Exploring EngineeringChapter 3, Part 1

Engineering Problem Solving

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.

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!

The “Need-Know-How-Solve” Method

Need: The 1st step is obvious: read the problem

very carefully. 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.

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.

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.

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] = 8.8 106 N/m2

– Hopefully less than the breaking stress in the cable.

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0.5 tons0.5 ton engine

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Example

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)

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.

More

The “Need-Know-How-Solve” Method

Summary:1) Engage the mind before the pencil!2) Delay solution until you have all in the facts.3) Allow for a traceable solution for other

members of an team (warts and all!).4) 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.)