1. Exploring Engineering Chapter 3, Part 1 Engineering Problem Solving

2. What You Will Learn • A formal technique to help you solve problems - “Need-Know-How-Solve” method breaks the problem done into four constituent parts that are easier to formulate than overall problem.

3. Problem Solving • Engineers of all disciplines are often challenged with unfamiliar problems • By breaking them down into a systematic methodology, many “impossible” problems can be solved • By systematizing your approach, you will leave an “auditing” trail for all those who later work on the same project. • As a huge bonus, the suggested method really helps in getting a good grade!

4. The “Need-Know-How-Solve” Method • Need: • The 1st step is obvious: read the problem verycarefully. Look for what is being sought. Don’t try to solve it now. Just write down what you are seeking. • Know: • Look at what you have been given (or look it up in available resources if not explicit in the statement of the problem). Again, don’t try to solve it now. Just write down what you know as the 2nd step.

5. How: • The 3rd step formulates your intended approach. It may be trivial (e.g., how many apples for $1?) or it may be an equation (e.g., E = mc2) or it the need for a spreadsheet analysis etc. Still don’t try to solve it now. • Solve: • The 4th and last step does what your instincts told you (incorrectly) to try as step 1: go ahead and get to a solution.

6. Example • Stress is defined as the force/area. Calculate the stress in SI units in an 0.50 inch diameter cable supporting a 1,000. lbm truck engine. • Need: Stress, symbol Greek sigma , in cable • Know: Force, F = 1,000. lbf = 4450. N (Convert.exe) and diameter is 0.50 inch • How: Stress F/A, where A = R2 = D2/4 • Solve:A = 3.14  (0.50  0.0254)2/4 • [in]2[m/in]2 = 5.07  10-4 m2 • Hence  = F/A = 4450./5.07  10-4 [N]/[m2] = 8.8  106N/m2 • Hopefully less than the breaking stress in the cable.

7. The “Need-Know-How-Solve” Method • On a single lane highway, you measure that there are 3140 cars/hr passing under a bridge. What is the separation between cars in seconds? • Need: Spacing in time between cars • Know: 3140 cars/hr • How: Dimensional analysis based on […] units • Solve: If 3140 cars/hr, time in s = 3600/3140 [s/hr][hr/car] = 1.15 s/car (to 3 significant figures)

8. More • If, in the previous example, the cars are traveling at 69 mph, what is their separation in m? In approximate car lengths? Does this meet a 1 car length per 10 mph spacing? • Need: Spacing between cars • Know: 3140 cars/hr, interval = 1.15 s and v = 69 mph • How: […] method. v = 69 mph = 30.8 m/s. Assume average car is ~4. m long. • Solve: Since t = 1.15 s/car and v = 30.8 m/s, distance/car = 30.8  1.15 [m/s][s/car]= 35.3 m of which 4. m. is car length. • Spacing = 31 m or 31/4 [m/car][car/m] = 7.8 ~ 8 (car lengths), which is greater than the recommended 7 car lengths.

9. The “Need-Know-How-Solve” Method Summary: • Engage the mind before the pencil! • Delay solution until you have all in the facts. • Allow for a traceable solution for other members of an team (warts and all!). • As a practical matter, you can get most of the grade for the same wrong answer if you follow this methodology! • E.g, write just the answer as “84.7” and may get you “0” grade but not for a clear development to a solution that said T = 8670/10.0 = 84.7! You would still get most of the grade. (This mimics the auditing trail required of a practicing engineer.)