lecture 03 design of rc members for flexure and axial
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 1
Lecture-03
Design of Reinforced Concrete
Members for Flexure and Axial
Loads
By: Prof. Dr. Qaisar Ali
Civil Engineering Department
UET Peshawar
drqaisarali@uetpeshawar.edu.pk
www.drqaisarali.com
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 2
Topics Addressed
� General
� Reinforced Concrete Members Subjected to Flexure Load only
� Reinforced Concrete Members Subjected to Axial Compressive
Load only
� Reinforced Concrete Members Subjected to Axial Compressive
Load with Uniaxial Bending
� Reinforced Concrete Members Subjected to Axial Compressive
Load with Biaxial Bending
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 3
General
� While transmitting load from floors and roof to the foundations,
frame members (beams and columns) of a RC frame structure are
subjected to one or more of the following load effects :
� Axial Load (compression or tension), Flexure, Shear and Torsion
� If all of these effects exist together in a RC frame member, Axial
and Flexure loads are considered as one set of effects in the
design process; whereas Shear and Torsion are considered as
another set of load effects.
� It means that the design for Axial+ Flexure is not affected by Shear +
Torsion and vice versa.
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 4
� When frame members are designed for the effects of Axial and
Flexure loads (with or without shear+ torsion) , following cases are
possible
� Members subjected to Flexure Load only
� In this case normal beam design procedures are followed.
� Members subjected to Axial Load only
� Pure compression member design procedures are used
� Members subjected to Combined Axial and Flexure Loads
� Interaction diagram procedures, considering Axial and Flexure effects together, are used.
� The Provisions of Chapter 10 shall apply for design of members subjected to
flexure or axial loads or to combined flexure and axial loads.
� These cases will be discussed one by one in the next slides
General
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Reinforced Concrete
Members Subjected to
Flexure Load only
5
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Contents
� Loading Stages Before Collapse
� Design of Solid Rectangular Members
� Design of Solid T Members
� Design of Hollow Rectangular Members
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 7
Loading Stages Before Collapse
� Beam Test
In order to clearly understand the behavior of RC members subjected
to flexure load only, the response of such members at three different
loading stages is discussed.
BEAM TEST VIDEO
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 8
Loading Stages Before Collapse
1. Un-cracked Concrete – Elastic Stage:
� At loads much lower than the ultimate, concrete remains un-
cracked in compression as well as tension and the behavior of
steel and concrete both is elastic.
2. Cracked Concrete (tension zone) – Elastic Stage
� With increase in load, concrete cracks in tension but remains un-
cracked in compression. Concrete in compression and steel in
tension both behave in elastic manner.
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 9
Loading Stages Before Collapse
3. Cracked Concrete (tension zone) – Inelastic (Ultimate Strength)
Stage
� Concrete is cracked in tension. Concrete in compression and steel
in tension both enters into inelastic range. At collapse, steel yields
and concrete in compression crushes.
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Stage-1: Behavior
ft = frM = Mcr
fc = ft << fc'
fc
ft
h
b
d
Compression zone
Tension Zone
Strain DiagramStress Diagram
Tensile Stress
Compressive Stress
fc'
ft = fr = 7.5 √fc'
Concrete stress-strain diagram
• This is a stage where concrete is at theverge of failure in tension
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Loading Stages Before Collapse
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
bfc = ft = Mc/Igwhere c = 0.5h
Ig = bh3/12
fc = ft = 6M/(bh2)
OR
C = T ; fc = ftM = 0.5fc × (b × 0.5h) × (2/3 h)
= 1/6 fc × b × h2
fc = ft = 6M/(bh2)
At ft = fr , where modulus of rupture, fr = 7.5 √fc′
Cracking Moment Capacity, Mcr = fr × Ig/(0.5h) = (fr × b × h2)/6
ft
hd
Compression zonefc
C= 0.5fc × (b × 0.5h)
T=0.5ft × (b × 0.5h)
2/3 h
1/2 h
1/2 h
M
Stage-1: Calculation of Forces
The contribution of steel isignored for simplification.
If there is no reinforcement,member will fail in tension.
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Loading Stages Before Collapse
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
εc
εt
fy
0.5fy
εc < 0.003
εs = fs/Es
0.45fc'
h
b
d
Compressionzone
Tension Zone Concrete Cracked
Strain Diagram Stress Diagram
Compressive Stressfc'
fs = 0.5 fy
Es
0.003
ft > frM > Mcr
fc = 0.45fc'fs =0.5 fy
fc = 0.45fc'
Stage-2: Behavior
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Loading Stages Before Collapse
7
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
h
In terms of moment couple (∑M = 0)
M = Tla = Asfs (d – c/3)
As = M/fs(d – c/3)
C = T (∑Fx = 0)
(½)fcbc = Asfs
c = 2Asfs / fcb {where fs = nfc and n =Es/Ec}
c = 2Asn/b
C = 0.5fc × (bc)
b
d
Compressionzone
Stress Diagram
c
T= Asfs
la = d – c/3
M
fc
Stage-2: Calculation of Forces
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Loading Stages Before Collapse
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
ft > >frM > >Mcr
fs = fyfc = αfc′, where α < 1 h
b
d
Compression zone
Tension Zone Concrete Cracked
Strain Diagram Stress DiagramCompressive Stress
fc'
T = Asfy
εt
fy
Stress-Strain Diagram for Concrete in Compression
Stress-Strain Diagram for Reinforcing Steel in Tension
εc = 0.003
εs = fy/Es
Es0.003εc
fc
Stage-3: Behavior
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Loading Stages Before Collapse
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
In terms of moment couple (∑M = 0)
M = Tla = Asfy (d – a/2)
As = M/fy(d – a/2)
C = T (∑Fx = 0)
0.85fc ′ab = Asfy
a = Asfy/ 0.85fc ′ b
T = Asfy
C = 0.85fc′ab
la = d – a/2h
b
d
Stress Diagram
T = Asfy
εc = 0.003
εs = fy/Es
M
a = β1c
0.85fc′
Equivalent Stress Diagram
fc
Stage-3: Calculation of Forces
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Loading Stages Before Collapse
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Stage-3: Calculation of Forces
• According to the strength design method (ACI 10.3.3), the nominal
flexural capacity of RC Members shall be calculated from the
conditions corresponding to stage 3.
• ACI code, R10.3.3 — The Nominal Flexural Strength (Mn) of a RC member
is reached when the strain in the extreme compression fiber reaches the
assumed strain limit of 0.003, (i.e. strains at stage 3.)
• In other words, the member finally fails by crushing of concrete, even if
steel in tension has yielded well before crushing of concrete.
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Loading Stages Before Collapse
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Stage-3: Calculation of Forces
• When concrete crushes at εc =0.003, depending on the amount of
steel (As) present as tension reinforcement, following conditions are possible
for steel strain (εs)
1. εs = εy Balanced Failure Condition, Brittle Failure
2. εs < εy Over reinforced condition, brittle failure
3. εs > εy Under Reinforced Condition, Ductile Failure
• For relative high amount of tension reinforcement, failure may occur
under conditions 1 & 2, causing brittle failure. It is for this reason that ACI
code restricts maximum amount of reinforcement in member subjected to
flexural load only.
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Loading Stages Before Collapse
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Stage-3: Calculation of Forces
• To ensure ductile failure & hence to restrict the maximum amount of
reinforcement, the ACI code recommends that for tension controlled sections
(Beams) εs = εt = 0.005
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Loading Stages Before Collapse
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 19
� Singly Reinforced:
� Flexural Capacity
� Mn = Asfy (d – a/2) [Nominal capacity]
� ΦMn = ΦAsfy(d – a/2) [Design capacity]
� To avoid failure, ΦMn ≥ Mu
� For ΦMn = Mu; ΦAsfy(d – a/2) =Mu ;
� As = Mu/ {Φfy (d – a/2)} and a = Asfy/0.85fc′b
T = Asfy
C = 0.85fc′ab
la = d – a/2h
b
d
Stress Diagram
T = Asfy
εc = 0.003
εs = fy/Es
M
a = β1c
0.85fc′
Equivalent Stress Diagram
fc
Design of Solid Rectangular Members
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 20
� Singly Reinforced:
� Maximum reinforcement (Asmax):
� From equilibrium of internal forces,
� ∑Fx = 0 → C = T
� 0.85fc′ab = Asfy …………(a)
� From similarity of triangles, in strain diagram
at failure condition,
� c/εu = (d – c)/εs
� c = dεu/(εu + εs)
� substituting a = β1c , As = ρmax b d and εs = εt , in equation (a) yields;
� ρmax = 0.85 β1(fc′/fy) εu/ (εu + εt)
Design of Solid Rectangular Members
11
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 21
� Singly Reinforced:
� Maximum reinforcement (Asmax):
� For ductility in Tension Controlled sections (Beams)
� εs = εt = 0.005 (ACI 10.3.5)
� and at failure εu = 0.003 (ACI R10.3.3),
� c = dεu/(εu + εs)→ c = 0.375d and, a = β1c = β10.375d
� Therefore, when a = β10.375d, As = Asmax in equation (a). Hence equation (a)becomes,
� 0.85fc′β10.375db = Asmaxfy
� Asmax = 0.31875β1bd fc′/fy … (b)
10.2.7.3 — Factor β1 shall be taken as 0.85 for concrete strengths fc′ up to and including 4000 psi.For strengths above 4000 psi, β1 shall be reduced continuously at a rate of 0.05 for each1000 psi ofstrength in excess of 4000 psi, but β1 shall not be taken less than 0.65.
Design of Solid Rectangular Members
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 22
� Singly Reinforced:
� Maximum reinforcement (Asmax):
� Asmax = 0.31875 β1bd fc′/fy … (b)
� For β1 = 0.85; fc′ = 3 ksi ; and fy = 40 ksi
� Asmax = 0.0203 bd; which means 2 % of effective area of concrete
� β1 = 0.85; fc′ = 3 ksi ; and fy = 60 ksi
� Asmax = 0. 0135 bd; which means 1.35 % of gross area of concrete
Design of Solid Rectangular Members
12
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 23
� Singly Reinforced:
� Maximum flexural capacity ( Mnmax):
Table 1: Maximum factored flexural capacity (M n in in-kips) of singly reinforced RC rectangular bea m for specified material strength and dimensions
fc′ = 3 ksi b (in)fy = 40 ksi 12 15 18
h (in)Assuming
distance from Centre of the main bar to
outer tension fiber=2.5”
12 740 (2.32) 925(2.90) 1110(3.47)
18 1970(3.78)2462(4.72)
2955(5.67)
20 2511(4.27)3139(5.33) 3767(6.40)
24 3790(5.24) 4738(6.55)5685(7.86)
30 6201(6.71)7751(8.38) 9301(10.06)
Note: The values in brackets represents A smax in in 2.
Design of Solid Rectangular Members
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 24
� Singly Reinforced:
� Flexural capacity at other strains
� We know that the ductility requirement of ACI code does not allow us to utilize
the beam flexural capacity beyond ΦMnmax. The code wants to ensure that
steel in tension yield before concrete crushes in compression.
� However, if we ignore ACI code restriction, let see what happens.
� We know that
� c = dεu/(εu + εs) ; a= 0.85c ; As = 0.85fc′ab/ fs; Mn = Asfs(d – a/2) ; fs = Eεs ≤ fy;
� For εu = 0.003 and assuming various values of εs , we can determine As and Mn
Design of Solid Rectangular Members
13
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 25
� Singly Reinforced:
� Flexural capacity at other strains
Table 2: Flexural Capacity (Mn) of 12 x 24 inch [d=2 1.5”] RC beam at different tensile strain condition
εs (in/in) 0.0005 0.001 0.00137* 0.0021 0.003 0.004 0.005** 0.007
c (in)18.43 16.13 14.76 12.65 10.75 9.21 8.06 6.46
As (in2)33.06 14.46 9.66 8.22 6.99 5.99 5.24 4.19
fs (ksi)14.5 29 39.73 40 40 40 40 40
Mn (in-kips)6551 6143 5846 5304 4734 4214 3790 3147
*Yield strain for grade 40 steel**ACI limit
Design of Solid Rectangular Members
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 26
� Singly Reinforced:
� Flexural nominal capacity at other strains
� Conclusions
� At balance condition, Yield strain = 0.00137, M = 5856; we see no substantial
increase in capacity beyond this point i.e. with further increase in steel reinforcement,
or decrease in strain there is no appreciable increase in flexural capacity.
� At ACI code limit of strain = 0.005, M = 3790; we see that there is considerable gap
between moment capacity at balance and moment capacity at ACI limit. Therefore if
ductility is not required, beam capacity can be further increased up to capacity at
balanced point.
� However if ductility is also required, we can increase moment capacity (without
changing dimensions) only if we provide reinforcement in compression (doubly
reinforced).
Design of Solid Rectangular Members
14
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 27
� Singly Reinforced:
� Minimum reinforcement (Asmin):
� According to ACI 10.5.1, at every section of a flexural member
where tensile reinforcement is required by analysis, the area As
provided shall not be less than that given by ρminbwd where, ρmin is
equal to 3√ (fc′)/fy and not less than 200/fy.
Design of Solid Rectangular Members
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 28
� Doubly Reinforced
� Background
� We have seen that we can not provide tensile reinforcement in excess of
Asmax = 0.31875β1bd fc′/fy , so there is a bar on maximum flexural capacity.
� We can increase Asmax if we increase b, d, fc′ or decrease fy .
� If we can’t do either of these and provide reinforcement in excess of Asmax ,
concrete in compression may crush before steel in tension yields.
� However if we provide this excess reinforcement also on compression side so
that the compression capacity of concrete also increases, we would be able to
increase the flexural capacity of the member. In this case the member is called
doubly reinforced.
� In other words the range of Asmax is increased. In such a case
� Asmax = 0.31875β1bd fc′/fy + compression steel.
Design of Solid Rectangular Members
15
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 29
� Doubly Reinforced
� Flexural Capacity
� Consider figure d and e, the flexural capacity of doubly reinforced beamconsists of two couples:
� The forces Asfy and 0.85fc′ab provides the couple with lever arm (d – a/2).
� Mn1 = Asfy (d – a/2) ……..………………… (c)
� The forces As′fy and As′fs′ provide another couple with lever arm (d – d′).
Mn2 = As′fs′ (d – d′) ………………………………….. (d)
Design of Solid Rectangular Members
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 30
� Doubly Reinforced
� Flexural Capacity
� The total nominal capacity of doubly reinforced beam is therefore,
� Mn = Mn1 + Mn2 = Asfy (d – a/2) + As′fs′ (d – d′)
Design of Solid Rectangular Members
16
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 31
� Doubly Reinforced
� Flexural Capacity
� Factored flexural capacity is given as,
ΦMn = ΦAsfy (d – a/2) + ΦAs′fs′ (d – d′) …………….. (e)
� To avoid failure, ΦMn ≥ Mu. For ΦMn = Mu, we have from equation (e),
Mu = ΦAsfy (d – a/2) + ΦAs′fs′ (d – d′) ……………..… (f)
� Where, ΦAsfy (d – a/2) is equal to ΦMnmax (singly) for As = Asmax
� Therefore, Mu = ΦMnmax (singly) + ΦAs′fs′ (d – d′)
� {Mu – ΦMnmax (singly)} = ΦAs′fs′ (d – d′)
� As′ = {Mu – ΦMnmax (singly)}/ {Φfs′ (d – d′)} ……….….... (g) ; where, fs′ ≤ fy
Design of Solid Rectangular Members
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 32
� Doubly Reinforced
� Maximum reinforcement
� Cc + Cs = T [ ∑Fx = 0 ]
� 0.85fc′ab + As′fs′ = Astfy
� For Amax ; a = β1c = 0.85 × 0.375d
� Ast will become Astmax
� 0.85fc′β10.375db + As′fs′ = Astmaxfy
� Astmax = β10.31875bdfc′/fy + As′fs′/fy
� Astmax = Asmax (singly) + As′fs′/fy
Cc = Compression forcedue to concrete incompression region,Cs = Compression forcein steel in compressionregion needed tobalance the tensionforce in addition to thetension force providedby Asmax (singly).
Design of Solid Rectangular Members
17
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 33
� Doubly Reinforced
� Maximum reinforcement
� Astmax = Asmax (singly) + As′fs′/fy
� The total steel area actually provided Ast as tension reinforcement
must be less than Astmax in above equation i.e. Ast ≤ Astmax
� Astmax (singly ) is a fixed number, whereas As′ is steel area actually
placed on compression side. (For more clarification, see example)
� Note that Compression steel in the above equation may or may not yieldwhen tension steel yields.
Design of Solid Rectangular Members
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 34
� Doubly Reinforced
� Conditions at which fs′ = fy when tension steel yields.
� By similarity of triangle (fig b), compression steel strain (εs′) is,
� εs′ = εu (c – d′)/ c …………………………….. (h)
� For tensile steel strain (εs) = εt = 0.005 (for under reinforced behavior):
� c = 0.375d
� Substituting the value of c in eqn. (h), we get,
� εs′ = εu (0.375d – d′)/ 0.375d = (0.003 – 0.008d′/d) …………….….. (i)
� Equation (i) gives the value of εs′ for the condition at which reinforcement on
tension side is at strain of 0.005 ensuring ductility.
Design of Solid Rectangular Members
18
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 35
� Doubly Reinforced
� Conditions at which fs′ = fy when tension steel yields.
� εs′ = {0.003 – 0.008d′/d} ……..……………….. (i) OR
� d′/d = (0.003 - εs′)/0.008 ………………………. (j)
� Substituting εs′ = εy,in equation (j).
� d′/d = (0.003 - εy)/0.008 …………..………..…. (k)
� Equation (k) gives the value of d′/d that ensures that when tension steel is at a
strain of 0.005 (ensuring ductility), the compression steel will also be at yield.
� Therefore for compression to yield, d′/d should be less than the value given by
equation (k).
Design of Solid Rectangular Members
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
� Doubly Reinforced
� Conditions at which fs′ = fy when tension steel yields.
� Table 3 gives the ratios (d′/d) and minimum beam effective depths (d) for
compression reinforcement to yield.
� For grade 40 steel, the minimum depth of beam to ensure that
compression steel will also yields at failure is 12.3 inch.
36
Design of Solid Rectangular Members
Table 3: Minimum beam depths for compression reinfor cement to yield
fy, psi Maximum d'/dMinimum d for d' =
2.5" (in.)
40000 0.2 12.3
60000 0.12 21.5
75000 0.05 48.8
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 37
� Doubly Reinforced
� Example
� Design a doubly reinforced concrete beam for an ultimate flexural
demand of 4500 in-kip. The beam sectional dimensions are restricted.
Material strengths are fc′ = 3 ksi and fy = 40 ksi.
d = 20″
b = 12″
Design of Solid Rectangular Members
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 38
� Doubly Reinforced
� Solution:
� Step No. 01: Calculation of ΦMnmax (singly)
ρmax (singly) = 0.0203
Asmax (singly) = ρmax (singly)bd = 4.87 in2
ΦMnmax (singly) = 2948.88 in-kip
� Step No. 02: Moment to be carried by compression steel
Mu (extra) = Mu – ΦMnmax (singly)
= 4500 – 2948.88 = 1551.12 in-kip
Design of Solid Rectangular Members
20
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 39
� Doubly Reinforced
� Solution:
� Step No. 03: Find εs′ and f s′
From table 2, d = 20″ > 12.3″, and for d′ = 2.5″, d′/d is 0.125 < 0.20 for grade 40
steel. So compression steel will yield.
Stress in compression steel fs′ = fy
Alternatively,
εs′ = (0.003 – 0.008d′/d) ………………….. (i)
εs′ = (0.003 – 0.008 × 2.5/20) = 0.002 > εy = 40/29000 = 0.00137
As εs′ is greater than εy, so the compression steel will yield.
Design of Solid Rectangular Members
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 40
� Doubly Reinforced
� Solution:
� Step No. 04: Calculation of A s′ and A st.
As′ = Mu(extra)/{Φfs′(d – d′)}=1551.12/{0.90×40×(20–2.5)}= 2.46 in2
� Total amount of tension reinforcement (Ast) is,
Ast = Asmax (singly) + As′= 4.87 + 2.46 = 7.33 in2
� Using #8 bar, with bar area Ab = 0.79 in2
No. of bars to be provided on tension side = Ast/ Ab= 7.33/ 0.79 = 9.28
No. of bars to be provided on compression side = As′/ Ab=2.46/ 0.79 = 3.11
Provide 10 #8 (9.7 in 2 in 3 layers) on tension side and 4 #8 (3.16 in 2 in 1
layer) on compression side.
Design of Solid Rectangular Members
21
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 41
� Doubly Reinforced
� Solution:
� Step No. 05: Ensure that d ′/d < 0.2 (for grade 40) so that selection of bars
does not create compressive stresses lower than yield.
With tensile reinforcement of 10 #8 bars in 3 layers and compression
reinforcement of 4 #8 bars in single layer, d = 19.625″ and d′ = 2.375
d′/d = 2.375/ 19.625 = 0.12 < 0.2, OK
Design of Solid Rectangular Members
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 42
� Doubly Reinforced
� Solution:
� Step No. 06: Ductility requirements: Ast ≤ Astmax
� Ast , which is the total steel area actually provided as tension reinforcement must
be less than Astmax .
� Astmax = Asmax (singly) + As′fs′/fy
� Astmax (singly ) is a fixed number for the case under consideration and As′ is
steel area actually placed on compression side.
� Asmax (singly) = 4.87 in2 ; As′ = 4 × 0.79 = 3.16 in2; Astmax= 4.87 + 3.16 = 8.036 in2
Ast = 7.9 in2
Therefore Ast = 7.9 in2 < Astmax OK.
Design of Solid Rectangular Members
22
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 43
� Difference between T-beam and T-beam Behavior
Design of Solid T Members
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 44
� Flexural Capacity
Design of Solid T Members
23
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Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 45
� Flexural Capacity
� Asf =0.85fc′(b – bw)hf/fy
� Asf, is the steel area which when stressed to fy, is required to balance the longitudinal
compressive force in the overhanging portions of the flange that are stressed uniformly at
0.85fc′.
� ΦMn1 = ΦAsffy (d – hf/2)
� As = ΦMn2 /Φ fy (d – a/2) = (Mu – ΦMn1)/Φ fy (d – a/2)
� a = Asfy/ (0.85fc′bw)
� As represents the steel area which when stressed to fy, is required to balance the
longitudinal compressive force in the rectangular portion of the beam.
� Total steel area required (Ast) = Asf + As
Design of Solid T Members
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 46
� Flexural Capacity (Alternate Formulae)
� ΦMn = Mu= ΦAstfy (d – x)
� Ast = Mu/ {Φfy (d – x)}
� x = {bwa2/2 + (b – bw)hf2/2}/ {bwa + (b – bw)hf}
� a = {Astfy – 0.85fc′ (b – bw)hf}/0.85fc′bw
Design of Solid T Members
24
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 47
� Ductility Requirements
� T = C1 + C2 [ ∑Fx = 0 ]
Astfy = 0.85fc′abw + 0.85fc′(b – bw)hf
Astfy = 0.85fc′abw + Asffy
� For ductility εs = εt = 0.005 (ACI 10.3.5),
� a = amax = β1c = β10.375d, and Ast will become Astmax, Therefore,
Astmaxfy= 0.85fc′β10.375dbw + Asffy
Astmaxfy= 0.85fc′β10.375dbw + Asf
Astmax = 0.31875 β1(fc′/fy)dbw + Asf OR Astmax = Asmax (singly) + Asf
� So, for T-beam to behave in a ductile manner Ast, provided ≤ Astmax
Design of Solid T Members
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 48
� Effective Flange width for T and L beam (ACI 8.10)
Design of Solid T Members
25
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Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 49
� Example 03
� Design a beam to resist a factored moment equal to 6500 in-kip. The
beam is 12″ wide, with 20″ effective depth and supports a 3″ slab. The
beam is 25′ long and its c/c distance to next beam is 4 ft. Material
strengths are fc′ = 3 ksi and fy = 40 ksi.
Design of Solid T Members
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 50
� Example Solution:
� Span length (l) = 25′
� d = 20″; bw = 12″; hf = 3″
� Effective flange width (b) is minimum of,
� l/4 = 25 × 12/4 = 75″
� 16hf + bw = 16 × 3 + 12 = 60″
� c/c distance to next beam = 4 × 12 = 48″
� Therefore, b = 48″
Design of Solid T Members
26
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Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 51
� Example Solution:
� Check if the beam behaviour is T or rectangular.
� Let a = hf = 3″
As = Mu/Φfy(d – a/2) = 6500/{0.90 × 40 × (20 – 3/2)} = 9.76 in2
a = Asfy/(0.85fc′b) = 9.76 × 40/ (0.85 × 3 × 48) = 3.20″ > hf
� Therefore, design as T beam.
Design of Solid T Members
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 52
� Example Solution:
� Design:
� We first calculate Asf, the steel area which, when stressed to fy, is required
to balance the longitudinal compressive force in the overhanging portions
of the flange that are stressed uniformly at 0.85fc′.
Asf = 0.85fc′ (b – bw) hf/fy
= 0.85 × 3 × (48 – 12) × 3/40 = 6.885 in2
� The nominal moment resistance (ФMn1), provided by Asf is,
ФMn1 = ФAsffy {d – hf/2} = 0.9 × 6.885 × 40 × {20 – 3/2} = 4585.41 in-kip
Design of Solid T Members
27
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 53
� Example Solution:
� Design:
� The nominal moment resistance (ФMn2), provided by remaining steel As is,ФMn2 = Mu – ФMn1 = 6500 – 4585.41 = 1914.45 in-kip
� Let a = 0.2d = 0.2 × 20 = 4″
As = ФMn2/ {Фfy(d – a/2)} = 1914.45/ {0.9 × 40 × (20 – 4/2)}= 2.95 in2
a = Asfy/(0.85fc′bw) = 2.95 × 40/(0.85 × 3 ×12) = 3.90″
� This value is close to the assumed value of “a”. Therefore,
Ast = Asf + As = 6.885 + 2.95 = 9.84 in2 (13 #8 Bars)
Design of Solid T Members
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 54
� Example Solution:
� Ductility requirements, (Ast = As + Asf) ≤ Astmax
Astmax = Asmax (singly) + Asf
= 4.87 + 6.885 = 11.76 in2
Ast = As + Asf = 13 × 0.79 = 10.27 in2 < 11.76 O.K.
Design of Solid T Members
28
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 55
� Example Solution:
� Ensure that Ast > Asmin
Ast = 10.27 in2
� Asmin = ρminbwd
� ρmin = 3√(fc′)/fy ≥ 200/fy
3√(fc′)/fy = 3 × √(3000)/60000 = 0.004
200/fy = 200/40000 = 0.005
ρmin = 0.005 ; Asmin = 0.005 × 12 ×20 = 1.2 in2 < Ast, O.K.
Design of Solid T Members
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 56
� Example Solution:
� Design:
� We design the same beam by alternate method.
� Trial 01:
� Assume a = hf = 3″
x = {bwa2/2 + (b – bw)hf2/2}/ {bwa + (b – bw)hf}
= {12×32/2+(48 – 12)×32/2}/ {12×3+ (48 – 12)×3} = 1.5″
Ast = Mu/ {Φfy (d – x)} = 6500/ {0.90 × 40 ×(20 – 1.5) = 9.76 in2
Design of Solid T Members
29
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 57
� Example Solution:
� Design:
� Trial 02:
� a = {Astfy – 0.85fc′ (b – bw)hf}/0.85fc′bw
= {9.76 × 40 – 0.85×3×(48 – 12)×3}/ (0.85×3×12)= 3.76″
x = {12×3.762/2+(48 – 12)×32/2}/ {12×3.76+ (48 – 12)×3} = 1.61″
Ast = 6500/ {0.90 × 40 × (20 – 1.61)} = 9.81 in2
Design of Solid T Members
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 58
� Example Solution:
� Design:
� Trial 03:
a = {9.81 × 40 – 0.85×3×(48 – 12)×3}/ (0.85×3×12)= 3.83″
x = {12×3.832/2+(48 – 12)×32/2}/ {12×3.83+ (48 – 12)×3} = 1.62″
Ast = 6500/ {0.90 × 40 × (20 – 1.62)} = 9.83 in2, O.K.
� This is same as calculated previously for T-beam.
Design of Solid T Members
30
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 59
� Flexural Capacity
Design of Hollow Rectangular Members
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 60
� Flexural Capacity
� As1 =0.85fc′bohf/fy
� As1 represents the steel area which when stressed to fy, is required to balance the
longitudinal compressive force in the rectangular portion of the area bohf that is stressed
uniformly at 0.85fc′.
� ΦMn1 = ΦAs1fy (d – hf/2)
� As2 = ΦMn2 /Φ fy (d – a/2) = (Mu – ΦMn1)/Φ fy (d – a/2)
� a = As2fy/ {0.85fc′(b - bo)}� As2 is the steel area which when stressed to fy, is required to balance the longitudinal
compressive force in the remaining portion of the section that is stressed uniformly at
0.85fc′.
� Total steel area required (Ast) = As1 + As2
Design of Hollow Rectangular Members
31
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 61
� Flexural Capacity (Alternate Formulae)
� ΦMn = Mu = ΦAstfy (d – x)
� Ast = Mu/ {Φfy (d – x)}
� x = {bohf2/2 + (b – bo)a2/2}/ {(b –bo)a + bohf}
� a = {Astfy – 0.85fc′bohf}/0.85fc′(b –bo)
Design of Hollow Rectangular Members
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 62
� Ductility Requirements
� For summation of internal forces,
� Astfy = 0.85fc′ba – 0.85fc′bo(a – hf)
� For εt = 0.005, a = β1 × 0.375d, we have Ast = Astmax, hollow, therefore,
� Astmax, hollow = {0.85fc′bβ1 × 0.375d – 0.85fc′bo(0.375d – hf)}/ fy
� Astmax, hollow = 0.319(fc′/fy)β1bd – 0.85(fc′/fy)bo(0.375d – hf)
� Astmax, hollow = Asmax (singly) – 0.85(fc′/fy)bo(0.375d – hf)
� So, for hollow beam to behave in a ductile manner:
Ast, provided ≤ Astmax, hollow
Design of Hollow Rectangular Members
32
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Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 63
� Example
� Design a beam to resist a factored moment equal to 2500 in-kip. The
beam has a hollow section with 12″ width and overall depth of 24″. The
hollow part inside the section is 3″ wide and 16″ deep. Material strengths
are fc′ = 3 ksi and fy = 60 ksi.
24″
12″
3″
16″
Design of Hollow Rectangular Members
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 64
� Example Solution
� h = 24″; d = 21.5″ (assumed)
� b = 12″
� bo = hf = 4″
� Check if the beam behaviour is rectangular or hollow rectangular.
� Let a = hf = 4″
As = Mu/Φfy(d – a/2) = 2500/{0.90 × 60 × (21.5 – 4/2)} = 2.37 in2
a = Asfy/(0.85fc′b) = 2.37 × 60/ (0.85 × 3 × 12) = 4.65″ > hf
� Therefore, design as hollow beam.
Design of Hollow Rectangular Members
33
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 65
� Example Solution
� First calculate As1,
As1 = 0.85fc′bohf/fy
= 0.85 × 3 × 3 × 4/60 = 0.51 in2
� The nominal moment resistance (ФMn1), provided by As1 is,
ФMn1 = ФAs1fy {d – hf/2} = 0.9 × 0.51 × 60 × {21.5 – 4/2} = 537.03 in-kip
� The nominal moment resistance (ФMn2), provided by remaining steel As2
is,
ФMn2 = Mu – ФMn1 = 2500 – 537.03 = 1962.97 in-kip
Design of Hollow Rectangular Members
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 66
� Example Solution
� Let a = 4″
As2 = ФMn2/ {Фfy(d – a/2)} = 1962.97/ {0.9 × 60 × (21.5 – 4/2)} = 1.86 in2
a = As2fy/ {0.85fc′ (b – bo)} = 1.86 × 60/ {0.85 × 3 × (12 – 3)} = 3.65″
� This value is close to the assumed value of “a”. Therefore,
Ast = As1 + As2 = 0.51 + 1.86 = 2.37 in2
� Using #8 bar, with bar area Ab = 0.79 in2
# of bars = Ast/ Ab = 2.37/ 0.79 = 3 bars
Design of Hollow Rectangular Members
34
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 67
� Example Solution
� Ductility requirements, (Ast) provided <Astmax, hollow
� Astmax, hollow = Asmax (singly) – 0.85(fc′/fy)bo(0.375d – hf)
= 3.48 – 0.85 × (3/60) × 3 × (0.375 × 21.5 – 4) = 2.96 in2
� Therefore, Ast = 2.37 in2 < 2.96 in2 O.K.
Design of Hollow Rectangular Members
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 68
� Example Solution
� Design the same beam by alternate approach.
� Trial 01:
Assume a = hf = 4″
� x = {bohf2/2 + (b – bo)a2/2}/ {(b –bo)a + bohf}
= {3×42/2+(12 – 3)×42/2}/ {(12–3)×4 + 3×4} = 2″
� Ast = Mu/ {Φfy (d – x)} = 2500/ {0.90 × 60 × (21.5 – 2)} = 2.37 in2
Design of Hollow Rectangular Members
35
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 69
� Example Solution
� Design the same beam by alternate approach.
� Trial 02:
� a = {Astfy – 0.85fc′bohf}/0.85fc′(b –bo)
= {2.37 × 60 – 0.85×3×3×4}/ {0.85×3×(12 – 3)} = 4.87″
x = {3×42/2+(12 – 3)× 4.872/2}/ {(12–3)×4.87 + 3×4} = 2.34″
Ast = 2500/ {0.90 × 60 × (21.5 – 2.34)} = 2.41 in2, O.K.
� This is close to the value calculated previously for hollow-beam.
Design of Hollow Rectangular Members
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Reinforced Concrete
Members Subjected to Axial
Compressive Loads
70
36
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Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Contents
71
� Axial Capacity
� Maximum Reinforcement Ratio
� Example
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Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 72
Axial Capacity
� Consider a Rectangular Section
� Nominal Axial Capacity is given as
� Cs1+ Cs2+Cs3+ Cc = Pn
� Cs1 = As1 * fs1
� Cs2 = As2 * fs2
� Cs3 = As3 * fs3
� Cc = Ac * fc
� As1 * fs1 + As2 * fs2 + As3 * fs3 + Ac * fc = Pn
37
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Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 73
Axial Capacity
� The section will reach its axial capacity when strain in concrete reaches
a value of 0.003.
� The yield strain values of steel for grade 40 and 60 are 0.00138 and
0.00207 respectively. Therefore steel would have already yielded at
0.003 strain. Hence fs1 = fs2 = fs3 = fs4 = fy and fc = 0.85 fc′
� Let As1 + As2 + As3 = Ast and Ac = Ag – Ast , Then
� Ast fy + 0.85 fc′(Ag – Ast) = Pn
� where Ag = gross area of column section,
� Ast = total steel area
� ΦPn = Pu
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Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 74
Axial Capacity
� As per ACI code (10.3.6 and 10.3.7 ), the axial capacity for
� Spiral Columns
� ΦPn (max) = 0.85Φ [0.85fc′(Ag − Ast) + fy Ast] ; Φ = 0.70
� Tied Columns
� ΦPn (max) = 0.80Φ [0.85fc′(Ag − Ast) + fy Ast] ; Φ = 0.65
� The ACI factors are lower for columns than for beams, reflecting
their greater importance in a structure.
� The additional reduction factors of 0.80 and 0.85 are used to
account for accidental eccentricities not considered in the analysis
that may exist in a compression member, and to recognize that
concrete strength may be less than fc′ under sustained high loads.
R10.3.6 and R10.3.7
38
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Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 75
Maximum Reinforcement Ratio
� 1 % ≤ Ast /Ag ≤ 8 %
� Practically, however reinforcement more than 6 % is seldom used.
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 76
Example
� Design a 18″ × 18″ column for a factored axial compressive load
of 300 kips. The material strengths are fc′ = 3 ksi and fy = 40 ksi.
18″
18″
39
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Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 77
Example
� Solution
� Nominal strength (ΦPn) of axially loaded column is:
� ΦPn = 0.80Φ{0.85fc′(Ag–Ast) + Astfy}
� Let Ast = 1% of Ag
� ΦPn = 0.80 × 0.65 × {0.85 × 3 × (324 – 0.01 × 324) + 0.01 × 324 × 40}
= 492 kips > (Pu = 300 kips), O.K.
Therefore, Ast = 0.01 × 324 = 3.24 in2
� Using 3/4″ Φ(#6) {# 19, 19 mm}, with bar area Ab =0.44 in2
� No. of bars = As/Ab = 3.24/0.44 = 7.36 ≈ 8 bars
� Use 8 #6 bars {8 #19 bars, 19 mm}
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 78
Reinforced Concrete Members
subjected to Axial Compressive
Load with Uniaxial Bending
40
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Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 79
Contents
� Behavior of Columns subjected to Uniaxial Bending
� Axial Capacity
� Flexural Capacity
� Design by Trial and Success Method
� Alternative Approach
� Interaction Diagram
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Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 80
Behavior of Columns subjected to
Uniaxial Bending
� Shown in figure, is a vertical rectangular RC
member subjected to axial compressive load
Pu at some eccentricity ex along x-axis of the
cross section causing moment Muy.
� Such a column is called uniaxial column.
� The bending is called uniaxial bending
because the bending exists only about one
of the centroidal axis of the cross section.
xy
41
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Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 81
� Pu = ΦPn = Φ (Cc + Cs – T) [ ∑F = 0 ]
= Φ (0.85fc′ab + As1fs1 – As2fs2)
� Pu = Φ{0.85fc′ab+As1fs1 – As2fs2} …..(1)
� fs1 = Eεs1 = 0.003E (c – d′)/c ≤ fy
� fs2 = Eεs2 = 0.003E (d – c)/c ≤ fy� Note: Negative sign with As2 shows that
steel layer As2 is under tensilestresses.
For εs1:εs1/(c - d′) = εu/c
For εs2:εs2/(d - c) = εu/c
Axial Capacity
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Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 82
� Mu = ΦMn [ ∑M = 0 ] (about geometric center),
� Mu = Φ [Cc× {(h/2) – (a/2)} + As1fs1 × {(h/2) – d′} + As2fs2 × {d – (h/2)}]
� With (d – h/2) = {h – d′ – h/2} = {(h/2) – d′}
� Mu = Φ [Cc× {(h/2) – (a/2)} + As1fs1 × {(h/2) – d′} + As2fs2 × {(h/2) – d′}] ………(2a)
� Note: All internal forces are in counter clockwise sense to resist flexural demand caused by Pu.
Flexural Capacity
42
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Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 83
� Mu = Φ [Cc× {(h/2) – (a/2)} + As1fs1 × {(h/2) –
d′} + As2fs2 × {(h/2) – d′}] ………(2a)
� With, Cc = 0.85fc′ab ; As1 = As2 = As
The equation (2a) becomes (2b) as:
� Mu=Φ[0.425fc′ab(h–a)+As{(h/2)–d′}(fs1+fs2)] .…(2b)
� Where, fs1 = Eεs1 = 0.003E (c – d′)/c ≤ fy &
� fs2 = Eεs2 = 0.003E (d – c)/c ≤ fy
Flexural Capacity
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 84
� It is important to note that equation (1) & (2b)
are valid for 2 layers of reinforcements only.
� Pu=Φ{0.85fc′ab+As1fs1 – As2fs2}………(1)
� Mu=Φ[0.425fc′ab(h–a)+As{(h/2)–d′}(fs1+fs2)]…(2b)
� For intermediate layers of reinforcement, the
corresponding terms with “As” shall be added in
the equations.
Flexural Capacity
43
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Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 85
� As discussed in previous lectures, the singly reinforced flexural
member can be designed by trial and success method using
following formulae:
� As = Mu/ {Φfy (d – a/2)} & a = Asfy/0.85fc′b
� In the same way, equations (1) and (2b) may be used for design
of RC member subjected to compressive load with uniaxial
bending
� Pu=Φ{0.85fc′ab+As1fs1 – As2fs2} …………………………………(1)
� Mu=Φ[0.425fc′ab(h – a) + As{(h/2) - d′}(fs1 + fs2)] …………… .(2b)
Design by Trial and Success Method
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 86
� However unlike equations for beam where fs = fy, here we don’t
know values of fs1 and fs2 . But we do know that steel stress shall
be taken equal to or less than yield strength. Therefore
� fs1 = Eεs1 = 0.003E (c – d′)/c ≤ fy
� fs2 = Eεs2 = 0.003E (d – c)/c ≤ fy
� Equation (1) can be now written in the following form
� Pu = Φ {0.85fc′β1cb + AsE × 0.003(c – d′)/c – AsE × 0.003(d – c)/c)}---(1)
Design by Trial and Success Method
44
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 87
� Equation (1) can be transformed into a quadratic equation to
obtain the value of “c” for a particular demand Pu and assumed
As:
� Φ0.85fc′ β1bc2 + (Φ174As – Pu)c – Φ87As (d – d′) = 0
� However such approach will not be convenient because the
check that stresses in reinforcement layers fs1 and fs2 shall not
exceed fy can not be applied in the above equation.
Design by Trial and Success Method
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 88
� As an example, with Mu = 40 ft-kip, Pu = 145 kips, As = 0.88 in2, fc′ = 3
ksi, b = h = 12″, d = 9.5″ and d′ = 2.5″, c comes out to be 6.08″ from
quadratic equation.
� For c = 6.08″, now fs1 and fs2 shall be ≤ fy
� fs1 = Eεs1 = 0.003E (c – d′)/c = 51 ksi ; greater than 40 ksi
� fs2 = Eεs2 = 0.003E (d – c)/c = 49 ksi ; greater than 40 ksi
� It means that every time when we obtain value of c, we have to checkstresses in steel and only that value of c will be used when fs1 and fs2are ≤ fy .
� Therefore this method of trial and success will not work in members subjected to
axial load and flexure together. We now look at another approach.
Design by Trial and Success Method
45
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 89
� Instead of calculating c, we assume c and calculate ФPn and
ФMn for a given set of data such as follows:
� ФPn =Φ{0.85fc′ab+ AsE × 0.003(c – d′)/c – AsE × 0.003(d – c)/c)}
� ФMn = Φ [0.425fc′β1c b (h – a) + As {(h/2) – d′} (fs1 + fs2)]
� For As = 0.88 in2, fc′ = 3 ksi, b = h = 12″, d = 9.5″ and d′ = 2.5″ ,
all values in the above equations are known except “c”.
Alternative Approach
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 90
� ФPn and ФMn are calculated for various values of “c” from 0 to h,
with the check that during calculations fs1 and fs2 do not exceed
fy for both eqns.
Table 4
c (in)0 ≤ c ≤ (h = 12)
ФPn (kips) ФMn (kip-ft)
3.69 0 36.255 64.6 41.597 133 43.099 185.3 36
12 252.64 19.44Axial capacity 281 0
Alternative Approach
46
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 91
� Plot the values and check the
capacity of the column for the
demand equal to Mu = 40 ft-kip
and Pu = 145 kips
Demand point(40,145)
Alternative Approach
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 92
� General:
� For a column of known dimensions
and reinforcement, several pairs of P
and M from various values of “c”
using equations 1 and 2b can be
obtained and plotted as shown.
Such a graph is known as capacity
curve or interaction diagram.
Nominal and Design diagram are
given in the figure.
Interaction Diagram
47
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 93
� General:
� If the factored demand in the form of
Pu and Mu lies inside the design
interaction diagram, the given
column will be safe against that
demand.
Interaction Diagram
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 94
� Important Features of InteractionDiagram
� Horizontal Cutoff: The horizontal
cutoff at upper end of the curve
at a value of αΦPnmax represents
the maximum design load
specified in the ACI 10.3.5 for
small eccentricities i.e., large
axial loads.
Interaction Diagram
48
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 95
� Important Features of InteractionDiagram:
� Linear Transition of Φ from 0.65
to 0.90 is applicable for εt ≤ fy/Es
to εt = 0.005 respectively.
Interaction Diagram
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 96
� Development of InteractionDiagram:
� Interaction diagram can be
developed by calculation of
certain points as discussed
below:
� Point 01: Point representing
capacity of column when
concentrically loaded.
� This represents the point for
which Mn = 0.
Interaction Diagram
49
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 97
� Development of InteractionDiagram:
� Point 02: c = h
� Point 2 corresponds to
crushing of the concrete at
the compression face of the
section and zero stress at
the other face.
Interaction Diagram
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 98
� Development of InteractionDiagram:
� Point 03: c = (h-d′)
� At Point 3, the strain in the
reinforcing bars farthest
from the compression face
is equal to zero.
Interaction Diagram
50
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 99
� Development of InteractionDiagram:
� Point 04: c = 0.68d (Grade 40)
c = 0.58d (Grade 60)
� Point representing capacity of
column for balance failure
condition (εc = 0.003 and εt = εy).
Interaction Diagram
c = d {εc/ (εc + εy)}
εc = 0.003
εy = 0.0013 (Grade 40)
εy = 0.0021 (Grade 60)
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 100
� Development of InteractionDiagram:
� Point 05: c = 0.375d
� Point in tension controlled
region for net tensile strain
(εt) = 0.005, and Φ = 0.90,
(εc = 0.003).
Interaction Diagram
c = d {εc/ (εc + εt)}
εc = 0.003
εt = 0.005
51
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 101
� Development of InteractionDiagram:
� Point 06: c = 0.23d
� Point on capacity curve for
which εt >> 0.005 and
εc = 0.003.
εt >> 0.005
Interaction Diagram
c = d {εc/ (εc + εt)}
εc = 0.003
εt >> 0.005
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 102
� Example: Develop interaction diagram for the given column.
The material strengths are fc′ = 3 ksi and fy = 40 ksi with 4 no.
6 bars.
12″
12″
Interaction Diagram
52
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 103
� Solution:� Design interaction diagram will be developed by plotting (06)
points as discussed earlier.
� Point 1: Point representing capacity of column when
concentrically loaded: Therefore
� ΦPn = Φ [0.85fc′(Ag − Ast) + fyAst]
= 0.65 × [0.85×3×(144 – 1.76) + 40 × 1.76] = 281.52 kip
� ΦMn = 0
Interaction Diagram
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 104
� Solution:� Point 2: c = h
� c = 12 ″ (c = h); a = β1c = 0.85 × 12 = 10.2″
� fs1 = 0.003E (c – d′)/c = 0.003×29000(12 – 2.25)/12 = 70.69 ksi > fy,
use fy = 40 ksi.
� fs2 = 0.003E (d – c)/c = 0.003×29000(9.75 – 12)/12 = -16.31 ksi< fy
� Therefore, ΦPn = Φ {0.85fc′ab + As fs1 – Asfs2}
= 0.65{0.85×3×10.2×12 +0.88×40+0.88×16.31} = 235.09 kip
� ΦMn = Φ [0.425fc′ab (h – a) + As {(h/2) – d′} (fs1 + fs2)]
= 0.65[0.425×3×10.2×12×(12–10.2)+0.88×{(12/2) – 2.25}(40-16.31)]
= 233.41 in-kip = 19.45 ft-kip
Interaction Diagram
53
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 105
� Solution:� Point 3: c = (h-d ′)
� c =12-2.25=9.75; a = β1c = 0.85 × 9.75 = 8.29″
� fs1 = 0.003E (c – d′)/c = 0.003×29000(9.75 – 2.25)/9.75 = 66.92 ksi > fy,
use fy = 40 ksi.
� fs2 = 0.003E (d – c)/c = 0.003×29000(9.75 – 9.75)/9.75 = 0 ksi< fy
� Therefore, ΦPn = Φ {0.85fc′ab + As fs1 – Asfs2}
� = 0.65{0.85×3×8.29×12 +0.88×40} = 187.77 kip
� ΦMn = Φ [0.425fc′ab (h – a) + As {(h/2) – d′} (fs1 + fs2)]
= 0.65[0.425×3×8.29×12×(12–8.29)+0.88×{(12/2) – 2.25}(40)]
= 391.67 in-kip = 32.64 ft-kip
Interaction Diagram
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 106
� Solution:� Point 4: Point representing balance failure: The neutral axis for
the balanced failure condition is easily calculated from
c = d {εu/ (εu + εy)} with εu equal to 0.003 and εy = 40/29000 =
0.001379, c = 0.68d
� cb = d {εu/ (εu + εy)} = 9.75 × 0.003/ (0.003 + 0.001379)
= 0.68d = 6.68″ giving a stress-block depth;
ab = β1cb = 0.85 × 6.68 = 5.67″
Interaction Diagram
54
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 107
� Solution:� Point 4: Balance failure: For the balanced failure condition, fs = fy.
� fs1 = 0.003E (c – d′)/c = 0.003×29000(6.68–2.25)/6.68= 57.69 ksi > fy,
� fs2 = 0.003E (d – c)/c = 0.003×29000(9.75 – 6.68)/6.68 = 40 ksi = fy
� Therefore, ΦPb = Φ {0.85fc′ab + Asfs1 – Asfs2}
= 0.65{0.85×3×5.67×12 +0.88×40–0.88×40} = 112.77 kip
� ΦMb = Φ [0.425fc′ab (h – a) + As {(h/2) – d′} (fs1 + fs2)]
= 0.65[0.425×3×5.67×12×(12–5.67)+0.88×{(12/2) – 2.25}(40 + 40)]
� = 528.54 in-kip = 44.05 ft-kip
Interaction Diagram
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 108
� Solution:� Point 5: This point is in tension controlled region for which εt = 0.005, Φ = 0.90:
� For εt = 0.005; c = d {εu/ (εu + εt)} = 9.75× {0.003/ (0.003 + 0.005)}= 0.375d = 3.66″
� a = β1c = 0.85 × 3.66 = 3.11″
� fs1 = 0.003E (c – d′)/c = 0.003×29000(3.66 – 2.25)/3.66 = 33.51 ksi < fy
� fs2 = 0.003E (d – c)/c = 0.003×29000(9.75 – 3.66)/3.66 = 144.76 ksi > fy,
use fy = 40 ksi.
� Therefore, ΦPn = Φ{0.85fc′ab + Asfs1 – Asfs2}
= 0.90{0.85×3×3.11×12 +0.88×33.51–0.88×40}= 80.50 kip
� ΦMn = Φ [0.425fc′ab (h – a) + As {(h/2) – d′} (fs1 + fs2)]
� = 0.90[0.425×3×3.11×12×(12–3.11)+0.88×{(12/2) – 2.25}(33.51+40)]
= 599 in-kip = 49.91 ft-kip
Interaction Diagram
55
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 109
� Solution:� Point 6: Point on capacity curve for which εt >> 0.005:
� Let εt = 2 × 0.005 = 0.01; c = d {εu/ (εu + εt)} = 9.75× {0.003/ (0.003 + 0.01)}= 0.23d = 2.25″
� a = β1c = 0.85 × 2.25 = 1.91″
� fs1 = 0.003E (c – d′)/c = 0.003×29000(2.25 – 2.25)/2.25 = 0 < fy
� fs2 = 0.003E (d – c)/c = 0.003×29000(9.75 – 2.25)/2.25 = 290 ksi > fy,use fy = 40 ksi.
� Therefore, ΦPn = Φ{0.85fc′ab + Asfs1 – Asfs2
= 0.90{0.85×3×1.91×12 +0.88×0 – 0.88×40} = 20.90 kip
� ΦMn = Φ [0.425fc′ab (h – a) + As {(h/2) – d′} (fs1 + fs2)]
= 0.90[0.425×3×1.91×12×(12–1.91)+0.88×{(12/2) – 2.25}(0 +40)
= 384.16 in-kip = 32.01 ft-kip
Interaction Diagram
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 110
� Solution:
Nominal Interaction Curve
Design Interaction Curve
0.80φPo
0
50
100
150
200
250
300
350
400
450
500
0 20 40 60 80
P (
kip)
M (kip-ft)
M vs Pd'
dh
b
Layer 0
1
Layer 0
2
Interaction Diagram
56
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
� Use of Design Aids:
� The uniaxial columns can be designedusing design aids e.g, normalizedinteraction diagrams such as given ingraph A5-A16 (Nilson). These diagramsrequire the calculation of a dimensionlessconstant γ.
� h = γh+2d′
γ = (h-2d′)/h
� Once γ is calculated, the interactiondiagram corresponding to the value of γ isselected & then column can be designedusing steps given on the next slides.
111
Reference: Design of Concrete Structures 13th Ed. by Nilson, Darwin and Dolan.
Interaction Diagram
b
h
X
Y
γh d′d′
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
� Use of Design Aids: Graph A.5 toA.16 (Nilson)
� Calculate γ = (h − 2 × d′) / h, selectthe relevant interaction diagram.
� Given Pu, e, Ag, fy, and fc′
� Calculate Kn = Pu/(Φfc′Ag)
� Calculate Rn = Pue/( Φfc′Agh)
� From the values of Kn & Rn, find ρfrom the graph as shown.
� Ast = ρAg
112
Reference: Design of Concrete Structures 13th Ed. by Nilson, Darwin and Dolan.
Interaction Diagram
Rn
Kn
57
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Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 113
� Example: Using design aids, design a 12″ square column to
support factored load of 145 kip and a factored moment of 40
kip-ft. The material strengths are fc′ = 4 ksi and fy = 60 ksi.
12″
12″
Interaction Diagram
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 114
� Solution: Design Aids (using fc′ = 4 ksi andfy = 60 ksi)
� With d′ = 2.5 in, γ = (12 − 2 × 2.5)/12 = 0.60.
� Kn = Pu/(Φfc′Ag) = 145/(0.65 × 4 × 144) = 0.40
� Rn = Pue/( Φfc′Agh) = (40 × 12)/ (0.65 × 4 ×144 × 12) = 0.11
� ρ = 0.007
� Ast = 0.007 × 144 = 1.0 in2. < 1 % of Ag =1.44
� Using #6 bar,
No. of bars = Ast/Ab = 1.44/0.44 ≈ 4 bars
Interaction Diagram
58
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 115
Reinforced Concrete Members
subjected to Axial Compressive
Load with Biaxial Bending
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Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 116
Contents
� Behavior of Columns subjected to Biaxial Bending
� Difficulties in Constructing Biaxial Interaction Surface
� Approximate Method for Converting Biaxial case to Uniaxial case
� Bresler’s Approximate Methods for Design of Biaxial Columns
� Reciprocal Load Method
� Load Contour Method
� Circular Columns
59
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Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 117
� Column section subjected to
compressive load (Pu) at
eccentricities ex and ey along
x and y axes causing
moments Muy and Mux
respectively.
Behavior of Columns subjected to
Biaxial Bending
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 118
� The biaxial bending
resistance of an axially
loaded column can be
represented as a surface
formed by a series of
uniaxial interaction curves
drawn radially from the P
axis.
Behavior of Columns subjected to
Biaxial Bending
60
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 119
� (a) uniaxial bending
about y axis.
� (b) uniaxial bending
about x axis.
� (c) biaxial bending
about diagonal axis.
Behavior of Columns subjected to
Biaxial Bending
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 120
� Force, strain and stress
distribution diagrams for
biaxial bending
Behavior of Columns subjected to
Biaxial Bending
61
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 121
� The triangular or
trapezoidal compression
zone.
� Neutral axis, not in
general, perpendicular to
the resultant eccentricity.
Difficulties in Constructing Biaxial
Interaction Surface
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 122
� For rectangular sections with
reinforcement equally distributed on
all faces.
� Biaxial demand can be converted to
equivalent uniaxial demand using
following equations: (reference PCA)
� Mnxo = Mnx + Mny (h/b)(1 – β)/β
� Mnyo = Mny + Mnx (b/h)(1 – β)/β
Approximate Method for Converting
Biaxial Case to Uniaxial Case
62
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 123
� 0.55 ≤ β ≤ 0.7
� A value of 0.65 for β is generally a good
initial choice in a biaxial bending
analysis.
� For a value of β = 0.65, the equations
can be simplified as below:
� Mnxo = Mnx + 0.54Mny (h/b)
� Mnyo = Mny + 0.54 Mnx(b/h)
� Pick the larger moment for onward
calculations
Approximate Method for Converting
Biaxial Case to Uniaxial Case
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 124
� Design Example
� Using equations for converting bi-axial column to uni-axial
column, design a 12″ square column to support factored load of
190 kip and factored moments of 35 kip-ft about x axis and 50
kip ft about y axis. The material strengths are fc′ = 4 ksi and
fy = 60 ksi.
Approximate Method for Converting
Biaxial Case to Uniaxial Case
b=12″
h =12″
X
Y
63
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 125
� Design Example
� Solution:
� Assuming compression controlled behavior (Φ = 0.65), the requirednominal strengths are:
� Mnx = Mux/ Φ = 35/ 0.65 = 53.84 ft-kip
� Mny = Muy/ Φ = 50/ 0.65 = 76.92 ft-kip
� Mnxo = Mnx + 0.54Mny (h/b)
= 53.84 + 0.54 × 76.92 × 1 = 95 ft-kip
Similarly,
� Mnyo = Mny + 0.54 Mnx (h/b)=76.92+0.54×53.84 × 1 = 105.9 ft-kip
� Muy = 0.65 × 105.9 = 68.84 ft kip. The biaxial column can now bedesigned as an equivalent uni-axial column with moment about y-axis.
Approximate Method for Converting
Biaxial Case to Uniaxial Case
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 126
� Design Example
� Solution:
� Note: In the original equations developed by PCA, they have used
nominal values of moments because the resultant Moment was
supposed to be used on the nominal interaction diagram. However if
we have factored interaction diagram, the equation can be directly
applied on factored moments without any difference in the final
output, as follows:
� Mux = 35, Muy= 50 ;
� Mu = Mux + 0.54Muy (h/b) = 35 + 0.54 × 50 = 62 ft-kip
� Mu = Muy + 0.54Mux (h/b) = 50 + 0.54 × 35 = 68.9 ft-kip
Approximate Method for Converting
Biaxial Case to Uniaxial Case
64
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 127
� Design Example
� Solution:
� Pu = 190 kip and Mu = 68.84 fi-kip
� With 2.25 in. d′, γ = (12 − 2 × 2.25)/12 = 0.63
≈ 0.60.
� Kn = Pu/(Φfc′Ag) = 190/(0.65 × 4 × 144) = 0.51
� Rn = Pue/( Φfc′Agh) = 68.84 × 12/ (0.65 × 4 ×
144 × 12) = 0.18
� From the graph, with the calculated values of
Kn and Rn, ρg = 0.031. Thus,
� Ast = 0.031 × 144 = 4.46 in2.
� Using #6 bar, # of bars = Ast/Ab = 4.46/ 0.44 =
10.33 ≈ 12 bars
Approximate Method for Converting
Biaxial Case to Uniaxial Case
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 128
• Design ExampleSolution:
Alternatively, we can design the Column from the uniaxial interaction
diagram developed for 12 x 12 inch column having 12 no. 6 bars, fc′ = 4 ksi
and fy = 60 ksi. The red dot shows that column is safe for the given values
of Pu = 190 kips and Mu = 68.9
Approximate Method for Converting
Biaxial Case to Uniaxial Case
65
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 129
� Reciprocal Load Method
� For Pn ≥ 0.1fc′Ag
Where Pn = Pu/ Ф
� Load Contour Method
� For Pn < 0.1fc′Ag
Where Pn = Pu/ Ф
Bresler’s Approximate Methods for
Design of Biaxial Columns
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 130
� Bresler's reciprocal load equation derives from the geometry of the
approximating plane. It can be shown that:
� {(1/Pn) = 1/ (Pnxo) +1/ (Pnyo) − (1/Pno)
� If ФPn ≥ Pu O.K.
Where,
� Pn = approximate value of nominal load in biaxial bending witheccentricities ex and ey.
� Pnyo = nominal load when only eccentricity ex is present (ey = 0),
� Pnxo = nominal load when only eccentricity ey is present (ex = 0),
� Pno = nominal load for concentrically loaded column
Reciprocal Load Method
66
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 131
� Steps
� Step 1:
� Mnx = Mux/Ф
� Mny = Muy/Ф
� Check if Pn ≥ 0.1 fc′Ag
� Reciprocal Load Method
applies
� Step 2:
� γ= (h−2d′)/h
� Assuming As, ρ = As/ bh
� Pno can be determined
Reciprocal Load Method
Pno ρ
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 132
� Steps� Step 3:
� ex/h = (Mny/Pn)/ h
� Pnxo can be determined
� Step 4:
� ey/b = (Mnx/Pn)/ b
� Pnyo can be determined
� Step 5:
� Using the equation;
{(1/Pn) = 1/ (Pnxo) +1/ (Pnyo) − (1/Pno)
� If ФPn ≥ Pu O.K.
� Note: All values determined from graph shall be
multiplied with fc′Ag
Reciprocal Load Method
Pnxo
Pno
ex/h
ρ
ey/b
Pnyo
67
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 133
� Design Example
� Using Reciprocal Load Method, design a 12″ square column to
support factored load of 190 kip and factored moments of 35 kip-ft
each about x and y axis respectively. The material strengths are fc′ =
4 ksi and fy = 60 ksi.
Reciprocal Load Method
b=12″X
Y
h =12″
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 134
� Design Example
� Solution:
� Design using approximate methods (Reciprocal Load Method):
� Given demand: Mnx = Mux/Ф = 35/0.65 = 53.84 ft-kip
� Mny = Muy/Ф = 35/0.65 = 53.84 ft-kip;
� Pu = 190 kips
� Check if Pn ≥ 0.1 fc′Ag
� Pn = 190/ 0.65 = 292.31 kip
� 0.1fc′Ag = 0.1 × 4 × 12 × 12 = 57.6 kip
� As Pn > 0.1 fc′Ag, therefore reciprocal load method applies.
Reciprocal Load Method
68
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 135
Reciprocal Load Method
� Design Example
� Solution:
� With d′=2.5 in., γ= (12 − 2 × 2.5)/12
= 0.60; Graph A.5 of Nilson 13th Ed
applies
� Assuming the column to be
reinforced with 4 #6 bars, therefore,
ρ = As/ bh = 4 × 0.44/ (12 × 12)
= 0.012
� Pno/fc′Ag = 1.09
Pno = 1.09 × fc′Ag
Pno = 1.09 × 4 ×144= 628 kips
Pno
ρ =0.012
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 136
Reciprocal Load Method
� Design Example
� Solution:
� Consider bending about Y-axis
� ex/h = 0.18
� Kn = 0.68
� Pnyo/fc′Ag = 0.68
� Pnyo = 0.68 × fc′Ag
Pnyo = 0.68 × 4 ×144= 391
kips
Pno ρ
Pnyo
ex/h
69
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 137
Reciprocal Load Method
� Design Example
� Solution:
� Consider bending about X-axis
� ey/b = 0.18
� Kn = 0.68
� Pnxo/fc′Ag = 0.68
� Pnxo = 0.68 × fc′Ag
Pnxo = 0.68 × 4 ×144= 391
kips
Pno ρ
Pnxo
ey/b
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Reciprocal Load Method
138
� Design Example
� Solution:
� Design using approximate methods (Reciprocal Load Method):
� Now apply reciprocal load equation,
� (1/Pn) = 1/ (Pnxo) +1/ (Pnyo) − 1/ ( Pno)
� (1/Pn) = 1/ (391) +1/ (391) − 1/ (628) = 0.00372
� Pn = 284 kip, and the design load is:
� ΦPn = 0.65 × 284 = 184 kips ≈ 190 kips, O.K.
70
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Reciprocal Load Method
139
� Design Example
� Solution:
� Instead of using Nelson charts, the Interaction diagramdeveloped earlier for 12 x 12 inch column with 4 no 6 bars isused in the next steps of Reciprocal Load Method. .
� Pn = 190/ 0.65 = 292.31 kip
� Mnx = Mny = 53.84 ft-kip
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
� Solution:
� Design using Approximate methods:
140
This interaction curve is for both x and y axes as the column is square
Reciprocal Load Method
71
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 141
� Design Example
� Solution:
� Design using Reciprocal Load Method :
� From nominal interaction curve,
� Pno = 590 kip,
� For Mnx = 53.84 ft-kip, Pnxo = 450 kip
� For Mny = 53.84 ft-kip, Pnyo = 450 kip
� Now apply reciprocal load equation,
� (1/Pn) = 1/ (Pnxo)+1/ (Pnyo)−1/ ( Pno)= 1/ (450) +1/ (450) − 1/ (590) = 0.00285
� Pn = 344.50 kip, and the design load is:
� ΦPn = 0.65 × 344.50 = 223.92 kips > 190 kips, O.K.
Reciprocal Load Method
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Reciprocal Load Method
142
� Design Example
� Solution:
� Designing the same column by
converting bi-axial case to uni-
axial case.
� Mu = 35 + 0.54 *20 = 53.9kip-ft
� Pu = 190 kip
� Interaction diagram for 12 x 12
inch column with 4 no 6 bars is
given in the figure.
� The blue dot shows that column
is safe under the given demand.
Nominal Interaction
Curve
Design Interaction
Curve0.80φPo
050
100150200250300350400450500550600650
0 20 40 60 80 100
P (
kip)
M (kip-ft)
72
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Load Contour Method
143
� The load contour method is based on representing the failuresurface of 3D interaction diagram by a family of curvescorresponding to constant values of Pn.
� (Mnx/Mnxo)α1 + (Mny/Mnyo)α2 ≤ 1
� Where,
� Mnx = Pney ; Mnxo = Mnx (when Mny = 0),
� Mny = Pnex ; Mnyo = Mny (when Mnx = 0)
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
� When α1 = α2 = α, the shapes of such
interaction contours are as shown for
specific α values. For values of Mnx/Mnx
and Mny/Mny , α can be determined
from the given graph.
144
Load Contour Method
73
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Load Contour Method
145
� Calculations reported by Bresler indicate that α falls in the range
from 1.15 to 1.55 for square and rectangular columns. Values
near the lower end of that range are the more conservative.
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Load Contour Method
146
� Steps:
� Step 1:
� Mnx = Mux/Ф
� Mny = Muy/Ф
� Check if Pn < 0.1 fc′Ag
� Load contour method applies
� Step 2:
� γ = (h− 2d′)/h
� Assuming As, ρ = As/ bh
ρ
74
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Load Contour Method
147
� Steps:
� Step 3:
� ex/h = (Mny/Pn)/ h
� Mnyo can be determined
� Step 4:
� ey/b = (Mnx/Pn)/ b
� Mnxo can be determined
� Step 5:
� (Mnx/Mnxo)α1 + (Mny/Mnyo)α2 ≤ 1
� Note: All values determined from graph should be multiplied with
fc′Agh Mnxo
ρ ey/b
Mnyo
ex/h
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Load Contour Method
148
� Design Example
� Using Load Contour Method, design a 12″ square column to
support factored load of 30 kip and factored moments of 20 kip-ft
each about x axis and 30 kip-ft about y axis. The material strengths
are fc′ = 4 ksi and fy = 60 ksi.
b=12″
h =12″
X
Y
75
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Load Contour Method
149
� Design Example
� Solution:
� Design using Load Contour Method:
� Given demand: Mnx = Mux/Ф = 20/0.65 = 30.76 ft-kip
� Mny = Muy/Ф = 30/0.65 = 46.15 ft-kip;
� Pn =Pu/Ф = 30/ 0.65 = 46.15 kips
� Check if Pn < 0.1 fc′Ag
� 0.1fc′Ag = 0.1 × 4 × 12 × 12 = 57.6 kip
� As Pn < 0.1 fc′Ag, therefore load contour method applies.
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 150
Load Contour Method
� Design Example
� Solution:
� With d′=2.5 in., γ = (12 − 2
× 2.5)/12 = 0.60 (graph
A.5 of Nilson 13th Ed
applies)
� Assuming the column to
be reinforced with 4 #6
bars, then, ρ = As/ bh
= 4 × 0.44/ (12 × 12)
= 0.012
ρ
76
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 151
Load Contour Method
� Design Example
� Solution:
� Consider bending about Y-axis
� ex/h = 1
� Rn = 0.12
� Mnyo/fc′Agh= 0.12
� Mnyo = 0.12 × fc′Agh
� Mnyo = 0.12 × 4 ×144 ×12
= 830 in-kip
ρ
Mnyo
ex/h
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 152
Load Contour Method
� Design Example
� Solution:
� Consider bending about X-axis
� ey/b = 0.65
� Rn = 0.14
� Mnxo/fc′Agh= 0.14
� Mnxo = 0.14 × fc′Agh
� Mnxo = 0.14 × 4 ×144 × 12
= 968 in-kip
ρ
Mnxo
ey/b
77
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Load Contour Method
153
� Design Example
� Solution:
� Design using Load Contour Method:
� Now apply load contour equation,
� (Mnx/Mnxo)α1 + (Mny/Mnyo)α1 = 1
� For α ≈ 1.15
� (30.76×12/968)1.15+(46.15×12/830)1.15=
0.95 < 1, OK
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Load Contour Method
154
� Design Example
� Solution:
� Design by converting bi-axialcase to uni-axial case.
� Mu= 30 + 0.54 *20 = 40.8 kip-ft
� Pu = 30 kip
� Interaction diagram for 12 x 12inch column with 4 no 6 bars isgiven in the figure.
� The blue dot shows thatcolumn is safe under the givendemand.
Nominal Interaction
CurveDesign
Interaction Curve
0.80φPo
050
100150200250300350400450500550600650
0 20 40 60 80 100
P (
kip)
M (kip-ft)
78
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Circular Columns
155
� Behavior
� Strain distribution at ultimate
load.
� The concrete compression zone
subject to the equivalent
rectangular stress distribution
has the shape of a segment of a
circle, shown shaded.
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Circular Columns
156
� Design Example
� Design a circular column, using approximate methods, for a factored
load of 60 kips and a factored moment of 20 ft-kips about x axis and
30 kip-ft about y axis. The diameter of column is 16″. Material
strengths are fc′ = 4000 psi and fy = 60000 psi.
16″ diameter
79
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Circular Columns
157
� Design Example
� Solution:
� Check that which method applies?
� Pn = Pu /Ф = 60/0.65 = 90.30 kips
� Mnx = Mux /Ф = 20/0.65 = 30.76 ft-kips
� Mny = Muy /Ф = 30/0.65 = 46.15 ft-kips
� Check if Pn ≥ 0.1 fc′Ag
� 0.1fc′Ag = 0.1 × 4 × π × 162/4= 80.42 kip; 92.30 kip > 80.42 kip
� Therefore, reciprocal load method applies.
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
� Design Example
� Solution
� With d′=2.5 in., γ = (16 − 2 ×
2.5)/16 = 0.70 (graph A.5 of
Nilson 13th Ed applies)
� Take 6 #6 bars, ρ =As/(Ag) =
(6 × 0.44)/(π × 162/4) = 0.013
� Pno/fc′Ag = 1.04
� Pno = 1.04 × fc′Ag
� Pno= 1.04 × 4 × 201= 836 kips
158
Circular Columns
ρ
Pno
80
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
� Design Example
� Solution
� Consider bending about Y-axis
� ex/d = 0.75
� Kn = 0.15
� Pnyo/fc′Ag = 0.15
� Pnyo = 0.15 × fc′Ag
� Pnyo = 0.15 × 4 × 201= 121 kips
159
Circular Columns
ρ
Pno
ex/d
Pnyo
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
� Design Example
� Solution
� Consider bending about X-axis
� ey/d = 0.50
� Kn = 0.25
� Pnxo/fc′Ag = 0.25
� Pnxo = 0.25 × fc′Ag
� Pnxo = 0.25 × 4 × 201= 201 kips
160
Circular Columns
ey/dPno ρ
Pnxo
81
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Circular Columns
161
� Design Example
� Solution:
� Apply reciprocal load equation:
� (1/Pn) = 1/ (Pnxo) +1/ (Pnyo) − 1/ ( Pno)
� (1/Pn) = 1/ (201) +1/ (121) − 1/ (836) = 0.0012
� Pn = 83 kip, and the design load is:
� ΦPn = 0.65 × 83 = 54 kips ≈ 60 kips, O.K.
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
References
162
� Design of Concrete Structures (13th Ed.) by Nilson, Darwin and
Dolan.
� Reinforced Concrete - Mechanics and Design (4th Ed.) by James
MacGregor.
� ACI 318.
� PCA notes 2002
82
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
The End
163
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