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Calculation of a beam’s moment capacity:

The triangular beam indicated is made of 4ksi concrete and carries three

#9 rebars of fy=60ksi.

Note that geometry offers itself for calculations regarding the Compression, the

lever-arm, and the depth of a.

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Calculation of a beam’s moment capacity:

The triangular beam indicated is made of f`c=4ksi concrete and

carries three (3) #9 rebars of fy=60ksi.

Note that geometry offers itself for calculations regarding the Compression,

the lever-arm, and the depth of α.

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We begin by the fundamental assumption that fs=fy (we verify this at

a later step) and we calculate the area of steel As:

3 #9 rebars give us a total area of 3 sq. in.

T=As(fy) = (3in^2)(60ksi) = 180 kips

fs=Calculated stress of reinforcement at service loads, fy=specified yield strength of non prestressed reinforcement.

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We compute the compression block so that we have C=T:

Given C=T = 180 kips

The specific geometry of the beam allowed us to generate

the formula for C. That can be reversed to solve for a:

…which gives us a value of 10.29 in.

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Verifying our assumption that fs=fy:

As mentioned earlier,

ε

cu will carry the value of 0.003

c=α/β1…thus c=10.29in/0.85…..= 12.11in

Using the similar triangles method as seen above…

thus

ε

s is 0.00394 >

ε

y which is f`c/fy that is 60ksi/29000ksi yielding a

value of 0.00207, proving that fs=fy

ε

s=strain in steel,

ε

y= yield strain.

fs=Calculated stress of reinforcement at service loads, fy=specified yield strength of non prestressed reinforcement.

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Computing the Mn:

We can derive from the geometry that jd=d-(2a/3)

Thus it would be safe to claim that

..which yields the result of 285.4 k`

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That wonderful property of structural materials to bend, crack and yet not

break, is one of the possible characteristics of RC.

When flexural forces surpass the limit My, steel reinforcement continues to

elongate. Resistance increases slightly, related to the increase of distance

between C and T. That distance increases as the depth of the concrete

stress block decreases until the concrete fractures. Although the stress of

the steel remains constant, the strain at the point of failure is several times

greater than the steel yield strain

ε

y

, ..approximately ε

y=fy/Es ≈.002

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On a section that fractures when the strain of steel is 0.006, if the As

was multiplied by a certain factor (let’s say doubled), then the Whitney

block would take a similar magnification (double in this case). The strain

at the tension could only be 0.003. The stretching of the steel at the

range between yield and beam failure would only be 0.001 instead of

0.004 as it was for the section with half as much tension reinforcement.

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There are three methods

of flexural failure of

concrete members:

Tension

Compression

Balanced

Source: http://www.shef.ac.uk/content/1/c6/04/71/91/fig32_3_concrete_crushing.jpg, Sept.20/09

Source: http://www.tfhrc.gov/structur/pubs/06115/images/fig29.jpg, Sept.20/09

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The section in the left carries four rebars to counterbalance the

compression force. The middle section indicates a T-beam with a

larger cross sectional area in compression, and it is set to

equilibrium through the application of more tensile rebars. Inversely,

the notched section on the right carries less tensile reinforcement.

Note that values of â1c are independent of the section’s shape.

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The capacity reduction

“φ” factor also reflects

the relative ductility of

the cross section at

failure expressed in

terms of the strain εt

as presented by ACI.

To determine the

value of εt the

proportional triangle

method can be applied

as seen in the

diagram.

εt=0.003[(d/c)-1]

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The capacity reduction

“φ” factor also reflects

the relative ductility of

the cross section at

failure.

For tension controlled,

ductile, flexural failure at

εt>0.005, (i.e 2.5 times

larger than εy for Grade

60), φ= 0.90

Failure at εt≤fy/Es is

compression controlled,

non-ductile, and φ=0.65

For further details and info, please refer to in class reading: pp 40-

41.

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ACI recognizes that the

magnitude of moments at

critical locations of a flexural

member, estimated through

elastic analysis, can not be

totally precise. Therefore,

designers are allowed to

“redistribute” moment values

(from support regions to span)

provided that:

At failure, the extreme fiber

tension strain εt exceeds

0.0075

The relative amount of

moment value that will be

redistributed is 1000(εt)

percent<20% of the moment

at the end of the beam.

Source: R.W. Furlong: Basic Decisions for Designing Reinforced Concrete Structures, Morgan Printing, Austin , TX, Sept. 2003

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ACI recognizes that the

magnitude of moments at

critical locations of a flexural

member, estimated through

elastic analysis, can not be

totally precise. Therefore,

designers are allowed to

“redistribute” moment values

(from support regions to span)

provided that:

At failure, the extreme fiber

tension strain εt exceeds

0.0075

The relative amount of

moment value that will be

redistributed is 1000(εt)

percent<20% of the moment

at the end of the beam.

Source: R.W. Furlong: Basic Decisions for Designing Reinforced Concrete Structures, Morgan Printing, Austin , TX, Sept. 2003

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Source: R.W. Furlong: Basic Decisions for Designing Reinforced Concrete Structures, Morgan Printing, Austin , TX, Sept. 2003

Equilibrium of forces at limit

load are maintained by insuring

that the avg of required end

moments plus the max +ve

moment is never reduced.

Midspan moments are

increased when end moments

are reduced, or else end

moments are increased if

midspan moment is reduced.

Practically, designing for

redistributed moments transfers

more reinforcement on midspan

and less at supports.

Moment redistributed should

not exceed 10(εt) multiplied by

the moment at support.

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Source: R.W. Furlong: Basic Decisions for Designing Reinforced Concrete Structures, Morgan Printing, Austin , TX, Sept. 2003

To facilitate the redistribution process

as stated earlier, a minimum

ε

t

value of 0.01 should be set.

Given an fy value of 60ksi, that

translates to a stretch

approximately 5 times that of the

strain experienced at fy.

When

ε

t is 0.01 (see lower

configuration in figure) the value

“c” can be deduced geometrically

to be 0.003d/(0.003+0.001)

which returns: c=0.231d

By substituting this value we can

deduce that

As(fy)=0.196(f’c)b(â1)d

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*Note: The min effective depth suggests that after As is determined, a cover per ACI guidelines should be determined.

To recap and bring to surface a few standardized

values:

A recommended ρ_Max will be applied when

the objective is to minimize the depth of a

beam. The εt will be 0.004 and φ will be 0.81.

As example you can use the following formula for minimum effective

depth.*

The ρ_Min is a threshold value that we shall

never cross.

The ρ_lim is the limit that keeps us still within an

εt of .005

The ρ_10 is an ideal condition that we should

always aim for in order to have a comfortable

condition for our designed element and where

the εt is 0.010

OR

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Select flexural bars for the section and the required moment and determine the ΦMn, the b1

ratio (ratio of depth of rectangular stress block, a, to the depth to neutral axis c) the distance c

and the strain on the extreme fiber of reinforcement εt.

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Assignments will be received at the beginning of class period.

Reading: Furlong Chapt. 4.1 & 4.2

Assignment 2 will be due one week

from today.

BEAM DEFLECTION FORMULAE

BEAM TYPE

SLOPE AT FREE END

DEFLECTION AT ANY SECTION IN TERMS OF x

MAXIMUM

DEFLECTION

1. Cantilever Beam – Concentrated load P at the free end

2

2

Pl

E

I

θ=

()

2

3

6

Px

yl

x

EI

=

−

3

max

3

Pl

E

I

δ=

2. Cantilever Beam – Concentrated load P at any point

2

2

Pa

E

I

θ=

()

2

3f

o

r

0

6

Px

ya

x

x

a

EI

=

−<

<

()

2

3f

o

r

6

Pa

yx

a

a

x

l

EI

=

−<

<

()

2

max

3

6

Pa

la

EI

δ

=−

3. Cantilever Beam – Uniform

ly distributed load ω (N/m)

3

6

l

E

I

ω

θ=

()

2

22

64

24x

yx

l

l

x

EI

ω

=+

−

4

max

8

l

E

I

ω

δ=

4. Cantilever Beam – Uniform

ly varying load: Maximum intensity ω

o

(N/m)

3

o

24

l

E

I

ω

θ=

()

2

32

2

3

o

10

10

5

120

x

yl

l

x

l

x

x

lEI

ω

=−

+

−

4

o

max

30

l

E

I

ω

δ=

5. Cantilever Beam – Couple moment M at the free end

M

l

E

I

θ=

2

2

M

x

y

E

I

=

2

max

2

M

l

E

I

δ=

BEAM DEFLECTION FORMULAS

BEAM TYPE

SLOPE AT ENDS

DEFLECTION AT ANY SECTION IN TERMS OF x

MAX

IMUM

AND CENTER

DEFLECTION

6. Beam Simply Supported at Ends – Concentrated load P at the center

2

12

16

Pl

E

I

θ=

θ

=

2

2

3

for

0

12

4

2

Px

l

l

yx

x

EI

⎛⎞

=

−<

<

⎜⎟

⎝⎠

3

max

48

Pl

E

I

δ=

7. Beam Simply Supported at Ends – Concentrated load P at any point

22

1

()

6

Pb

l

b

lEI

−

θ=

2

(2

)

6

Pab

l

b

lEI

−

θ=

()

222

for

0

6

Pbx

yl

x

b

x

a

lEI

=

−−

<

<

()

()

3

22

3

6

for

Pb

l

yx

a

l

b

x

x

lEI

b

ax

l

⎡

⎤

=−

+

−

−

⎢

⎥

⎣

⎦

<

<

(

)

32

22

max

93

Pb

l

b

lEI

−

δ=

at

()

22

3

xl

b

=−

()

2

2

at the center, if

34

48

Pb

lb

EI

δ=

−

ab

>

8. Beam Simply Supported at Ends – Uniformly distributed load ω (N/m)

3

12

24l

E

I

ω

θ=

θ

=

()

32

3

2

24

x

yl

l

x

x

EI

ω

=−

+

4

max

5

384l

E

I

ω

δ=

9. Beam Simply Supported at Ends – Couple momen

t M at the right end

1

6

M

l

E

I

θ=

2

3

M

l

E

I

θ=

2

2

1

6

M

lx

x

y

E

Il

⎛⎞

=−

⎜⎟

⎝⎠

2

max

93

M

l

EI

δ=

at

3

l

x

=

2

16

M

l

E

I

δ=

at the center

10. Beam Simply Supporte

d at Ends –

Uniformly varying load: Maxim

um intensity ω

o

(N/m)

3

o

1

7

360

l

E

I

ω

θ=

3

o

2

45

l

E

I

ω

θ=

()

42

2

4

o

71

0

3

360

x

yl

l

x

x

lEI

ω

=−

+

4

o

max

0.00652

l

E

I

ω

δ=

at

0.519

x

l

=

4

o

0.0065

1

l

E

I

ω

δ=

at the center

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