reinforced concrete analysis and flexural design – factors ... · 0.016in if the concrete is...

Reinforced Concrete Analysisand Flexural Design – Factors for

Safety and Strain Comfort

MORGAN STATE UNIVERSITY

SCHOOL OF ARCHITECTURE AND PLANNING

LECTURE III

Dr. Jason E. Charalambides

Copyright J. CharalambidesCopyright J. Charalambides

– As we design RC, it will prove convenientto use the following:

! A ratio of the volume of steel within thevolume of concrete: Of course, we caneliminate the dimension of length andcalculate the cross sectional area of thesteel and of the concrete, and get the sameresult. That will be our first targetedparameter.

! A section shape coefficient, which willinvolve the ultimate moment and the beam'sdimensions:

Density And SelectionCoefficient

ρ=Αsbw⋅d

Note: The above formulae do not incorporate safety factors that shall apply at a later stage in design exercises.

k=Mu

bw⋅d2


– Given the target value of 0.010 for thestrain in tensile steel (εt), then the value“c” can be deduced geometrically to be:

! From there on we can continue with the

following:

and by substituting the nominal C value innominal M...

Flexural Design, Factors:

c= 0.003⋅d0.003+0.010

=0.231⋅d

Try a simple in class example of a concrete beam of bw=12” and h=22” at f'c=4ksi, fy=60ksi.

Cn=0.85⋅ f ' c⋅bw⋅β1⋅c=T=As⋅f y

Mn=Cn [d−β1⋅c2 ]=[0.195⋅ f ' c⋅β1][1−0.115⋅β1]⋅bw⋅d

2

therefore a= β1⋅c≈20%d


– Let's name the “ρ” (rho) value and “k” for this specific scenario ofεt=0.010 as “ρ10” and “k10.” We can generate a standard formula foreach of these within which we can also incorporate the “φ” factor ofsafety at the value of 0.9:

! You can always refer to the following table for a set of standardized values of

the above, based on selected steel reinforcement strengths f'c and fy respectively.

Density And SelectionCoefficient

ρ10=0.196⋅ f ' c⋅β 1

f y

Try the previous example again but estimate the As based on the ρ10

k 10=0.177⋅f ' c⋅β 1⋅(1−0.115⋅β1)


Conditions of “ñρ”


Analyzing Simplified Formulae

We presented the simplified formula for initialestimation of the steel area:However, what is the factor that is missinghere? We use this formula for initialestimation based on the fact that the mostcommon steel strength is fy=60ksi and theassumption that steel will be in tension. Thusthe φ=0.9 is incorporated.But what if our steel was not 60ksi but asignificantly stronger 75ksi? Shall we need asmuch steel?...Likely not! In fact given therelation of 60/75 we can approximate thesame equation to ….So what if we have grade 40 steel?

As60Mu4d

As75Mu5d


Analyzing Simplified Formulae

Just for the sake of understanding the formulas and nottaking them for “face value” let’s just see the process ofhow they derive:

– We all recognize the “force” multiplied by “moment-arm”function giving us the moment:

– Can we substitute values from the Cn formula in the oneabove?

– The above allows us to solve for a/2

Mn=As⋅ f y⋅[d−a2 ]

Cn=0.85⋅ f ' c⋅bw⋅a thus a=1.1765⋅Cnf ' c⋅bw

a2=

As⋅f y1.701( f ' c⋅bw)


Analyzing Simplified Formulas

If we approximate the quantity (d-0.5*a) to 0.9* d andconsider the value of “φ” to be 0.9 for tensile ductileflexure, we can derive the formula….

– ...which can be simplified to ….. or if we apply grade60 steel

As=12( inch

ft)⋅Mu(k⋅ft )

φ⋅f y( kinch2 )⋅0.9⋅d (inch)

=12⋅Mu

0.81⋅ f y⋅d

As=15⋅Mu

f y⋅dAs≈

Mu

4d


A Basic Formula forRectangular Beam Dimensioning

If we approximate that a good height to base width ratiowould be roughly between 1.95 and 2.15 we can derive thefollowing relation:

...which can followed by

Those numbers can be rounded and adjusted

bw≈d

2.1

drecomm= 3√ −2.326Mu

ρ⋅ f y [ 0.59 ( ρ⋅ f y )f ' c

−1]


In Class Example cont


T-Beams in Flexure

In most cases, slabs and beams are cast together, integrating the action of thebeam to that of the slab. The compressive forces that are generated due toflexure extend from the width of the beam into the slab, and thus we have aform that reasonably is named T-section. It is reasonable to anticipate that the Whitney block will most often be confinedwithin that upper area. Nevertheless, it is necessary to point out that there is alimit to how far this upper area extends horizontally away from the edge of thebeam. Please see diagrams of limits of effective width deriving from the ACIcode:


T-Beams in Flexure

Since the effective width of the Whitney block can be very large, thedepth a=β1*c shall be rather shallow, and the value of εt very large.Therefore, ductility will likely not be of major concern in +ve momentregions for T-beams.

Please note that for defining the steel density “ρ” and thereinforcement limits, we apply the base width bw of the stem, not ofthe flange.


T-Beams in Flexure

Longitudinal bars in compression can improve ductility by reducingthe area of concrete required to equilibrate As*fy. Whenever:

OR

Use in compression longitudinal bars with an area A`s: A`s>As-As10

ΦM n

bw⋅d2<k 10 As>As10 As10= ρ10⋅bw⋅d


T-Beams in Flexure

As we see in the diagram, -ve moment areas are limited to the stem width bw.The depth of “a” in such cases can be large enough to limit the value of εt to notallow moment redistribution or even to force a value of “φ” to be less than 0.90.Reinforcement can be applied in the compression zone to equilibrate therequired tension forces much more effectively than concrete.

Bar placement requirements insure that longitudinal reinforcement at the bottomof beams at joints will always be available.


Determining The “hs”

There is a certainminimum thickness aconcrete slab can haveaccording to the ACI 318code and that is given inthe following tableextruded from theaforementioned:

Source: American Concrete Institute code 318

Table 7.3.1.1


Determining The “hs”

One more table allowsdesigners to determine theminimum thickness of a slabthat has no interior beams.

This is only provided in thiscase for information although itis not applicable to the subjectof study of T-Beams.

Source: American Concrete Institute code 318

Table 8.3.1.1


Longitudinal Reinforcement inCompression

If the compression steel yields before concrete reaches a compression strainof 0.003, the expressions below each diagram provide correct values for thecompression reinforced section. The strain at the level of compression steelmust exceed fy/Es – about .002 for Grade 60 – if the compression steel yields.

}


Longitudinal Reinforcement inCompression

Generally a depth above 28” will allow comp. steel to yield at failure. If the steeldoes not yield when concrete reaches 0.003, proportional triangles can producethe relationship between steel strain and the surface failure strain of 0.003. Thatstrain value multiplied by Es and As’ gives the force in compression steel. Thatforce together with C=.85f`c*bw*a must equal the T=As*fy


In Class Example: Analysis ofa Beam to Flexural Stress (f's)


In Class Example:


Serviceability – Crack Control

! Where fs is stress in deformed reinforcement, can be calculated as 2/3*fy.! For fy=60ksi and Cc=2.0”, then smax=10” for beams! Cc=.75”, then smax=12” for slabs! Allowing 2.5” from corner bars to edge of section, this clause suggests

minimum number of bars nmin=(5+bw)/10

smax≤15( 40,000f s )−2.5⋅Cc<12inch

! Cracks that form before the tensional rebars action takes place, are notnoticeable to the untrained eye of a passer-by. Their width would be about0.016in if the concrete is joined together by bars that are not particularly large.

! ACI318-14 ¶10.6.4 requires that bar spacing “s” be limited to the lesser of thebelow indicated formulae:

smax≤12( 40,000f s )<12inch


Rebar Maximum Arrangementin Single Layer Per ACI Code

! Maximum rebar, or minimum rebar spacing should fulfill the following criteria:

• Rebars should not be closer than an inch from one another (edge to edge)

• If the largest rebar used is larger than 1” in diameter, the minimum distancebetween rebars should exceed that distance

• The largest size aggregates should not exceed ¾ of that distance asdetermined in the above parameters.


Table for Rebar MaximumArrangement in Single Layer

Source: Goswami, I: All in One, Civil Engineering PE Breath & Depth Exam Guide


Serviceability - Deflection

! The stiffness of a beam serves as a deflection control factor. That flexuralstiffness is primarily a function of the member’s thickness, and we can referto ACI318 Table 9.5 (a) (Member thickness for deflection control), which is atabulation of thickness requirements for members with various conditions ofcontinuity at the ends.


Serviceability - Deflection

! If the thickness of a member does not exceed the limits as given in table 9.5(a) it is necessary to show computationally, that deflections under appliedloads are within the limits given in table 9.5 (b) (Deflection limits for beamsand slabs).


Serviceability – Computing Deflection

1. Use unfactored service loads with LL located at points where itwill cause maximum deflection.

2. Sufficient accuracy can be obtained from downward deflection,using formula for simply supported span, minus upwarddeflection from the average of end moments

3. Formula:

4. Deflection from loading indefinitely can be taken as twice theinitial deflection generated by a sustained loading. Deflectiondue to load of up to 3 months sustained is to be taken as equalto initial deflection.

5. The value of Ec= 57000*√f`c with psi units for E and for f`c.

Δ=[ 5⋅w⋅l2

384−Mavg

8 ]⋅l2E⋅I

RC Beam with compressive reinforcement Select grade 75 rebars required for a section 20” wide and 26” deep (including cover)to resist service load moments Md=180k` and Ml=275k`. Use f'c=5ksi. After barsare selected calculate FMn to show that the section and the bars are adequate:

fy 75ksi:= f'c 5.ksi:= bw 20in:= Md 170.85k':= Ml 275k':= Φ .9:=

h 26in≡ cover 2.5in≡ d' cover≡ d h cover−:= Es 29000ksi:=

Mu 1.2 Md⋅ 1.6 Ml⋅+:= Mu 645.02 k'⋅= d 23.5 in⋅= β1 0.8=

AS_estimated15 Mu⋅

12d fy:= AS_estimated 5.49 in2

⋅=

Bar Designation Number

Weight per foot (lbf) Diameter db Area As Perimeter

3 0.376 0.375 0.11 1.1784 0.668 0.500 0.20 1.5715 1.043 0.625 0.31 1.9636 1.502 0.750 0.44 2.3567 2.044 0.875 0.60 2.7498 2.670 1.000 0.79 3.1429 3.400 1.128 1.00 3.54410 4.318 1.270 1.27 3.99011 5.304 1.410 1.56 4.43014 7.650 1.693 2.25 5.31918 13.600 2.257 4.00 7.091

We can take two #8s and four # 9s for tensile reinforcement

Abar1 .79in2≡ Abar2 1in2

≡

nbar1 2:= nbar2 4:=

bar1dia 1.0in:= bar2dia 1.128in:=

Verifying that spacing is good: We calculate how much spacing remains after subtracting the side covers and the sum ofdiameters of all selected rebars. Then we divide the remaining space by the number of spaces that will remain betweenrebars. Spacing needss to be equal or larger than 1" or largest bar diameter:

Assuming that the smallest bars are at the ends...

spacingbw nbar1 bar1dia⋅ nbar2 bar2dia⋅+( )− 2 cover⋅ min bar1dia bar2dia, ( )−( )−

nbar1 nbar2+ 1−( ):=

Min_Spacing_Condition if spacing max 1in bar1dia, bar2dia, ( )< "redesign", "Ok", ( ):=

spacing 1.898 in⋅= Min_Spacing_Condition "Ok"=

Calculating Area of Steel:As Abar1 nbar1⋅( ) Abar2 nbar2⋅( )+:= As 5.58 in2

⋅=

T As fy⋅:= T 418.5 kip⋅=

aT

0.85 f'c⋅ bw⋅:= a 4.924 in⋅=

ρAs

d bw⋅( ):= ρ 0.01187= ρmin

200psify

:= ρmin 0.0027=

caβ1

:= c 6.15441 in⋅=ad

0.20951=

εt .003dc

1−

⋅:= εt 0.00846= εtyfyEs

:= εty 0.00259=

Φ 0.9 .65 0.25εt εty−( )

0.005 εty−( )⋅+≥ 0.65≥= Φ 0.9=

ΦMn 0.9T da2

−

⋅:= ΦMn 660.338 k'⋅=

Strength if ΦMn Mu≥ "OK", "Not acceptable", ( ):= Strength "OK"=

Considering that up to +5%strength can be acceptableEfficiency if ΦMn Mu 1.05⋅≥ "Overdesigned", "Not Overdesigned", ( ):=

Efficiency "Not Overdesigned"=

ρ100.196 f'c⋅ β1⋅( )

fy:= ρ10 0.01045=

Examining the "k" section shape coefficient with respect to the "k10"

k10 .177 f'c⋅ β1⋅ 1 .115 β1⋅−( )⋅:= k10 642.864 psi⋅=

k Φ.85 f'c⋅ β1⋅ c⋅ d⋅( )

bw d2⋅

⋅ dβ1 c⋅

2−

⋅:= k 842.9842 psi⋅=k

k101.311=

Let's assume that the flexural stress in the compressed steel (f's) is equal to the yield strength (fy):

A's As ρ10 bw⋅ d⋅−:= A's 0.66693 in2⋅=

Assume two #7 rebars (Need at least two bars and we like to keep size of not more than 2 # sizedifference not for addressing maximum stresses but more for assuring that we maintain ease of onsite process. Also the 2 bars of that size are to be extended from midspan into the support asrequired to satisfy Code Clause 18.3.2):

Abar3 .60in2≡ nbar3 2:= bar3dia .875in:=

A's nbar3 Abar3⋅:= A's 1.2 in2⋅=

T' A's fy⋅:= T' 90 kip⋅=

aT T'−

0.85 f'c⋅ bw⋅:= a 3.865 in⋅=

caβ1

:= c 4.83088 in⋅=

εt .003dc

1−

⋅:= εt 0.01159=

Φ if εt .005> .9, "calculate", ( ):= Φ 0.9=

εs .003c d'−

c

⋅:= εs 0.00145=

f's εs Es⋅:= f's 41.98 ksi⋅=

ΦMn Φ T T'−( ) da2

−

⋅ T'( ) d cover−( )⋅[ ]+

⋅:= ΦMn 8077.475 k''⋅= Or ΦMn 673.123 k'⋅=


k10 .177 f'c⋅ β1⋅ 1 .115 β1⋅−( )⋅:= k10 642.864 psi⋅=

k Φ.85 f'c⋅ β1⋅ c⋅ d⋅( )

bw d2⋅

⋅ dβ1 c⋅

2−

⋅:= k 678.3484 psi⋅=k

k101.055=

Let's assume that the flexural stress in the compressed steel (f's) is unequal to the yield strength (fy):

Equilibrium requires that Concrete compression and Steel Compression will equal to Steel Tension.Thus:

0.85 f'c⋅ b⋅ β1⋅ c⋅( ) A's f's⋅( )+ As fy⋅=c 5.33814 in⋅=

εs .003c d'−

c

⋅:= εt 0.01159=

f's εs Es⋅:= f's 46.26 ksi⋅=

a c β1⋅:= a 4.271 in⋅=

εt .003dc

1−

⋅:= εt 0.01021=

Φ if εt .005> .9, "calculate", ( ):= Φ 0.9=

ΦMn Φ 0.85 f'c⋅ bw⋅ a⋅( ) da2

−

⋅ A's f's⋅( ) d d'−( )+

⋅:= ΦMn 8028.81 k''⋅= Or ΦMn 669.067 k'⋅=


k10 .177 f'c⋅ β1⋅ 1 .115 β1⋅−( )⋅:= k10 642.864 psi⋅=

k Φ.85 f'c⋅ β1⋅ c⋅ d⋅( )

bw d2⋅

⋅ dβ1 c⋅

2−

⋅:= k 742.5251 psi⋅=k

k101.155=

reinforced concrete analysis and flexural design – factors ... · 0.016in if the concrete is...

Documents