1. Thinking Like an Engineer 2e Instructor Slides
SOLVEM
Thinking Like an Engineer
An Active Learning Approach, 2e
Stephan, Bowman, Park, Sill, Ohland
Copyright © 2013 Pearson Prentice-Hall, Inc.
SOLVEM
SOLVEM
Thinking Like an Engineer 2e
Instructor Slides
Instructor Slides

2. Thinking Like an Engineer 2e
SOLVEM
SOLVEM Methodology
S = Sketch
O = Observations and / or Objectives
L = List
V = Variables and constants
E = Equations
M = Manipulate
SOLVEM
Thinking Like an Engineer 2e
Instructor Slides

3. Thinking Like an Engineer 2e
SOLVEM
Example SOLVEM_1
What is the area of the largest equilateral triangle that can be cut from a circular piece of material 50 centimeters in diameter?
SOLVEM
Thinking Like an Engineer 2e
Instructor Slides

4. Example, continued SOLVEM_1
Thinking Like an Engineer 2e
SOLVEM
Example, continued SOLVEM_1
What is the area of the largest equilateral triangle that can be cut from a circular piece of material 50 cm in diameter?
SOLVEM H L
60°
120°
60°
Thinking Like an Engineer 2e
Instructor Slides

5. Example, continued SOLVEM_1
Thinking Like an Engineer 2e
SOLVEM
Example, continued SOLVEM_1
Objective:
Determine the area of an equilateral
triangle inscribed in a circle
Observations:
Each angle of the triangle is 60°
Each side of the triangle is a chord of the circle
The angle subtended by one of the chords is 120° (360°/3)
60°
120° H L
SOLVEM
Thinking Like an Engineer 2e
Instructor Slides

6. Example, continued SOLVEM_1
Thinking Like an Engineer 2e
SOLVEM
Example, continued SOLVEM_1
D Diameter of circle 50 cm
R Radius of circle
φ Internal angle of equilateral triangle 60°
θ Angle subtended by chord 120°
L Length of chord (base of triangle)
H Height of triangle
A Area of triangle
SOLVEM
Thinking Like an Engineer 2e
Instructor Slides

7. Example, continued SOLVEM_1
Thinking Like an Engineer 2e
SOLVEM
Example, continued SOLVEM_1
Radius of circle:
R = 0.5 D
Area of triangle:
A = 0.5 L H
Length of chord:
L = 2 R sin(θ/2)
tan(φ) = length of opposite side / adjacent side
H/(L/2)
for the right triangle defined by half of the equilateral triangle
SOLVEM
Thinking Like an Engineer 2e
Instructor Slides

8. Example, continued SOLVEM_1
Thinking Like an Engineer 2e
SOLVEM
Example, continued SOLVEM_1
Length of chord:
L = 2 R sin(60°) = 1.73 R
Height of triangle:
H = L tan(φ)/2
H = (1.73 R) tan(60°)/2
H = 1.5 R
Area of triangle:
A = 0.5 (1.73 R) (1.5 R)
A = 1.30 R2
SOLVEM
Thinking Like an Engineer 2e
Note that by solving for the general case in terms of R (A = 1.30 R2), we can now easily calculate the area of an inscribed equilateral triangle in any size circle.
Instructor Slides

9. Example, continued SOLVEM_1
Thinking Like an Engineer 2e
SOLVEM
Example, continued SOLVEM_1
NOW, plug in numbers!
R = 0.5 D
R = 0.5 (50 cm) = 25 cm
A = 1.30 R2
A = 1.30 (25 cm)2
A = 811 cm2
SOLVEM
Thinking Like an Engineer 2e
Reasonableness check: Visual inspection leads one to think that the triangle area is about half the area of the circle. The area of the circle is 1960 cm2, so 811 seems reasonable.
Instructor Slides

10. Example SOLVEM_2
You purchase a truck having wheels and tires such that each wheel-tire combination has a radius of 14 inches.
The speedometer, which reads accurately, is calibrated for these tires. Inspired by a monster-truck event, you decide to jack up your truck such that the wheel-tire radius is 20 inches.
Identify the effect this has on your speedometer reading.
Specifically, if your speedo-meter indicates that you are traveling at 60 miles per hour, what is your actual speed?
SOLVEM
Thinking Like an Engineer 2e

11. Example, continued SOLVEM_2
Thinking Like an Engineer 2e

12. Example, continued SOLVEM_2
Observations:
Assuming the tires do not slip, for one revolution, the tire circumference will be the distance traveled by the truck.
The number of wheel revolutions per time determines the speedometer reading.
Since more distance is covered per revolution for a bigger tire, the truck will be moving faster than the speedometer says it is in this case.
Since we are comparing one tire to another, I should be able to solve this problem as a ratio, without having to worry about unit conversions.
SOLVEM
Thinking Like an Engineer 2e

13. Example, continued SOLVEM_2
R wheel radius 14 inches, 20 inches
Vtire the truck velocity for a particular tire
t time
revtire revolution distance for a particular tire
tire tire radius
SOLVEM
Thinking Like an Engineer 2e

14. Example, continued SOLVEM_2
Circumference of wheel = 2 π R
V14 = (rev14 / t)
V20 = (rev20 / t)
rev14 = circumference of wheel14
rev20 = circumference of wheel20
SOLVEM
Thinking Like an Engineer 2e

15. Example, continued SOLVEM_2
Thinking Like an Engineer 2e
SOLVEM
Example, continued SOLVEM_2
Step 1: Solve Equations 2 & 3 for time (t), which is unimportant since we are finding a rate.
t = K (rev14 / V14)
t = K (rev20 / V20)
Step 2: Since unit time is the same in each case, equate the two expressions:
(rev14 / V14) = (rev20 / V20)
V20 = V14 (rev20 / rev14)
SOLVEM
Thinking Like an Engineer 2e
Instructor Slides

16. Example, continued SOLVEM_2
Step 3: Substitute for circumference
(rev20 /rev14) = (2 π R20) / (2 π R14)
= (R20 / R14)
Step 4: Now we can use ratio of radii instead of ratio of velocities
V20 = V14 (rev20 / rev14)
= V14 (R20 / R14)
SOLVEM
Thinking Like an Engineer 2e

17. Example, continued SOLVEM_2
NOW plug in number!
V20 = V14 (R20 / R14)
= (60 mph) (20 in / 14 in)
V20 = 86 mph
SOLVEM
Thinking Like an Engineer 2e