Analysis and Design of Reinforced Concrete Structure-I
CEN 323
Ultimate Strength Design (USD) of Beam:
Mu ϕMn
Pu ϕPn
Vu ϕVn
The nominal strength of a proposed member is calculated based on the best current knowledge of member and material behavior. That nominal strength is modified by a strength reduction factor ϕ, less than unity, to obtain the design strength.
The required strength, should the hypothetical overload stage actually be realized, is found by applying load factors γ, greater than unity.
Flexural Design
ϕMn Mu
The distribution of concrete compressive stresses at or near ultimate load have no well-defined shape –parabolic, trapezoidal, or other shape. For this and other reasons, wholly rational flexural theory for reinforced concrete has not yet been developed. Present methods of analysis, therefore, are based in part on known laws of mechanics and are supplemented, where needed, by extensive test information.
It has been mentioned before that an exact criterion for concrete compression failure is not yet known, but that for rectangular beams, strains of 0.003 to 0.004 have been measured immediately preceding failure. If one assumes, usually slightly conservatively, that the concrete is about to crush when the maximum strain reaches ε = 0.003.
Figure 1: Stress distribution at ultimate load
Figure 1: Stress distribution at ultimate load
(3)(2)(1)Let
(3)
(2)
(1)
α=favfc'
Then
C=αfc'bc
For a given distance c to the neutral axis, the location of C can be defined as some fraction β of the distance. Thus, as indicated in Fig. 1 for a concrete of given strength it is necessary to know only α and β to completely define the effect of the concrete compressive stresses.
Extensive direct measurements, as well as indirect evaluations of numerous beam tests, have shown that the following values for α and β are satisfactorily accurate.
Figure 2: Variation of α and β with concrete strength
Figure 2: Variation of α and β with concrete strength
If this experimental information is accepted, the maximum moment can be calculated from the laws of equilibrium and from the assumption that plane cross sections remain plane. Equilibrium requires that
C=T or αfc'bc=Asfs
Also, the bending moment, being the couple of the forces C and T, can be written as either
(4)
(4)
(5)M=Tz=Asfsd-βc
(5)
M=Cz=αfc'bcd-βc
(7)(6)For failure initiated by yielding of the tension steel, fs=fy. Substituting this value in Eq. (3), one obtain the distance to the neutral axis
(7)
(6)
c=Asfyαfc'b
Alternatively, using As=ρbd, the neutral axis distance is
c=ρfydαfc'
(9)(8)giving the distance to the neutral axis when tension failure occurs. The nominal moment Mnis then obtained from Eq. (4), with the value for c just determined, and fs=fy; that is
(9)
(8)
Mn= ρfybd21-βfyραfc'
With the specific, experimentally obtained values for α and β given previously, this becomes
Mn= ρfybd21-0.59fyρfc'
The above equation is valid only for a steel ratio less than the balance steel ratio, ρb
(10)ρb= αfc'fy ϵuϵu+ϵy
(10)
Equivalent Rectangular Stress Distribution
A equivalent rectangular stress distribution of concrete stress was proposed by C. S. Whitney and was subsequently elaborated and checked experimentally by others and accepted in many countries are presented below
Figure 3: Actual and equivalent stress distribution at ultimate load
Figure 3: Actual and equivalent stress distribution at ultimate load
Where C= αfc'bc=γfc'ab from which γ=αca
fc' psi
4000
5000
6000
7000
8000
α
0.72
0.68
0.64
0.60
0.56
β
0.425
0.400
0.375
0.350
0.325
β1=2β
0.85
0.80
0.75
0.70
0.65
γ=αβ1
0.85
0.85
0.85
0.86
0.86
Figure 4: Concrete stress block parameter
Figure 4: Concrete stress block parameter
(11)The concrete compression force at failure in a rectangular beam of with b is
(11)
C=0.85 fc' ab
From of Eq. (10),
(12)ρb= 0.85β1fc'fy ϵuϵu+ϵy
(12)
Overerreinforced Beams
A compression failure in flexure, should it occur, gives little if any warning of distress.
Why Underreinforced Beams desirable
In Underreinforced beam failure initiated by yielding of the steel, typically is gradual.
Distress is obvious from observing the large deflections and widening of concrete cracks associated with yielding of the steel reinforcement, and measures can be taken to avoid total collapse.
Strain-hardening of the reinforcing steel is not considered, which provide substantial strength to the beam.
That is why to design a beam, which fail by yielding of steel, reinforcement ration, ρ is chose less the balance reinforcement ration ρb.
ACI provision for Underreinforced beam
To ensure underreinforced behavior, ACI Code 10.3.4 establishes a minimum net tensile strain ϵt at the nominal strength of 0.005 for members subjected to axial loads less than 0.10fc'Ag . By way of comparison ϵy, the steel strain at the balanced condition, is 0.00207 for fy=60,000 psi.
Using ϵt= ϵy=0.005 in Eq. (12) provides the maximum reinforcement ratio allowed by ACI code for beam
ρmax= 0.85β1fc'fy ϵuϵu+0.005
The value of reduction factor depend upon the net tensile strain of steel of which can be determined by following figure.
Figure 5: Net tensile strain
Figure 5: Net tensile strain
Minimum Steel Ratio
Minimum steel ratio according to ACI 10.5 is ρmin=200fy for flexural member.
Example 1
A rectangular beam has the dimensions b=10 in., h=25 in., and d=23 in., and is reinforced with three No. 8 (No. 25) bars. The concrete cylinder strength fc'=4000 psi. and yield point of steel is fy=60,000 psi. Compute the nominal and design strength of the section.
Solution:
For this beam reinforcement ratio, ρ=Asbd=3×0.7910×23=0.0103
Max reinforcement ratio, ρ0.005= 0.85β1fc'fy 0.0030.003+0.005=0.85×0.85×460×0.0030.008=0.0181
As ρ<ρ0.005 the beam is Under reinforced beam
Nominal Moment, Mn= ρfybd21-0.59fyρfc'=0.0103×60,000×10×2321-0.590.0103×604
=2,970,000 in-lb=248 ft-kips
Design Moment, ϕMn=0.9×248=223.2 ft-kips
Example 2
A rectangular beam must carry a distributed live load of 680 plf and support the dead load of a wall weighing 380 plf, in addition to its own weight, on a simple span of 24 ft. Design the beam for flexure, using fy=60ksi and fc'=4 ksi.
Solution 1:
The total load acting on the beam includes its own weight, which must be estimated. As a trial a total depth equal to 1 in. per ft of span and a width of one-half that amount are assumed; the beam weight is estimated as (12×24144)150=300 plf,
Dead load=self weight+wall dead load=300+380=680 plf
d=21 in.d=21 in.b=12 in.h=24 in.Figure: Assumed sectionLive load=680 plf
d=21 in.
d=21 in.
b=12 in.
h=24 in.
Figure: Assumed section
Factored load =1.2×DL+1.6×LL
=1.2×680+1.6×680
=1904 plf
Factored Moment, Mu=18ωl2
=18×1904×242 ft-lb
=137088 ft-lb
=1645056 in-lb
Now we have, ϕMn=ϕρfybd21-0.59fyρfc'
1645056=0.9×ρ×60,000×12×2121-0.59ρ×604
1645056=285.768×106ρ1-8.85ρ
1645056=285.768×106ρ-2.529×109ρ2
2.529×109ρ2-285.768×106ρ+1645056=0
by solving this two degree equation, ρ1=0.1069, ρ2=0.00608
assume, ρ=0.00608
Max reinforcement ratio, ρ0.005= 0.85β1fc'fy 0.0030.003+0.005=0.85×0.85×460×0.0030.008=0.0181
Minimum steel ratio according to ACI 10.5 is ρmin=200fy=0.003
ρmin<ρ<ρ0.005
with this reinforce ration the beam would be underreinforced one.
As=ρbd=0.00608×12×21=1.53 in2
This could be provided by two No. 6 (No.19) and two No.5 (No.16) bars in a row which provide a steel area 1.50 in2 , which is sufficiently close to the required steel.
d=21 in.b=12 in.h=24 in.Figure: Design section2#6 & 2#5(1.50 in2)
d=21 in.
b=12 in.
h=24 in.
Figure: Design section
2#6 & 2#5
(1.50 in2)
Solution 2:
The total load acting on the beam includes its own weight, which must be estimated. As a trial a total depth equal to 1 in. per ft of span and a width of one-half that amount are assumed; the beam weight is estimated as (12×24144)150=300 plf,
Dead load=self weight+wall dead load=300+380=680 plf
Live load=680 plf
Factored load =1.2×DL+1.6×LL
=1.2×680+1.6×680
=1904 plf
Factored Moment, Mu=18ωl2
=18×1904×242 ft-lb
=137088 ft-lb
=1645056 in-lb
Max reinforcement ratio, ρ0.005= 0.85β1fc'fy 0.0030.003+0.005=0.85×0.85×460×0.0030.008=0.0181
Taking ρ=0.0181
Now we have, ϕMn=ϕρfybd21-0.59fyρfc'
1645056=0.9×0.0181×60,000×bd21-0.590.0181×604
820.84bd2=1645056 in3
bd2=2004.11in3
Assume b=10 in.
Then 10d2=2004.11in3
d2=200.41in2
d=200.41 in.
d=14.15 in.
d 14.5 in.
Assume a total depth,h=d+3.5=14.5+3.5=18 in.
As=ρbd=0.0181×10×14.5=2.62 in2
This could be provided by six No. 6 (No.19) bars in two rows which provide a steel area 2.64 in2 , which is slightly greater to the required steel.
d=14.5 in.b=10 in.h=18 in.Figure: Design section6#6 bars(2.64 in2)
d=14.5 in.
b=10 in.
h=18 in.
Figure: Design section
6#6 bars
(2.64 in2)
An improved economy may be possible, refining the steel area based on the actual larger effective depth and smaller concrete section.
Now self weight can be estimated =(10×18144)150=187.5 190 plf
Dead load=self weight+wall dead load=190+380=570 plf
Live load=680 plf
Factored load =1.2×DL+1.6×LL
=1.2×570+1.6×680
=1772 plf
Factored Moment, Mu=18ωl2
=18×1772×242 ft-lb
=127584 ft-lb
=1531008 in-lb
Now we have, ϕMn=ϕρfybd21-0.59fyρfc'
1531008=0.9×ρ×60,000×10×14.521-0.59ρ×604
1531008=113.535×106ρ1-8.85ρ
1531008=113.535×106ρ-1.004×109ρ2
1.004×109ρ2-113.535×106ρ+1531008=0
solving this two degree equation ρ1=0.09743, ρ2=0.015650
assume, ρ=0.015650
As=ρbd=0.0156×10×14.5=2.26 in2
This could be provided by three No. 6 (No.19) and three No.5 (No.16) bars in two row which provide a steel area 2.25 in2, which is sufficiently close to the required steel.
d=14.5 in.b=10 in.h=18 in.Figure: Alternate Design section3#6 bars & 3#5(2.25 in2)
d=14.5 in.
b=10 in.
h=18 in.
Figure: Alternate Design section
3#6 bars & 3#5
(2.25 in2)
Overreinforced Beams
According to the ACI Code, all beams are to be designed for designed for yielding of the tension steel with εt not less than 0.004 and, thus, ρ ρmax. Occasionally, however, such as when analyzing the capacity of existing construction, it may be necessary to calculate the flexural strength of an overreinforced compression controlled member, for which fs is less than fy at flexural failure.
In this case, the steel strain, ϵs=ϵud-cc
From the equilibrium requirement that C=T, one can write
0.85β1fc'bc=ρϵsEsbd
Substituting the steel strain from in the last equation, and defining ku=cd, one obtains a quadratic equation in ku as follows:
ku2+mρku-mρ=0
Here, ρ=Asbd as usual and m is a material parameter given by
m=Esϵu0.85β1fc'
Solving the quadratic equation for ku,
ku=mρ+mρ22-mρ2
Neutral axis depth for the overreinforced beam, c=kud, after the stress-block depth a=β1c. With steel stress fs=Esϵs the nominal flexural strength is
Mn=Asfsd-a2
The strength reduction factor ϕ will be equal 0.65 for beams in this range
Practical Consideration in the design of beams
To focus attention initially on the basic aspect of flexural design, the preceding examples were carried out with only minimum regard for certain practical considerations that always influence the actual design of beams. These relate to optimal concrete protection, rounding of dimension, standardization of dimension and selection of bar combination.
Concrete protection for reinforcement
ACI Code 7.7
Not exposed directly to ground or weather
Exposed to weather or in contact with ground
Concrete is cast in direct contact with the ground
Slab
and
wall
34 in.
2 in.
(112 in. for No. 5 and smaller bar)
3 in.
Beam
and
column
112 in.
2 in.
(for No. 5 and smaller bar)grounddentof bar combinationnsion,mum ning nsion steel (112 in. for No. 5 and smaller bar)
3 in.
In general, the center of main flexural bars in beams should be placed 212 to 3 in. from the top or bottom surface to furnish at least 112 in. of clear cover
In slabs, 1 in. to the center of the bar is ordinarily sufficient to give the required
34 in. cover.
To simplify construction and there by to reduce costs, the over all dimension of beams, b and h, are almost always rounded to the nearest inch, and often to the next multiple of 2 in.
For slab, the total depth h, is generally rounded to the nearest 12 in. up to 6 in. and to the nearest in above that thickness.
Selection of Bars and Bar Spacing
No. 14 and No. 18 bars are used mainly in column
Should be arranged symmetrically about the vertical centerline
Many designers limit the variation in diameter of bars in a single layer to two bar sizes
According to ACI Code 7.6, the minimum clear distance between adjacent bars in beam shall not be less than the nominal diameter of the bars, or 1 in.
When reinforcement are placed two or more layers, the clear distance between layers must not be less than 1 in., and the bar in the upper layer should be placed directly above those in the bottom layer.
Rectangular Beams with Tension and Compression Reinforcement
If a beam cross section is limited because of architectural or other consideration, it may happen that the concrete cannot develop the compression force required to resist the given bending moment. In that case, reinforcement is added in the compression zone, resulting in a so-called doubly reinforced beam.
Tension and Compression Steel Both at Yield Stress
Mn1=As'fyd-d'
Mn2=(As-As')fyd-a2 or
With the definition ρ=Asbd and ρ'=As'bd
Mn2=(ρ-ρ')fybd21-0.59fy(ρ-ρ')fc'
The total nominal resisting moment is then
Mn=Mn1+Mn2=As'fyd-d'+(As-As')fyd-a2
Or
Mn=Mn1+Mn2=As'fyd-d'+(ρ-ρ')fybd21-0.59fy(ρ-ρ')fc'
In accordance with the safety provision of the ACI Code, the net tensile strain is checked, and if ϵt 0.005, this nominal capacity is reduced by the factor ϕ=0.9 to obtain the design strength. For ϵt between 0.005 to 0.004, ϕ must be adjusted.
In doubly reinforced beam, maximum reinforcement ratio should be limited by
ρmax=ρmax+ρ'
Compression Steel below Yield Stress
Minimum tensile reinforcement ratio ρcy that will ensure yielding of the compression steel at failure
ρcy=0.85β1fc'fyd'dϵuϵu-ϵy+ρ'
To be underreinforced beam the reinforced ratio should be less than
ρmax=ρ0.005+ρ'fs'fy
Where fs'= Esϵs'=Esϵu-d'd(ϵu+0.005) fy
Then value of c could be found out by solving
Asfy=0.85β1fc'bc+As'ϵuEsc-d'c
Then a=β1c
Nominal capacity of the section would be
Mn=0.85fc'abd-a2+As'fs'd-d'
7 ft.DL: 1 k/ftLL: 5 k/ft10"20"Example 3
7 ft.
DL: 1 k/ft
LL: 5 k/ft
10"
20"
A rectangular must carry a dead and live load as shown as figure and its self weight. For Architectural reason the section is fixed by 10 in. width and 20 in total depth. If fy=60 ksi and fc'=4 ksi, what steel must be provided?
Solution:
Self weight can be estimated =(10×20144)0.150=0.21 klf
Dead load=self weight+dead load=0.21+1=1.21 klf
Live load=5 klf
Factored load =1.2×DL+1.6×LL
=1.2×1.21+1.6×5
=9.452 klf
Factored Moment, Mu=12ωl2
=12×9.452×72 ft-kips
=231.57 ft-kips
=2778.88 in-kips
Assuming the tensile steel centroid will be 4 in. below the top face of the beam and that compression steel, if required, will be placed 2.5 in. above the beam's bottom surface. Then d=16 in.
First, check the capacity of the section if singly reinforced with reinforcement ratio
ρ0.005= 0.85β1fc'fy 0.0030.003+0.005=0.85×0.85×460×0.0030.008=0.0181
The maximum nominal moment that the section can provide is
Mn= ρfybd21-0.59fyρfc'=0.0181×60×10×1621-0.590.0181×604
=2334.82 in-kips
Because corresponding design moment, ϕMn=2101.33 in-kips, is less than the required capacity, 2779 in-kips, compression steel is needed as well as additional tension steel.
The remaining moment to be carried by the compression steel couple is
Mn1=27790.9-2334.82=752.95 in-kips
As d is less then the value required to develop the compression reinforcement yield stress (Table 3.2, page 99, Nilson 13th edition), a reduced stress in the compression reinforcement will be used.
fs'= Esϵs'=Esϵu-d'd(ϵu+0.005)
= 29000×.00175 ksi
= 29000×.00175 ksi
= 50.75 ksi
For the compression reinforcement to obtain the area of steel
As'=Mn1fs'd-d'=752.95 50.75 (16-2.5)=1.12 in2
The total area of tensile reinforcement at 60 ksi is
As=ρ×bd+As'fs'fy
=0.0181×10×16+1.12 50.7560
=3.84 in2
Four No. 9 bars in two rows will be placed in the tension zone and two No. 7 bar provide slightly grater than the required steel.
20 "14.75 "212 "10 "2 # 74 # 9212 "234 "
20 "
14.75 "
212 "
10 "
2 # 7
4 # 9
212 "
234 "
A final check is made to ensure that the section has sufficient capacity
Asfy=0.85β1fc'bc+As'ϵuEsc-d'c
4×60=0.85×0.85×4×10×c+1.20×0.003×29000c-2.5c
4×60=0.85×0.85×4×10×c+1.20×0.003×29000c-2.5c
28.9 c2-135.6 c-261=0
Solving the equation c1=6.16 and c2=-1.47
c -1.47 , c=6.16
fs'= Esϵs'=ϵs'=29000×0.0036.16-2.56.16=51.69 ksi
ϵt=0.00317.25-6.166.16=0.0054
For which ϕ=0.9 and a=β1×c=0.85×6.16=5.236
Then ϕMn=0.90.85×4×5.236×1016-5.2362+1.2×51.69×16-2.5
= 2897.7 in-kips
This is greater than Mu, so all right