P11215
Analysis Documents
Anthony Poli
Wheel Speed / Kinematic Drive Analysis
Motor
Transmission
Wheel
Gear Ratios:
Motor to Transmission:
Transmission to Wheel:
Wheel Diameter:
10:72
24:40
12 in.
.1388
.6
ωm = Motor Speed
ωT = Transmission Speed
ωw = Wheel Speed
V = Vehicle Velocity
𝜔𝑇 =
10
𝜔
72 𝑚
𝜔𝑤 =
24
𝜔
40 𝑇
𝑉=
1[𝑓𝑡]
12[𝑖𝑛]
𝑓𝑡
∗ 12 [𝑖𝑛]𝜋𝜔𝑤 [ 𝑠 ]
Combine to get:
𝑉=
1[𝑓𝑡]
24 10
𝑓𝑡
∗ 12 [𝑖𝑛]𝜋
∗ 𝜔𝑚 [ ]
12[𝑖𝑛]
40 72
𝑠
So at the lowest motor revolutions obtained by the previous group, 22.5 [°/s] or 3.75rpm,
V = 11.78 [ft./min]
The old goal was 10 [ft./min] so this is acceptable.
For lowest speed,
Which is the autonomous
Operation speed
ωm = 22.5 [°/s] or 3.75rpm
ωT = 0.52 rpm
ωw = 0.3125 rpm
P11215
Analysis Documents
Anthony Poli
P11215
Analysis Documents
Anthony Poli
Force on Pin:
Motor Torque
Hole
Motor
Transmission
Max torque output of the motor:
Radius of 10T Drive Cog:
Radius of 72T Cog:
Radius to pin hole:
Diameter of pin hole:
Radius to hole center
τm
d1
d2
d3
21 [N*m]
0.6896 in.
4.405 in.
1 in.
0.5 in.
or
185.87 [in.*lb.]
𝜏 =𝐹∗𝑑
τ = Torque
F = Force
d = Distance from pivot
𝐹𝑐ℎ𝑎𝑖𝑛 =
𝜏𝑚
𝑑1
𝜏𝑡𝑟𝑎𝑛𝑠. = 𝐹𝑐ℎ𝑎𝑖𝑛 ∗ 𝑑2
𝐹ℎ𝑜𝑙𝑒 =
𝜏𝑡𝑟𝑎𝑛𝑠.
𝑑3
Combine to get:
𝐹ℎ𝑜𝑙𝑒
𝜏𝑚
∗ 𝑑2
𝜏𝑚 𝑑2
𝑑1
=
=
𝑑3
𝑑1 𝑑3
Sub to get:
𝐹ℎ𝑜𝑙𝑒 =
21[𝑖𝑛.∗ 𝑙𝑏. ] ∗ 4.405[𝑖𝑛. ]
= 1187 𝑙𝑏𝑓.
1[𝑖𝑛. ] ∗ 0.6896[𝑖𝑛. ]
P11215
Analysis Documents
Pin Stress:
Force on Pin
Direction of Motion
½“ gap
The Pin will receive most of its stress in the form of shear.
Max shear force
Ey of 10060 Steel
Poisson’s Ratio
Fp
E
ν
Shear Modulus
𝐸 = 2𝐺(1 + 𝜈)
For Steel 𝐺 =
24000[𝑝𝑠𝑖]
2(1+.3)
1200 [lbf]
24[kpsi] (lowest grade I could find in Shigley p.1020)
0.3 (for most Structural Metals)
𝐺=
𝐸
2(1+𝜈)
= 9230.77
For a pin:
𝜏𝑚𝑎𝑥 =
𝐹
𝐴
τ = Shear Stress
F = Force Applied
A = Cross-sectional Area
𝐴=
𝜋 2 𝜋
𝑑 = 0.5[𝑖𝑛. ]2 = 0.19635[𝑖𝑛.2 ]
4
4
Sub:
𝜏𝑚𝑎𝑥 =
G = 9231[psi]
τmax = 6112[psi]
G > τmax therefore the pin won’t break.
1200[𝑙𝑏𝑓]
= 6111.55 [𝑝𝑠𝑖]
0.19635[𝑖𝑛.2 ]
Anthony Poli
P11215
Analysis Documents
Anthony Poli
Gear Stress:
The Stress on the hole in the 24tooth gear was performed in ANSYS by Chris.
A view of the mesh
A close up of the stress concentration at the hole.
The max stress experienced by the gear hole is 23000ksi
A conservative yield strength for steel is 34000ksi
This yields a factor of safety of 1.7 which is within tolerable limits of the part.
Also this is stress concentration was simulated with the max stall force the motor can output. This force
is much greater than the force required to drive the robot.
The last team found it took 9[in.*lbs.] of motor output to move the robot and this simulation was run
with a motor output of over 185[in.*lbs.]
P11215
Analysis Documents
Anthony Poli
Transmission Axle Stress
Cantilevered Beam
Weld joint for axle
To determine if the axle is strong enough the max force applied at the end of the axle will be found. If no
forces exceed that force the beam will hold.
Max. Force at
furthest distance
Axle
L
P
Ey
G
Length
Load
Yield Strength
Shear Modulus
4[in.]
Find.
218000 [psi]
11900000 [psi]
The properties are for O1 Hardened Tool Steel (from Matweb.com) which is the steel to be ordered for
this axel. This specification was based on workability and price for the length and diameter needed.
For Bending:
𝜎𝑥 =
Where:
σx = Bending Stress
M = Bending Force
y = distance from axis
I = Area Moment of Inertia
𝑀𝑦
𝐼
[psi]
[in. * lb]
[in.]
[in4]
I for a cylinder:
𝐼=
𝜋 4
𝜋
𝑑 =
0.254 = 0.000192[𝑖𝑛.4 ]
64
64
Max Bending Force:
𝜎𝑥 𝐼
𝐸𝐼
218000[𝑝𝑠𝑖] ∗ 0.000192[𝑖𝑛.4 ]
𝑀=
=
𝑠𝑢𝑏: =
= 334.408 [𝑖𝑛.∗ 𝑙𝑏]
𝑦
𝑦
. 125[𝑖𝑛. ]
Max Force:
𝑀 334
𝑃= =
= 83[𝑙𝑏]
𝐿
4
P11215
Analysis Documents
For Shear:
𝜏𝑚𝑎𝑥 = 𝐺 =
𝐹
𝐴
τ = Shear Stress
F = Force Applied
A = Cross-sectional Area
𝐴=
𝜋 2 𝜋
𝑑 = 0.75[𝑖𝑛. ]2 = 0.049087[𝑖𝑛.2 ]
4
4
Max Shear Force:
𝐹 = 𝐺𝐴 = 11900000 [psi] ∗ 0.049087[𝑖𝑛.2 ] = 584140 [𝑙𝑏]
No worries here.
Anthony Poli
P11215
Analysis Documents
Anthony Poli
Axle Stress
Cantilevered Beam
To determine if the axle is strong enough the max force applied at the end of the axle will be found. If no
forces exceed that force the beam will hold.
Max. Force at
furthest distance
Axle mount
Axle
L
P
Ey
G
Length
Load
Yield Strength
Shear Modulus
6[in.]
Find.
34100 [psi]
11600000 [psi]
The properties are for Ultra-Machinable 12L14 Carbon Steel (from Matweb.com) which is the steel to be
ordered for this axel. This specification was based on workability and price for the length and diameter
needed.
For Bending:
𝜎𝑥 =
Where:
σx = Bending Stress
M = Bending Force
y = distance from axis
I = Area Moment of Inertia
𝑀𝑦
𝐼
[psi]
[in. * lb]
[in.]
[in4]
I for a cylinder:
𝐼=
𝜋 4
𝜋
𝑑 =
0.754 = 0.015532[𝑖𝑛.4 ]
64
64
Max Bending Force:
𝜎𝑥 𝐼
𝐸𝐼
34100[𝑝𝑠𝑖] ∗ 0.015532[𝑖𝑛.4 ]
𝑀=
=
𝑠𝑢𝑏: =
= 1412.34[𝑖𝑛.∗ 𝑙𝑏]
𝑦
𝑦
. 375[𝑖𝑛. ]
Max Force:
𝑀 1412
𝑃= =
= 235[𝑙𝑏]
𝐿
6
P11215
Analysis Documents
Anthony Poli
Since the Robot is supported by 4 wheels and may weigh a total of 200 pounds this should be
acceptable. If not a Steel with a higher yield modulus can be used.
For Shear:
𝜏𝑚𝑎𝑥 = 𝐺 =
𝐹
𝐴
τ = Shear Stress
F = Force Applied
A = Cross-sectional Area
𝐴=
𝜋 2 𝜋
𝑑 = 0.75[𝑖𝑛. ]2 = 0.441786[𝑖𝑛.2 ]
4
4
Max Shear Force:
𝐹 = 𝐺𝐴 = 11600000 [psi] ∗ 0.441786[𝑖𝑛.2 ] = 5124723.02 [𝑙𝑏]
No worries here.