The document analyzes the failure of a bike pedal spindle. It determines the location of fatigue crack origin and calculates various forces, moments, and stresses acting on the spindle. The largest stress was found to be torque stress. This agrees with observations that the spindle failed in a perpendicular direction to the torque stress. Based on the images and fracture mechanics analysis, the critical crack length and size of the plastic zone were estimated. The maximum pedal force required to cause final fracture was also calculated based on the material's fracture toughness.

1. MTRL 485 – Case 5
Spindle Failure
Group 4
Masseir, Alex (55761118)
May, Michael (47957105)
McGrew, Kennan (55398101)
Mei, Helen (31494115)
Muhammad, Harith (18204115)

2. Task 1
Figure 1: Fatigue Surface
The 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 98.5
2 34.3 0.6 0.1 65.0
3 37.7 0.5 61.8
4 - - 100
5 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.

3. Task 2
Forces
The 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° ∗ si𝑛2 ( 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 forces
Σ𝑁𝑖 = 0
𝑁𝑥 + 0 = 0 => 𝑁𝑥 = 0
𝑁𝑦 + 𝐹𝑟 = 0 => 𝑁𝑦 = −𝐹𝑟
𝑁𝑧 − 𝐹𝑛 = 0 => 𝑁𝑧 = 𝐹𝑛

4. 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
Moments
The 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 = 0.055⁡𝑏 = 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.

5. Figure 2
Graph 2: Moment and Reaction Forces vs. Degrees
-300.000
-200.000
-100.000
0.000
100.000
200.000
300.000
400.000
500.000
600.000
0 50 100 150 200
MomentandReactionForces
Degrees
Moment and ReactionForces Vs. Degrees
Mx
My
Mz
Nx
Ny
Nz

6. Stresses
The bending, shear, and torque stresses were calculated using the equations below and previous
information.
𝜎𝐵𝑒𝑛𝑑𝑖𝑛𝑔 =
𝑀𝑜𝑚𝑒𝑛𝑡
𝐼
∗ 𝑍
𝜎𝑆ℎ𝑒𝑎𝑟 =
𝑆ℎ𝑒𝑎𝑟⁡𝑓𝑜𝑟𝑐𝑒
𝐴𝑟𝑒𝑎
𝜎𝑆ℎ𝑒𝑎𝑟−𝑇𝑜𝑟𝑞𝑢𝑒 =
𝑇𝑜𝑟𝑞𝑢𝑒
𝐽
∗ 𝑟
𝐴𝑟𝑒𝑎 = ⁡𝜋 ∗ 𝑟2
𝐼 =
𝜋
4
∗ 𝑟4
𝐽 =
𝜋
2
∗ 𝑟4
𝑟 = 0.006
The 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.
Graph 3: Stresses vs. Degrees
-200.000
-150.000
-100.000
-50.000
0.000
50.000
100.000
150.000
200.000
250.000
300.000
0 20 40 60 80 100 120 140 160 180 200
Stresses
Degrees
Stresses vs. Degrees
Bending y=R, z=0
Bending y=-R, z=0
Bending y=0, z=R
Bending y=0, z=-R
Shear xy
Shear xz
Torque

7. Task 3
The 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 tothe 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 ac that is critical for brittle
fracture. The following formula is used to calculate the critical crack length of the material.

8. 𝑎 𝑐 =
1
𝜋
(
𝐾𝑐
𝐹𝜎 𝑚𝑎𝑥
)2
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 stress⁡𝜎𝑓𝑎𝑖𝑙 =
𝐾𝐼𝐶
√ 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

9. 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 = 60MPam0.5
KIC = 𝑌 ∗ 𝜎 ∗ √ 𝜋 ∗ 𝑎 𝑐⁡𝑎𝑠⁡𝑖𝑓⁡𝑎 ≪ 𝑏
Y = geometric factor
𝜎 = applied stress
ac = critical crack length
Area = 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 = 565MPa
We know that 𝜎max = 565 MPa based on the material properties. The formula below can be used to
calculate the critical crack length:
𝑎 𝑐 =
1
𝜋
(
𝐾𝐼𝐶
𝐹𝜎 𝑚𝑎𝑥
)2
Where F is the geometrical factor and 𝜎max is the maximum stress.
𝜎 𝑚𝑎𝑥 =
𝐹𝑜𝑟𝑐𝑒
𝐴𝑟𝑒𝑎

10. By assuming 𝜎max = 565 MPa, we calculate the maximum force to cause fracture by using the formula
above.
𝐹𝑜𝑟𝑐𝑒 = 𝜎 𝑚𝑎𝑥 ∗ 𝐴𝑟𝑒𝑎
= 565 ∗ 106
∗ 0.000113097⁡
= 63899.8⁡𝑁
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.

11. Appendix
Appendix Table 1
Appendix Table 2

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