9. Shear Lug Design. 8.2 Calculating Resisting Friction Force The resisting friction force, V f, may be computed as follows: V f.

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1 TECHNICAL CORRECTION October 2006 PIP STE05121 Anchor Bolt Design Guide will develop friction even when the column or vertical vessel is uplift. This downward load can be considered calculatg frictional resistance. Care shall be taken to assure that the downward load that produces frictional resistance occurs simultaneously with the shear load. In resistg horizontal loads, the friction resistance attributable to downward force from overturng moment may be used. The frictional resistance can also be used combation with shear lugs to resist the factored shear load. The frictional resistance should not be used combation with the shear resistance of anchors unless a mechanism exists to keep the base plate from slippg before the anchors can resist the load (such as weldg the washer to the base plate). Note: If the PIP design STE05121 requires weldg the washer to the base plate, pla washers Anchor steel Bolt Design plate (rather Guide than hardened washers) must be specified PIP - to Oct ensure 2006 that a good weld can be produced. 8.2 Calculatg Resistg Friction Force The resistg friction force, V f, may be computed as follows: V f P μ = μp 9. Shear Lug Design Shear Lug Verification Example = normal compression force = coefficient of friction The materials used and the embedment depth of the base plate determe the value of the coefficient of friction. (Refer to Figure F for a pictorial representation.) a. μ = 0.90 for concrete placed agast as-rolled steel with the contact plane a full plate thickness below the concrete surface. b. μ = 0.70 for concrete or grout placed agast as-rolled steel with the contact plate cocidental with the concrete surface. c. μ = 0.55 for grouted conditions with the contact plane between grout and as-rolled steel above the concrete surface. Normally, friction and the shear capacity of the anchors used a foundation adequately resist column base shear forces. In some cases, however, the engeer may fd the shear force too great and may be required to transfer the excess shear force to the foundation by another means. If the total factored shear loads are transmitted through shear lugs or friction, the anchor bolts need not be designed for shear. A shear lug (a plate or pipe stub section, welded perpendicularly to the bottom of the base plate) allows for complete transfer of the force through the shear lug, thus takg the shear load off of the anchors. The bearg on the shear lug is applied only on the portion of the lug adjacent to the concrete. Therefore, the engeer should disregard the portion of the lug immersed the top layer of grout and uniformly distribute the bearg load through the remag height. Process Industry Practices Page 19 of 24

2 PIP STE05121 TECHNICAL CORRECTION Anchor Bolt Design Guide October 2006 The shear lug should be designed for the applied shear portion not resisted by friction between the base plate and concrete foundation. Grout must completely surround the lug plate or pipe section and must entirely fill the slot created the concrete. When usg a pipe section, a hole approximately 2 ches diameter should be drilled through the base plate to the pipe section to allow grout placement and spection to assure that grout is fillg the entire pipe section. 9.1 Calculatg Shear Load Applied to Shear Lug The applied shear load, V app, used to design the shear lug should be computed as follows: V app = V ua - V f 9.2 Design Procedure for Shear Lug Plate Design of a shear lug plate follows (for an example calculation, see Appendix Example 3, this Practice): a. Calculate the required bearg area for the shear lug: A req = V app / (0.85 * φ * fc ) φ = 0.65 b. Determe the shear lug dimensions, assumg that bearg occurs only on the portion of the lug below the grout level. Assume a value of W, the lug width, on the basis of the known base plate size to fd H, the total height of the lug, cludg the grout thickness, G: H = (A req /W) + G c. Calculate the factored cantilever end moment actg on a unit length of the shear lug: M u = (V app /W) * (G + (H-G)/2) d. With the value for the moment, the lug thickness can be found. The shear lug should not be thicker than the base plate: t = [(4 * M u )/(0.9*f ya )] 0.5 e. Design weld between plate section and base plate. f. Calculate the breakout strength of the shear lug shear. The method shown as follows is from ACI , Appendix B, section B.11: V cb = A Vc *4*φ*[f c ] 0.5 where A Vc = the projected area of the failure half-truncated pyramid defed by projectg a 45-degree plane from the bearg edges of the shear lug to the free edge. The bearg area of the shear lug shall be excluded from the projected area. φ = concrete strength reduction factor = 0.85 Page 20 of 24 Process Industry Practices

3 TECHNICAL CORRECTION October 2006 PIP STE05121 Anchor Bolt Design Guide Example 3 - Shear Lug Plate Section Design PLAN V = 40 K u (ULTIMATE) SECTION Process Industry Practices Page A-25

4 TECHNICAL CORRECTION October 2006 PIP STE05121 Anchor Bolt Design Guide EXAMPLE 3 - Shear Lug Plate Section Design Design a shear lug plate for a 14-. square base plate, subject to a factored axial dead load of 22.5 s, factored live load of 65 s, and a factored shear load of 40 s. The base plate and shear lug have f ya = 36 and f c ' = 3. The contact plane between the grout and base plate is assumed to be 1. above the concrete. A 2-ft 0-. square pedestal is assumed. Ductility is not required. V app = V ua V f = 40 (0.55)(22.5) = 27.6 s Bearg area = A req = V app / (0.85 φ f c ') = 27.6 s / (0.85*0.65*3 ) = On the basis of base plate size, assume the plate width, W, will be 12. Height of plate = H = A req / W + G = / = Use 3. Ultimate moment = M u = (V app / W) * (G + (H G)/2) = (27.6 s / 12.) * (1. + (3.-1.)/2) = 4.61 k-. /. Thickness = t = [(4 * M u )/(φ* f ya )] ½ = ((4* )/(0.9*36 )) ½ = Use This 12-. x 3-. x plate will be sufficient to carry the applied shear load and resultg moment. Design of the weld between the plate section and the base plate is left to the engeer. Check concrete breakout strength of the shear lug shear. Distance from shear lug to edge of concrete = ( ) / 2 = A V = 24 * ( ) (12 * 2) = V cb = A Vc *4*φ*[f c '] 0.5 = 303 * 4 * 0.85 * [3000] 0.5 = lb = 56.4 s > 27.3 s Process Industry Practices Page A-26

5 Engeer: Javier Encas, PE GEOMETRY Column Section... W8X31 Width Length Column... Plate... Concrete Support Wp1 Wp2 Rod Offset... Thickness of Grout Lp Lp FACTORED LOADS (LRFD) Vertical Load P... Bendg Moment M... Horizontal Load V... Design Eccentricity e Design Eccentricity Is < L/6 MATERIALS Plate Steel Strength Fy... Pier Concrete Strength f'c k-ft Cantilever Model Bearg Stress fp... Critical Long m Critical Short n Plate Thickness tp AXIALLY LOADED PLATES Thornton Model Bearg Strength ϕfp... Critical Int λn'. Design Plate... Plate Thickness tp k-/ Blodgett Method Max. Bearg Stress fp... Critical Section Critical Section Moment due to Rod Tension Design Plate... Plate Thickness tp BASE PLATES WITH MOMENT k-/ k-/ k-/ DeWolf Method Max. Bearg Stress fp... Critical Section Critical Section Moment due to Rod Tension Design Plate... Plate Thickness tp k-/ k-/ k-/ 1

6 Engeer: Javier Encas, PE ANCHORAGE DESIGN Rod Material Specification... F (4) Rods, fya = 36.0, futa = 58.0 Anchor Rod Size.. 1" diam. x 12.0 emb. Concrete Is Uncracked at Service Load Level Tension Analysis () Total Tension Force Nug... Tension Force per Rod Nui... No Reforcg Bars Provided Failure Mode ϕ Nn Nu / ϕnn Steel Strength Nsa Rebars Strength Nrg 0.75 N.A. N.A. Conc. Breakout Ncbg Pullout Strength Npn Side Blowout Nsbg 0.70 N.A. N.A. Nu / ϕnn Tension Design Ratio... 0 SUMMARY OF RESULTS Design Plate... Plate Thickness tp... Max. Bearg Stress fp... Bearg Strength ϕfp... fp / ϕfp Design Ratio... DESIGN IS DUCTILE k-/ 4 Shear Analysis () Shear Taken by Shear Lug + Friction Total Shear Force V... 4 Compression Force C... Friction Coefficient... Friction Strength ϕfr... V / ϕfr Shear Friction Ratio... Shear Force Lug... Shear Lug Width Wl... Shear Lug Height Hl... Shear Lug Thickness tl Failure Mode ϕ Vn Vu / ϕvn Conc. Bearg Vcbr Conc. Breakout Vcb V / ϕvn Shear Design Ratio Steel design... Base plate design... Anchorage design... DESIGN CODES AISC (14th Ed.) AISC Design Series # 1 ACI Appendix D 2

7 Engeer: Javier Encas, PE Tension Breakout Shear Breakout 3

8 Engeer: Javier Encas, PE Column Section... Vertical Load P... Width Length Bendg Moment M... k-ft Column... Horizontal Load V... Plate... Design Eccentricity e... Concrete Support Wp1 Wp2 Lp1 Lp2 Design Eccentricity Is < L/6 Rod Offset... Thickness of Grout... Plate Steel Strength Fy... Pier Concrete Strength f'c Bearg stress 22.5 / (14.0 * 14.0) = 0.1 Bearg strength = 0.85 * 3.0 * = 4.4 ACI Under-strength factor ϕ = 0.65 ACI Bearg strength ratio = = * 2.8 = 4 < 1.0 Critical section m = Critical section n = [ 0.5 * ( *8.0) = 3.2 AISC-DG# * ( *8.0) = * 8.0 * 8.0 ] * 4 = 4 AISC-DG# ( )² = + - = 0.20 = = 2.0 Controllg section Plate moment Max (3.2, 3.8, 0.20 * 2.0) = * 3.8² / 2 = 0.8 k-/ Plate thickness = 3.8 * = 0.32 AISC-DG#

9 Engeer: Javier Encas, PE Eccentricity * 12 / 22.5 = < L / 6 = 14.0 / 6 = 2.3 Max bearg stress = * * * * 14.0² = 0.1 M bearg stress = = * * * * 14.0² = 0.1 Bearg at critical section / 14.0 * ( ) = 0.1 Moment due to bearg Mb = 0.1 * 3.2² / 2 * ( ) * 3.2² / 3 = 0.6 k-/ Plate thickness = = 0.27 AISC-DG#

10 Engeer: Javier Encas, PE Rod Material Specification... Anchor Rod Size... 1" diam. x 12.0 emb., Total tension force Nu =, F , Use (4) Rods, fya = 36.0, futa = 58.0 Ase = 0.61 ², Abrg = 1.50 ² # of tension rods = 0, Tension force per rod Nui = ACI D.5 - Steel strength of anchors tension ACI D.5.1 Steel strength * 58.0 = 35.1 ACI Eq. (D-2) Under-strength factor ϕ = 0.75 Steel strength ratio = = 0.75 * 35.1 ACI D.4.3 = 0 < 1.0 ACI D Concrete breakout strength of anchors tension ACI D.5.2 No Reforcg Bars Provided Effective embedment / 1.5 = ACI D Anchor group area Anc = ( ) * ( ) = ² ACI D Sgle anchor area 9 * (11.7)² = ² Eq. (D-5) Sgle anchor strength = 24 = 52.4 Eq. (D-6) Eccentricity factor 1.00 (No eccentric load) ACI D Edge effects factor = = 0.81 ACI D * 11.7 Crackg factor 1.25 (Uncracked concrete at service load level) ACI D Breakout strength * 0.81 * 1.25 * 52.4 = 25.0 Eq. (D-4) Under-strength factor ϕ = 0.70 Breakout strength ratio = = 0.70 * 25.0 ACI D.4.3 = 0 < 1.0 ACI D Breakout strength ratio controls (0 < 0) ACI D Concrete pullout strength of anchors tension ACI D.5.3 Sgle anchor strength 8 * 1.50 * 3.0 = 36.0 ACI Eq. (D-14) Crackg factor 1.25 (Uncracked concrete at service load level) ACI D Pullout strength 1.25 * 36.0 = 50.4 ACI Eq. (D-13) Under-strength factor ϕ = 0.70 Pullout strength ratio = = 0.70 * 50.4 ACI D.4.3 = 0 < 1.0 ACI D

11 Engeer: Javier Encas, PE - Concrete side-face blowout strength of anchors tension ACI D.5.4 Side-face blowout Nsbg = N.A. (Embed < 2.5 Ca₁, 12.0 < 2.5 * 6.5 = 16.3) ACI D Tension Design Ratio = = 0 < 1.0 ACI D Shear resisted by Shear Lug + Friction Total shear force Vu = 4, Compression force C = 22.5, Friction coeff. = 0.20 ACI D.5 Friction strength Friction strength ratio = = 0.75 * 22.5 * 0.20 = = 1.00 <= * 3.4 Shear lug width Wl = 12.0, Shear lug height Hl = 3.0, Shear lug thickness tl = Steel strength of lug flexure Lug moment 36.6 * (1.0 + ( ) / 2) = 73.3 k- Lug flexural strength 12.0 * 36 * 1.0 ² / 4 = k- AISC F.11 Under-strength factor ϕ = 0.90 AISC F.1 Flexural strength ratio = = - Steel strength of lug shear Shear force lug * = 36.6 = 0.75 < 1.0 AISC B3.4 Lug shear strength 0.6 * 36 * 12.0 * 1.0 = AISC Eq. (G2.1) Under-strength factor ϕ = 0.90 AISC G.1 Shear strength ratio = = - Weld strength shear lug * = 0.16 < 1.0 AISC B3.4 Shear lug fillet weld size a = ) (M size = ) NG AISC J2.2a Shear per unit width Tension per unit width 36.6 / (2 * 12.0) = / [( * / 3) * 12.0] = 5.2 Resultant per lug width R = = = Weld stress 0.6 * 70 * ( ) = 63.0 AISC Eq. (J2.5) Weld strength Under-strength factor ϕ = 0.75 Weld strength ratio = = - Concrete bearg strength of lug shear Lug bearg area 63.0 * * * 2 * 12.0 = * AISC J3b.4 = 0.65 < 1.0 AISC B3.3 ( ) * = 24.0 ² Lug bearg strength 1.3 * 3.0 * 24.0 = 93.6 ACI 349 D.4.6 Under-strength factor ϕ = 0.65 ACI Bearg strength ratio = = * 93.6 = 0.60 < 1.0 ACI

12 Engeer: Javier Encas, PE - Concrete breakout strength of lug shear ACI 349 D.11.2 Lug breakout area Avc = ( ) * ( ) = ² Breakout strength = * 4 = 57.8 Under-strength factor ϕ = 0.75 Breakout strength ratio = = * 57.8 ACI D.4.3 = 0.84 < 1.0 ACI D Shear Design Ratio = = 0.84 < 1.0 ACI D Anchorage design is ductile Steel design... Base plate design... Anchorage design... AISC (14th Ed.) AISC Design Series # 1 ACI Appendix D 5