load combination - ubc 97 and aci

1

Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan

Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011

Lecture-15

Introduction to Earthquake

Resistant Design of RC

Structures (Part II)

By: Prof Dr. Qaisar Ali

Civil Engineering Department

UET Peshawar

[email protected]



Topics Addressed

Load Combinations

UBC-97 Load Combinations

Load Combinations in other codes

Compatibility Issue in BCP and ACI codes

Application of Load Combinations

Study on Results of Analysis using UBC and ACI Load

Combinations.

Analysis using software

2

mailto:[email protected]

2



Topics Addressed

ACI Special Provisions for Seismic Design

General Requirements

ACI Provisions for SMRF

ACI Provisions for IMRF

Miscellaneous Considerations

Example on SMRF Requirements

3




UBC-97 adopts load combinations and strength reductions

factors of ACI 318-99.

According to UBC-97, earthquake combinations shall be

multiplied by 1.1 for concrete structures.

4

Load Combinations

UBC-97 (section 1612.2.1)

Load Combinations Strength Reduction Factors

1.4D 0.9 (flexure)

1.4D + 1.7L 0.85 (Shear)

1.2D + 0.5L ± 1.0E 0.70 (Tied )

0.9D ± 1.0E 0.75 (Spiral)

3




Definition of E in UBC-97

According to section 1630.1.1 of UBC-97 (section 5.30.1.1 of BCP SP-2007),

E is given as:

E = ρEh + Eν

Eh = Horizontal component of the earthquake load (storey Force).

Eν= Vertical component of the earthquake ground motion.

In most of the case, ρ ≈ 1, so,

E = Eh + Eν

Now, Eν = 0.5CaID, therefore,

E = Eh + 0.5CaID

5

Load Combinations




Therefore, the following load combinations of UBC-97 are

generated:

1.1[1.2D + 0.5L ± 1.0 (Eh + 0.5CaID)} ]…..……(i)

1.1[0.9D ± 1.0 (Eh + 0.5CaID)] …………………(ii)

NOTE: D, L and Eh represents load effects axial force, shear, and

bending moment due to respective loads.

6

Load Combinations

4




As an example let’s write UBC Load Combinations for

following seismic zone data:

Seismic Zone: 2B

Soil type: SD

Importance factor (I)= 1

For seismic zone 2B and soil type SD, Seismic coefficient Ca = 0.28

With this data, following UBC-97 load combinations are

obtained:

7

Load Combinations




1.1[1.2D + 0.5L ± 1.0 (Eh + 0.5CaID)]…..……(i)

1.1[1.2D + 0.5L ± 1.0 (Eh + 0.5 × 0.28 × 1.00 ×D)]

1.32D + 0.55L ± 1.1Eh ± 0.154D

1.474D + 0.55L ± 1.1Eh …………….. (ia)

1.166D + 0.55L ± 1.1Eh …………….. (ib)

8

Load Combinations

5




Similarly,

1.1[0.9D ± 1.0 (Eh + 0.5CaID)] …………………(ii)

1.1[0.9D ± 1.0 (Eh + 0.5 × 0.28 × 1.00 ×D)]

0.99D ± 1.1Eh ± 0.154D

1.14D ± 1.1Eh …………….. (iia)

0.84D ± 1.1Eh ……………. (iib)

9

Load Combinations




Finally, six load combinations are used for analysis of structure

1.4D

1.4D+1.7L

1.474D + 0.55L ± 1.1Eh …………….. (ia)

1.166D + 0.55L ± 1.1Eh …………….. (ib)

1.14D ± 1.1Eh …………….. (iia)

0.84D ± 1.1Eh ……………. (iib)

10

Load Combinations

6



Load Combinations in Other Codes

ACI 318-02 & 05

1.4D

1.2D+1.6L

1.2D + 1.0L ± 1.0E

0.9D ± 1.0E

Note: This “E” must be calculated using IBC code.

BCP SP-2007

BCP has same combinations as UBC-97.

11

Load Combinations



Compatibility of BCP (UBC) and ACI

BCP chapter 7 can be used for earthquake resistant design

of RC structures using load combination and Strength

Reduction Factors of chapter 5 of BCP (UBC 97 load

combinations).

To maintain compatibility in the usage of BCP code, analysis

is done using load combinations of UBC 97. Design can be

done using:

UBC 97 design procedure of chapter 19 which is ACI 318-99.

ACI 318-05 using load combinations and strength reduction factors of

UBC 97.

12

Load Combinations

7




Following steps are followed to apply load combinations:

The structure is analyzed for unamplified load cases separately e.g.,

Analysis for unamplified dead load (1.0D),

Analysis for unamplified live load (1.0L)

Analysis for unamplified lateral storey load cases (1.0Eh).

Load effects obtained for each load case are multiplied with amplification

factors and combined as per code load combination requirements.

With this approach, the structure has to be analyzed only for each load case.

After analysis, any load combinations can be performed with load cases.

13

Load Combinations




Example 1: Apply Load combinations to analysis results of the encircled bay of

the given structure.

14

Load Combinations

20 ft 20 ft 20 ft 20 ft

15 ft

15 ft

15 ft

fc′ = 3 ksi

fy = 40 ksi

SDL = 40 psf

LL = 60 psf

SDL = 40 psf

LL = 60 psf

SDL = 40 psf

LL = 60 psf

Seismic Zone: 2B

Soil: SD

Slab: 6″

Beams:12″ × 18″

Columns: 12″

BAY

8




Example 1

Following load combinations for zone 2B and soil SD are used:

1.4D+1.7L

1.474D + 0.55L ± 1.1Eh …………….. (ia)

1.166D + 0.55L ± 1.1Eh …………….. (ib)

1.14D ± 1.1Eh …………….. (iia)

0.84D ± 1.1Eh ……………. (iib)

15

Load Combinations




Example 1

16

Load Combinations

E

-17 -17

9

-0.5

1.0

22

-22 -39

37

-37

33

1.0D 1.0L 1.0E −1.0E

39

-37

37

-33

-22

22

1.4D + 1.7L

-89 -89

52

-2.8

4.5

1.474D + 0.55L + 1.1E

-43 -90

35

-46

44

-41

35

1.166D + 0.55L − 1.1E

-79 -29

29

41

-37

41

-37

1.14D + 1.1E

-20 -68

23

-45

43

-41

36

0.84D − 1.1E

-57 -8

20

42

-38

41

-36

-89 -90 52

-46 -41

Envelop

44 -37

For higher zone, this value might

become positive

E

Bent left (+) Bent right (−)

BM Sign convention for column

Bent down (+) Bent up (−)

BM Sign convention for beam

-43 -43

26

-1.4

2.0

9




Case Study 1: Study on Results of Analysis using UBC and ACI

Load Combinations.

17

Load Combinations

20 ft 20 ft 20 ft 20 ft

15 ft

15 ft

15 ft

fc′ = 3 ksi

fy = 40 ksi

SDL = 40 psf

LL = 60 psf

SDL = 40 psf

LL = 60 psf

SDL = 40 psf

LL = 60 psf

Seismic Zone: 2B

Soil: SD

Slab: 6″

Beams:12″ × 18″

Columns: 12″

Portion of frame

considered




Case Study 1: Following load combinations for zone 2B and soil

SD are used:

In ACI load combination, “E” is as per IBC. In this example E is

taken as per UBC with vertical component ignored.

18

Load Combinations

UBC 97 Load Combinations:

1.4D+1.7L

1.474D + 0.55L + 1.1E

1.166D + 0.55L − 1.1E

1.14D + 1.1E

0.84D − 1.1E

ACI 318-05 Load Combinations:

1.2D+1.6L

1.2D + 1.0L + 1.0E

1.2D + 1.0L − 1.1E

0.9D + 1.0E

0.9D − 1.0E

10




Case Study 1

19

-17 -17

9

-0.5

1.0

22

-22 -39

37

-37

33

1.0D

1.0L

1.0E

−1.0E

Analysis Results for Unamplified Individual load Cases

39

-37

37

-33





Load Combinations

-22

22

-43 -43

26

-1.4

2.0




Case Study 1

20

-89 -89

52

-2.8

4.5

1.4D + 1.7L (UBC-97) 1.2D + 1.6L (ACI-318-05)

-75 -75

37

-3.3

4.3

Load combination 1

Load Combinations

11




Case Study 1

21

1.474D + 0.55L + 1.1E (UBC-97) 1.2D + 1.0L + 1.0E (ACI-318-05)

-42 -85

33

-42

40

-37

33

-43 -90

35

-46

44

-41

35

-85

-42

33

1.2D + 1.0L − 1.0E (ACI-318-05)

36

-33

36

-33

1.166D + 0.55L − 1.1E (UBC-97)

-79

-29

29

41

-37

41

-37

Load combination 2 & 3

Load Combinations




Case Study 1

22

1.14D + 1.1E (UBC-97) 0.9D + 1.0E (ACI-318-05)

-85

-42

33

0.9D − 1.0E (ACI-318-05)

36

-33

36

-33

-13 -56

19

-41

39

-36

33

-20 -68

23

-45

43

-41

36

-57

-8

20

42

-38

41

-36

Load combination 4 & 5

Load Combinations

0.84D − 1.1E (UBC-97)

12




Case Study 1: Conclusions

For the given frame, except gravity load combination 2, there is no significant

difference between UBC-97 and ACI 318-05 load combination results.

However in some cases it may be more.

Note that in each case, strength reduction factors are different. Therefore,

there will be difference in reinforcement as well.

Caution about use of load combination in SAP2000: Before designing in SAP2000,

make sure that combinations used are the same as used in the relevant code.

23

Load Combinations




Example 2: Complete Example using Approximate Analysis

24

Load Combinations

20 ft 20 ft 20 ft 20 ft

15 ft

15 ft

15 ft

fc′ = 3 ksi

fy = 40 ksi

SDL = 40 psf

LL = 60 psf

SDL = 40 psf

LL = 60 psf

SDL = 40 psf

LL = 60 psf

Seismic Zone: 2B

Soil: SD

Slab: 6″

Beams:12″ × 18″

Columns: 12″

13




Example 2: E-W Frame to be analysed. Lateral load from Static

Lateral Force Procedure are shown.

25

F3 =23 kips

l1=20 ft l2=20 ft l3=20 ft

h=10.5 ft

F2 =15.5 kips

F1 = 7.74 kips

l4=20 ft

h=10.5 ft

h=10.5 ft

Load Combinations




Example 2: Analysis Results for 1.0D

26

Column

Moments

Beam

Moments

Load Combinations

14




Example 2: Analysis Results for 1.0L

Column

Moments

Beam

Moments

Load Combinations

27



Applications of Load Combinations

Example 2: Analysis Results for 1.0E

Load Combinations

28

15




Example 2: Analysis Results for −1.0E

Load Combinations

29




Example 2: Analysis Results for 1.2D+1.6L

Column

Moments

Beam

Moments

Load Combinations

30

16



Applicationn of Load Combinations

Example 2: Analysis Results for 1.2D+1.0L+1.0E

Column

Moments

Beam

Moments

Load Combinations

31




Example 2: Analysis Results for 1.2D + 1.0L − 1.0E

Column

Moments

Beam

Moments

Load Combinations

32

17




Example 2: Analysis Results for 0.9D + 1.0E

Column

Moments

Beam

Moments

Load Combinations

33




Example 2: Analysis Results for 0.9D − 1.0E

Column

Moments

Beam

Moments

Load Combinations

34

18



Application of load Combinations

Example 2: Envelop Used For Design & Comparison with SAP

Column

Moments

Beam

Moments

Load Combinations

74 52 31 52 74

55 62 52 62 55

91 133 123 133 91

-74 -109 -88 -88 -88 -88 -109 -74

78 58 58 78

-100 -135 -114 -114 -114 -114 -135 -100

78 58 58 78

-146 -181 -160 -160 -160 -160 -181 -146

78 58 58 78

10 10 2 2 2 2

56 56 48 48 48 48 37 37

54 22 21 22 54

56 42 37 42 56

85 81 77 81 85

-38 -60 -80 -71 -71 -80 -60 -38

84 36 36 84

-72 -93 -86 -85 -85 -86 -93 -72

50 37 37 50

-97 -123 -109 -105 -105 -109 -123 -97

54 37 37 54

6 6

46 10 10 46 3 10 10 3

35

Approximate Analysis Envelop SAP2000 Envelop



Case Study 2

Comparative study between gravity and earthquake load

analysis for various zones of the given structure using

SAP2000.

36

Gravity vs. Earthquake

Loading

20 ft 20 ft 20 ft 20 ft

15 ft

15 ft

15 ft

The study has been done

using SAP2000 using ACI

318-05 load combinations

and is done for all

seismic zones. The study

has been done on the

same 4 by 3 (20′×15′)

panel building.

19



Case Study 2

The objective of this study is to determine for the given 3D

structure:

Bending moment due to gravity loads

Bending moments due to earthquake loads from zone 1 to 4.

Compare the bending moments to see the variation in bending

moments due to change in loading.

Compare the reinforcement requirement due to change in loading.

37


Loading



Case Study 2

Gravity Load Analysis (1.2D+1.6L) for all seismic zones

38

34 -50

51

-80 -75 -73

37

-36

1.2D+1.6L

44

-38

-2 -38

54

-85 -79 -71

36


Loading

20



Case Study 2

Combinations for Zone 1

39

11 -21

35

-65 -41 -59

25

-16

1.2D+1.0L+1.0E

1.2D+1.0L−1.0E

34 -46

31

-43 -60 -39

25

-32

36

-31

22

-19

-35

32

-5

54

-95 -51 -80

31

-60

41

-51 -86 -42

33

0.9D+1.0E

0.9D−1.0E

4 -10

27

-48 -25 -44

25

-8

13

-12

-32 9

32

-61 -19 -52

17

27 -36

21

-27 -44 -24

18

-25

27

-23

32 -45

22

-18 -55 -15

19


Loading



Case Study 2

Combinations for Zone 2A

40

8 -19

48

-89 -47 -80

33

-16

1.2D+1.0L+1.0E

25

-22

-60

50 -66

40

-49 -81 -45

33

-46

51

-43

64

17

59

-113 -30 -95

33

-83

41

-33 -100 -27

37

-5 0

28

-57 -18 -52

18

-2

0.9D+1.0E

8

-7

-61 32

40

-79 -5 -67

19

37 -46

21

-18 -52 -16

19

-32

33

-28

63 -68

23

0 -69 0

22

0.9D−1.0E 1.2D+1.0L−1.0E


Loading

21



Case Study 2

Combinations for Zone 2B

41

2 -13

50

-94 -42 -85

33

-12

1.2D+1.0L+1.0E

21

-19

-77

1.2D+1.0L−1.0E

56 -72

40

-44 -86 -40

34

-50

54

-45

81

0.9D+1.0E

0.9D−1.0E

-11 6

30

-63 -13 -56

19

-2

4

-4

-80

46

-90 3 -77

20

42 -53

22

-12 -57 -12

20

-36

37

-30

80 -82

24

10 -78 9

24

-31

63

-123 -21 -105

33

-97

41

-22 -109 -18

38


Loading



Case Study 2


42

1.2D+1.0L+1.0E

1.2D+1.0L−1.0E

64 -81

39

-37 -92 -34

36

-56

59

-49

104

-16

-5 -4

53

-101 -36 -91

33

-6

16

-100 49

70

-138 -15 -117

36

-116

42

-8 -121 -5

40

0.9D+1.0E

0.9D−1.0E

0

-18 14

32

-70 -7 -63

20

7

0

-101 64

49

-105 15 -89

23

50 -61

22

-5 -63 -5

21

-31

42

-34

104 -101

26

24 -90 22

28


Loading

22



Case Study 2


43

1.2D+1.0L+1.0E

1.2D+1.0L−1.0E

71 -89

40

-29 -97 -27

35

-61

64

-53

127

-11

-13 3

54

-108 -29 -98

33

-1

12

-122 68

78

-152 -4 -130

37

-134

42

6 -132 6

44

0.9D+1.0E

2

-26 23

35

-77 0 -69

20

12

-5

-123 83

58

-119 26 -102

21

0.9D−1.0E

58 -70

23

1 -69 0

22

-47

46

-38

126 -119

29

39 -101 34

34


Loading



Case Study 2

Beam Moment Comparison for all zones

44

A

B

(A) (B)


Loading

Tension on opposite face

for higher seismic zone No significant effects of

Lateral loads

23



Case Study 2


45

A

B

(A) (B)


Loading


for higher seismic zone

No significant effects of

Lateral loads



Case Study 2


46

C

(C) (F)

F


Loading

Marginal effect of lateral

loads

Marginal effect of lateral

loads

24



Case Study 2

Comparison for all zones

47

C F

(C) (F)


Loading







Case Study 2


48

E

(E)


Loading

No significant effects of

Lateral loads

25



Case Study 2

Comparison for all zones

49

E


Loading

(E) No significant effects of

Lateral loads



Case Study 2

Column Moment Comparison for all zones

50

Top moment


Loading

Increase in BM for higher

seismic zone

26



Case Study 2

Conclusions

Lower storey positive end moment in beams may become significant in

higher seismic zones.

There is no significant change in beam mid span positive moments for

all zones.

The column moments increases with increase in seismic zone.

Generally, the moment due to lateral loads in beams and columns both

increases from top to bottom stories and is maximum at the aground

storey

Within a storey, exterior negative moment in a beam increases more

than that of interior negative moments in the same seismic zone.

51


Loading



ETABS

Following slides present broad steps required to perform analysis.

52

Analysis & Design Using

Software

27



1. After completing the modeling, goto

Option → Preferences → Concrete Frame Design

ETABS

Select UBC-97 so that UBC load combinations can be generated

3. Goto Define → Special Seismic Load Effects and select from two

options based on requirement:

(i) Include Special Seismic Design Data

(ii) Do Not Include Special Seismic Design Data

Option (i) is normally selected.

Vertical component of earthquake loads can be

included from this option.

Vertical component of earthquake loads can also be neglected

if acceptable. Goto Define Load Add Default Design

Combos to Generate load combinations Select Concrete Frame Design to

generate load combinations for concrete

material

Goto Define → Load Combinations to see the

generated Load combinations

Click Modify Button to see details of

load combinations Load Combination details

(Without inclusion of vertical

component of earthquake load)

Load combination including vertical

component of earthquake load



SAP2000

Following slides present broad steps required to perform analysis.

54

Analysis & Design Using

Software

28



SAP2000

After completing the structural model, go to Define Load cases

and all required load patterns.

Earthquake load pattern should be defined as discussed in

previous lecture

Defined Load patterns

Note: UBC 97 used for earthquake load definition.

Go to Design Concrete Frame Design View/ Revise Preferences

to select code for design.

To generate load combinations select the code. As UBC is used for

earthquake definition therefore select UBC.

Go to Define load Combinations to generate Load

combinations as per UBC-97

Sometimes the desired load combinations might not match the required code

combinations, therefore all combinations should be checked before final design.

After this, the model is ready for analysis and design as per UBC-97.

In a second option, To perform analysis and design as per BCP, manually define all load

combinations as per UBC-97 in Define Load Combination section.

Then go to design Concrete Frame Design View/ Revise Preferences and change design code

to ACI 318-05.

Change the strength reduction factors as per UBC i.e., Φ for shear and torsion = 0.85 instead

of 0.75.

Option 1 Option 2



ACI Special Provisions for Seismic

Design

56

29



The principal goal of the Special Provisions is to ensure adequate

toughness under inelastic displacement reversals brought on by

earthquake loading.

The provisions accomplish this goal by requiring the designer to

provide for concrete confinement and inelastic rotation capacity.

No special requirements are placed on structures subjected to low or

no seismic risk.

Structural systems designed for high and moderate seismic risk are

referred to as Special and Intermediate respectively.

57

ACI Special Provisions for

Seismic Design



Based on moment resisting capacity, there are three types of RC

frames,

SMRF (Special Moment Resisting Frame),

IMRF (Intermediate Moment Resisting Frame),

OMRF (Ordinary Moment Resisting Frame).

Some general requirements will be presented first, which are

common to all frames. Specific requirements for each type of frame

are presented later on.

58


Seismic Design

30




Concrete in members resisting earthquake induced forces

Min f’c = 3000 Psi (cylinder strength) for all types

No maximum limit on ordinary concrete

5000 psi is maximum limit for light weight

59


Seismic Design




Reinforcement in members resisting earthquake induced forces

Grade 60, conforming to ASTM A 706 (low alloy steel)

Grade 40 or 60, conforming to ASTM A 615 (billet steel) provided that

Fy (actual) – Fy (specified) ≤ +18 Ksi

Actual Ultimate / Actual Fy ≥ 1.25

60


Seismic Design

31




Hoops, Ties and Cross Ties

Confinement for concrete is provided by transverse reinforcement

consisting of stirrups. hoops, and crossties.

To ensure adequate anchorage, a seismic hook (shown in figure) is used

on stirrups, hoops and crossties .

61


Seismic Design

(Seismic Hook)




Hoops, ties and Crossties: Advantages

Closely spaced horizontal closed ties in column help in three ways:

i. they carry the horizontal shear forces induced by earthquakes, and thereby

resist diagonal shear cracks,

ii. they hold together the vertical bar and prevent them from excessively

bending outwards (in technical terms, this bending phenomenon is called

buckling), and

iii. they contain the concrete in the column. The ends of the ties must be bent

at 135° hooks. Such hook ends prevent opening of hoops and

consequently buckling of concrete and buckling of vertical bars.

62


Seismic Design

32



Provisions for Flexural Members

These provision applies to flexural members with:

Factored axial compressive force Agf’c/10.

Note: These provisions generally apply to beams because axial load on beams

is generally less than Agfc′/10.

However they are also applicable to columns subjected to axial load less

than Agfc′/10.

63

ACI Provisions for Special

Moment Resisting Frames (SMRF)




1. Size: The members must have:

a. clear span-to-effective-depth ratio of at least 4, (Ln/d ≥ 4)

e.g., for Ln = 15 ft, d = 16″, Ln/d = 15 × 12/16 = 11.25 > 4, O.K.

b. width-to-depth ratio of at least 0.3, b/d ≥ 0.3

e.g., for width (b) = 12″ and depth (h) = 18″, b/h = 12/18 = 0.67 > 0.3, O.K.

c. web width of not less 10 inches.

64



33




2. Flexural Reinforcement

65



Asl−

Asl+ (Asl−)/2

As− or As+ (at all section) (maximum of As at either joint)/4

Asr−

Asr+ (Asr−)/2

rmin = 3f’c/fy, 200/fy (at critical sections)

rmax = 0.025 (at critical sections) Min. 2 bars continuous at all sections




3. Transverse Reinforcement

66



s

d/4

8 smallest longitudinal bar diameter

24 hoop bar diameter

12” 2”

≥ 2h ≥ 2h s d/2

Column

Column

34




4. Lap Splice

67



Lap splice length =1.3 ld = 1.3 0.05 (fy/ √fc′)db

50 db for fc′ 3 and fy 40 ksi


Spacing of stirrups Least of d/4 or 4 inches

Lapping prohibited in regions where longitudinal bars can yield in tension

Lapping of Longitudinal bars

≥ 2h ≥ 2h


Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011 68


Mechanical Splice of Longitudinal Reinforcement

Mechanical Splices shall conform to 21.2.6.

Section 21.2.6 says that welded splice shall conform to 12.14.3.2

which states “A full mechanical splice shall develop in tension or

compression, as required, at least 125 % of the specified yield

strength (fy) of the bar.

68



35




Welded Splice of Longitudinal Reinforcement

Welded Splices shall conform to 21.2.7.

Section 7.3.6 says that welded splice shall conform to 12.14.3.4

which states “ A full welded splice shall develop at least 125 % of

the specified yield strength (fy) of the bar.

69





Provision for Frame Members Subjected to Bending and Axial Load

The provision applies to members with:

Factored axial compressive force > Agf’c/10

70



36




1. Size

a) Each side at least 12 inches

b) Shorter to longer side ratio ≥ 0.4.

i.e. 12/12, 12/18, 12/24 OK; but 12/36 not O.K

71





Provision for Frame Members

Subjected to Bending and Axial

Load

2. Longitudinal Reinforcement

72



0.01 rg 0.06

Clear span, hc

37



Provision for Frame

Members Subjected

to Bending and

Axial Load

3. Trans. Rein.

73



h2

h1

s 0.25 (smaller of h1 or h2)

6 long. bar dia.

so

s 6 long. bar dia.

6”

lo

lo Larger of h1 or h2

Clear span/6

18”




3. Trans. Rein.

74



4” ≤ so = 4 + [(14 – hx)/3] ≤ 6”

hx = max. value of hx on all column faces

hx 14”

hx hx hx

hx

hx

6db 3” 6db extension

Alternate 90-deg hooks

Provide add. trans. reinf. if thickness > 4”

38



Provision for Frame Members Subjected

to Bending and Axial Load

4. Lap Splice

75



Tension lap splice within center half of member length

Lap splice length =1.3 ld = 1.3 0.05 (fy/ √fc′)db



Spacing of ties in lap splice not more than smaller of d/4 or 4″



Joints of Special Moment Frame



Beam

Column

Beam Column Joint

Column ties (with 135o)

hook continued through joint

(ACI 21.5.2)

76

39




Successful seismic design of frames require that the

structures be proportioned so that hinges occur at locations

that least compromise strength. For this, “weak Beam-strong

column” approach is used.

After design, the member capacities are calculated based

on designed section.

Column flexural capacity > Beam flexural capacity

77






Minimum Flexural Strength of Column at Joint

78

M+nc,t

M-nc,b

M+nb,r

M-nb,l

M+nc,b + M-

nc,t 6(M+nb,l + M-

nb,r)/5

M-nc,t

M+nc,b

M+nb,l M-

nb,r

M-nc,b + M+

nc,t 6(M-nb,l + M+

nb,r)/5



40




To prevent beam column joint cracking, ACI Code 21.5.1

requires that the column dimension parallel to the beam

reinforcement must be at least 20 times the diameter of the

largest longitudinal bar.

79



Beam longitudinal

reinforcement with

diameter (db)

20db

Beam

Column




Beam longitudinal reinforcement that is terminated within a

column. must be extended to the far face of the column core.

The development length (ldh) of bars with 90° hooks must be not

less than 8db, 6 inch, Or fydb/ (65 √ fc′).

80



Beam

Column

Beam longitudinal

reinforcement

ldh

41



Provision for Flexural Members

1. Size: No special requirement (Just as ordinary beam

requirement).

2. Flexural steel: Less stringent requirement as discussed next.

3. Transverse steel: Same as for SMRF.

4. Lap: No special requirement (Just as ordinary beam

requirement).

81

ACI Provisions for Intermediate

Moment Resisting Frames (IMRF)




2. Flexural Reinforcement

82



Asl−

Asl+ (Asl−)/3

As− or As+ (at all section) (maximum of As at either joint)/5

Asr−

Asr+ (Asr−)/3

rmin = 3f’c/fy, 200/fy (at critical sections)

t ≥ 0.004

42



Provision for Columns

1. Size: No special requirement (Just as ordinary column

requirement)

2. Flexural steel: No special requirement (Just as ordinary column

requirement)

3. Transverse steel: Less Stringent requirement as given next.

4. Lap: No special requirement (Just as ordinary column

requirement)

83






84



h2

h1

so/2

Trans. reinf. based on Mn and factored tributary gravity load

so

8 smallest long. bar dia.

24 tie bar dia.

0.5 min. (h1 or h2)

12”

lo

lo

Larger of h1 or h2

Clear span/6

18”

s ≤ d/2 (As per ACI 11.5.4)

43



IMRF are not allowed in regions of high seismic risk, however,

SMRF are allowed in regions of moderate seismic risk.

Unlike regions of high seismic risk, two way slab system without

beams are allowed in regions of moderate seismic risk.

In regions of low or no seismic risk ordinary moment resisting

frames OMRF are allowed but IMRF and SMRF may also be

provided.

85

Miscellaneous Considerations



Detail the selected frame of E-W interior frame of the given

structure as per SMRF requirements. The structure is already

designed for the following seismic zone data.

Seismic zone: 4

Magnitude of earthquake ≥ 7.0

Slip rate ≥ 5.0

Closest distance to known seismic source > 15 km.

Soil type: SD (stiff).

Concrete compressive strength = 3 ksi,

Steel yield strength = 40 ksi

Modulus of elasticity of concrete = 3000 ksi.

86

Design Example

44



Given 3D structure:

87

Design Example

20 ft 20 ft 20 ft 20 ft

15 ft

15 ft

15 ft

10.5 ft

10.5 ft

10.5 ft (floor to floor)

SDL = 40 psf

LL = 60 psf

SDL = 40 psf

LL = 60 psf

SDL = 40 psf

LL = 60 psf

fc′ = 3 ksi

fy = 40 ksi

Zone = 4

Ca = 0.44

Cv = 0.64

I = 1

W = 2002

V = 259 kip

Slab-Beam

Frame

Structure

Beams: 15″ × 24″

Columns: 15″ square



Load Combinations

ACI 318-05 load combinations have been used.

1.2D+ 1.6L

1.2D + 1.0L + 1.0E

1.2D + 1.0L − 1.0E

0.9D + 1.0E

0.9D − 1.0E

88

Design Example

45



Design Example

Analysis

Analysis has been done using SAP2000. SAP2000 develops

envelop of maximum bending moments automatically for the

given load combinations.

During the design, the software automatically checks SMRF

requirements at each and every section.

89



Design Example

Analysis

Analysis results for shown portion of E-W interior frame is

shown next.

90

20 ft 20 ft 20 ft 20 ft

15 ft

15 ft

15 ft

10.5 ft

10.5 ft

10.5 ft (floor to floor)

46



Analysis Results

91

E

-43 -43

26

-17.5 -17.5

11

-0.5

1.0

-1.4

2.0

43

-43 -72

67

-68

61

1.0D 1.0L 1.0E −1.0E

72

-67

68

-61

-43

43

1.2D + 1.6L

-65 -65

40

-2.2

3.2

1.2D + 1.0L + 1.0E

-13 -98

40

-74

70

-68

61

1.2D + 1.0L − 1.0E

-98 -12

40

70

-64

68

-61

1.14D + 1.1E

+14

-70

26

-73

68

-68

61

0.84D − 1.1E

-70

+15 26

71

-66

68

-61

-98 -98

+40

-74 -68

Envelop

70 61

+14 +15

E





Design Example



Design Example

Selected portion of E-W interior Frame:

Bending Moment Envelop

As the software checks SMRF moment capacity requirements and ACI

minimum moment capacity requirements at critical sections. Therefore

final shape of the bending moment envelop of the beam is as shown:

92

20 ft 20 ft 20 ft 20 ft

57

−98 −98

48 48 Values used in design -98 -98

+40

-74 -68

Analysis Envelop

70 61

+14 +15

47



Design Example

Calculation of number of bars:

93

20 ft 20 ft 20 ft 20 ft

# of bars = As/Ab

Use No. 5 bar,

Negative reinforcement at joint:

Left joint:

1.61/0.31= 5.19 (take 6 bars in 2 layers)

Right joint:


Positive bars (mid span):

1.21/0.31 = 3.9 (take 4 bars in 1 layer)

Positive bars (at joint):


Column reinforcement:

2.25/ 0.31 = 7.25 (take 8 bars for even distribution

of bars at all faces of column)

1.61 1.61

1.02 1.02 1.21

2.25 2.25

Reinforcement in in2



Design Example

Calculation of number of bars:

94

20 ft 20 ft 20 ft 20 ft

# of bars = As/Ab

Use No. 5 bar,

Negative reinforcement at joint:

Left joint:


Right joint:


Positive bars (mid span):


Positive bars (at joint):


Column reinforcement:

2.25/ 0.31 = 7.25 (take 8 bars for even distribution

of bars at all faces of column)

6 bars 6 bars

4 bars 4 bars 4 bars

8 bars 8 bars

No. of #5 bars

48



SMRF Requirements Checklist

Provisions for Beams

Sizes

ln/d = 20 × 12/21 = 11.4 > 4 (ACI 21.3.1.2 satisfied)

Width/ depth = 15/24 = 0.625 > 0.3 (ACI 21.3.1.3 satisfied)

Width = 15″ > 10″, O.K.

Therefore 15″ × 24″ deep beams is OK.

95

Design Example





Flexural Reinforcement

96

Design Example

As+ (at joints) ≥ ½ As− (at joints)

4 #5 bars ≥ ½ (6 #5 bars)

As (any section) ≥ ¼ Max. As at joints

2 #5 bars ≥ ¼ (6 #5 bars)

As (at all critical sections) ≥ Asmin

As Asmin = 6 #5 bars

As (at any section) ≤ Asmax

Asmax = 0.025bd = 25 #5 bars

OK

OK

Provide at least 6 bars at critical sections)

OK

Asl− = 6 #5 Asr− = 6 #5

Asmid+ = 4 #5 Asl+ = 4 #5 Asr+ = 4 #5

N.G

49





Flexural Reinforcement

97

Design Example

Asl− = 6 #5 Asr− = 6 #5

Asmid+ = 6 #5 Asl+ = 6 #5 Asr+ = 6 #5





Transverse Reinforcement

98

Design Example

s

d/4 = 21/4 = 5.25″

8 smallest long. bar dia.= 8 × 5/8= 5″

24 hoop bar diameter = 24 × 3/8= 9″

12” 2”

2h = 48″ s d/2 = 21/2 = 11″

Column

2h = 48″

50





Lap Splice: If required then,

Not to be provided within joints. Not to be provided within 2h region from face of

the support.

Spacing of hoops within lap = least of d/4 or 4″ c/c = 4″ c/c

Lap splice length =1.3 ld = 1.30.05 (fy/ √fc′)db ≈ 30″ = 2.5′

50 db = 50 (5/8) = 31.25″ ≈ 2.5′ for fc′ 3 and fy 40 ksi

99

Design Example

2h=48″ 2h=48″




Provisions for Columns

Size: All columns are 15″ square, which is more than least required for

SMRF (i.e., 12″).

Flexural Reinforcement: All columns are reinforced with 8 #5 bars

which gives ρg = 0.011, within the specified range 0.01 ≤ ρg ≤ 0.06.

100

Design Example

51



Design Example

SMRF Requirements

Checklist


Transverse Reinforcement:

lo = max (larger column

dimension, hc/6, 18″) = 18″

Spacing of ties in lo region is

least of = smaller column

dimension/4, 6 long bar

dia = 3.75″

Spacing in the remaining

region will be least of 6

long bar dia or 6″ = 3.75″

hc = 8.5′

lo

lo

15” × 15” column

8, #5 bars

101



Design Example


Provisions for Columns

Lap Splice:

Tension lap splice within center

half of member length.

Spacing of ties in lap splice not

more than smaller of d/4 or 4″

Lap length = 1.3 0.05 (fy/ √fc′ )db=

30″ ≈ 2.5′

And from 50db = 50(5/8) =

31.25″

hc = 8.5′

102

52



Design Example


Provision for Joints

To prevent beam column joint cracking, ACI Code 21.5.1 requires that

the column dimension parallel to the beam reinforcement must be at

least 20 times the diameter of the largest longitudinal bar.

20 × 5/8 = 12.5″

Column dimension parallel to beam long bar = 15″, OK

103



Design Example



6 #5 bars

6 #5 bars

2″

Column

Beam Joint

Interior joint

104

53



Design Example



For exterior columns, the column dimension parallel to beam longitudinal

bar must be greater than the development length of beam bars in

columns with 90° hooks is not to be less than largest of:

The development length of beam bars in columns with 90° hooks is not

to be less than largest of:

8db = 8 × 5/8 = 5″

6″

ldh = fydb/(65 √fc′) = 40000 × (5/8)/ {65 × √(3000)} = 7″

Therefore, development length = 7″. The column dimension is 15″ which

satisfies this requirement.

105



Design Example



2″

Development of beam

reinforcement in column

= 13″ > ldh = 7″

Development of beam

reinforcement in column

= 22″ > ldh = 7″

Column

Beam Joint

Exterior Joint 106

54



References

ACI 318

Design of Concrete Structures by Nilson, Darwin and

Dolan.

UBC-97

BCP

107



The End

108

load combination - ubc 97 and aci

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