective chain ring bolt because it produces a distance that is in the range of 0.8 and 1.5 and has the highest factor of safety with respect to bearing failure of the bolts meeting the constraint for distance.