Example 2 Seismic Design of RC Column ACI 318-19 PDF

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CME 400–Column Design EXAMPLE 2: Determine the required reinforcement details for the column above the third floor. The columns of an interior bay of the special moment-resisting frame shown in Figure 1 below have a clear height of 10 feet. The structure has a redundancy factor of  = 1.0 and SDS = 0.826. The proposed beam and column sections are shown in Figure 1. The service level gravity loads, bending moments, and the moments due to the design level seismic forces are shown in Figure 2. The bending moments are shown acting at the face of the joint. Second-order effects may be neglected and the axial force due to seismic loads and the bending moments due to dead and live load are negligible. f'c = 4,000 psi, fy = 60,000 psi

Determine the required reinforcement details for the column above the third floor.

Beam Cross Section

Column Cross Section

Figure 1 Beam and Column Cross Sections

1

sidesway to the left Figure 2 Applied Loads 2

Solution: The longitudinal reinforcement to resist the factored loads is determined first. Load combinations:

For dead load and live load the applicable load combination for the column above the third floor is given by ACI Eq. (5.3.1b), which is U = 1.2D + 1.6L The factored column axial load for dead and live load is then Pu3, D+L = 1.2 x 400 + 1.6 x 100 = 640 kips

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When the effects of dead load and seismic load are additive, the applicable loading cases are given by ACI Eq. (5.3.1e), as U

= 1.2D + 1.0E + 0.5L + 0.2S = (1.2 + 0.2SDS)D + QE + 0.5L + 0.2S

The corresponding factored column axial load is Pu3 = l.2D + 1.0E + 0.5L + 0.2S = (1.2 + 0.2 x 0.826) x 400 + 1.0 x 0 + 0.5 x 100 + 0 = 596 kips The corresponding factored column moment is Mu3 = (1.2 + 0.2SDS)D + QE + 0.5L + 0.2S = (1.2 + 0.2 x 0.826) x 0 + 1.0 x 176 + 0.5 x 0 + 0 = 176 kip-ft For the column below the third floor, the applicable factored loads for the dead load plus seismic load combination are Pu2 = l.2D + 1.0E + 0.5L + 0.2S = (1.2 + 0.2 x 0.826) x 470 + 1.0 x 0 + 0.5 x 120 + 0 = 702 kips The corresponding factored column moment is Mu2

= (1.2 + 0.2SDS)D + QE + 0.5L + 0.2S = (1.2 + 0.2 x 0.826) x 0 + 1.0 x 224 + 0.5 x 0 + 0 = 224 kip-ft

Column reinforcement Providing eight #8 bars gives a reinforcement area of As = 6.32 in2 Gross area of the column is and 4

Ag = 23 x 23 = 529 in2 g = As /Ag = 6.32/529 = 0.012 = 1.2% 0.01Ag = 5.29 < As  satisfies ACI Section 18.7.4 0.06 Ag = 31.74 > As  satisfies ACI Section 18.7.4 18.7.4 Longitudinal reinforcement 18.7.4.1 Area of longitudinal reinforcement, Ast, shall be at least 0.01Ag and shall not exceed 0.06Ag. 18.7.4.2 In columns with circular hoops, there shall be at least six longitudinal bars. Column axial and flexural capacity For zero applied moment on the column above the third floor, the design capacity in axial compression is obtained from the appropriate interaction diagram shown in Figure 3, (ACI section 11.5.1.1) as

 Pn3

= 1110 kips > P u3 = 640 kips, D+L ... satisfactory

For a factored applied load of 596 kips on the column above the third floor, the design flexural capacity from the column-interaction diagram is:  Mn3 = 400 kip-ft > Mu3 = 176 k.ft... satisfactory For a factored applied load of 702 kips on the column below the third floor, the design flexural capacity from the column-interaction diagram is: Mn2 = 370 kip-ft > Mu2 = 224 k.ft ... satisfactory For f'c = 4,000 psi, fy = 60,000 psi, and  = h/h =(23-2.5-2.5)/23=0.78 From the column-interaction diagrams shown below, interpolate for  = 0.78, for  = 0.70 and  = 0.80. You will find the same numbers for Mn values. 5

Figure 3 Details for Example 2

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Strong Column-Weak Beam The formation of plastic hinges at both ends of the columns in a given story, due to seismic loads, may produce a sides-way mechanism that causes the story to collapse. To prevent this, a strong column-weak beam design is required by ACI Section 18.7.3. A column forming part of 8

the lateral-force-resisting system and with factored axial force exceeding 0.1ƒ'cAg shall be designed to satisfy ACI Equation (18.7.3.2), which is: 𝟔

Mnc  𝟓 Mnb Mnc

= sum of the nominal flexural strengths of columns at the face of a joint calculated for the applicable factored axial force resulting in the lowest flexural strength.

Mnb

= sum of the nominal flexural strengths of beams at the face of the joint and in the same plane as the columns.

In T-beam construction, slab reinforcement within an effective width of the flange is assumed to contribute to the negative flexural strength. The effective flange width is defined in ACI Section 6.3.2.1 as the lesser of: be = 16hf + bw or

be

= s+ bw or be = s1/2 + s2/2 where s1 & s2 are centerline to centerline of webs 9

or

be

= l /4

Where

l

= beam span = flange thickness = width of stem = clear distance between stems

hf bw s

one-quarter of the span length of the beam

In determining the negative moment strength of the beam framing into the column, the reinforcement within an effective width of the flange is assumed to contribute to the negative flexural strength. The effective flange width is defined in ACI Section 6.3.2.1 as the lesser of: be = 16hf + bw = 16 X 0.5 + 21/12 = 9.75 ft or

be

= s+ bw = 12 ft

or

be

= l /4

one-quarter of the span length of the

beam

= 24/4 = 6 ft ... governs The distribution reinforcement in this width of flange is Asf = 0.0018 x 6 x 6 x 12 = 0.78 in2 Hence, the total area of reinforcement in the top of the beam at the left face of the column is: As = 3.38 + 0.78 = 4.16 in2 The nominal negative moment strength of the beam framing into the right-hand-side face of the joint is determined from the expression: MnR = Asfyd (d-a/2, from C = T  a = Asfy/0.85f c b) 10

= Asfy (d- Asfy/(2x85f c b))= Asfyd (1-0.59fy/fc), where  = As/bd = 4.16 x 60 x 21.5(1 -0.59 x 0.0092 x 60/4)/12 = 411 kip-ft where  = As/bd = 4.16/(21x21.5) = 0.0092 The nominal positive moment strength of the beam framing into the lefthand-side face of the joint is obtained from Example 1 (Beam Design) as: For the bottom reinforcement in the beam span, provide three #7 bars to give a reinforcement area of: As = 1.80 in2

 = As/bd = 1.8/(21 x 21.5) = 0.004 The design flexural strength provided is given by Mu

= Asfy d (1-0.59 fy/fc), where  = As/bd = 0.90 x1.80 x 60 x 21.5(1 -0.59 x 0.004 x 60/4)/12 = 168 kip-ft

The design flexural strength provided is given by: MnL = 168/ = 168/0.9 = 187 kip-ft Hence,

1.2 Mg = 1.2(411 + 187) = 717 kip-ft

The sum of the nominal flexural capacities of the columns framing into the joint at the third floor is ΣMC = ( Mn3 +  Mn2)/ = (400 + 370)/0.65 = 1185 > l.2ΣMg ... satisfies ACI Equation (18.7.3.2) 11

Column shear 18.6.5.1 Design forces The design shear force Ve shall be calculated from consideration of the forces on the portion of the beam between faces of the joints. It shall be assumed that moments of opposite sign corresponding to probable flexural strength Mpr, act at the joint faces and that the beam is loaded with the factored gravity and vertical earthquake loads along its span. The maximum factored column shear is obtained from Figure 4 as: Vu = l.2D + 1.0E + 0.5L + 0.2S = 1.2 x 0 + 1.0 x 35 + 0.5 x 0 + 0 = 35 kips In accordance with ACI Sections 18.6.5.1 the design shear force for the column above the third floor may be calculated from the probable moment strengths at the top and bottom of the column. The column probable moment strength is determined by assuming  = 1.0 and a tensile reinforcement stress of, fS = 1.25ffy. The maximum design moment, at both the top and bottom of the column, occurs at the balanced strain condition and is obtained from the appropriate column-interaction diagram as:

 Mn = 412 kip-ft (for the balanced condition) The strength reduction factor for this condition of axial compression with flexure is given by ACI Section 21.2.1 as  = 0.65

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The probable moments are assumed to occur at the axial load producing a balanced strain condition that gives rise to the maximum moments. The probable moment strength may be determined from the column design moment strength as: Mpr   Mn x 1.25/0.65 = 412 x 1.25/0.65 where  Mn =  Mnb (balanced moment) = 412 k-ft = 792 kip-ft Where Mpr is based on  = 1.0 and fs = 1.25 fy The clear height of the column is Hn = 10 ft The design shear force at the top and bottom of the column shown in Figure 4 is then:

Figure 4 Column shear due to probable flexural strength

Ve

= 2MprlHn = 2 x 792/10 = 158 kips

However, in accordance with ACI Section 18.6.5.1, the maximum design shear force in the column need not exceed that determined from the probable flexural strengths of the beams that frame into either side of the joint. 14

The probable moment strength at the left end of the beam, assuming a strength reduction factor of  = 1.0 and a tensile reinforcement stress fs=1.25fy, were derived in Example 1 as: Mpr1 = Asfyd (1-0.59fy/f c), where  = As/bd = 3.38 x 60 x 21.5(1.25 -0.92 x 0.0075 x 60/4)/12 = 417 kip-ft where  = As/bd = 3.38/(21x21.5) = 0.0075

The probable flexural strength at the right end of the beam is given by: Mpr2 = Asfyd (1-0.59fy/fc), where  = As/bd = 1.80 x 60 x 21.5(1.25 -0.92 x 0.004 x 60/4)/12 = 231 kip-ft where  = As/bd = 1.80/(21x21.5) = 0.004

Mpr1 Mpr2

= Mpr3 = 417 kip-ft = Mpr4 = 231 kip-ft

However, the column design shear need not exceed the value determined from the probable moment strengths of the beams framing into the top and bottom of the column. As shown in Figure 5, the design shear force for this condition, provided that the column stiffness is the same in all stories, is given by:

Figure 5 Column shear due to beam probable flexural strength 15

Ve

Then:

= (Mpr1 + Mpr2 + Mpr3 + Mpr4)/ 2Hn = (417 + 231 + 417 + 231)/(2 x 10) = 65 kips < 158 kips Column shear due to probable flexural strength > Vu = 35 kips due to seismic force Ve

= 65 kips ... governs

In addition, the transverse reinforcement must also be designed to resist the maximum shear caused by factored loads. The compressive force value given by Ag fc /20 = 529 x 4/20 = 106 kips < Pu+ = 640 kips 18.6.5.2 Transverse reinforcement Transverse reinforcement over the length identified in 18.6.4.1 shall be designed to resist shear assuming Vc = 0 when both (a) and (b) occur: (a) The earthquake-induced shear force calculated in accordance with 18.6.5.1 represents at least one-half of the maximum required shear strength within those lengths. (b) The factored axial compressive force Pu including earthquake load effects is less than Agf’c/20. Hence, in accordance with ACI Section 18.6.5.2 the design shear strength provided by the concrete may be utilized (Vc  0), then shear reinforcement shall be provided to resist the total design shear force and this is given by ACI Table 22.5.5.1 as:

 𝑽𝒄 =  𝑥 2 √𝑓𝑐′ 𝑥 𝑏𝑤 𝑥 𝑑 where d = h-2,50 = 23-2.5 = 21.5 in. 2

= 0.75 x 2 (4000)0·5 x 23 x (21.50)/1000 = 45 kips < Ve = 65 kips The design shear strength required from shear reinforcement is given by ACI Table 22.5.1.1 as: Vs = Ve - Vc 16

Vs = 65 - 45 Vs = 20 kips < 4 x Vc •• satisfies ACI Section 22.5.1.2 ACI 318-19 Code

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Size effect factor, λs Measured shear strength, attributed to concrete (Vc) does not increase in direct proportion with member depth. Beam that is twice as deep, may fail at less than twice the shear of shallower beam, even though ACI 318-14 implies that shear should scale with depth. (Sneed and Ramirez, 2010). •

Consistent with fracture mechanics theory for RC

•

Proposed by ACI Committee 446 (fracture mechanics)

•

Power law, proportionality to d-1/2

• General form verified for many geometries and quasi-brittle materials 18

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18.7.5.5 Beyond the length ℓo given in 18.7.5.1, the column shall contain spiral reinforcement satisfying 25.7.3 or hoop and crosstie reinforcement satisfying 25.7.2 and 25.7.4 with spacing s not exceeding the least of 6 in., 6db of the smallest Grade 60 longitudinal column bar, and 5db of the smallest Grade 80 longitudinal column bar, unless a greater amount of transverse reinforcement is required by 18.7.4.4 or 18.7.6. s s or

= 6db = 6 x 1.0 = 6 in = 6 in

Hence a spacing of 6 inches is appropriate, with the exception of confinement reinforcement at the ends of the column and at lap splices.

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The area of shear reinforcement that is required, at a spacing of 6 inches, to provide shear strength of  Vs is specified by ACI Equation (22.5.10.5.4) as Av

=  Vs s/  d f y = 20 x 6/(0.75 x 20.5 x 60) = 0.13 in2

The minimum size of crosstie required for a #8 longitudinal bar is specified, by ACI Section 9.7.6.4.2, as a #3 bar and at least one crosstie is required to satisfy the lateral support requirements of ACI Section 9.6.3.3. The minimum area of shear reinforcement, which may be provided at a spacing of 6 inches, is given by ACI Table 9.6.3.3 as: Av(min) = 50bws/ f y ... governs for fc < 4444 psi = 50 x 23 x 6/60,000 = 0.12 in2 < 0.13 in2 Providing a #4 hoop and one #4 crosstie (3 legs) gives an area of Av

= 3 x 0.2 21

= 0.6 in2 > 0.13 in2 ... satisfactory Confinement reinforcement The point of inflection of the column lies within the center half of the column clear height. Hence, in accordance with ACI Section 18.7.5.1 confinement reinforcement is required for a distance from each joint face given by the greater of lo: 18.7.5.1 Transverse reinforcement required in 18.7.5.2 through 18.7.5.4 shall be provided over a length ℓo from each joint face and on both sides of any section where flexural yielding is likely to occur as a result of lateral displacements beyond the elastic range of behavior. Length ℓo shall be at least the greatest of (a) through (c): (a) The depth of the column at the joint face or at the section where flexural yielding is likely to occur. (b) One-sixth of the clear span of the column (c) 18 in. = Hn/6 lo = 10 x 12/6 = 20 in or

lo

= 18 in

or

lo

= h = 23 in ... governs

18.7.5.3 Spacing of transverse reinforcement shall not exceed the least of (a) through (d): (a) One-fourth of the minimum column dimension (b) For Grade 60, 6db of the smallest longitudinal bar (c) For Grade 80, 5db of the smallest longitudinal bar (d) so, as calculated by:

𝑠𝑜 = 4 + [

14−ℎ𝑥 ] 3

(18.7.5.3)

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The value of 𝒔𝒐 from Eq. (18.7.5.3) shall not exceed 6 in. and need not be taken less than 4 in. 18.7.5.4 Amount of transverse reinforcement shall be in accordance with Table 18.7.5.4. The concrete strength factor kf and confinement effectiveness factor kn are calculated according to Eq. (18.7.5.4a) and (18.7.5.4b): 𝑘𝑓 =

𝑓𝑐′

25,000

𝑘𝑛 =

+ 0.6 ≥ 1.0

(18.7.5.4a)

𝑛𝑙

(18.7.5.4b)

𝑛𝑙 −2

where nl is the number of longitudinal bars or bar bundles around the perimeter of a column core with rectilinear hoops that are laterally supported by the corner of hoops or by seismic hooks. The spacing of the confinement reinforcement is limited by ACI Section 18.7.5.3 to the minimum value given by so

= hmin/4 =23/4 = 5.75 in

or for Grade 60, so = 6db, so = 6db = 6 x 1.0 =6 in or

so

= 6in = 4 + (14 - hx)/3 ... ACI Equation (18.7.5.3) = 4 + (14 -9.75)/3 = 5.4 in ... governs

Using #4 hoop reinforcement bars at 4 inches on center, and providing 1- inches clear cover to the bars, gives a core dimension, measured center to center of the hoop reinforcement, of: bc = hc = 23 -3 - 0.5 = 19.5 in 23

The dimension measured out-to-out of the confining bars is, in both directions, h c = hc + 0.5 = 19.5 + 0.5 = 20 in The area, calculated out-to-out of the confining bars, is Ach = (hc)2 =202 = 400 in2 The required area of confinement reinforcement is given by the greater value obtained from ACI Table 18.7.5.4, which are

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Ash

= 0.3sbc (Ag/Ach -1) f c/ f y Table 18.7.5.4 (a) = 0.3 x 4 x 19.5 (529/400 -1)4/60 = 0.50 in2

Ash = 0.09sbc fc/ f y = 0.09 x 4 x 19.5 x 4/60 = 0.47 in2

or

Table 18.7.5.4 (b)

A #4 hoop with one #4 crosstie provides an area of confinement reinforcing of: Ash = 0.60 in2 > Ash ... satisfactory To conform to ACI Section 18.7.4.2, a tension splice is required within the center half of the clear column height. Hoop reinforcement, at a spacing of 4 inches, is provided over the splice length in accordance with ACI Section 18.7.5.3. The lap length required for a Class B splice is specified by ACI Section 25.5.3.1 as being equal to 1.3 times the tensile development length. The development length is given by ACI section 25.4.2.3 as: 25.4.9 Development of deformed bars and deformed wires in compression 25.4.9.1 Development length l dc for deformed bars and deformed wires in compression shall be the greater of (a) and (b) (a) Length calculated in accordance with 25.4.9.2 (b) 8 in. 25.4.9.2 l dc shall be the greater of (a) and (b), using the modification factors of ACI section 25.4.9.3 or Table 25.4.9.3 𝑓𝑦 𝑟

(a)

𝐥𝑑𝑐 =

(b)

0.0003fy db

50 √𝑓𝑐′

𝑥𝑑𝑏

25.4.9.3 For the calculation of l dc modification factors shall be in accordance with Table 25.4.9.3, except r shall be permitted to be taken as 1.0. 25

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And using Table 25.4.2.5 from ACI 318-19 𝜓𝑔 = 𝜓𝑠 = 𝜓𝑒 = 𝜓𝑡 = 1 therefore:

l dc

= 0.075db f y [(f c )0.5 cb + Ktr)/db]

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For the reinforcement layout indicated, (cb + Ktr)/db equals its maximum permissible value of 2.5 and 1.3 l d = 1.3 x 0.075 x 1.0 x 60,000/2.5(4000)0.5 = 37 in Details of the column reinforcement are shown in Figure 6 below.

Splices For tension Splices length can be found in ACI 318-19 25.5.2

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For compression, Splices length can be found in ACI 318-19 25.5.5.1

Compression lap splice length 𝑙𝑠𝑐 of No. 11 or smaller deformed bars in compression shall be calculated in accordance with (a), (b), or (c):

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