This document contains calculations for various mechanical, electrical, and computer systems for a device called MARTHA. The calculations are organized by system and include analyses of materials like PVC, steel, aluminum, and various fasteners used. Stress, strain, load, and separation number calculations are shown for components like the chassis, motor mount, screws, and wheels to ensure the design meets strength and safety requirements.

1. APPENDIX E
CALCULATIONS

2. E1
Introduction:
The following are all calculations, mechanical, computer and electrical for the MARTHA
device. The calculations are organized by what system they correspond to. All common
assumptions are to be made, unless otherwise stated (i.e. g=9.81 m/s2).

3. E2
PVC Analysis
PVC Material Specifications:
𝑆 𝑢𝑡 = 10500 𝑝𝑠𝑖
𝑆 𝑦 = 8500 𝑝𝑠𝑖
E = 490 ksi
Chain Material Specifications: 4130 Normalized alloy steel
𝑆 𝑢𝑡= 97200 psi
𝑆 𝑦 = 63100 𝑝𝑠𝑖
E = 29700 ksi
Screw Specifications: #6-32 ¾” Machine Screw
d = 0.138”
𝑑 𝑟 = 0.0974”
N = 32
𝐴𝑡 = 0.0119 𝑖𝑛2
𝑆 𝑢𝑡 = 25000 psi
𝑆 𝑝 = 22479 psi
𝑆 𝑦 = 22000 psi
E = 1490 ksi
𝐴 𝐵 = 0.138(0.5) = 0.069 𝑖𝑛2
l = 0.5”
P = 15 lb
𝐹𝑖 = .9𝑆 𝑝 𝐴𝑡 = (.9)(0.0119)(22479) = 241 lb
𝐾𝑏 = 𝐸
𝐴𝑡
𝑙
+ 𝐸
𝐴 𝑏
𝑙
=
0.0119(490000)
0.5
+
0.069(490000)
0.5
= 135000
𝐾 𝑚 = 𝐴 𝑚
𝐸 𝑚
𝑙
=
0.069(490000)
0.5
= 67600
𝐶 =
𝐾𝑏
𝐾 𝑚 𝐾𝑏
=
135000
(135000)(67600)
= 0.67
𝑃 𝑏 = 𝐶𝑃 = (. 67)(15) = 10.05 𝑙𝑏
𝑃𝑚 = (1 − 𝐶) 𝑃 = (1 − .67)15 = 4.95 𝑙𝑏
𝐹𝑏 = 𝐹𝑖 + 𝑃 𝑏 = 241 + 10.05 = 251.05 𝑙𝑏
𝐹𝑚 = 𝐹𝑖 − 𝑃𝑚 = 241 − 4.95 = 236.05 𝑙𝑏
𝜎𝑏 =
𝐹𝑏
𝐴𝑡
=
251.05
0.0119
= 21100 𝑝𝑠𝑖
𝜎 𝑚 =
𝐹𝑚
𝐴 𝑏
=
236.05
0.069
= 3421 𝑝𝑠𝑖
𝑁 𝑦 =
𝑆 𝑦
𝜎𝑏
=
22000
21100
= 1.04

4. E3
𝑃0 =
𝐹𝑖
(1 − 𝐶)
=
241
(1 − .67)
= 730.3 𝑙𝑏
𝑁𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 =
𝑃0
𝑃
=
730.3
15
= 49
Front wheel: (6/15) 13 = 5.2 lb
Back wheel: (11/15) 13 = 7.8 lb
Front:
5.2lb/2 wheels = 2.6 lb
σ = F/A = 2.6/0.036 = 72.2 psi
Back:
7.8lb/2 wheels = 3.9 lb
σ = F/A = 3.9/0.036 = 108.3 psi

5. E4
Chassis Analysis
Chassis material: 6061-Aluminum
Ρ = 0.0975 lb/𝑖𝑛2
𝑆 𝑢𝑡 = 45000 𝑝𝑠𝑖
𝑆 𝑦 = 40000 𝑝𝑠𝑖
E = 10000 ksi
Screw: #8-32 1” machine screw
d = 0.164”
N = 32
𝑑 𝑟 = 0.1234"
𝐴𝑡 = 0.0140 𝑖𝑛2
𝑆 𝑢𝑡 = 25000 𝑝𝑠𝑖
𝑆 𝑦 = 22000 𝑝𝑠𝑖
E = 1490 ksi
𝑆 𝑝 = 22500 𝑝𝑠𝑖
l = 0.8”
𝐴 𝑏 = (0.164)(0.8) = 0.1312𝑖𝑛2
Chassis-Screw Connection:
P = 12 lb
𝐹𝑖 = .9𝑆 𝑝 𝐴𝑡 = 283.5 𝑙𝑏
𝐾𝑏 = 𝐸
𝐴𝑡
𝑙
+ 𝐸
𝐴 𝑏
𝑙
= 88935
𝑙𝑏
𝑖𝑛
𝐾 𝑚 =
𝐴 𝑚 𝐸 𝑚
𝑙
=
(0.044)(100000000)
0.8
= 440000
𝑙𝑏
𝑖𝑛
𝐴 𝑚 = (
𝜋
4
) ( 𝐷 − 𝑑)2
= (
𝜋
4
) (0.4 − 0.164)2
= 0.044𝑖𝑛2
𝑃 𝑏 = 𝐶𝑃 = (12)(0.17) = 2.04 𝑙𝑏
𝑃𝑚 = (1 − 𝐶) 𝑃 = 9.96 𝑙𝑏
𝑃0 =
𝐹𝑖
(1 − 𝐶)
=
283.5
(1 − 0.17)
= 341.57 𝑙𝑏
𝑁𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 =
𝑃0
𝑃
= 28.5
𝐹𝑏 = 𝐹𝑖 + 𝑃 𝑏 = 285.55 𝑙𝑏
𝐹𝑚 = 𝐹𝑖 − 𝑃𝑚 = 273.54 𝑙𝑏
𝐶 =
𝐾𝑏
𝐾 𝑚 + 𝐾𝑏
= 0.17
𝜎𝑏 =
𝐹𝑏
𝐴𝑡
=
285.55
0.014
= 20396 𝑝𝑠𝑖 𝑥28 𝑠𝑐𝑟𝑒𝑤𝑠 = 728.43 𝑝𝑠𝑖 𝑝𝑒𝑟 𝑠𝑐𝑟𝑒𝑤
𝜎 𝑚 =
𝐹𝑚
𝐴 𝑏
= 2084.9 𝑝𝑠𝑖

6. E5
𝑁 𝑦 =
𝑆 𝑦
𝜎𝑏
=
22000
728.43
= 30.2
Shear on side screws:
𝐹 = 𝑚𝑔 = (12𝑙𝑏)(
9.81𝑚
𝑠2
) = 117.72 𝑁
𝐴 = 0.014 𝑖𝑛2
𝜎𝑠ℎ𝑒𝑎𝑟 =
117.72
0.014
= 8408.6 𝑝𝑠𝑖 𝑥 8 𝑠𝑐𝑟𝑒𝑤𝑠 = 1051.075 𝑝𝑠𝑖 𝑝𝑒𝑟 𝑠𝑐𝑟𝑒𝑤

7. E6
MotorMount Analysis (Screw Connections)
Material Specifications: 6061-Aluminum
𝜌 = 0.0975
𝑙𝑏
𝑖𝑛3
𝑆 𝑢𝑡 = 45000 𝑝𝑠𝑖
𝑆 𝑦 = 40000 𝑝𝑠𝑖
𝐸 = 10000 𝑘𝑠𝑖
Bolt Material: ¼” x 1” 307 Grade A Steel – SAE Grade 5, medium carbon, cold drawn
Head width: 3/8”
𝑙 𝑡 = 0.75"
𝑆 𝑦 = 92 𝑘𝑠𝑖
𝑆 𝑢𝑡 = 120 𝑘𝑠𝑖
𝑆 𝑝 = 85 𝑘𝑠𝑖
𝐸 = 30000 𝑘𝑠𝑖
𝐴𝑡 = 0.032 𝑖𝑛2
𝐴 𝑏 = 𝜋(0.25)2
= 0.2
Connections Relating to Drawing:
1. L-Bar to Chassis
2. L-Bar to Motor Face Plate
3. L-Bar to Bearing
4. Top L-bar to Plate with ABS Plastic
5. Motor Face Plate to Motor
L-Bar to Motor Al face connection:
Same Preload Specs as #8-32 x 1” screw listed above
C = 0.17
P = 1.8 lb
𝐹𝑖 = 283.5 𝑙𝑏
𝐴𝑡 = 0.014 𝑖𝑛2
𝐴 𝑏 = 0.1312 𝑖𝑛2
𝑃 𝑏 = 𝐶𝑃 = 0.306 𝑙𝑏
𝑃𝑚 = (1 − 𝐶) 𝑃 = 1.494 𝑙𝑏
𝐹𝑏 = 𝐹𝑖 + 𝑃 𝑏 = 283.806 𝑙𝑏
𝐹𝑚 = 𝐹𝑖 − 𝑃𝑚 = 282.006 𝑙𝑏
𝜎𝑏 =
𝐹𝑏
𝐴𝑡
= 20271.9 𝑝𝑠𝑖 𝑥 2 𝑠𝑐𝑟𝑒𝑤𝑠 = 10135.95 𝑝𝑠𝑖 𝑝𝑒𝑟 𝑠𝑐𝑟𝑒𝑤
𝜎 𝑚 =
𝐹𝑚
𝐴 𝑏
= 2149.44 𝑝𝑠𝑖
𝑁 𝑦 =
𝑆 𝑦
𝜎𝑏
= 1.08

8. E7
𝑃0 =
𝐹𝑖
(1 − 𝐶)
= 341.57
𝑁𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 =
𝑃0
𝑃
= 190
Shear on screw head:
F = 1.8 lb x 9.8m/𝑠2
= 17.66 𝑁
A = 0.014 𝑖𝑛2
𝜎𝑠ℎ𝑒𝑎𝑟 =
17.60
0.014
= 1261.4 𝑝𝑠𝑖 𝑥2 𝑠𝑐𝑟𝑒𝑤𝑠 = 630.7 𝑝𝑠𝑖 𝑝𝑒𝑟 𝑠𝑐𝑟𝑒𝑤
L-Bar to bearing connection:
Material: SAE 1840 Bronze
Bearing housing: 6061-Aluminum
Same Preload Specs as #8-32 x 1” screw listed above
P = 1 lb
C = 0.95
𝑃 𝑏 = 𝐶𝑃 = 0.95 𝑙𝑏
𝑃𝑚 = (1 − 𝐶) 𝑃 = 0.05 𝑙𝑏
𝐹𝑏 = 𝐹𝑖 + 𝑃 𝑏 = 2448.95 𝑙𝑏
𝐹𝑚 = 𝐹𝑖 − 𝑃𝑚 = 2447.95 𝑙𝑏
𝜎𝑏 =
𝐹𝑏
𝐴𝑡
= 76530 𝑝𝑠𝑖 𝑥 2 𝑠𝑐𝑟𝑒𝑤𝑠 = 38265 𝑝𝑠𝑖 𝑝𝑒𝑟 𝑠𝑐𝑟𝑒𝑤
𝜎 𝑚 =
𝐹𝑚
𝐴 𝑏
= 12239.75 𝑝𝑠𝑖
𝑁 𝑦 =
𝑆 𝑦
𝜎𝑏
= 1.2
𝑃0 =
𝐹𝑖
(1 − 𝐶)
= 37210 𝑙𝑏
𝑁𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 =
𝑃0
𝑃
= 48960
L-Bar to Chassis bolt connection:
Specifications for bolt preload listed above
𝐹𝑖 = 0.9𝑆 𝑝 𝐴𝑡 = 2448 𝑙𝑏
𝑙 𝑡ℎ𝑟𝑒𝑎𝑑 = 2𝑑 + 0.25 = 0.75"
𝑙 𝑠 = 𝑙 𝑏𝑜𝑙𝑡 − 𝑙 𝑡ℎ𝑟𝑒𝑎𝑑 = 1 − 0.75 = 0.25"
𝑙 𝑡 = 𝑙 − 𝑙 𝑠 = 0.75"
𝐾𝑏 = 𝐸
𝐴𝑡
𝑙
+ 𝐸
𝐴 𝑏
𝑙
= 2.53𝑥107
𝑙𝑏
𝑖𝑛
𝐾 𝑚 =
𝐴 𝑚 𝐸 𝑚
𝑙
= 1.25𝑥106
𝑙𝑏
𝑖𝑛
𝐴 𝑚 =
1
8
𝑥 0.25 = 0.09375 𝑖𝑛2
𝐶 =
𝐾 𝑏
𝐾 𝑚+𝑘 𝑏
= 0.95
P = 2 lb

9. E8
𝑃 𝑏 = 𝐶𝑃 = 1.9 𝑙𝑏
𝑃𝑚 = (1 − 𝐶) 𝑃 = 0.1 𝑙𝑏
𝐹𝑏 = 𝐹𝑖 + 𝑃 𝑏 = 2449.5 𝑙𝑏
𝐹𝑚 = 𝐹𝑖 − 𝑃𝑚 = 2447.5 𝑙𝑏
𝜎𝑏 =
𝐹𝑏
𝐴𝑡
= 76547 𝑝𝑠𝑖 𝑥 2 𝑠𝑐𝑟𝑒𝑤𝑠 = 38273.5 𝑝𝑠𝑖 𝑝𝑒𝑟 𝑠𝑐𝑟𝑒𝑤
𝜎 𝑚 =
𝐹𝑚
𝐴 𝑏
= 26107 𝑝𝑠𝑖
𝑁 𝑦 =
𝑆 𝑦
𝜎𝑏
= 1.2
𝑃0 =
𝐹𝑖
(1 − 𝐶)
= 48960 𝑙𝑏
𝑁𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 =
𝑃0
𝑃
= 24480
L-Bar to Aluminum Motor face connection with ABS Plastic Spacer:
ABS Specifications:
𝑆 𝑢𝑡 = 5500 𝑝𝑠𝑖
𝑆 𝑦 = 3000 𝑝𝑠𝑖
E = 270 ksi
𝜎𝑎𝑏𝑠 =
𝐹 𝑚
𝐴
= 6216.8 𝑝𝑠𝑖 ∶ 𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑣𝑒
#8-32 Screws
P = 1 lb
𝐹𝑖 = .9𝑆 𝑝 𝐴𝑡 = 283.5 𝑙𝑏
𝐾𝑏 = 𝐸
𝐴𝑡
𝑙
+ 𝐸
𝐴 𝑏
𝑙
= 88935
𝑙𝑏
𝑖𝑛
𝐾 𝑚 =
𝐴 𝑚 𝐸 𝑚
𝑙
=
(0.044)(100000000)
0.8
= 23760
𝑙𝑏
𝑖𝑛
𝐴 𝑚 = (
𝜋
4
) ( 𝐷 − 𝑑)2
= 0.044𝑖𝑛2
𝑃 𝑏 = 𝐶𝑃 = 0.79 𝑙𝑏
𝑃𝑚 = (1 − 𝐶) 𝑃 = 0.21 𝑙𝑏
𝑃0 =
𝐹𝑖
(1 − 𝐶)
=
283.5
(1 − 0.17)
= 1350 𝑙𝑏
𝐹𝑏 = 𝐹𝑖 + 𝑃 𝑏 = 284.29 𝑙𝑏
𝐹𝑚 = 𝐹𝑖 − 𝑃𝑚 = 283.29 𝑙𝑏
𝜎𝑏 =
𝐹𝑏
𝐴𝑡
= 6401.1 𝑝𝑠𝑖 𝑥 2 𝑠𝑐𝑟𝑒𝑤𝑠 = 3230.55 𝑝𝑠𝑖 𝑝𝑒𝑟 𝑠𝑐𝑟𝑒𝑤
𝑁 𝑦 =
𝑆 𝑦
𝜎𝑏
= 4.65
𝑁𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 =
𝑃0
𝑃
= 1350
Aluminum face to Motor connection:

10. E9
#10-24 1.5” screws:
d = 0.19”
𝑑 𝑟 = 0.1359"
N = 24
𝐴𝑡 = 0.0175 𝑖𝑛2
𝑆 𝑦 = 22000 𝑝𝑠𝑖
𝑆 𝑢𝑡 = 25000 𝑝𝑠𝑖
E = 490 ksi
𝐴 𝑏 = 𝛱( 𝑑2) = 0.028 𝑖𝑛2
𝐴 𝑚 =
𝜋 𝐷𝑒𝑓𝑓
2
4
= 0.042 𝑖𝑛2
𝐷 𝑒𝑓𝑓 = 0.23"
𝐹𝑖 = .9𝑆 𝑝 𝐴𝑡 = 354.375 𝑙𝑏
𝐾𝑏 = 𝐸
𝐴𝑡
𝑙
+ 𝐸
𝐴 𝑏
𝑙
= 17150
𝑙𝑏
𝑖𝑛
𝐾 𝑚 =
𝐴 𝑚 𝐸 𝑚
𝑙
=
(0.042)(100000000)
1.3
= 319567
𝑙𝑏
𝑖𝑛
𝐶 =
𝐾 𝑏
𝐾 𝑚+𝑘 𝑏
= 0.05
𝑃 𝑏 = 𝐶𝑃 = 0.09 𝑙𝑏
𝑃𝑚 = (1 − 𝐶) 𝑃 = 1.71 𝑙𝑏
𝑃0 =
𝐹𝑖
(1 − 𝐶)
= 373 𝑙𝑏
𝐹𝑏 = 𝐹𝑖 + 𝑃 𝑏 = 354.465 𝑙𝑏
𝐹𝑚 = 𝐹𝑖 − 𝑃𝑚 = 352.665 𝑙𝑏
𝜎𝑏 =
𝐹𝑏
𝐴𝑡
= 20250 𝑝𝑠𝑖 𝑥 2 𝑠𝑐𝑟𝑒𝑤𝑠 = 10125 𝑝𝑠𝑖 𝑝𝑒𝑟 𝑠𝑐𝑟𝑒𝑤
𝑁 𝑦 =
𝑆 𝑦
𝜎𝑏
= 1.1
𝑁𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 =
𝑃0
𝑃
= 207.2

11. E10
MotorMount Analysis (Bearing & Motor)
Motor Specs: 48:1 ratio
Load speed = 33 rpm
Load torque = 280 oz-in = 17.5 lb-in.
0.5” diameter, 1.75” long motor shaft, 𝐷 𝑜 = 0.5", 𝐷𝑖 = 0.25"
𝜏 =
𝑇𝜌
𝐽
=
17.5(0.25)
3.8 𝑥 10−4
= 11513.2 𝑙𝑏𝑓 = 𝜏 𝑚𝑎𝑥
𝐽 =
𝜋( 𝐷𝑜 − 𝐷𝑖)4
32
= 3.8𝑥10−4
Set Screws: 10-32 set screws ½”
𝑑 𝑠𝑒𝑡 𝑠𝑐𝑟𝑒𝑤 = 0.165"
𝐴𝑡 = 0.0175𝑖𝑛.2
𝑑 𝑝 = 𝑑 − (
0.64519
2
) = 0.163"
𝑆 𝑦 = 22000 𝑝𝑠𝑖
Alloy steel: F = 311.376 N
𝜎𝑡 =
𝐹
𝐴𝑡
= 1.78𝑥104
𝑝𝑠𝑖
𝑁 𝑦 =
𝑆 𝑦
𝜎𝑡
= 1.24
𝐽 =
𝜋(0.165)4
32
= 7.3𝑥10−5
𝑆𝑒𝑡 𝑆𝑐𝑟𝑒𝑤 #1: 𝜏 =
17.5 (
0.165
2
)
7.3𝑥10−5
= 19777.4 𝑝𝑠𝑖 − 𝑖𝑛
𝑆𝑒𝑡 𝑆𝑐𝑟𝑒𝑤 #2: 𝜏 =
17.5(1)
7.3𝑥10−5
= 2.4𝑥105
𝑝𝑠𝑖 − 𝑖𝑛
Hydrodynamic Bearing Analysis:
𝑙
𝑑
= 0.75
Lube thickness: h = 0.0017”
Abs. viscosity = η = Uρ = 3.5E-6
n = 33 rpm = 1980 rps
A = π (0.25)(0.25) = 0.19𝑖𝑛2
U = πdn = 1555 in/sec
𝐶 𝑑 = 𝑑𝑛 = 0.000425
𝐶𝑟 =
𝐶 𝑑
2
= 0.0004 = 𝑒
𝑂 𝑛 = 20
𝜀 =
𝑒
𝐶𝑟
= 0.747
𝜀 𝑥 = 0.21394+ 0.385170(𝑂𝑛 − 0.0008( 𝑂𝑛 − 60) = 0.747
𝐾𝜀 =
𝑂 𝑛
4𝜋
= 1.592

12. E11
𝜏 𝑥 = 𝜂
𝑈
ℎ
= 3.2 𝑝𝑠𝑖 𝐹 = 𝐴𝜏 𝑥 = 6.08 𝑙𝑏
Average oil pressure: 𝑃𝑎𝑣𝑔 =
𝑃
𝑙𝑑
= 17 𝑝𝑠𝑖
𝑆𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑟𝑦: 𝜏𝑠 =
𝜂𝑑3
𝑙𝑛𝜋2
𝐶 𝑑(1 − 𝜀2)0.5
= 0.45 𝑙𝑏 − 𝑖𝑛
𝜃 𝑚𝑎𝑥 = cos−1
1 − √(1 + 24𝜀^2)
4𝜀
= 159 𝑑𝑒𝑔𝑟𝑒𝑒𝑠
𝑃 =
𝜂𝑈
𝑟𝐶𝑟
2
𝑙2
4
3𝜀𝑠𝑖𝑛𝜃
(1 + 3𝑐𝑜𝑠𝜃)3
= 1670.3 𝑝𝑠𝑖
𝜙 = tan−1
(
𝜋√1 − 𝜀2
4𝜀
) = 35 𝑑𝑒𝑔𝑟𝑒𝑒𝑠
𝑅𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑔: 𝜏 𝑟 = 𝜏𝑠 + 𝑃𝑒 𝑠𝑖𝑛𝜙 = 0.8332 𝑙𝑏 − 𝑖𝑛
𝑃𝑜𝑤𝑒𝑟 𝑙𝑜𝑠𝑡 𝑖𝑛 𝑏𝑒𝑎𝑟𝑖𝑛𝑔: 𝛷 = 2𝜋𝜏 𝑟(925) = 0.006 𝐻𝑃
𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛: 𝜇 =
2𝜏 𝑟
𝑃 𝑑
= 0.004
𝑁 =
0.8
0.25
= 3.2

13. E12
Gear Assembly
Bending Stress on Drive Gear:
𝜎𝑏 =
𝑊𝑡 𝑃 𝑑
𝐹𝐽
𝐾𝑎 𝐾 𝑚
𝐾𝑣
𝑥 𝐾𝑠 𝐾𝑏 𝐾𝑙
For series 116-4 DC Motor: w = 33 rpm, T = 280 in-oz = 17.5 lb-in
F = 0.25”
𝑁 𝑝 = 8
a = 0.3”
𝑑 =
𝑁 𝑝
𝑃 𝑑
𝐷 𝑜 =
𝑁 + 2
𝑃
= 𝑃 𝑑 + 2𝑎
3.5 = 𝑃 𝑑 = 2(.3) 𝑃𝑑 = 2.9"
𝑑 =
8
2.4
= 2.76
𝑉𝑡 =
𝑑𝑤
2
=
(2.76)(33𝑟𝑝𝑚)
2
2𝜋
12
= 23.85
𝑓𝑡
𝑚𝑖𝑛
𝑄𝑣 = 6 𝑡𝑜 8 = 7
𝐾𝑣 = 0.96 𝐽 = 0.2 𝐾 𝑚 = 1.6 𝐾𝑎 = 1.25( 𝑎𝑠𝑠𝑢𝑚𝑒 𝑚𝑜𝑑𝑒𝑟𝑎𝑡𝑒 𝑠ℎ𝑜𝑐𝑘 𝑡𝑒𝑟𝑟𝑎𝑖𝑛)
𝐾𝑠 = 𝐾𝑙 = 1( 𝑛𝑜𝑛 − 𝑖𝑑𝑙𝑒𝑟)
𝑊𝑡 =
2𝑇𝑝
𝑑 𝑝
=
2(17.5)
2.76
= 12.68 𝑙𝑏𝑓
𝜎𝑏 =
(12.68)(2.9)
(. 25)(.2)
(1.25)(1.6)
(0.96)
= 1532.16 𝑝𝑠𝑖
Al-6061: 𝑆𝑓𝑏′ = 14𝑥103
𝑝𝑠𝑖
𝑆𝑓𝑏 =
𝐾𝐿
𝐾 𝑇 𝐾 𝑅
𝑆𝑓 𝑏′
𝐾𝑟 = 1.25 𝐾𝐿 = 1.6831( 𝑁)−0.0323
= 1.073 𝐾𝑡 = 1(𝑟𝑜𝑜𝑚 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒)
𝑁 = 30𝑟𝑝𝑚 (
60𝑚𝑖𝑛
ℎ𝑟
)(3
ℎ𝑟𝑠
𝑑𝑎𝑦
) (
4𝑑𝑎𝑦𝑠
𝑤𝑒𝑒𝑘
)(
52𝑤𝑒𝑒𝑘𝑠
𝑦𝑒𝑎𝑟
)(1𝑦𝑒𝑎𝑟) = 1.123𝑥106
𝑐𝑦𝑐𝑙𝑒𝑠
𝑆𝑓𝑏 =
1.073
(1)(1.25)
(14𝑥103) = 12017.6 𝑝𝑠𝑖
𝑁𝑓𝑏 =
𝑆𝑓𝑏
𝜎𝑏
= 7.84

14. E13
Gear Shaft Analysis
Steel 1050: 𝑆 𝑢𝑡 = 90 𝑘𝑠𝑖 𝑆 𝑦 = 50 𝑘𝑠𝑖
𝑇 𝑚 = 17.5 𝑙𝑏 − 𝑖𝑛 𝑇𝑎 = 0
𝑀 𝑎 = 𝑀 𝑚𝑎𝑥
𝐹 =
𝑇
𝑟
=
17.5
2.76
2
= 12.68 𝑙𝑏𝑓
𝛴𝐹𝑦: 𝑅1 − 𝐹 − 𝐹 = 0, 𝑅1 = 2𝐹 = 25.36 𝑙𝑏𝑓
𝑉 = −𝑅1 < 𝑥 − 0 >0
+ 𝐹 < 𝑥 − 1 >0
+ 𝐹 < 𝑥 − 1.5 >0
𝑀 = ∫ 𝑉 = −𝑅1 < 𝑥 − 0 >1
+ 𝐹 < 𝑥 − 1 >1
+ 𝐹 < 𝑥 − 1.5 >1
= 31.7 𝑙𝑏𝑓
𝑆 𝑒′ = .5𝑆 𝑢𝑡 = 45 𝑘𝑠𝑖
𝐶𝑙𝑜𝑎𝑑 = 1( 𝑏𝑒𝑛𝑑𝑖𝑛𝑔) 𝐶𝑠𝑖𝑧𝑒 = 0.869( 𝑑)−0.097
= 0.994 𝐶𝑠𝑢𝑟𝑓 = 𝐴( 𝑆 𝑢𝑡) 𝑏
= 0.569
𝐶𝑡𝑒𝑚𝑝 = 1 𝐶𝑟𝑒𝑙𝑖𝑎𝑏 = 0.814 ( 𝑎𝑠𝑠𝑢𝑚𝑒 99% 𝑟𝑒𝑙𝑖𝑎𝑏𝑖𝑙𝑖𝑡𝑖𝑦)
𝑆 𝑒 = (45)(0.994)(0.569)(0.814)(1)(1) = 20.7188 𝑘𝑠𝑖
𝑑 =
(
(
32𝑁𝑠𝑓
𝜋
)
(
√(𝐾𝑓 𝑀 𝑎)
2
+
3
4
(𝐾𝑓𝑠 𝑇𝑎)
2
20.7188𝑘
+
√(𝐾𝑓𝑚 𝑀 𝑎)
2
+
3
4
(𝐾𝑓𝑠𝑚 𝑇𝑎)
2
𝑆 𝑢𝑡
)
1
3
)
𝑑 = 0.25" 𝑆 𝑢𝑡 = 90𝑘 𝑁𝑠𝑓 = 9.199
𝜎 𝑚𝑎𝑥 =
𝑀 𝑚𝑎𝑥 𝐶
𝐼
𝐼 =
1
12
𝑏ℎ2
= 3.255𝑥10−4
𝑖𝑛4
𝜎 𝑚𝑎𝑥 = 12.2 𝑘𝑠𝑖 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 𝑆 𝑢𝑡 𝑠𝑜 𝑖𝑡′
𝑠 𝑠𝑎𝑓𝑒
𝜏 𝑚𝑎𝑥 =
𝑇𝑝
𝐽
𝐽 = 𝑏ℎ
1
12
( 𝑏2
+ ℎ2) = 6.51𝑥10−4
𝜏 𝑚𝑎𝑥 = 3.5 𝑘𝑠𝑖

15. E14
Surface Fatigue Internal Square
𝜎𝑐 = 𝐶 𝑝√
𝑤 𝑏
𝐹𝐼𝑑
𝐶 𝑎 𝐶 𝑚
𝐶 𝑣
𝐶𝑠 𝐶𝑓
𝐹 = 0.25" 𝑊𝑡 = 12.68 𝑙𝑏𝑓
𝐼 = (
𝑠𝑖𝑛𝜙𝑐𝑜𝑠𝜙
2
) (𝑁 𝑝)
1
𝑁𝑔 𝑁 𝑝
= 0.08
𝐶 𝑝 = 1950 𝑝𝑠𝑖( 𝑠𝑡𝑒𝑒𝑙 𝑜𝑛 𝑎𝑙𝑢𝑚𝑖𝑛𝑢𝑚) 𝐶 𝑚 = 𝐾 𝑚 = 1.6 𝐶 𝑎 = 1
𝐶 𝑣 = (𝐴/(𝐴 + √𝑉𝑡 )
𝐵
𝐴 = 50 + 56(1 − 𝐵) 𝐵 = .25(12− 𝑄𝑣).667
= 0.7314
𝐴 = 65.04 …. 𝐶 𝑣 = 1.27
𝐶𝑠 = 𝐾𝑠 = 𝐶𝑓 = 1 ( 𝑐𝑜𝑛𝑣𝑒𝑛𝑡𝑖𝑜𝑛𝑎𝑙 𝑚𝑒𝑡ℎ𝑜𝑑𝑠)
𝜎𝑐 = 33727 𝑝𝑠𝑖
𝑁𝑓𝑠 = (
𝑆𝑓𝑐
𝜎𝑐
)
2
𝑆𝑓𝑐 =
𝐶 𝐿 𝐶 𝐻
𝐶 𝑇 𝐶𝑅
𝑆𝑓𝑐′
𝐶 𝐿 = 2.466( 𝑁)−.056
= 1.13 𝐶𝑅 = 𝐾 𝑅 = 1.25(99% 𝑟𝑒𝑙𝑖𝑎𝑏𝑖𝑙𝑖𝑡𝑦)
𝐶 𝐻 = 𝐶 𝑇 = 1 ( 𝑅𝑜𝑜𝑚 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒)
𝑆𝑓 𝑐′ = 60 𝑡𝑜 70 𝑘𝑠𝑖 = 65 𝑘𝑠𝑖
𝑆𝑓𝑐 = 58760 𝑝𝑠𝑖
𝑁𝑓𝑠 = 3.035
𝐵𝑎𝑐𝑘𝑙𝑎𝑠ℎ: 𝑔𝑎𝑝 = 0.001"
% 𝑇 𝑖𝑛𝑐𝑟𝑒𝑠𝑒 𝑜𝑛 𝑠ℎ𝑎𝑓𝑡:
0.001
0.25
= .005 = 4%
𝜏 =
𝑇𝑝
𝐽
𝑝 𝑛𝑒𝑤 =
. 251
2
= .1255 𝐽 𝑛𝑒𝑤 = (
. 2512
12
) (. 2512
+. 2512) = 0.003969
𝜏 = 33200𝑝𝑠𝑖, 𝜏 𝑜𝑙𝑑 = 3360 𝑝𝑠𝑖
3360
3320
= 101.2%

16. E15
Spring Analysis
Spring Constant:
19.11 lb = limit
Robot max weight = 20 lb
𝐹 = −𝑘𝑥 𝑥 = 3.5𝑚𝑚 𝐹 = 50𝑁 𝑘 =
50𝑁
3.5𝑚𝑚
=
14.28𝑁
𝑚𝑚
𝑇𝑦𝑝𝑒: 𝑒𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑠𝑝𝑟𝑖𝑛𝑔𝑠
Shear on Axle:
𝑀𝑎𝑥 𝑇 𝑖𝑛 𝑠𝑝𝑟𝑖𝑛𝑔 = 19.11 𝑙𝑏𝑓
𝜎 =
𝑀𝑦
𝐼
𝛴𝑀 = 19.1(7)+ 19.1(11)− 𝐹2(18) 𝐹2 = 19.47 𝑙𝑏𝑓
𝛴𝐹𝑦 = −𝐹1 + 19.1 + 19.1 − 19.47 𝐹1 = 18.73 𝑙𝑏𝑓
𝑀 𝑚𝑎𝑥 = 131.11
Figure 1
Figure 2
-30
-20
-10
0
10
20
30
0 5 10 15 20
V(shear)
Length (inches)
Shear Diagram
-140
-120
-100
-80
-60
-40
-20
0
0 5 10 15 20
Moment
Length (inches)
Moment Diagram

17. E16
𝜎 =
(131.11)(
. 25
2
)
𝐼
𝐼 = 1.917𝑥10−4
𝑖𝑛4
𝜎 𝑚𝑎𝑥 = 85.47 𝑘𝑠𝑖

18. E17
Wheels & Axles Analysis
𝐴𝐵𝑆 𝑃𝑙𝑎𝑠𝑡𝑖𝑐: 𝑆 𝑢𝑡 = 4000 𝑝𝑠𝑖 𝑆 𝑦 = 1400 𝑝𝑠𝑖
20𝑙𝑏𝑓
4
= 5𝑙𝑏𝑓 𝑝𝑒𝑟 𝑤ℎ𝑒𝑒𝑙
𝜎 =
𝐹
𝐴
𝜋(5)2
4
= 𝐴 = 19.63𝑖𝑛2
𝜎 = 2546𝑝𝑠𝑖 𝑝𝑒𝑟 𝑤ℎ𝑒𝑒𝑙

19. E18
Wire Cutting Device Analysis
Motor Specs:
𝑃𝑟𝑜𝑑𝑢𝑐𝑒𝑠 2300 𝑟𝑝𝑚/𝑣𝑜𝑙𝑡
𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑔𝑖𝑣𝑒𝑛: 12 𝑉
𝐴𝑚𝑝𝑠 𝐷𝑟𝑎𝑤𝑛: 5 𝐴
2300 𝑟𝑝𝑚/𝑉 𝑥 12 𝑉 = 27,600 𝑟𝑝𝑚
𝑃𝑜𝑤𝑒𝑟 = 𝑉𝑥𝐼 = 12𝑥5 = 60 𝑤𝑎𝑡𝑡𝑠 = 0.0805 𝐻𝑃
𝑅𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝐶𝑢𝑡𝑡𝑖𝑛𝑔 𝐵𝑙𝑎𝑑𝑒 = 1.5𝑖𝑛
𝑇𝑜𝑟𝑞𝑢𝑒 = 𝑟𝑥𝐹 =
33000( 𝐻𝑃)
2𝜋( 𝑟𝑝𝑚)
= 0.1535 𝑓𝑡 − 𝑙𝑏 = 1.838 𝑙𝑏 − 𝑖𝑛.
𝐹 =
𝑇
𝑟
=
1.838
1.5
= 1.225 𝑙𝑏𝑓
Wire Specs:
𝐴 𝑤𝑖𝑟𝑒 = 0.0075𝑖𝑛2
𝑆𝑡𝑟𝑒𝑛𝑔𝑡ℎ =
18𝑙𝑏
𝑖𝑛2
𝐹𝑜𝑟𝑐𝑒 𝑡𝑜 𝑐𝑢𝑡 𝑤𝑖𝑟𝑒 = 𝑆𝑡𝑟𝑒𝑛𝑔𝑡ℎ 𝑥 𝐴 𝑤𝑖𝑟𝑒 = 0.135 𝑙𝑏𝑓

20. E19
Soil/Air CollectorAnalysis
Servo Specs:
Torque = 38 oz.-in
Force required closing lid of air collector:
Lid = 2.1 oz.
𝐴𝑙𝑖𝑑 = 𝜋( 𝑟)2
= 𝜋(. 5)2
= 0.785 𝑖𝑛2
Servo Arm = 4” long
𝑇𝑜𝑟𝑞𝑢𝑒 𝑢𝑠𝑒𝑑 𝑡𝑜 𝑐𝑙𝑜𝑠𝑒 𝑙𝑖𝑑 = 𝜏 = 38𝑜𝑧.
𝑖𝑛2
4𝑖𝑛
= 9.5𝑜𝑧. −𝑖𝑛.
𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝜏 𝑡𝑜 𝑐𝑙𝑜𝑠𝑒 𝑙𝑖𝑑 = 1.05 𝑜𝑧.−𝑖𝑛.

21. E20
MATLAB Code Calculations
%%
%Tyler Maydew - LAW OF COSINES
clc
close all
%%
%This section calculates the angle. This does not change regardless of
%position.
a=sqrt((n1-n2)^2+(w1-w2)^2);
b=sqrt((n2-n3)^2+(w2-w3)^2);
c=sqrt((n3-n1)^2+(w3-w1)^2);
A=acos((b^2+c^2-a^2)/(2*b*c));
theta=180-(A*180/pi);
%%
%This set of code is to determine whether we need to spin clockwise or
%counterclockwise.
dn32=n3-n2;
dn12=n1-n2;
dw32=w3-w2;
dw12=w1-w2;
alpha=atan(dn32/dw32)*180/pi;
beta=atan(dn12/dw12)*180/pi;
%This will find theta for line 3 off horizontal of 2
if dn32>=0
if dw32<=0
theta2=abs(alpha);
else
theta2=180-alpha;
end
else
if dw32>=0
theta2=180+abs(alpha);
else
theta2=360-alpha;
end
end
%this will find theta1 for line 1 off horizontal of 2
if dn12>=0
if dw12<=0
theta1=abs(beta);
else
theta1=180-beta;
end
else
if dw12>=0
theta1=180+abs(beta);
else
theta1=360-beta;
end
end
if theta1>=theta2

22. E21
direc=true;
else
direc=false;
end
%%
%Distance Calc
n=n1-n3;
nd=n*0.000277777778;
nm=111028.2811574018*nd;
w=w1-w3;
wd=w*0.000277777778;
wm=wd*85803.10402713598;
d=sqrt(nm.^2+w.^2);