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Exam #1 Prep. -&- Structures (Ch. 4)

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Exam #1 Prep.-&- Structures (Ch. 4)

ENGR B36 - Statics

Pat Aderhold

9/29/2014

- Notes
- More feedback from Quiz #2
- Logistics for Exam #1 (this Wednesday!)

- Review Quiz #3
- Last chance Ch. 2-3 questions
- Start Ch. 4

- Use the units you’re given
- Define your axes
- Bold print means vector
- Know momentsandcouples
- Add diagrams and labels
- Some people got the “wrong” answer and full credit (because the right answer was in their work)
- I’m not sure I’m being as lenient as I promised, there may be bonus points opportunities / throw out low grades, etc. in the future

- I’ll start handing them out 5 min. before class starts
- Same as quiz: answers on front page and all work shown on blank pages
- I’ll have extra blank pages
- Make it clear and you’re more likely to get “free” points

- Organization
- Each problem starts on new page
- Make a big mistake? Cross out with thin line and move on

- Calculators acceptable (recommended)
- Read the directions carefully
- I’ll give you a few extra minutes after to finish up but there is a class meeting after this one

A barge is pulled by a horse along a canal as seen in the image below. The tension on the connecting rope is 1000 N in magnitude and in the direction of the unit vector n = 0.913i + 0.365j - 0.138k. Find the amount of the force that acts along the centerline of the boat, which has the position vector r = 10mi + 10mj.

Replace the two forces and one couple acting on the rigid pipe frame by their equivalent resultant force R acting at point O and a couple MO. What is the component of MOthat acts in the same direction as R?

The world-famous “Bakersfield” sign is undergoing maintenance during a wind storm. There is a 100 kg worker painting the letter “K”, located 5 m from the left column. The weight of the span is 1000 kg and its weight is uniformly distributed, meaning that the force can be modeled as concentrated and acting at its center of mass (the center of the span). The force of the wind is 1 kN, which also acts as a concentrated force at the center of the span (going into the page, i.e. in the negative x direction). Assume that this structure is fixed at the left end (near point “A”) and on rollers at the opposite end. Neglect the width and weight of the columns, draw a free-body diagram and calculate the force and moment acting at point “A” of the structure: (Note: the instructor is not trying to be “tricky” with this problem. If something seems inaccurate or vague, ask for clarification).

- “Practice Problems”?
- Quizzes?
- Unassigned problems from text?
- Concepts?

Chapter 4Structures

Meriam, JL and Kraige, LG. Statics 7th Ed. Wiley 2012. p.172

- Why bother?
- Is there a difference (to us)?

(Read the book for all the details)

- In real life, many advantages
- Weight, money, modularity, reparability

- To us, responds the sameto external loading
- Now we can observe internal forces

- “Simple”: built from a basic triangles
- Members can’t stretch => doesn’t deform

- “Plane”: back to 2D for a while
- As usual: doesn’t exist but useful construct

“Method of Joints”

“Method of Sections”

They mean exactly what they think you mean

Meriam, JL and Kraige, LG. Statics 7th Ed. Wiley 2012. p.176

- Solve at one location => move on to next
- “Two-force members” have equal, opposite, collinear forces... always!

Meriam, JL and Kraige, LG. Statics 7th Ed. Wiley 2012. p.175

Meriam, JL and Kraige, LG. Statics 7th Ed. Wiley 2012. p.177

- Only using two pieces of information(ΣFX=0 & ΣFY=0)
- Why not use third? ΣMO=0

Meriam, JL and Kraige, LG. Statics 7th Ed. Wiley 2012. p.188

Nothing magic about FBDs. Can chop right through a structure if we like. Still get rid of the rest of the universe. Still adding in all necessary forces.

Meriam, JL and Kraige, LG. Statics 7th Ed. Wiley 2012. p.188

- Focus on Exam #1
- We’ll pick up the pace after
- Read up to 4/4 for the next Monday
- We’ll be going through practice problems

Exam #1 on Wednesday!