L6-Introduction-to-EQ-Resistant-Design-RC Structures

Department of Civil Engineering, N-W.F.P UET Peshawar

A Monograph on Earthquake Resistant Design of R C Structures

Earthquake Resistant Design Of Reinforced Concrete Structures By

Professor Dr. Qaisar Ali Earthquake Engineering Center, Civil Engineering Department N-W.F.P University of Engineering and Technology Peshawar, Pakistan.

Prof. Dr. Qaisar Ali (http://www.eec.edu.pk)

Page 1 of 56



1. Introduction:

Earthquake results from the sudden movement of the tectonic plates in the earth’s crust, figure 01. The movement takes place at the fault lines, and the energy released is transmitted through the earth in the form of waves, figure 02, that cause ground motion many miles from the epicentre, figure 03. Regions adjacent to active fault lines are the most prone to experience earthquakes. As experienced by structures, earthquakes consist of random horizontal and vertical movements of the earth’s surface. As the ground moves, inertia tends to keep structures in place, figures 04, resulting in the imposition of displacements and forces that can have catastrophic results, figure 05. The purpose of seismic design is to proportion structures so that they can withstand the displacements and the forces induced by the ground motion.

Figure 01: Earth’s tectonic plates.


Page 2 of 56



Figure 02: Motions caused by body and surface waves.

Figure 03: Epicentre and focus.


Page 3 of 56



Figure 04: Effect of inertia in a building when shaken at its base.

Figure 05: Inertia force and relative motion within a building.

Figure 06: Arrival of seismic waves at a site.


Page 4 of 56



Historically, seismic design has emphasized the effects of horizontal ground motion, because the horizontal components of an earthquake usually exceed the vertical component and because structures are usually much stiffer and stronger in response to vertical loads than they are in response to horizontal loads. Experience has shown that the horizontal components are the most destructive. For structural design, the intensity of an earthquake is usually described in terms of the ground acceleration as a fraction of the acceleration of gravity, i.e., 0.1, 0.2, or 0.3g. Although peak acceleration is an important design parameter, the frequency characteristics and duration of an earthquake are also important; the closer the frequency of the earthquake motion is to the natural frequency of a structure and the longer the duration of the earthquake, the greater the potential for damage. Based on elastic behaviour, structures subjected to a major earthquake would be required to undergo large displacements. However, recent design practices require that structures be designed for only a fraction of the forces associated with those displacements. The relatively low design forces are justified by the observations that the buildings designed for low forces have behaved satisfactorily and that structures dissipate significant energy as the material yield and behave in-elastically. This nonlinear behaviour, however, usually translates into increased displacements, which may result in major non-structural damage and require significant ductility. Displacements may also be of such a magnitude that the strength of the structure is affected by stability considerations. Designers of structures that may be subjected to earthquakes, therefore, are faced with a choice: (a) providing adequate stiffness and strength to limit the response of structures to the elastic range or (b) providing lower-strength structures, with presumably lower initial costs, that have the ability to withstand large inelastic deformations while maintaining their load-carrying capability.

2. Structural Response:

The safety of a structure subjected to seismic loading rests on the designer’s understanding of the response of the structure to ground motion. For many years, the goal of earthquake design has been to construct buildings that will withstand


Page 5 of 56



moderate earthquakes without damage and severe earthquakes without collapse. Building codes have undergone regular modification as major earthquakes have exposed weaknesses in existing design criteria. Design for earthquakes differs from design for gravity and wind loads in the relatively greater sensitivity of earthquake-induced forces to the geometry of the structure. Without careful design, forces and displacements can be concentrated in portions of a structure that are not capable of providing adequate strength or ductility. Steps to strengthen a member for one type of loading may actually increase the forces in the member and change the mode of failure from ductile to brittle. a. Structural consideration: The closer the frequency of the ground motion is to one of the natural frequencies of a structure, the greater the likelihood of the structure experiencing resonance, resulting in an increase in both displacement and damage. Therefore, earthquake response depends strongly on the geometric properties of a structure, especially height. Tall buildings respond more strongly to long-period (low frequency) ground motion, while short buildings respond more strongly to short period (high frequency) ground motion. Figure 07 shows the shapes for the principal modes of vibration of a three storey frame structure. The relative contribution of each mode to the lateral displacement of the structure depends on the frequency characteristics of the ground motion. 3 t h ig 2 e H y e r o t S1

0 Displacement (a)

(b)

(c)

Figure 07: Modal shapes for a three storey building (a) first mode; (b) second mode; (c) third mode.


Page 6 of 56



The first mode, figure 07a, usually provides the greatest contribution to lateral displacement. The taller a structure, the more susceptible it is to the effects of higher modes of vibration, which are generally additive to the effects of the lower modes and tend to have the greatest influence on the upper stories. Under any circumstances, the longer the duration of an earthquake, the greater the potential of damage. The configuration of a structure also has a major effect on its response to an earthquake. Structures with a discontinuity in stiffness or geometry can be subjected to undesirably high displacements or forces. For example, the discontinuance of shear walls, infill walls or even cladding at a particular story level, will have the result of concentrating the displacement in the open, or “soft,” story, figure 08, 09. The high displacement will, in turn, require a large amount of ductility if the structure is not to fail. Such a design is not recommended, and the stiffening members should be continued to the foundation.

Figure 08: Upper storeys of open ground storey move together as single block.


Page 7 of 56



Figure 09: Ground storey of reinforced concrete building left open to facilitate parking.

Similarly, any kind of horizontal or vertical mass or stiffness irregularity in structures places them in undesirable position against earthquake forces. Buildings with simple geometry in plan, figure 10a, perform well during strong earthquakes. Buildings with re-entrant corners, like those U, V, H and + shaped in plan, figure 10b, have sustained significant damage in past earthquakes. Many times, the bad effects of these interior corners in the plan of buildings are avoided by making the buildings in two parts. For example, an L-shaped plan can be broken up into two rectangular plan shapes using a separation joint at the junction, figure 10c. Figure 11 shows buildings with one of their sizes much larger or much smaller than the other two. Such shapes do not perform well during the earthquakes. Buildings with vertical setbacks (like the hotel buildings with a few storeys wider than the rest) cause a sudden jump in earthquake forces at the level of discontinuity, figure 12. Within a structure, stiffer members tend to pick up a greater portion of the load. When a frame is combined with a shear wall, this can have the positive effect of reducing the displacements of the structure and decreasing both structural and non-structural damage. However, when the effects of higher stiffness members,


Page 8 of 56



such as masonry infill walls, are not considered in the design, unexpected and often undesirable results can occur. Finally, any discussion of structural considerations would be incomplete without emphasizing the need to provide adequate separation between structures. Lateral displacements can result in structures coming in contact during an earthquake, resulting in major damage due to hammering, figure 13. Spacing requirements to ensure that adjacent structures do not come into contact as a result of earthquake induced motion are specified in relevant codes.

Figure 10: Simple plan shape buildings do well during earthquake.

Figure 11: Buildings with one of their overall sizes much larger or much smaller than the other two.


Page 9 of 56



Figure 12: Buildings with setbacks.

Figure 13: Hammering or Pounding.

b. Member Considerations: Members designed for seismic loading must perform in a ductile fashion and dissipate energy in a manner that does not compromise the strength of the structure. Both the overall design and the structural details must be considered to meet this goal. The principal method of ensuring ductility in members subject to shear and bending is to provide confinement for the concrete. This is accomplished through the use of closed hoops or spiral reinforcement, which enclose the core of the beams and columns. When confinement is provided, beams and columns can undergo nonlinear cyclic bending while maintaining their flexural strength and


Page 10 of 56



without deteriorating due to diagonal tension cracking. The formation of ductile hinges allows reinforced concrete frames to dissipate energy. Successful seismic design of frames requires that the structures be proportioned so that hinges occur at locations that least compromise strength. For a frame undergoing lateral displacement, such as shown in figure 14a, the flexural capacity of the members at a joint, figure 14b, should be such that the columns are stronger than the beams. In this way, hinges will form in the beams rather than the columns, minimizing the portion of the structure affected by nonlinear behaviour and maintaining the overall vertical load capacity. For these reasons, the “weak beam-strong column” approach is used to design reinforced concrete frames subject to seismic loading.

M C1

M

-

M

M C2

(a)

(b)

Figure 14: Frame subjected to lateral loading (a) deflected shape; (b) Moments acting on beam-column joint.

3. Seismic loading criteria:

In Pakistan, the design criteria for earthquake loading are based on design procedures presented in chapter 5, division II of Building Code of Pakistan, seismic provision 2007 (BCP, SP 2007), which have been adopted from chapter 16, division II of UBC97 (Uniform Building Code), volume 2, attached for reference in Appendix A of this document. The total design seismic force imposed by an earthquake on the structure at its base is referred to as base shear “V” in the UBC. The UBC-97 calculates the base shear from


Page 11 of 56

+



the total structure weight and then appropriates the base shear in accordance with dynamic theory. The design seismic force can be determined based on the UBC-97 static lateral force procedure [sec. 1630.2, UBC-97 or Sec. 5.30.2, BCP 2007] and/or the dynamic lateral force procedure [sec. 1631, UBC-97 or sec. 5.31, BCP-2007]. The static lateral force procedures (section 1630 of the UBC-97) may be used for the following structures: 1. All structures, regular or irregular, in Seismic Zone 1 and in Occupancy Categories 4 and 5 in Seismic Zone 2. 2. Regular structures under 240 feet (73152 mm) in height with lateral force resistance provided by systems listed in Table 16-N, except where section 1629.8.4, Item 4, applies. 3. Irregular structures not more than five stories or 65 feet (19812 mm) in height. 4. Structures having a flexible upper portion supported on a rigid lower portion where both portions of the structure considered separately can be classified as being regular, the average story stiffness of the lower portion is at least 10 times the average story stiffness of the upper portion and the period of the entire structure is not greater than 1.1 times the period of the upper portion considered as a separate structure fixed at the base. The dynamic lateral force procedure of section 1631 shall be used for all other structures including the following: 1. Structures 240 feet (73152 mm) or more in height, except as permitted by Section 1629.8.3, Item 1. 2. Structures having a stiffness, weight or geometric vertical irregularity of Type 1, 2 or 3, as defined in Table 16-L, or structures having irregular features not described in Table 16-L or 16-M, except as permitted by Section 1630.4.2. 3. Structures over five stories or 65 feet (19812 mm) in height in Seismic Zones 3 and 4 not having the same structural system throughout their height except as permitted by Section 1630.4.2. 4. Structures, regular or irregular, located on Soil Profile Type S F, which have a period greater than 0.7 second. The analysis shall include the effects of the soils at the site and shall conform to Section 1631.2, Item 4.


Page 12 of 56



3.1. Static lateral force procedure:

The static procedure is also referred to as the equivalent static lateral force procedure. UBC-97 sec. 1630.2 provides the provisions for determining base shear by static lateral force procedure as follows: The total design base shear in a given direction can be determined from the following formula: V = (CνI/RT) W Where, Cν = Seismic coefficient (Table 16-R of UBC-97 given below in Table 1 of this document). I = Seismic importance factor (Table 16-K of UBC-97 given in Appendix A) R = numerical coefficient representative of inherent over strength and global ductility capacity of lateral force-resisting systems (Table 16-N or 16-P given in Appendix A of this document). W = the total seismic dead load defined in Section 1630.1.1 as follows: Seismic dead load, W, is the total dead load and applicable portions of other loads listed below. i) In storage and warehouse occupancies, a minimum of 25 percent of the floor live load shall be applicable. ii) Where a partition load is used in the floor design, a load of not less than 10 psf (0.48 kN/m2) shall be included. iii) design snow loads of 30 psf (1.44 kN/m2) or less need not be included. 2

Where design snow loads exceed 30 psf (1.44 kN/m ), the design snow load shall be included, but may be reduced up to 75 percent where consideration of siting, configuration and load duration warrant when approved by the building official. iv) total weight of permanent equipment shall be included.

The total design base need not exceed the following: V = (2.5CaI/R) W Ca = Seismic coefficient (Table 16-Q of UBC-97 given below in Table 2)


Page 13 of 56



The total design base shear shall not be less than the following: V = 0.11CaIW In addition for seismic zone 4, the total base shear shall also not be less than the following: V = (0.8ZNνI/R) W Nν = near source factor (Table 16-T of UBC-97 given below in Table 3) Z = Seismic zone factor (Table 16-I of UBC-97 given below in Table 4) Note: Table for soil profile type is given in appendix A of this document.

Table 1: Seismic Coefficient C ν Seismic Zone Factor, Z Soil Profile Type

SA SB SC SD SE SF 1

Z = 0.075 0.06

Z = 0.15 0.12

0.08 0.13 0.18 0.26

0.15 0.25 0.32 0.50

Z = 0.3 0.24

Z = 0.4 0.32Nv

0.20 0.32 0.40 0.64

0.30 0.45 0.54 0.84

0.40Nv 0.56Nv 0.64Nv 0.96Nv

See Footnote 1

Site Specific geotechnical investigation and dynamic site response analysis shall be performed to determine seismic coefficients for Soil Profile Type SF.

Table 2: Seismic Coefficient C a Seismic Zone Factor, Z

Soil Profile Z = 0.075 0.06 0.08

Type SA SB SC SD SE SF 1

Z = 0.2 0.16

Z = 0.15 0.12 0.15

Z = 0.2 0.16 0.20

Z = 0.3 0.24 0.30

Z = 0.4 0.32Na 0.40Na

0.09 0.12

0.18 0.22

0.24 0.28

0.33 0.36

0.40Na 0.44Na

0.19

0.30

0.34 See Footnote 1

0.36

0.36Na

Site Specific geotechnical investigation and dynamic site response analysis shall be performed to determine seismic coefficients for Soil Profile Type SF.

Table 3: Near source factor N

ν

Closest Distance To Known Seismic Source Seismic Source Type

≤

A

2 km 2.0

5 km 1.6

10 km 1.2

≥

15 km 1.0

B

1.6

1.2

1.0

1.0

C

1.0

1.0

1.0

1.0

Table 4: Seismic zone factor Z Zone

1

2A

2B

3

4

Z

0.075

0.15

0.20

0.30

0.40


Page 14 of 56



Table 5: Near source factor N a Closest Distance To Known Seismic Source Seismic Source Type

A

≤

2 km 1.5

5 km 1.2

≥

10 km 1.0

B

1.3

1.0

1.0

C

1.0

1.0

1.0

Table 6: Seismic source type. 2 Seismic Source Definition

Seismic Source Type

A

Seismic Source Description

Maximum Moment Slip Rate, SR Magnitude, M (mm/year)

Faults that are capable of producing large magnitude events and that have a high rate of seismic activity All faults other than Types A and C

B

C

Faults that are not capable of producing large magnitude earthquakes and that have a relatively low rate of seismic activity

M ≥ 7.0

SR ≥ 5

M ≥ 7.0 M < 7.0 M ≥ 6.5

SR < 5 SR > 2 SR < 2

M < 6.5

SR ≤ 2

Figure 15: Seismic zoning map of Pakistan.


Page 15 of 56



Structural period: The value of T shall be determined from one of the following

methods: a. Method A: For all buildings, the value T may be approximated from the

following formula: T = Ct (hn)3/4 Where, Ct = 0.035 (0.0853) for steel moment-resisting frames. Ct = 0.030 (0.0731) for reinforced concrete moment-resisting frames and eccentrically braced frames. Ct = 0.020 (0.0488) for all other buildings. hn = Actual height (feet or meters) of the building above the base to the nth level. Alternatively, the value of Ct for structures with concrete or masonry shear walls may be taken as 0.1/ √Ac (For SI: 0.0743/√Ac for Ac in m2). The value of Ac shall be determined from the following formula: Ac = ∑Ae[0.2 + (De/hn)2] The value of De/hn used in formula above shall not exceed 0.9. Where, Ac = the combined effective area, in square feet (m 2), of the shear walls in the first story of the structure. Ae = the minimum cross-sectional area in any horizontal plane in the first story, in square feet (m2) of a shear wall. De = the length, in feet (m), of a shear wall in the first story in the direction parallel to the applied forces. b. Method B: The fundamental period T may be calculated using the

structural properties and deformational characteristics of the resisting elements in a properly substantiated analysis. The analysis shall be in accordance with the requirements of Section 1630.1.2. The value of T from Method B shall not exceed a value 30 percent greater than the value of T obtained from Method A in Seismic Zone 4, and 40 percent in Seismic Zones 1, 2 and 3.


Page 16 of 56



The fundamental period T may be computed by using the following formula:

Where,

wi = that portion of W located at or assigned to Level i.

δi = horizontal displacement at Level i relative to the base due to applied lateral forces, f.

g = acceleration due to gravity. fi = lateral force at Level i. The values of fi represent any lateral force distributed approximately in accordance with the principles of Formulas (30-13), (30-14) and (30-15) in UBC-97 or any other rational distribution. The elastic deflections,

δi, shall be

calculated using the applied lateral forces, fi.

Example: Calculate the base shear and storey forces of a five storey building

given in figure 16. The structure is constructed on stiff soil which comes under soil type SD of table 16-J of UBC-97. The structure is located in zone 3.

w5 = 700 kip w4 = 800 kip w3 = 800 kip

60'-0"

w2 = 800 kip w1 = 800 kip

12'-0"

Figure 16: Two storey frame structure.


Page 17 of 56



Solution: i) Base shear: According to static lateral force procedure the total design base

shear in a given direction can be determined from the following formula: V = (CνI/RT) W From table 16-R, Cν = 0.54 From table 16-K, I = 1.00, standard occupancy structures. From table 16-N, R = 8.5, Concrete SMRF (will be discussed later). T = Ct (hn)3/4 = 0.030 × (60)3/4 = 0.646 sec. W = w1 + w2 + w3 + w4 + w5 = 4 × 800 + 700 = 3900 kip Therefore, V = {(0.54 × 1.00)/ (8.5 × 0.646)} × 3900 = 383 kip The total design base need not exceed the following: V = (2.5CaI/R) W From table 16-Q, Ca = 0.36 Therefore, V = (2.5CaI/R) W = {(2.5 × 0.36 × 1.00)/ (8.5)} × 3900 = 413 kip The total design base shear shall not be less than the following: V = 0.11CaIW V = 0.11CaIW = 0.11 × 0.36 × 1.00 × 3900 = 154.44 kip Therefore, V = 383 kip ii) Vertical distribution of base shear to storey:

The joint force at a level x of the structure is given as: Fx = (V – Ft)ωxhx/∑ωihi

{i ranges from 1 to n, where n = number of stories}

Ft = Additional force that is applied to the top level (i.e., the roof) in addition to the Fx force at that level. Ft = 0.07TV {for T > 0.7 sec} Ft = 0 {for T ≤ 0.7 sec}

∑ωihi = 800×12+800×24+800×36+800×48+700×60 = 138000 kip Therefore for the case under consideration, Force for storey 1 is: F1 = (383 – 0) × 800 × 12/ {(138000)} = 27 kip Storey forces for other stories are given in table 6 below.


Page 18 of 56



Table 7: Storey shears. Level x

hx (ft)

5 4 3 2 1

60 48 36 24 12

wx (kip) 700 800 800 800 800

wxhx (ft-kip)

wxhx /(Σwihi)

Fx (kip)

42000 0.304 117 38400 0.278 107 28800 0.209 80 19200 0.139 53 9600 0.070 27 Σwihi 138000 Check ΣFx =V = 383 kip OK =

w5 = 700 kip

F5 = 117 kip

w4 = 800 kip

F4 = 107 kip

w3 = 800 kip

F3 = 80 kip

60'-0"

w2 = 800 kip

F2 = 53 kip

w1 = 800 kip

F1 = 27 kip

12'-0"

V = 383 kip Figure 17: Base shear and storey forces.

3.2. Dynamic lateral force procedure:

UBC-97 section 1631 include information on dynamic lateral force procedures that involve the use of (a) response spectra, or (b) time history analyses of the structural response based on a series of ground motion acceleration histories that are representative of ground motion expected at the site. The details of these methods are presented in sections 1631.5 and 1631.6 of the UBC-97.


Page 19 of 56



4. Gravity vs. Earthquake Loading in a Reinforced Concrete Building:

Gravity loading (due to self weight and contents) on the buildings causes RC frames to bend resulting in stretching and shortening at various locations. Tension is generated at surfaces that stretch and compression at those that shorten (figure 18b).

Figure 18: Earthquake shaking reverses tension and compression in members. Reinforcement is required on both faces of members.

Under gravity loads, tension in the beams is at the bottom surface of the beam in the central location and is at the top surface at the ends. On the other hand, earthquake


Page 20 of 56



loading causes tension on the beam and column faces at locations different from those under gravity loading (figure 18c); the relative levels of this tension (in technical terms, bending moment) generated in members are shown in figure 18d. The level of bending moment due to earthquake loading depends on severity of shaking and can exceed that due to gravity loading. Thus, under strong earthquake shaking, the beam ends can develop tension on either of the top and bottom faces. Since concrete cannot carry this tension, steel bars are required on both faces of beams to resist reversals of bending moment. Similarly, steel bars are required on all faces of columns too.

5. ACI special provision for seismic design:

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. Based on moment resisting capacity, there are three types of RC frames, i) SMRF (Special Moment Resisting Frame), ii) IMRF (Intermediate Moment Resisting Frame), iii) OMRF (Ordinary Moment Resisting Frame). Section 5.1 of this document describes some general requirements which are common to all frames. Specific requirements for each type of frame are presented in Section 5.2. 5.1. General requirements: (i) Materials:

To ensure adequate ductility and toughness under inelastic rotation, ACI Code 21.2.4 sets a minimum concrete strength of 3000 psi. The ACI Code 21.2.5 allows the use of Grades 40 and 60 reinforcement meeting the requirements of ASTM A615, provided that the actual yield strength does not exceed the specified yield by more than 18 ksi and that the actual tensile strength exceeds the actual yield strength by at least 25 percent.


Page 21 of 56



(ii) Lateral Reinforcement:

Confinement for concrete is provided by transverse reinforcement consisting of stirrups, hoops, and crossties, figure 19. To ensure adequate anchorage, a seismic hook [with a bend not less than 135° and a 6 times bar diameter (but not less than 3 in.) extension that engages the longitudinal reinforcement and projects into the interior of the stirrup or hoop] is used on stirrups, hoops and crossties, figure 20.

Hoops are closed ties that can be made up of several reinforcing elements, each having seismic hooks at both ends, or continuously wound ties with seismic hooks at both ends, figure 20. A crosstie is a reinforcing bar with a seismic hook at one end and a hook with not less than a 90° bend and at least a 6 bar diameter extension at the other end. The hooks or crossties must engage peripheral longitudinal reinforcing bars, figure 20.

Figure 19: Confinement in concrete using transverse reinforcements.


Page 22 of 56



Closed Hoops

Figure 20: Example of transverse reinforcement in column. 5.2. ACI provisions for special moment resisting frames (SMRF): 5.2.1. Provisions for beams: a. Size: The members must have a:

Clear span-to-effective-depth ratio of at least 4, ( ln/d ≥ 4) or (d ≤ ln/4). A width-to-depth ratio of at least 0.3, (b/d

≥ 0.3) or (b ≥ 0.3d).

A web width of neither less than 10 ″ nor more than the support width plus three-quarters of the flexural member depth on either side of the support. (bw ≥ 10″), bw ≤ width of support + ¾ × depth of beam + ¾ × depth of beam, figure 21.

Width of beam 3 4

depth of beam Support (Column)

Figure 21: Limitation on beam web width.

b. Flexural Reinforcement:

Neither positive nor negative moment strength at any section in a member may be less than one-fourth of the maximum moment strength at either end of the member.


Page 23 of 56



The positive moment capacity at the face of columns must be at least onehalf of the negative moment strength at the same location, figure 22. Minimum two reinforcing bars top and bottom, throughout the member.

ρmax ≤ 0.025.

Figure 22: Location and amount of longitudinal steel bars in beams.

c. Lap splices:

Not within the joints. Not within twice the member depth “2h”, from the face of a joint or at other locations where flexural yielding is expected. Lap splices must be enclosed by hoops or spirals with a maximum spacing of one-fourth of the effective depth or 4 ″.

Lap splice length =1.3 ld = (1.3 x 0.05 f y/ √fc′)db = 50 d b for fc′ 3 ksi and f y 40 ksi. =70 db for fc′ 3 ksi and fy 60 ksi.

Figure 23: Details of lapping steel reinforcement in seismic beams.


Page 24 of 56



d. Transverse reinforcement:

Transverse reinforcement in the form of hoops, figure 24 must be used over a length equal to twice the member depth measured from the face of the supporting member toward mid span at both ends, as per calculation, but fulfilling following conditions, figure 25. (i)

The first hoop must be located not more than 2 in from the face of the supporting member.

(ii)

Max spacing of the hoops over the length must not exceed: a. ¼ of effective depth of beam, b. 8 times dia of smallest longitudinal steel, c. 24 dia of hoop bar d. 12 in.

(iii) Elsewhere spacing not to exceed d/2.

Figure 24: Steel reinforcement in seismic beams.


Page 25 of 56



Figure 25: Location and amount of vertical stirrups in beams.

5.2.2. Provisions for columns: a. Size:

Each side at least 12 in. Shorter to longer side ratio

≥ 0.4.i.e. 12/12, 12/18, 12/24 OK; but 12/36 not

O.K. b. Flexural reinforcement:

0.01 ≤ ρg ≤ 0.06 c. Lap splices:

Within the middle half as shown in figure 26. Tie spacing at lap splice is d/4 or 4 inch; whichever is less where d is effective depth along least dimension of column, figure 26.


Page 26 of 56



Lap splice length =1.3 ld =(1.3 × 0.05 fy/ √fc′)db =50 db for fc′ 3 ksi and fy 40 ksi =70 db for fc′ 3 ksi and fy 60 ksi

Figure 26: Placing vertical bars and closed ties in columns.


Page 27 of 56



d. Transverse reinforcement:

ACI Code 21.4.4 specifies the use of minimum transverse reinforcement over length lo from each joint face. The length lo may not be less than (i)

the depth “D” of the member at the joint face or at the section where flexural yielding is likely to occur.

(ii)

one-sixth of the clear span (h c)of the member; or

(iii) 18 in. Max spacing of ties within length lo. (i)

Least lateral dimension of column /4,

(ii)

6 times the diameter (d b) of longitudinal reinforcement in column,

(iii) 4 + (14 – hx) /3; but not more than 6 inch and not less than 4 inch. Where hx = maximum horizontal spacing (inches) of hoop or crosstie legs on all faces of the column. Elsewhere spacing of ties is least of 6 d b or 6 inch.

Figure 27: Spacing of tie reinforcement.


Page 28 of 56



5.2.3. Beam-column joints:

In RC buildings, portions of columns that are common to beams at their intersections are called beam-column joints, figure 28. Since their constituent materials have limited strengths, the joints have limited force carrying capacity. When forces larger than these are applied during earthquakes, joints are severely damaged. Repairing damaged joints is difficult, and so damage must be avoided. Thus, beam-column joints must be designed to resist earthquake forces. ACI recommendations: To

provide

adequate

confinement

within

a

joint,

the

transverse

reinforcement used in columns must be continued through the joint in accordance with ACI 21.5.2, figure 29 and 30. To provide adequate development of beam reinforcement passing through a joint, ACI 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, figure 31. Beam longitudinal reinforcement that is terminated within a column must be extended to the far face of the column core. The development length of bars with 90° hooks must be not less than largest of 8d b, 6″, or ldh = fydb/(65 √fc′), figure 31. In interior joints, the beam bars (both top and bottom) need to go through the joint without any cut in the joint region. Also, these bars must be placed within the column bars and with no bends, figure 32.


Page 29 of 56



Figure 28: Beam column joints are critical parts of a building. Push pull forces on joint cause (a) loss of grip on beam bars in joint region; (b) distortion of joints. These result in irreparable damage in joints under strong seismic shaking.

Figure 29: Closed loop steel ties in beam-column joints. Such ties with 135 o hook resist the ill effects of distortion of joints.


Page 30 of 56



Figure 30: Providing horizontal ties in the joint --- three stage procedure is required.


Page 31 of 56



Figure 31: Anchorage of beam bars in beam column joint.

Figure 32: Anchorage of beam bars in interior joint. Diagram (a) and (b) show cross sectional views in plan of a joint region.


Page 32 of 56



5.3. ACI provisions for Intermediate moment resisting frames (IMRF): 5.3.1. Provision for beams: a. Sizes: No special requirement (Just as ordinary beam requirement). b. Transverse steel: Same as for SMRF. c. Lap: No special requirement (Just as ordinary beam requirement). d. Flexural Reinforcement: Less Stringent requirement as discussed below.

Neither positive nor negative moment strength at any section in a member may be less than one-fifth of the maximum moment strength at either end of the member. The positive moment capacity at the face of columns must be at least one-

third of the negative moment strength at the same location, fig. 33. Min two reinforcing bars top and bottom, throughout the member.

Figure 33: Location and amount of longitudinal steel bars in beams. These resist tension due to flexure. 5.3.2. Provision for columns: a. Sizes: No special requirement (Just as ordinary column requirement). b. Flexural steel: No special requirement (Just as ordinary column requirement). c. Lap: No special requirement (Just as ordinary column requirement). d. Transverse steel: For columns, within length lo from the joint face, the tie spacing, in accordance with ACI Code 21.12.5.2 may not exceed:


Page 33 of 56



• 8 times the diameter of the smallest longitudinal bar, • 24 times the diameter of the tie bar, • one-half of the smallest cross-sectional dimension of the column, • 12 in. Miscellaneous considerations:

•

IMRF are not allowed in regions of high seismic risk (zone 3 and 4), however, SMRF are allowed in regions of moderate seismic risk.

•

In regions of low or no seismic risk (zone 0 or 1) ordinary moment resisting frames OMRF are allowed but IMRF and SMRF may also be provided, ACI 21.2.1.2.

•

Unlike regions of high seismic risk, two way slab system without beams are allowed in regions of moderate seismic risk (zone 2a, 2b), ACI 21.12.6.

Shear Wall is a wall designed to resist lateral forces parallel to the plane of the wall. It is

sometimes referred to as vertical diaphragm or structural wall. To ensure adequate ductility, ACI 21.7.2 requires that structural walls have minimum shear reinforcement ratios in both longitudinal and transverse directions ρν and

ρn of 0.0025 and a maximum

reinforcement spacing of 18″. For details on shear walls, please refer to Seismic design of reinforced concrete and masonry structures by Priestley, page 362.

Figure 34: Cross sections of structural walls with boundary elements.


Page 34 of 56



6. Load combinations:

Broadly speaking, there are two types of codes in practice for design of structures. The codes like UBC, IBC etc. provide guidelines on the overall performance of structural systems, which also include procedures for determination of wind and seismic demand on structures. Other codes such as ACI, AISC provide material specific recommendations on the design of structures. Therefore designers use first type of codes for evaluating seismic forces and second type for design and detailing of earthquake resistant structures. Occasionally the two codes combined in this way may not be compatible. To avoid any mismatch resulting from such merger, every code explicitly mentions its counterpart code or codes. For example UBC 97 in its chapter 19 on concrete reproduces the complete code of ACI 318-95, and wherein it categorically states that UBC 97 shall be used in conjunction with ACI 318-95 for design of all reinforced concrete structures. Similarly IBC 2000 is compatible with ACI 318-02/05 and not with ACI 318-95/99 due to many reasons but mainly due to difference in load amplification and strength reduction factors. The chapter 7 of BCP SP-2007 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).The chapter is compatible with ACI 318-05.

Table 8: UBC-97 Load combinations and strength reduction factors (section 1612.2.1) Load Combinations

Strength Reduction Factors

1.4D

0.9 (flexure)

1.2D + 1.6L

0.85 (Shear)

1.1 (1.2D + 0.5L ± 1.0E)

0.70 (Tied )

1.1 (0.9D ± 1.0E)

0.75 (Spiral)

The amalgamation of the codes shall be done carefully when softwares such as SAP2000 or ETABS are used for analysis and design of structures. Though these softwares allow the use of UBC 97 for calculation of earthquake forces and ACI for design, when ACI 318-02/05 is selected for design, the software automatically loads the combination of ACI against the recommendation of BCP which uses the


Page 35 of 56



combination of UBC 97. Therefore to conform to recommendations of BCP, the designers should manually adjust the default ACI 318-05 load combinations and strength reduction factors according to UBC-97. Now according to the definition of E as given in section 1630.1.1 of UBC-97 and section 5.30.1.1 of BCP SP-2007, the final load combinations will come out to be as follows: As E = ρEh + Eν Eh represents the forces associated with the horizontal component of the earthquake load. While Eν represents loads resulting from the vertical component of the earthquake ground motion. In most of the case, ρ ≈ 1, therefore, E = E h + Eν Eν = 0.5CaID Therefore, E = Eh + 0.5CaID Therefore, in table 8, load combinations of UBC-97 including E become, 1.1{1.2D + 0.5L ± 1.0 (Eh + 0.5CaID)} D (1.32 ± 0.55CaI) + 0.55L ± 1.1Eh …..……(i) and 1.1{0.9D ± 1.0 (Eh + 0.5CaID)} D (0.99 ± 0.55CaI) ± 1.1Eh …………………(ii) Hence all UBC-97 can be reproduced as follows: U = 1.4D U = 1.2D + 1.6L U = (1.32 + 0.55CaI)D + 0.55L + 1.1Eh U = (1.32 + 0.55CaI)D + 0.55L − 1.1Eh U = (1.32 − 0.55CaI)D + 0.55L + 1.1Eh U = (1.32 − 0.55CaI)D + 0.55L − 1.1Eh U = (0.99 + 0.55CaI)D + 1.1Eh

−

U = (0.99 + 0.55CaI)D 1.1Eh U = (0.99 − 0.55CaI)D + 1.1Eh U = (0.99 − 0.55CaI)D − 1.1Eh


Page 36 of 56



Design Pb: Design the structure given in figure 35 as SMRF. The structure is located in

seismic zone 4 of UBC-97. The seismic source in the locality is such that magnitude of earthquake and slip rate may exceed 7.0 and 5.0 respectively. The closest distance of structure to the known seismic source is greater than 15 km. The soil type is stiff. The structure is 10′ high. Concrete compressive strength = 3 ksi, steel yield strength = 40 ksi and Modulus of elasticity of concrete = 3000 ksi. 20'

R.C.C Column

Slab

Beam

N

15'

PLAN

B

C

10'

F

G

E

H

7'

A

D

20'

15'

ELEVATION

Figure 35: 15′ x 20′ RCC frame structure.

Data Given: Material properties:

Concrete: fc′ = 3000 psi. Reinforcement: fy = 40,000 psi. Loads: Live load = 30 psf

Superimposed dead load = 3″ mud layer.


Page 37 of 56



Seismic design data:

The structure is located in a region of high seismic risk (seismic zone 4). Soil type = Stiff soil, SD (Table 16-J of UBC-97) Magnitude of earthquake ≥ 7.0; Slip rate ≥ 5.0, i.e, seismic source type = A (Table 16-U of UBC-97)

Solution: Step No 1: Sizes. 1. Columns:

As mentioned in section 5.2.2 of this document, the shortest dimension of column shall not be less than 12″, therefore assume 12″ × 12″ column. 2. Beam:

Assuming beam cross-sectional dimensions = 12″ × 24″. According to ACI 9.5.2.1, table 9.5 (a): Minimum thickness of simply supported beam = h min = l/16

l = clear span (ln) + depth of member (beam) [ACI 8.7] l = 20 – (12/12) + (24/12) = 21′ Depth (h) = (21/16) × (0.4 + 40000/100000) × 12 = 12.6″ (this is minimum requirement by ACI 9.5.2.1). Therefore, for h = 24″; d = h – 3 = 21″

Checks on limitation of section dimensions: ln/d = 14 × 12/21 = 8 > 4 (ACI 21.3.1.2 satisfied) Width/ depth = 12/24 = 0.5 > 0.3 (ACI 21.3.1.3 satisfied) Width = 12″ > 10″, O.K. Take 12″ × 24″ deep beams. 3. Slab:

lb/la = 20/15 = 1.33 < 2 “two way slab” hmin = perimeter/180 = 2 × (20 + 15) × 12/180 = 4.66 in

″ Assume hf = 6


Page 38 of 56



Step No 2: Loading. 1. Gravity loads: a. Slab:

Service dead load of/on slab = (6/12)×0.15+ (3/12)×0.12=0.105 ksf or 105 psf Service Live load over slab = 30 psf = 0.03 ksf

b. Beam:

B1

RCC Column B2

B2

RCC Beam

15'-0" Tributary area of slab on beam

45° 20'-0"

B1

Figure 36: Tributary area of slab on beam.

Load on beam B1: Service dead load =Service DL on slab × AB1/lcB1 + self wt of beam B1 = 0.105×{0.5(5+20)×7.5}/20+0.15×(12/12) x (24/12) = 0.524 k/ft Where, AB1 = tributary area of slab on beam B1.

lcB1 = c/c span of beam B1. Service live load = Service live load on slab x A/ lc

= 0.03 × {0.5(5+20)×7.5}/20= 0.141 k/ft Load on beam B2: Service dead load =Service DL on slab × AB2/lcB2 + self wt of beam B2 = 0.105× {0.5×7.5×15}/15+0.15×(12/12) x (24/12) = 0.425 k/ft Service live load = Service live load on slab x A/ lc

= 0.03 × {0.5×7.5×15}/15= 0.1125 k/ft


Page 39 of 56



c. Column: All columns in the system are biaxial. The service loads on column

will come from analysis of the frame.

2. Earthquake Load:

Static Lateral force procedure: Also referred to as equivalent static lateral force procedure. The total design base shear is: V = (CνI/RT) W Where, for the case under consideration, I = 1.00 {Standard occupancy structures, table 16-K of UBC-97} R = 8.5 {for SMRF, table 16-N of UBC-97} Cν = 0.64Nν {seismic zone 4/soil type SD, Table 16-Q of UBC-97} Nν = 1.0 (For seismic source type A

≥ 15 km, Table 16-T of UBC-97)

Therefore, Cν = 0.64 T = Ct (hn)3/4 Where, Ct = 0.030 (for reinforced concrete structures) hn = total height of the structure = 10 ′ Therefore, T = 0.030 × (10)3/4 = 0.169 sec Total seismic weight of the structure (W) is: W = wslab + wbeams + wcolumns + wmud = [(6/12)×15×20+(2×20+2×15)×(12/12)×(24/12)+4×(12/12)×(12/12)×10]×0.15 +(3/12) ×15×20×0.12 = 58.5 kip Therefore, V = (CνI/RT) W = {(0.64 x 1.00)/ (8.5 x 0.169)} x 58.5 = 26.06 kip

The total design base need not exceed the following: V = (2.5CaI/R) W = (2.5× 0.44Na× 1.00/8.5) × 58.5 = (2.5 × 0.44 × 1.00/8.5) × 58.5 = 7.57 k Ca = Seismic coefficient = 0.44Na (Soil type SD/zone 4, table 16-Q, UBC-97) Na = near source factor = 1.00 (Table 16-S of UBC-97)


Page 40 of 56



The total design base shear shall not be less than the following: V = 0.11CaIW = 0.11 × 0.44 × 1.00 × 58.5 = 2.83 kip In addition for seismic zone 4, the total base shear shall also not be less than the following: V = (0.8ZNνI/R) W = (0.8 × 0.4 × 1.00 × 1.00/8.5) × 58.5 = 2.20 kip Therefore design base shear is V = 7.57 kip The base shear shall be converted to storey forces in order to be used in the analysis. As the structure is single storey, the storey forces assigned to storey level in each direction will be as follows:

Figure 37: Base shear converted to storey forces in two mutually orthogonal directions of structure.

Load Combinations:


Page 41 of 56



According to UBC-97, various load combinations of gravity and EQ loads are: U = 1.4D …………………………………….(i) U = 1.2D + 1.6L……………………………..(ii) U = 1.1 (1.2D + 0.5L ± 1.0E)………………..(iii) U = 1.1(0.9D ± 1.0E)………………………...(iv) As discussed earlier, according to UBC-97, sec. 1630.1.1: E = ρ Eh + Eν In most of the case, ρ ≈ 1, therefore, E = Eh + Eν Eν = 0.5CaID Ca = 0.44Na; Na = 1.0 Therefore, Ca = 0.44, E = Eh + 0.5 × 0.44 × D = Eh + 0.22D Hence all load combinations of UBC-97 can be reproduced as follows: U = 1.4D U = 1.2D + 1.6L U = 1.56D + 0.55L ± 1.1Eh U = 1.08D + 0.55L ± 1.1Eh U = 1.23D ± 1.1Eh U = 0.75D ± 1.1Eh

Step No 3: Analysis.

Any structural analysis method may be used to analyze the frames shown above. However, for the sake of clarity, only one of the frames will be analyzed manually using moment distribution method while the rest of the frames will be analyzed by finite element method based software (SAP2000). Analysis for service dead load

Analysis is performed using Moment distribution method. Only analysis of N-S frame (15 ft wide) is presented here.


Page 42 of 56



Step a: Data required.

Dead load on frame = 0.425 k/ ft Modulus of elasticity of Concrete (E) = 3000 ksi Width of frame (LBeam) = 15′-0″ Height of frame (LColumn) = 10′-0″ IColumn = 12×123/12 = 1728 in4 IBeam = 12×243/12 = 13824 in4

Step b: Relative stiffnesses for frame members.

KAB = KBA = (EI/L)AB = EIcolumn/Lcolumn = 3000 × 1728/ (10 × 12) = 43200 k-in KBC = KCB = (EI/L)BC = EIBeam/LBeam = 3000 × 13824/ (15 × 12) = 230400 k-in KCD = KDC = (EI/L)CD = (EI/L)AB = 43200 k-in

Step c: Distribution factors.

DFAB = 1 (Hinge support) DFBA = KBA/ (KBA + KBC) = 43200/ (43200 + 230400) = 0.158 DFBC = KBC/ (KBC + KBA) = 230400/ (230400 + 43200) = 0.842 DFCB = KCB/ (KCB + KCD) = 230400/ (230400 + 43200) = 0.842 DFCD = KCD/ (KCD + KCB) = 43200/ (43200 + 230400) = 0.158 DFDC = 1 (Hinge support)

Step d: Fixed end moments.

MFAB = 0 (No applied transverse load) MFBA = 0 (No applied transverse load) MFBC = − wdl2/12 = − 0.425×152/12 = − 7.97 k-ft = − 95.625 k-in MFCB = wdl2/12 = 0.425×152/12 = 7.97 k-ft = 95.625 k-in MFCD = 0 (No applied transverse load) MFDC = 0 (No applied transverse load)


Page 43 of 56



Step e: Moment Distribution table. Joint Members K Cycles DF FEM 1 Balancing 2 CO Balancing 3 CO Balancing 4 CO Balancing 5 CO Balancing 6 CO Balancing CO 7 Balancing CO 8 Balancing CO 9 Balancing 10 CO 11 12

Balancing CO Balancing CO Balancing

Table 09: Moment distribution table A B C AB BA BC CB CD 43200 43200 230400 230400 43200 1 0.157895 0.842105 0.842105 0.157895 0 0 -177.75 177.75 0 0 28.06579 149.6842 -149.684 -28.0658 14.03 0.00 -74.84 74.84 0.00 -14.03 11.82 63.02 -63.02 -11.82 5.91 -7.02 -31.51 31.51 7.02 -5.91 6.08 32.45 -32.45 -6.08 3.04 -2.95 -16.22 16.22 2.95 -3.04 3.03 16.15 -16.15 -3.03 1.51 -1.52 -8.07 8.07 1.52 -1.51 1.52 8.08 -8.08 -1.52 0.76 -0.76 -4.04 4.04 0.76 -0.76 0.76 4.04 -4.04 -0.76 0.38 -0.38 -2.02 2.02 0.38 -0.38 0.38 2.02 -2.02 -0.38 0.19 -0.19 -1.01 1.01 0.19 -0.19 0.19 1.01 -1.01 -0.19 0.09 -0.09 -0.50 0.50 0.09 -0.09 0.09 0.50 -0.50 -0.09 0.05 -0.05 -0.25 0.25 0.05 -0.05 0.02 -0.02 0.01 -0.01 0.00

0.05 -0.02 0.02 -0.01 0.01 39.02

0.25 -0.13 0.13 -0.06 0.06 -39.02

-0.25 0.13 -0.13 0.06 -0.06 39.02

D DC 43200 1 0 0 -14.03 14.03 -5.91 5.91 -3.04 3.04 -1.51 1.51 -0.76 0.76 -0.38 0.38 -0.19 0.19 -0.09 0.09 -0.05

-0.05 0.02 -0.02 0.01 -0.01 -39.02

These values are compared with FE method based software SAP2000, shown in figure 38.


Page 44 of 56

0.05 -0.02 0.02 -0.01 0.01 0.00



Figure 38: Bending moment analysis of N-S frame in SAP2000.

The remaining analysis (for service load) of N-S frame and E-W frame is done in SAP2000 and complete analysis results are given in tables 10 and 11 respectively.

Table 10: Service values of load effects for N-S frame (15 ft) of the structure. Moment (kip-in) Member Axial (kip) Shear (kip) Positive (mid) (supports) (supports) Dead

AB BC CD

5.61 0.31 5.61

0.31 5.61 0.31

n/r 215 n/r

0 - 36 36

36 - 36 0

Live

AB BC CD

0.84 0 0.84

0 0.84 0

n/r 32.45 n/r

0 -5.52 5.52

5.52 -5.52 0

EQ

AB BC CD

2.53 -1.89 -2.53

1.89 2.53 1.8

n/r 0 n/r

0 227.57 -227.57

-227.57 -227.57 0


Page 45 of 56



Table 11: Service values of load effects for E-W frame (20 ft) of the structure. Moment (kip-in) Member Axial (kip) Shear (kip) Positive (mid) (supports) (supports) Dead

Live

EQ

EF FG GH

6.9 0.62 6.9

0.62 6.9 0.62

n/r 339 n/r

0 -75 75

75 -75 0

EF FG

1.41 Nil

Nil 1.41

n/r 69.31

0 -15.29

15.29 -15.29

GH

1.41

Nil

n/r

15.29

0

EF FG GH

1.89 -1.89 -1.89

1.89 1.89 1.89

n/r 0 n/r

0 227.61 -227.61

-227.61 -227.61 0

The factored values according to the calculated load combinations for all frame members (beams and columns) are summarized in tables 12 through 15. Table 12: Factored axial forces and shear forces in N-S frame (15 ft). Axial Force (kip) 1.4D 1.2D + 1.6L 1.56D + 0.55L + 1.1Eh 1.56D + 0.55L - 1.1Eh 1.08D + 0.55L + 1.1Eh 1.08D + 0.55L - 1.1Eh 1.23D + 1.1Eh 1.23D - 1.1Eh 0.75D + 1.1Eh 0.75D + 1.1Eh Max + Max -


AB 7.85 8.08 12.00 6.43 9.30 3.74 9.68 4.12 6.99 6.99 12.00 0.00

BC 0 0 0 0 0 0 0 0 0 0 0 0

CD 7.85 8.08 6.43 12.00 3.74 9.30 4.12 9.68 1.42 1.42 12.00 1.42

Shear Force (kip) AB 0 0 0 0 0 0 0 0 0 0 0 0

BC 7.85 8.08 12.00 6.43 9.30 3.74 9.68 4.12 6.99 6.99 12.00 0.00

CD 0 0 0 0 0 0 0 0 0 0 0 0

Page 46 of 56


1.4D 1.2D + 1.6L 1.56D + 0.55L + 1.1Eh 1.56D + 0.55L - 1.1Eh 1.08D + 0.55L + 1.1Eh 1.08D + 0.55L - 1.1Eh 1.23D + 1.1Eh 1.23D - 1.1Eh 0.75D + 1.1Eh 0.75D + 1.1Eh Max + Max -


Table 13: Factored moments in N-S frame (15 ft). Bending Moment AB BC CD AB BC CD Support Support Support Positive Positive Positive A B C n/r 301.00 n/r 0 -50.40 50.40 n/r 309.92 n/r 0 -52.03 52.03 n/r 353.25 n/r 0 191.13 -191.13 n/r 353.25 n/r 0 -309.52 309.52 n/r n/r n/r n/r n/r n/r n/r n/r

250.05 250.05 264.45 264.45 161.25 161.25 353.25 0.00

n/r n/r n/r n/r n/r n/r n/r n/r

0 0 0 0 0 0 0 0

208.41 -292.24 206.05 -294.61 223.33 223.33 223.33 - 3 0 9 .5 2

-208.41 292.24 -206.05 294.61 -223.33 -223.33 309.52 - 223. 33

AB Support B 50.40 52.03 -191.13 309.52

BC CD Support Support C D -50.40 0 -52.03 0 -309.52 0 191.13 0

-208.41 292.24 -206.05 294.61 -223.33 -223.33 309.52 -223.33

-292.24 208.41 -294.61 206.05 -277.33 -277.33 208.41 -309.52

Table 14: Factored axial forces and shear forces in E-W frame (20 ft).


Axial Force Shear Force EF FG GH EF FG GH 9.66 0 9.66 0 9.66 0 10.536 0 10.536 0 10.536 0 13.6185 0 9.4605 0 13.6185 0 9.4605 0 13.6185 0 9.4605 0 10.3065 0 6.1485 0 10.3065 0 6.1485 0 10.3065 0 6.1485 0 10.566 0 6.408 0 10.566 0 6.408 0 10.566 0 6.408 0 7.254 0 3.096 0 7.254 0 7.254 0 3.096 0 7.254 0 13.6185 0 13.6185 0 13.6185 0 0 0 3.096 0 0 0


Page 47 of 56

0 0 0 0 0 0 0 0




Table 15: Factored moments in E-W frame (20 ft). Bending Moment EF FG GH EF FG GH EF FG GH Support Support Support Support Support Support E F G F G H Positive Positive Positive n/r 474.6 n/r 0 -105 105 105 -105 0 n/r 517.696 n/r 0 -114.46 114.464 114.464 -114.46 0 n/r 566.961 n/r 0 124.962 -124.96 -124.96 -375.78 0 n/r 566.961 n/r 0 -375.78 375.781 375.781 124.962 0 n/r n/r n/r n/r n/r n/r n/r n/r

404.241 404.241 416.97 416.97 254.25 254.25 566.961 0

n/r n/r n/r n/r n/r n/r n/r n/r

0 0 0 0 0 0 0 0

160.962 -160.96 -160.96 -339.78 -339.78 339.781 339.781 160.962 158.121 -158.12 -158.12 -342.62 -342.62 342.621 342.621 158.121 194.121 -194.12 -194.12 -306.62 194.121 -194.12 -194.12 -306.62 194.121 375.781 375.781 160.962 -375.78 -194.12 -194.12 -375.78

Step No 4: Design.

From the above analysis results, the beams and columns are designed for the following demands. Table 16: Maximum demand on members. Axial Force (kips)

Shear Force (kip) 12 13.6185

Beams

BC FG

0 0

Columns

AB,CD EF, GH

12 13.618

0 0

Bending moment (in-kip) + 353 309 566.96 375 309 375

223 194

From above data, the moment demand of 566.96 corresponds to 0.76 in2, quite below the Asmin mark of 1.26 in2. This means that all moment demands will give steel area less than Asmin.


Page 48 of 56

0 0 0 0 0 0 0 0



i) Design of Beams:

Beam

E-W

Table 17: Flexural design of beams Flexural design steel area from Design steel area from analysis Asmax Bar 2 Positive Negative (in ) Positive Negative used steel steel steel steel (in2) (in2) (in2) (in2)

# of bars Positive bars

0.47

0.41

5.12

1.36

1.26

#5

5

0.76

0.50

5.12

1.26

1.26

#5

5

5 (curtailing 3 bars) 5 (curtailing 3 bars)

Beam N-S Beam

Negative bars

Note: Asmin = 1.26 in2

Beam E-W Beam N-S Beam

Table 18: Shear design of beam Design shear ΦVc analysis Stirrup spacing (kip) (kip) 12 20.7 as per SMRF requirements 13. 6185 20.7 as per SMRF requirements

ii) Design of column:

All columns are biaxial in behavior. However, all columns are designed using SAP 2000. Column flexural reinforcement from SAP = 2.10 in2 Using #5 bars, with bar area Ab = 0.30 in2 No. of bars = 2.10/0.31 = 6.77 ≈ 8 bars.

Figure 39: Design results of SAP2000.


Page 49 of 56



ACI Check lists for SMRF: Check list for beams: Flexural Reinforcement:

Positive bars at any section ≥ ¼ maximum bars at either ends of beam: ¼ maximum bars at either ends of beam = ¼ (5 #5 bars) Positive bars at any section = 5 #5 bars > ¼ maximum bars at either ends of beam

O.K. Negative bars at any section ≥ ¼ maximum bars at either ends of beam: ¼ maximum bars at either ends of beam = ¼ (5 #5 bars) Negative bars at any section = 2 #5 bars > ¼ max bars at either ends of beam

O.K. Positive bars at face of column ≥ ½ Negative bars at face of column: ½ Negative bars at face of column = ½ × 5# 5 bars Positive bars at face of column = 5# 5 bars ≥ ½ Negative bars at face of column

O.K. Minimum two reinforcing bars top and bottom, throughout the member , O.K.

ρmax ≤ 0.025: ρmax = 5 × 0.30/ (12 × 21) = 0.0059 < 0.025, O.K. At any section, the

Lap splices:

Not to be provided within the joints. Not to be provided within 2h = 2 × 24 = 48″, from the face of the joint or at other locations where flexural yielding is expected. Maximum spacing of hoops on Lap splices is least of: d/4 = 21/4 = 5.25″ 4″ Therefore, maximum spacing of hoops on Lap splice = 4″

Lap splice length =(1.3 × 0.05 f y/ √fc′)×db = 30″


Page 50 of 56



Transverse reinforcement:

Transverse reinforcement shall be provided over a length of 2h = 2 × 24 = 48 ″, measured from the face of the supporting member toward mid span at both ends. The first hoop shall be located at a distance of 2″ from the face of the supporting member. Max spacing of the hoops over the length must not exceed least of: a. ¼ × d = ¼ × 21 = 5.25″ b. 8db = 8 × 5/8 = 5″ c. 24dhoop bar = 24 × 3/8 = 9″

d. 12″ Therefore, maximum spacing of hoops over a region of 48 ″ from both ends of beam ≤ 5″. Finally, using spacing of 5 ″.

Elsewhere spacing shall not exceed d/2 = 21/2 = 10.5″. Checklist for columns: Flexural reinforcement:

0.01 ≤ ρg ≤ 0.06: Therefore, for 8 #5 bars, As/bh = 8 × 0.30/ (12 × 12) = 0.0166, O.K.

Lap splices:

To be provided within the middle half of column height. Maximum spacing of ties on Lap splices is least of: d/4 = {12 – 1.5 – (3/8) – (5/8)/2}/4 = 2.45 ″ 4″ Therefore, maximum spacing of ties on Lap splices = 2.45 ″ ≈ 3″

Lap splice length =(1.3 × 0.05 fy/ √fc′)db = 30″

Transverse reinforcement:

Length lo from each joint face is least of: (i) Depth of member at joint face = 12″.


Page 51 of 56



(ii) hc/6 = 10 × 12/6 = 20″ (iii)18″ Therefore lo = 12″

Maximum spacing of ties within length lo is least of: (i)

Least lateral dimension of column/4 = 12/4 = 3″

(ii)

6db = 6 × 5/8 = 3.75″,

(iii) 4″ ≤ 4 + (14 – hx) /3 ≤ 6″ = 4 + [14 – {12 – 2 × 1.5 – 2 × (3/8)}]/3 = 5.92″ Therefore, maximum spacing of ties within length lo = 3″

Elsewhere spacing of ties shall be least of: 6 db = 6 × 5/8 = 3.75″ 6″ Finally, however, 3″ of spacing will be provided throughout the column height.

Beam-column joints:

The transverse reinforcement used in columns shall be continued through the joint in accordance with ACI 21.5.2. Column dimension parallel to beam longitudinal bar ≥ 20 × Beam long bar: Column dimension parallel to beam longitudinal bar = 12″ 20 × 5/8 = 12.5″ ≈ 12″, O.K.

The development length of beam bars in columns with 90° hooks is not to be less than largest of: a. 8db = 8 × 5/8 = 5″ b. 6″ c. ldh = fydb/(65 √fc′) = 40000 × (5/8)/ {65 × √(3000)} = 7″ Therefore, development length = 7″


Page 52 of 56



Step No 5: Drafting.

5bars A #5

#5 2bars

B

C

#5 5bars

C A #3 ties @3" c/c

B

C

#3, 2 legged stirrups



@ 5" c/c upto 2h = 4'

@ 10.5" c/c

@ 5" c/c upto 2h = 4'

8 #5 bars E

E

D

D

Lap splice 3'-0"

Figure 40: Typical detail of frame.

5 #5 bars

Development length of beam bars in columns = 10" > 7" Development length of column bars in beam = 22" > 7" 2"

5 #5 bars

#3 ties @3" c/c


Page 53 of 56



Figure 41: Details at C. 5 #5 Bars

2 #5 Bars

6"

#3, 2 legged @ 5" c/c

18"

2 #5 Bars

6"

#3, 2 legged @ 5" c/c

18"

5 #5 Bars

#3, 2 legged @ 10.5" c/c

18"

5 #5 Bars

12"

5 #5 Bars

12"

SectionA-A

12"

SectionB-B

SectionC-C

Figure 42: Beam sections.

8 #5 Bars

#3 ties @ 3" c/c

12"

8 #5 Bars

#3 ties @ 3" c/c

12"

12"

12"

Section D-D

Section E-E

Figure 43: Column sections.

Table 11: Minimum Clear Cover S.No

LOCATION

MINIMUMCOVER

1.

FOOTING

3"

2.

COLUMN

1-1/2"

3.

SLAB

3/4"

4.

BEAM

1-1/2"


6"

Page 54 of 56



Table 12: SPLICE LENGTH, ACI 21.3.2.4, class B splice.

MINIMUM SPLICE LAP LENGTH (INCHES) fc' =3,000psi

BAR DIA

fy = 40,000psi

SLAB

BEAM / COLUMN

3/8"

16

23

1/2"

21

30

5/8"

26

38

3/4"

-

45

Table 13: STANDARD HOOKS, ACI 7.1. 2 / B

BAR DIA

A(in.)

3/8"

2.5

2.5

4.5

1/2"

2.5

3.0

6.0

5/8"

2.5

4.0

7.5

3/4"

3.0

4.5

9

B(in.)

C(in.)

A

B

B/2

C

K O O H ° 0 8 1

K O O H ° 0 9

Table 14: DEVELOPMENT LENGTH IN TENSION, ACI 21.5.4.1.

DEVELOPM ENT LEN GTH (INCHES ) fc' =3,000psi

BAR DIA

STRAIGHT BARS

fy = 40,000psi WITH STANDARD HOOK

3/8"

18

6

1/2"

24

6

5/8"

30

8

3/4"

36

9


Page 55 of 56

Earthquake Engineering Center, N-W.F.P UET Peshawar


References



Design of Concrete Structures by Nilson, Darwin and Dolan (13 th ed.)



Learning Earthquake Design and Construction (Earthquake Tips) by C. R Murty, Indian Institute of Technology Kanpur.



UBC-97, Volume 2 (Uniform Building Code)



BCP SP-2007 (Building Code of Pakistan Seismic Provisions-2007)



ACI 318-02/05


Page 56 of 56

L6-Introduction-to-EQ-Resistant-Design-RC Structures

Recommend Documents