Earthquake Resistant Design of RC Structures

7/29/2019 Earthquake Resistant Design of RC Structures

1/56

Department of Civil Engineering, UET Peshawar Notes on Earthquake Resistant Design of R C Structures

Prof. Dr. Qaisar Ali (http://www.drqaisarali.com) Page 1 of 56

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.


2/56



1. Introduction:Earthquake results from the sudden movement of the tectonic plates in the earths

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 earths

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: Earths tectonic plates.


3/56



Figure 02: Motions caused by body and surface waves.

Figure 03: Epicentre and focus.


4/56



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.


5/56



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 designers

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


6/56



0

1

2

3

Displacement

(a) (b) (c)

StoreyHeight

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.

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

(c) third mode.


7/56


8/56


9/56


10/56


11/56


12/56


13/56



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 = (CI/RT) W

Where,

C = 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.

Where design snow loads exceed 30 psf (1.44 kN/m2), 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)


14/56


15/56


16/56


17/56



12'-0"w = 800 kip1

w = 800 kip2

w = 800 kip3

w = 800 kip4

w = 700 kip5

60'-0"

The fundamental period Tmay be computed by using the following formula:

Where,

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

i = 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, i, 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.

Figure 16: Two storey frame structure.


18/56



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 = (CI/RT) W

From table 16-R, C = 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 levelx of the structure is given as:

Fx = (VFt)xhx/ihi {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}

ihi = 80012+80024+80036+80048+70060 = 138000 kip

Therefore for the case under consideration, Force for storey 1 is:

F1 = (3830) 800 12/ {(138000)} = 27 kip

Storey forces for other stories are given in table 6 below.


19/56


20/56



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


21/56


22/56


23/56


24/56


25/56


26/56


27/56


28/56


29/56



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 transversereinforcement 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 ajoint, 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 beextended to the far face of the column core. The development length of bars

with 90 hooks must be not less than largest of 8db, 6, or ldh = fydb/(65 fc),figure 31.

In interior joints, the beam bars (both top and bottom) need to go throughthe joint without any cut in the joint region. Also, these bars must be placed

within the column bars and with no bends, figure 32.


30/56


31/56


32/56


33/56



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:


34/56


35/56


36/56



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 = Eh + E

Eh represents the forces associated with the horizontal component of the earthquake

load. While E represents loads resulting from the vertical component of the

earthquake ground motion.

In most of the case, 1, therefore,

E = Eh + E

E = 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


37/56


38/56


39/56


40/56


41/56


42/56


43/56


44/56


45/56


46/56


47/56



Table 13: Factored moments in N-S frame (15 ft).

Bending Moment

AB BC CD AB BC CD AB BC CD

Positive Positive Positive

Support

A

Support

B

Support

C

Support

B

Support

C

Supp

D

1.4D n/r 301.00 n/r 0 -50.40 50.40 50.40 -50.40 0

1.2D + 1.6L n/r 309.92 n/r 0 -52.03 52.03 52.03 -52.03 0

1.56D + 0.55L + 1.1Eh n/r 353.25 n/r 0 191.13 -191.13 -191.13 -309.52 0

1.56D + 0.55L - 1.1Eh n/r 353.25 n/r 0 -309.52 309.52 309.52 191.13 0

1.08D + 0.55L + 1.1Eh n/r 250.05 n/r 0 208.41 -208.41 -208.41 -292.24 0

1.08D + 0.55L - 1.1Eh n/r 250.05 n/r 0 -292.24 292.24 292.24 208.41 0

1.23D + 1.1Eh n/r 264.45 n/r 0 206.05 -206.05 -206.05 -294.61 0

1.23D - 1.1Eh n/r 264.45 n/r 0 -294.61 294.61 294.61 206.05 0

0.75D + 1.1Eh n/r 161.25 n/r 0 223.33 -223.33 -223.33 -277.33 0

0.75D + 1.1Eh n/r

161.25

n/r

0

223.33

-223.33

-223.33

-277.33

0Max + n/r 353.25 n/r 0 223.33 309.52 309.52 208.41 0Max - n/r 0.00 n/r 0 -309.52 -223.33 -223.33 -309.52 0

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

Axial Force Shear Force

EF FG GH EF FG GH

1.4D 9.66 0 9.66 0 9.66 0

1.2D + 1.6L 10.536 0 10.536 0 10.536 0

1.56D + 0.55L + 1.1Eh 13.6185 0 9.4605 0 13.6185 0

1.56D + 0.55L - 1.1Eh 9.4605 0 13.6185 0 9.4605 0

1.08D + 0.55L + 1.1Eh 10.3065 0 6.1485 0 10.3065 0

1.08D + 0.55L - 1.1Eh 6.1485 0 10.3065 0 6.1485 0

1.23D + 1.1Eh 10.566 0 6.408 0 10.566 0

1.23D - 1.1Eh 6.408 0 10.566 0 6.408 0

0.75D + 1.1Eh 7.254

0

3.096

0

7.254

0

0.75D + 1.1Eh 7.254 0 3.096 0 7.254 0

Max + 13.6185 0 13.6185 0 13.6185 0Max - 0 0 3.096 0 0 0


48/56


49/56


50/56


51/56



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 2from the face of the supportingmember.

Max spacing of the hoops over the length must not exceed least of:a. d= 21 = 5.25b. 8db= 8 5/8 = 5c. 24dhoop bar= 24 3/8 = 9

d. 12Therefore, maximum spacing of hoops over a region of 48 from both ends of

beam 5. Finally, using spacing of5.

Elsewhere spacing shall not exceedd/2= 21/2 = 10.5.Checklist for columns:

Flexural reinforcement:

0.01 g 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 = {121.5(3/8)(5/8)/2}/4 = 2.45 4Therefore, 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.


52/56



(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) 44 + (14hx) /3 6= 4 + [14{122 1.52 (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 6Finally, 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 jointin 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 = 5b. 6c. ldh = fydb/(65 fc) = 40000 (5/8)/ {65 (3000)} = 7Therefore, development length = 7


53/56


54/56



4. BEAM 1-1/2"

2.

3.

COLUMN

SLAB

1. FOOTING

1-1/2"

3/4"

3"

S.No LOCATION MINIMUM COVER

Table 11: Minimum Clear Cover

18"

12"

6"

5 #5Bars

5 #5Bars

#3, 2 legged

@ 5" c/c 18"

12"

6"

2 #5Bars

5 #5Bars

#3, 2 legged

@ 5" c/c 18"

12"

6"

2 #5Bars

5 #5Bars

#3, 2 legged

@ 10.5" c/c

Section A-A Section B-B Section C-C

12"

12"

8 #5Bars

#3 ties @ 3" c/c

Section D-D

12"

12"

8 #5Bars

#3 ties @ 3" c/c

Section E-E

Figure 41: Details at C.

Figure 42: Beam sections.

Figure 43: Column sections.


55/56



MINIMUM SPLICE LAP LENGTH (INCHES)

26

-

5/8"

3/4"

SLAB

DIA

16

211/2"

3/8"

BAR

BEAM / COLUMN

38

45

23

30

fc' = 3,000 psi fy = 40,000 psi

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

6.0

4.5

C(in.)

3.02.51/2"

2.55/8"

3/4" 3.0

4.0

4.5

7.5

9

2.5

DIA

BAR

3/8"

A(in.)

2.5

B(in.)

B/2 C

B

A

B/2

Table 13: STANDARD HOOKS, ACI 7.1.

90

HOOK

180HOOK

DEVELOPMENT LENGTH (INCHES)

30

36

5/8"

3/4"

STRAIGHT BARS

DIA

18

241/2"

3/8"

BAR

WITH STANDARD HOOK

8

9

6

6

fc' = 3,000 psi fy = 40,000 psi

Table 14: DEVELOPMENT LENGTH IN TENSION, ACI 21.5.4.1.


56/56

Date post:	04-Apr-2018
Category:	Documents
Upload:	prabhu81
View:	233 times
Download:	2 times

Earthquake Resistant Design of RC Structures

Documents