By:
Noemi Y. Rodriguez
Antonio Galloza
Levi L Vargas
Julio Alvarez
Objectives

In this project we were told to design the
shaft that joints the two pedals of a ministepper that could support a maximum
weight of 250 lb. by analyzing combined
loads and stress concentrators. We should
take in consideration the type of material to
be selected, deflections and deformations
in the material. We will be also considering
finite life, the critical section and safety
factors of the components.
Mini-Stepper
Design Details

Why do we design this device, what is
its application?
We design this device because it’s
something that we have in our daily use.
A mini-stepper helps people maintain a
healthy life exercising themselves.

What are the most important
engineering considerations in our
design?
 Material
 Safety
 Costs
Design
Stress Analysis (Static Load)

Bendings:
 σx=Mc/I
 σy=Mc/I

Shears:
 τx=4V/3A
 τy=4V/3A

Torsions:
 Due to the sleeve the:
Sleeve
Torsion→0
Stress Analysis (Static Load)
Bending Stresses
X 
Mc
I

1.0in.

340lb. in


2

X 
4
4
 1.0in. 0.90in  

  
 
4  2   2  
Y 
 X  10070.4029Psi.
 Y  17442.449Psi.

Mc
I

1.0in.

588.92lb. in


2

Y 
4
4
 1.0in. 0.90in  

  
 
4  2   2  
Stress Analysis (Static Load)
Shear Stresses
4V
X 
3A
X 
 YZ
485lb.
2
3
0.14922565in.
 

4V

3A
4147.53lb.

 YZ 
2
3
0.14922565in.
 

 YZ  1315.4514Psi.
 X  759.47622Psi.



Stress Analysis (Static Load)

Von Misses Stress
 X   Y    Y   Z    Z   X 
2
 VM 
2
2

 6  XY   YZ   ZX
2
VM  24253.3751Psi.
2
2
2

Stress Analysis (Static Load)

Safety Factor
Y
n
 VM
60,200Psi.
n
24253.3751Psi.
n  2.48
Deflection Analysis




EIV=F
EIM=EI∫Vdx
EIθ=EI∫Mdx
Eiv=EI∫θdx
Deflection Analysis

Force component Fy

Force component Fx
EIV  147.22lb
EIV  85lb
EIM  147.22 * x  C1
EIM  85 * x  C1
x  L, M  0  C1  147.22 L
x  L, M  0  C1  85 L
EIM  147.22 * x  147.22 L
EIM  85 * x  85 L
x2
EIangulo  147.22  147.22 Lx  C 2
2
x  0, angulo  0  C 2  0
x2
EIangulo  85  85 Lx  C2
2
x  0, angulo  0  C2  0
x2
EIangulo  147.22  147.22 Lx
2
3
x
x2
EIv  147.22  147.22 L
 C3
6
2
x  0, v  0  C 3  0
x2
EIangulo  85  85 Lx
2
x3
x2
EIv  85  85 L  C3
6
2
x  0, v  0  C3  0
x3
x2
EIv  147.22  147.22 L
6
2
v max  0.006414in
x3
x2
EIv  85  85 L
6
2
vmax  0.003703in
Deflection Analysis
Fy  250  80  170lb

Overall
EIV  170lb
EIM  170 * x  C1
x  L, M  0  C1  170 L
EIM  170 * x  170 L
x2
EIangulo  170  170 Lx  C 2
2
x  0, angulo  0  C 2  0
x2
EIangulo  170  170 Lx
2
x3
x2
EIv  170  170 L
 C3
6
2
x  0, v  0  C 3  0
x3
x2
EIv  170  170 L
6
2
v max  0.007407in
Material Selection

Material Indexes
 M=Cvr/E^(1/2)
 M=Cvr/σf^(2/3)
Has to be minimized
Material Selection
Material Selection
Material Selection
Material
Modulus E GPa
Fracture Strength MPa
Relative cost Cvr
Al Alloys
70
190
.9
Steel 1040
200
450
.75
Cast Iron
150
690
.65
For Al Alloy
For Steel 1040
For Cast Iron
Dynamic Load Analysis

For the Stress concentration

Endurance Limit
 Se=ksize*ksurface*kload*ktemperature*
kreliability*Se’
Dynamic Load Analysis

Endurance Limit
 Se=(0.957092)*(0.819629)*(0.7532)*(44.95)
=26.5519ksi

Component live: 10^6 Cycles
Dynamic Loads Analysis

Stress concentration factor
Kt,bend=1.62
Kt,shear=1.35
Kf,bend=1.52
Kf,shear=1.29
Dynamic Load Analysis

Fmax,Fmin
Dynamic Loads

Fa,Fm
Dynamic Load Analysis
Dynamic Load Analysis
Point A
Point B
Dynamic Load Analysis

Safety Factor (Modified Goodman)

What is the unique part of our project?
 In our project we analyzed a shaft for a mini-
stepper, the unique part comes when four
minds merge to achieve the goal of a design
of a safety shaft.

What is the most challenging part in our
project?
 Determine the outer and inner diameters of
our shaft to obtain an appropriate safety
factor without overdesign

What is the weakness of our project?
 The weakness of our project is that in a long
period of time, the corrosion will affect our
shaft due to the selection of an inexpensive
material.

What we learn from this project?
 Team work
 Design is not just stetic, but is a more
complex analysis on which men life depends
QUESTIONS?