[Code]ACI 349.2R-97 Embedment Design Examples(ACI,1997)

ACI 349.2R-97 (Reapproved 2002) Embedment Design Examples Reported by ACI Committee 349 Charles A. Zalesiak Chairman Hans G. Ashar Ranjit Bandyopadhyay * Ronald A. Cook* Jack M. Daly Arobindo Dutt Branko Galunic Dwaine A. Godfrey Herman L. Graves III * Major contributor to the report † Deceased Gunnar A. Harstead Christopher Heinz Charles J. Hookham Richard E. Klingner Timothy J. Lynch Frederick L. Moreadith Dragos A. Nuta Richard S. Orr* Robert B. Pan Julius V. Rotz † Robert W. Talmadge Chen P. Tan Richard E. Toland Donald T. Ward Albert Y. C. Wong Appendix B of ACI 349 was developed to better define the design requirements for steel embedmnts revisions are periodically made to the code as a result of on-going research and testing. As with other concretebuilding codes, the design of embedments attempts to assure a ductile failure mode so that the reinforcement yields before the concrete fails. In embedments designed for direct loading, the concrete pullout strength must be greater than the tensile strength of the steel. This report presents a series of design examples of ductile steel embedments. These examples have been updated to include the revision incorparated in Appendix B of ACI 349-97. Keywords: Anchorage (structural); anchor bolts; anchors (fasteners); embedment; inserts; loads (forces); load transfer; moments; reinforced concrete; reinforcing steels; shear strength; structural design; studs; tension. PART A—Examples: Ductile single embedded element in semi-in nite concrete. . . .p. 349.2R-3 Example A1 Single stud, tension only Example A2 Single stud, shear only Example A3 Single stud, combined tension and shear Example A4 Anchor bolt, combined tension and shear Example A5 Single rebar, combined tension and shear PART B—Examples: Ductile multiple embedded elements in semi-in nite concrete. .p. 349.2R-10 Example B1 Four-stud rigid embedded plate, tension only Example B2(a) Four-stud rigid embedded plate, combined shear and uniaxial moment Example B2(b) Four-stud flexible embedded plate, combined shear and uniaxial moment Example B2(c) Four-bolt rigid surface-mounted plate, combined shear and uniaxial moment Example B3(a) Four-stud rigid embedded plate, combined tension, shear, and uniaxial moment Example B3(b) Four-stud flexible embedded plate, combined tension, shear, and uniaxial moment Example B4 Four-stud rigid embedded plate in thin slab, tension only APPENDIX A—Projected area (Acp) for four studs. . . . . . . . . . . . . . . . . . .p. 349.2R-26 ACI 349.2R-97 became effective October 16, 1997. Copyright © 2002, American Concrete Institute. All rights reserved including rights of reproduction and use in any form or by any means, including the making of copies by any photo process, or by electronic or mechanical device, printed, written, or oral, or recording for sound or visual reproduction or for use in any knowledge or retrieval system or device, unless permission in writing is obtained from the copyright proprietors. CONTENTS Introduction. . . . . . . . . . . . . . . . . . . . . . .p. 349.2R-2 Notation. . . . . . . . . . . . . . . . . . . . . . . . . . p. 349.2R-2 ACI Committee Reports, Guides, Standard Practices, and Commentaries are intended for guidance in planning, designing, executing, and inspecting construction. This document is intended for the use of individuals who are competent to evaluate the significance and limitations of its content and recommendations and who will accept responsibility for the application of the material it contains. The American Concrete Institute disclaims any and all responsibility for the stated principles. The Institute shall not be liable for any loss or damage arising therefrom. Reference to this document shall not be made in contract documents. If items found in this document are desired by the Architect/Engineer to be a part of the contract documents, they shall be restated in mandatory language for incorporation by the Architect/ Engineer. 349.2R-1 349.2R-2 MANUAL OF CONCRETE PRACTICE INTRODUCTION This report has been prepared by members of the ACI 349 Sub-Committee on Steel Embedments to provide examples of the application of the ACI 349 Code to the design of steel embedments. The ACI 349 Committee was charged in 1973 with preparation of the code covering concrete structures in nuclear power plants. At that time, it was recognized that design requirements for steel embedments were not well defined and a special working group was established to develop code requirements. After much discussion and many drafts, Appendix B was approved and issued in the 1978 Supplement of ACI 349 covering the design of steel embedments. Subsequently, the Sub-Committee has continued to monitor on-going research and testing and to incorporate experience of applying the Code. Periodic revisions have been made to the Code and Appendix B. The underlying philosophy in the design of embedments is to attempt to assure a ductile failure mode. This is similar to the philosophy of the rest of the concrete building codes wherein, for example, flexural steel for a beam is limited to assure that the reinforcement steel yields before the concrete crushes. In the design of an embedment for direct loading, the philosophy leads to the requirement that the concrete pull-out strength must be greater than the tensile strength of the steel. This report includes a series of design examples starting with simple cases and extending to more complex cases for ductile embedments. The format for each example follows the format of the Strength Design Handbook, SP-17, and provides a reference back to the code paragraph for each calculation procedure. NOTATION depth of equivalent stress block, in. effective stress area defined by the projected area of the 45 degree stress cone radiating towards the attachment from the bearing edge of the anchor, sq. in. effective stress area of anchor, sq. in. area of anchor head, sq. in. area of steel, sq. in. area of steel required to resist tension, sq. in. area of steel required to resist shear, sq. in. Ar = A vf = b = B c C db dh ds Fy fc ′ f ut fy h k tr ld Ld Mn Mu My n Pd Pn Pu R S t T Th Vn Vu α β γ µ φ = = = = = = = = = = = = = = a = Acp = Ac = Ah = As = Ast = Asv = = = = = = = = = = = = = = = = = = = = reduction in effective stress area to account for limited depth of concrete beyond the bearing surface of the embedment, sq. in. area of shear friction reinforcement, sq. in. width of embedded or surface mounted plate, or width of an anchor group, measured out to out of bearing edges of the outermost anchor heads, in. overlapping stress cone factor (see Appendix A) spacing or cover dimension, in. compressive reaction nominal diameter of reinforcing bar, in. diameter of anchor head or reinforcing bar, in. diameter of tensile stress component, in. specified yield strength of steel plate, psi specified compressive strength of concrete, psi specified tensile strength of steel, psi specified yield strength of steel, psi overall thickness of concrete member, in. transverse reinforcement index development length, in. embedment depth of anchor head measured from attachment of anchor head to tensile stress component, to the concrete surface, in. nominal moment strength factored moment load on embedment elastic moment capacity of steel plate number of threads per inch design pullout strength of concrete in tension nominal axial strength factored external axial load on the anchorage radius of 45 degree stress cone, in. (see A cp ) spacing between anchors, in. thickness of plate, in. tension force thickness of anchor head, in. nominal shear strength factored shear load on embedments reinforcement location factor coating factor reinforcement size factor coefficient of friction strength reduction factor combined tension and shear Single rebar. shear only Single stud. combined tension and shear Anchor bolt.EMBEDMENT DESIGN EXAMPLES 349.2R-3 PART A EXAMPLES: Ductile single embedded element in semi-infinite concrete Example A1 Example A2 Example A3 Example A4 Example A5 Single stud. tension only Single stud. combined tension and shear . 79 in.38 = 5.196 × 60 = 11.312 in. diameter stud 1-5/16 in.5. Ah = π(d h /2 )2 = 0.5 OK Th = 0. OK STEP 3: Determine required embedment length for the stud to prevent concrete cone failure B. must exceed the minimum specified tensile strength (As fut ) of the tensile stress component. 2 .4.2 (per manufacturer’s data.2R-4 MANUAL OF CONCRETE PRACTICE Example A1—Single stud.8 Pd = φ4 f c Acp ′ Acp = π[(Ld + dh /2)2 – ( dh /2) 2] Compute Ld from the equation: π[Ld + dh /2)2 – ( dh /2) 2]φ4 f c ≥ As fut ′ L d + Ld – 22.) Ah / As = 0.5. Pd . Pd > Asfut As fut = 0. CODE SECTION P u = 8 kips DESIGN PROCEDURE STEP 1: Determine required steel area of the stud Assume that the load is applied directly over the stud and that a plate size of 3 in.196 in. Th = 0.25 in.2 Use one 1/2 in.8 kips φ4 f c = 0.65 × 4 × ( 4000 ) ′ = 165 psi (see Note 2) π[(Ld + 0.1 B.8 Ld (Ld + 1.312 > 0.1. As = 0.87 + 0.30 in. Pu = φ P n = φ Asfy As = 8/ [(0.349.1(a) B.79 / 0.18 in.0 times the greatest dimension from the outer most bearing edge of the anchor head to the face of the tensile stress component. b) Thickness of the anchor heat (Th) is at least 1. tension only Design an embedment using a stud welded to an embedded plate.165 ≥ 11. which has an effective length of 4. Use 1/2 in.1 Equate the external (required strength) and internal (design strength) forces and solve for the required steel area for the stud.5 times the area of the tensile stress component. (per manufacturer’s data) (dh – ds )/2 = 0.25 in.2 of the Code.2 a) Area of the anchor head (Ah) (including the area of the tensile stress component) is at least 2.9)(50)] = 0. diameter stud.1.18 in.000 psi f ut = 60.5 2]0.25 OK Head and tensile stress component are concentric.5.87 in.6.5)2 – 0.196 = 4 > 2. × 3 in.2 The design pullout strength of the concrete. has been established by requirements of the attachment.8 ≥ 0 Ld ≥ 4. long. × 3/8 in. d h = 1 in.4.2 OK STEP 2: Check anchor head bearing B. dh Th Ld Given: f c = 4000 psi ′ f y = 50. giving Ld = 4. c) Bearing area of head is approximately evenly distributed around the perimeter of the tensile stress component. CALCULATION B.000 psi Pu = 8 kips where Pu is the required factored external load as defined in Section 9.5.0) ≥ 22.2 > 0. .2R-5 Example A1. then the embedment length would exclude the thickness of the embedded plate.65 per Category (d) of Section B. continued CODE SECTION DESIGN PROCEDURE STEP 4: Check plate thickness Since the load is applied directly over the stud. 2) In all design examples.4. diameter studs is acceptable on 3/8 in. the strength reduction factor φ for concrete pullout is taken as 0. thick plate per manufacturer. the embedment length Ld is taken to the face of the concrete. the only requirement on plate thickness is that it satisfy the minimum thickness required for stud welding. If the plate were larger than the stress cone.2.EMBEDMENT DESIGN EXAMPLES 349. NOTE: Stud welding of 1/2 in. OK CALCULATION 1) In the above example. “Shear Strength of Thin Flange Composite Sections.2.4.185 3/8 in. Equate the external (required strength) and internal (design strength) forces and solve for the required steel area for the stud.196 = 4 > 2.4.” G.7 with φ = 0.25 OK Head and tensile stress component are concentric.5/2.5 OK Th = 0.2 of the Code.1. Th = 0.: .7* Use 1/2 in.25 in. Goble.000 psi Vu = 6 kips where Vu is the required factored external load as defined in Section 9. As = 0. NOTE: The provisions of Section 11. µ = 0. 1968. long (see calculation in Example A1) t > 0.2 a) Procedure is identical to that in Example A1 b) Procedure is identical to that in Example A1 Ah / As = 0. thick plate is OK. April.2 B.2 > 0.16 in.5 on shear strength are not applicable at the surface between the steel plate and the concrete. STEP 2: Check anchor head bearing Vu = φVn = φµ A vf f y A vf = Vu /(φµ f y ) A vf = 6/(0.3.2 Use one 1/2 in.2 OK B. OK c) Procedure is identical to that in Example A1 STEP 3: Determine required embedment length for the stud to prevent concrete cone failure B. STEP 4: Check plate thickness Select plate thickness such that ds / t < 2.2R-6 MANUAL OF CONCRETE PRACTICE Example A2—Single stud.85. Shear loads at this interface are carried by local bearing and wedge action as described in commentary Section B.312 > 0. Given: f c = 4000 psi ′ f y = 50.312 in. Th Ld dh V u = 6 kips CODE SECTION DESIGN PROCEDURE STEP 1: Determine required steel area of the stud CALCULATION B.1 Procedure is identical to that in Example A1 since tensile capacity of the stud must be developed. G. diameter stud 5-3/16 in.5.85 × 0.349.5.9.6.9 × 50) = 0. * Ref. AISC Engineering Journal.4. (per manufacturer’s data) (d h – ds )/2 = 0. diameter stud.5.5.196 in.79 / 0.7 = 0.7.000 psi f ut = 60. shear only Design an embedment using a stud welded to an embedded plate.16 in.2 Use the shear friction provision of Section 11. 7 Eq.2 11.000 psi Pu = 4 kips Vu = 2 kips where Pu and Vu are the required factored external loads as defined in Section 9.2. (11-26) B.EMBEDMENT DESIGN EXAMPLES 349.5. thick plate is OK 8 The provisions of Section 11. Given: f c = 4000 psi ′ f y = 50. OK c) Procedure is identical to that in Example A1 STEP 3: Determine required embedment length for the stud to prevent concrete cone failure B.14 in.9.9 × 50) = 0.2 of the Code.2 > 0.79 / 0. Sum the area of steel required for tension with the area of steel required for shear.85.25 in.14 in.4.6.1. Shear loads at this interface are carried by local bearing and wedge action as described in commentary Section B. Goble.5.2 Vu = φVn = φµ Asv fy Asv = Vu /(φµ f y ) Asv = 2/(0.25 OK Head and tensile stress component are concentric.7 with φ = 0. µ = 0.09 + 0. Th Ld dh P u = 4 kips V u = 2 kips CODE SECTION DESIGN PROCEDURE STEP 1: Determine required steel area of the stud CALCULATION B. long (see calculation in Example A1) STEP 4: Calculate minimum plate thickness Select plate thickness such that ds / t < 2. Use the shear friction provision of Section 11.” G. April.5/2. As = 0.6.1 Equate the external (required strength) and internal (design strength) tension forces and solve for the required steel area for tension.000 psi f ut = 60.312 in. Th = 0.5.05 in.2 OK B.09 in. Equate the external (required strength) and internal (design strength) forces and solve for the required steel area for shear.05 = 0.2 B. diameter stud. Total Area As = Ast + Asv STEP 2: Check anchor head bearing Pu = φ Pn = φ A st f y A st = 4/(0.196 = 4 > 2. AISC Engineering Journal.2 Use one 1/2 in.185 3/ in.312 > 0.196 in. . 1968.4.2 a) Procedure is identical to that in Example A1 b) Procedure is identical to that in Example A1 Ah / As = 0.5.3. G.5 OK Th = 0.5.9 × 50) = 0. “Shear Strength of Thin Flange Composite Sections. (per manufacturer’s data) (dh – ds )/2 = 0. diameter stud 5-3/16 in.2R-7 Example A3—Single stud.1 Procedure is identical to that in Example A1 Use 1/2 in.85 × 0.7.6.2 B.4.7 = 0.7* NOTE: * Ref.: t > 0.5 on shear strength are not applicable at the surface between the steel plate and the concrete.2 As = 0. combined tension and shear Design an embedment using a stud welded to an embedded plate.3. 4. diameter bolt width across flats = 2.000 psi Pu = 40 kips Vu = 20 kips where Pu and Vu are the required factored external loads as defined in Section 9.78 > 0. Use provision for contact surface of the base plate flush with the surface of the concrete.46 in.5.95 in. Tensile stress area = 0.125 + 196.2 Vu = φ Vn = φ (0. 7 threads per inch. Total Area As = Ast + Asv As = 0.6.1.2 Sum the area of steel required for tension with the area of steel required for shear.2 Ah / As = 3.2 > 0.1 B.65 × 4 × ( 4000 ) = 165 psi [(Ld + 1.96 in. Equate the external (required strength) and internal (design strength) forces and solve for the required steel area for shear.46 / 0.7)(φ f y )] A sv = 20/(0.1 Equate the external (required strength) and internal (design strength) tension forces and solve for the required steel area for tension.97 = 3. Ah = (1. diameter bolt.5.2 of the Code.97 × 105 = 102 kips φ4 f c′ = 0.4.2.1 Procedure is identical to that in Example A1 A s fut = 0.2 B.55 + 0.85.2 OK STEP 2: Check anchor head bearing B..6. OK OK OK c) Procedure is identical to that in Example A1 STEP 3: Determine required embedment length for the bolt to prevent concrete cone failure B.7 × 0.2R-8 MANUAL OF CONCRETE PRACTICE Example A4—Single bolt.2 Use one 1-1/4 in.85 × 81) = 0.97 in.165 ≥ 102 Ld (Ld + 2.2 B. combined tension and shear Design an embedment using a high strength bolt (A 325). P u = 40 kips CODE SECTION DESIGN PROCEDURE STEP 1: Determine required steel area of the stud CALCULATION B.5. Th = 0.000 psi f ut = 105.2 a) Procedure is identical to that in Example A1 A 325 Heavy Hex Head for 1-1/4 in.57 > 2.8 ) = 12.41= 0.6. thickness = 0.96 in. Pu = φ Pn = φ A st f y A st = 40/(0.9 × 81) = 0. 2 .7 f y A sv ) A sv = Vu / [(0.349.125 + ( 1.97 in.8 Ld ≥ –1.0)2 × 2 × 3 = 3.25)/2 = 0.5.3.5.0 in.53 in.5 b) Procedure is identical to that in Example A1 (dh – ds )/2 = (2 × 2 3 – 1.125 2 ]0.41 in.2 As = 0.25) ≥ 196.78 in.55 in. φ = 0.53 Head and tensile stress component are concentric.125) 2 – 1. Given: f c = 4000 psi ′ f y = 81. 7 Eq. Equate the external (required strength) and internal (design strength) forces and solve for the required steel area for shear. 6 bar (γ = 0.28 + 0.3.11 in. Given: f c = 4000 psi ′ f y = 60.28 in.5. (11-26) B.2.39 in.1(b) 12.4 .0 × 0.5.5.1 Equate the external (required strength) and internal (design strength) forces and solve for the required steel area for tension.2 Use No. µ = 0. combined tension and shear Design an embedment using a straight reinforcing bar welded to an embedment plate.3.2.6. As = 0.4 ld 3 fy αβγ ----.85.2 Vu = φVn = φ µ A sv f y Asv = Vu /(φµ f y ) Asv = 5/(0.2 Provide full penetration weld between bar and plate per AWS D1.2 11. max.5.). Use 5/16 in.3 12.2 As = 0. Use ld = 24 inches t ≥ 0.7 with φ = 0.11 = 0. of fresh concrete to be cast below the anchor (α = 1.EMBEDMENT DESIGN EXAMPLES 349.3) f ut = 90.14. STEP 3: Calculate minimum plate thickness Select the plate thickness as shown for Example A2.9 × 60) = 0. thick plate 12.39 in. more than 12 in.2R-9 Example A5—Single rebar.= ----.75 = 22.44 in.3.85 × 0.2.3 × 1.7 = 0.---------. 6 Grade 60 reinforcing bar. Total Area As = Ast + Asv STEP 2: Calculate required embedment length Pu = φ Pn = φ A st f y Ast = 15/(0.000 OK per Code Section 3.8).5.2 > 0. uncoated anchor (β = 1.2 in.0). based on attachment configuration and welding requirements.6.2 OK B. no adjacent anchors or edges ([c + k tr ]/db = 2.5. Use the shear friction provision of Section 11.8 ) ⁄ ( 2.9. CODE SECTION P u = 15 kips V u = 5 kips DESIGN PROCEDURE STEP 1: Determine required steel area of the stud CALCULATION B.3).6.28 in.5 ) ] × 0.75/2.000 psi (≤ 60. STEP 4: Connection of reinforcing bar to plate l d = [ ( 3 ⁄ 40 ) ] × [ ( 60000 ⁄ 4000 ) ] × [ ( 1.2 B. No. Sum the area of steel required for tension with the area of steel required for shear.-----------------------------------db 40 f ′ [ ( c + k tr ) ⁄ d b ] c Assume no transverse reinforcement (k tr = 0).1.2 of the Code.000 psi (based on typical test results) Pu = 15 kips Vu = 5 kips where Pu and Vu are the required factored external loads as defined in Section 9.9 × 60) = 0. and uniaxial moment Four-stud flexible embedded plate. combined tension. tension only Four-stud rigid embedded plate. combined tension. combined shear and uniaxial moment Four-bolt rigid surface-mounted plate. combined shear and uniaxial moment Four-stud flexible embedded plate. shear. and uniaxial moment Four-stud rigid embedded plate in thin slab. shear. combined shear and uniaxial moment Four-stud rigid embedded plate.349.2R-10 MANUAL OF CONCRETE PRACTICE PART B EXAMPLES: Ductile multiple embedded element in semi-infinite concrete Example B1 Example B2(a) Example B2(b) Example B2(c) Example B3(a) Example B3(b) Example B4 Four-stud rigid embedded plate. tension only . 1 Equate the external (required strength) and internal (design strength) forces and solve for the required steel area for the stud. Given: f c = 4000 psi ′ f y = 50.2 . diameter studs.6 in.1 The design pullout strength of the concrete. Pd .25 OK Head and tensile stress component are concentric.79 / 0.442 in.5.2R-11 Example B1—Four-stud rigid embedded plate.2 > 0.400 /165 = 160. As = 0.000 psi (studs) f ut = 60. must exceed the minimum specified tensile strength of the tensile stress components.4 kips φ4 f c′ = 0.5.65 × 4 × ( 4000 ) = 165 psi Pd = φ4 f c′ Acp Where Acp = the projected area of the 45 deg stress cones radiating toward the attachment from the bearing edge of the anchors.5 OK Th = 0. Pu = φ Pn = φµ A s f y As = 18 /45 = 0. This area must be limited by overlapping stress cones and by the bearing area of the anchor heads.196 = 4 > 2.000 psi Fy = 36.40 in.312 in.1.4.2 a) Procedure is identical to that in Example A1 b) Procedure is identical to that in Example A1 Ah / As = 0.312 > 0. Pd ≥ As fut As fut = 4 × 0. (per manufacturer’s data) (dh – ds )/2 = 0. P u = 18 kips CODE SECTION DESIGN PROCEDURE STEP 1: Determine required steel area of the stud CALCULATION B. OK c) Procedure is identical to that in Example A1 STEP 3: Determine required stud spacing and embedment length to prevent concrete pullout B.25 in. A cp min = As fut /(φ4 f c′ ) = 26.EMBEDMENT DESIGN EXAMPLES 349.5.2 of the Code.40 in. Th = 0.6. A 501 structural tube attachment.2 Use four 3/8 in. tension only Design an embedment with four welded studs and a rigid embedded plate for a 3 × 3 × 3/16 in.2 OK STEP 2: Check anchor head bearing B.110 × 60 = 26.000 psi (plate) Pu = 18 kips where Pu is the required factored external load as defined in Section 9. 09 in.2 6 in..5 in. For the stud configuration selected above (R = 4.2 290.4 in. the projected area is (see Appendix A) The projected area of the stress cones may be calculated for each standard stud length (Ld) and a range of stud spacings.09 in. an approximate calculation is sufficient.71 Spacing S Radius R.).2 5 in. spacing.2R-12 MANUAL OF CONCRETE PRACTICE Example B1. All values greater than: CALCULATION Acp = (4π – 2B )R 2 – 4 Ah Acp = (3π – B)R 2 + S 2 – 4 Ah (2R > S > 2 R) ( 2 R > S) Acp min = As fut /(φ4 f c′ ) are then satisfactory. NOTE: The above calculation utilizes an exact calculation of the projected area.2 7 in. inches 3.2 This compares with the exact value of 175. stud with effective length of 3.09 + π × 4.4 in.09′′ R 6′′ R Table B1-1—Projected areas (Acp ) for varying Ld and S Development length Ld. inches 4. continued CODE SECTION DESIGN PROCEDURE STEP 3: (continued) For a four-stud plate with studs at spacing S and radius R of the projected stress cone.0 in. 152.2 226.2 4 in. Such a method is used in Example B2. S = 6 in.0 in.4 in. and Ld + dh /2 + t at the outer surface of the concrete and plate.349.2 in.09 in. In many cases.8 in.375 2 = 184.2 calculated in Table B1-1 R 4.2 324.71 5. 175.2 6.2 . S (see Table B1-1). the approximate method would give: R 6′′ Acp = 6 × 6 + 4 × 6 × 4. Conservatively neglect the thickness of the plate t.09 2 = 4 × π × 0. × 4-1/8 in. The radius of the projected stress cone is Ld + dh /2 at the underside of the embedded plate.2 258. Select 3/8 in. at 6 in. 129.71 in.4 in.9 in. 195. Evaluate plate sections to determine minimum load capacity. On diagonal (b-b): M = 4.5 in.2R-13 Example B1.5 in. continued CODE SECTION DESIGN PROCEDURE STEP 4: Calculate required plate thickness Try an 8 in.1 B.5 2 = 9.6.-kips 0.5′′ 1.5 = 13.56 in. × 8 in. × 8 in. × 5/8 in.5 ⁄ 43.5′′ b Calculate the bending strength of the plate based on the yield moment capacity using yield stress. embedded plate . plate The plate must transmit to the studs all loads used in the design of the attachment.9 × 36 = 43.2t 2 t min = ( 9. The design strength for embedments shall be based on a maximum steel stress of φf y .9 × My = 1/6 × 8 × t 2 × 0.50 in.2 b a 1. Use 8 in. At face of tube (a-a): M = 9 × 1.5 × 1.9 × 36 = 38.-kips My = 1/6 × 5 2 × t 2 × 0.2 ) = 0.2 t 2 t min = ( 13.2 ) = 0.3. a CALCULATION B.5 ⁄ 38.EMBEDMENT DESIGN EXAMPLES 349. 3 Capacity reduction factor for shear Design shear strength must be greater than the required strength . × 7 in.1 Equate the internal forces and solve for a : 0.6.4 kips where Mu and Vu are the required factored external loads as defined in Section 9.9 kips) φV n = 19.9 Ae (50)(6 – 2.2 of the Code. stud spacing CALCULATION T = Aefy 10. The stud area not used for moment is available for shear transfer by shear friction.2 is applicable.85 φV n = 0.85f c′ ab = Ae f y a = Ae f y /(0.2 0. combined shear and uniaxial moment Design an embedment using welded studs and a rigid embedded plate for a 3 × 3 × 1/4 in.2.10Ae Equate the external (required strength) and internal moment (design strength) and solve the resulting quadratic equation for Ae : For 2 studs in tension.4 kips 7′′ 5′′ M u = 70 in. Section B. A sv = 2 (0.000 psi (studs) f ut = 60.10Ae /2) Ae = 0. Given: f c = 4000 psi ′ f y = 50.5.85 (4)(7)] a = 2. diameter studs STEP 2: Design for shear Mu = φ Mn 70.196) A sv = 0.9 kips φ = 0.4 kips OK B.2 9.2. 6′′ C 0. plate with 5 in.5 > 12.6.5.2.2.90 V n = 0.2 Ae /stud = 0.2R-14 MANUAL OF CONCRETE PRACTICE Example B2(a)—Four-stud rigid embedded plate.000 psi (plate) Mu = 70 in.85 f c′ b ) a = Ae 50 / [ 0.2 B.508)(50) V n = 22.000 psi Fy = 36.-kips 3 × 3 × 1/4 ′′ Tube 5′′ DESIGN PROCEDURE STEP 1: Design for moment Try a 7 in.85 (22. CODE SECTION 7′′ V u = 12.2 Since this is an embedded base plate.138) + 2 (0.2 9.275 in.7 Assume a uniform stress block for concrete compression zone and the two top studs as the tension components.0 = 0.196 – 0.90Av f f y V n = 0.2 Ae /stud = 0. Ae per stud is: Try 1/2 in.90 (0.138 in.349.2 9.6. A 501 structural tube attachment.-kips Vu = 12.196 in.85 f c′ a M u = 70 9.6.508 in. × 5 in.2 Shear-friction coefficient Nominal shear strength B.3.5. A φ factor of 0. Note that the plate thickness is calculated using the nominal strength of the anchors.5 in. With multiple rows of anchors in tension (e.EMBEDMENT DESIGN EXAMPLES 349.-kips ← controls Moment on compression side Note that C = T . CALCULATION T = Aefy dt No Yield dc c t Determine minimum base plate thickness to prevent base plate yielding. both the moment on the tension side and the moment on the compression side need to be determined. NOTE: For this example. continued CODE SECTION DESIGN PROCEDURE STEP 3: Design for rigid base plate In order to ensure rigid base plate behavior.0) M = 13.g.1(0. use 3/4 in. The moment in the plate at the edge of the attached member is used for sizing the base plate thickness. a middle row of anchors).. The φ factor is not included since the calculation of plate thickness should be based on the maximum nominal tensile force in the anchor rather than the design force. The larger of the moment on the tension side or the compression side will control the design of the base plate.75 in.-kips M = Cd c M = A e f y dc M = (0.9 is used in calculating the required area of the anchor in Step 1 . it is only necessary to show that d c is greater than d t . it is essential that the base plate not yield on either the compression or tension side of the connection.275 (50)[ 2 – 2. plate Determine minimum base plate thickness to prevent yielding of plate.275)(50)(1.75 in. Moment on tension side a /2 M = Tdt or M = Cd c whichever is greater M = Tdt M = Ae fy dt M = (0.275)/2 ] M = 23.5 t = 0. Nominal moment capacity of base plate Mn = F y S Mn = F y ( bt 2/6) = (36)(7)t 2 /6 Mn = 42t 2 42 t 2 = 23.2R-15 Example B2(a).275)(50)(6 – 4 – a /2) M = 0. This general procedure is shown in this example. and then calculate the moment as M = (d c)(C ). R = Ld stud + plate thickness + stud head radius = 5.94 in.0) = 51.5 = 6.69 + 0. NOTE: Although slightly unconservative.349.4.79) = 312.65)* ( 4000 ) (312. the embedment radius will usually exceed the anchor spacing.0 in.94) + (6.2R-16 MANUAL OF CONCRETE PRACTICE Example B2(a). Acp = 5 (5) + 4 (5)(6.75 + 0.300 lb Pd = 51. This assumption of all four studs in tension assures ductility even in the event of a pure tension load. the assumption above is reasonable for embedments where the embedment radius (R ) exceeds the spacing between individual anchors.0 kips See Example A1 Pd = φ4 ( f c′ ) Acp where: φ = 0. If the designer can assure that such a tensile condition cannot occur. continued CODE SECTION DESIGN PROCEDURE STEP 4: Embedment length B.0 kips .196)(60) = 47. particularly those in the “rigid” plate category.3 kips > 47.2 Calculate capacity of concrete CALCULATION Put = As f ut = 4(0. Assume the outer boundaries of the stress cones are connected by tangents. Check bearing requirements of stud head B.1 Calculate design load assuming all studs may resist concurrent tensile loads.5. For most embedments. Pd = 4(0.2 Therefore. long having an effective length.94) 2 π – 4(0.65 Acp = projected area of concrete R = radius of projected cones Try 1/2 in. diameter stud 6-1/8 in.69 in.1. Ld = 5. it is sufficient only to develop two of the studs at a time since the other two studs are in compression. pp.000 psi (plate) Mu = 70 in. and Klingner. R. thick base plate and determine yield capacity Mn of plate. American Society of Civil Engineers.2 of the Code.000 psi (studs) f ut = 60. This will cause the compressive resultant to move inward toward the attached member. June. 1992.2. stud spacing CALCULATION T = Aefy 10.-kips c = Mn / C c = F y S / Ae f y c = 16. A.-kips Vu = 12. × 7 in. No. plate with 5 in. Physically.EMBEDMENT DESIGN EXAMPLES 349. 1645-1665.” Journal of Structural Engineering. 6. NOTE: For simplicity. Mn = F y S Mn = (36)(7)(0. a hinge will form on the compression side of the base plate at the edge of the attached member. C c M u = 70 Reference: Cook. “Ductile Multiple-Anchor Steel-to-Concrete Connections. 4′′ If the base plate is not stiff enough to obtain rigid base plate behavior..4 kips 7′′ 5′′ M u = 70 in. V. Given: f c = 4000 psi ′ f y = 50. Assume a 5/8 in.4 kips where Mu and Vu are the required factored external oads as defined in Section 9. × 5 in.4/ Ae (50) c = 0. 118.7 Determine the amount of tensile steel required for the applied moment. NOTE: From summation of forces T = C = Ae f y .4 in.-kips 3 × 3 × 1/4 ′′ Tube 5′′ DESIGN PROCEDURE STEP 1: Design for moment Try a 7 in.. CODE SECTION 7′′ V u = 12. R. E. it may also be assumed that the compressive reaction is located at the edge of the attached member.625)2 /6 Mn = 16.000 psi Fy = 36. A 501 structural tube attachment.328 / Ae Determine c. the nearest the compressive reaction can be to the edge of the attached member is a distance “c ” equal to the yield moment of the plate divided by the compressive reaction. combined shear and uniaxial moment Design an embedment using welded studs and a flexible embedded plate for a 3 × 3 × 1/4 in.2R-17 Example B2(b)—Four-stud flexible embedded plate. 2 For 2 studs in tension.349.90 Vn = 0. M = Td t M = Ae f y d t M = (0.478 in. plate is OK 8 Nominal moment capacity of base plate. Shear-friction coefficient Nominal shear strength Asv = 2 (0.-kips Mn = F y S Mn = F y ( bt 2 /6) Mn = (36)(7)(0. .2 Ae /stud .4 in.196) Asv = 0.2 0.2 Ae /stud = 0.2 B.328/Ae ) 70.3 Capacity reduction factor for shear Design shear strength must be greater than the required strength STEP 3: Design for flexible base plate In order to ensure that prying action does not occur on the tension side of the base plate.196 – 0.0.3 kips > 12.9 Ae (50)(4 + c ) 70.2 is applicable.90 is already included.0) M = 15.5.307 in.85 φ Vn = 0.2R-18 MANUAL OF CONCRETE PRACTICE Example B2(b).-kips > 15.2.2.0 = 0.6.6. T = Aefy dt No Yield Yield OK C t Determine minimum base plate thickness to prevent base plate yielding and possible prying action on tension side. Section B. continued CODE SECTION DESIGN PROCEDURE STEP 1: continued 9.2 Since this is an embedded base plate.2.478) (50) Vn = 21.0 = 45 (4 Ae + 0.307)(50)(1.90 (0. diameter studs STEP 2: Design for shear B.625) 2/6 Mn = 16.-kips 5/ in.196 in.5. Ae per stud is: Try 1/2 in.5. The stud area not used for moment is available for shear transfer by shear friction.6. it is essential that the base plate not yield on the tension side of the connection.3.2 9.5 kips) φ Vn = 18.153) + 2 (0.4 kips OK B.35 in.0 = 0.6.6.1 Equate the external (required strength) and internal (design strength) moment and solve the resulting linear equation for Ae : CALCULATION Mu = φ Mn 70.5 kips φ = 0.35 in.2 9.85(21.5.3 B.328) Ae = 0.153 in. Note that a φ = 0.9 Ae (50)(4 + 0.90 Asv f y Vn = 0.2 9. EMBEDMENT DESIGN EXAMPLES 349. continued CODE SECTION DESIGN PROCEDURE STEP 4: Embedment length See Example B2(a) NOTE: As can be seen from this flexible base plate example (5/8 in. base plate) and Example B2(a) with a rigid base plate (3/4 in. base plate).2R-19 Example B2(b). The real advantages of flexible base plate analysis become apparent with multiple rows of anchors. CALCULATION . the compressive reaction becomes so large that the assumption of rigid base plate behavior results in excessively thick plates. In the case of multiple rows of anchors. For multiple row connections. there is very little difference between the two analyses for a typical two row connection. the flexible base plate procedure will result in more reasonable base plate thicknesses. taken as 0.2 Since this is a surface mounted plate. threaded bolts A e = 0.41A e /2 ) A e = 0.2 A e / bolt = 0.000 psi (plate) Mu = 70 in.2R-20 MANUAL OF CONCRETE PRACTICE Example B2(c)—Four-bolt rigid surface-mounted plate.70 A vs f y + 0.2.85 f c′ M u = 70 Equate the internal forces and solve for a : 0.4 in this example.2 B. T = Aefy 6′′ a C 0. 7′′ V u = 12.7 Try a 7 in. diameter bolts (for 3/8 in.4 kips 7′′ 5′′ M u = 70 in. bolt spacing Determine the amount of tensile steel required for the applied moment.6.41A e 9. Section B.2.2 A e / bolt = 0.4 kips where Mu and Vu are the required factored external loads as defined in Section 9.85 f c′ ab = A e f y a = A e f y /(0. × 5 in.1 Equate the external (required strength) and internal (design strength) moments and solve the resulting quadratic equation for A e : For 2 bolts in tension.5. × 7 in. plate with 5 in.000 psi (bolts) Fy = 36. Mu = φ Mn 70. Given: f c = 4000 psi ′ f y = 105.078 in. combined shear and uniaxial moment Design an embedment using cast-in-place bolts and a rigid surface-mounted plate for a 3 × 3 × 1/4 in.1 is applicable.065 in.6.2 9.349.0 = 0. Assume standard concrete beam compression block.2 ) STEP 2: Design for shear B.000 psi (bolts) f ut = 125.130 in.078 in.6.40 C .85 (4)(7) ] a = 4. The nominal shear strength is the sum of the shear strength provided by the anchors and the friction force between the base plate and concrete due to the compressive reaction.9 A e (105)(6 – 4.-kips Vu = 12.-kips 3 × 3 × 1/4 ′′ Tube 5′′ CODE SECTION DESIGN PROCEDURE STEP 1: Design for moment CALCULATION 10.5.2 of the Code.3.1 Vn = 0. A e per bolt is: Try 3/8 in.5.85 f c′ b ) a = A e 105 / [0. A 501 structural tube attachment.2. 130)(105)(6 – 4 – a /2) M = 0. the actual design values used should be based on the appropriate structural steel code.182)(105) + (5.46 kips CALCULATION Vn = 0. and then calculate the moment as M = (dc )(C).70(0. Determine minimum base plate thickness to prevent base plate yielding.3 The anchor area not used for moment is available A vs = 2(0.2 Shear contribution from friction between the base plate and concrete due to the compressive reaction.40 C Vn = 0. .EMBEDMENT DESIGN EXAMPLES 349. This general procedure is shown in this example. both the moment on the tension side and the moment on the compression side need to be determined..130 (105)[ 2 – 4.2 9.130)/2 ] M = 23.5.130)(105)(1. Assume threads in shear plane. NOTE: For this example.4 kips OK B.0) M = 13. it is only necessary to show that d c is greater than dt . With multiple rows of anchors in tension (e.85(18. a middle row of anchors).7 in.-kips ← controls Moment on compression side: Note that C = T.40 T = 0.065) + 2(0.6.2.-kips M = Cd c M = A e f y dc M = (0. A vs = 0.40(0.70 A vs f y + 0.40 C = 0. STEP 3: Design for rigid base plate In order to ensure rigid base plate behavior. Nominal shear strength from anchors and friction between the base plate and concrete 0. The moment in the plate at the edge of the attached member is used for sizing the base plate thickness. Moment on tension side T = Aefy dt No Yield dc C a /2 M = Tdt or M = Cd c whichever is greater M = Td t M = Ae fy dt M = (0.130)(105) 0.078) for shear transfer.g. Actual design of the base plate is not covered by ACI 349 Appendix B.40( A e f y ) = 0. it is essential that the base plate not yield on either the compression or tension side of the connection.182 in. NOTE: This step in the example is for information only. Although the design procedure shown is appropriate for base plate design.46) Vn = 18.4 in.078 – 0.3 Capacity reduction factor for shear Design shear strength must be greater than the required strength.40 C = 5. continued CODE SECTION DESIGN PROCEDURE STEP 2: continued B.6.41(0.8 kips) φ Vn = 16.2 9.2R-21 Example B2(c).8 kips φ = 0. The larger of the moment on the tension side or the compression side will control the design of the base plate.85 φVn = 0.0 kips > 12. use 3/4 in. . Note that the plate thickness is calculated using the nominal strength of the anchors. It is not necessary to include a φ factor in this calculation of plate thickness. A φ factor of 0.74 in.9 is used in calculating the required area of the anchor in Step 1. continued CODE SECTION DESIGN PROCEDURE STEP 3: continued Nominal moment capacity of base plate CALCULATION Mn = F y S Mn = F y ( bt 2 /6) = (36)(7)t 2 /6 Mn = 42 t 2 42 t 2 = 23.4 t = 0. STEP 4: Embedment length The calculation of the required embedment length is similar to that in Example B2(a).349.2R-22 MANUAL OF CONCRETE PRACTICE Example B2(c). plate Determine minimum base plate thickness to prevent yielding of plate. 000 psi (plate) Mu = 70 in.4 kips Pu = 11.85 STEP 3 and STEP 4 See Example B2(a) A e = 0.466) × (7)[6 – 2.46/2 = 0.85(4)(2.2 A sv = 2(0.85(33. × 7 in. Given: f c = 4000 psi ′ f y = 50.2. Pu .466)] Solving the quadratic equation: Steel / Stud Provide 5/8 in.78 = 50 A e (2. and Vu are the required factored external loads as defined in Section 9.0 = 0.85 f c′ See Example B2(a). the attachment plate is assumed rigid.46 in.85 f c′ ab ( d – 2.1 Sum external (required strength) and internal (design strength) forces.EMBEDMENT DESIGN EXAMPLES 349. Sum moments about the center line of base plate (line of axial load).5 – 0.3.74)(0.466 70. and uniaxial moment Design an embedment using welded studs and a rigid embedded plate for a 3 × 3 × 1/4 in. plate with 5 in.3) = 28.1 kips where Mu .2 9.23) + 2(0. CODE SECTION 7′′ V u = 12.74 in.30) = 0.9) = 33.2R-23 Example B3(a)—Four-stud rigid embedded plate.3 Shear capacity φ = 0.1 A e – 0.1A e – 0. C a 0.2 9.85(4)(a )(7) = 11.4 kips.2 0.000 psi (studs) f ut = 60.000 psi Fy = 36.9[50A e (2.-kips Vu = 12.2 A e = 0.1 a = 2.5 –a /2)] 77.30 – 0. diameter studs STEP 2: Design for shear Assume the total stud area not used for moment and tension is available for shear transfer by shear friction.-kips 3 × 3 × 1/4 ′′ Tube 5′′ DESIGN PROCEDURE STEP 1: Design for moment and tension Try a 7 in.9 9.3 kips > 12.7 Assume a uniform stress block for concrete compression zone and the top two studs as the tension components. Nominal shear capacity µ = 0. combined tension.5) + 0.2 of the Code. A 501 structural tube attachment.9 for flexure M u = 70 50A e – 0.5(2. OK . stud spacing CALCULATION T = Aefy P u = 11. φ = 0.23 in. 9. shear.2 V n = 50(0.4 kips P u = 11.1 A e – 0.5) + 0.30 in.3 kips φ V n = 0. × 5 in.1 7′′ 5′′ M u = 70 in.1 6′′ 10. and Vu are the required factored external loads as defined in Section 9.2 V n = 0.1 7′′ 5′′ M u = 70 in.1(1.2.2 of the Code.3 B. 7′′ V u = 12. and uniaxial moment Design an embedment using welded studs and a flexible embedded plate for a 3 × 3 × 1/4 in. it is assumed that the compressive reaction is located at the edge of the attached member.5.4) = 27. shear.000 psi (studs) f ut = 60.000 psi Fy = 36.85 φ V n = 0.24 in.72 in.5)/4 φ T = 21.4 kips Pu = 11.2 A e provided = 0. Lever arm for moment = 4 in.6.000 psi (plate) Mu = 70 in.1 6′′ C 0.24) + 2 (0.-kips 3 × 3 × 1/4 ′′ Tube 5′′ CODE SECTION DESIGN PROCEDURE STEP 1: Tension in top studs Try a 7 in.72)(50) = 32.4 kips P u = 11. combined tension. A 501 structural tube attachment.9 Tension in each stud φ T = 70/4 + 11. φ = 0.2 9.4 kips.0 kips A e .0 kips 24. × 5 in.30 – 0.85 f c′ a M u = 70 For simplicity. STEP 3 and STEP 4 See Example B2(b) A e = 12.6.6. diameter stud STEP 2: Design for shear B. Given: f c = 4000 psi ′ f y = 50.2 A v f = 2 (0.85(32.30) = 0.5 kips > 12.349.9 Capacity reduction factor for shear Design shear strength must be equal to or greater than the required strength.0 / 2 = 12.3 9.40 kips φ = 0. OK .-kips Vu = 12.2 B.7 kips T = 24. × 7 in.2R-24 MANUAL OF CONCRETE PRACTICE Example B3(b)—Four-stud flexible embedded plate.1 kips where Mu .2.0 / 50 = 0. area for each stud Try 5/8 in.5. Nominal shear strength µ = 0.9(0..3 The stud area not used for moment and tension is available for shear transfer by shear friction.30 in. plate with 5 in. stud spacing CALCULATION T = Aefy P u = 11. Pu .2 9. all sides of the rectangular stress reduction area are equal and.8 sq.4 sq.2 Total stress reduction area = (b + 2 L d + 2 t – 2 h ) 2 = [ 6.625) – 2(6. therefore.EMBEDMENT DESIGN EXAMPLES 349.8 in. OK Ductility requirements met. only one side needs to be found. .6 in.4. rather than the length of 4. P u = 18 kips Given: fc = ′ fy = f ut = 4000 psi 50. Due to biaxial symmetry of the assembly in Example B1. in.71′′ 3. in.33 in. tension only Determine the reduction of projected stress area due to limited concrete thickness for the embedment of Example B1. COMMENT: Example B1 describes an embedment assembly with a projected area of 175.000 psi h = 6′′ 5/ ′′ 8 thick plate Thickness of concrete slab = 6 in. to the face of concrete.86 in.71) + 2 (0.2 < 14. conservatively neglects the thickness of the plate and uses L d = 3.75 + 2 ( 3. available for reduction in projected area before the minimum requirements for concrete strength are no longer met.2R-25 Example B4—Four-stud rigid embedded plate in thin slab.5)] 2 = 5. L d = 3.2 area still available for stress reduction.000 psi (studs) 60. when there was a minimum required area of 160.75′′ CODE SECTION DESIGN PROCEDURE STEP 1: Determine area available for stress reduction CALCULATION B. Note that the projected area calculated in Example B1.71 in. this leaves approximately 14.2.42′′ b + 2 (L d + t – h ) b = 6. the angle α of the common segment is given by: cos(1/2 α) = S /(2 R ) α = 2cos –1 [ S /(2 R )] The area of the common segment A seg equals the area of the two sectors minus the area of the triangles: R α A seg = 2 [α R 2 /2 – R 2 sin(α/2) cos(α/2)] = (α – sinα) R 2 S /2 CASE I The projected area is equal to the area of four full cones minus the area of the four overlapping portions minus the area of the four heads.349. The studs are located at the corners of a rectangle with spacing S x and S y in each direction. The radius of the projected stress cones (45 deg cone angle) is R = L d + d h / 2. CASE I A cp2 = S x S y + (4)( 3/4 )π R 2 – (α x – sinα x) R 2 – (α y – sin α y ) R 2 – 4 A h SUMMARY R Sx A cp1 = (4π – 2 B ) R 2 – 4 A h A cp2 = (3π – B ) R 2 + S x S y – 4 A h where B = (α x – sin α x ) + (α y – sin α y ) αx = 2cos –1 (S x Sy / 2R) α y = 2cos –1 (S y /2 R ) CASE II . R A cp1 = 4πR 2 – 2(αx – sinαx )R 2 – 2(α y – sin α y)R 2 – 4 A h Sx Sy CASE II The projected area is equal to the area of the central rectangle plus the area of the four three quarter cones minus the area of the four overlapping portions outside the rectangle minus the area of the four heads.2R-26 MANUAL OF CONCRETE PRACTICE APPENDIX A—Projected area (Acp ) for four studs This appendix develops the projected area of four stress cones at the surface of the concrete. There are two cases of overlapping stress cones. For two overlapping stress cones of radius R and spacing S. In Case I. In Case II. all four stress cones overlap. The projected area for the two cases is formulated below. there is no overlap at the center of the rectangle since R < 1 / 2 ( S x 2 + S y 2 ) 1/2 .