MTRL 485 – Case 5

Spindle Failure

Group 4

Masseir, Alex (55761118)May, Michael (47957105)

McGrew, Kennan (55398101)Mei, Helen (31494115)

Muhammad, Harith (18204115)

Task 1

Figure 1: Fatigue SurfaceThe location for the fatigue crack origin was determined by the colour difference of the sample as well as the beach marks surrounding the origin. Beach marks are the curved lines on the sample which indicates fatigue propagation and they are caused by the change of cyclic loads in a sample. Using these markings, we can determine the fatigue crack origin. The dark coloured surface is suspected to be the origin of the fatigue crack because it most likely grew over periods of days or months, leading to oxidation on the surface. The region that appears to be brightest is the fast fracture region where the sample failed rapidly due to the sufficiently large increase of the crack. The fast fracture region is also verified by the rough section of the sample. It is presumed that the stress intensity exceeded the critical value, creating the rapid failure. Figure 1 also shows the hardened layers at the edges of the sample which are formed from case hardening.

Run # C, wt% Mn, wt% Si, wt% Fe, wt%1 - 1.1 0.4 98.52 34.3 0.6 0.1 65.03 37.7 0.5 61.84 - - 1005 37.7 - 62.3

Table 1: EDX analysis of the fracture surface

Referring to Table 1, the composition of the samples can be determined however the data is rather scattered especially for the carbon and iron content. Therefore, it is assumed that the analysis have been performed incorrectly.

Task 2ForcesThe three main forces in the bike pedal spindle example were calculated using the three equations below and F0 = 500N

Fθ=F °sin (θ )

Fn=F°∗si n2 (θ )=¿Fn=Fθ∗cos (θ)

Fr=F°∗sin (θ )∗cos (θ )=¿Fr=Fθ∗sin (θ)

Its was expected that Fθ and Fn would be equal at θ=90° because at this point the forces are pointing in the same direction as Fθ is perpendicular at this point like Fn always is. The assumption that the two forces would also be greatest at point are correct because when θ = 90° both Fn and Fθ are pointing in the same downward direction as the initial for applied by the foot of the biker F°. Fθ is also larger at all time except 90° because it is always counting Fo a downward force and Fn is always perpendicular to the spot of the pedal compared to the rotation of the crank. Fr becomes negative at any angle above 90° because sin(θ) is negative between 90°-180° and because below 90° Fr is pointing down and 0° - just above 90° it points up. Fn and Fθ never become negative because between 0° - 180° in this example they are always pointing downward. A graph of the force vs. degrees in below. All values can be found in the appendix

Graph 1: Forces vs. Degrees

Reaction Forces

The equations below were used to calculate the reaction forcesΣ∋¿0

Nx+0=0=¿Nx=0

Ny+Fr=0=¿Ny=−Fr

Nz−Fn=0=¿Nz=Fn

It is expected that Nx will equal zero because there are no forces in the Nx direction. As we can see looking at chart 2 below and chart one above reaction force Ny is the opposite if Fr because the Ny reaction forces are acting in the Y direction and are equal and opposite. Figure one will gives a good visual representation of the relation ship between Ny and Fr. Also from looking at the two charts and figure one we see is acting in the Z direction and has the same for as values as Fn. All values can be found in the appendix

Figure 1

MomentsThe Mx, My, and Mz moments were calculated using the formulas below.

ΣMi=0

Mx−Fn∗b=0=¿Mx=Fn∗b

My−Fn∗a=0=¿My=Fn∗a

Mz=Fr∗a=0=¿Mz=Fr∗a

a=0.055b=0.17

The moments were calculated using a = 0.055 m and b = 0.17 m. Looking at figure two it is easy to see the moments present on the crank. We can see the moment in the X direction will in general be the largest because it is the only one that has the large b distance in its equation. Moments in the Y and Z direction will in general be small because their distance used in their momentum equations is a, which is significantly smaller. Mz moment is smaller than My and becomes negative because it is dependant on Fr which is smaller and becomes negative past 90°. All values can be found in the appendix.

Figure 2

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Moment and Reaction Forces Vs. Degrees

MxMyMzNxNyNz

Degrees

Mom

ent

and

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ctio

n F

orce

s

Graph 2: Moment and Reaction Forces vs. Degrees

StressesThe bending, shear, and torque stresses were calculated using the equations below and previous information.

σ Bending=Moment

I∗Z

σ Shear=Shear force

Area

σ Shear−Torque=Torque

J∗r

Area=π∗r2

I=π4∗r4

J= π2∗r 4

r=0.006The bending stresses were calculated at y = ± R & z = 0 and at y = 0 and z = ± R. When y = R and z = 0 the shear stress was calculated using the Mz moment as the moment and the first moment of inertia I, as calculated below, z in the bending stress equation is r (.006). When y = -R and z = 0 the answer is the same but negative cause z is negative. When y = 0 and z = R the bending stress is calculated using the My moment as well as I and r the same as previously. When y = 0 and z = -R the answer is the same but negative. Doing this leads to the bending stresses being greatest when y = 0 and z = ± R. The shear stresses xy and xy were calculated using above equations for shear stress and area. In the xy shear reaction force Ny was used as the shear force, in the xz shear stress the Nz reaction force was used. This leads to the xz shear being larger. Finally the torque stress was found using the second moment of inertia and the equation above. Moment Mx was used as a value for torque and this lead to the largest about of stress coming by means of torque.All values can be found in the appendix.

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Bending y=R, z=0Bending y=-R, z=0Bending y=0, z=RBending y=0, z=-RShear xyShear xzTorque

Degrees

Stre

sses

Graph 3: Stresses vs. DegreesTask 3The material properties presented and measured show a large variability. Try to bound each property between an upper and lower bound.

What does case hardening do to a material and its properties?Case hardening increases the hardness of the outer layer while the core remains relatively soft. The combination of a hard shell with a soft core allows the material to withstand higher stresses and fatigue. The shell would have a higher wear resistance and fatigue strength while the core would have a higher impact strength.

Where would you expect yielding to initiate? Failure always occurs perpendicular to the direction of the largest stress. In this case this is where we would expect yielding to initiate. In our results the largest stress occurs in torque. Since torque stress is the largest stress observed in our analysis we would expect yielding to occur perpendicular to the torque stress. We expect the yielding to happen at a 45-degree angle as it did in the example.

Fatigue cracks tend to initiate and grow perpendicular to the maximum principal stress. Do your analysis agree with experimental observations? In our analysis we found that the largest stress in the spindle is from torque. For this analysis the direction that the spindle would fail would be in a perpendicular direction of the torque, this is in a 45-degree angle into the depth of the spindle. Looking at the images of the actual failure we see that this is the way that the spindle cracked and eventually broke.

Perform a fracture analysis based on the crack length at final fracture from the images in this report. a) Crack Growth

Rate of crack growth depends on the stress intensity at the head of our fatigue crack. Hypothesis can be made that the higher the stress intensity, the higher the crack growth rate. When Kmax (maximum stress intensity factor) reaches the value fracture toughness Kc of the material and thickness of the spindle, crack propagation is expected to occur when the length of crack is equivalent to a c that is critical for brittle fracture. The following formula is used to calculate the critical crack length of the material.

ac=1π

¿

In addition, crack growth causes a loss of cross-sectional area, and thus an increase in the stress on the remaining uncracked (net) area. Depending on the material and the member geometry and size, fully plastic yielding occurs before Kmax = Fracture toughness value.

b) Plastic Zone Size

Based on the crack length at final fracture from the images and fracture mechanics, it is estimated that the

critical length is 9.1765 mm. The material starts to fail at stressσ fail=K IC

√πa.

For the material that deforms plastically, the local stresses eventually exceed the yields strength. The stress cannot continue to rise according to linear elastic behavior. We can estimate the size of the plastic zone, the region denoted rp where the stresses are predicted based on the elastic solution to be above the yield stress. This plastic zone depends on the ratio of the applied stress to the yield stress as well on the crack size. With this, we can define the effective stress intensity factor where the crack size is assumed to be larger by a factor equal to the plastic zone size.

Plastic Zone Size is obtained by this formula:

c) Fracture

Before the point at the material maintains its shape and undergoes the plastic deformation. However, it reaches the fracture limit as the curve drops exponentially according to the fracture toughness equation shown as it weakens the materials and causes the crack propagation to occur and finally fails. Having this plastic zone shows us that the material yields when the crack length is very small before reaching the value of critical crack length.

According to the Paris Law, region 3 represents the period where the unstable crack growth leads the material to fracture.

How large has the pedal force has to be to cause final fracture based on the KIC = 60 MPam0.5?The question asks for the maximum pedal force to initiate the crack propagation thus leads to the fracture. The KIC given is 60MPam0.5. A given material can resist a crack without brittle fracture occurring as long as this K is below a critical value called as facture toughness This fracture toughness is dependent on the thickness of the material. Thicker material has lower KIC. KIC is defined as a measure of given ability to resist fracture in the presence of crack.

Method 1: Using KIC = 60MPam 0.5 KIC = Y∗σ∗√π∗acas if a≪b

Y = geometric factorσ = applied stressac = critical crack lengthArea = 0.000113097 m^2

Initially we solve the applied stress to obtain the maximum force. To do this, you need to have the critical crack length of the material. Based on the image given in the notes, the critical crack length is about 9.1765 mm.

Solving for stress and maximum force gives us that the maximum force is about 39.966 kN.

Method2: Assuming σmax = 565MPaWe know that σmax = 565 MPa based on the material properties. The formula below can be used to calculate the critical crack length:

ac=1π

¿

Where F is the geometrical factor and σmax is the maximum stress.

σ max=ForceArea

By assuming σmax = 565 MPa, we calculate the maximum force to cause fracture by using the formula above.

Force=σmax∗Area¿565∗106∗0.000113097

¿63899.8N

How are modern spindles designed?Modern spindles are design as a square taper because this is a simple and durable system for modern bikes. The square spindle can be made from various materials; however aluminum and stainless steel are the materials being used on the market.

Other spindles that are found for modern spindles designs are the Octalink Interface and the ISIS bottom bracket interface. The Octalink interface and the ISIS bottom bracket interface are eight grooved splines that are in lined with eight inverse splines on the crank [1]. The two designs provide more power transfer from the legs of the rider into the mechanical system of the bike [1]. For high end bikes, the spindle is designed as an external bottom bracket to improve the performance of the bike performance while maintaining its durability. The power and the stiffness of the bottom bracket are improved by increasing the diameter and the size of the spindle [1]. However, with this design, the lifespan of the product will decrease.

What parameters do we know well and what parameters don’t we know well in this problem?The parameters that we are aware of are the dimensions of the spindle and the crank system. We can also calculate the forces and moments applied on the system. The parameters that we are unaware of are the number of cycles which caused the failure and the environmental effects on the spindle.

Appendix

Appendix Table 1

Appendix Table 2

Reference

[1] Sohner, Brad. Bottom Bracket Tech Breakdow.Bike Rumor. Retrieved from http://www.bikerumor.com/2010/02/17/bottom-bracket-tech-breakdown/