aisc lrfd base plates

Beam-Column Base Plate Design—LRFD Method

ENGINEERING JOURNAL / FIRST QUARTER / 1999

Richard M. Drake is Principal Structural Engineer, FluorDaniel, Irvine, CA.

Sharon J. Elkin is Structural Engineer, Fluor Daniel, Irvine,CA.

29

1

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4

INTRODUCTION

RICHARD M. DRAKE and SHARON J. ELKIN

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Fig. 1. Base Plate Design Variables

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t is c mm n design practice t design a building r struc-base plate width perpendicular t m ment direc-ture beam-c lumn with a m ment-resisting r fixed base.ti n, in.Theref re the base plate and anch r r ds must be capablebase plate length parallel t m ment directi n, in.f transferring shear l ads, axial l ads, and bending m -c lumn flange width, in.ments t the supp rting f undati n.verall c lumn depth, in.Typically, these beam-c lumn base plates have been

anch r r d distance fr m c lumn and base platedesigned and/ r analyzed by using service l ads r bycenterline parallel t m ment directi n, in.appr ximating the stress relati nship assuming the c m-base plate bearing interface cantilever directi npressi n bearing l cati n. The auth rs present an therparallel t m ment directi n, in.appr ach, using fact red l ads directly in a meth d c nsis-

tent with the equati ns f static equilibrium and the LRFD 0 95(1)Specificati n. 2

The m ment-resisting base plate must have designbase plate bearing interface cantilever perpendic-strengths in excess f the required strengths, flexural ( ),ular t m ment directi n, in.axial ( ), and shear ( ) f r all l ad c mbinati ns.

A typical beam-c lumn base plate ge metry is sh wn0 80in Figure 1, which is c nsistent with that sh wn n page (2)211-61 f the LRFD Manual.

base plate tensi n interface cantilever parallel tm ment directi n, in.

(3)2 2

c lumn flange thickness, in.

The pr gressi n f beam-c lumn l adings, in rder f in-creasing m ments, is presented in f ur l ad cases.

Case A is a l ad case with axial c mpressi n and shear,with ut bending m ment. This case results in a full lengthunif rm pressure distributi n between the base plate andthe supp rting c ncrete. This case is summarized in theLRFD Manual beginning n page 11-54 and is summa-rized herein f r c mpleteness.

Case B ev lves fr m Case A by the additi n f a smallbending m ment. The m ment changes the full lengthunif rm pressure distributi n t a partial length unif rmpressure distributi n, but is n t large en ugh t cause sepa-rati n between the base plate and the supp rting c ncrete.

Case C ev lves fr m Case B by the additi n f a spe-cific bending m ment such that the unif rm pressure dis-tributi n is the smallest p ssible length with ut separati n

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CASE A: NO MOMENT—NO UPLIFT

CASE B: SMALL MOMENT WITHOUT UPLIFT

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Fig. 2. No Moment - No Uplift

Fig. 3. Small Moment Without Uplift

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between the base plate and the supp rting c ncrete. This 1. Assume that the resultant c mpressive bearing stressc rresp nds t the c mm n elastic limit where any addi- is directly under the c lumn flange.ti nal m ment w uld initiate separati n between the base 2. Assume a linear strain distributi n such that the an-plate and the supp rting c ncrete. ch r r d strain is dependent n the bearing area

Case D ev lves fr m Case C by the additi n f suffi- strain.cient bending m ment t require anch r r ds t prevent 3. Assume independent strain distributi n.separati n between the base plate and the supp rting c n-

All three meth ds summarized by AISC assume a lin-crete. This is a c mm n situati n f r fixed base plates

ear triangular distributi n f the resultant c mpressivein structural ffice practice. That is, a rigid frame with a

bearing stress. This implies that the beam-c lumn basefixed base plate will usually attract en ugh bending m -

plate has n additi nal capacity after the extreme fiberment t require anch r r ds t prevent uplift f the base

reaches the c ncrete bearing limit state. The auth rs pr -plate fr m the supp rting c ncrete.

p se that a unif rm distributi n f the resultant c mpres-sive bearing stress is m re appr priate when utilizingLRFD.

If there is n bending m ment r axial tensi n at the base Case B, a beam-c lumn with a small m ment and nf a beam-c lumn, the anch r r ds resist shear l ads but uplift at the base plate elevati n, is sh wn in Figure 3.

are n t required t prevent uplift r separati n f the base The m ment is expressed as l cated at s me ec-plate fr m the f undati n. Case A, a beam-c lumn with centricity ( ) fr m the beam-c lumn neutral axis.n m ment r uplift at the base plate elevati n, is sh wnin Figure 2.

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(4)0

06If the magnitude f the bending m ment is small relative

t the magnitude f the axial l ad, the c lumn anch r0r ds are n t required t restrain uplift r separati n f 6

the base plate fr m the f undati n. In service, they nly2resist shear. They are als necessary f r the stability f

the structure during c nstructi n.(5)AISC addresses three different variati ns f the elastic 2

meth d when using an ultimate strength appr ach f r thewhere:

design f beam-c lumn base plates subjected t bendingbearing length, in.m ment.

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ENGINEERING JOURNAL / FIRST QUARTER / 1999 31

3

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CASE C: MAXIMUM MOMENT WITHOUTUPLIFT

CONCRETE BEARING LIMIT STATE

LRFD Specification Requirements

CASE D: MOMENT WITH UPLIFT

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Fig. 4. Maximum Moment Without Uplift

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shear. Case D, a beam-c lumn with sufficient m ment tcause uplift at the base plate elevati n, is sh wn in Figure5. This is the m st c mm n case in design practice, espe-The maximum m ment with ut base plate uplift is as-cially f r rigid frames designed t resist lateral earthquakesumed t ccur when the c ncrete bearing limit state isr wind l adings n the building r structure.reached ver a bearing area c ncentric with the applied

l ad at its maximum eccentricity. If the eccentricity ex-

ceeds , the tendency f r uplift f the plate is assumed t6

ccur. This assumes a linear pressure distributi n in acc r-dance with elastic the ry and n tensi n capacity betweenthe base plate and supp rting c ncrete surfaces. Case C, abeam-c lumn with the maximum m ment with ut upliftat the base plate elevati n, is sh wn in Figure 4.

(4)

06

(4) (7)6

06 T satisfy static equilibrium at the c ncrete bearing limit

state, the centr id f the c ncrete bearing reacti n ( )must be aligned with the line- f-acti n f the applied axial6l ad.

2 26

2 The LRFD Specificati n defines the c ncrete bearing(6)3 limit state in Secti n J9.

(8)

When the m ment at the beam-c lumn base plate exceeds On the full area f a c ncrete supp rt:, anch r r ds are designed t resist uplift as well as

0 85 (LRFD J9-1)6

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Case B: Small Moment Without Uplift

Practical Design Procedure—Required Area

Case C: Maximum Moment Without Uplift

Case D: Moment with Uplift

Case A: No Moment - No Uplift

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On less than the full area f a c ncrete supp rt:

0 85 (LRFD J9-2)( 2 )

2 (0 60)(0 85) (0 60)(0 85) (2)

where:

c mpressi n resistance fact r = 0.60 ( 2 ) (12)specified c ncrete c mpressive strength, ksi

N te that equati n 12 is n t a cl sed f rm s luti n be-area f steel c ncentrically bearing n a c ncretecause;supp rt, in.

maximum area f the p rti n f the supp rting is a functi n f ,surface that is ge metrically similar t and c n- is a functi n f ,centric with the l aded area, in. is a functi n f , and

is a functi n f .

H wever, if is defined as s me fixed distance r asSelect base plate dimensi ns such that: s me percentage f , the c rresp nding maximum values

f and can be determined directly.(8)

And n ting that:

(9) As previ usly stated, Case C is the situati n where uplift

is imminent and .F r c nvenience, define a new variable, , the c ncrete6bearing strength per unit width (K/in).

0 85 0 85 (2) 2(6)

3

(0 60)(0 85) (0 60)(0 85) (2)(0 60)(0 85) (0 60)(0 85) (2)

0 51 1 02 (10) 2 20 51 1 02

23 3F r m st c lumn base plates bearing directly n a c n- 3crete f undati n, the c ncrete dimensi n is much greaterthan the base plate dimensi n, and it is reas nable t 0 667 (13)

assume that the rati 2. F r m st c lumn( )

6base plates bearing n gr ut r a c ncrete pier, the c n-crete (gr ut) dimensi n is equal t the base plate dimen-

0 111 (14)si n, and it is reas nable t c nservatively take the rati

1.

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(4)(11)

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ANCHOR ROD SHEAR AND TENSION LIMITSTATES


Required Strength

Practical Design Procedure—Rod Sizes

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Tw equati ns will be needed t s lve f r the tw un-kn wns, the required tensile strength f the anch r r ds,

, and bearing length, .T maintain static equilibrium, the summati n f verti-

cal f rce must equal zer : The LRFD Specificati n defines the anch r r d (b lts)shear and tensi n limit states in Secti ns J3.6 and J3.7,

0 and Tables J3.2 and J3.5.

0 (21)

(16) (22)

F r ASTM A307 b lts:T maintain static equilibrium, the summati n f m mentstaken ab ut the f rce must equal zer : 59 1 9 45 (Table J3.5)

F r ASTM A325 b lts, threads excluded fr m the shear( ) 0 plane:2 2

117 1 5 90 (Table J3.5)

( ) 0 (17) where:2 2required anch r r d shear strength, kipsanch r r d resistance fact r 0 75

( ) 0n minal shear strength, ksi2 2anch r r d n minal (gr ss) area, in.required anch r r d tensile strength, kips( ) 0 (18)

2 2 n minal tensile strength, ksianch r r d shear stress, ksi

This is in the f rm f a classic quadratic equati n, withunkn wn . (23)

0 (19) F r A307 b lts:

24 ksi (Table J3.2)4F r A325 b lts when threads are excluded fr m the shear2plane:

60 ksi (Table J3.2)4 [ ( )]

2The shear stress ( ) is calculated c nsidering the requiredshear strength f the c lumn base.

2 ( )(20) (24)2 2

where:T determine the ther unkn wn, , substitute the value

number f r ds sharing shear l ad, unitlessf r int the equati n:

N te that all the base plate anch r r ds are c nsidered(16)effective in sharing the shear l ad.

As a check, back substitute the value f r int theequati n:

( ) 0 (17) 0 75 (25)2 2

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BASE PLATE FLEXURAL YIELDING LIMITSTATE


Required Strength—Tension Interface

Required Strength—Bearing InterfaceNominal Strength

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On secti n parallel t c lumn flanges:59 1 9 45 (26)

(29)2

0 75 (27)On secti n parallel t c lumn web:

where:

(30)number f r ds sharing tensi n l ad, unitless 2

N te that all f the base plate anch r r ds are n t c n- where:sidered effective in sharing the tensi n l ad. F r m st base

c ncrete bearing stress, ksiplate designs, nly half f the anch r r ds are required tresist tensi n f r a given l ad c mbinati n. The bearing pressure may cause bending in the base plate

The embedment, edge distances, and verlapping shear in the area between the flanges, especially f r lightly l adedc nes f the anch r r ds int the c ncrete must be checked c lumns. Yield line the ry is used t analyze this c n-t assure that the design tensile strength als exceeds the siderati n.required tensile strength. This check sh uld be in acc r-dance with the appr priate c ncrete design specificati n, (31)

4and is bey nd the sc pe f this paper.It sh uldben ted thatbaseplateh lesare ften versized ( )

(32)withrespect t theanch rr ds. In thiscase, s me“slippage”2may be necessary bef re the anch r r d shear limit state

is reached. F r large shear l ads, the designer may ch se Let the larger f , , and :t investigate alternate shear transfer limit states inv lvingpretensi ned b lts, fricti n and/ r shear lugs.

(33)2

where:

yield line the ry cantilever distance fr m c lumnThe entire base plate cr ss-secti n can reach the specifiedweb r c lumn flange, in.yield stress ( ).largest base plate cantilever, in.

N te that f r m st base plate ge metries, the cantileverdimensi n ( ) is very small and “c rner bending” f the

TheLRFDSpecificati n defines theflexuralyielding limitbase plate is neglected. When the dimensi n is large t

state in Secti n F1.acc mm date m re anch r r ds r m re bearing surface,c rner bending plate m ments sh uld be c nsidered and(28)used in the base plate thickness calculati ns.

(LRFD F1-1)

where:The tensi n n the anch r r ds will cause bending in the

required base plate flexural strength, in-K base plate f r the cantilever distance .flexural resistance fact r = 0.90 F r a unit width f base plate:n minal flexural strength, in-Kplastic bending m ment, in-K

(34)

The bearing pressure between the c ncrete and the baseF r a unit width f base plate:plate will cause bending in the base plate f r the cantilever

distances and . The bearing stress, (ksi), is calculated(35)c nsidering the required axial and flexural strength f the 4

c lumn base, and respectively.

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Practical Design Procedure—Bearing Interface Case D: Moment with UpliftBase Plate Thickness

Practical Design Procedure—Tension Interface BasePlate Thickness

DESIGN EXAMPLE 1

Case A: No Moment—No Uplift

Case B: Small Moment Without Uplift

Case C: Maximum Moment Without Uplift

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Fig. 6. Design Example 1

Required:

Solution:

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(44)Setting the design strength equal t the n minal strengthand s lving f r the required plate thickness ( ): F r all cases:

(28)2 11 (45)

(LRFD F1-1)

If :0 90

2 41 49 (46)

1 49 (36)If :

2 11 (47)

Setting the design strength equal t the n minal strengthand s lving f r the required plate thickness:

(28)

0 904

2 11 (37)

(38)

1 49 (39)

(40)( 2 )

1 49 (41) a) Design anch r r ds( 2 )b) Determine base plate thickness

1. Dimensi ns:1 5(42)

22.0 in. 0.95(12.12 in.)5 24 in. (1)

2

1 5 16.0 in. 12.12 in. 0.605 in.1 49 (43) 2 24 in. (3)

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Select: Base Plate 2 20 1’-10

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DESIGN EXAMPLE 2

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Select: 4 - 3/4 in. Diameter Anchor Rods

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Fig. 7. Design Example 2

Required:

Solution:

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2. Eccentricity:

6. Check bearing n c ncrete bel w gr ut layer120 ft-K(12 in./ft)11 08 in. (4) The gr ut is 2 in. thick. Assume that the c ncrete130K

extends at least 2 in. bey nd gr ut in each directi n.22.0 in.

3 67 in. 11 08 in. (7)6 6 (24 in.)(6.67 in.)

(0 51)(4 ksi)(20.0 in.) (10)(20 in.)(2.27 in.)

3. C ncrete bearing:76.6 K/in. 61.2 K/in. used in designAssume the bearing n gr ut area will g vern.

(0 51)(6 ksi)(20.0 in.) 1 61.2 K/in. (10)

16.0 in. 22.0 in.19 0 in.

2 2 2

16.0 in.11 08 in. 19 08 in.

2

2(130)(19 08)19 0 (19 0) (20)

61 2

19 0 16 73 2 27 in.

61 2 K/in.(2.27 in.) 130 K 8.92 K (16)

4. Anch r r d shear and tensi n:Check 4 in. dia. anch r r ds

30 0 K7.50 K (25)

4

0 75(24 ksi)(0.4418 in. )

7.96 K 7.50 K

7 50 K59 1 9 26 7 ksi (26)

0.4418 in.

8.92 K4.46 K (27)

2

a) Determine required tensile strength0 75(26.7 ksi)(0.4418 in. )b) Determine base plate thickness

8.85 4.46 K

N te that this pr blem is Example 16 fr m the AISCC lumn Base Plate Steel Design Guide Series.5. Base plate flexural yielding:

1. Required strength: (LRFD A4-2)2 27 in. 5 24 in. and n t applicable

1.2(21K) 1.6(39K) 87.6K(8 92 K)(2.24 in.) 1.2(171 in.-K) 1.6(309 in.-K) 700 in.-K2 11 0.35 in. (45)(20.0 in.)(36 ksi)

2. Dimensi ns:14.0 in. 0.95(7.995 in.)

3 20 in. (1)(130 K) 5.24 in. 22 11 (47)(20.0 in.)(36 ksi) 11.0 in. 7.995 in. 0.435 in.

1 72 (3)2 2 21.82 in.

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SUMMARY AND CONCLUSIONS

Required Tensile Strength 17.3 KREFERENCES

Select: Base Plate 1 14 1 -2

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Design Of Welded Structures

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. Man-t .ual Of Steel Construction, Load & Resistance FactorDesign

controls Col-umn Base Plates

Engineering Journal,

Design.Of Anchor Bolts In Petrochemical Facilities

.Engineering Journal

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3. Eccentricity: f r the design f the anch r r ds is slightly smallerbecause the centr id f the c mpressi n reacti n is700 in.-K

7.99 in. (4) a greater distance fr m the anch r r ds.87 6 K

14.0 in.2 33 in. 7 99 in. (7)

6 6 A meth d l gy has been presented that summarizes the4. C ncrete bearing: design f beam-c lumn base plates and anch r r ds using

fact red l ads directly in a manner c nsistent with the(0 51)(3 ksi)(14 in.) 4 42.8 K/in. (10) equati ns f static equilibrium and the LRFD Specifi-

cati n. Tw design examples have been presented. A11.0 in. 14.0 in.12 5 in. direct c mparis n was made with a pr blem s lved by2 2 2

an ther AISC meth d.11.0 in. The step-by-step meth d l gy presented will be benefi-7 99 in. 13.49 in.

2 cial in a structural design ffice, all wing the design prac-titi ner t use the same fact red l ads f r the design f the

2(87 6)(13 49)steel structure, base plate, and anch r r ds. In additi n the12 5 ( 12 5) (20)

42 8 unif rm “rectangular” pressure distributi n will be easiert design and pr gram than the linear “triangular” pressure12 5 10 05 2 45 in.distributi n utilized in all wable stress design and ther

42 8 K/in.(2.45 in.) 87.6 K 17.3 K (16) published LRFD f rmulati ns.

5. Base plate flexural yielding: 1. Bl dgett, Omer W., ,1966.2 45 in. 3 20 in. and n t applicable

2. Smith, J. C., ,2nd Editi n, 1996.(17 3 K)(1.72 in.)

2 11 0 51 in. (45) 3. American Institute f Steel C nstructi n (AISC),(14.0 in.)(36 ksi)“L ad and Resistance Fact r Design Specificati n f rStructural Steel Buildings”, December 1, 1993.

(87 6 K) 3.20 in. 4. American Institute fSteelC nstructi n (AISC),2 11 (47)(14.0 in.)(36 ksi)

, 2nd Editi n, V lume 2, 1994.1.24 in. 5. American Institute f Steel C nstructi n (AISC),

, Steel Design Guide Series, 1990./6. Shipp, J.G., and Haninger, E.R., “Design Of Headed

6. C mparis n: Anch r B lts,” V l 20, N . 2,AISC s luti n f r this pr blem: (2nd Qtr.), pp 58-69, AISC, 1983.

7. American S ciety f Civil Engineers (ASCE),Required Anch r R d Tensile Strength 21 2 K, pp 4-3 t

Select: Base Plate 1 / 14 1 -2 4-8, 1997.Length f triangular c mpressi n bl ck 5 1 in. 8. Th rnt n, W. A., “Design f Small Base plates f r

Wide-Flange C lumns,” , V l 27,Auth r’s s luti n f r this pr blem:

N . 3, (3rd Qtr.), pp 108-110, AISC, 1990a.Required Anch r R d Tensile Strength 17 3 K 9. Th rnt n, W. A., “Design f Small Base plates f r

Wide-FlangeC lumns-AC ncatenati n fMeth ds,”Select: Base Plate 1 / 14 1 -2, V l 27, N . 4, (4th Qtr.), pp 108-

Length f rectangular c mpressi n bl ck 110, AISC, 1990b.2 45 in.

Remarks:

area f steel c ncentrically bearing n a c ncreteThe auth rs’ s luti n yields the identical basesupp rt, in.plate size and thickness. Required tensile strength

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maximum area f the p rti n f the supp rting c lumn verall depth, in.surface that is ge metrically similar t and c n- axial eccentricity, in.centric with the l aded area, in. anch r r d distance fr m c lumn and base plateanch r r d n minal (gr ss) area, in. centerline parallel t m ment directi n, in.base plate width perpendicular t m ment direc- specified c ncrete c mpressive strength, ksiti n, in. c ncrete bearing stress, ksin minal tensile strength, ksi anch r r d shear stress, ksin minal shear strength, ksi base plate bearing interface cantilever parallelspecified minimum yield stress, ksi t m ment directi n, in.n minal flexural strength, in.-K base plate bearing interface cantilever perpen-plastic bending m ment, in.-K dicular t m ment directi n, in.required base plate flexural strength, in.-K yield line the rycantileverdistancefr mc lumn

web r c lumn flange, in.required flexural strength, in.-Kc ncrete ( r gr ut) bearing strength per unitbase plate length parallel t m ment directi n,width, kips/in.in.c lumn flange thickness, in.n minal bearing l ad n c ncrete, kipsbase plate thickness, in.required axial strength, kipsbase plate tensi n interface cantilever parallel trequired tensile strength, kipsm ment directi n, in.required anch r r d tensile strength, kipsanch r r d resistance fact r = 0.75required shear strength, kipsflexural resistance fact r = 0.90required anch r r d shear strength, kipsc mpressi n resistance fact r = 0.60bearing length, in.number f r ds sharing tensi n l ad, unitlessc lumn flange width, in.number f r ds sharing shear l ad, unitlesslargest base plate cantilever, in.

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