Lecture 03 Design of RC Members for Flexure and Axial

1

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

[email protected]

www.drqaisarali.com



Topics Addressed

� General

� Reinforced Concrete Members Subjected to Flexure Load only

� Reinforced Concrete Members Subjected to Axial Compressive

Load only


Load with Uniaxial Bending


Load with Biaxial Bending

2



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.



� 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

3


Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures

Reinforced Concrete

Members Subjected to

Flexure Load only

5



Contents

� Loading Stages Before Collapse

� Design of Solid Rectangular Members

� Design of Solid T Members

� Design of Hollow Rectangular Members

6

4



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




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.

5




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.



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

10


6



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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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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7



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


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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

14


8



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


15





• 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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9




• 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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• 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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10



� 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




� 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)


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� 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.






� 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


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� 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.





� 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


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� 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





� 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).


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� 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.




� 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.


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� 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)






� The total nominal capacity of doubly reinforced beam is therefore,

� Mn = Mn1 + Mn2 = Asfy (d – a/2) + As′fs′ (d – d′)


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� 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





� 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).


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� Maximum reinforcement


� 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.





� 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.


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� 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).






� 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.

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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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� 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″





� 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


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� 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.





� 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.


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� 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





� 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 (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.


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� Difference between T-beam and T-beam Behavior

Design of Solid T Members





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� 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




� 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


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� 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




� Effective Flange width for T and L beam (ACI 8.10)


25



� 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.




� 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″


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� 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:

� 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


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� 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)





� 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.


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� 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:

� 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


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� 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:

� 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.


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Design of Hollow Rectangular Members




� 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


31



� 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)




� 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


32



� 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″




� 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.


33




� 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





� 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


34




� 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 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


35




� 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.




Reinforced Concrete

Members Subjected to Axial

Compressive Loads

70

36



Contents

71

� Axial Capacity

� Maximum Reinforcement Ratio

� Example



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



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



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



Maximum Reinforcement Ratio

� 1 % ≤ Ast /Ag ≤ 8 %

� Practically, however reinforcement more than 6 % is seldom used.



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



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}



Reinforced Concrete Members

subjected to Axial Compressive

Load with Uniaxial Bending

40



Contents

� Behavior of Columns subjected to Uniaxial Bending

� Axial Capacity


� Design by Trial and Success Method

� Alternative Approach

� Interaction Diagram



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



� 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



� 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



� 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



� 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



� 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



� 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)


44



� 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.




� 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.


45



� 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



� 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


46



� 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)




� 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



� 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



� 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



� 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



� 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




� 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




� 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




� 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)




� 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




� 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



� 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



� 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



� 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



� 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


= 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



� 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



� 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



� 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


� = 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



� 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


= 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



� 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



� 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′



� 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



� 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



� 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



Reinforced Concrete Members

subjected to Axial Compressive

Load with Biaxial Bending



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



� Column section subjected to

compressive load (Pu) at

eccentricities ex and ey along

x and y axes causing

moments Muy and Mux

respectively.


Biaxial Bending



� 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.


Biaxial Bending

60



� (a) uniaxial bending

about y axis.

� (b) uniaxial bending

about x axis.

� (c) biaxial bending

about diagonal axis.


Biaxial Bending



� Force, strain and stress

distribution diagrams for

biaxial bending


Biaxial Bending

61



� The triangular or

trapezoidal compression

zone.

� Neutral axis, not in

general, perpendicular to

the resultant eccentricity.

Difficulties in Constructing Biaxial

Interaction Surface



� 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



� 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





� 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.



b=12″

h =12″

X

Y

63



� 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.





� 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



64



� 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





• 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



65




� 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



� 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



� Steps

� Step 1:

� Mnx = Mux/Ф

� Mny = Muy/Ф

� Check if Pn ≥ 0.1 fc′Ag


applies

� Step 2:

� γ= (h−2d′)/h

� Assuming As, ρ = As/ bh

� Pno can be determined


Pno ρ



� 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


Pnxo

Pno

ex/h

ρ

ey/b

Pnyo

67



� 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.


b=12″X

Y

h =12″



� 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


� 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.


68




� 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




� 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




� 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




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




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



� Solution:

� Design using Approximate methods:

140

This interaction curve is for both x and y axes as the column is square


71



� 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.





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



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)



� 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



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.



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



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



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



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.



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



Load Contour Method

� Design Example

� Solution:


� 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



Load Contour Method

� Design Example

� Solution:


� 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



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



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



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.



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



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


� 0.1fc′Ag = 0.1 × 4 × π × 162/4= 80.42 kip; 92.30 kip > 80.42 kip

� Therefore, reciprocal load method applies.



� 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



� Design Example

� Solution


� 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



� Design Example

� Solution


� 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



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.



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



The End

163

Documents

Lecture 03 Design of RC Members for Flexure and Axial