1. Bending Forces Or Beam Me Up Scotty
(Credit for many illustrations is given to McGraw Hill publishers and an array of internet search results)

2. Parallel Reading Chapter 6 Section 6.1 Introduction
Section 6.2 Strain Displacement Analysis
Section 6.3 Flexural Stress in Linear Elastic Beams
(Do Reading Assignment Problem Set 6A)

3. Consider the Case of a Beam With a Load in the Middle
The members on top are in a
Squeeze Play, while the
members on the bottom are in
Tension.

4. Lets Take a Look at That If we are in tension on one side of the
Bend and compression on the other,
Somewhere there must be a neutral plain

5. As We Move Away from the Neutral Axis the Strain Varies Linearly

6. Hooke’s Law Now Tells Us About Stress in Beam Bending
Where E is Young’s Modules

7. Now We Get A Tip from Statics
The forces (stress * area) above and below the neutral plane have
to be equal.

8. Only One Way that is True
That neutral plain has to go right through the centroid of the beam

9. So What is a Centroid? (we hope the heck this is a review)
The physical center
The center of mass
for the beam
So someone tell me where the centroid is
for the 4 X 6 beam?
Where is it for the 3 X 8 beam?

10. So How do We Find the Centroid When its’ not Stupid Obvious
90 mm
20 mm
40 mm
30 mm

11. Someone's bound to come up with Integrating Over the Area
90 mm
90 mm
20 mm
40 mm
A less painful option is usually
Available.
30 mm
Most of the objects we work with break-down into
Simple parts where the centroid is obvious

12. Lets Peg the Obvious Centroids
90 mm
20 mm
10 mm
45 mm
15 mm
40 mm
20 mm
30 mm

13. Next We’ll Weight Each Obvious Centroid by the Area of It’s Object

14. Now We’ll Divide Through by Total Area
We just nailed ourselves the Centroid of a T beam
(Of course finding centroids is not a key topic
of this course, but if we can’t do it, it will make
our lives miserable for this course).

15. Back to Bent Beams When our beam deflects it bends
along the arc of a circle of radius ρ
through an angle of θ. The radius
extends from the center point of the
arc of the bend to the neutral plane.

16. If We Take a Closer Look at Deformation in a Cross-Section
The plain of our cross-section remains
A nice flat cross-section – But
On the compression side our nice former
Rectangle puffs out increasingly toward
The top (prob not a surprise if we
Remember Poisson’s ratio)
And get increasingly skinny on the bottom
As we into higher tension areas away
From the neutral plain.

17. The Amount of Deflection is Related
To a bunch of terms including the
Bending moment on the beam,
Young’s Modulus (a material
Property), and something called I
That comes from the geometry of
The beam.

18. Similarly, the amount of thinning or thickening is proportional to material properties
Any given cross section stays a plain but
The Compression size fattens up
The Tension size skinnies down
Not surprisingly the amount of plumping out or
Skinnieing down is proportional to Poisson’s
Ratio

19. Lets Review Our Materials Properties
Young’s Modulus is the slope of the line in a stress
Strain plot. It relates change in length from a tension
Or compression load to the stress

20. When Things Stretch in One Direction – They Skinny Up in the other

21. The Proportion is Called Poisson’s Ratio

22. So Let Make Sure We Understand the Terms for the amount of Deflection
There is Young’s Modulus
(We know what that is)
It makes sense that we might not want
Excessive deflection

23. Checking Out More Terms
M is that bending moment couple that
Is deflecting the beam.

24. And Then There is I
Right now I’m not seeing what that is

25. Let’s Review that Moment Term
Note its just the
Force * lever arm y

26. Handily a Look at that Last Equation gives us I
Obviously
We call this term the Moment of Inertia
(Yes we do hope this is a review for you from Statics)

27. Section Modulus is a Closely Related Term

28. That Inertia Term is a Measure of the Ability of a Beam to Resist Bending
There are
Precalculated
Tables of these
Values available
For most
Structural steel
shapes

29. Lets Try Doing Something With This Stuff
The allowable Tensile
Stress is 12 Ksi
The allowable Compressive
Stress is 16 Ksi
What is the largest Moment Couple
I can put on this thing?
6 in
4 in

30. Our Basic Equation Limiting stress 3 in Neutral Axis or Plane
Since our beam is symmetric our most limiting stress will be tension

31. Obviously We Need S
6 in =
For a rectangle
4 in

32. Going for the Answer

33. Assignment 13
Do Problems and

34. What if We Tried a Different Shape6 4
For our rectangle
A 24 section modulus allowed us to put
A 288,000 in*lb moment on our beam while
Staying in allowable tensile stress

35. Use a Table of Standard Wide Flange Beams
First term after W
(in this case 12 inches)
Second term is the weight in lbs per foot
If I pick a weight of 22 lbs/ft I will get
A Section Modulus of 25.4 > 24
A wide W12X22 wide flange beam will carry slightly more load than our
6X4 beam.

36. Here’s the Kicker The area of my wide flange beam is 6.48 in^2
Instead of 24 in^2 for my rectangular beam
I get more from only 25% of the material by
Using a wide flange beam!

37. Is there a Down Side? 4 in 6 in S = 16
Maximum Moment = 12,000 * 16 = 192,000 in*lbs

38. So Could Anything Go Wrong with Our Wide Flange Beam?NA
S for a W12X22 beam about
The weak axis
S= 2.31 < < 16 for our 4X6
Things really go to crap around
The weak axis.

39. Assignment 14
Do problems and

40. Controlling Cost With Beams
We might want to consider a less expensive
material

41. The Problem with Concrete Beams
Like most brittle rock materials – they
Have little tensile strength
We already saw in our last problem that
Tensile strength can form our design
Limit Lk

42. The Practical Trick Put steel reinforcing Rebar near the tensile
Edge of the beam

43. Theory of Reinforced Concrete
Compression load area
Neutral Plane or axis
(Which is not at the centroid)
Concrete holds
The rods out at
A distance to
Maximize their
Inertial value
Tensile load rides entirely on
the steal reinforcing rods

44. Convert the Steel Cross Sectional Area to an Equivalent Concrete Area
Here is our concrete
Compression area
Hear is the equivalent concrete area
To replace the rebar

45. Lets Walk Through This We have a concrete beam
E for Concrete is 25 GPa
It has steel rebar
Reinforcement.
E for Steel is 200 GPa
This is the area of
The steel

46. Conversion to an Equivalent Concrete area is Proportional to Young’s Modulus
So an equivalent concrete area is *

47. We Now Need to Find the Neutral Axis (Which is not at the Centroid this time)b
We exploit the fact that the moment
Of the top part must be equal to the
Moment of the steel equivilent

48. Set Up Our Quadratic Equation
Moment of our
Equivalent concrete
(steel) section about axis
Moment of Concrete
About neutral axis

49. Solving Our Quadratic
177.87mm
302.13mm
Let us now assume the bending moment on this beam is 175 KN*m
lets check out the resulting stresses

50. We Know the Forces Above and Below the Neutral Plain are Equal and Opposite
Looks like we need the value of I

51. Get the Inertia of the Upper Compression Block around Neutral Axis
Contribution to
Inertia of Beam
From Compression
Concrete

52. Now the Inertia of Our Equivalent Concrete Area

53. Solve I for Our Equivalent Beam System

54. Now Let Get the Stress in Our Concrete Compression Area
Our given Moment Load
(from our previous calculation)

55. Going for the Stress on the Steel
Stress in equivalent concrete
Is the same.
But we remember the ratio of
Our Young’s Modulus
* 8 =