Seismic Design

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 Em

Eh

Ev o Eh

NSCP eq. 208-1

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 dead 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.

Page 132/7



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.

Page 132/7



Structural Design and Detailing for Earthquake E arthquake Resistance E xample 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 de design ax axial lo load an and mo moment at at co column to top C. C.

 Distance

A

B 8000

C

D

8000

 Soil

8000

I =  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 E arthquake Resistance S olution olution and discussion: F ind 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 Apply load combinations involving earthquake. The basic load combinations for strength design 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 =

Mu − Mu 1

where

l

2

+ 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 (1− k )d E s 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

u

is

∈cu ; from a = β c c β ∈cu φ u = a φ u =

1

1

0.85 f ' c b

and

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 a c= and

β 1

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

c=

A' in which ρ ' = s bd

( 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. A 3 * 804 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

c=

Hence, the ultimate curvature is

A s f y β 1 * 0.85f ' c b

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

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

5

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




5.2.5 Seismic Shear Forces in Beams and Columns. Columns Shear failure in reinforced concrete members is regarded as brittle failure. Therefore, in designing earthquake-resistant structures, it is important to provide excess shear capacity over and above that corresponding to flexural failure. The code requirements are based on the strong columnweak beam concept. Hence plastification of the critical regions at the ends of the beams will have to be considered as a possible loading condition. The shear force is then computed based on the moment resistances in the developed plastic hinges, labeled as probable moment resistance M PR , developed when the longitudinal flexural steel enters into the hardening stage. Consequently, the computation of the probable moment resistance, 1.25f y , is used as the stress in the longitudinal reinforcement. In order to absorb the energy that can cause plastic hinging, the earthquake-resistant frame has to be ductile in part through confinement of the longitudinal reinforcement of the columns and the beam-column joints and in part through the provision of the excess shear capacity. Figure 5.6 shows the deformed geometry of and the seismic moment and shear forces for a beam subjected to gravity loading and reversible sidesway. (a) sidesway to the left; (b) sidesway to the right.

Page 151/7



Structural Design and Detailing for Earthquake Resistance The seismic shear forces are

V L = V R =

− + + MPR MPR

l

M +M + PR

− PR

l

+ −

1.4D + 1.7L 2 1.4D + 1.7L 2

where = span, L and R subscripts = left and right ends, and M PR = probable moment strength at the end of the beam based on steel reinforcement tensile strength of 1.25 f y and strength reduction factor = 1.0. These instantaneous moments M PR should be computed on the basis of equilibrium of moments at the joint where the beam moments are equal to the probable moments of resistance. The shear forces in the columns are computed in a similar manner, so the horizontal V e at top and bottom of the column is M + MPR 2 V e = PR1 h except that end moments for columns (M PR 1 and M PR 2) need not be greater than the moments generated by the M PR of beams framing into the beam-column joint, where h = column height, and the subscripts 1 and 2 indicate the top and bottom column end moments, respectively, as seen in Figure 5.7.

Page 152/7




Figure 5.7 Seismic moments and shears at column ends: (a) joint moments; (b) sway to right; (c) sway to left.

Page 153/7




5.2.6 Strong Column - Weak Weak Beam Concept . As previously discussed in 5.2.2, the strong column-beam concept is ensured by the following inequality:

 6  ΣMcol ≥  ΣMbeam  5 

For a joint subjected to reversible base shear forces, as shown in Figure 5.8, the above equation becomes

 6  (φ Mn+ + φ Mn− )col ≥  (φ Mn+ + φ Mn− )beam  5  where = 0.90 for beams, 0.65 for tied columns, and 0.70 for spiral columns. For beam-columns, = 0.90 to 0.65.

Figure 5.8 Seismic moment summation at beam-column joint: (a) sidesway to the left; Page 154/7 (b) sidesway to the right.




5.2.7 Design of Confining Reinforcement for Beam - Column Column Connection. Connection Transverse reinforcement in the form of closely spaced hoops (ties) or spirals has to be adequately provided. The aim is to produce adequate rotational capacity within the elastic hinges that may develop as a result of the seismic forces. 

For column spirals, the minimum volumetric ratio of the spiral hoops needed for the concrete core confinement cannot be less than the larger of: 0.12 f ' c ρ s ≥ or f yh

 Ag

 f ' − 1 c  Ach  f yh

ρ s ≥ 0.45

whichever is greater, where s

= ratio of volume of spiral reinforcement to the core volume measured out-to-out.

Ag = gross area of the column section. Ach = core area of section measured to the outside of the transverse reinforcement. f yh = specified yield strength of transverse reinforcement.

Page 155/7




For column rectangular hoops, the cross-sectional area within spacing s cannot be less than the larger of: f ' A sh ≥ 0.09 shc c or f yh

 Ag

 f ' − 1 c  Ach  f yh

A sh ≥ 0.3 shc  whichever is greater, where

Ash = total cross-sectional area of transverse reinforcement (including cross ties) within spacing s and perpendicular to dimension hc . hc = cross-sectional dimension of column core measured center-to-center of confining reinforcement. h x = maximum horizontal spacing of hoops or cross-ties on all faces of the column. Ach = cross-sectional reinforcement.

area

of

structural

member,

s = spacing of transverse reinforcement within length 150mm and need not be taken less than 100mm.

measured o.

out-to-out

transverse

Whose value should not exceed

smax = one-quarter of the smallest cross-sectional dimension of the member, 6 times diameter of longitudinal reinforcement. Page 156/7




Additionally, if the thickness of the concrete outside the confining transve exceeds 100mm, additional transverse reinforcement has to be provided a exceed 300mm. The concrete cover on the additional reinforcement 100mm. 

The confining transverse reinforcement in columns should be placed o potential hinge over a distance o. The largest of the following three cond length o: (a) depth of member at joint face (b) one-sixth of the clear span (c) 450mm

Increase o by 50% or more in locations of high axial loads and flexural de base of a building.

When transverse reinforcement is not provided throughout the column len




For beam confinement, the confining transverse reinforcement at beam ends should be placed over a length equal to twice the member depth h from the face of the joint on either side or of any other location where plastic hinges can develop. The maximum hoop spacing should be the smallest of the following four conditions: (a) one-fourth effective depth d (b) eight times diameter of longitudinal bars (c) 24 times diameter of the hoop (d) 300mm however, the Code requires that confining reinforcement spacing need not exceed 100mm. Figure 5.9 summarizes typical detailing requirements for a confined column.

Page 158/7




Figure 5.9 Typical detailing of seismically reinforced column: (a) spirally confined; (b) confined with rectangular hoops; (c) cross-sectional detailing of ties. X ≤ 350mm. Consecutive cross ties have 90° hooks on opposite sides. Page 159/7




Reduction in confinement at joints: a 50% reduction in confinement and an increase in the minimum tie spacing to 150mm are allowed by the code if a joint is confined on all four faces by adjoining beams with each beam wide enough to cover three-quarters of the adjoining face.



The yield strength of reinforcement in seismic zones (particularly zone 4) should not exceed 410 MPa.

Horizontal Shear in Beam - Column Column Connection Connection Test of joints and deep beams shave shown that shear strength is not as sensitive to joint (shear) reinforcement as for that along the span. On this basis, the code has assumed the joint strength as a function of only the compressive strength of the concrete and requires a minimum amount of transverse reinforcement in the joint. The effective area A j in Figure 5.10 should in no case be greater than the column cross-sectional area. The minimal shear strength of the joint should not be taken greater than the forces V n specified below for normal-weight concrete.

Page 160/7




Figure 5.10 Seismic effective area of joint. Page 161/7




Reduction in confinement at joints: a 50% reduction in confinement and an increase in the minimum tie spacing to 150mm are allowed by the code if a joint is confined on all four faces by adjoining beams with each beam wide enough to cover three-quarters of the adjoining face.



The yield strength of reinforcement in seismic zones (particularly zone 4) should not exceed 410 MPa.

Horizontal Shear in Beam - Column Column Connection Connection Test of joints and deep beams shave shown that shear strength is not as sensitive to joint (shear) reinforcement as for that along the span. On this basis, the code has assumed the joint strength as a function of only the compressive strength of the concrete and requires a minimum amount of transverse reinforcement in the joint. The effective area A j in Figure 5.10 should in no case be greater than the column cross-sectional area. The minimal shear strength of the joint should not be taken greater than the forces V n specified below for normal-weight concrete. 

Confined on all faces by beams framing into the joint:

V n ≤ 1.66 f ' c A j

Page 162/7




Confined on three faces or on two opposite faces:

V n ≤ 1.25 f ' c A j 

All other cases:

V n ≤ 1.0 f ' c A j A framing beam is considered to provide confinement to the joint only if at least three-quarters of the joint is covered by the beam. The value of allowable V n should be reduced by 25% if lightweight concrete is used. Some test data indicate that the value of V n for all other cases is unconservative when applied to corner joints. A j = effective cross-sectional area within a joint in a plane parallel to the plane of reinforcement generating shear at the joint. The code assumes that the horizontal shear in the joint is determined on the basis that the stress in the flexural tensile steel = 1.25 f y .

Page 163/7




5.2.8 Design of Confining Reinforcement for Beam - Column Column Connection. Connection E xample Problem 5.3. Design the transverse confining reinforcement of joint A in a ductile moment-resisting frame of a building as shown in the figure below. The structure is situated in seismic zone 4. The following design criteria applies to the building frame as: All beams are 300mm x 600mm with 4- 25 longitudinal bars top and bottom and columns are 400mm x 600mm. Stirrup size is 12.

Joint B

Joint A

0 0 6

Column size 400mm x 600mm

All beams are 300mm x 600mm with 0 0 6

600

4-Ø25 bars top and bottom.

wD =

21 kN/m

wL =

36 kN/m

MPR =

460 kN-m

f’c =

27.6 MPa

fy =

410 MPa

0 0 6 3

7500

FRAME ELEVATION

600

Page 164/7



Structural Design and Detailing for Earthquake Resistance Check the web shear reinforcement along beam span outside the inelastic zone. Consider the figure of isolated joint A below showing schematic of the lines of action of the beam-column joint forces. 0 0 6

Column size

All beams are 300mm x 600mm with 0 0 6

4-Ø25 bars top and bottom.

0 0 6 3

0 0 8 1 =

600

7500

FRAME ELEVATION

600

0 2 0 / 8 2 1 h = 2 / 1

h

col Joint A

Vu

MEQ

Vcol

Shear forces at beam-column joint. Page 165/7



Structural Design and Detailing for Earthquake Resistance depth of reinforcement, d = 600-(40+12+25/2) = 535.5 mm reinforcement, As = 4*491 = 1964 mm2 longitudinal steel ratio, ρ =

A s = bd

wD = 21 kN/m wL = 36 kN/m

:

1964 300( 535.5)

M A

ρ = 0.0122 < 0.025

 

Mn = 1.25 A s f y  d −

a 

 ≤ MPR

2 

1.25 A s f y 0.85 f ' c b

V A

VB

Beam AB Equilibrium

where a=

MB

Ln = 7500

=

1.25(1964)(410) 0.85( 27.6)(300)

a = 143 mm, then 143    /1000000 2   Mn = 467.039 kN - m > MPR = 460 kN - m

Mn = 1.25(1964)(410) 535.5 −

therefore, 4- 25 bars at top and bottom are sufficient.

Page 166/7



Structural Design and Detailing for Earthquake Resistance Determine the beam transverse confining reinforcement in the inelastic zone of plastic hinging. Using the following equations for seismic shear forces:

V L =

M A + MB

+

   φ  460 + 460 1.4(157.5) + 1.7( 270)  1  V L = +   7.5 2  0.75  V L = 575.667 kN ln

wD = 21 kN/m

1.4D + 1.7L  1  2

Computing shear strength provided by the concrete beam,

V c = (1/ 6) f ' c bw d =

(1/ 6)( 27.6 )(300)(535.5)

wL = 36 kN/m M A MB

Ln = 7500

V A

VB

Beam AB Equilibrium

1000

V c = 140.664 kN Calculate the nominal shear force at a distance d from the

VDA = 21(7.5/2) = 78.75 kN VLA = 36(7.5/2) = 135 kN

face of the column support,

V n =

575.667(7.5 / 2 − 0.5355)

(7.5 / 2)

V n = 493.462 kN Page 167/7



Structural Design and Detailing for Earthquake Resistance Then, the nominal shear strength to be resisted by the reinforcement,

V s = V n − V c = 493.462 − 140.664 V s = 352.798 kN Using 12 hoops, Av = 2(113) = 226 mm2, the required spacing is

s =

Av f y d V s

=

(226)(410)(535.5) 352.798(1000)

s = 140 mm These confining hoops shall be placed over beam within a distance of and shall be spaced not to exceed the least value of

o

= 2h = 2(600) = 1200 mm

(d /4) = 535.5/4 = 133 mm……….. Governs, say 125 mm (8*smallest longitudinal bar d b) = 8(25) = 200 mm (24*hoop diameter) = 24(12) = 288 mm or (maximum spacing of ) = 300 mm Therefore, within

o

= 1200mm, use 12 hoops and crossties at 125 mm c-c over this distance.

Further, use 12 closed hoops at 150 mm c-c beyond critical section, then increase spacing to d /2 = 535.5/2 = 267 mm, say 250 mm approaching midspan and stop stirrups at V c /2. Page 168/7



Structural Design and Detailing for Earthquake Resistance Determine the confining reinforcement in the column in beam-column joint. Calculate the joint shear strength. Column shear forces should not exceed those base on the probable end moment strengths M PR of the beams framing into the joint. 460 MPR = V col = h1 / 2 + h2 / 2 3.60 / 2 + 3.60 / 2

V col = 127.778 kN then

V n = A s f y − V col =

(1964)(410) 1000

− 127.778

V n = 677.462 kN and this V n ≤ 1.25 f ' c A j where A j = 400(600) = 240,000 mm 2, then allowable

V n =

1.25 27.6 (240,000) 1000

V n = 1576.071kN > actual V n = 677.462 kN Hence, the confined column joint is adequate to resist the seismic shear.

Page 169/7



Structural Design and Detailing for Earthquake Resistance Determine the column confinement in the inelastic zone. column d = 600 - (40+12+25/2) = 535.5 mm At the A j plane, the nominal shear strength provided by concrete is given also as

V c = (1/ 6) f ' c bd =

(1/ 6)(

27.6 )(400)(535.5) 1000

V c = 187.552 kN then, the nominal shear strength to be resisted by confinement is

V s = V n − V c = 677.462 − 187.552 V s = 489.91kN Using 12 hoops, Av = 2(113) = 226 mm2, the required spacing is

s =

Av f y d V s

=

(226)(410)(535.5) 489.91(1000)

s = 101mm

Page 170/7



Structural Design and Detailing for Earthquake Resistance Determine the greater value of the following expressions. where hc = column core dimension measured c-c of confining reinforcement hc = 600 – 2(40+12) = 496 mm

A sh ≥ 0.09 shc

f ' c or f yh

 A  f ' A sh ≥ 0.3 shc  g − 1 c  Ach  f yh

try spacing s = 90 mm

A sh ≥ 0.09(90)(496)

27.6 410

= 270 mm2 or

 400 * 600  27.6 − 1 = 570 mm2 - -- > controls A sh ≥ 0.3(90)(496)  264 * 496  410 Check with the maximum spacing, the least value of (smaller column dimension/4) = 400/4 = 100 mm…….. governs (6*longitudinal bar diameter) = 6(25) = 150 mm (maximum spacing of ) = 100 mm or 350 − h x < 150 mm s x = 100 + 3 > 100 mm

s x = 100 +

350 − ( 496 − 64.5) / 2 3

= 144 mm Page 171/7




hc = 496

296

Determine the distance o over which these confinements shall be placed in the column of both sides of potential hinge and shall be the largest of

600

400

8- 25

12 @ 90mm

(depth of the member h) = 600 mm (beam clear span over 6) = 7500/6 = 1250 mm or ……… governs

14 spaces @ 90mm = 1260mm

(minimum of ) = 450 mm

4- 25

Hence, provide 12 hoops and 12 crossties at 100 mm c-c over the distance of say o = 1250 mm.

12 @ 100mm

4- 25 12 spaces @ 100mm = 1200mm

150

50

12 @ 90mm

Page 172/7

150




5.2.9 In- situ situ Concrete Detailing – General Requirements. Requirements The following notes and associated detail drawings have been compiled to enable the elements of reinforced concrete structures to be detailed in a consistent and satisfactory manner for earthquake resistance. These details should be satisfactory in regions of medium and high seismic risk in so far as they reflect the present stateof-the-art. However considerable uncertainty exists regarding effective details for some members, particularly columns and beam-columns connections. In low risk regions, relaxations may be made to the following requirements, but the principles of lapping, containment and continuity must be retained if adequate ductility is to be obtained. 

Laps. Laps Laps in earthquake resisting frames must continue to function while the members or joints undergo large deformations. As the stress transfer is accomplished through the concrete surrounding the bars, it is essential that there be adequate space in a member to place and compact good quality concrete. Laps should preferably not be made in regions of high stress, such as near beam-to-column connections, as the concrete may become cracked under large deformations and thus destroy the transfer of stress by bond. In regions of high stress, laps should be considered as an anchorage problem rather than a lap problem, i.e. the transfer of stress from one bar to another is not considered; instead the bars required to resist tension should be extended beyond the zone of expected large deformations in order to develop their strength by anchorage. Laps should preferably be staggered but where this is impracticable and large numbers are lapped at one location (I.e. columns) adequate links or ties must be provided to minimize the possibility of splitting in concrete. In columns and beams even when laps are made in regions of low stress at least two links should be provided as shown in the details. Code provisions on laps are given in NSCP Section 412.15 to 412.20.

Page 173/7



Structural Design and Detailing for Earthquake Resistance 

Anchorage. Anchorage Satisfactory anchorage may be achieved by extending bars as straight lengths, or by using 90° and 180° bends, but anchorage efficiency will be governed largely by the state of stress of the concrete in the anchorage length. Tensile reinforcement should not be anchored in zones of high tension. If this cannot be achieved, additional reinforcement in the form of links should be added, especially where high shear exists, to help to confine the concrete in the anchorage length. It is especially desirable to avoid anchoring bars in the ‘panel’ zone of beam-column connections. Large amounts of the reinforcement should not be curtailed at any one location. See NSCP Section 412 for development and splices of reinforcement.



Bar bending. bending The code has adopted standardization of bar shapes but due attention must be made to the bearing stresses in bends. The bearing stress inside a bend in a bar which does not extend or is not assumed to be stressed beyond a point four times the bar size past the end of the bend need not be checked, as the longitudinal stresses developed in the bar at the bend will be small. See NSCP Section 407.2 through 407.407.4 for details of reinforcement. The bearing stress inside a bend in any other bar should be calculated as

bearing stress f p = where

1.5 f ' c Ft ≤ r φ 1+ 2φ / ab

F t =

tensile force due to ultimate loads in a bar or group of bars, N

r =

internal radius of the bend, mm

=

diameter of the bar or, in bundle, the diameter of a bar of equivalent area, mm Page 174/7



Structural Design and Detailing for Earthquake Resistance ab = center-to-center distance perpendicular to the plane of the bend between bars or groups of bars for a particular bar or group of bars in contact, respectively, mm 

Concrete cover cover.

Minimum cover to reinforcement as set forth in NSCP Section 407.8.1.



Concrete quality quality. The minimum recommended 28-day compressive strength, f’ c for structural concrete is 20 N/mm 2. The use of lightweight aggregates for structural purposes in seismic zones should be very cautiously proceeded with, as many lightweight concretes prove very brittle in earthquakes. Appropriate advice should be sought in order to obtain a suitably ductile concrete. It cannot be over-emphasized that quality control, workmanship and supervision are of the utmost importance in obtaining earthquake-resistant concrete.



Reinforcement quality quality. For adequate earthquake resistance, suitable quality of reinforcement must be ensured by both specification and testing. As the properties of reinforcement vary greatly between manufacturers, much depends on knowing the source of the bars, and on applying the appropriate tests. The following points should be observed: - Adequate minimum yield stress may be ensured by specifying steel to an appropriate standard (PNS 49 or ASTM A615). - Grades of steel with f’ c in excess of 410 N/mm 2 may not be permitted in areas of high seismic risk, but slightly greater strengths may be used if adequate ductility is proven by Page 175/7 tests.

Seismic Design

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