# File

```Exam #1 Prep.
-&amp;- Structures (Ch. 4)
ENGR B36 - Statics
9/29/2014
Agenda
• Notes
 More feedback from Quiz #2
 Logistics for Exam #1 (this Wednesday!)
• Review Quiz #3
• Last chance Ch. 2-3 questions
• Start Ch. 4
Quiz #2 Feedback
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Use the units you’re given
Bold print means vector
Know moments and couples
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
Exam #1 Logistics
• 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)
• I’ll give you a few extra minutes after to finish up but there
is a class meeting after this one
Quiz #3 - Problem 1
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.
Quiz #3 - Problem 2
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 MO that acts in
the same direction as R?
Quiz #3 - Problem 3
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
Lingering Ch. 2 &amp; 3 Questions?
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“Practice Problems”?
Quizzes?
Unassigned problems from text?
Concepts?
Chapter 4
Structures
Meriam, JL and Kraige, LG. Statics 7th Ed. Wiley 2012. p.172
Rigid Bodies vs. Structures
• Why bother?
• Is there a difference (to us)?
Structures
(Read the book for all the details)
• In real life, many advantages
 Weight, money, modularity, reparability
• Now we can observe internal forces
“Simple” and “Plane”
• “Simple”: built from a basic triangles
• Members can’t stretch =&gt; doesn’t deform
“Simple” and “Plane”
• “Plane”: back to 2D for a while
• As usual: doesn’t exist but useful construct
Two Ways to Go
“Method of Joints”
“Method of Sections”
They mean exactly what they think you mean
Method of Joints
Meriam, JL and Kraige, LG. Statics 7th Ed. Wiley 2012. p.176
Method of Joints
• Solve at one location
=&gt; 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
Method of Joints
Meriam, JL and Kraige, LG. Statics 7th Ed. Wiley 2012. p.177
Method of Sections
• Only using two pieces of information
(ΣFX=0 &amp; 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
Wrap-Up
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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!
```