samir shehada_ aci 05_rc design 2

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ECIV 4316 Design of Reinforced Concrete Structures (II)

1st Semester, 2013-2014 The Islamic University of Gaza

Instructors Name: Prof. Samir M. Shihada

E-Mail: [email protected]

Office: Administration Building, Room B243

Office Phone: 2860700 - Ext. No. 2814

Office Hours: 11:00-12:00, Saturdays through Wednesdays Course Description:

Design for torsion, beam-columns, two-way slabs, serviceability requirements, and design of some shallow footing systems.

Objectives:

1. Students will learn advanced design topics that are essential for reinforced concrete design of structures.

2. Students will ultimately learn how to use these design skills in analyzing and

designing a comprehensive reinforced concrete project.

Instructional Methods:

1. Three lecture hours per week covering theoretical background and solution of numerical examples.

2. One discussion hour per week focusing mainly on the comprehensive design project. This design project will be completed throughout the semester. Design groups will be assigned to work on the project by the teaching assistant.

Textbooks: 1. Reinforced Concrete Design , by S.M. Shihada, 3rd Ed., (first draft).

References:

2. Building Code Requirements for Reinforced Concrete, ACI 318-08, Farmington Hills, MI, USA.

3. Reinforced Concrete, Mechanics and Design, by J.G. MacGregor and J.K. Wight, 4th Ed., Prentice Hal, 2005.

4. Design of Reinforced Concrete , by J.C. McCormac and J. Nelson, 6th Ed., Wiley, 2005.

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Course Outline:

1. Design of Beam-Columns: 1.1 Short columns 1.2 Slender columns 1.3 Joints of moment resisting frames

2. Design for Torsion

3. Design of Two-Way Slabs: 3.1 Coefficient Methods 3.2 Direct Design Method

4. Serviceability Requirements 4.1 Deflection 4.2 Crack width control 4.3 Skin reinforcement

5. Design of Some Footing Systems:

5.1 Eccentrically loaded isolated footings 5.2 Wall footings 5.3 Combined footings 5.4 Continuous footings 5.5 Strap footings

6. Applications

6.1 Comprehensive design project

Attendance:

Regular attendance is encouraged for maintaining pace with the lectures. Grading Policy: The students will be evaluated by two examinations and an assigned comprehensive design project. The final grades for this course will be based on the following tentative table: Midterm Exam. 25 % Final Exam. 55 % Project 20 % Total 100 %

CHAPTER NINE COLUMNS 1

9 CHAPTER 9: COLUMNS

9.1 First-Order versus Second-Order Analysis

A first-order analysis is based on the initial geometry of the structure, assuming elastic behavior. On the other hand, second-order analysis considers the influence of lateral drift, cracking, member curvature, shrinkage and creep on the forces in the structure.

9.2 Sway and Nonsway Frames

9.2.1 Nonsway Frames

Structural frames whose joints are restrained against lateral displacement by attachment to rigid elements or bracing are called nonsway frames, shown in Figure 9.1.

According to ACI Code 10.10.5.1 a column in a structure is nonsway if the increase in column end moments due to second-order effects does not exceed 5 percent of the first-order end moments. Moreover, ACI Code 10.10.5.2 assumes a story within a structure is nonsway if:

cus

ulV

PQ = oD (9.1)

is less than or equal to 0.05, where Q is the stability index which is the ratio of secondary

moment due to lateral displacement and primary moment, uP is the total factored vertical load in the story, usV is the factored horizontal story shear, cl is length of column

measured center-to-center of the joints in the frame, and oD is the first-order relative

deflection between the top and bottom of that story due to usV .


Figure 9.1: Nonsway frame

9.2.2 Sway Frames

Structural frames, not attached to an effective bracing element, but depend on the bending stiffness of the columns and girders to provide resistance to lateral displacement are called sway frames, shown in Figure 9.2.

Figure 9.2: Sway frame

9.3 Effective Length Factor k of Columns of Rigid Frames

The effective length of column is the length of a column hinged at both ends and having the same buckling load as the column being considered. Thus, the effective length factor k, is the ratio of the effective length to the original length of column.


The ACI Code R10.10.6.3 recognizes the Jackson and Moreland Alignment Charts, shown in Figure 9.3, to estimate the effective length factor k for a column of constant cross section in a multibay frame. The effective length factor k is a function of the relative stiffness at each end of the column. In these charts, k is determined as the intersection of a line joining the values of y at the two ends of the column. The relative stiffness of the beams and

columns at each end of the column y is given by Eq. (9.2)

=

bbb

ccc

lIElIE

//

y (9.2)

where,

cl = length of column center-to-center of the joints

bl = length of beam center-to-center of the joints

cE = modulus of elasticity of column concrete

bE = modulus of elasticity of beam concrete

cI = moment of inertia of column cross section about an axis perpendicular to the plane of

buckling being considered.

bI = moment of inertia of beam cross section about an axis perpendicular to the plane of

buckling being considered.


(a) (b)

Figure 9.3: Alignment chart; (a) nonsway frames; (b) sway frames

indicates a summation of all member stiffnesses connected to the joint and lying in the plane in which buckling of the column is being considered.

Consider the two-story frame shown in Figure 9.4. To determine the effective length factor k for column EF,

Figure 9.4: Two-story frame


)/()/()/()/(

21

21

LIELIEHIEHIE

EHbBEb

EFcDEcE +

+=y

and

)/()/()/(

21

2

LIELIEHIE

FIbCFb

EFcF +

=y

ACI Code 10.10.6.3 specifies that for columns in nonsway frames, the effective length factor k should be taken as 1.0, unless analysis shows that a lower value is justified.

The ACI Code 10.10.4.1 specifies that the influence of cracking along the length of the member, presence of axial loads, and effects of duration of loads be taken into consideration when calculating k by using reduced values of moment of inertia as follows:

Beams -------------------------------- gI35.0

Columns------------------------------ gI70.0

Uncracked Walls -------------------- gI70.0

Cracked Walls ----------------------- gI35.0

where gI is the gross moment of inertia.

As an alternate to using alignment charts to determine k, the following simplified equations are used for computing the effective length factors for nonsway and sway frame members.

For columns in nonsway frames, the smaller of Eq. (9.3) and Eq. (9.4) is used

( ) 0.105.07.0 ++= BAk yy (9.3)

0.105.085.0 min += yk (9.4)

where Ay and By are the values of y at the two ends of the column, and miny is the

smaller of the two values.

For columns in sway frames restrained at both ends, k is taken as

for 2


mk y+= 19.0 (9.6)

where my is the average of y values at the two ends of the column.

For columns in sway frames hinged at one end, k is taken as

y3.00.2 +=k (9.7)

where y is the values at the restrained end of the column.

9.4 The ACI Procedure for Classifying Short and Slender Columns According to ACI Code 10.10.1, columns can be classified as short when their effective slenderness ratios satisfy the following criteria:

For nonsway frames

( ) 0.40/1234 21 - MMrlk u (9.8)

or

For sway frames 22/ rlk u (9.9)

Furthermore, compression members are considered braced against sidesway when bracing elements have a total stiffness, resisting lateral movement of that story, of at least 12 times the gross stiffness of the columns within the story.

where k = effective length factor

ul = unsupported length of member, defined in ACI Code 10.10.1.1 as clear distance between floor slabs, beams, or other members capable of providing lateral support, as shown in Figure 9.5.

Figure 9.5: Unsupported length of member


r = radius of gyration associated with axis about which bending occurs. For rectangular cross sections r = 0.30 h, and for circular sections, r = 0.25 h as specified by ACI Code

10.10.1.2.

h = column dimension in the direction of bending.

1M = smaller factored end moment on column, positive if member is bent single curvature,

negative if bent in double curvature.

2M = larger factored end moment on column, always positive.

Chart 9.1 summarizes the process of column design as per the ACI Code.

Chart 9.1: Column Ddesign

Example (9.1): The frame shown in Figure 9.6 consists of members with rectangular cross sections, made of the same strength concrete. Considering buckling in the plane of the figure, categorize column bc as long or short if the frame is:

Column Design

Sway frame Non-sway frame

22r

klu

10022 r

klu

)M/M(1234r

kl21

u -

->

2

1uMM

1234r

kl100

100>r

klu

Neglect slenderness

(short)

Moment magnification

method (slender)

Exact P-D analysis (slender)


a. Nonsway

b. Sway.

Figure 9.6: Frame and loads on column bc

Solution: a. Nonsway: For a column to be short,

( ) 0.40/1234 21 - MMrlk u

cmlu 3403030400 =--=

k is conservatively taken as 1.0.

( )( ) 38.32353.0

3401/ ==rlk

( ) ( ) 38.3240 astaken 1.4240/271234/1234 21 >=--=- MM i.e., column is classified as being short.

b. Sway:

The column is classified as being short when 22/ rlk u

[ ( )( ) ( )][ ( )( ) ( )] [ ( )( ) ( )]

406.075012/603035.090012/603035.0

40012/35307.033

3

=+

=cy


[ ( )( ) ( ) ] [ ( )( ) ( )][ ( )( ) ( )] [ ( )( ) ( )]

945.075012/603035.090012/603035.045012/40307.040012/35307.0

33

33

=++

=by

Using the appropriate alignment chart, k = 1.14, and ( )( ) 2291.36353.034014.1

>==rlk u

i.e., column is classified as being slender, or long.

y could have been evaluated using Eq. (9.5)

675.02

945.0406.0=

+=my

25.1675.0120

675.020=+

-=k

9.5 Short Columns Subjected to Axial force and Bending

Generally, columns are subjected to axial forces in addition to some bending moments. These moments are generally due to:

Unsymmetrically placed floors, as shown in Figure 9.7.

Figure 9.7 Unsymmetrically placed floors

Lateral loading such as wind or earthquake loads, shown in Figure 9.8.


Figure 9.8: Rigid Frame

Loads from eccentric loading such as crane loads acting on corbels.

End restraints resulting from monolithic action between floor beams and columns.

Accidental eccentricity resulting from column misalignment or other execution deficiencies.

9.5.1 Interaction Diagram

Unlike pure axial or bending loading, where unique strength exists for a particular section, combined axial load and bending result in an infinite number of strength combinations evaluated by using equilibrium equations and compatibility of strains. When these strength combinations are plotted, a strength curve called Interaction Diagram is produced. To plot an interaction diagram for a particular cross section, at least five strength combinations are required, as shown in Figure 9.9. All strength combinations located in the area under the curve are possible safe strength capacities while combinations located outside the curve are failure cases.

Point A:

This point on the curve represents pure axial compression capacity of the column cross section, where the eccentricity e is equal to zero. The nominal axial capacity nAP is given

by Eq. (9.10)

( ) ysgsggcnA fAAAfP +-= 85.0 (9.10)

where sgA is total column reinforcement.


Figure 9.9: Strength interaction diagram

Point C:

This point on the curve represents pure flexural capacity of the column cross section analyzed as doubly reinforced section, where the eccentricity e is equal to infinity. The nominal flexural capacity nCM is given by Eq. (9.11)

( ) ( )ddCadCM scnC -+-= 2/ (9.11)

Point B:

This point is characterized by its maximum bending strength and represents a balanced failure of the column section. Crushing of the concrete occurs simultaneously with yielding of the reinforcement, or

003.0== cuc ee , and yt ee = .

Point D:


Points along the curve between points A and B are characterized by compression failure of the section. Failure is initiated by crushing of the concrete before the initiation of yielding of the reinforcement, or

003.0== cuc ee , and yt ee < .

The eccentricity e is smaller than the eccentricity at balanced failure be , where the eccentricity increases when moving from point A towards point B along the interaction curve.

Point E:

Points along the curve between points B and C are characterized by tension failure of the section. Failure is initiated by yielding of the reinforcement, or

003.0== cuc ee , and 005.0t e .

The eccentricity e is larger than the eccentricity at balanced failure be , where the eccentricity increases when moving from point B towards point C along the interaction curve.

Point F:

This point on the curve represents pure axial tension capacity of the column cross section where the eccentricity e is equal to zero. The nominal axial capacity nFP is given by Eq.

(9.12)

ysgnF fAP = (9.12)

where sgA is total column reinforcement.

9.5.2 Modes of Failure

Three modes of failure are possible for columns subjected to axial force plus bending.

9.5.2.1 Balanced Failure

Figure 9.10 shows a rectangular section subjected to an axial load with eccentricity adjusted so as to produce a balanced failure using the principle of static equivalence.


Figure 9.10: Balanced failure

The distance to the neutral axis from the extreme compression fiber is given as

df

xy

b

+=

61206120 (9.13)

The compressive force resisted by concrete is given by

bxfC bccb 185.0 b=

The tensile force resisted by reinforcement in the tension side is

ysb fAT =

The compressive force resisted by reinforcement on the compression side of the cross section is

( )cyssb ffAC -= 85.0


In the last of the previous equations, it is assumed that compression reinforcement does

yield, or ys ee . This can be easily checked from similar triangles, or

-=

b

bs x

dx003.0e

If yielding of the reinforcement does not occur, where ys ee


Figure 9.11: Tension failure

The compressive force resisted by concrete is given by bxfC cc 185.0 b=

where dx

005.0003.0003.0

=+

and d375.0x =

The tensile force resisted by reinforcement in the tension side is ys fAT =

The compressive force resisted by reinforcement on the compression side of the cross section is ( )cyss ffAC -= 85.0 when ys ee . If yielding of the reinforcement does not occur, where ys ee


Substituting ,, sc CC and T in Eq. (9.17), one gets

( ) yscyscn fAffAbxfP --+= 85.085.0 1b (9.18) From equilibrium of moments,

( ) ( ) ( )'ddC2/adC''deP scn -+-=+ is used to evaluate e

The nominal flexural strength ePM nn = .

9.5.2.3 Compression Failure

When the nominal compression strength nP exceeds the balanced nomial compression

strength nbP , or when the eccentricity e is less than be or when te at the extreme layer

of steel at the face opposite the maximum compression force is less than ye , the section is

compression controlled.

Figure 9.12 shows a rectangular section subjected to an axial load with eccentricity chosen smaller than be .

Figure 9.12: Compression failure

The compressive force resisted by concrete is given by


bafC cc = 85.0

Since the reinforcement does not yield, the tensile force resisted by reinforcement in the tension side is

( )

-=

-===

aadA

xxdEAEAfAT scusssssss 16120

bee

The compressive force resisted by reinforcement on the compression side of the cross section is

( )cyss ffAC -= 85.0 for ys ee . From equilibrium of forces in the vertical direction

TCCP scn -+= (9.19)


( ) ( )

---+=

aad

AffAbafP scyscn1612085.085.0

b (9.20)

From equilibrium of moments, by taking the moments about the axial load ( ) ( ) ( ) 02/ =++---+--- edTddedCdeadC sc

( ) ( ) ( ) 02/ =+-+++-++++- edTddedCdeadC sc (9.21)


( ) ( )( )( ) ( ) 06120

85.02/85.0

=+

--

+++--++++-

dea

adA

ddedffAdeadbaf

s

cysc

b (9.22)

Eq. (9.22) which is a cubic equation in terms of a , can be written in the following form:

023 =+++ DaCaBaA (9.23)

where,

bfA c= 425.0

( )ddebfB c -+= 85.0

( )( ) ( )( )edAdddeffAC scys ++-++-= 612085.0


( ) ( )dedAD s +-= 16120 b Eq. (9.23) can be solved using Newton-Raphson iteration method, or any available mathematical software. When this iteration method is used,

( ) DaCaBaAaf +++= 23 (9.24)

and the first derivative of this function is given by

( ) CaBaAaf ++= 23 2 (9.25)

Assume a trial value for a , named oa

The first iteration value 1a is given as

( )( )o

oo af

afaa

-=1 (9.26)

A second iteration is evaluated using 1a evaluated from Eq. (9.26). Repeat for more

iterations until you reach a converged value for a . Then, a is substituted in Eq. (9.20) to get nP . The nominal flexural strength is evaluated by multiplying nP by e .

Example (9.6): For the column cross section shown in Figure 9.13, plot a nominal strength interaction diagram, using five strength combinations at least.

Use 2/250 cmkgfc = , 2/4200 cmkgf y = , and ( ) 26 /1004.2 cmkgE s = .

F

Figure 9.13: Column cross section

Solution:


d = 55 6.25 = 48.75 cm

d = 6.25 cm

d = (55.0 6.25 6.25)/2 = 21.25 cm

( )002.0

1004.24200

6 ==ye

Point A:

This is a case of pure axial compression load, the nominal axial load is

( ) ( ) ( )[ ] tonsPn 65.565420027.3927.39355525085.010001

=+-=

Point B:

The strength combination at this point corresponds to a balanced failure.

( ) cmxb 91.2875.48420061206120

=

+

=

( ) cmab 573.2491.2885.0 ==

( )( )( ) tonsCcb 76.1821000/35573.2425085.0 ==

tonsTb 49.821000420064.19 =

=

ys ee >=

-= 0023.0

91.2825.691.28

003.0

( ) tonsC sb 31.7825085.04200100064.19

=-=

From equilibrium of forces in the vertical direction,

tonsPnb 58.17849.8231.7876.182 =-+=

From equilibrium of moments,

( ) ( )

( ) mt

Mnb

.97.6125.21100

49.82

25.2125.675.48100

31.7825.212/57.2475.48100

76.182

=+

--+--=


The eccentricity causing balanced failure meb 347.058.17897.61

==

Point E:

005.0003.0003.0

dx

+= and

( ) cm281.1875.48375.0d375.0x ===

ys 00197.0281.1825.6281.18003.0 ee @=

-=

0254.066.0

25.666.0003.0 -=

-=se

( )( )( )( ) tons57.1151000/35281.1885.025085.0Cc ==

tonsT 49.821000420064.19 =

=

( ) tonsC s 31.7825085.04200100064.19

=-= , assuming that ys ee


TCCP scn -+=

tons39.11349.8231.7857.115Pn =-+=

From equilibrium of moments,

( ) ( ) ( )'ddC2/adC''deP scn -+-=+

( ) ( ) ( )25.675.4831.782

281.1885.075.4857.11525.21e39.113 -+

-=+

and cm15.51e=

m.t97.56100

15.5139.111Mn =

=

Point D:

The eccentricity is set at 0.25 m (smaller than 0.347 m) in order to locate a compression failure strength combination.


( )( ) cmkgA /75.371835250425.0 == ( )( )( ) kgB 75.1859375.4825.21253525085.0 -=-+= ( )( )( )( ) kg.cm

C 5852781.372525.21612064.19

75.4825.2125.62525085.0420064.19=++

-++-=

( )( )( )( ) ( ) 28 .103035529.275.482525.2185.0612064.19 cmkgD -=+-= ( ) ( )823 103035529.237.585278175.1859375.3718 -+-= aaaaf ( ) 37.58527815.3718725.11156 2 +-= aaaf

Let oa = 25 cm as a first trial

( ) 75.3755138025 -=f ( ) 12.1189575025 =f

The first iteration value 1a is given as

( )( ) cmf

fa 15.28

12.1189575075.3755138025

2525

251 =+=-=

( ) 5.261941615.28 =f ( ) 1364641715.28 =f

( )( ) cmf

fa 96.27

136464175.261941615.28

15.2815.28

15.282 =-=-=

( ) 75.3723796.27 =f ( ) 1353454796.27 =f

( )( ) cmf

fa 96.27

1353454775.3723796.27

96.2796.27

96.273 =-=-=

O.K0024.085.0/96.27

25.685.0/96.27003.0 ys ee >=

-=

( )( )( ) ( )

( )tons

Pn

32.22896.27

96.2775.4885.01000612064.19

25085.042001000

64.191000/3596.2725085.0

=

-

-

-+=


( ) mtM n .08.5725.032.228 ==

Point F:

This is a case of pure axial tension load, the nominal axial load is

( ) tonsfAP ysyn 93.1641000/420027.39 === .

Point C:

This is a case of pure bending moment where the nominal flexural strength is calculated as follows:

( )( )( )( ) tonsxxCc 32.61000/3585.025085.0 ==

tonsT 49.821000420064.19 =

=

( ) tonsC s 31.7825085.04200100064.19

=-= , assuming that ys ee


049.8231.7832.6 =-+x , and x = 0.66 cm

0254.066.0

25.666.0003.0 -=

-=se , which means that the compression reinforcement is

stressed in tension, a contradiction to the equilibrium equation.

tonsxx

xCs

-=

-

-=

23.751023.11625085.025.661201000

64.19

From equilibrium of forces,

49.8223.751023.11632.6 =-+x

x

023.751533.3332.6 2 =-+ xx

( ) ( )( )( )

64.12829.141533.33

32.6223.75132.64533.33533.33 2

-=

+-=

x

x

x = 8.568 cm, and the other root is negative (rejected)


mt

M n

.472.36100

25.675.48568.8

23.751023.116

2568.885.075.48

100568.832.6

=

-

-+

-

=

Figure 9.14 shows the resulting strength interaction diagram for the given cross section.

Figure 9.14: Strength interaction diagram

9.5.3 Design Interaction Diagrams

The design interaction diagram for a tied or spirally-reinforced column is evaluated by carrying out three modifications on the nominal strength interaction curve, shown in Figure 9.15 as follows:

a. All nominal strength combinations on the curve are multiplied by the strength reduction factor F . This factor is equal to 0.65 for tied columns and 0.75 for spirally reinforced columns.

CHAPTER NINE COLUMNS

24

b. The modified axial strength in compression is reduced to account for accidental eccentricity. The magnitude of axial force evaluated in step (a) is multiplied by 0.80 in case of tied columns, and by 0.85 for spirally reinforced columns.

c. ACI Code 9.3.2.2 specifies that for sections in which the net tensile strain in the extreme tension steel at nominal strength, te is between the limits for compression-controlled

and tension-controlled sections,j shall be permitted to be linearly increased from that

for compression-controlled sections to 0.90 as te increases from the compression-

controlled strain limit to 0.005.

Figure 9.15: Adjusted interaction diagram

9.1.1 Non-dimensional Design Interaction Diagrams

Non-dimensional design interaction diagrams are independent of column dimensions. In this text, ACI 340 prepared diagrams are used in design of rectangular and circular columns ( a sample of which is shown in Figure 9.16). The vertical coordinate ( )gcnn A'f/PK = represents the non-dimensional form of the nominal axial load capacity of the section. The horizontal coordinate ( )hA'f/MR gcnn = represents the non-dimensional nominal bending moment capacity of the section. These diagrams could be used with any system of units. The strength reduction factor () was considered to be 1.0 so that the nominal values contained in the interaction diagrams could be used with any set of factors. In order to use these diagrams, it is necessary to compute the value of , which is equal to the distance from the center of the bars on one side of the column to the center of the bars on the other side of the column divided by the depth of the column, both taken in the direction of bending.


25

Figure 9.16: Nominal load-moment strength interaction diagram

Example (9.7): Design reinforcement for a 40 cm 50 cm tied column. The column, which is part of a nonsway frame (bent into single curvature), has an unsupported length of 3.0 m. It is subjected to a factored axial load of 240 tons in addition to a factored bending moment of 50 t.m at both column ends.

Use 2c cm/kg280'f = and 2

y cm/kg4200f = .

Solution: For the column to be short,


26

( ) 0.40/1234 21 - MMrlk u

( )( ) 0.20503.0

3001/ ==rlk u

( ) ( ) 0.200.2250/501234/1234 21 >=-=- MM i.e., the column is classified as being short.

hdderh bstirrup ---=

)(2)(cov2g

Assuming mm25f for bars,

( ) ( ) 75.050

5.2124250=

---=g

( )( )( ) 659.05040280

65.0/1000240A'f

PKgc

nn ===

( )( )( )( ) 275.0505040280

65.0/10000050hA'f

ePRgc

nn ===

Using strength interaction diagram L4-60.7, %.67.4g =r , while using interaction diagram

L4-60.8, %.25.4g =r .

Interpolating between the two values of g , one gets %.46.4g =r

( )( ) 2s cm2.8950400446.0A == , use mm2520 f Clear distance between bars in the short direction

( ) ( ) ( ) ( ) cm0.4cm5.25.1cm0.43

5.26154240=>=

---= (O.K)

Spacing of f 10 mm ties is the smallest of:

48 (1) = 48 cm

16 (2.5) = 40 cm

40 cm

Use f 10 mm ties @ 40 cm as shown in Figure 9.17.


27

Figure 9.17: Designed cross section

Example (9.8): Design a short, square tied column to carry the following service loads: tonsPD 75= ,

tonsPL 53= , mtM D .5.6= , and mtM L .5.3= .

Use 2c cm/kg280'f = , 2/4200 cmkgf y = , and %1=r .

Solution: ( ) ( ) tons8.174536.1752.1Pu =+=

( ) ( ) tons4.135.36.15.62.1Mu =+= In the next table, different cross sections are assumed and corresponding reinforcement ratios are then evaluated using strength interaction diagrams L4-60.6, L4-60.7 and L4-60.8.

40 cm x 40 cm 50 cm x 50 cm 60 cm x 60 cm

gc

nn A'f

PK = ( )( )( ) 60.0404028065.0/10008.174

= ( )( )( ) 384.05050280

65.0/10008.174=

( )( )( ) 267.06060280

65.0/10008.174=

hA'feP

Rgc

nn =

( )( )( )( ) 115.0404040280

65.0/1000004.13= ( )( )( )( ) 059.0505050280

65.0/1000004.13= ( )( )( )( ) 034.0606060280

65.0/1000004.13=

g ( ) ( ) 687.040

5.2124240=

--- ( ) ( ) 750.050

5.2124250=

--- ( ) ( ) 792.060

5.2124260=

---

gr 0.01 0.01 0.01


28

Choosing 40 cm 40 cm cross section.

( )( ) 2s cm0.16404001.0A == , use mm168 f . Clear distance between bars

= ( ) ( ) ( ) ( ) cm0.4cm6.15.1cm53.73

6.14134240>>=

---


48 (1) = 48 cm

16 (1.6) = 25.6 cm

40 cm

Use f 10 mm ties @ 25 cm as shown in Figure 9.18.


Example (9.9): Design a short spirally reinforced column using minimum reinforcement to support a factored axial load of 24 tons and a factored bending moment of 5 t.m. Use 2/280 cmkgfc = , and

2/4200 cmkgf y = ..

Solution: tons0.24Pu =

m.t0.5Mu =

( ) ( )D

0.21242D ---=g


29

In the next table, different cross sections are assumed and corresponding reinforcement ratios are then evaluated using interaction diagrams C4-60.6, C4-60.7 and C4-60.8.

Try a 30 cm diameter cross section.

Now, let us check the assumed value of g ,

( )( ) 22s cm06.7304/01.0A == p , use mm127 f .

Trying mm8f spirals, the pitch S is given by

( )

( ) ( )( )( )( )

cm

ff

AA

D

aS

sy

c

c

gc

s 52.3

42002801

224/304/2245.0

5.04

145.0

4

2

2=

-

=

-

=

pp

Clear pitch of spiral cmSc 72.280.052.3 =-=

Use f 8 mm spiral with a pitch of 3.5 cm as shown in Figure 9.19.


30 cm 35 cm 40 cm 45 cm

gc

nn A'f

PK = 0.16 0.12 0.09 0.07

hA'feP

Rgc

nn =

0.11 0.07 0.05 0.03

g 0.61 0.67 0.71 0.74

gr 0.01 0.01 0.01 0.01


30

9.2 Design of Long (Slender) Columns

The total moment in along column is the sum of the secondary moment caused by deformations resulting from buckling, and the primary moment caused by transverse loads and column end moments, as shown in Figure 9.20. Once the deflected shape of the column, shown in Figure 9.21, is determined, the secondary moment can be evaluated at any section. One way to determine the deflected shape is to integrate the basic equation

(a) (b)

(c) (d) (e) Figure 9.20: (a) beam-column; (b) free body diagram; (c) primary moment; (d) secondary moment; (e) total moment

( )IExM

xdyd

=22

(9.27)

The stiffness IE can not be determined accurately due to change of the modulus of

elasticity of concrete with compressive stress level and an unknown extent of cracking of the concrete sections.


31

ACI Code 10.10.1 suggests that the moments in long columns be determined using second-order analysis that accounts for the influence of cracking and creep of concrete. If second-order analysis software is not available, approximate analysis using the moment magnification method can be used. The moment magnification method is based on magnifying the primary moments, determined from first-order analysis, by a certain factor in an attempt to reach approximate value of the total moment originally caused by primary and secondary moments.

9.2.1 Moment Magnification

In his book Theory of Elastic stability, Timoshenko studied the elastic stability of structures. One of the problems he discussed was a beam-column subjected to an axial load P in addition to a concentrated load Q applied at midspan, shown in Figure 9.22. He proved that the maximum deflection maxd and the maximum bending moment maxM are given by the

following equations

( )

-= 3

3

maxtan3

48 uuu

EILQ

d (9.28)

=u

uLQM

tan4max

(9.29)

where EIPLu

2=

Figure 9.21: Deflected shape of column


32

(d)

Figure 9.22: (a) Beam-column and loads; (b) primary moment (P = 0); (c) secondary moment (Q = 0); (d) primary + secondary moments

Looking carefully at the two previous equations, one can easily notice that the maximum

deflection maxd is equal to EILQ

48

3

which is the maximum deflection due to the load Q in the

absence of P, multiplied by the factor ( )

3

tan3u

uu - which accounts for the presence of the

force P . Similarly, the maximum bending moment maxM is 4LQ , which is the maximum

moment due to the load Q in the absence of the load P, multiplied by the factor u

utan,

which accounts for the presence of the force P . These factors are called amplification factors, and their values are unity when P is equal to zero.

(a)

(b)

(c)


33

Timoshenko derived an approximate expressions for the amplification factor as crPP /1

1-

for crPP / less than 0.60, as shown in Figure 9.23.

Figure 9.23: Amplification factors

Eq. (9.28) and (9.29) can be written as

-

=crPPEI

LQ/1

148

3

maxd (9.30)

-

=crPP

LQM

/11

4max (9.31)

where crP is Eulers critical buckling load given as

( ) 22

ucr Lk

EIP = (9.32)

The previous equations can be used for beam-columns bent in single curvature in nonsway frames and with the maximum values of primary and secondary moments coincide near midspan.


34

9.2.2 ACI Moment Magnification Method for Nonsway Frames

Figure 9.24: (a) nonsway column bent in single curvature; (b) nonsway column bent in double curvature

If a beam-column in nonsway frame is loaded by unequal end moments, without transverse loads, the maximum design moment will occur either at one of the column ends when the secondary moment is small or between the two ends when the secondary moment is large, as shown in Figure 9.24. Since the maximum primary and secondary moments do not

coincide, the amplification factor crPP /1

1-

can not be applied directly. To deal with this

(a)

(b)


35

situation, the maximum end moment 2M is multiplied by an equivalent moment correcting

factor mC . This factor is used to convert 2M into an equivalent uniform moment which

gives the same total moment due to actual primary and secondary moments when

multiplied by the amplification factor crPP /1

1-

, as shown in Figure 9.25.

(a) (b) Figure 9.25: (a) Actual moments Mmax = primary moment+secondary moment; (b) Equivalent Uniform moment Mmax = CmM2

According to ACI Code 10.0.6, slender columns in nonsway frames are designed for the factored axial load uP and the factored moment amplified for the effects of member

curvature, cM where

min,2ns2nsc MMM dd = (9.33)

and

2M = columns larger-end moment, not taken less than min,2M ,where

( )hPM u 03.00.15min,2 += , where the units within the brackets are given in millimeters.

nsd = moment magnification factor for nonsway frames , given by

0.1

P75.0P

1

C

c

um

ns -

=d (9.34)

where


36

mC = factor relating actual moment diagram to an equivalent uniform moment diagram. For

members without transverse loads between the supports, mC is taken as

2

1m M

M4.060.0C += (9.35)

where 21 M/M is positive if the column is bent in single curvature, and negative if the

member is bent into double curvature.

For columns with transverse loads between supports, mC is taken as 1.0. For members for

which min,2M exceeds 2M , mC is either taken equal to 1.0, or be based on the ratio of the

computed end moments 21 M/M .

cP = Eulers critical buckling load given by

( )2u

2c

lkEIP p=

In the previous equation, the flexural stiffness EI is to account for cracking, and creep given by either of the two following equations

( )dns

sesgc

1IEIE2.0

EIb+

+= (9.36)

dns

gc

1IE4.0

EIb+

= (9.37)

cE = modulus of elasticity of concrete

sE = modulus of elasticity of steel reinforcement

gI = moment of inertia of gross concrete section about centroidal axis, neglecting

reinforcement

seI = moment of inertia of reinforcement about centroidal axis of member cross section

dnsb = creep effect factor equals the ratio of maximum factored axial sustained load to

maximum factored axial load associated with the same load combination, but shall not be taken greater than 1.0.


37

Example (9.10): Design the reinforcement for a 50 cm 55 cm column that carries a factored axial load of 250 tons, a service dead load of 72.9 tons, a smaller-end moment of 30 t.m, and a larger-end moment of 40 t.m, shown in Figure 9.26.a. The column is considered nonsway, its effective length klu = 8.0 m, and is bent into single curvature.

Use 2/280 cmkgfc = and 2/4200 cmkgf y = .

Figure 9.26.a: Given cross section

Solution: 1- Check whether the column is short or long:

( ) 48.48553.0800

==rlk u

( ) ( ) 0.2540/301234/1234 21 =-=- MM i.e., the column is classified as being long, thus columns larger-end moment needs to be magnified.

2- Evaluate the equivalent moment correction factor mC :

( ) ( ) 9.040/304.06.0/4.06.0 21 =+=+= MMCm

3- Evaluate the critical buckling load cP :

( ) 35.0250

9.722.1dns ==b

232825267128015100 kg/cm.Ec ==


38

( ) ( )( )( ) ( )

2103

10195350112

555032825267140kg.cm.

...

EI =+

=

( )( )( ) ( )

tons36.8001000800

1019.5P2

102c ==

p

4- Evaluate the magnification factor nsd :

( )

0.154.1

36.80075.02501

90.0>=

-=nsd

5- Evaluate the design moment cM :

( )mtmtM .40.875.7

100055003.015

250min,2 =>=--- O.K


48 (1) = 48 cm


39

16 (2.5) = 40 cm

50 cm

Use two sets of f 10 mm ties @ 40 cm as shown in Figure 9.26.b.

Example (9.11): Design a 7.0 m high column that carries a service dead load of 55 tons, and a service live load of 45 tons, shown in Figure 9.27.a.


Figure 9.27: (a) Column and eccentricities; (b) end moments


40

Solution: 1- Compute column end moments 1M and 2M :

( ) ( ) tons138456.1552.1Pu =+=

( ) mtM .28.8100/61381 == ( ) mtM .8.13100/101382 ==

2- Estimate the column size:

For an assumed reinforcement ratio of 1%, gA may be assumed as follows:

( )( )

( )238.952

420001.028045.01000138

45.0cm

ffPA

ygc

ug =+

=+

=r

Try a 40 cm 40 cm cross section

3- Check whether the column is short or long:

( ) 10033.58403.0700


41

( )

0.125.2

83.29375.01381

84.0ns >=

-=d

7- Evaluate the design moment maxM :

( )mtmtM .8.13.73.3

100040003.015

138min,2 =

-=d

11- Evaluate the design moment cM :


42

( )mtmtM .8.13.14.4

100050003.015

138min,2 =--- O.K.


48 (0.8) = 38.4 cm

16 (1.8) = 28.8 cm

50 cm

Use two sets of f 8 mm ties @ 25 cm as shown in Figure 9.27.c.


43

Figure 9.27.c: Designed cross section


44

9.1.1 Moment Magnification for Sway Frames

The design moments 1M and 2M at the ends of a compression member are taken as

s1sns11 MMM d+= (9.38.a)

s2sns22 MMM d+= (9.38.b)

where

nsM1 = factored end moment at the end 1M acts due to loads that cause no sway calculated

using a first-order elastic frame analysis

nsM 2 = factored end moment at the end 2M acts due to loads that cause no sway calculated

using a first-order elastic frame analysis

sM1 = factored end moment at the end 1M acts due to loads that cause substantial sway

calculated using a first-order elastic frame analysis

sM 2 = factored end moment at the end 2M acts due to loads that cause substantial sway

calculated using a first-order elastic frame analysis

sd = moment magnification factor for sway frames to reflect lateral drift resulting from lateral and gravity loads

The magnified sway moments ss Md are computed in one of three methods.

1. A second-order elastic frame analysis may be used.

2. The moment magnification factor sd is calculated from Eq. (9.39)

1

P75.0P

1

1

c

us

-=

d (9.39)

where

uP = the summation of all vertical loads in a story

cP = the summation of critical buckling loads for all sway resisting columns in a story

3. The moment magnification factor is calculated from Eq. (9.40)

11

1

-=

Qsd (9.40)


45

where Q is the stability index, defined in the previous sections.

If sd calculated from Eq. (9.40) exceeds 1.5, it has to be calculated using one of the two

methods described in 1 and 2.

When sustained lateral loads are present, I for compression members is to be divided by )1( dsb+ . The term dsb is taken as the ratio of maximum factored sustained shear within

a story to the maximum shear in that story associated with the same load combination, but shall not be greater than 1.0.

Example (9.12):

For the frame shown in Figure 9.28.a, design column BC given the following: service dead load including own weight = 6 t/m, service live load = 4 t/m, concentrated wind load at I = 3.0 t acting to the left and a concentrated wind load at H = 6.0 t acting to the left.


Note that all frame members are 30 60 cm in cross section.

Figure 8.28.a: Frame dimensions

Solution:

1- Evaluate internal forces in second story members

Effective moments of inertia are given by

( ) ( ) 43 18900012/603035.0 cmI beam == ,


46

and

( ) ( ) 43 37800012/60307.0 cmI column == The modules of elasticity of concrete is given by

2/328.25267128015100 cmKgEc ==

Using STAAD-III structural analysis software, the normal forces and bending moments for service dead load, live load and wind loads are shown in ! . .b.


47

! . .b: Normal forces and bending moments

1- Check whether columns on the second floor are sway or nonsway: Case (1): D + L + W

For this case, the horizontal displacement at point I = 6.2 cm, and at point H = 3.94 cm, as evaluated from STAAD-III.

mtWu /40.11)]4(7.1)6(4.1[75.0 =+=

tVu 825.3)]3(7.1[75.0 ==

The stability index ( ) ( ) ( )

( ) 05.0269.00.5825.3100/94.32.6204.11

>=-

=Q

i.e., second story is unbraced (sway).

Case (2): D + W

For this case, the horizontal displacement at point I = 6.21 cm, and at point H = 4.11 cm, as evaluated from STAAD-III.

mtWu /4.5)6(9.0 ==

tVu 90.3)3(3.1 ==

The stability index ( ) ( ) ( )

( ) 05.0116.00.59.3100/11.421.6204.5

>=-

=Q

i.e., second story is unbraced (sway).

2- Check whether the column is short or long:


48

For column BC,

[ ( )( ) ( )][ ( )( ) ( )]

0.4100012/603035.050012/60307.0

3

3

==Cy

[ ( )( ) ( )]( )[ ( )( ) ( )]

0.8100012/603035.0

250012/60307.03

3

==By

Using the appropriate alignment chart, k = 2.33, and

( )( ) 2295.56603.044033.2

>==rlk u , i.e column is long.

Sway and nonsway moments:

Case (1) : D + L

( ) ( ) mtM ns .77.10457.277.136.414.1 =+=

mtM s .0.0=

( ) ( ) tPu 87.7218.197.176.284.1 =+=

Case (2) : D + L + W

( ) mtM ns .58.7877.10475.0 ==

( )( ) mtM s .28.693.47.175.0 ==

( ) ( ) ( ) tPu 87.5596.07.175.087.7275.0 =+=

Case (3) : D + W

( ) mtM ns .22.3736.4190.0 ==

( ) mtM s .41.693.43.1 ==

( ) ( ) tPu 13.2796.03.176.2890.0 =+= 2/328.25267128015100 cmkgEc ==

( ) ( ) ( )( )( ) ( )210

3

.10457700685.50.0112

6030328.2526714.0cmkgEI eff =+

=

( )( )( )

tonsPcr 5.512100044033.210457700685.5

2

102

=

=p

For column EF,


49

[ ( )( ) ( )][ ( )( ) ( )]( ) 0.22100012/603035.0

50012/60307.03

3

==Fy

[ ( )( ) ( )]( )[ ( )( ) ( )]( ) 0.42100012/603035.0

250012/60307.03

3

==Ey

Using the appropriate alignment chart, k = 1.79

( )( )( )

tonsPcr 36.868100044079.110457700685.5

2

102

=

=p

For dead, live, and wind loads,

( )

19.1

5.5125.51236.86875.0)20(4.111

1=

++-

=sd

For dead plus wind loads,

( )

08.1

5.5125.51236.86875.0)20(4.51

1=

++-

=sd

3- Evaluate the magnified moments: Case (1) : D + L

mtM .77.104max =

tPu 87.72=

Case (2) : D + L + W

( ) mtM .05.8628.619.158.78max =+=

tPu 87.55=

Case (3) : D + W

( ) mtM .14.4441.608.122.37max =+=

tPu 13.27=

4- Design the reinforcement: Case (1) seems to be the most critical of the three loading cases

( ) ( )80.0

600.2124260

=---

=g


50

( )( )

24840603010008772

kg/cm..

AP

AP

g

u

g

n ===

( )( )( )

20197606030

10000077104kg/cm.

.hA

MhA

M

g

u

g

n ===

Using the interaction diagram given for 80.0=g , the reinforcement ratio 07.0=r .

( )( ) 2126603007.0 cmAs == , use mm3216 f , as shown in ! . .c. Clear distance between bars

= ( ) cmcmcm 0.45.25.16.5)8.0(2)2.3(4)1(2)4(230 >>=---- O.K. Use 4-bar bundles at the corners of the section as shown in Figure 9.28.c to satisfy clear spacing requirement.

Spacing of ties is the smallest of:

48 (1) = 48 cm

16 (2.5) = 40 cm

30 cm

Use f 10 mm ties @ 30 cm.

(c)

Figure 9.28: (continued); Designed cross section; (c) using mm3216f bars

Note that the given beam and column dimensions in this example are rather small resulting in high reinforcement ratios. The small section dimensions are advertently selected to demonstrate the design procedure of slender columns which are part of way frames.


51

9.2 Problems

P9.17.1 In Example (9.12), design columns AB and DE for ACI load combinations and check overall structural stability under gravity loading.

P9.17.2 Design a short, spirally reinforced column to support a factored axial load of 250 tons.


P9.17.3 Design a short tied column to support a factored axial load of 250 tons.

Use 2/350 cmkgfc = and 2/4200 cmkgfy = .

P9.17.4 Using the strength interaction diagrams, design a short, spirally reinforced column to support a factored axial load of 200 tons and a factored moment of 15 ton.m.


P9.17.5 Check whether the given cross section shown in Figure P9.17.5, is adequate for resisting a factored axial force of 250 tons and a factored moment of 20 ton.m.


Figure P9.17.5


52

P9.17.6 A hinged end column 6.0 m tall supports a service dead load of 80 tons, and service live load of 40 tons. The loads are applied at an eccentricity of 6 cm at the bottom and 8 cm at the top. The top eccentricity is to the right of the centerline and the bottom eccentricity is to the left of the centerline. Design a tied column, rectangular in cross section to support the given loads.


1

Example:

For the frame shown in Figure 1.a, design column BC given the following: service dead load including own weight = 6 t/m, service live load = 4 t/m, concentrated wind load at I = 3.0 t acting to the left and a concentrated wind load at H = 6.0 t acting to the left.


Note that all frame members are 30 60 cm in cross section.

Figure 1.a: Frame dimensions

Solution:

Evaluate internal forces in second story members

Effective moments of inertia are given by

( ) ( ) 43 18900012/603035.0 cmI beam == , and

( ) ( ) 43 37800012/60307.0 cmI column == The modules of elasticity of concrete is given by

2/328.25267128015100 cmKgEc ==

Using SAP-2000 structural analysis software, the normal forces and bending moments for service dead load, live load and wind loads are shown in Figure 1.b.

2

Figure1.b: Normal forces and bending moments

Check whether columns on the second floor are sway or non-sway:

3

Case (1): D + L + W (1.2 D + 1.6 W + 1.0 L)

For this case, the horizontal displacement at point I = 7.46 cm, and at point H = 5.02 cm, as evaluated from SAP-2000.

m/t20.11)4(0.1)6(2.1Wu =+=

t8.4)3(6.1Vu ==

The stability index ( ) ( ) ( )( ) 05.023.00.58.4100/02.546.7202.11Q >=-=

i.e., the second story is considered as sway.

Case (2): D + W (0.9 D + 1.6 W)

For this case, the horizontal displacement at point I = 7.43 cm, and at point H = 5.04 cm.

m/t4.5)6(9.0Wu ==

t8.4)3(6.1Vu ==



Case (3): D + W (1.2 D + 0.8 W)

For this case, the horizontal displacement at point I = 3.74 cm, and at point H = 2.50 cm.

m/t2.7)6(2.1Wu ==

t4.2)3(8.0Vu ==



Case (4): D + L (1.2 D + 1.6 L)

Due to symmetry of loading and geometry, the second story is considered as non-sway.

Check whether the column is short or long:

For column BC,

4

[ ( )( ) ( )][ ( )( ) ( )]

0.4100012/603035.050012/60307.0

3

3

==Cy

[ ( )( ) ( )]( )[ ( )( ) ( )]

0.8100012/603035.0

250012/60307.03

3

==By

Using the appropriate alignment chart (for sway frames), k = 2.33, and

( )( ) 2295.56603.044033.2

>==rlk u , i.e column is

long (applies to cases 1, 2 and 3).

For case 4, k is taken as 1.0 (non-sway)

( )( ) 44.24603.04400.1

rlk u ==

( ) ( ) m.t11.9468.276.152.412.1M2 =+=

( ) ( ) m.t89.7373.216.160.322.1M1 =+=

42.4311.9489.731234

MM1234

2

1 =

--=

- , taken as 40 ( > 24.44) and column is considered

short.

Sway and non-sway moments:

Case (1) : (1.2 D + 1.6 W + 1.0 L)

( ) ( ) m.t50.7768.270.152.412.1Mns =+=

( ) m.t81.788.46.1Ms ==

( ) ( ) ( ) t18.5516.190.195.06.175.282.1Pu =++= Case (2) : (0.9D + 1.6 W)

( ) m.t37.3752.419.0Mns ==

( ) m.t81.788.46.1Ms ==

( ) ( ) t40.2795.06.175.289.0Pu =+=

Case (3) : (1.2 D + 0.8 W)

( ) m.t82.4952.412.1Mns ==

( ) m.t9.388.48.0Ms ==

( ) ( ) t26.3595.08.075.282.1Pu =+= Case (4) : (1.2 D + 1.6 W + 1.0 L)

Not required, since the moments need not be magnified for short columns.

5

Design moments (magnified):

For cases 1, 2 and 3 2/328.25267128015100 cmkgEc ==

( ) ( ) ( )( )( ) ( )210

3

.10457700685.50.0112

6030328.2526714.0cmkgEI eff =+

=

( )( )( )

tonsPcr 5.512100044033.210457700685.5

2

102

=

=p

For column EF,

[ ( )( ) ( )][ ( )( ) ( )]( ) 0.22100012/603035.0

50012/60307.03

3

==Fy

[ ( )( ) ( )]( )[ ( )( ) ( )]( ) 0.42100012/603035.0

250012/60307.03

3

==Ey

Using the appropriate alignment chart, k = 1.79

( )( )( )

tonsPcr 36.868100044079.110457700685.5

2

102

=

=p

For case 1,

( )

19.1

5.5125.51236.86875.0)20(2.111

1s =

++-

=d

( ) m.t80.8681.719.15.77Mu =+=

t18.55Pu =

For case 2,

( )

08.1

5.5125.51236.86875.0)20(4.51

1s =

++-

=d

( ) m.t80.4581.708.137.37Mu =+=

t40.27Pu =

For case 3,

( )

11.1

5.5125.51236.86875.0)20(2.7

1

1s =

++-

=d

( ) m.t15.549.311.182.49Mu =+=

t26.35Pu =

6

For case 4,

m.t11.94Mu =

( ) ( ) t16.6516.196.175.282.1Pu =+=

Reinforcement ratios:

( ) ( )80.0

600.2124260

=---

=g

Using ACI L4-60.8 design interaction diagram, the reinforcement ratios for the four cases are tabulated.

Reinforcement ratios:

Loading case

Case (1)

Case (2)

Case (3)

Case (4)

cg

u'fA65.0

P 0.17 0.08 0.11 0.20

cg

u'fhA65.0

M 0.44 0.23 0.28 0.48

r (%) 6.25 3.25 3.75 6.75

The column is to be provided with reinforcement ratio not less than 6.75 %.

1

Moment Frames

Based on ACI 2.1, 21.1 and 21.2, Moment Frames are defined

as frames in which members and joints resist forces through

flexure, shear, and axial force. Moment frames are categorized as

follows:

Ordinary Moment Frames Concrete frames complying

with the requirements of Chapters 1 through 18 of the ACI

Code. They are used in regions of low-seismic risk.

Intermediate Moment Frames Concrete frames

complying with the requirements of 21.2.2.3 and 21.12 in

addition to the requirements for ordinary moment frames.

They are used in regions of moderate-seismic risk.

Special Moment Frames Concrete frames complying

with the requirements of 21.2 through 21.5, in addition to

the requirements for ordinary moment frames. They are

used in regions of moderate and high-seismic risks.

2

Beam-Column Joints

A- Corner Joints:

A-1 Opening: If a corner joint of a rigid frame tends to be opened by the applied moments it is called opening joint.

Measured Efficiency of Opening Joints

3

A-2 Closing: If a corner joints tends to be closed by the applied moments it is called closing joints.

B- T- Joints:

4

Exterior Beam Column Joint

C- Cross- Joints:

(a) Forces due to gravity loads (b) Forces due to lateral loads

Chapter 3

Short Column Design

By Noel. J. Everard1 and Mohsen A. Issa2 3.1 Introduction The majority of reinforced concrete columns are subjected to primary stresses caused by flexure, axial force, and shear. Secondary stresses associated with deformations are usually very small in most columns used in practice. These columns are referred to as "short columns." Short columns are designed using the interaction diagrams presented in this chapter. The capacity of a short column is the same as the capacity of its section under primary stresses, irrespective of its length. Long columns, columns with small cross-sectional dimensions, and columns with little end restraints may develop secondary stresses associated with column deformations, especially if they are not braced laterally. These columns are referred to as "slender columns". Fig. 3-1 illustrates secondary moments generated in a slender column by P- effect. Consequently, slender columns resist lower axial loads than short columns having the same cross-section. This is illustrated in Fig. 3-1. Failure of a slender column is initiated either by the material failure of a section, or instability of the column as a member, depending on the level of slenderness. The latter is known as column buckling. Design of slender columns is discussed in Chapter 4. The classification of a column as a short column or a slender column is made on the basis of its Slenderness Ratio, defined below. Slenderness Ratio: rk u /l where, l u is unsupported column length; k is effective length factor reflecting end restraint and lateral bracing conditions of a column; and r is the radius of gyration reflecting the size and shape of a column cross-section. A detailed discussion of the parameters involved in establishing the slenderness ratio is presented in Chapter 4. Columns with slenderness ratios less than those specified in Secs. 10.12.2 and 10.13.2 for non-sway and sway frames, respectively, are designed as short columns using this chapter. 1 Professor Emeritus of Civil Engineering, the University of Texas at Arlington, Arlington, Texas. 2 Professor, Department of Civil and Materials Engineering, University of Illinois at Chicago, Illinois.

Non-sway frames are frames that are braced against sidesway by shear walls or other stiffening members. They are also referred to as braced frames. Sway frames are frames that are free to translate laterally so that secondary bending moments are induced due to P- effects. They are also referred to as unbraced frames. The following are the limiting slenderness ratios for short column behavior:

Non-sway frames: )/M12(M34r

k21

u l (3.1)

Sway frames: 22r

k u l (3.2) Where the term [ )/M12(M34 21 ] 40 and the ratio 21/MM is positive if the member is bent in single curvature and negative if bent in double curvature.

Fig. 3-1 Failure Modes in Short and Slender Columns

3.2 Column Sectional Capacity In short columns the column capacity is directly obtained from column sectional capacity. The theory that has been presented in Section 1.2 of Chapter 1 for flexural sections, also applies to reinforced concrete column sections. However, column sections are subjected to flexure in combination with axial forces (axial compression and tension). Therefore, the equilibrium of internal forces changes, resulting in significantly different flexural capacities and behavioral modes depending on the level of accompanying axial load. Fig. 3-2 illustrates a typical column section subjected to combined bending and axial compression. As can be seen, different combinations of moment and accompanying axial force result in different column capacities and corresponding strain profiles, while also affecting the failure modes, i.e., tension or compression controlled behavior. The combination of bending moment and axial force that result in a column capacity is best presented by column interaction diagrams. Interaction diagrams are constructed by computing moment and axial force capacities, as shown below, for different strain profiles.

ssscn TCCCP ++= 21 (3-3)

3112 xTxCxCM sscn ++= (3-4)

Compressioncontrolled

b

h

Cross-Section

Balanced section

0.003

Tension controlledn.a.

Strain Distribution Stress Distribution

t= y

t=0.005

Tran

sition

zone

x1x2

x3

Cs1

Cs2

CcPn

Ts

Mn

Fig. 3-2 Analysis of a column section

3.2.1 Column Interaction Diagrams The column axial load - bending moment interaction diagrams included herein (Columns 3.1.1 through Columns 3.24.4) conform fully to the provisions of ACI 318-05. The equations that were used to generate data for plotting the interaction diagrams were originally developed for ACI Special Publication SP-73. In addition, complete derivations of the equations for square and circular columns having the steel arranged in a circle have been published in ACI Concrete International4. The original interaction diagrams that were contained in SP-7 were subsequently published in Special Publication SP-17A5. The related equations were derived considering the reinforcing steel to be represented as follows:

(a) For rectangular and square columns having steel bars placed on the end faces only, the reinforcement was assumed to consist of two equal thin strips parallel to the compression face of the section.

(b) For rectangular and square columns having steel bars equally distributed along all four faces of the section, the reinforcement was considered to consist of a thin rectangular or square tube.

(c) For square and circular sections having steel bars arranged in a circle, the reinforcement was considered to consist of a thin circular tube.

The interaction diagrams were developed using the rectangular stress block, specified in ACI 318-05 (Sec. 10.2.7). In all cases, for reinforcement that exists within the compressed portion of the depth perpendicular to the compression face of the concrete (a = c), the compression stress in the steel was reduced by 0.85 /cf to account for the concrete area that is displaced by the reinforcing bars within the compression stress block. The interaction diagrams were plotted in non-dimensional form. The vertical coordinate [ )/( / gcnn AfPK = ] represents the non-dimensional form of the nominal axial load capacity of the 3 Everard and Cohen. Ultimate Strength Design of Reinforced Concrete Columns, ACI Special Publication SP-7, 1964, pp. 152-182. 4 Everard, N.J., Axial Load-Moment Interaction for Cross-Sections Having Longitudinal Reinforcement Arranged in a Circle, ACI Structural Journal, Vol. 94, No. 6, November-December, 1997, pp. 695-699. 5 ACI Committee 340, Ultimate Strength Design Handbook, Volume 2, Columns, ACI Special Publication 17-A, American Concrete Institute, Detroit, MI, 1970, 226 pages.

section. The horizontal coordinate [ )/( / hAfMR gcnn = ] represents the non-dimensional nominal bending moment capacity of the section. The non-dimensional forms were used so that the interaction diagrams could be used equally well with any system of units (i.e. SI or inch-pound units). The strength reduction factor () was considered to be 1.0 so that the nominal values contained in the interaction diagrams could be used with any set of factors, since ACI 318-05 contains different factors in Chapter 9, Chapter 20 and Appendix C. It is important to point out that the factors that are provided in Chapter 9 of ACI 318-05 are based on the strain values in the tension reinforcement farthest from the compression face of a member, or at the centroid of the tension reinforcement. Code Section 9.3.2 references Sections 10.3.3 and 10.3.4 where the strain values for tension control and compression control are defined. It should be note that the eccentricity ratios ( PMhe // = ), sometimes included as diagonal lines on interaction diagrams, are not included in the interaction diagrams. Using that variable as a coordinate with either nK or nR could lead to inaccuracies because at the lower ends of the diagrams the e/h lines converge rapidly. However, straight lines for the tension steel stress ratios ys ff / have been plotted for assistance in designing splices in the reinforcement. Further, the ratio 0.1/ =ys ff represents steel strain syy Ef /= , which is the boundary point for the factor for compression control, and the beginning of the transition zone for linear increase of the factor to that for tension control. In order to provide a means of interpolation for the factor, other strain lines were plotted. The strain line for 005.0=t , the beginning of the zone for tension control has been plotted on all diagrams. For steel yield strength 60.0 ksi, the intermediate strain line for 035.0=t has been plotted. For Steel yield strength 75.0 ksi, the intermediate strain line for 038.0=t has been plotted. It should be noted that all strains refer to those in the reinforcing bar or bars farthest from the compression face of the section. Discussions and tables related to the strength reduction factors are contained in two publications in Concrete International6,7. In order to point to designs that are prohibited by ACI 318-05, Section 10.3.5, strain lines for

004.0=t have also been plotted. Designs that fall within the confines of the lines for 004.0=t and nK less than 0.10 are not permitted by ACI 318-05. This includes tension axial loads, with nK

negative. Tension axial loads are not included in the interaction diagrams. However, the interaction diagram lines for tension axial loads are very nearly linear from 0.0=nK to 0.0=nR with [ )/( / gcystn AffAK = ]. This is discussed in the next section.

6 Everard, N. J., Designing With ACI 318-02 Strength Reduction Factors, Concrete International, August, 2002, Vol. 24, No. 8, pp 91-93. 7 Everard, N. J., Strain Related Strength Reduction Factors () According to ACI 318-02, Concrete International, August, 2002, Vol. 34, No. 8, pp. 91-93.

Straight lines for maxK are also provided on each interaction diagram. Here, maxK refers to the maximum permissible nominal axial load on a column that is laterally reinforced with ties conforming to ACI 318-05 Section 7.10.5. Defining 0K as the theoretical axial compression capacity of a member with 0.0=nR , 0max 80.0 KK = , or, considering ACI 318-05 Eq. (10-2), without the factor, 80max .Pn, = [ stystg/c Af)A(Af. +850 ] (3-5) Then, g

/c A/fPK maxmax = (3-6)

For columns with spirals conforming with ACI 318-05 Section 7.10.4, values of maxK from the interaction diagrams are to be multiplied by 0.85/0.80 ratio. The number of longitudinal reinforcing bars that may be contained is not limited to the number shown in the illustrations on the interaction diagrams. They only illustrate the type of reinforcement patterns. However, for circular and square columns with steel arranged in a circle, and for rectangular or square columns with steel equally distributed along all four faces, it is a good practice to use at least 8 bars (and preferably at least 12 bars). Although side steel was assumed to be 50 percent of the total steel for columns having longitudinal steel equally distributed along all four faces, reasonably accurate and conservative designs result when the side steel consists of only 30 percent of the total steel. The maximum number of bars that may be used in any column cross section is limited by the maximum allowable steel ratio of 0.08, and the conditions of cover and spacing between bars. 3.2.2 Flexure with Tension Axial Load Many studies concerning flexure with tension axial load show that the interaction diagram for tension axial load and flexure is very nearly linear between Ro and the tension axial load value ntK , as is shown in Fig. 3-3. Here, 0R is the value of nR for 0.0=nK , and )/( / gcystnt AffAK =

Fig. 3.3 Flexure with axial tension

Design values for flexure with tension axial load can be obtained using the equations: [ ]001 /RR.KK nntn = (3-7) [ ]ntnon /KK.RR = 01 (3-8) Also, the tension side interaction diagram can be plotted as a straight line using 0R and ntK , as is shown in Fig. 3.3. 3.3 Columns Subjected to Biaxial Bending Most columns are subjected to significant bending in one direction, while subjected to relatively small bending moments in the orthogonal direction. These columns are designed by using the interaction diagrams discussed in the preceding section for uniaxial bending and if required checked for the adequacy of capacity in the orthogonal direction. However, some columns, as in the case of corner columns, are subjected to equally significant bending moments in two orthogonal directions. These columns may have to be designed for biaxial bending. A circular column subjected to moments about two axes may be designed as a uniaxial column acted upon by the resultant moment;

2ny2nxn

2uy

2uxu MMMMMM +=+= (3-9)

For the design of rectangular columns subjected to moments about two axes, this handbook provides design aids for two methods: 1) The Reciprocal Load (1/Pi) Method suggested by Bresler8, and 2) The Load Contour Method developed by Parme, Nieves, and Gouwens9. The Reciprocal Load Method is more convenient for making an analysis of a trial section. The Load Contour Method is more suitable for selecting a column cross section. Both of these methods use the concept of a failure surface to reflect the interaction of three variables, the nominal axial load Pn and the nominal biaxial bending moments Mnx and Mny, which in combination will cause failure strain at the extreme compression fiber. In other words, the failure surface reflects the strength of short compression members subject to biaxial bending and compression. The bending axes, eccentricities and biaxial moments are illustrated in Fig. 3.4.

8 Bresler, Boris. Design Criteria for Reinforced Columns under Axial Load and Biaxial Bending, ACI Journal Proceedings, V. 57, No.11, Nov. 1960, pp. 481-490. 9 Parme, A.L. Nieves, J. M. and Gouwens, A. Capacity of Reinforced Rectangular Columns Subjected to Biaxial Bending. ACI Journal Proceedings, V. 63, No. 9, Sept. 1966, pp.911-923.

xx

y

y

ex

eyPn

Mnx = Pn eyMny = Pn ex

Fig. 3.4 Notations used for column sections subjected to biaxial bending

A failure surface S1 may be represented by variables Pn, ex, and ey, as in Fig. 3.5, or it may be represented by surface S2 represented by variables Pn, Mnx, and Mny as shown in Fig. 3.6. Note that S1 is a single curvature surface having no discontinuity at the balance point, whereas S2 has such a discontinuity. (When biaxial bending exists together with a nominal axial force smaller than the lesser of Pb or 0.1 fc Ag, it is sufficiently accurate and conservative to ignore the axial force and design the section for bending only.)

Fig. 3.5 Failure surface S1 Fig. 3.6 Failure surface S2

3.3.1 Reciprocal Load Method In the reciprocal load method, the surface S1 is inverted by plotting 1/Pn as the vertical axis, giving the surface S3, shown in Fig. 3.7. As Fig. 3.8 shows, a true point (1/Pn1, exA, eyB) on this reciprocal failure surface may be approximated by a point (1/Pni, exA, eyB) on a plane S3 passing through Points A, B, and C. Each point on the true surface is approximated by a different plane; that is, the entire failure surface is defined by an infinite number of planes.

Point A represents the nominal axial load strength Pny when the load has an eccentricity of exA with ey = 0. Point B represents the nominal axial load strength Pnx when the load has an eccentricity of eyB with ex = 0. Point C is based on the axial capacity Po with zero eccentricity. The equation of the plane passing through the three points is;

onynxni P

1P1

P1

P1 += (3-10)

Where: Pni: approximation of nominal axial load strength at eccentricities ex and ey Pnx: nominal axial load strength for eccentricity ey along the y-axis only (x-axis is axis of bending) Pny: nominal axial load strength for eccentricity ex along the x-axis only (y-axis is axis of bending) Po: nominal axial load strength for zero eccentricity

Fig. 3.7 Failure surface S3,, which is reciprocal Fig. 3.8 Graphical representation of Reciprocal of surface S1 Load Method For design purposes, when is constant, the 1/Pni equation given in Eq. 3.9 may be used. The variable Kn = Pn / (f c Ag) can be used directly in the reciprocal equation, as follows:

onynxni K1

K1

K1

K1 += (3-11)

Where, the values of K refer to the corresponding values of Pn as defined above. Once a preliminary cross section with an estimated steel ratio g has been selected, the actual values of Rnx and Rny are calculated using the actual bending moments about the cross section X and Y axes, respectively. The corresponding values of Knx and Kny are obtained from the interaction diagrams presented in this Chapter as the intersection of appropriate Rn value and the assumed steel ratio curve for g. Then, the

value of the theoretical compression axial load capacity Ko is obtained at the intersection of the steel ratio curve and the vertical axis for zero Rn. 3.3.2 Load Contour Method The load contour method uses the failure surface S2 (Fig. 3.6) and works with a load contour defined by a plane at a constant value of Pn, as illustrated in Fig. 3.9. The load contour defining the relationship between Mnx and Mny for a constant Pn may be expressed nondimensionally as follows:

1MM

MM

noy

ny

nox

nx =

+

(3-12)

For design, if each term is multiplied by , the equation will be unchanged. Thus Mux, Muy, Mox, and Moy, which should correspond to Mnx, Mny, Mnox , and Mnoy, respectively, may be used instead of the original expressions. This is done in the remainder of this section. To simplify the equation (for application), a point on the nondimensional diagram Fig. 3.10 is defined such that the biaxial moment capacities Mnx and Mny at this point are in the same ratio as the uniaxial moment capacities Mox and Moy; thus

oy

ox

ny

nx

MM

MM = (3-12)

or; oynyoxnx MMandMM == (3-13)

Fig. 3.10 Load contour for constant Pn on failure surface

In physical sense, the ratio is the constant portion of the uniaxial moment capacities which may be permitted to act simultaneously on the column section. The actual value of depends on the ration Pn/Pog as well as properties of the material and cross section. However, the usual range is between 0.55 and 0.70. An average value of = 0.65 is suggested for design. The actual values of are available from Columns 3.25. The load contour equation given above (Eq. 3-10) may be written in terms of , as shown below:

1MM

MM

0.5/log log

noy

ny0.5/log log

nox

nx =

+

(3-14)

A plot of the Eq. 3-12 appears as Columns 3.26. This design aid is used for analysis. Entering with Mnx/Mox and the value of from Columns 3.25, one can find permissible Mny/Moy. The relationship using may be better visualized by examining Fig. 3.10. The true relationship between Points A, B, and C is a curve; however, it may be approximated by straight lines for design purposes. The load contour equations as straight line approximation are:

i) For ox

oy

nx

ny

MM

MM

+=

1

MM

MMMox

oynxnyoy (3-13)

ii) For ox

oy

nx

ny

MM

MM

+=

1

MMMMM

oy

oxnynxox (3-14)

For rectangular sections with reinforcement equally distributed on all four faces, the above equations can be approximated by;

+=

1

hbMMM nxnyoy (3-15)

For ox

oy

nx

ny

MM

MM or

hb

MM

nx

ny where b and h are dimensions of the rectangular column section parallel to x and y axes, respectively. Using the straight line approximation equations, the design problem can be attacked by converting the nominal moments into equivalent uniaxial moment capacities Mox or Moy. This is accomplished by;

(a) assuming a value for b/h (b) estimating the value of as 0.65 (c) calculating the approximate equivalent uniaxial bending moment using the appropriate one of

the above two equations (d) choosing the trial section and reinforcement using the methods for uniaxial bending and axial

load. The section chosen should then be verified using either the load contour or the reciprocal load method.

3.4 Columns Examples COLUMNS EXAMPLE 1 - Required area of steel for a rectangular tied column with bars on four

faces (slenderness ratio found to be below critical value) For a rectangular tied column with bars equally distributed along four faces, find area of steel. Given: Loading Pu= 560 kip and Mu= 3920 kip-in. Assume = 0.70 or, Nominal axial load Pn = 560/0.70 = 800 kip Nominal moment Mn = 3920/0.70 = 5600 kip-in. Materials Compressive strength of concrete /cf = 4 ksi Yield strength of reinforcement fy = 60 ksi Nominal maximum size of aggregate is 1 in. Design conditions Short column braced against sidesway.

Procedure Calculation ACI

318-05 Section

Design Aid

Determine column section size. Given: h = 20 in. b = 16 in.

Determine reinforcement ration g using known values of variables on appropriate interaction diagram(s) and compute required cross section area Ast of longitudinal reinforcement.

Pn= 800 kip Mn = 5600 kip-in. h = 20 in. b = 16 in. Ag = b x h = 20 x 16 = 320 in.2

A) Compute gc

n

AfP

nK '= ( )( ) 625.03204800 ==nK

B) Compute hAf

M

gc

nnR '= ( )( )( ) 22.0203204

5600 ==nR

C) Estimate h

5 - h 0.75 = 205 - 20

D) Determine the appropriate interaction diagram(s)

For a rectangular tied column with bars along four faces, /cf = 4 ksi, fy = 60 ksi, and an estimated of 0.75, use R4-60.7 and R4-60.8. For kn= 0.625 and Rn= 0.22

10.2 10.3

E) Read g for kn and Rn values from appropriate interaction diagrams

Read g = 0.041 for = 0.7 and g = 0.039 for = 0.8 Interpolating; g = 0.040 for = 0.75

F) Compute required Ast from Ast=g Ag

Required Ast = 0.040 320 in.2 = 12.8 in2

Columns 3.2.2 (R4-60.7) and 3.2.3 (R4-60.8)

COLUMNS EXAMPLE 2 - For a specified reinforcement ratio, selection of a column section size for a rectangular tied column with bars on end faces only

For minimum longitudinal reinforcement (g= 0.01) and column section dimension h = 16 in., select the column dimension b for a rectangular tied column with bars on end faces only. Given: Loading Pu= 660 kips and Mu= 2790 kip-in. Assume = 0.70 or, Nominal axial load Pn = 660/0.70= 943 kips Nominal moment Mn = 4200/0.70= 3986 kip-in. Materials Compressive strength of concrete /cf = 4 ksi Yield strength of reinforcement fy = 60 ksi Nominal maximum size of aggregate is 1 in. Design conditions Slenderness effects may be neglected because kl u/h is known to be below critical value


318-05 Section

Design Aid

Determine trial column dimension b corresponding to known values of variables on appropriate interaction diagram(s).

Pn= 943 kips, Mn = 3986 kip-in. h = 16 in. /cf = 4 ksi, fy= 60 ksi g = 0.01

A) Assume a series of trial column sizes b, in inches; and compute Ag=bh , in.2 24 384

26 416

28 448

B) Compute gc

n

AfP

nK '= ( )( )61.03844

943

= ( )( )

57.04164

943

= ( )( )

53.04484

943

=

C) Compute hAf

M

gc

nnR '= ( )( )( )16.0

1638443986

=

( )( )( )14.0

1641643986

=

( )( )( )14.0

1644843986

=

D) Estimate h

5 - h 0.7 0.7 0.7 D) Determine the appropriate interaction diagram(s)

For a rectangular tied column with bars along four faces, /cf = 4 ksi, fy = 60 ksi, and an estimated of 0.70, use Interaction Diagram L4-60.7

0.018 0.014 0.011 E) Read g for kn and Rn values For = 0.7, select dimension corresponding to g nearest desired value of g = 0.01 Therefore, try a 16 x 28-in. column

10.2 10.3

Columns 3.8.2 (L4-60.7)

COLUMNS EXAMPLE 3 - Selection of reinforcement for a square spiral column (slenderness ratio is below critical value)

For the square spiral column section shown, select reinforcement. . Given: Loading Pu= 660 kips and Mu= 2640 kip-in. Assume = 0.70 or, Nominal axial load Pn = 660/0.70= 943 kips Nominal moment Mn = 2640/0.70= 3771 kip-in. Materials Compressive strength of concrete /cf = 4 ksi Yield strength of reinforcement fy = 60 ksi Nominal maximum size of aggregate is 1 in. Design conditions Column section size h = b = 18 in Slenderness effects may be neglected because kl u/h is known to be below critical value


318-05 Section

Design Aid

Determine reinforcement ration g using known values of variables on appropriate interaction diagram(s) and compute required cross section area Ast of longitudinal reinforcement.

Pn= 943 kips Mn = 3771 kip-in. h = 18 in. b = 18 in. Ag=bh= 1818=324 in.2

A) Compute gc

n

AfP

nK '= ( )( ) 73.03244943 ==nK

B) Compute hAf

M

gc

nnR '= ( )( )( ) 16.0183204

3771 ==nR

C) Estimate h

5 - h 218

18 0.7 = 5 - D) Determine the appropriate interaction diagram(s)

For a square spiral column, /cf = 4 ksi, fy = 60 ksi, and an estimated of 0.72, use Interaction Diagram S4-60.7 and S4-60.8

For kn= 0.73 and Rn= 0.16 and, = 0.70: g = 0.035 = 0.80: g = 0.031

for = 0.72: g = 0.034

E) Read g for kn and Rn values.

Ast = 0.034 320 in.2 = 12.8 in2

10.2 10.3

Columns 3.20.2 (S4-60.7) and 3.20.3 (S4-60.8)

COLUMNS EXAMPLE 4 - Design of square column section subject to biaxial bending using resultant moment

Select column section size and reinforcement for

samir shehada_ aci 05_rc design 2

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