Seismic Design and Detailing_pp131-150

Adam C Abinales f.asep, pice

Seismic Design of Concrete Structures 5.

Structural Design and Detailing for Earthquake E arthquake Resistance

5.1 5.1

Eart Earthq hqua uake ke Loa Load d Comb Combin inat atio ions ns:: Stre Streng ngth th Des Desig ign n

5.1.1 Earthquake Loads and Modeling Requirements . Structures shall be designed for ground motion producing structural response and seismic forces in any horizontal direction. The earthquake loads that shall be used in the load combinations combinations (set forth forth in NSCP NSCP Section 203) shall be be in accordance with the requirements of NSCP Section 208.5.1.1. E = ρ Eh + Ev

NSCP eq. 208-1

Em = Ω o Eh

NSCP eq. 208-2

where E

= the earthqu earthquake ake load load on an elemen elementt of the structur structure e resulting resulting from from the combinati combination on of the horizontal component E h and the vertical component E v .

E h = the earthq earthquak uake e load load due due to to the the base base shea shearr V or V or the design lateral force F p. E m = the estimated estimated maximu maximum m earthquak earthquake e force that that can be be developed developed in the struct structure ure and and used in the design of specific elements of the structure. E v = the load load effect effect resulti resulting ng from the the vertical vertical componen componentt of the earthqu earthquake ake ground ground motion motion and is equal to an addition of 0.5 C a*I*D to the deal load effect, D, for strength design method, and may be taken as zero for allowable (or working) stress design method. o

= the seismic seismic force force amplificat amplification ion factor factor that that is required required to accoun accountt for structur structure e overstrength. (Section 208.5.3.1). Page 131/7



Structural Design and Detailing for Earthquake E arthquake Resistance = Reliab Reliabili ility/ ty/Red Redund undanc ancy y Factor Factor determ determin ined ed as: ρ = 2 −

6.1

NSCP eq. 208-3

r max AB

r max = the maximum maximum element-s element-story tory shear shear ratio; ratio; the the ratio ratio of the design design story story shear shear in the most most heavily loaded single element to the total design story shear. AB = the groun ground d floor floor area of the the structu structure re expres expresses ses in in m 2.

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Structural Design and Detailing for Earthquake Resistance E xample Problem 5.1. A four-storey concrete building of special moment resisting frame system has been analyzed. Beam A-B and column C-D are elements of SMRF. Structural analysis yielded the following results due to dead load, office building live load and lateral seismic forces: Find the following:

 Structure

1.

Strength design moment at beam end A.

 Seismic

2.

Strength design axial load and moment at column top C.

 Distance

A

B 8000

C

D

8000

 Soil

8000

 I =

Roof

 0 0 0 4

4th 0 0 0 4

3rd 0 0 0 4

A

B

C

2nd

0 0 0 4

D

GF

is located in Zone 4;

source type: A to seismic source = 10 km

profile type: S D

1.0

=1.1; f 1 = 0.5

Member/ Stress

Dead Load D

Live Load L

Lateral Seismic Eh

Beam moment at A

135 kN-m

65 kN-m

165 kN-m

Column C-D axial load

400 kN

180 kN

490 kN

Column moment at C

55 kN-m

30 kN-m

Page 133/7

220 kN-m



Structural Design and Detailing for Earthquake Resistance S olution and discussion: F ind the strength design moment at beam end A. T o determine strength design moments for design, the earthquake component E must be combined with the dead and live load components D and L, as illustrated below. Determine earthquake load E E = ρ Eh + Ev Sect. 208.5.1.1 where, the moment due to vertical earthquake force is Ev = 0.5CaID; in which Ca = 0.44Na = 0.44(1.0) Ev = 0.5(0.44)(1.0)(135) Ev = 29.7 kN - m

while the moment due to horizontal earthquake force is Eh = 165 kN - m

then, E = 1.1(165) + 29.7 = 211.2 kN - m

Apply load combinations involving earthquake. The basic load combinations for strength design A per Section 203.3.1 is 1.2D + 1.0E 0.9D ± 1.0E

+ 1.0 f L 1

NSCP eq. 203-5 NSCPPage eq.134/7 203-6



Structural Design and Detailing for Earthquake Resistance F or reinforced concrete frame, the above equations shall be multiplied by 1.1 per Section 409.3.3 and become 1.32D + 1.10E + 1.10 f L 1

0.99D ± 1.10E

therefore, strength design moment at beam end A M A = 1.32MD + 1.10ME + 1.10 f 1ML

M A = 0.99MD ±1.10ME

M A = 1.32(135) + 1.10(211.2) + 1.10(0.5)(65)

M A = 0.99(135) ± 1.10(211.2)

M A = 446.27 kN - m and

M A = 365.97 kN - m or − 98.67 kN - m

F ind the strength design axial load and moment at column top C. Determine the earthquake load E E = ρ Eh + Ev

where, for the axial load E = 1.1(490) + 0.5(0.44)(1.0)(400) = 627 kN

for the moment at top E = 1.1(220) + 0.5(0.44)(1.0)(55) = 254.1kN - m

Page 135/7



Structural Design and Detailing for Earthquake Resistance Apply load combinations involving earthquake. A for the axial load P c

PC = 1.32PD + 1.10PE + 1.10 f 1 pL PC = 1.32( 400) + 1.10(627) + 1.10(0.5)(180) PC = 1316.7 kN

and PC = 0.99P D ±1.10PE PC = 0.99(400) ± 1.10(627) PC = 1085.7 kN or − 293.7 kN

therefore,

PC = 1316.7 kN or − 293.7 kN

for the moment M c

MC = 1.32MD + 1.10ME + 1.10 f 1ML MC = 1.32( 55) + 1.10( 254.1) + 1.10(0.5)(30) Mc = 368.61kN - m

and MC = 0.99MD ±1.10ME MC = 0.99(55) ± 1.10(254.1) MC = 333.96 kN - m or − 225.06 kN - m

Note that the column section capacity must be designed for the interaction of P c = 1316.7 kN compression and M c = 368.61 kN-m (for D+L+E ), and the interaction of P c = 293.7 kN tension and Page 136/7 M c = -225.06 kN-m (for D+E ).



Structural Design and Detailing for Earthquake Resistance

5.2

In-situ Reinforced Concrete Design and Detail Reinforced concrete for most structures is generally desirable because of its availability and economy, and its stiffness can be used to advantage to minimize seismic deformations and hence reduce the damage to non-structure. Difficulties arise due to reinforcement congestion when trying to achieve high ductility in framed structures, and at the time of writing the problem of detailing beam-column joints to withstand strong cyclic loading had not been resolved. It should be recalled that no amount of good detailing will enable an ill-conceived structural form to survive a strong earthquake.

5.2.1 Seismic Response of Reinforced Concrete. Concrete Even in well-designed reinforced concrete members, the root cause of failure under earthquake loading is usually concrete cracking. Degradation occurs in the cracked zone under cyclic loading. Cracks do not close up properly when the tensile stress drops because of permanent elongation of reinforcement in the crack, and aggregate interlock is destroyed in a few cycles. In hinge and joint zones, reversed diagonal cracking breaks down in the concrete between the cracks completely, and sliding shear failure occurs. Refer to Figure 5.1.

Figure 5.1. Progressive failure of reinforced concrete hinge zone under seismic loading. Page 137/7




5.2.2 Principles of Earthquake - Resistant Resistant Design. Design In reinforced concrete structures, the essential features of earthquake-resistance are embodied in ensuring the following:



Beams should fail before columns. “Strong Column - Weak Beam” Concept. Design codes require that earthquake-induced energy be dissipated by plastic hinging of the beams, rather than the columns. This hypothesis is due to the fact that compression members such as columns have lower ductility than flexure-dominant beams. If columns are not stronger than beams framing to a joint, inelastic action can develop in the column. Furthermore, the consequence of a column failure is far more severe than a local beam failure. This concept is ensured by the following inequality:

where

 6  ΣMcol ≥  ΣMbeam  5 

M col = sum of moments at the faces of the joint corresponding to the nominal flexural strength of the columns framing to that joint; M beam = sum of moments at the faces of the joint corresponding to the nominal flexural strengths of the beams framing into that joint. In T-beam construction, where the slab is in tension under moments at the face of the joint, slab reinforcement within the effective slab width has to be assumed to contribute to flexural strength is the slab reinforcement is developed at the critical section for flexure. Page 138/7



Structural Design and Detailing for Earthquake Resistance 

Failure should be in flexure rather than in shear. To prevent shear failure occurring before bending failure, it is good practice to design that the flexural steel in a member yields while the shear reinforcement is working at a stress less than yield (say normally 90%). In beams, a conservative approach to safety in shear is to make the shear strength equal to the maximum shear demands which can be made on the beam in terms of its bending capacity. Referring to Figure 5.2, the shear strength of the beam should correspond to V max =

Mu1 − Mu 2

where

l

+ V DL

V DL is the dead load shear force M u is the factored moment, determined as

Mu = A s f y z

As is the steel area in the tension zone f y is the maximum steel strength after hardening, say 95% z is the lever arm

Page 139/7



Structural Design and Detailing for Earthquake Resistance M u1

M u2 l

f y (+)

95 percentile

Figure 5.2. Shear strength consideration for reinforced concrete beams.

h

As

(-)

b

ε 

Premature failure of joints between members should be prevented. Joints between members such as beam-column joints are susceptible to failure earlier than the adjacent members due to destruction of a joint zone, in a manner similar to that shown in Figure 5.1. This is particularly true mostly to exterior columns.



Ductile rather than brittle failure should be obtained. In earthquake engineering, the effect of material behavior on the choice of the method of analysis is a much greater issue than in non-seismic engineering. The problem can be divided into two categories depending on whether the material behavior is brittle or ductile, i.e. whether it can be considered linear elastic or inelastic. The normal analytical and design methods of dealing with these two states are summarized in the following table. See next page. Page 140/7



Structural Design and Detailing for Earthquake Resistance Material Behavior

Method of Analysis Equivalentstatic

Linear elastic (brittle)

Linear dynamic

Equivalentstatic Inelastic (ductile) Linear dynamic

Seismic Loading

Design Provisions

Arbitrarily reduced

1. Working stress or factored ultimate stress design, plus imposed nominal ductility

Arbitrarily reduced

2. Working stress or factored ultimate stress design, plus imposed nominal ductility

Full

3. Ultimate stress design, plus imposed nominal ductility

Arbitrarily reduced

4. Working stress or factored ultimate stress design, plus imposed arbitrary ductility*

Arbitrarily reduced

5. Working stress or factored ultimate stress design, plus imposed arbitrary ductility*

Arbitrarily reduced

6. Working stress or factored ultimate stress design, plus imposed arbitrary ductility* 7. Structure intended to remain elastic, but nominal ductility imposed

Full Inelastic dynamic

Full

8. Ductility demands found from plastic hinge rotations Page 141/7




5.2.3 Available Ductility for Reinforced Concrete Members . The available section ductility of a reinforced concrete member is most conveniently expressed as the ratio of its curvature at ultimate moment from its first u to its curvature at its first yield y. The expression u / y may be evaluated principles, the answers varying with the geometry of the section, the reinforcement arrangement, the loading and the stress-strain relationships of the steel and the concrete. 

Single reinforced sections. sections Consider conditions at first yield and ultimate moment as shown in Figure 5.3. ε ce d'

ε cu

f ce

A s' k d

f c m = 0.85f'c a

c

d

Figure 5.3. Reinforced concrete section in flexure.

A s b

ε sy = f y /E s strain

f y stress

(a) a t first yield

ε sy > f y /E s strain

f y stress

(b) at ultim ate

Assuming an under-reinforced section, first yield will occur in the steel, and the curvature f ∈ sy φ y = ; in which ∈ sy = y E s (1− k )d Note that the formula for k is f y true for linear elastic φ y = where k = ( ρ n) + 2( ρ n) − ρ n behavior only, while for E s (1− k )d higher concrete stresses the A E 20000 in which ρ = s andn = s = true non- linear linear concrete bd Ec 4700 f'c Page 142/7 stress block shall be used. 2



Structural Design and Detailing for Earthquake Resistance T he ultimate curvature, where

a=

A s f y 0.85 f ' c b

u

is

and

φ u =

∈cu

; from a = β 1c c β ∈ φ u = 1 cu a

β 1, which describes the depth of the equivalent

rectangular stress block, may be taken as β 1 = 0.85 for f ' c ≤ 30 N/mm2 ,

otherwise β 1 = 0.85 −

0.05 7

( f ' c −30) ≤ 0.65

F rom the above derivation, the available section ductility may be written as

φ u

=

∈cu (1− k )dE s

φ y cf y The ultimate concrete strain cu may be taken as equal to 0.004 representing the limit of useful concrete strain, for estimating the ductility available for reinforced concrete in a strong earthquake.



Doubly reinforced sections sections. The ductility of doubly reinforced sections (Figure (d)) may be determined from the curvature in the same way as for singly reinforced sections. U sing the same expression for available section ductility as

φ u φ y

=

∈cu (1− k )dE s cf y Page 143/7



Structural Design and Detailing for Earthquake Resistance but to allow for the effect of compression steel ratio ’, the expressions for c and k become c=

and

k = [( ρ + ρ ' )n]2 + 2[( ρ + ρ ' )n] − ( ρ + ρ ' )n

c=

A' in which ρ ' = s bd

a β 1

( ρ − ρ ' ) f y d 0.85f ' c β 1

The above equations assume that the compression steel is yielding, but if this is not so, the actual value of the steel stress should be used f y . And as k has been found assuming linear elastic behavior in concrete, the qualifications mentioned for singly reinforced members also apply. d'

As'

d

Figure 5.4. Doubly reinforced concrete section.

As b

Page 144/7



Structural Design and Detailing for Earthquake Resistance 

Effect of confinement on ductility. ductility The ductility and strength of concrete is greatly enhanced by confining the compression zone with closely spaced lateral steel ties. In order to quantify the ductility of confined concrete, a number of stress-strain curves for confined concrete have been derived. It is known that rectangular all-enclosing links are moderately effective on small columns, but are of little use in large columns. In large columns, this is remedied to some extent by the use of intermediate lateral ties anchored to the all-enclosing links. The procedure for calculating the section ductility u / y is the same as that for unconfined concrete as described herein, the only difference being in determining an appropriate value of ultimate concrete strain cu for use in the expression for fu/fy. It is therefore recommended that a lower bound for the maximum concrete strain for concrete confined with rectangular links may be used.

∈cu = 0.003 + 0.02 where b

lc

r v

b lc

 ρ f  +  v yv   138 

2

= ratio of the beam width to the distance from the critical section to the point of contraflexure = ratio of volume of confining steel (including compression steel) to volume of concrete confined

f yv = yield stress of the confining steel in N/mm 2 Page 145/7



Structural Design and Detailing for Earthquake Resistance E xample Problem 5.2. Given a singly reinforced concrete beam section with 3- 32 reinforcing bars at the bottom. The confining steel consists of 12 mild steel bars ( f y = f yv = 275 N/mm2) at 75 mm centers and the concrete strength is f’ c = 25 N/mm2. Estimate the section ductility u / y. Assume b lc

= 1/ 8

n.a.

c

500

A s = 3 - Ø 3 2 b a r s 250

Page 146/7



Structural Design and Detailing for Earthquake Resistance S olution and discussion: T o find the curvature at first yield effectively singly reinforced. 3 * 804 A ρ = s = evaluate bd (250)(500)

y,

first estimate the depth of the neutral axis, the section being

ρ = 0.0193 and

n= n=

E s 200000 = Ec 4700 f'c 200000

= 8.511

4700 25 then, ρ n = 0.164

φ y =

k = ( ρ n)2 + 2( ρ n) − ρ n k = ( 0.164) + 2(0.164) − 0.164

f y E s (1− k )d

2

φ y =

k = 0.432

275 200000(1− 0.432)500

= 4.84 x10 − radian/mm 6

Although this implies a computed maximum concrete stress greater than 0.85 f’ c , the triangular A stress block gives a reasonable approximation. Thus, the curvature at first yield

∈cu = 0.003 + 0.02

b lc

 ρ f  +  v yv   138 

2

Page 147/7



Structural Design and Detailing for Earthquake Resistance T o find the ultimate curvature for the confined section, first determine the ultimate concrete strain cu . 113 * 2( 488 + 170) ρ v = where, consider link size, 488x170 (488)(170)(75)

then

 0.0239 * 275  ∈cu = 0.003 + 0.02(1/ 8) +   138   ∈cu = 0.00777

2

ρ = 0.0239

c=

Next, find the depth of the neutral axis at ultimate from

Hence, the ultimate curvature is

∈cu

c=

A s f y β 1 * 0.85f ' c b

(3 * 804)(275) 0.85 * 0.85( 25)( 250)

0.00777 c = 146.9 mm 146.9 c −5 φ u = 5.29 x10 radian/mm φ u =

=

Therefore, the available section ductility is

φ u φ y

=

5.29 x10

−5

4.84 x10

−6

= 10.9

It is of interest to observe that the ultimate strain cu = 0.00777 is about more than twice the value of 0.004 normally assumed for unconfined concrete. Hence the available section ductility has been roughly doubled by the use of confinement steel. Page 148/7




5.2.4 Ductility of reinforced concrete members with flexure and axial load . Axial load unfavorably affects the ductility of flexural members. It is therefore imperative that for practical levels of axial load, columns must be provided with confining reinforcement. For rectangular columns with closely spaced links, and in which the longitudinal steel is mainly concentrated in two opposite faces, the ratio u / y may be estimated from Figure 5.5.

Figure 5.5. u / y for columns of confined concrete.

Page 149/7



Structural Design and Detailing for Earthquake Resistance where As = and

β h = Ah =

area of tension reinforcement, mm 2 1.2 Ah f yh

shh f ' c cross-sectional area of the links, mm2

f yh =

yield stress of the link reinforcement, N/mm 2

s =

spacing of the link reinforcement, mm

hh =

the longer dimension of the rectangle of concrete enclosed by the links, mm

Page 150/7

Seismic Design and Detailing_pp131-150

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