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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.


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.


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.


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.


Verifying our assumption that fs=fy: As mentioned earlier, εcu 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 εs is 0.00394 > εy which is f`c/fy that is 60ksi/29000ksi yielding a value of 0.00207, proving that fs=fy

εs=strain in steel, εy= yield strain.

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


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 εy, ..approximately εy=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.20/09Source: 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.

jçãÉåí=oÉÇáëíêáÄìíáçå 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

jçãÉåí=oÉÇáëíêáÄìíáçå 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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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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To facilitate the redistribution process as stated earlier, a minimum εt 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 εt 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 DEFLECTION1. Cantilever Beam – Concentrated load P at the free end

2

2PlEI

θ = ( )2

36Pxy l xEI

= − 3

max 3PlEI

δ =

2. Cantilever Beam – Concentrated load P at any point

2

2Pa

EIθ =

( )2

3 for 06Pxy a x x aEI

= − < <

( )2

3 for6Pay x a a x l

EI= − < <

( )2

max 36Pa l a

EIδ = −

3. Cantilever Beam – Uniformly distributed load ω (N/m)

3

6lEIω

θ = ( )2

2 26 424

xy x l lxEI

ω= + −

4

max 8lEIω

δ =

4. Cantilever Beam – Uniformly varying load: Maximum intensity ωo (N/m)

3o

24lEI

ωθ = ( )

23 2 2 3o 10 10 5

120xy l l x lx xlEI

ω= − + −

4o

max 30lEI

ωδ =

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

MlEI

θ = 2

2Mxy

EI=

2

max 2Ml

EIδ =

BEAM DEFLECTION FORMULAS

BEAM TYPE SLOPE AT ENDS DEFLECTION AT ANY SECTION IN TERMS OF x MAXIMUM AND CENTER DEFLECTION

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

2

1 2 16Pl

EIθ = θ =

223 for 0

12 4 2Px l ly x xEI

⎛ ⎞= − < <⎜ ⎟

⎝ ⎠

3

max 48Pl

EIδ =

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

2 2

1( )6

Pb l blEI−

θ =

2(2 )

6Pab l b

lEI−

θ =

( )2 2 2 for 06Pbxy l x b x alEI

= − − < <

( ) ( )3 2 2 3

6for

Pb ly x a l b x xlEI b

a x l

⎡ ⎤= − + − −⎢ ⎥⎣ ⎦< <

( )3 22 2

max 9 3

Pb l b

lEI

−δ = at ( )2 2 3x l b= −

( )2 2 at the center, if 3 448

Pb l bEI

δ = − a b>

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

3

1 2 24lEI

ωθ = θ = ( )3 2 32

24xy l lx xEI

ω= − +

4

max5

384lEI

ωδ =

9. Beam Simply Supported at Ends – Couple moment M at the right end

1 6MlEI

θ =

2 3MlEI

θ =

2

216Mlx xy

EI l⎛ ⎞

= −⎜ ⎟⎝ ⎠

2

max 9 3Ml

EIδ = at

3lx =

2

16Ml

EIδ = at the center

10. Beam Simply Supported at Ends – Uniformly varying load: Maximum intensity ωo (N/m)

3o

17360

lEI

ωθ =

3o

2 45lEI

ωθ =

( )4 2 2 4o 7 10 3360

xy l l x xlEI

ω= − +

4o

max 0.00652 lEIω

δ = at 0.519x l=

4o0.00651 l

EIω

δ = at the center

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