Example 1 - Single Headed Anchor in Tension Away from Edges BORRADOR. Calculations and Discussion

Transcription

1 Example 1 - Single Headed Anchor in Tension Away from Edges Check the capacity of a single anchor, 1 in. diameter, F1554 Grade 36 headed bolt with heavy-hex head installed in the top of a foundation without edge effects to resist a factored load of 17,000 lb tension (determined from ACI ). The foundation is located in an area of high seismic risk. Assume normal weight concrete and that a crack forms in the plane of the anchor. ' f c = 6,000 psi 17,000 lb h ef = 8 in. ote: Foundation reinforcement not shown for clarity 1. The factored design load is: ua = 17,000 lb 2. General requirement for anchor strength φ n ua where φ n is the lowest design strength from all appropriate failure modes in tension as determined from consideration of φ sa, φ pa, φ sb, and φ cb. In regions of moderate or high seismic risk, the design strength of anchors is limited to 0.75 φ n φ n ua 3. Check strength of steel anchor and requirements for ductility Since the anchor is located in a region of moderate or high seismic risk, the anchor strength shall be governed by tensile or shear strength of a ductile steel element, ACI 318, D unless the other parts of the attachment are designed to yield (D.3.3.5). ASTM F1554 Grade 36 bolt material meets the requirements of the ductile steel element definition in D.1 [tensile elongation of at least 14 percent and reduction of area of at least 30 percent]. F1554 material has 23 % elongation in 2 in. of length with 40 % reduction in area (See Table A). 8 in. D Eq. (D-1) D D D The basic design strength for the steel anchor in regions of moderate or high seismic risk is: 0.75 φ sa ua and φ sa < capacity of concrete breakout, pullout, side-face blowout, or splitting failure in tension to satisfy ductility requirements, where: 1-1

2 φ = 0.75 and the steel bolt strength is: sa = n A se (f uta ) where the minimum value for f uta = 58,000 psi is given for F1554, Grade 36 material. See Table A of this design guide. ote, f uta shall not be taken greater than the smaller of 1.9 f ya and 125,000 psi (See D.5.1.2). n = 1 (single anchor) D.4.4(a) Eq. (D-3) ASI / ASME B1.1 defines A se : A se π = 4 d o nt 2 substituting 1 in. for d o and 8 (number of threads) for n t, A se = in. 2 sa = (1) (0.606 in. 2 ) (58,000 psi) = 35,148 lb φ sa = (0.75) 35,148 lb = 26,361 lb Reduction for regions of moderate or high seismic risk 0.75 (φ sa ) = 0.75 (26,361 lb) = 19,771 lb 19,771 lbs > 17,000 lbs (requirement of 0.75φ sa > ua is met) 4. Using the embedment depth, (h ef = 8 in.), check the various capacities of the concrete The basic capacity for nominal concrete breakout strength in tension for a single anchor in regions of moderate or high seismic risk is: 0.75 φ cb ua and φ cb > φ sa (to ensure that the steel failure mode governs) where: φ = 0.70 (for concrete breakout), φ = 0.75 (for ductile steel in tension) RD D D.4.4 Determine the concrete breakout strength assuming condition B (no supplementary reinforcement has been provided) A c cb = ψ ed, ψ c, ψ cp, Aco b Eq. (D-4) 1-2

3 where: Ac the = 1.0 for single anchor remote from edges (a complete breakout prism Aco is developed and no reduction is required) RD D ψ ed, = 1.0 since anchor is remote from edges, (c a,min 1.5 h ef ) ψ c, = 1.0 since for this example it has been assumed cracking occurs in the concrete at service load levels ψ cp, = 1.0 (for cast-in anchors) The basic concrete breakout strength of a single anchor in tension is: = k b c f ' c h 1.5 ef where k c = 24 for cast-in anchors For this example h ef = 8 in. which is less than 11 in. so equation D-8, ' 5 / 3 ( = f h ), does not apply b 16 c ef Substituting: 1.5 b = 24 6,000 (8) = 42,065 lb and φ cb = 0.7(42,065 lb) = 29,445 lb Reduction for regions of moderate or high seismic risk 0.75 φ cb = 0.75 (29,445 lb) = 22,084 lb 22,084 lb ua and 22,084 lb > 0.75φ sa (or 19,771 lb) D D (D-7) D therefore the anchor strength remains governed by ductile steel failure 5. Check the anchor bolt pullout strength To prevent a pullout failure mode, the requirement for pullout strength checks the bearing stress under bolt head. This failure mode is initiated by crushing of the concrete: 0.75 φ pn ua and φ pn > φ sa (to ensure ductile steel failure mode governs) where: φ = 0.70 D.4.4(c)(ii) 1-3

4 pn = ψ c, P p Eq. (D-14) where: =8 A f p brg c ψ c,p = 1.0 considering concrete cracking at service loads Eq. (D-15) D For 1 in. diameter heavy hex bolt head the bearing area of the head, A brg = in. 2 (See Table B1 of this design guide) Substituting: 2 p = 8(1.501in. )(6,000 psi) = 72, 048lb and φ pn = 0.7 (72,048 lb) = 50,433 lb Reduction for regions of moderate or high seismic risk 0.75 φ pn = 0.75 (50,433 lb) = 37,825 lb 37,825 lb ua and 37,825 lb > 0.75φ sa (or 19,771 lb) therefore the anchor strength remains governed by ductile steel failure 6. Evaluate concrete side-face blowout failure mode Since this anchor is remote from a free edge of concrete (c a1 0.4 h ef ), concrete side-face blowout failure mode is not applicable. 7. Evaluate splitting failure: This type of failure can occur in thin slabs where the anchors have torque applied. The minimum edge distance for cast-in anchors with torque applied is 6 (d o ) where d o is the bolt diameter. This failure mode is not applicable since the anchor is remote from an edge. 8. Summary: D D.5.4 D.8.2 Steel 0.75φ sa 19,771 lb (Controls) Concrete Breakout 0.75 φ cb 22,084 lb Concrete Pullout 0.75 φ pn 37,825 lb Side-Face Blowout A The lowest design strength for considering all failure modes 19,771 lb for steel failure (controlled by the seismic reduction design capacity equation 0.75φ sa ); therefore the ASTM F1554 Grade 36, 1-in. diameter anchor bolt with a heavyhex head and an 8-in. embedment is adequate to resist the 17,000 lb tension factored load in a moderate or high seismic region. 1-4

5 1-5

6 Example 2 Single Hooked Anchor in Tension Away from Edges Check the capacity of a single anchor, 1 in. diameter, F1554 Grade 36 standard hooked bolt (L-bolt) installed in the top of a foundation without edge effects to resist a factored load of 17,000 lb tension (determined from ACI ). The foundation is located in an area of high seismic risk. Assume normal weight concrete and that a crack forms in the plane of the anchor. ' f c = 6,000 psi 17,000 lb ote: Foundation reinforcement not shown for clarity 1. The factored design load is: ua = 17,000 lb 2. General requirement for anchor strength: φ n ua e h = 3 in. where φ n is the lowest design strength from all appropriate failure modes in tension as determined from consideration of φ sa, φ pa, φ sb, and φ cb. In regions of moderate or high seismic risk, the design strength of anchors is limited to 0.75 φ n. 3. Check strength of steel anchor and requirements for ductility Since the anchor is located in a region of moderate or high seismic risk, the anchor strength shall be governed by tensile or shear strength of a ductile steel element, ACI 318 D ASTM F1554 Grade 36 material meets the requirements of the ductile steel element definition in D.1 [tensile elongation of at least 14 percent and reduction of area of at least 30 percent]. F1554 material has 23 % elongation in 2 in. of length with 40% reduction in area (See Table A). h ef = 8 in. 8 in. D Eq. (D-1) D D D The basic design strength for the steel anchor in regions of moderate or high seismic risk is: 0.75 φ sa ua and φ sa < the capacity of concrete breakout, pullout, side-face blowout, or splitting failure in tension to satisfy ductility requirements, where: φ = 0.75 D.4.4(a) and the steel bolt strength is: 2-1

7 sa = n A se (f uta ) Eq. (D-3) where the minimum value for f uta = 58,000 psi is given for F1554 Grade 36 material. See Table A of this design guide. ote, f uta shall not be taken greater than the smaller of 1.9 f ya and 125,000 psi (See D.5.1.2). n = 1 (single anchor) ASI / ASME B1.1 defines A se : A se π = 4 d o nt 2. Substituting 1 in. for d o and 8 (number of threads) for n t A se = in. 2 sa = (1) (0.606 in. 2 ) (58,000 psi) = 35,148 lb φ sa = (0.75)35,148 lb = 26,361 lb Reduction for regions of moderate or high seismic risk 0.75 (φ sa ) = 0.75 (26,361 lb) = 19,771 lb 19,771 lbs > 17,000 lbs (requirement of 0.75φ sa ua is met) 4. Using the embedment depth (h ef = 8 in.) check the various capacities of the concrete The basic capacity for nominal concrete breakout strength in tension for a single anchor in regions of moderate or high seismic risk is: 0.75 φ cb ua and φ cb > φ sa (to ensure that the steel failure mode governs) where: φ = 0.70 (for concrete breakout), φ = 0.75 (for ductile steel in tension) RD D D.4.4 Determine the concrete breakout strength assuming condition B (no supplementary reinforcement has been provided) A c cb = ψ ed, ψ c, ψ cp, Aco b Eq. (D-4) where: Ac the = 1.0 for single anchor remote from edges (a complete breakout prism is Aco developed and no reduction is required) RD

8 ψ ed, = 1.0 since anchor is remote from edges, (c a,min 1.5 h ef ) D ψ c, = 1.0 since for this example it has been assumed cracking occurs in the concrete at service load levels ψ cp, = 1.0 (for cast-in anchors) D D The basic concrete breakout strength of a single anchor in tension is: = k b c f ' c h 1.5 ef where k c = 24 for cast-in anchors For this example h ef = 8 in. which is less than 11 ' 5 / 3 in. so equation D-8, ( = f h ), does not apply Substituting: b 16 c ef 1.5 b = 24 6,000 (8) = 42,065 lb and φ cb = 0.7 (42,065 lb)= 29,445 lb Reduction for regions of moderate or high seismic risk 0.75 (φ cb ) = 0.75 (29,445 lb) = 22,084 lb 22,084 lb ua and 22,084 lb > 0.75φ sa (or 19,771 lb) therefore the anchor strength at this point remains governed by ductile steel failure. 5. Check the anchor bolt pullout strength To prevent a pullout failure mode, the requirement for pullout strength of the hooked bolt checks the bearing stress inside the hook. This failure mode is initiated by crushing of the concrete: Eq. (D-7) D D φ pn ua and φ pn > φ sa (to insure ductile steel failure mode governs) where: φ = 0.70 pn = ψ c, P p D.4.4(c)(ii) Eq. (D-14) where: p = 0.9 f ' c e h d o Eq. (D-16) 2-3

9 ψ c,p = 1.0 considering concrete cracking at service loads. e h = 3 in., d o = 1 in. and the hook length falls within the range 3d e 4.5 d Substituting: o h o D D ( 6000 psi)( 3" )( 1" ) 16, lb p = 0.9 = 200 and φ pn = 0.7 (16,200 lb) = 11,340 lb Reduction for regions of moderate or high seismic risk 0.75 (11,340 lb) = 8,505 lb 0.75 φ pn < ua and φ pn < φ sa strength requirements are not met. Anchor pullout strength is less than the factored load and less than the steel capacity due to pullout capacity of the hook. Extending the hook will increase the capacity but the hook length is limited to 4.5 times anchor diameter or 4 ½ in. (D.5.3.5) Try extending hook to the maximum allowed amount. Substituting 4.5 in. for e h, recalculate p and φ pn : = psi 4.5" 1" = 24,300 lb p ( )( )( ) and φ pn = 0.7 (24,300 lb) = 17,010 lb Reduction for regions of moderate or high seismic risk 0.75 (φ pn ) = 0.75 (17,010 lb) = 12,757 lb Anchor pullout strength with seismic reduction is less than factored load, 17,000 lb, and less than the steel capacity, 19,771 lb. 6. Evaluate concrete side-face blowout failure mode Since this anchor is located far from a free edge of concrete (c a1 0.4 h ef ), concrete side-face blowout failure mode is not applicable, however, Section D.5.4 for sideface blowout applies only to headed anchors. 7. Evaluate splitting failure: This type of failure can occur in thin slabs where the anchors have torque applied. The minimum edge distance for cast-in anchors with torque applied is 6 (d o ) where d o is the bolt diameter. This failure mode is not applicable since the anchor is remote from an edge. D D D.5.4 D

10 8. Summary: Steel 0.75φ sa 19,771 lb Concrete Breakout 0.75 φ cb 22,084 lb Concrete Pullout 0.75 φ pn 12,757 lb Controls Side-Face Blowout A The ASTM F1554 Grade 36, 1 in. diameter hooked bolt with an 8 in. embedment using the maximum allowed effective hook length of 4 ½ in. with the 0.75 seismic reduction is inadequate to resist the 17,000 lb tension factored load and cannot be used in a moderate to high seismic region since the ductile steel failure mode does not govern the capacity. In a non-seismic or low seismic region the hooked bolt with 4 ½ in. hook is adequate to resist the factor load of 17,000 lbs. with the pullout strength (17,010 lb) being the governing mode of failure. 2-5

11 Example 3 Single Post-Installed Anchor in Tension Away from Edges Determine the minimum diameter post-installed torque-controlled expansion anchor for installation in the bottom of an 8-in. slab with a concrete compressive strength of f c = 4,000 psi to support a 3,000 lb service dead load. The anchor will be in the tension zone (cracking at service load level is assumed), away from edges and other anchors in normal weight concrete. See Table C of this document for sample anchor installation and performance data. The data in Table C is not from any specific anchor and should not be used for design in accordance with ACI , Appendix D. However it is similar to what would be expected from testing and an evaluation report prepared by an independent testing and evaluation agency for the manufacturer in accordance with ACI Example 3 1. Determine factored design load: ua = 1.4 D ua = 1.4 (3,000) = 4,200 lb 2. Determine fastener material: 3,000 lb 8-in. Select a qualified post-installed torque-controlled expansion anchor with a bolt conforming to the requirements of ASTM F 1554 Grade 55. Design information resulting from anchor prequalification testing according to ACI for the selected torque-controlled expansion anchor is given in Table C of this document. h ef ACI , Table 4.2 For ASTM F 1554 Grade 55: f uta = 75,000 psi f ya = 55,000 psi Elongation at 2 in. = 21 % min. Reduction of area = 30 % min. ACI 318 Appendix D requires 14 % min. elongation and 30 % min. reduction of area to qualify as a ductile steel anchor. D.1 3-1

12 Example 3 (cont d) 3. Steel strength requirement under tension loading: φ n ua = 4,200 lb where φ n is the lowest of φ sa, φ pn, φ sb, and φ cb sa = na se f uta f uta the smaller of 1.9f ya and 125,000 psi. φ na se f uta ua = 4,200 lb For ductile steel, φ = 0.75 n = 1 (single anchor) Substituting and solving for A se : 4,200 Ase φ futa A 4, 200 se , 000 A se in. 2 (minimum effective cross-sectional area of anchor) Bolt diameter 3/8-in. (A se = in. 2 ) 4. Concrete breakout strength requirement for tension loading: φ cb ua = 4,200 lb Ac cb = ψ ed ψ c, Aco where: ' 1.5 = k f h, ψ cp, b c substituting: c ef Ac 1.5 φ ψ ed, ψ c, ψ cp, kc f c hef = 4, 200 lb A co where: Ac = 1 for a single anchor A co b k c = 17 for a post-installed anchor where cracking is expected ψ ed, = 1 for no edge effects ψ cp, = 1 assuming cracking at service loads for a category 1 anchor, φ = 0.65 (condition B, no supplemental reinforcement) for a category 2 anchor, φ = 0.55 (condition B, no supplemental reinforcement) D.5.1 D and D Eq. (D-3) D D.4.4 D.5.2 Eq. (D-1) Eq. (D-4) Eq. (D-7) D D D.4.4(c)ii D.4.4(c)ii 3-2

13 Example 3 (cont d) [ote that an anchor category is established independently for each anchor diameter from the reliability tests of ACI Smaller diameter anchors are, in general, more sensitive to the conditions established for reliability tests, including reduced torque, variations in hole diameter and cracks in the concrete.] Assuming a category 1 anchor, solve for the required embedment, h ef, corresponding to the desired concrete breakout strength: 4,200 h ef = 3.3 in , If we instead assume a category 2 anchor, the required embedment becomes: h ef 4, ,000 for a category 2 anchor for breakout strength. 2 3 = 3.7 in. Substituting the data from anchor categories and the embedment depths from Table C into the combined expression: Ac ' 1.5 φ cb = φ ψ ed, ψ c, nψ cp, kc fc h ef yields the following table; A co Acceptable anchor diameters and embedment depths for concrete breakout strength in tension (for this example) are: Diameter (in.) Suitable embedments (in.) Category Minimum h ef (in.) 3/ ½ /8 4.5, /4 3.5, 5, Pullout strength: For static loading, pullout strength is established by reference tests in cracks and by the crack movement reliability test of ACI For postinstalled anchors, data from the anchor prequalification testing must be used. φ pn ua = 4,200 lb = ψ pn c, p p where: ψ c.p = 1 φ = 0.65 for a Category 1 and Condition B φ = 0.55 for a Category 2 and Condition B D.5.3 Eq. (D-1) Eq. (D-14) D D.4.4(c)ii D.4.4(c)ii From the evaluation report, the following 5 percent fractile capacities were established for the embedments that satisfy concrete breakout strength using the above equation: 3-3

14 Example 3 (cont d) diameter (in.) embedment (in.) anchor pullout capacity (lb) 3/8 4.5 φ pn = ,583 = 3,070 < 4,200 lb.g. 1/2 5.5 φ pn = ,544 = 4,149 < 4,200 lb.g. 5/8 4.5 φ pn = ,211 = 5,337 > 4,200 lb OK 5/8 6.5 φ pn = ,254 = 9,265 > 4,200 lb OK 3/4 3.5 φ pn = ,632 = 3,661 < 4,200 lb.g. 3/4 5 φ pn = ,617 = 6,251 > 4,200 lb OK 3/4 8 φ pn = ,463 = 12,651 > 4,200 lb OK Try a 5/8 in. diameter anchor with h ef = 4.5 in. from the selected anchor system. 6. Check all failure modes: For a post-installed torque-controlled expansion anchor with a 5/8 in. diameter and embedment depth of 4.5 in.: a) For steel strength: φ sa = φ na se f uta ua = 4,200 lb ,000 = 12,712 lb > 4,200 lb OK b) For concrete breakout strength: Ac ' 1.5 φ cb = φ ψ ed, ψ cp, kc fc h ef ua = 4,200 lb A co φ cb = c) For pullout strength: , = 6,671 lb > 4,200 lb OK φ pn ua = 4,200 lb (value obtained from Table C, manufacturer s evaluation report) ,211 = 5,337 lb > 4,200 lb OK 7. Check for minimum concrete thickness: In order to prevent splitting, the thickness of the concrete in which the anchor is embedded must be at least 1.5 h ef. D.5.1 D.5.2 D.5.3 D = 6-3/4 in. < 8 in. OK 8. Summary: Strength mode Strength (lb) Steel strength φ sa 12,712 Concrete breakout strength φ cb 6,671 Concrete pullout strength φ pn 5,337 Therefore pullout strength controls. 3-4

15 Example 3 (cont d) 9. Final Recommendation: Use a post-installed torque-controlled expansion anchor with a 5/8-in. diameter and an embedment depth of 4.5-in. meeting the requirements of ASTM F 1554, Grade

16 Example 4 - Group of Headed Studs in Tension ear an Edge Design a group of four welded, AWS D1.1 Type B welded headed studs spaced 6-in. on center each way and concentrically loaded with a 10,000 lb service dead load. The anchor group is to be installed in the bottom of an 8-in. thick normal weight concrete slab made with the centerline of the connection 6-in. from a free edge of the slab. ' f c = 4,000 psi 8 in. h ef 6 in. 6 in. ote: Reinf. not shown for clarity 1. Determine factored design load ua = 1.4 (10,000) = 14,000 lb 2. Determine anchor diameter Using AWS D1.1 Type B welded, headed studs. Assume tension steel failure controls. The basic requirement for the anchor steel is: φ sa ua where: φ = 0.75 ote: Per the Ductile Steel Element definition in D.1, AWS D1.1 Type B studs qualify as a ductile steel element (20% minimum elongation in 2-in. which is greater than the 14% required and a minimum reduction in area of 50% that is greater than the 30% required. See Table A for this information. sa = n A se f uta 10,000 lb ½ in. plate 6 in. 9.2 D.5.1 Eq. (D-1) D D.4.4(a)i Eq. (D-3) For design purposes, Eq. (D-1) with Eq. (D-3) may be rearranged as: A se φnf ua uta 4-1

17 Example 4 (cont d) where: ua = 14,000 lb φ = 0.75 n = 4 f uta = 65,000 psi ote: Per Section D5.1.2, f uta shall not be taken greater than 1.9 f ya or 125,000 psi. For AWS D1.1 headed studs, 1.9 f ya = 1.9 (51,000 psi) = 96,900 psi, therefore use the specified minimum f uta of 65,000 psi. Substituting: 14,000 Ase = in ( 4)( 65,000) Per Table B1, ½-in. diameter welded, headed studs will satisfy this requirement (A se = in. 2 ). ote: Per AWS D1.1 Table 7.1, Type B welded studs applies to studs in ½- in., 5/8-in., ¾-in., 7/8-in., and 1 in. diameters. Although individual manufacturers may list smaller diameters they are not explicitly covered by AWS D1.1 The total design steel strength of four ½-in. headed studs: φ sa ( 4)( 65,000)( 0.196) 38, 220 = 0.75 = 3. Determine the required embedment length (h ef ) based on concrete breakout in tension: Two different equations are given for calculating concrete breakout strength; for single anchors Eq. (D-4) applies, and for anchor groups Eq. (D-5) applies. An anchor group is defined as: lb. a number of anchors of approximately equal effective embedment depth with each anchor spaced at less than three times its embedment depth from one or more adjacent anchors. D D.5.2 D.1 Since the spacing between anchors is 6-in., the anchors must be treated as a group if the embedment depth exceeds 2-in.. Although the embedment depth is unknown, at this point it will be assumed that the provisions for an anchor group will apply. The basic requirement for embedment of a group of anchors is: φ cbg ua where: Eq. (D-1) D

18 Example 4 (cont d) φ = 0.70, for anchors governed by concrete breakout and Condition B applies since no supplementary reinforcement has been provided (e.g., hairpin type reinforcement that ties the failure prism into the structural member). = A A ψ ψ ψ ψ c cbg ec, ed, c, cp, b co D.4.4(c)ii RD.4.4 Eq. (D-5) Since this connection is likely to be affected by both group effects and edge effects, the embedment length h ef is difficult to solve for directly. In this case, an embedment length must be assumed at the outset and then be proven to satisfy the requirement of Eq. (D-1). ote: Welded studs are generally available in fixed lengths. Available lengths may be determined from manufacturers' catalogs. For this example, one manufacturers product line indicates a length of 4-in. (after welding) for a standard ½- in. headed stud. The effective embedment depth, h ef = 4-in in. = 4.5-in. ote: The effective embedment length, h ef, for the welded stud anchor is the shank length of the stud (4-in.) plus the thickness of the embedded plate (0.5-in.) as shown in the following figure. 0.5 in. Evaluate the terms in Eq. (D-5) with h ef = 4.5 in. Determine A c and A co for the anchor group: A c is the projected area of the failure surface as approximated by a rectangle with edges bounded by 1.5h ef (1.5 x 4.5 = 6.75 in. in this case) and free edges of the concrete from the centerlines of the anchors. A c shall not be taken greater than n A co where n = number of anchors in a group. 4 in. h ef = 4.5 in. D " 6" A c 6.75" 3 " 6 " 6.75" 4-3

19 Example 4 (cont d) A c = ( ) ( ) = 307 in. 2 A co = 9 h 2 ef = 9 (4.5) 2 = 182 in. 2 edge effects) (projected area of failure for one anchor without Eq. (D-6) Check: A c na co 307 in. 2 < 4(182 in. 2 ) = 728 in. 2 - OK Determine ψ ec, : D D ψ ec, = 1.0 (no eccentricity in the connection) Determine ψ ed, since c a1 = c a,min < 1.5h ef (note: c a1 = 3 in.) ca,min ψ ed, = hef 3.0 ψ ed, = = (4.5) Determine ψ c, : ψ c, = 1.0 for locations where concrete cracking is likely to occur (i.e., the bottom of the slab) and cracking is controlled by flexural or confining reinforcement. Determine ψ cp, : ote, this factor only applies to post-installed anchors. Therefore: ψ cp, = 1.0 for cast-in-place anchors. Determine b : = k b c f ' c h 1.5 ef where k c = 24 for cast-in anchors D Eq. (D-11) D D D Eq. (D-7) b ' 1.5 = kc f c hef = 24 4, ( 4.5) = 14, 490 lb Substituting into Eq. (D-5): cbg 307 = (1.0)(0.83)(1.0)(14,490) = 20, lb Check if: φ cbg ua Eq. (D-1) 4-4

20 Example 4 (cont d) Substituting: φ cbg =(0.70) (20,287) = 14,201 lb 14,201 lb > 14,000 lb OK Specify a 4-in. length for the welded, headed studs with the ½- in. thick base plate. 4. Determine if welded stud head size is adequate for pullout φ pn ua where: φ = 0.70 ote: Condition B applies in all cases when pullout strength governs. = ψ pn c, P p where: = p ' 8Abrg f c ψ c,p = 1.0 for locations where concrete cracking is likely to occur (i.e., the bottom of the slab) and cracking is controlled by flexural or confining reinforcement. For design purposes Eq. (D-1) with Eq. (D-14) and Eq. (D-15) may be rearranged as: A brg ua φψ 8 f ' c, P c For the group of four studs the individual factored tension load ua on each stud is: D.5.3 Eq. (D-1) D D.4.4(c)ii Eq. (D-14) Eq. (D-15) D ua 14,000 = = 3,500 lb 4 Substituting: 3,500 Abrg ( required ) = in (1.0) (8) (4,000) 4-5

21 Example 4 (cont d) The bearing area of welded, headed studs should be determined from manufacturers' catalogs. Figure 7.1 from AWS D1.1 lists the diameter of the head for a ½-in. diameter stud is 1 in. AWS states that alternate head configurations may be used with proof of full-strength development of design. Procedures in ACI 318 Appendix D can be used to confirm alternate head configurations. Abrg = π ( 1.0) ( 0.5) ) ( provided ) = in 2 > in 2 OK The total design pullout strength of four ½-in. headed studs: φn pn ( 4)( 8)( 1.0)( 0.589) (4,000) 52, 774 = 0.7 = 5. Evaluate side-face blowout Side-face blowout needs to be considered when the edge distance from the centerline of an anchor to the nearest free edge is less than 0.4 h ef. For this example: 0.4 h ef = 0.4 (4.5) = 1.8 in. < 3 in. actual edge distance The side-face blowout failure mode is not applicable. 6. Required edge distances, spacing, and thickness to preclude splitting failure Since a welded, headed anchor is not torqued the minimum cover requirements of ACI 318, Section 7.7 apply. Per Section 7.7, the minimum clear cover for a ½-in. bar not exposed to earth or weather is ¾-in. which is less than the 2 ¾-in. cover provided - OK 7. Summary of connection strength: Steel Concrete breakout Pullout Side-face blowout φ sa 38,220 lb. φ cbg 14,201 lb. Controls φ n pn 52,774 lb. φ sb ot Applicable lb D.5.4 D (c) Use four ½-in. diameter welded studs meeting AWS D1.1 Type B with an effective embedment of 4 ½- in. 4-6

22 Example 5 Single Headed Bolt in Shear ear an Edge Determine the reversible service wind load shear capacity for a single ½ in. diameter cast in hex headed bolt meeting ASTM F1554 Grade 36. The headed bolt is installed in a normal weight continuous concrete foundation with a 7 in. embedment and a 2¾ in. edge distance. o supplemental reinforcing is present. ote: This is the minimum anchorage requirement at the foundation required by IBC 2003 Section for conventional light frame wood construction. The 2¾ in edge distance represents a typical connection at the base of framed walls using wood 26 sill members. V service ' f c 4,000 psi 18 in 7 in ote: Foundation reinforcement not shown for clarity 1. This problem provides the anchor diameter, embedment depth, and anchor material type. The designer is required to compute the maximum reversible service shear load due to wind. In this case, it is best to first determine the controlling shear design strength,v n, based on the smaller of the steel strength and/or concrete strength. Step 6 of this example provides the conversion of the controlling factored shear load, V ua, to a service load due to wind using factored loads of Section Determine V ua as controlled by the steel strength of the anchor in shear V sa V ua where: = 0.65 Per the Ductile Steel Element definition in D.1, ASTM F1554 Grade 36 steel qualifies as a ductile steel element(see Table A). V sa = n 0.6 A se f uta 2.75 in D.6.1 Eq. (D 2) D D.4.4 Eq. (D 20) To determine V ua for the steel strength, Eq. (D 2) can be combined with Eq. (D 20) to give: V V n 0.6 A f ua sa se uta where: = 0.65 n = 1 A se = in. 2 for a 1/2 in. threaded bolt (See Table B1) f uta = 58,000 psi 5 1

23 Example 5 (cont d) Per ASTM F1554, Grade 36 has a specified minimum yield strength (f ya )of 36 ksi and a specified tensile strength (f uta )of 58 ksi (see Table A). For design purposes, the minimum tensile strength of 58 ksi should be used. ote: Per Section D.6.1.2, f uta shall not be taken greater than 1.9 f ya or 125,000 psi. For ASTM F1554 Grade 36, 1.9 f ya = 1.9 (36,000) = 68,400 psi, therefore use the specified minimum f uta of 58,000 psi. D Substituting, V ua as controlled by steel strength is: ,000 3, 212 V V lb ua sa 3. Determine V ua governed by concrete breakout strength with shear directed toward a free edge. V cb V ua. where: = 0.70 Condition B no supplementary reinforcement has been provided. V A V Vc cb ed, V c, V b AVco where: AVc 3(2.75)(1.5)(2.75) A 4.5(2.75) Vco ed,v = 1.0 In this instance, the member thickness is greater than 1.5 c a1, and the distance to an orthogonal edge, c a2, is greater than 1.5 c a1. c,v = 1.0 for locations where concrete cracking is likely to occur (i.e., the edge of the foundation is susceptible to cracks) and no supplemental reinforcement is provided or edge reinforcement is smaller than a o. 4 bar. D.6.2 Eq. (D 2) D D.4.4(c)i Eq. (D 21) D D D V b 0.2 c 1. 5 e 7 do f' c a1 d0 Eq. (D 24) where: e = load bearing length of the anchor for shear, not to exceed 8d 0 e = 8 d o = 8 (0.5) = 4.0 in. < 7.0 in For this problem 8d o will control since the embedment depth h ef is 7 in. D To determine V ua for the given embedment depth governed by concrete breakout strength, Eq. (D 2) can be combined with Eq. (D 21) and Eq. (D 23) to give: 5 2

24 Example 5 (cont d) V ua V cb A A Vc Vco 0.2 c 1. 5 e ed, V c, V7 o ' c a1 d0 d f Substituting, V ua as controlled by concrete breakout strength is: Vua Vcb , , 514 lb Determine V ua governed by concrete pryout strength ote: The pryout failure mode typically controls only for stiff anchors installed at shallow embedments where the concrete breakout occurs behind the anchor in a direction opposite the shear force (See Figure RD.4.1(ii)). For this example, where the shear may be directed either toward the free edge or away from the free edge, the small edge distance may be the controlling value for pryout strength. V cp V ua where: = 0.70 Condition B applies in all cases when pryout strength governs V k cp cp cb where: k cp = 2.0 for h ef A 2.5 in. c cb ed, c, cp, Aco Evaluate the terms of Eq. (D 4) for this problem: b A c is the projected concrete failure area on the surface as approximated by a rectangle with edges bounded by 1.5h ef (1.5(7)= 10.5 in. in this case) in a direction perpendicular to the shear force and the free edge of the concrete from the centerline of the anchor. D.6.3 Eq. (D 2) D D.4.4(c)i Eq. (D 29) Eq. (D 4) A c = ( )( ) = 278 in. 2 A co = 9 h ef 2 = 9 (7.0) 2 = 441 in. 2 Eq. (D 6) 5 3

25 Example 5 (cont d) 10.5 in A c 10.5 in Determine ed, : ca,min ed, h ef ed, Determine c, : 0.78 c, = 1.0 for locations where concrete cracking is likely to occur (i.e., the edge of the foundation is susceptible to cracks) with cracking controlled by flexural reinforcement or confining reinforcement. Determine cp, : cp, = 1.0 for cast in place anchors. Determine b for the anchor: k b c f ' c h 2.75 in 1.5 ef where, k c = 24 for cast in anchors 10.5 in D Eq. (D 11) D D D Eq. (D 7) b ,112 lb Substituting into Eq. (D 4): cb 278 (0.78)(1.0)(1.0)(28,112) 13, lb To determine V ua for the given embedment depth governed by pryout strength, 5 4

26 Example 5 (cont d) Eq. (D 2) can be combined with Eq. (D 29) to give: V V k ua cp cp cb Substituting, V ua for the embedment length governed by pryout strength is: V V 0.7(2.0)(13,822) 19,350 lb ua cp 5. Required edge distances, spacings and thickness to preclude splitting failure: Since a headed bolt used to attach wood frame construction is not likely to be torqued significantly, the minimum cover requirements of Section 7.7 apply. Per Section 7.7 the minimum clear cover for a 1/2 in. bar is 1 1/2 in. when exposed to earth or weather. The clear cover provided for the bolt is 2 1/2 in. (2 3/4in. to bolt centerline less one half bolt diameter). ote that the bolt head will have slightly less cover (2 3/16 in. for a hex head) OK 6. Summary: The factored shear load (V ua = V n ) based on the governing strength (steel, concrete breakout, and concrete pryout) can be summarized as: Steel strength V sa 3,212 lb Concrete Breakout V cb 1,514 lb Controls Concrete pryout V cp 19,350 lb In accordance with Section 9.2 the load factor for wind load is 1.6: V service Vcb 1.6 1, lb 1.6 The reversible service load shear strength from wind load of the IBC 2003 Section minimum foundation connection for conventional wood frame construction (1/2 in. diameter bolt embedded 7 in.) is 946 lb per bolt. The strength of the attached member (i.e., the 2x6 sill plate) also needs to be evaluated. D ote that this embedment strength is only related to the anchor being installed in concrete with a specified compressive strength of 4,000 psi. In many cases, concrete used in foundations such as this is specified at 2,500 psi, the minimum strength permitted by the code. Since the concrete breakout strength controlled the strength of the connection, a revised strength based on using 2,500 psi concrete rather than the 4,000 psi concrete used in the example can be determined as follows: 2,500 Vservice, f ' 2,500 psi lb c 4,

27 Example 6 Single Headed Bolt in Tension and Shear ear an Edge Determine if a single ½ in. diameter cast in hex headed bolt installed with a 7 in. embedment depth and a 2 ¾ in. edge distance in a normal weight, continuous concrete foundation is adequate for a service tension load from wind of 1,000 lb and reversible service shear load from wind of 600 lb. o supplemental reinforcing is present. ote: This is an extension of Example 5 that includes a tension load on the anchor as well as a shear load. f c = 4,000 psi ASTM F1554 Grade 36 hex head anchor bolt material 1,000 lb 600 lb 18 in. 7 in. ote: Foundation reinforcement not shown for clarity 1. Determine the factored design loads ua = 1.6 (1,000) = 1,600 lb V ua = 1.6 (600) = 960 lb 2. This is a tension/shear interaction problem where values for both the design tensile strength ( n ) and design shear strength ( V n ) will need to be determined. n is the smallest of the design tensile strengths as controlled by steel ( sa ), concrete breakout ( cb ), pullout ( pn ), and side face blowout ( sb ). V n is the smallest of the design shear strengths as controlled by steel ( V sa ), concrete breakout ( V cb ), and concrete pryout ( V cp ). 3. Determine the design 2.75 in. tensile strength ( n ) a. Steel strength, ( sa ): sa = n A se f uta 9.2 D D.5 D.5.1 Eq. (D 3) where: = 0.75 D.4.4(a)i n = 1 (for one anchor) Per the Ductile Steel Element definition in D.1, ASTM F1554 Grade 36 steel qualifies as a ductile steel element. 6 1

28 Example 6 (cont d) A se = in. 2 (for ½ anchor bolt, see Table B1) f uta = 58,000 psi (see Table A) ote: Per Section D.6.1.2, f uta shall not be taken greater than 1.9 f ya or 125,000 psi. For ASTM F1554 Grade 36, 1.9 f ya = 1.9 (36,000) = 68,400 psi, therefore use the specified minimum f uta of 58,000 psi. Substituting: sa = 0.75 (1) (0.142) (58,000) = 6,177 lb b. Concrete breakout strength ( cb ): Since no supplementary reinforcement has been provided, = 0.70 In the process of calculating the pryout strength for this fastener in Example o. 5, Step 4, cb for this anchor was found to be 13,822 lb. Substituting: cb = 0.70 (13,822) = 9,675 lb c. Pullout strength ( pn ) pn = c,p p where: = 0.70 Condition B applies in all cases when pullout strength governs c,p = 1.0, cracking may occur at the edges of the foundation P 8A brg f ' c A brg = in. 2, for a ½ in. hex head bolt (see Table B1) Pullout strength ( pn ): D.5.2 D.4.4(c)ii D.5.3 Eq. (D 14) D.4.4(c)ii D Eq. (D 15) 0.7(1.0)(8)(0.291)(4,000) 6,518 lb pn d. Concrete side face blowout strength ( sb ) The side face blowout failure mode must be investigated when the edge distance c a1 is less than 0.4 h ef 0.4 h ef = 0.4 (7) = 2.80 in. > 2.75 in., therefore, the side face blowout strength must be determined sb 160 c A f ' a1 brg c D.5.4 D Eq. (D 17) 6 2

29 Example 6 (cont d) where: = 0.70, no supplementary reinforcement has been provided c a1 = 2.75 in. D.4.4(c)ii A brg = in. 2, for a ½ in. hex head bolt (see Table B1) Substituting: 160 (2.75) , lb sb e. Summary: Steel strength sa 6,177 lb Controls Concrete breakout strength cb 9,675 lb Concrete pullout strength pn 6,518 lb Side face blowout strength sb 10,508 Check if n ua : 6,177 lb > 1,600 lb OK Therefore: n = 6,177 lb 4. Determine the design shear strength ( V n ) Summary of steel strength, concrete breakout strength, and pryout strength for anchor in shear from Example Problem o. 5: Steel strength V sa 3,212 lb Concrete Breakout V cb 1,514 lb Controls Concrete pryout V cp 19,350 lb Check V n V ua : 1,514 lb > 960 lb OK Therefore: D.5.1 D.5.2 D.5.3 D.5.4 Eq. (D 1) D.6 D.6.1 D.6.2 D.6.3 Eq. (D 2) V n = 1,514 lb 5. Check tension and shear interaction If V ua 0.2 V n then the full tension design strength is permitted D.7 D.7.1 V ua = 960 lb 0.2 V n = 0.2 (1,514) = 302 lb < 960 lb 6 3

30 Example 6 (cont d) V ua exceeds 0.2 V n, so the full tension design strength is not permitted If ua 0.2 n then the full shear design strength is permitted D.7.2 ua = 1,600 lb 0.2 n = 0.2 (6,177) = 1,235 lb < 1,600 lb ua exceeds 0.2 n, so the full shear design strength is not permitted The interaction equation must be used ua n Vua V n 1.2 1, OK 6,177 1, Required edge distances, spacings, and thickness to preclude splitting failure: Since a headed bolt used to attach wood frame construction is not likely to be torqued significantly, the minimum cover requirements of Section 7.7 apply. Per Section 7.7 the minimum clear cover for a 1/2 in. bar is 1 ½ in. when exposed to earth or weather. The clear cover provided for the bolt is 2 ½ in. (2 ¾ in. to bolt centerline less one half bolt diameter). ote that the bolt head will have slightly less cover (2 3/16 in. for a hex head) OK 7. Summary A single ½ in. diameter cast in hex headed bolt installed with a 7 in. embedment depth and a 2 ¾ in. edge distance in a concrete foundation is adequate for a service tension load from wind of 1,000 lb and reversible service shear load from wind of 600 lb. D.7.3 Eq. (D 31) D

31 Example 7 Single Post Installed Anchor in Tension and Shear ear an Edge Determine if a single 1/2 inch diameter post installed torque controlled expansion anchor with a minimum 5 1/2 inch effective embedment installed 3 inches from the edge of a continuous normal weight concrete footing (cast against the earth) is adequate for a service tension load of 1,000 lb for wind and a reversible service shear load of 350 lb for wind. The anchor will be installed in the tension zone and the concrete is assumed to be cracked. 1,000 lb 350 lb ote: Reinf. not shown for clarity. See Table C for a sample table of post installed anchor data from manufacturer (fictitious for example purposes only) as determined from testing in accordance with ACI Determine the factored tension and shear design loads ua = 1.6W = 1.6 1,000 = 1,600 lb. V ua = 1.6W = = 560 lb. 2. Design considerations This is a tension/shear interaction problem where values for both n and V n need to be determined. n is the lesser of the design tension strength controlled by: Steel ( sa ), Concrete Breakout ( cb ), Concrete Side Face Blowout ( sb ), or Concrete Pull Out ( pn ). V n is the lesser of the design shear strength controlled by: Steel (V sa ), Concrete Breakout (V cb ), or Pryout (V cp ). Concrete Side Face Blowout requirements usually apply to cast in anchors (see RD.5.4) and are not considered here. 3. Evaluate anchor material: 5 1/2 in. 3 in. f c =3,000 psi For this example consider that the post installed torque controlled expansion anchor stud is manufactured from carbon steel material conforming to the material requirements of ASTM F 1554 grade 55 which is a headed bolt ASTM specification. The data from anchor prequalification testing according to ACI is shown in Table C. 9.2 D

32 Example 7 (cont d) For ASTM F1554 Grade 55 material (See Table A): f uta = 75,000 psi f ya = 55,000 psi Elongation at 2 in. = 21 % min. with a reduction of area = 30 % min. Appendix D requires 14 % min. elongation and 30 % min. reduction of area for an anchor to be considered as a ductile steel element. D.1 Anchor steel is ductile. 4. Steel strength under tension loading sa ua sa = na se f uta na se f uta ua = 1,600 lb For ductile steel as controlling failure mode, = 0.75 n = 1 (single anchor) Calculating for sa : ,000 = 7,988 lb > 1,600 lb OK sa 1/2 in. diameter anchor steel strength is adequate under tension loading. 5. Minimum Edge Distance Requirements The minimum edge distance for post installed anchors shall be based on the greater of the minimum cover requirements in 7.7, minimum edge distance requirements for the products as determined by tests in accordance with ACI (See Table C), and shall not be less than 2 times the maximum aggregate size. D.5.1 D Eq. (D 3) D.4.4(a)i D.8.3 c a,min = (3 in. maximum cover requirements for concrete cast permanently against the earth; 2.5 in. Product requirement; or 2(0.75) = 1.5 in.) Assuming ¾ in. maximum aggregate size) c a,min = 3 in. 6. Concrete breakout strength under tension loading cb ua D.5.2 D

33 Example 7 (cont d) A c cb ed, c, cp, Aco b Eq. (D 4) where: k b c f ' c h 1.5 ef Eq. (D 7) Substituting: A c 1.5 cb ed, c, cp, k c f ' c h ef ua =1,600 lb A co where: k c = k cr (See Table C) = 17 ca,min ed, = when c a,min < 1.5h ef 1.5h by observation, c a,min < 1.5h ef ed, = ef c, = 1.0 assuming cracking at service loads (f t > f r ) cp, = 1.0 assuming cracking at service loads (f t > f r ) for a category 2 anchor, = 0.55 (o supplemental reinforcement provided) 2 A co = 9 h ef = 9 (5.5) 2 A co = in. 2 A c = (c a h ef ) (2 x 1.5h ef ) = ( (5.5)) (2 x (1.5)(5.5)) A c = in. 2 Eq. (D 11) D D D.4.4(c)ii Eq. (D 6) D Fig. RD.5.2.1(b) A A c co Calculating cb : = 0.68 cb = , cb = 0.55(6,615 lb) = 3,638 lb > ua =1,600 lb OK = 6,615 lb 7 3

34 Example 7 (cont d) 7. Pullout Strength D Pullout strength, p, for post installed anchors is established by reference tests in cracked and uncracked concrete in accordance with ACI Data from the anchor prequalification testing must be used. p = p,cr = 7,544 lb (See Table C ). pn ua D pn = c, P p for a category 2 anchor, = 0.55 (Condition B applies in all cases when pullout strength governs) c,p = 1.0 assuming cracking at service loads (f t > f r ) pn = ,544 = 4,149 > 1,600 lb OK 8. Check All Failure Modes under Tension Loading Summary: Steel Concrete breakout Pullout sa cb pn 7,988 lb. 3,638 lb. Controls 4,149 lb. n = 3,638 lb as concrete breakout strength controls. 9. Steel strength under shear loading V sa V ua V sa = n0.6a se f uta (Anchor does not have sleeve to extend through shear plane) n0.6a se f uta V ua = 560 lb For ductile steel as controlling failure mode, = 0.65 Eq. (D 14) D.4.4(c)ii D D D.6.1 D Eq. (D 20) D.4.4(a)ii n = 1 (single anchor) Calculating for V sa : V sa ,000 = 4,153 lb > 560 lb OK 1/2 in. diameter anchor steel strength is adequate under shear loading. 7 4

35 Example 7 (cont d) 10. Concrete breakout strength under shear loading V cb V ua A Vc V cb ed, V c, V AVco V b D.6.2 D Eq. (D 21) where: V b 0.2 c 1. 5 e ' 7 do fc a1 do Substituting: V cb where: A A Vc Vco e ed V c V,, 7 d o 0.2 d o f ' c 1.5 c a Vua 1 = 560 lb = 0.7 for anchors governed by concrete breakout due to shear (Condition B no supplemental reinforcement is provided) A A Vc Vco = 1.0, and ca1 = 3 in. (edge distance) c a2 is the distance from the center of the anchor to the edge of concrete in the direction orthogonal to c a1 (not specified in this example but consider this distance greater than 1.5 h ef ) ed,v = 1.0 since c a2 1.5c a1 c,v = 1.0 assuming cracking at service loads (f t > f r ) d o = 0.5 in. Eq. (D 24) D.4.4(c)ii Eq. (D 27) D e = h ef = 5.5 in., but e = 4 in. e 8d o ; 8d o = 8(0.5) = 4 in. D V cb ,000 = 1,495 lb > 560 lb OK Concrete Pryout Strength: V cp V cp V ua = k cp cb D.6.3 D

36 Example 7 (cont d) where: Eq. (D 29) k cp = 2.0, since h ef = 5.5 in. > 2.5 in. cb = 6,615 lb (See section 6 of this example) = 0.7 (Condition B applies) D D.4.4(c)i V cp ,615 = 9,261 lb > 560 lb OK 12. Check All Failure Modes under Shear Loading: Summary: Steel Concrete breakout Pryout V sa V cb V cp 4,153 lb. 1,495 lb. Controls 9,261 lb. V n = 1,495 lb from concrete breakout strength controls 13. Check Interaction of Tension and Shear Forces If V ua < 0.2 V n, then the full strength in tension is permitted: n ua 0.2 V n = 0.2(1,495 lb) = 299 lb < 560 lb Requirement not met If ua < 0.2 n, then the full strength in shear is permitted: V n V ua 0.2 n = 0.2(3,638 lb) = 728 lb < 1,600 lb Requirement not met Since V ua > 0.2 V n ua n V V ua n 1.2 and ua > 0.2 n, then: D D.7 D.7.1 D.7.2 D.7.3 Eq. (D 31) 1,600 3, Summary 560 1,495 = = 0.81 < 1.2 OK The Post Installed Torque Controlled Expansion Anchor, 1/2 in. diameter at a 5 1/2 in. effective embedment depth is adequate to resist the applied service tension and shear loads of 1,000 lb and 350 lb, respectively. 7 6

37 Example 8 Group of Anchors in Tension and Shear with Two Free Edges, and Supplemental Reinforcement 4,000 lb Check the capacity of a fastener group with four ¾ diameter ASTM F1554 Grade 55 cast-in anchor rods embedded 12 using hex nuts into the 2-0 thick, f c = 3000 thickened slab made of normalweight concrete to support a factored shear of 4 kips shear and simultaneous uplift of 12 kips. The plate is symmetrically placed at the corner. Seismic forces are not a consideration. 60 ksi reinforcement #5 continuous perimeter bar 12,000 lb ¾ x 1-0 x1-0 Base Plate with oversized holes 1 ½ " non-shrink, min 3000psi grout Hairpins each waysee example text #5 continuous perimeter bar slab edge reinforcement. Calculation and Discussion Because both shear and tension are to be considered the capacity of this detail will be based on the interaction equation given in D

38 Example 8 (cont d) Calculation and Discussion 1. Factored design load ua = 12,000 lb V ua = 4,000 lb 2. Design considerations The differences between Condition A and Condition B, and the associated requirements for supplemental reinforcement are the subject of different interpretations. The disagreement centers around whether Condition A supplemental reinforcement must be proportioned, detailed, and developed to carry the full design load, thereby essentially eliminating the breakout failure mode, or if Condition A is an intermediate condition between Condition B and a condition in which the supplementary reinforcement is proportioned, detailed, and developed to completely restrain the breakout prism and carry the full load. In the first interpretation, Condition A supplemental reinforcement must be proportioned, detailed, and developed to carry the full design load. In the second interpretation, Condition A supplemental reinforcement is not required to be proportioned, detailed, and developed to carry the full load. This second interpretation is consistent with three possible conditions: Condition A Supplemental reinforcement is present and is oriented and detailed to help restrain the breakout prism, but is not specifically designed to completely restrain the breakout prism. Because this reinforcement increases the ductility of the system a small increase in the phi factor is provided by the code. Condition B Either reinforcement is present near the anchor but is not oriented to aid in restraining the breakout prism or no reinforcement is present. Fully Developed Condition Reinforcement is proportioned, detailed, and developed to hold the breakout prism in place and to support a design load greater than the breakout strength. The authors of this problem consider the second interpretation correct and have considered Condition B for tension and Condition A for shear. It is hoped that this issue will be clarified in ACI In the following calculations, the supplementary reinforcement (#4s, #5s, hairpin) is not considered to directly restrain the shear breakout prism. It is considered to enhance the ductility of the system in shear as noted in the explanation of the second interpretation above for Condition A. Condition B of the second interpretation is assumed for all tension capacity calculations. The anchor system must be designed as follows: φ n ua Eq. (D-1) where φ n is the lowest design strength in tension from all appropriate failure 8-2

39 Example 8 (cont d) modes. φ V n V ua Calculation and Discussion Eq. (D-2) Where φ V n is the lowest design strength in shear from all appropriate failure modes. 3. Anchor material ASTM F1554 Grade 55 rod material meets the requirements of the ductile steel element definition in D.1 (i.e. elongation of at least 14% and reduction in area of at least 30%. F1554 material has 21 % elongation in 2" of length with 30 % reduction in area (see Table A of this design guide). Tensile Capacity Calculations 1. Determine the steel strength of anchors in tension, φ sa. φ sa = φ n A se f uta φ = 0.75 A se = See Table B1 of this design guide. φ na f = 0.75(4)(0.334)(75, 000) = 75,150 lb se uta 2. Determine the breakout capacity of the anchor group in tension, φ cbg. φ cbg A c = φ ψec, ψed, ψc,ψ cp, Aco (ѱ cp, = 1.0 for cast-in anchors) φ = 0.70 We select Condition B, no supplementary reinforcement is provided to resist tension breakout. There is no reinforcement provided which is oriented to restrain the tension breakout prism. The reinforcement present is oriented in such a way that it will aid in enhancing the ductility of the system in shear more on that in the shear section (See Section 7 in shear calculations) and not the tension breakout prism. b D.1 D Eq. (D-3) D.4.4 D.5.2 Eq. (D-5) D D.4.4 RD h ef =18 Check the basic concrete breakout strength. A c = ( )( ) = 900 in

40 Example 8 (cont d) Calculation and Discussion A co = 9 h = 9(12) 2 = 1,296 in. 2 2 ef ψ ec, = 1.0 as no eccentricity is exists Check to see if the edge distance modifier, ψ ed,, is other than 1.0 c a,min = 4 c a, min < 1.5 h ef = 18, therefore ψ ed, will be less than one. ca,min 4.0 ψ ed, = = = h 1.5(12) ef Check if the cracked concrete, ψ, modifier is other than one. c, Unless an analysis is done to show no cracking at service loads cracking must be assumed, therefore ψ c, = 1.0. h ef = 12" > 11", therefore Eq. (D-8) shall be applied Hatched Area Is A C Eq. (D-6) D D Eq. (D-11) D D Eq. (D-8) 5/3 16 ' 5/3 b = fc hef = 16 3, 000 (12) = 55,120 lb 900 φ cbg = 0.7 (1.0)(0.767)(1.0)(55,120) = 20,551 lb 1, Determine the pullout strength in tension, φ pn. φ pn = φ ψ c,p p D.5.3 Eq. (D-14) D

41 Example 8 (cont d) Calculation and Discussion φ = 0.70 D Unless an analysis is done to show no cracking is present at service loads cracking must be assumed, therefore ψ c,p = 1.0. Eq. (D-15) ' = 8 A f = (8)0.654 (3,000) = 15,696 lb p brg c See Table B1 for bearing area of hex nut. There are four rods, thus the total pullout capacity is (4)15,696 = 62,784lb. 4. Determine concrete side-face blowout capacity, φ sbg. Evaluate if side-face blowout is a consideration. The smallest edge distance, c a1, must be less than 0.4 h ef = 0.4(12) = 4.8 and the spacing of the anchors must be less than 6c a1 = 6(4) = 24. Both requirements are met, so side face blowout must be considered. s φsbg = φ ca1 φ = 0.70 s = 8 sb ' = 160c A f = 160(4) ,000 = 28,348 lb φ sb a1 brg c sbg 5. Summary 8 = ,348 = 26,458 lb 6(4) D.5.4 D Eq. (D-18) D.4.4 Eq. (D-17) φ sa = 75,150 lb (steel strength) φ cbg = 20,551 lb (concrete breakout) φ pn = 62,784 lb (concrete pullout) φ sbg = 26,458 lb (side face blowout) Concrete breakout controls the tension capacity at a capacity of kips. 8-5

42 Example 8 (cont d) Shear Capacity Calculation Calculation and Discussion 6. Determine the steel strength of anchor in shear, φ Vsa. φ V =φ n0.6a f sa φ = 0.65 se uta D Eq. (D-20) D.4.4 Ase = See Table B1 of this design guide The steel strength of each anchor is: φ n0.6 A f = 0.65 (0.6)(0.334)(75,000) = 9.77kips se uta Where anchors are used with built-up grout pads, the design strength must be multiplied by 0.8. φ V sa = 0.8 (9.77) = 7.82 kips per anchor 7. Determine the breakout strength in shear, φ Vcbg. AVc φvcbg = φ ψ ψ ψ A φ = 0.75 Vco ec,v ed,v c,v V b Condition A. The combination of the continuous #5 perimeter bars and #4 bent slab bars qualify as supplementary reinforcement for resisting a shear breakout, as do the hairpins. The hairpins could potentially be designed to directly restrain the concrete breakout thereby eliminating breakout as a failure mode. However, at this time the combination of the continuous #5 perimeter bars and #4 bent slab bars and hairpins are considered as supplementary reinforcement in terms of Condition A (in other words, the hairpin is not considered to directly restrain the breakout prism and does not eliminate breakout as a failure mode. We are acknowledging the fact that there is reinforcement present that will help retrain the breakout prism and improve ductility allowing an increase in the capacity by changing the phi factor to the Condition A value). D D.6.2 Eq. (D-22) D V = 7 d f (c ) e ' 1.5 b o c a1 do can not exceed 8 d = 8(3/ 4) = 6" e o Eq. (D-24) D c a1 can be either from the first line of fasteners or the second. The base plate will be fabricated with oversized holes. It is therefore possible the two fasteners closest to the edge will engage before those at the line farthest from the edge. Unless steps are taken to ensure that the fasteners farthest from the edge will be RD6.2.1 Fig RD.6.2.1(b) 8-6

43 Example 8 (cont d) Calculation and Discussion engaged (such as welding plate washers with standard holes to the base plate in the field which will be considered later in this example), c a1 should conservatively be taken as the distance from the free edge to the first line of fasteners. c a1 = 4 A Vc = (c a2 +s+1.5 c a1 )(1.5 c a1 )=( )(6) = 108 in. 2 A Vco = 4.5 (c a1 ) 2 = 4.5(4) 2 = 72 in. 2 V b ,000(4) 4,026 lb = = 0.75 ψ ec,v = 1.0 no eccentric shear is applied. c a2 = 4 therefore c a2 < 1.5c a1 and ψ ed.v must be calculated ca2 4.0 ψ ed,v = = = c 1.5(4.0) a1 A continuous #5 bar is present around the edge of the slab. Therefore ψ c,v = φ V cbg = 0.75 (1.0)(0.9)(1.2)(4,026) = 4.89 kips 72 The breakout capacity of the detail is limited. The designer may design the hairpins in the slab to directly restrain concrete breakout per D and thereby eliminate breakout as a failure mode. Side ote: Consider the effect on the capacity if plate washers with standard holes were welded to the base plate at each anchor. In such a case the breakout prism could be assumed to form from the back anchors per RD c a1 = c a1 = 18 A Vc = ( )(18) = 540 in. 2 A Vco = 4.5 (c a1 ) 2 = 4.5(12) 2 = 648 in. 2 Fig. RD.6.2.1(a) D Eq. (D-23) Eq. (D-24) D D Eq. (D-28) D Eq. (D-22) Vb = ,000 ( 12) = 20,921 lb 0.75 c a2 = 4 therefore c a2 < 1.5c a1 and ψ ed.v must be calculated ψ ed,v = φ V cbg = 0.75 (1)(0.767)(1.2)(20,921) = 12.0kips 648 Thus, the addition of plate washers with standard holes welded to the base plate would allow the designer to assume that the breakout would occur from the back row of fasteners and to realize a substantial increase in breakout capacity. If this assumption is made all of the load is carried by the rear fasteners and the steel strength of the rear fasteners alone must be sufficient to support the full design load. Additionally, the bearing capacity of the thin edge of the steel washer against the bolt should be checked. 8-7

44 Example 8 (cont d) Calculation and Discussion D Determine the concrete pryout strength, φ V cp. φv = φ k cpg cp cbg φ= 0.7 k = 2.0 when h 2.5" cp cbg ef = A 900 = b (0.767)(1)(55,120) 29.3 kips 1, 296 = c ed, c, A ψ ψ co Eq. (D-30) D.4.4 D Eq. (D-4) φ = 0.7(2)(29.3) = 41 kips V cp 9. Summary φ = 7.82 kips per anchor V sa φ = 4.89 kips assuming no welded plate washers are provided. V cbg φ V cp = 41 kips Side ote: In the case where plate washers were welded to the farthest line of fasteners a substantial increase in concrete breakout capacity is realized. Then the steel strength of these two anchors must be compared to the concrete breakout capacity engaged by these anchors. The concrete breakout capacity of 12 kips with two anchors effective is less than the steel capacity 2(7.82) = 15.6 kips. Furthermore, the pryout capacity of the two rear fasteners alone must be checked: φv = φ k cpg cp cbg φ= 0.7 k = 2.0 when h 2.5" cp A c = ef 2 = ( )(4 + 8) 660in for the rear two anchors. 660 cbg = (0.767)(1)(55,120) = 21. 5kips 1,296 φ V cp 0.7(2)(21.5) = 30kips Concrete Breakout controls. Shear breakout severely limits the capacity of this connection. 8-8

45 Example 8 (cont d) Calculation and Discussion Interaction of Tensile and Shear Forces 10. First the provisions of D.7.1 and D.7.2 are checked. Vua = 4 kips > 0.2 φvn = 0.2(4.89) = 0.98 kips (assumes no plate washers welded to base plate and no hairpins provided) D.7 D.7.1 ua = 12 kips > 0.2 φ = 0.2(20.6) = 4.12 kips n D.7.2 Therefore the interaction of the two must be checked ua φ n Vua = + = 1.4 > 1.2 o Good φv n Add field welded plate washers and thus increase φ V n to 12.0 kips (see above, concrete breakout failure of the two rear anchors governs). V ua = 4 kips > 0.2 φ V n = 0.2 (12.0) = 2.4 kips Therefore the interaction must still be checked, u φ na + Vu φv = = OK na The four anchor rod group is adequate to resist the applied loading assuming plate washers are welded to the base plate at all anchor rod locations. Eq. (D-31) 8-9

46 Example 10 - Multiple Anchor Connection Subjected to Moment and Shear Design an embed plate beam to support the end reaction and end bending moment of a steel beam using a group of eight welded headed studs spaced as shown in Figure 1. The factored design bending moment (M ua ) is 30,000 ft-lb and the factored shear load coming from the beam end reaction (V ua ) is 20,000 lb factored shear load (ACI Section 9.2 load combinations used). The connection is located on the side face of a reinforced concrete spandrel girder (see Figure 1). 7/8" dia. welded headed studs h ef = 9" 1-1/2" 1-1/2" x 2'-0" sq. embed plate Girder beyond #6 typ. #5 4" o.c. Figure 1 Beam Support Consisting of Embed Plate with Welded Studs 1. It is often difficult to simultaneously determine the required anchor diameter and embedment length simultaneously due to the complexity of the load distribution to each anchor and the interaction of tensile and shear forces. Thus, the general approach from a practical design perspective is to assume some design values in advance and verify their acceptability in the design calculation. In this example, the connection detail shown in Figure 1 as based on preliminary design is to be checked for conformance with code requirements. 2. Design basis: A. basic assumptions - cracked concrete - elastic design B. materials ormal wt. concrete f'c = 4,500 psi 1'-6" V ua W14x53 M ua 4 - #10 Bottom soffit of spandrel girder 1-1/2" typ y 4 5 z x V ua 10-1/2" 10-1/2" 6" 10-1/2" 10-1/2" 6"

47 Example 10 (cont d) - embed plate: ASTM A /2" x 24" x 24" (t p = 1-1/2 in) - anchors: 7/8" x 8-3/16" AWS D1.1 Type B mild steel welded headed stud (standard length shear stud) specified yield strength of anchor steel f ya = 51 ksi specified tensile strength of anchor steel f uta = 65 ksi stud diameter d = in. stud head diameter d h = in. stud head thickness t h = 3/8 in. reduction in stud length due to welding ~ 3/16 in. effective anchor embedment: h ef = 8-3/16" 3/8" 3/16" + 1-1/2" = 9.125" 9 in. stud length d h Figure 2 Welded Headed Stud 3. Determine anchor reactions: D.3.1 A. Tension: The determination of the anchor reactions is not trivial. While the use of finite element modeling may provide the closest approximation, it is possible to use conservative assumptions and statics to derive an acceptable set of anchor forces. Simplified statically determinant analysis: Assume that the compression reaction is located directly beneath the toe of the W14 beam (conservative) and that the embed plate exhibits rigid-body rotation (see Figure 3). d hef th 10-2

48 Example 10 (cont d) T T a 10.5" 7" T M 360 in - k Figure 3 Tension Load Distribution T T = tension reaction of top anchors (6, 7 and 8) T M = tension reaction of middle anchors (4 and 5) M = ( T T ) + 7 ( T M ) 360 = 0 (1) Assume rigid-body rotation of embed plate and determine ratio of anchor reactions (K = single anchor elastic stiffness, Δ = anchor elastic displacement): Δ T = Δ T = 25. Δ Δ 7 M M Tension reaction of back 3 anchors: TT = 3Δ TK (3) Substituting Eq. (2) into Eq. (3): TT = 7.5Δ MK (4) Tension reaction of middle 2 anchors: TM = 2Δ MK (5) Substituting Eq. (5) into Eq. (4): TT = 3.75T (6) M Substituting into Eq. (1) above: TM = 4. 95k T4 = T5 = / 2 = k T = 186k. T = T = T = 186. / 3= 620k. T Total tension reaction: ua = ( 3)( 6.20 ) + ( 2)( 2.48 ) = 23.6 k Check assumption of plate rigidity using results from analysis above: C The assumption that the compression reaction is located at the compression flange of the steel beam as shown in Figure 3 is rational based on observations of tested assemblies. While the load distribution to the tension-loaded anchors will also be affected by their proximity to the connected wide flange shape, such effects are likely to occur at larger rotations. Per D.3.1, the analysis may be based on elastic (2) 10-3

49 Example 10 (cont d) anchor response. On the tension side, checking the moment at the beam face and compare it to the moment in the plate: ( ) 2 bh M =S f = f = 36ksi=324in-k y,pl x y,pl 6 y,pl 6 ( )( ) M = T T a = = 65.1in - k 324 in - k face ok B. Shear: The anchors are welded to the attachment. Per RD for anchors welded to a common plate the shear is carried by the back anchor row. critical anchor row 27" 10.5" 6" Figure 4 Assumption of Critical Anchor Row for Shear Loading 4. Determine design tensile strength, φ n, as follows: ominal steel strength in tension, sa : ca1 RD D check: f uta = 65 ksi < 1.9 f ya = (1.9)(51)= 96.9 k < 125 ksi o.k. D Effective cross-sectional area of anchor: 2 2 π d π A se = = = in y Tension, sa = A se f uta = (0.601)(65)= k for a single stud Eq.(D-3) z x 20 k 6" c a2 Concrete breakout strength in tension, cbg : D.5.2 A A ψ ψ ψ ψ = Eq.(D-5) c cbg ec, ed, c, cp, b co Determine A c for the group of tensioned anchors, T 4, T 5, T 6, T 7 and T 8 and A co for single anchor: 10-4

50 Example 10 (cont d) The projected concrete failure area, A c, is shown shaded in Figure 5 and is calculated as follows: h ef = 9 in. A = (1.5 h +c +2 s ) (2 1.5 h +s ) c ef a2 x ef y = ( ) ( ) = 1,519 in. 2 D " Figure 5 Determination of Projected Failure Area and e' for Tension-Loaded Anchors Determine A co for the single anchor: A = 9 h = 9( 9) = 729 in. Eq.(D-6) co 13.5" 10.5" 13.5" ef A 2 sy 5 c a " y Determine the eccentricity modification factor ψ ec, : s x z 7 x 8 a c a2 resultant of tension-loaded anchors (T 4 - T 8 ) e' sy centroid of tensionloaded anchors (T 4 - T 8 ) 1 ψ ec, = e h ef Determine the eccentricity, e, of the resultant tension load with respect to the centroid of the tension-loaded anchors. The location of the geometric centroid of tension-loaded anchors (T 4 - T 8 ) as measured from axis a-a (see Figure 5) is given by(2 5 )s. Summing moments about a-a, the location of the resultant of the tension-loaded anchors is given y Eq.(D-9) Fig. RD.5.2.4(b) 10-5

51 Example 10 (cont d) (T 4+T 5 )sy by. The eccentricity e between the centroid of the tension-loaded ua anchors and the tension resultant is thus given as: 2 s (T +T )s (2)(10.5) ( )(10.5) = y 4 5 y e - = - = 1.99 in. 5 ua ψ ec, = = (1.99) 1+ 3(9) c = 6 in.< 1.5h = 13.5in. a,min ef Determine the near-edge modification factor ψ ed, : c 6 a,min ψ ed, = = = hef 1.5(9) Determine the modification factor for cracked/uncracked concreteψ c, : Per D.5.2.6, unless an analysis is performed to show no cracking at service loads, the concrete is assumed to be cracked for the purposes of the determining the anchor design strength: ψ =1.0 for cracked concrete. Determine the splitting modification factorψ cp, : c, Eq.(D-11) D ψ cp, = 1.0 for cast-in anchors. D Determine the basic concrete breakout tensile strength for a single anchor, b : ' 1.5 b c c ef D =k f h ote: h = 9 in.< 11 in. Use of Eq. (D-7) required. Eq.(D-7) k c =24 ef = (24) 4,500 (9) b 1.5 = 43,469lb = 43.5 k Substituting into Eq.(D-5): 1,519 cbg = (0.87)(0.83)(1.0)(1.0)(43.5) = 65.5 k 729 Determine the pullout strength of anchors in tension, pn : D.5.3 = ψ Eq.(D-14) pn c,p p where: ψ c,p = 1.0 cracking assumed D

52 Example 10 (cont d) Pullout strength of single anchor in tension, p D ' p =8Abrg f c Eq.(D-15) Determine bearing area of the head of a single stud: π(dh - d ) π( ) A brg = = =0.884 in. 4 4 p = (8)(0.884)(4,500)= 31,824lb = 31.8 k Substituting into Eq. (D-14): pn = (1.0)(31.8)= 31.8 k for a single anchor Determine the side-face blowout tensile strength, sb : c = 6 in. 0.4h = 3.6 in. a,min ef 2 D.5.4 side-face blowout failure mode not applicable D The design tensile strengths are calculated with the calculated nominal strengths and strength reduction factors of D.4.4. Steel strength, φ =0.75 D.4.4 Per AWS D1.1 Type B studs qualify as a ductile steel element (20% minimum elongation in 2", see Table A.) sa ( ) φ = (0.75) = 29.3 k for a single anchor. D Strength reduction factor for concrete breakout, Condition B: φ =0.70 D.4.4 ote: Although the cantilever beam contains significant amounts of beam reinforcing located in the anchor vicinity, this reinforcing is not configured to provide the necessary restraint to the tension-induced concrete breakout surface to warrant the use of the strength reduction factor associated with Condition A. Therefore, Condition B is assumed (supplementary reinforcement not provided). cbg D.4.4 φ = (0.7)(65.5)= 45.9 k for the anchor group comprised of anchors T 4 - T 8. D Strength reduction factor for pullout φ =0.70 D.4.4 φ = φ = (0.70)(31.8) = 22.3 k for a single anchor. D.5.3 pn pn ote: Since this is an elastic analysis, the check for the controlling strength is dependent on the nature of the load distribution. In this case the anchor loads are not uniform, so the steel and pullout checks must be performed on the most 10-7

53 Example 10 (cont d) critically-loaded anchor. Since the CCD method predicts the concrete breakout resistance of the group, the concrete breakout check is always performed for the group of tension-loaded anchors. Summary of connection strength tension: D Failure mode Critical anchor(s) Associated strength, φ n Factored Load, ua Ratio φ ua n Steel 6, 7 or 8 φ sa 29.3 k 6.4 k 0.22 Concrete breakout Concrete pullout Side-face blowout group 4-8 φ 45.9 k 23.6 k 0.51 controls cbg 6, 7 or 8 φ pn 22.3 k 6.4 k φ sb 5. Determine design shear strength, φ V n, as follows: ominal steel strength in shear for cast-in headed studs, V sa : not applicable D V sa = nase f uta = (1)(0.601)(65)= 39.1 k Eq.(D-19) For a single anchor: V sa = (1)(0.601)(65)= 39.1 k Concrete breakout strength in shear, V cbg : Vc cbg ec,v ed,v c,v b Vco D A V = V A ψ ψ ψ Eq.(D-22) ote: Per RD for anchors welded to a common plate the shear is carried by the back anchor row (anchors 6, 7 and 8, see Figure 4) The shear load per anchor is thereforev ua = 20k /3= 6.7k. ote also that the contribution to the shear capacity of the fastening by the concrete located in front of the embedded plate will be neglected. RD

54 Example 10 (cont d) 40.5" 10.5" 10.5" 6" " 27" ca1 = 12" Figure 6 Determination of Projected Failure Area for Shear Loading A Vc #5 o.c Figure 7 Determination of Projected Failure Surface for Shear Loading View of Girder Soffit The edge distance from the edge to the farthest anchor row = 27 in. (see Figure 6). Determine the projected concrete failure areas, A Vco and A Vc : 45" ote that anchors are influenced by three edges (including the beam width), and are located less than 1.5 h ef, therefore the value of c a1 used in Eq. D-21 through D-28 will be reduced therefore check: c h s a2 a x,max c a1 = max ; ; = max ; ; = 12 in. D

55 Example 10 (cont d) A = 4.5 c = 4.5(12) = 648 in. Vco a1 Eq.(D-23) 2 A Vc = (1.5 c a1 + 2 s x + c a2 )(h)= (45)(18)= 810 in for anchors 6, 7 and 8 D Determine the eccentricity modification factor ψ ec,v : ψ ec,v = 1.0 for no eccentricity in the connection D Determine the near-edge modification factor ψ ed, V : c 6 a2 ψ ed,v = = = ca Determine the modification factor for cracked/uncracked concrete ψ cv, : Per D.6.2.7: ψ c,v = 1.4 for cracked concrete with supplementary reinforcement of a o. 4 bar or greater between the anchor and the edge, and with the supplementary reinforcement enclosed within stirrups spaced at not more than 4 inches The #10 girder flexural reinforcement substantially exceeds the requirement of a o. 4 bar, and the associated stirrups for this example will be considered effective in meeting the requirements of D One issue arises from the fact that the edge reinforcing, in this case a #10 bar, does not extend beyond the failure surface on the right-hand side. While the language of D does not specifically require this, it is a reasonable expectation that the edge reinforcing is intended to act as a tension tie across the face of the failure prism. The assumption is made here that the beam reinforcing is adequate to justify the increase provided by D A check for development of the #5 stirrups in the assumed breakout surface as shown below should be made. Refer to Figure 8. Per Section , for #6 and smaller straight deformed bars, the development length is given by: d f y ψt ψe λ 60, = d b= 0.625=22.3in. 25 f ' c 25 4,500 Eq.(D-28) D d,provided ( ) = 23in. The #5 stirrups in this case are also closed loop stirrups anchored by their interaction with the #10 longitudinal steel, so say ok. For this example the #5 stirrups will be considered as supplemental reinforcement in meeting the requirements of tying the shear failure prism to the structure

56 Example 10 (cont d) assumed failure surface #5 stirrups #10 Figure 8 Elevation of Girder Reinforcement at End Determine basic concrete breakout shear strength, V b, for cast-in anchors continuously welded to a steel attachment having a minimum thickness equal to the greater of 3/8 in. or half of the anchor diameter: 0.2 e ' 1.5 b = o c a1 d o V 8 d f c where: e D Eq.(D-25) = 8 d o = (8)(0.875) = 7 in. (8 d o controls) D Substituting into Eq.(D-25) development length 23 in. development length V b = (8) ,500 (12) = 31,629lb = 31.6 k Substituting into Eq.(D-22) 810 V cbg = (1.0)(0.80)(1.4)(31.6) = 44.2 k 648 Determine the concrete pryout strength V cpg D

57 Example 10 (cont d) V cpg =kcp cbg Eq.(D-30) Check the pryout capacity associated with the back three anchors. This is conservative. Determine the concrete breakout strength of rear three anchors (6, 7 and 8): A = ψ ψ ψ c cbg ec, ed, c, b Aco Eq.(D-5) Figure 9 Determination of Projected Failure Area for Pryout A c = (3h ef )(1.5h ef + 2 s x +c a2 )= (27)( )= 1,094 k cp =2.0 for hef 13.5" 13.5" 2.5in. A c 13.5" 1,094 cbg = (1.0)(0.83)(1.0)(43.5) = 54.1 k " 10.5" 6" ote: For the calculation of the pryout capacity, the tension resistance of the anchor group is calculated assuming uniform load distribution to the anchors, thereforeψ ec, = 1.0. V cpg = (2.0)(54.1)= k The design shear strengths are calculated with the calculated nominal strengths and strength reduction factors of D

58 Example 10 (cont d) Per AWS D1.1 Type B studs qualify as a ductile steel element (20% minimum elongation in 2 ) D.1 Determination of the number of anchors contributing to the steel shear capacity is dependent on the assumption for concrete edge failure. The contributions of anchors 1, 2, 3, 4 and 5 are neglected for the steel shear calculation, since these anchors are contained within the assumed breakout surface (see Figure 10). This is reflected in the calculation of V ua = 6.7k Figure 10 Determination of o. of Anchors Contributing to V sa Strength reduction factor for steel strength, φ =0.65 D.4.4 φ V = (0.65)(39.1)= 25.4 k D.6.1 sa Strength reduction factor for concrete breakout, Condition A: φ =0.75 D.4.4 ote: For the concrete breakout strength of the anchors in shear, the # 5 stirrups in this example will be considered as supplemental reinforcement (Condition A), since the girder reinforcement will substantially engage the breakout surface. cbg D.4.4 φ V = (0.75)(44.2)= 33.2 k D.6.2 Strength reduction factor for pryout, Condition B: φ =0.70 D.4.4 φ V = (0.70)(108.3) = 75.8 k D.6.3 cpg ote: As with the tension case, it is important to ensure that the most criticallyloaded anchor is identified for the steel check. Since the CCD method predicts the concrete breakout resistance of the group, the concrete breakout and pryout checks are performed for the anchor group only

59 Example 10 (cont d) Summary of connection strength shear: D Failure mode Critical anchor(s) Associated strength, φ V n Factored Load, V ua Ratio Vua φv n Steel 6, 7 or 8 φ Vsa 25.4 k 6.7 k 0.26 Concrete breakout group 4-8 φ Vcbg 33.2 k 20 k 0.60 controls Concrete pryout group 6-8 φ V 75.8 k 20 k 0.26 cpg 6. Check tension and shear interaction (concrete breakout): D.7 ua = 23.6 k > 0.2 φ n (= 9.18 k) V ua = 20 k > 0.2 φ V n (= 6.64 k) V ua ua + = φ + n φ V = n < Interaction for steel failure ok by inspection. ok D.7.2 D.7.1 Eq.(D-31) 10-14

60 Example 12 - Multiple Anchor Connection Subjected to Seismic Moment and Shear Check the design of an embedded plate with a group of eight welded headed studs spaced as shown to support a factored reversible bending moment, M ua of 30,000 lb-ft (360 kip-in ) and a factored reversible end reaction, V ua of 60,000 lbs (60 kips) (based on elastic analysis, see note below ). The loading results from an earthquake in a region of moderate or high seismic risk. The connection is located far away from any edges of the concrete member; assume it to be in a shear wall (see Figure 1). normal wt. concrete f'c = 6,000 psi welded headed studs 2'-0" h ef = 14" Figure 1 Beam Support Consisting of Embed Plate with Welded Studs 1. ote : As discussed in the ACI Commentary R21.2.1, the design forces may be less than those corresponding to linear response at the anticipated earthquake intensity. The design loads for this connection should be established consistent with the requirements of a code such as ASCE 7. Any reduction factor, R w, should consider the type of construction and the behavior required for the steel beam to concrete connection. 2. The connections shown in Figure 1 is to be checked for the given loading and ductility requirements of Appendix D for resisting seismic loads in a region of moderate or high seismic risk. 3. Design basis: (1) basic assumptions - cracked concrete - plastic design V ua 2" M ua W14x53 o.c., typ. 2" x 1'-0" x 1'-8" embed plate 2" typ. 5 6 y 3 x 4 z 1 2 4" 4" 8" 8"

61 Example 12 (cont d) (2) materials - embed plate: ASTM A 36, 2 x 12 x 20 (t p = 2 in.) - anchors: 3/4" diameter x 13-3/16" AWS D1.1 Type B mild steel welded headed stud specified yield strength of anchor steel f ya = 51 ksi specified tensile strength of anchor steel f uta = 65 ksi stud head diameter d h = 1.25 in. stud head thickness t h = 3/8 in. reduction in stud length due to welding ~ 3/16 in. effective anchor embedment: h ef = 13-3/16 3/8 3/ = Assume h ef = 14 in. 4. Requirements for design: Connection must be proportioned to satisfy D.3.3, i.e., either the anchor design is governed by steel strength or the attachment will undergo ductile yielding before the anchors reach design strength. Checking the plastic capacity of the attached beam (assume F y = 50 ksi): Z = 87.1 in 3, M pl = 87.1 x 50 = 4,355 kip-in. Clearly, it is not practical to develop the plastic strength of the beam. In this case, the anchor design must be controlled by steel strength of the anchors. D D Determine anchor forces: D.3.1 A. Tension: Anchor forces are determined on the basis of plastic response. Assume a simplified elastic statically determinate analysis: Assume that the compression reaction is located directly beneath the toe of the W14 beam (conservative) and that the embedded plate is stiff enough to exhibit rigid-body rotation (see Figure 2). 12-2

62 Example 12 (cont d) 8" 7" Figure 2 Tension Load Distribution Check assumption of plate rigidity: T T = tension force on top anchors (5 and 6) T M = tension force on middle anchors (3 and 4) M = 0 15 ( 2T T ) + 7 ( 2TM ) 360 = 30 T T + 14 T M = 360 Because plastic behavior is assumed, all tension anchors have equal force: T T = T M Substituting into the above equation: 360 kip in. TT = TM = = 8.18 k 44 = 4 ua ( 8.18) = k The 2-in. embed plate has a nominal yield moment capacity of: M y = S x f y bh = 6 2 f y 12 = ( 2.0) ksi = 288 kip in. However, the plastic moment capacity of the plate is given by: M pl = f pl M y = 1.5 M = kip in. = 432 kip in. y Since the yield capacity of the embed plate at the toe of the connected shape is close to the applied moment, the stiffness of the plate should minimize the prying forces in this connection. The assumption of the compression reaction at the toe is reasonable and conservative. While the load distribution to the tension-loaded anchors will also be affected by their proximity to the connected wide flange shape, such effects are likely to occur at larger rotations. Per D.3.1, plastic analysis is permitted. On the tension side, checking the moment in the plate at the top face of the beam flange: M T T T M C ( 8.18 k )( 1 in. ) = T a = 2 = kip in. 288 kip in ok face T. 6. Determine design tensile strength of the anchors, φ n : a 360 in - k Calculate the nominal steel strength in tension, sa : D check: f = 65 ksi 1.9 f = = 96.9 ksi < ksi OK D uta y 125 Effective cross-sectional area of anchor: 12-3

63 Example 12 (cont d) 2 2 π d π 0.75 A se = = = in 4 4 ominal Steel Tensile Strength ( 0.442)( 65) = 28. kips 2 sa = Ase futa = 73 for a single anchor Eq.(D-3) Calculate the nominal concrete breakout strength in tension, cbg : D.5.2 cbg Ac = ψ ec, ψ ed, ψ c, ψ cp, ( b ) Eq.(D-5) A co Determine the ratio of A c for the tension anchor group, T 3, T 4, T 5 and T 6, and A co for a single anchor. The projected concrete failure area, A c, shown as a shaded area on the figure below is calculated as follows (see Figure 3): h ef = 14 in. A c = (1.5h ef + s x + 1.5h ef )(1.5h ef + s y + 1.5h ef ) = ( )( ) = 2,500 in 2 D

64 Example 12 (cont d) 50" 21" 8" 21" Figure 3 Determination of Projected Failure Area and e' for Tension-Loaded Anchors Determine A co for single anchor A co = 9(h ef ) 2 = 9(14) 2 = 1,764 in 2 for a single anchor. Determine the modification factors ψ ec,, ψ ed,, ψ c, and ψ cp, : Determine the eccentricity modification factor, ψ ec, ψ 1 = ' 2 e h ec, ef 1.0 Because plastic analysis is being used, all tension anchors have equal force, and the resultant of the four tension anchors is concentric with respect to the center of ' gravity of the anchors group, and the eccentricity e is zero. ψ ec, = 1.0 s y 2 A c a 25" s x 6 4 a 25" resultant of tension-loaded anchors (T 3, T 4, T 5 and T 6 ) sy Eq.(D-6) Eq.(D-9) Fig. RD.5.2.4(b) Determine the edge modification factor, ψ ed, Connection is remote from edges: ψ ed, = 1. 0 Determine the modification factor for cracked concrete, Per D.5.2.6, unless an analysis is performed to show no cracking at service loads, the concrete is assumed to be cracked for the purposes of the determining ψ c, D

65 Example 12 (cont d) the anchor design strength (this is a prudent assumption for most seismic design cases): ψ c, = 1. 0 Determine the modification factor for post-installed anchor in uncracked concrete, ψ cp, Since it is not a post-installed anchor, ψ cp, = 1.0 Determine the concrete breakout strength, b, for a single anchor in tension : b c ' c 1.5 ef D = k f h Eq.(D-7) h ef = 14 in. k = 16 for cast-in anchors, 11 in. hef 25 in. c 5/ 3 ( 14) = 100. kips b = 16 6,000 8 Substituting into Eq.(D-5) for the group capacity: 2,500 cbg = 9 1,764 ( 1.0)( 1.0)( 1.0) (1.0)( 100.8) = 142. kips Determine the pullout strength of anchors in tension, pn : pn c, P where: p D.5.3 = ψ Eq.(D-14) ψ = 1.0 cracked concrete assumed D c, P Determine pullout capacity of single anchor in tension, p D = p A f ' 8 brg c Eq.(D-15) Determine bearing area of the head of a single anchor : π(dh - d ) π( ) A brg = = = in. 4 4 ( 0.785)( 6,000) 37. kips p = 8 = 6 Substituting into Eq. (D-14)

66 Example 12 (cont d) pn = (1.0)(37.6) = 37.6 kips for a single anchor Determine side-face blowout tensile capacity, sb (not applicable since there are no near edges) D.5.4 Determine strength reduction factors applicable for the conditions The design tensile strengths are determined with the calculated nominal strengths and strength reduction factors of D.4.4. Determine the design steel tensile strength: Strength reduction factor for ductile steel in tension, φ =0.75 Per AWS D1.1 Type B studs qualify as a ductile steel element (20% minimum elongation in 2 ) φ = φ n ( 4)( 28.73) 86. kips sa t sa = 0.75 = 2 Determine design concrete breakout strength in tension: Strength reduction factor for concrete breakout, Condition B: φ =0.70 ote: The wall element in which the connection is embedded contains normal orthogonal reinforcement, which will not significantly interact with the tension failure surface of the tension-loaded anchors. Therefore, Condition B is assumed (supplementary reinforcement not provided). ( 142.9) 100. kips D.4.4.a.i D D.4.4.c.ii D.4.4 φ cbg = 0.7 = 03 D Determine design concrete pullout strength in tension : Strength reduction factor for pullout φ =0.70 pn t pn ( )( ) D.4.4.c.ii φ = φn = = 105.3k ips D Summary of design anchor strength tension: D Steel Concrete breakout Concrete pullout φ sa 86.2 kips controls OK D φ cbg kips φ pn kips 12-7

67 Example 12 (cont d) ote: The design strength of steel is exceeded by the design strength of concrete breakout and concrete pullout. For plastic design, it may be prudent to require that the nominal strength of steel be exceeded by the design strength of concrete breakout and concrete pullout. This would require an increase in embedment from 14 in. to about 17 in. 7. Design Strength for Tension in a region of moderate or high seismic risk : ( 86.2) 64. kips 0.75 φ n = 0.75 = 65 D Determine design shear strength, φ V n, The contribution to the shear capacity of the anchorage by the concrete located in front of the embedded plate is neglected. ominal steel strength in shear, V sa : for six anchors V ( 0.442)( 65) 172. kips D = n A f = 6 = Eq.(D-19) sa se uta 4 ominal Concrete breakout strength in shear, V cbg : ot applicable, since there are no proximate edges. Determine the anchor group nominal pryout strength, V cpg Concrete pryout strength V cpg D D.6.3 V cpg =kcp cbg Eq.(D-30) Concrete breakout strength of four anchors (3,4,5 and 6): ote : As all six anchors are in shear, technically cbg of all six anchors need to be determined. However, conservatively, cbg for the anchors in tension are taken for the following calculation. cbg = kips k cp =2.0 for h ef 2.5 in V cpg = 2.0 x = kips Determine design shear strength of the anchors The design shear strength is determined with the calculated nominal strengths and strength reduction factors of D.4.4. Eq.(D-5) Per AWS D1.1 Type B studs qualify as a ductile steel element (20% minimum elongation in 2 ) D

68 Example 12 (cont d) Strength reduction factor for ductile steel in shear, φ =0.65 D.4.4.a.i ( 172.3) 112. kips φ V sa = 0.65 = 0 Eq.(D-19) Strength reduction factor for concrete pryout φ =0.70 D.4.4.c.i φ V cpg = 0.70(285.8) = kips D Summary of anchor strength shear: Steel Concrete pryout D φ V sa kips controls OK D V cpg φ kips 9. Design strength for Shear in a region of moderate or high seismic risk : φ n ( ) 0.75 V = = 84.0 k ips D Check tension and shear interaction: D.7 ua = 32.7 kips > 0.2 φ n = 0.2 (64.65) = kips V ua = 60 kips > 0.2 φ V n = 0.2(84) = 16.8 kips ua 0.75φ n Vua φ V n = + = 1.22 > Interaction is not met and design strength must be increased. It can be achieved by using larger diameter anchors. D.7.2 D.7.1 Eq.(D-31) 12-9

69 Example 16 - Group of Tension Anchors on a Pier with Shear Lug Design the anchors and shear lug for the steel column located atop the concrete pedestal shown in Figure A.1. Tension is resisted by the anchors, and all the shear is resisted by the shear lug. This is a common design situation encountered in industrial facilities. The pedestal, in this example, could just as easily have been a wall or pilaster. The challenge for this application is to design all of the elements to work properly together while making certain that the design is constructible. This example goes beyond the provisions of Appendix D and uses other provisions such as ACI 349 and AISC Column Base Plate Design Guide. These other provisions are used to evaluate the shear lug. The reader is cautioned to review these other documents to fully understand their limitations. Given: Pedestal geometry shown in Figure A.1 Pedestal vertical reinforcement 16 - #8 bars (designed per ACI 318 for tension plus bending at the top of footing) Pedestal transverse tie sets of #4 bars arranged as shown Concrete strength f c = 4000 psi, reinforcement strength f y = 60,000 psi W12x58 column with 16 square base plate 1-½ thick All plate steel is ASTM A36 Use 4 anchors, as shown in Figure A.1, are ASTM F 1554 Grade 55 material Use anchors with double heavy hex nut at the embedded end Anchors are to be a ductile design Service Dead Load (DL) = 200 kips Service Wind Load (WL) = 230 kips uplift and 40 kips shear (acting as shown in Figure A.1) 16-1

70 Example 16 (cont d) Figure A.1 Design Ultimate Loads Load combinations are from Section 9.2 of ACI 318. Combination (9-6) controls for tension and shear on the anchors. Tension T u = 0.9D + 1.6W + 1.6H = 0.9 (200 kip) (-230 kip) = -188 kips (upward) Shear V u = 0.9D + 1.6W + 1.6H = 1.6 (40 kip) = 64 kips Anchor Design 1. Select the anchor diameter to resist the factored load using Section D.5.1 of ACI 318. Try 1-¼ in. diameter anchor. ASTM F 1554 Grade 55 is ductile steel. Check the assumed anchor diameter by checking the tensile capacity. Therefore, φ equals 0.75 for tension sa se uta 16-2 D.4.4 φ = φ n A f Eq. (D-3) 2 A se = in From Table B1 of this design guide f uta = 75 ksi From Table A of this design guide 2 φ sa = 0.75(4)(0.969 in )(75 ksi) = 218 kip > 188 kips (looks ok) 2. Make Initial Estimate of Embedment Depth: Estimate the required embedded length h ef at 12 diameters. Given the 7 in. anchor spacing and 30 in. pier, 12 diameters equals h ef of ~15 in., which will include the entire area of the pier in the projected failure area A c. ote from Section RD.5.2 that the failure surface projects outward 1.5 units per every 1 unit of depth (~35 degrees to the concrete surface). Since the tension force in this example places the entire concrete pier section in tension it is doubtful that the concrete breakout capacity will be sufficient to resist the applied load (this will be verified by calculation in Section 3). In this case, the load path must be from the anchors to the vertical pier reinforcement. The transfer of force from anchors to reinforcement requires development of the reinforcing bars into the concrete breakout cone. The #8 vertical bars have a

71 Example 16 (cont d) tension development length of approximately 48 in. For now, provide h ef such that a #6 bar with hook could be developed into the failure cone as shown in Figure A.2within a distance of L d. Figure A.2 Allowing 10 in. to develop a #6 hooked bar (including the length reduction factor of 0.7 from ACI 318, (a)) and 2 in. for cover at the top of pier: h ef = = in Try 20 in. as the embedment depth for the anchors 3. Determine Concrete Breakout Capacity of Anchor Group: Use Section D.5.2 of ACI 318 to calculate the concrete breakout capacity. ote that this check is included in this example for illustrative purposes only, since it is apparent that the concrete breakout load will have to be resisted by supplemental reinforcement. According to Section D.4.4.a.i and D.4.4.c.i of ACI 318, the strength reduction factor φ will be 0.75 for shear and tension since Condition A applies. The group tension capacity is: A φ = Eq. (D-5) c cbg φ ψ ec, ψ ed, ψ c, ψ cp, Aco Determine the maximum embedment depth allowed: This determination is required because if the actual embedment depth is used the calculation is overly conservative. b 16-3

72 Example 16 (cont d) c a,max = (30-7) / 2 = 11.5 in. ; 1.5h ef = 1.5(20) = 30 in. ca, max = 11.5in < 1.5hef = 30in Per ACI 318, D.5.2.3, limit the embedment depth used in EQ. D-4 through D-11. h ef > s max /3 = 7/3 =2.33 in h = c /1.5 = 11.5 / in For use in D-4 through D-11 ef a, max = 67 h ef maximum of c a,max /1.5 or s max /3 h ef = 7.67 in Determine the concrete breakout areas to account for spacing - Determine concrete breakout area for anchor group A c = projected area includes the entire pier area. 2 2 = ( 30 in) 900in Since 1.5h ef = 30 in, by inspection the - Determine the concrete breakout area for a single anchor co 2 9h ef A = Eq. (D-6) A co = = 2 2 9(7.67in) 529in Determine the single anchor tension breakout capacity b c c 1.5 ef = k f h ote: Although the anchors have an embedment depth of about 20 in., D5.2.3 reduces the embedment depth because of the multiple side effects which requires the use of Eq. D-7 instead of D-8. Eq. ( D-7) k = 24 For cast-in anchors c 1.5 b = (7.67) = lb Determine the eccentricity modification factor, Ψ ec, ψ ec, = 1.0 Eq. (D-9) Factor equals 1, since there is no eccentric load. Determine the edge distance modification factor, Ψ ed, c a,min ψ ed, = Eq. (D-11) 1.5hef 16-4

73 Example 16 (cont d) Factor applies due to close proximity of edges. ψ ed 11.5 = (7.67), = 1.0 Determine the cracking modification factor, Ψ c, ψ D c, = 1.25 By inspection of the magnitude and direction of the applied factored loads (axial tension and shear only) it is concluded that at service loads there will be no cracking along the axis of the anchors. Since there is no cracking that would affect the concrete breakout capacity, The cracking modification factor equals Calculate the anchor group concrete tension breakout capacity 900 φ cbg = 0.75 (1.0)(1.0)(1.25)(32243) = lb = kips 529 Clearly, the concrete tension breakout capacity is not adequate to resist the factored tension load. Supplemental reinforcement is required. 4. Determine the anchor pullout capacity. D.5.3 φ = Eq. (D-14) n pn φ nψ c, P p brg c p = 8 A f Eq. (D-15) 2 A brg = 2.237in See Table B1 of this design guide Determine the cracking modification factor, Ψ c, ψ c, P = 1.4 D Factor equals 1.4, since there is no cracking that would affect bearing at the embedded head of the anchor. Calculate the single anchor pullout capacity, p p = 8 (2.237) (4000) = 71584lb = 71. 5kips Calculate the anchor group pullout capacity, pg φ n pn = 0.75(4)(1.4)(71584) = lb = kips 5. Check side-face blowout. D.5.4 c a1 = 11.5 in. Edge distance 16-5

74 Example 16 (cont d) 0.4h ef = 0.4(20) = 8 in. < c a1 = 11.5 in. Tension capacity is not affected by side-face blowout. Summary of anchor capacity for tension load: Anchor steel capacity = 218 kips Concrete breakout capacity = 51.4 kips Pullout capacity = kips Side-face blowout capacity = not applicable Since the concrete breakout capacity controls and is less than the factored tension of 188 kips, reinforcement is required to be anchored on both sides of the breakout planes. In addition, a ductile design is required as part of the problem statement. A ductile design requires that overall capacity be limited by the anchor s steel capacity, which is 218 kips. Reinforcement Design 1. There are many possible options for designing the reinforcement. The challenge is to make certain that the design is constructible. Often, the design will have to be sketched to scale in order to check for congestion and interferences. It is also advisable to coordinate with a member of the construction team. This example requires vertical reinforcement to constrain the concrete failure prism. The reinforcement must be developed on both sides of the failure plane. As previously discussed, the #8 vertical bars have an approximate tension development length of 48 in. (ACI 318, Section 12.2). Since the effective embedment h ef is equal to 20 in., a straight #8 bar cannot be developed. One option, which will not be examined in this example, would be to make the anchors long enough to develop the #8 bars; if there were an excess area of #8 bars, then the development length could be reduced in accordance with ACI 318, Section Several other options will be evaluated. The following is the area of reinforcement required to develop the anchor capacity of 218 kips: (RD.4.2.1) 218kips A s = 0.9(60 ksi) 2 = 4.04 in #8 bars equivalent to A s = 4.04/0.79 = 5.11 => 6 bars #6 bars equivalent to A s = 4.04/0.44 = 9.18 => 10 bars 16-6

75 Example 16 (cont d) Figure A.3 Figure A.3 shows three reinforcement options. One option not considered in this example is the use of headed reinforcing bars for the #6 and #8 bar options. A general discussion of the three options considered follows: A. Option (A) requires a 180 degree hook at the top of each #8 bar. The development length for a standard hook on a #8 bar is approximately 14 in. (including the length reduction factor of 0.7 from ACI 318, Section (a)). Since there is an excess of #8 bars for developing the anchors (6 bars required, 16 bars in the pier), the development length may be reduced according to ACI 318, Section (d), as follows: l As required 4.04 = ( 14 in) = (14 in) = 4. in (d) A provided 16(0.79) dh 47 s However, l dh shall not be less than 8 db or 6 in. l dh = 8 (1.0 in) = 8 in Since 10 in. was previously allotted for bar development, the #8 bars with hooks will work. The J dimension, or width of the hook, is 8 in., which puts the hooks in close proximity to the anchors. Also, the hooks would have to be rotated in order to avoid interfering with one another at the corners of the pier. As shown in Figure A.1, the #8 bars are continuous into the foundation below the pier; therefore, once the #8 bars are cast into the foundation, it will be difficult to adjust the hooks at the top in order to accommodate the anchors and the shear lug blockout. For this example, Option (A) is ruled out. 16-7

76 Example 16 (cont d) B. Option (B) requires #6 U-bars (or hairpins) to be installed adjacent to the anchors. Since 10 - #6 bars are required to develop the anchors, 5 U-bars would have to be placed. This option allows the #6 U-bars bars to be placed at the same time as the anchors and shear lug blockout. Unfortunately, the U-bars will have to compete for space with the anchors, shear key blockout, and lateral tie sets. In addition, construction will have to take additional measures to secure the U-bars during the casting of concrete. For this example, Option (B) is ruled out. C. Option (C) is the solution chosen for this example. This option makes use of hooked #6 bars that are lap spliced with the #8 pier reinforcement. The required number of #6 bars is 10, but 12 bars will be provided for convenience because there are three #8 bars between corners in the pedestal (3 on each side of the pier). The hooked bars may be placed at the same time as the anchors and shear key blockout, which will permit adjustment of the hooks to avoid congestion and interferences. Also, the hooked bars will not interfere with the interior legs of the lateral tie sets. The lap splice for the #6 bars must be Class B, and the required lap length is approximately 37 in., according to ACI 318, Section Therefore, the overall length of the hooked bars is 10 in. (the development length) plus 37 in. (the lap splice length) plus 7.67 in due to slope of shear cone (See Figure A.3.c), which equals 54 in. The 4 ft-6 in. (54 in.) pier height is sufficient to accommodate the hooked bars. 2. ACI 318, Section , requires lateral reinforcement to encompass the anchors in the top of a pier/column. As shown in Figure A.1, 2 sets of #4 ties will be located within the top 5 in. of the pier. Shear Lug Design A single cantilever type shear lug is proposed to transfer the entire shear load to the top of the concrete pier. For convenience, the lug will use the same 1-½ -thick plate used for the column base plate. The methodology used in this example is based on ACI , Appendix D, and Column Base Plates: Steel Design Guide Series 1, American Institute of Steel Construction, Inc., Chicago, Illinois, As a matter of good practice, the shear lug should be embedded a minimum of 2 in. into the concrete pier. The 1-½ in. thick grout pad between the base plate and top of pier is considered to be ineffective for transfer of shear. The shear lug must therefore be a minimum of

77 Example 16 (cont d) ½ in. deep. Try 2 in. of concrete embedment a lug length of 12 in. (see sketch in Figure A.4) Figure A.4 1. Check the lug for concrete bearing. (ote that φ factors From ACI 349 are different than those used in ACI 318 and that ACI 349 φ factors will be used for this design) φ P n = φ1. 3 f c A Bearing capacity lug according to ACI 349, Section D φ P n = 0.7(1.3)(4000)(2 12) = lb = kips > V u = 64 kips O.K. 2. Check for shear acting towards the edge of the pier. The methodology of ACI 349, Section D.11.2, is applied. The effective stress area is found by projecting 45-degree planes from the bearing edges of the shear lug as shown in Figure A.4. A eff = n 2 = ( ) (12 2) in ote, the area of the φ V = φ 4 f A φ = 0.8 c eff lug must be deducted. φ V n = 0.8(4) 4000(436.5) = lb = kips > V u = 64 kips O.K. 16-9

78 Example 16 (cont d) 3. Check the shear lug for bending and shear stresses. A uniform bearing pressure is assumed over the 2 in. of embedded depth into the top of the pedestal. The maximum moment will occur where the lug attaches to the baseplate. The plate capacity is based on the Load and Resistance Factor Design (LRFD) capacity for a plate in bending and shear. Determine the bending capacity of shear lug M u = ( 64)( ) = 160kip in Applied factored moment at the weld of 2 the lug to the baseplate, based on the shear load being applied at mid-depth of lug. φ M = φ Plastic moment capacity of the shear lug plate n ZF y AISCLRFD F1.1 2 (12)(1.5) φ M n = (0.9) (36) = kip in > M u = 160 kip-in O.K. 4 Determine the shear capacity of the shear lug φ V = φ 0. 6A F Shear capacity of the shear lug plate AISC LRFD J5.3 n g y φ V n = 0.9(0.6)(12 1.5)(36) = 350 kips > V u = 64 kips O.K. Figure A.5 4. Design the shear lug-to-baseplate weld using E70XX electrodes. The resultant forces on the welds are as shown in Figure A.5. The shear force is shared equally between the two fillets. The moment is resisted by a force 16-10

79 Example 16 (cont d) couple at the two fillets. The vertical force couple with distance s is taken between the centroids of the two fillet welds. For 1-½ in. plate, the minimum allowable fillet size is 5/16 in. f v = ( 64) /(12 2) = 2.67 kip / in Shear force per in. of weld 160 f t = = 7.8 kip / in Tension force per in. of weld (12)[1.5 + (2 1 5 )] R = f + f = 8.24 kip in Total force per in. of weld v t /.5 φ F φ 0.6F ( sin 1 θ ) Weld capacity AISC LRFD w = EXX θ = 90 degrees since the force is perpendicular to the axis of the weld. φ = 0.75 Solve for the required weld leg size, a : a = 8.24 = in < 5/16 in (0.6)(70)( sin (90))(0.707) Use the minimum allowable size of 5/16 in. Design Summary Anchors: 1-¼ in. diameter ASTM 1554 Grade 55 w/ heavy hex nuts 20 in. embedment depth (top of concrete to top of embedded head/nut) Supplementary reinforcement: 12 - #6 bars with standard 180 hooks at the top, lap splice with #8 pier reinforcement Shear lugs: 12 in. x 3-½ in. x 1-½ in. thick plate ASTM A36 5/16 in. fillet weld E70XX on each of shear lug side to baseplate Appendix J

80 Table A MATERIALS FOR HEADED ACHORS AD THREADED RODS Material Welded Studs AWS D1.1:2004 ASTM A29-05 / A Grade or Type B Diameter (in.) Tensile Strength, Min. (ksi) Yield Strength, Min. Elongation, Min. (ksi) Method (%) Length Reduction of Area, Min. (%) ½ to % 20 2 in. 50 ASTM F ¼ to % 23 2 in * % 21 2 in. 30 (H,HD, T) ¼ to % 15 2 in. 45 ASTM A (H,T) B7 2-½ % 16 4D 50 2-½ to % 16 4D 50 Over 4 to % 18 4D 50 ASTM A A ¼ to in. (Gr. A: HD) (Gr. C: H, T) C ¼ to in. ASTM A36-05 (H, T) ASTM A449-04b (H, HD, T) otes: - To in. 1 ¼ to % 14 4D 35 Over 1 to 1-½ % 14 4D 35 Over 1-½ to % 14 4D 35 1 Anchor type availabilities are denoted as follows: H = hooked bolt, HD = headed bolt, and T = threaded rod. Comments Structural Welding Steel, Section 7, covers welded headed or welded bent studs. AWS D1.1 requires studs to be made from cold drawn bar stock conforming to the requirements of ASTM A108. ASTM F1554, Standard Specification for Anchor Bolts, Steel, 36, 55, and 105-ksi Yield Strength is the preferred material specification for anchors. *Diameters larger than 2 in. (up to 4 in) are available, but the reduction of area will vary for Grade 55. ASTM A193, Standard Specification for Alloy-Steel and Stainless Steel Bolting Materials for High-Temperature Service: Grade B7 is an alloy steel for use in high-temperature service. ASTM A307, Standard Specification for Carbon Steel Bolts and Studs, PSI Tensile Strength: ACI 349 specifies that elements meeting ASTM A307 shall be considered ductile. ote that Grade C conforms to tensile properties for ASTM A36. ASTM A36, Standard Specification for Carbon Structural Steel: Since ACI 318 considers ASTM A307 to be ductile, A36 will also qualify since it is the basis for ASTM A307 Grade C. ASTM A449, Standard Specification for Quenched and Tempered Steel Bolts and Studs: This specification is for general high strength applications.

81 Table B1 Gross Area (A D ), Effective Area (A se ), and Bearing Area (A brg ) for cast-in-place anchors, threaded rod with nuts, and threaded rods with nuts and washers ominal Diameter otes: Gross Cross- Sectional Area d o (in.) A D (in. 2 ) A se (in. 2 ) Table B1 - Dimensional Characteristics of Cast-In-Place Bolts & Threaded Rods with uts Effective Cross- Sectional Area Square Head Bolt Hex Head Bolt Heavy Hex Head Bolt Dimensions taken from AISC Steel Design Manual 2 A se = (d 0 -(0.9743/n)) 2 3 A H = F 2 (Square Head Bolt/ut) or A H = 1.5F 2 tan30 (Hex/Heavy Hex Head Bolt/ut) 4 A brg = A H -A D Threaded Rod w/ Square ut A brg (in. 2 ) Threaded Rod w/ Heavy Square ut Threaded Rod w/ Hex ut Threaded Rod w/ Heavy Hex ut ote: All washers need to meet the minimum thickness requirements in Table B2, or a reduced A brg. needs to be calculated according to D