EFFECTIVE TEACHING OF REINFORCED CONCRETE

RICHARD MILLER, FPCIProfessor of Civil Engineering

University of CincinnatiChair, PCI R&D Council

1

2

Every once in a while, something changes how you think.

TAKING OVER SENIOR DESIGN

3

I started working with the Neihoff Studio at UC.

Engineering students worked with architects, planners and political scientists.

I had to THINK about: What IS an engineer.

4

This article really made me think, too.

Michael Speaks – Design Thinking

5

I saw a lecture by Michael Speaks, Dean College of Design, U Kentucky on Design Thinking.

It made me think, too.

http://www.theberlage.nl/events/details/2009_04_06_design_thinking

Design Thinking• Jeff Hawkins – Palm Pilot Inventor• On Intelligence - http://www.onintelligence.org/• Intelligence is really

– Storing Patterns– When the world does not fit the pattern, form a

new pattern.– Intelligence is the ability to predict patterns.– Knowledge is produced, not found.

6

Design Thinking• “Storing patterns” is what the Scientific

American article was about.– Chess masters store patterns.– Athletes store patterns.– Engineers store patterns.

7

www.houseofstaunton.com

Design Thinking

8

Engineering is pattern recognition, pattern construction and pattern prediction!!!!!!!!

Design Thinking

9

Example:

You recognize the pattern:

sin( )

?

sin

bx

h x

f x aedfdx

f x a g x

g x e

h x bx

I do this without thinking. I did not really think about it until I had to teach to my son (business major).

Pattern:

Design Thinking

10

sin( )

sin

' '

' '

' cos

cos

bx

h x

bx

f x ae

f x a g x

g x h x e

h x b bx

df ab bx edx

Constant “rule”

e to the f(x) rule

Trig function

Design Thinking

• Engineers do not just try to solve the problem, they REFRAME IT.

• Engineers ask “what if”.• Engineers try to learn all they can about

something.• Engineers then extend and extrapolate

that knowledge.

11

Design Thinking

• Speaks gives an example.– An architect wanted to use a new material for

a building.– The engineers studied the material,

characterized it and tested it.– In the end, the ENGINEERS gained

knowledge and became more valuable.

12

Design Thinking

• Learn patterns• Form new patterns• Extend the patterns• You can only ask “what if” if you

know “what is.”

13

WE HAVE A REAL OPPORTUNITY!

14

Technology presents us with an opportunity to change how we teach.

It goes to the heart of “What is an engineer.”

EXAMPLE

15

What really affects the performance of a concrete beam?

16

First, we have teach them the basics and then show the patterns.

17

M = (C or T) d

M = Cd = Td

Start with something they know!Simple linear theory. They know this pattern.

18

0

1000

2000

3000

4000

5000

6000

0 0.002 0.004 0.006 0.008 0.01 0.012

Strain

Stre

ss (p

si)

Now introduce something new – this is the stress strain curve for concrete in compression. They don’t know non-linear materials.

The beginning is linear!!!

19

Walk them through the sequence. Here is a pattern they know.

Uncracked

20

Cracked – Linear – Now we start to extend the pattern. This “doesn’t fit the mold”.

Remember, the beginning of the stress strain curve is linear!!!We ignore cracked concrete!

Strain is linear with “y”.Plane sections remain plane.

21

Now, we could introduce a new pattern and go this way:

Linear elastic approach to a cracked member.

22

But let’s go this way: Cracked – Non-linearNow we have a really different problem!pattern.

Strain is still linear with “y”.Plane sections remain plane.

Now, the stress strain curve is non linear!!!

23

Now introduce c = 0.003.

Tell them it is a definition. It is supported by research, but other values can be justified. The code writers had to choose a reasonable value and they chose 0.003. If justified by future research, this could change. But this is KNOWLEDGE for the “what if”.

24

Show a similarity – remind them that an integral is an area.

dVol = dA

2/

0

2/

0

2/

0

)(h

h

h

dA

Vold

dTC

VOLUME OF COMPRESSION WEDGE

Here, we EXTEND a known pattern!

25

Show that the stress block is a “convenience” for integration.

26

Notice that we still start with some theory. Using the baseball analogy: you have to start with the theory of “Hit the ball and run to first base!” But theory is introduced a little at a time, not all at once.

If we taught baseball like engineering, 2nd grade everyone would learn infield, 3rd grade everyone would learn outfield, 4th grade everyone would learn hitting in 5th grade everyone would learn pitching, in 6th grade everyone would learn baserunning. Maybe in 7th grade you would actually play a game. Who would stick around for THAT??

27

Consider the concrete beam shown which is designed for positive moment. The distance from the extreme compression fiber (top in this case) to the centroid of the reinforcing steel is called “d”, the effective depth.

fc’ = 5000 psify = 60 ksi

THIS EXAMPLE USES NUMBERS!!!!

28

Why use an example with NUMBERS??

29

When you learn a new subject, it is easier to start with a “concrete” example rather than a theoretical one!

(no pun intended … Or maybe just a little pun!)

30

Let us assume the concrete stress strain curve can be modeled as a parabola. The ACI code allows the concrete stress-strain curve to be modeled as a parabola, trapezoid, triangle or any shape which the engineer can justify. There is a lot of research which shows that the pre-peak part of the curve can be modeled as a parabola if the concrete is normal strength.

31

22

003.0003.0250002' cc

cf

c

cf

ccc psiff

Parabolic Stress-Strain

0

1000

2000

3000

4000

5000

6000

0 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035Strain

Stre

ss (p

si)

Again, the example uses numbers!

Note: we could do “what if’s with different values of extreme fiber strain and use numerical integration!

You may have to show them how to write the equation of a parabola to fit this curve.

32

The values of c and fs are not known. Assume that the steel yields and fs = fy .The integral is just a volume. The area of the parabola is (2/3)cfc’. Multiply by width, b, to get volume! Apply equilibrium C=T.

"4.5

600000.3"10500032 2

c

psiincpsi

33

ftkinkM

ksiinM

n

n

2553055

"4.583"19600.3 2

The centroid of a parabolic area is (3/8)c from the vertex.

Don’t Assume They RememberGeometry

Always use your units!

34

Does the steel yield?

C = 5.4”

ys

s

f19 5.40.003 0.0076 0.00215.4 E

YES

35

Whitney Stress Block

bffAa

fAbafTC

c

ss

ssc

'85.0

'85.0

Now show the stress block:

36

1

1

1

0.85 ' 4000

0.65 ' 8000

0.80 ' 5000

c

c

c

f psi

f psi

interpolate in betweenf psi

37

Assume steel yields fs = fy .

"30.580.0

"23.4

"23.4"10585.0

600.3

1

2

ac

ksiksiina

38

Assume steel yields fs = fy .

ftkinkM

ksiinM

n

n

25330402

"23.4"19600.3 2

39

Does the steel yield?

C = 5.3”

s19 5.30.003 0.0077 0.0021

5.3

YES

DifferenceParabola Stress Block

C inch 5.4 5.3

Mn k-in 3055 3040

40

Good place to point out that ANYTHING + 10% is good in concrete!

41

Wait a minute!! What happened to the k factors for the stress block???

At this point, you can introduce these factors and show how the stress block was “calibrated”.

42

However, ask yourself this about the k factors for the stress block:

Are they necessary for student understanding or just confusing?

Is it useful to the students at this point?

Will they even remember?

It is probably a good idea to make the aware of how the stress block was calibrated – but don’t dwell on it.

Effect of parameters on Mn

• Again, the tendency here is to do this “theoretically”.

• This is a good chance to get the students “doing something”.

• Have them bring lap tops and work in groups.

• They form patterns and ask “WHAT IF”.

43

44

fc’ = 5000 psify = 60 ksi

Mn = 3040 k-in = Minitial

Now we can explore changing the parameters.

We could do this theoretically or with a numerical example

Problem – what if we start to change things on this beam??

45

c d b es fs As f'c beta1 a C T M0.9 19 10 0.060333 60 0.5 5 0.8 0.72 30.6 30 559.2

1.78 19 10 0.029022 60 1 5 0.8 1.424 60.52 60 1097.282.65 19 10 0.018509 60 1.5 5 0.8 2.12 90.1 90 1614.63.53 19 10 0.013147 60 2 5 0.8 2.824 120.02 120 2110.564.43 19 10 0.009867 60 2.5 5 0.8 3.544 150.62 150 2584.2

5.3 19 10 0.007755 60 3 5 0.8 4.24 180.2 180 3038.46.2 19 10 0.006194 60 3.5 5 0.8 4.96 210.8 210 3469.2

7.05 19 10 0.005085 60 4 5 0.8 5.64 239.7 240 3883.27.95 19 10 0.00417 60 4.5 5 0.8 6.36 270.3 270 4271.48.85 19 10 0.003441 60 5 5 0.8 7.08 300.9 300 463810.6 19 10 0.002377 60 6 5 0.8 8.48 360.4 360 5313.6

11.55 19 10 0.001935 56.11688 7 5 0.8 9.24 392.7 392.8182 5648.72511.98 19 10 0.001758 50.97997 8 5 0.8 9.584 407.32 407.8397 5794.587

The students can make a simple spreadsheet program like this:

(or you can make it for them).

0

1000

2000

3000

4000

5000

6000

7000

0 1 2 3 4 5 6 7 8 9

Area Steel in^2

Mn

k-in

Tension Control

Compression Control

This shows the variation of Mn with As. This graph is NOT multiplied by

46

0

500

1000

1500

2000

2500

3000

3500

4000

0 1 2 3 4 5 6 7 8 9

Area Steel in^2

Mn

k-in

Tension Control

Compression Control

This shows the effect of reducing the moment by . Note that once the tension control limit is passed, there is little gain for more steel.

47

0

1000

2000

3000

4000

5000

6000

7000

0 5 10 15 20 25 30 35

d in

M k

-in

Ideal linearSteel yields at d greater than 8.5

This shows the effect of varying d.

48

0

1000

2000

3000

4000

5000

6000

7000

0 5 10 15 20 25 30 35

d in

M k

-in

Ideal linear

Steel yields at d greater than 8.5

Tension Control

Here is the effect of the variation of d, but the moment is multiplied by .

49

0

1000

2000

3000

4000

5000

6000

7000

0 1 2 3 4 5 6 7 8 9 10

Concrete Strength ksi

Mom

ent k

-in

This shows the effect of varying fc’. Note that the section is tension controlled in all cases.

50

0

1000

2000

3000

4000

5000

6000

7000

0 5 10 15 20 25

b in

M k

-in

This shows the effect of varying b. Again, all sections are tension controlled.

51

0

0.5

1

1.5

2

2.5

0 0.5 1 1.5 2 2.5 3

M/M

init

d Area of Steel

b, Concrete Strength

b and Concretestrength overlap

Here is a normalized graph. The factor is NOT applied. Note that As and d have the largest effect on moment. The concrete strength and b have almost no effect.

This graph is normalized. The point 1,1 is the beam with As = 3 in2, b = 10 in, d = 19 in and fc’ = 5ksi. The x axis is the fraction of the original parameter, e.g. 0.5 would be d = 9.5 inches. The y axis is the ratio of the moment capacity to the moment capacity of the original cross section.

52

53

Now bring in the theory:

bffAdfAM

adfAM

c

ssssn

ssn

'6.1

2

54

bffAfdAM

bffAdfAM

c

ssssn

c

ssssn

'6.1

'6.122

0

1000

2000

3000

4000

5000

6000

7000

0 1 2 3 4 5 6 7 8 9

Area Steel in^2

Mn

k-in

Tension Control

Compression Control

Clearly parabolic in As

55

bffAdfAMc

ssssn '6.1

Linear in d, as long as the steel yields (a/2 is constant). When the steel does NOT yield, fs depends on d, but there is only a slight deviation from linear.

0

1000

2000

3000

4000

5000

6000

7000

0 5 10 15 20 25 30 35

d in

M k

-in

Ideal linear

Steel yields at d greater than 12

56

0

1000

2000

3000

4000

5000

6000

7000

0 1 2 3 4 5 6 7 8 9 10

Concrete Strength ksi

Mom

ent k

-in

0

1000

2000

3000

4000

5000

6000

7000

0 5 10 15 20 25

b in

M k

-in

bffAdfAMc

ssssn '6.1

The variation of b and fc’ are the same. Note that these graphs all have the steel yielding.

1.57

An interesting note on tension control: Before this, we all remember ρmax =0.75 ρbal. To define tension control, Bob Mast calculated the ductility (basically δmax/ δy) and then found the extreme fiber strains that corresponded to the same ductility.

In the AASHTO LRFD Bridge Specification, there are values of ε for tension control of steel with yield strengths above 60ksi where the stress/strain curve is not elastic/perfectly plastic.

We used the computer to calculate the ductility for a number of different beam shapes with GR 60 steel. We then recalculated them with GR 100 steel and found values of ε that provided the same ductility.

This is a really good example of “what if” engineering!

58

Another Example:

A Square Column:

59

As d A barb= 18 layer 1 3 1.8 2 0.6 Plastic Centroid 0 rho 0.011111h= 18 0 0 9 0 y bar 9

layer 3 3 1.8 16 0.6fc' 4000beta1 0.85

c epst a eps1 eps2 fs1 fs2 Pc PS1 PS2 P M M P10000 -0.003 18 -0.003 -0.003 -60 -60 -1101.6 -101.88 -101.88 1305.36 1.82E-12 1.52E-13 1305.36

22 -0.00082 18 -0.00273 -0.00082 -60 -23.7273 -1101.6 -101.88 -36.5891 1240.069 457.0364 38.08636 1240.06918.17 -0.00036 15.4445 -0.00267 -0.00036 -60 -10.3902 -945.203 -101.88 -12.5824 1059.666 1832.817 152.7348 1059.666

18 -0.00033 15.3 -0.00267 -0.00033 -60 -9.66667 -936.36 -101.88 -11.28 1049.52 1898.286 158.1905 1049.5216 0 13.6 -0.00263 0 -60 0 -832.32 -101.88 6.12 928.08 2587.104 215.592 928.0813 0.000692 11.05 -0.00254 0.000692 -60 20.07692 -676.26 -101.88 36.13846 742.0015 3316.133 276.3444 742.001512 0.001 10.2 -0.0025 0.001 -60 29 -624.24 -101.88 52.2 673.92 3513.096 292.758 673.9211 0.001364 9.35 -0.00245 0.001364 -60 39.54545 -572.22 -101.88 71.18182 602.9182 3686.284 307.1904 602.9182

9.4 0.002106 7.99 -0.00236 0.002106 -60 60 -488.988 -101.88 108 482.868 3916.545 326.3787 482.8689 0.002333 7.65 -0.00233 0.002333 -60 60 -468.18 -101.88 108 462.06 3891.992 324.3326 462.068 0.003 6.8 -0.00225 0.003 -60 60 -416.16 -101.88 108 410.04 3799.656 316.638 410.047 0.003857 5.95 -0.00214 0.003857 -60 60 -364.14 -101.88 108 358.02 3663.104 305.2586 358.026 0.005 5.1 -0.002 0.005 -58 60 -312.12 -98.28 108 302.4 3457.134 288.0945 302.45 0.0066 4.25 -0.0018 0.0066 -52.2 60 -260.1 -87.84 108 239.94 3159.068 263.2556 239.94

4.5 0.007667 3.825 -0.00167 0.007667 -48.3333 60 -234.09 -80.88 108 206.97 2981.273 248.4394 206.974 0.009 3.4 -0.0015 0.009 -43.5 60 -208.08 -72.18 108 172.26 2780.244 231.687 172.263 0.013 2.55 -0.001 0.013 -29 60 -156.06 -46.08 108 94.14 2284.124 190.3436 94.14

2.079159 0.020086 1.767285 -0.00011 0.020086 -3.31233 60 -108.158 0.157805 108 6.34E-05 1632.743 136.0619 6.34E-05

A simple EXCEL Sheet will plot the interaction diagram.“Goal Seek” finds important points like zeros and yield.

60

0

200

400

600

800

1000

1200

1400

1600

1800

0 100 200 300 400 500 600

Axia

l Loa

d k

Moment k-ft

#7

#8

#9

#10

#11

Increasing ρ

We are all familiar with this graph from textbooks; the effect of increasing the amount of steel.

61

0

500

1000

1500

2000

2500

0 100 200 300 400 500 600

Axia

l Loa

d k

Moment k-ft

4000 psi

5000 psi

6000 psi

7000 psi

Increasingfc'

Big Effect on Axial Load

Almost no effecton moment!

But you almost NEVER see the effect of fc’!

Why this works better• Students still learn theory, just in a different

order.• Students learn in a way more understandable to

them.• Students get to DO something, not just listen to

you.• Students FORM PATTERNS• Students EXTEND PATTERNS• Students ask “WHAT IF”.

62

When to Change Up

63

Sometimes covering theory first works better.

Shear in prestressed beams:

db 'f.V

Vdb 'f0.6)(MM

VV

wcci

dwccrmax

ici

71

64

db 'f.V

Vdb 'f0.6)(MM

VV

wcci

dwccrmax

ici

71

The elements of this equation are very confusing.

If you derive the equation, the meaning an application of the terms becomes clear.

EFFECTIVE TEACHING• Remember the students are in unfamiliar

territory.• Teach to their style of learning, not yours.

– Students tend to like “concrete” examples first, theory later.

• Get them ACTIVE! They learn more by doing.• Teach INTELLIGENCE not just KNOWLEDGE.

65