More Applications of Linear Stress-Strain Relations (Credit for many illustrations is given to McGraw Hill publishers and an array of internet search results) - PowerPoint PPT Presentation

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More Applications of Linear Stress-Strain Relations (Credit for many illustrations is given to McGraw Hill publishers and an array of internet search results)

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Title: More Applications of Linear Stress-Strain Relations (Credit for many illustrations is given to McGraw Hill publishers and an array of internet search results)


1
More Applications of Linear Stress-Strain
Relations (Credit for many illustrations is
given to McGraw Hill publishers and an array
ofinternet search results)
2
Parallel Reading
  • 3.4 Statically Determinate Structures
  • 3.5 Statically Indeterminate Structures
  • 2.13 Generalized Hookes Law

3
I Mentioned That Sometimes in Statics Problems we
run out of equations before we have answers
A
Looks like no brainer statics
300600 A B But we are out of equations
to Break-down how much force is At A and how much
is at B
B
This is one of those Statically
Indeterminate Things.
4
Material Properties to the Rescue
Blow the bottom support out and let the loaded
bar Just hang there. Calculate how much
lengthening we will see.
5
Next Impose that the Ground Did not Disappear and
will Push Up as Necessary to Ensure 0 displacement
See how large the force B has to be to Cancel the
displacement Now you have B Now we can use our
statics equation 600300 A B
6
A General Comment On Solution Methods
  • Look at the Problem
  • Look at What You Know
  • Look at What You Want to Know
  • Look for What Equations Apply
  • Plan your solution strategy before you start
    number crunching
  • Some people just start trying equations hoping
    that some miracle will suddenly pop out (it
    usually doesnt)

7
Lets Do the Math
Chop our block into 4 pieces What force is
yanking on the bottom of Block 1 Well lets see
Nothing so P10 What force is yanking on the
bottom of Block 2 Looks like 600 KN so P2
600 What force is yanking on the bottom of Block
3 Well pretty clearly Block 2 is hanging
on Worth about 600 KN so P3 600
Throw in Block P4 has 600 KN From below plus
300 KN for a Total of 900 KN
8
Estimate some Deformations
A400X10-6 M
P2
0.15 M
d 6000.15/(400X10-6 E) 225000/E
600KN
9
Add Up All the Deformations(P2, P3, and P4)
10
Now We Will Have the Reaction at the Base Reverse
the Deformation
11
We Know the Supports Working Together Prevent
Stretching Out
We Used Material Properties to Determine RB
12
Finish It Off With Statics
13
Statics Folks Eat Your Heart Out
14
Assignment 5
  • Problem 3.5-2

15
What Happens if I Try to Pull a Block Apart in 3
Directions at Once?
Make it Easy to Solve If the deformations are
small Geometry from one force Wont change
anything for the Next force. We can just Put
them over the top of Each other. Actually thats
what we did When we printed physical Compression
over thermal Expansion or solved the Statically
indeterminate problem
16
Remember Each Force Stretches in Its Direction
and Thins things down in the other directions
Force in X direction stretched In X direction,
but the pulls In Y and Z thinned it down
Principle of Superposition
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