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1 Thank you for joining our live webinar today. We will begin shortly. Please standby. Thank you. Need Help? Call ReadyTalk Support: Today s audio will be broadcast through the internet. Alternatively, to hear the audio through the phone, dial (800)

2 Today s live webinar will begin shortly. Please standby. As a reminder, all lines have been muted. Please type any questions or comments through the Chat feature on the left portion of your screen. Today s audio will be broadcast through the internet. Alternatively, to hear the audio through the phone, dial (800) AISC is a Registered Provider with The American Institute of Architects Continuing Education Systems (AIA/CES). Credit(s) earned on completion of this program will be reported to AIA/CES for AIA members. Certificates of Completion for both AIA members and non- AIA members are available upon request. This program is registered with AIA/CES for continuing professional education. As such, it does not include content that may be deemed or construed to be an approval or endorsement by the AIA of any material of construction or any method or manner of handling, using, distributing, or dealing in any material or product. Questions related to specific materials, methods, and services will be addressed at the conclusion of this presentation. 8.

3 Copyright Materials This presentation is protected by US and International Copyright laws. Reproduction, distribution, display and use of the presentation without written permission of AISC is prohibited. The 016 The information presented herein is based on recognized engineering principles and is for general information only. While it is believed to be accurate, this information should not be applied to any specific application without competent professional examination and verification by a licensed professional engineer. Anyone making use of this information assumes all liability arising from such use. Course Description Columns usually transfer their load to a supporting foundation through base plates. This session will present the design process for base plates according to the AISC Specification, Manual, and Design Guide 1. Base plates that are required to carry compression, tension and bending will be presented. The stress distribution on base plates will be explained for different load cases, and alternate stress distributions will be considered. 8.3

4 Learning Objectives Gain an understanding of AISC Specification, Manual, and Design Guide 1 provisions that govern design of column base plates Determine the stress distribution on a column bearing plate Design column base plates concentrically loaded in compression and tension Design column base plates loaded with small and large moments based on AISC Specification for Structural Steel Buildings Lesson 8 Presented by Louis F. Geschwindner, Ph.D., P.E. Emeritus Professor at Penn State University Former Vice-President at AISC 8 8.4

5 based on AISC Specification for Structural Steel Buildings Night School 10 Lesson Lesson 8 This lesson will address column base plates Those subjected to concentric compressive and tensile forces will be addressed Small moment base plates will be treated Large moment base plates will be designed Alternate stress distributions will be considered

6 Column base plates are the interface between the steel column and the concrete foundation They may be called upon to transfer compression, tension, or moment Design resources include the AISC Manual Part 14 and Design Guide 1 The Specification has limited information 8.11 The Specification J8. Column Bases and Bearing on Concrete φ = 0.65 (LRFD) Ω =.31 (ASD) c The nominal bearing strength on the full area of concrete support P = P = 0.85 f A J8-1 np p c On less than the full area of concrete support P = P = 0.85 f A A A 1.7 f A c np p c 1 1 c 1 1 J

7 The Specification f c = compressive strength of concrete (ksi) A = area of steel concentrically bearing on a 1 concrete support (in. ) A = maximum area of supporting concrete geometrically similar and concentric (in. ) Concrete Column Base Plate 8.13 The Specification F11. Rectangular Bars and Rounds 1. Limit state of flexural yielding M = M = F Z 1.6M = 1.5M F11-1 n p y y y φ b = 0.9 (LRFD) Ω = 1.67 (ASD) b For a rectangular section, the shape factor is always 1.5. Limit state of lateral torsional buckling does not apply for bending about minor axis

8 The Manual Part 14 provides basic equations for determining the strength of a base plate when loaded in compression It also provides recommendations for details of construction and installation Tension and moment resisting base plates are briefly mentioned There are no design examples 8.15 Design Guide 1 Base Plate and Anchor Rod Design, second edition Discusses materials, fabrication, installation and repairs Addresses the base plate, the anchor rods, and the concrete to which it is all attached. Includes derivation of some equations and examples of design

9 Basis of design There are two parts to the design of the base plate 1. Determination of the plan dimensions (the area) which is a function of the bearing stress permitted on the concrete. Determination of the plate thickness as a flexural member based on actual bearing stress 8.17 We will consider several types of base plates Concentrically loaded in compression Concentrically loaded to resist tension Eccentrically loaded with small moment Eccentrically loaded with large moment

10 OSHA Safety Standards for Steel Erection requires 1. A minimum of four anchor rods in a column base plate. A minimum moment strength to resist a 300 pound eccentric gravity load 3. This load at minimum eccentricity of 18 in. from the extreme outer face of the column in each direction 8.19 OSHA These OSHA standards can be met with ¾ in. anchor rods on a 4 in. by 4 in. pattern 4 in. ¾ in. anchor rods A b = 0.44 in. F1554 Gr 36 4 in. F = 58 ksi F u nt = 0.75F = = 43.5 ksi u φ r =φ F A = = 14.4 kips n nt b Available moment strength M c = φ r 4.0 = = 115 in.-kips n Required moment strength M = = 19. in.-kips u J3-1 Half the depth of a W

11 Concentrically loaded b f P rp d N P rp = f BN rp B f = required bearing stress Pressure distribution for compression rp 8.1 Concentrically loaded Assumed bending lines 0.95d 0.95d N m 0.80b f B 0.80b f n Cantilever dimensions N 0.95d m = B 0.80b f n =

12 Concentrically loaded Assumed bending lines 0.95d N λ n =λ db f b f B For small base plates Cantilever dimensions 8.3 Concentrically loaded X λ= X If X is greater than 0.64, λ will be equal to 1.0 For all W-shapes 4db f P 4db r X f = ( d + b ) P f c ( d b ) + f required strength Pr If = 0.90 then λ = 1.0 available strength P c λ n =λ db f 4 To start, assume λ=1. If λn is the largest cantilever, then calculate and use λ Cantilever dimensions

13 Concentrically loaded Flexural strength of plate assuming a 1.0 in. width of plate from Eq. F t p Mn = M p = FyZ = F y For a uniform bearing pressure, the required moment strength is M = r f l rp 4 where l = max( m, n, λn ) f = required bearing stress rp l 8.5 Concentrically loaded Setting the available strength equal to the required strength LRFD t f l p rp P l u ; P P φ F u u y = = tp = l = l BN φfybn FyBN ASD This is my preferred format F t f l y p rp P l a ; ΩP P = = t a a p = l = l Ω BN FyBN FyBN

14 Example 1 Determine the required base plate on a concrete pedestal approximately the same size, for a W1x96 column which has a nominal compressive strength of P n = 360 kips. LRFD A d Pu = 0.9P f n up =φc ( 0.85 f c ) A1 = 0.9( 360) = ( 3.0) 1.0 = 34 kips = 1.66 ksi b f f c = 3.0 ksi Fy = 36 ksi N bf = 1. in. d = 1.7 in. B 8.7 Example 1 Required area (LRFD) A 1req Pu 34 = = = 195 in. f 1.66 up To get as close as possible to the minimum thickness, set m = n ( N 0.95d) ( B 0.80b f ) m= = n= N = B 0.80b d f = B = B

15 Example 1 Required area (LRFD) If the base plate were square, B = N = 195 = 14.0 So try B = 13, N = 15.5 d = 1. in. A b f = 1. in. B = 13.0 in. N = 15.5 in. 1 = = 0 in ( 1.7) m = = 1.7 in ( 1.) n = = 1.6 in. 1.7( 1.) λ n = ( 1.0) = 3.11 in. 4 Since λn is critical, determine λ 8.9 Example 1 For this W1x96 and base plate Therefore P = P = 34 kips r c u p P = P = = 334 kips 4db f P 4( 1.7)( 1.) r 34 X = = = 0.97 ( d b ) P ( ) c f X 0.97 λ= = = X Since X > 0.64 we should have expected that λ would be

16 Example 1 Determine base plate thickness (LRFD) t p Final design 34 Pu = l = 3.11 = 0.98 in. φ FBN b y PL1.0 in. x 13.0 in. x 1.0 ft 3½ in. Since no anchor rod forces exist, use the OSHA minimum four ¾ in.-diameter rods, F1544, Grade 36, rods in the 4 in. pattern we calculated earlier 8.31 Example 1 This column is a lightly loaded member, equivalent to a pin ended column with a length of 36 ft. and KL/r y = 140 Thus, a small base plate was required and the interior cantilever was the controlling factor

17 Example 1 ASD ASD For this example, required strength is based on the nominal strength of the column In all provisions of the Specification, available strength for LRFD is always 1.5 times the available strength for ASD Thus, this example will give the same results for ASD as it did for LRFD as shown in the next few slides 8.33 Example 1 ASD Determine the required base plate on a concrete pedestal approximately the same size, for a W1x96 column which has a nominal compressive strength of P n = 360 kips. ASD A d Pa = Pn Ω fap = ( 0.85 f c Ωc ) A1 = = 0.85( 3.0) = 16 kips = 1.10 ksi b f f c = 3.0 ksi Fy = 36 ksi N bf = 1. in. d = 1.7 in. B

18 Example 1 ASD Required area A 1req Pa 16 = = = 196 in. f 1.10 ap To get as close as possible to the minimum thickness, set m = n ( N 0.95d) ( B 0.80b f ) m= = n= N = B 0.80b d f = B = B Required area Example 1 ASD If the base plate were square, B = N = 196 = 14.0 So try B = 13, N = 15.5 d = 1. in. A b f = 1. in. B = 13.0 in. N = 15.5 in. 1 = = 0 in ( 1.7) m = = 1.7 in ( 1.) n = = 1.6 in. 1.7( 1.) λ n = ( 1.0) = 3.11 in. 4 Since λn is critical, determine λ

19 Example 1 ASD For this W1x96 and base plate Therefore P = P = 16 kips r c a p P = P = = kips 4db f P 4( 1.7)( 1.) r 16 X = = = 0.97 ( d b ) P ( ) c + + f X 0.97 λ= = = X Since X > 0.64 we should have expected that λ would be Example 1 ASD Determine base plate thickness t p Final design ( 1.67)( 16) ( 36)( 13)( 15.5) ΩPa = l = 3.11 = 0.98 in. FBN y PL13.0 in. x 1.0 ft 3½ in. x 1.0 in. Since no anchor rod forces exist, use four ¾ in.-diameter rods, F1544, Grade 36, L= 1 in. What we showed previously would satisfy OSHA

20 Example The column of Example 1 will now have a nominal compressive strength, P n = 1100 kips. Determine the required base plate on a concrete footing sufficient to permit the upper limit on bearing stress b f f c = 3.0 ksi LRFD Pu = 0.9Pn = 0.9( 1100) = 990 kips A d =φ ( 0.85 f ) A1 = = 3.3 ksi φ 1.7 f up c c f ( ) c c B N Fy = 36 ksi bf = 1. in. d = 1.7 in Example Required area (LRFD) A 1req Pu 990 = = = 98 in. f 3.3 up To get as close as possible to the minimum thickness, set m = n ( N 0.95d) ( B 0.80b f ) m= = n= N = B 0.80b d f = B = B

21 Example Required area (LRFD) If the base plate were square, B= N = 98 = 17.3 So try B = 16.5 in., N = 18.5 in. A b f = 1. in. 1 = = ( 1.7) d = 1. in. N = 18.5 in. B = 16.5 in. m = = 3. in ( 1.) n = = 3.37 in λ n = ( 1.0) = 3.11 in in. So n = 3.37 in. is critical Example For this W1x96 and base plate Pr = Pu = 990 kips P = P = = 1010 kips c p Determine base plate thickness (LRFD) t p 990 Pu = l = 3.37 = 1.51 in. φ FBN b y Use a PL1-3/4 in. x 16½ in. x 1.0 ft 6½ in. base plate on a minimum concrete pad to be determined next

22 Example The maximum permitted bearing stress is attained when where P = P= 0.85 fa A A 1.7 fa J8- np p c 1 1 c 1 A A which is A A A = maximum area of the portion of the supporting surface that is geometrically similar to and concentric with the loaded area 8.43 Example The question has always been, what is geometrically similar? For a square base plate it means a square of concrete, that s easy For an 18 in. x 18 in. base plate, A 1 =34 in. The supporting concrete must have 4 times the area so its dimensions are 36 in. x 36 in. and A =196 in. A A = 1 34 = =

23 Example However, when the base plate is rectangular, we have to think about how the load distributes out from the base plate through the concrete From ACI 318, use a rise:run of 1: in all directions until you hit an edge 1 Thus, increase the dimension equally in all directions 8.45 Example If we are to add the same dimension on both sides and the area is to be 4 times A 1 then B+ x N + x = 4BN Solving for x gives 3 ( ) x 3( 16.5)( 18.5) ( 5.457) x B N x BN x + + = + + = x =+ N x/ x/ x/ B x/

24 Example For our 16½ in. x 18 ½ in. base plate add 17½ in. to each dimension Thus the concrete will be 34 in. x 36 in. 1 1 A = = 305 in. A = = 14 in. A A 14 = = Tensile Base Plate A column that will be required to resist a net uplift force will need a base plate designed to resist tension The anchor rod tensile strength is a function of the rod material strength The pull out strength of the anchor rod is a function of the concrete strength, embedment length and rod placement

25 Tensile Base Plate We will not address the concrete side of this problem although there is guidance in DG 1 and others We will look at determining the base plate thickness and the column to plate weld As for the compression base plate, the flexural strength of the base plate is based on Eq. F Tensile Base Plates The flexural strength of the plate will be a function of the effective width resisting the moment, b eff From F11.1 Yielding beff tp Mn = M p = FyZ = F y For a single rod force, T r, the required moment strength is M r 4 F11-1 = Tl where l = the distance from the rod r to the face of the column flange or web l

26 Tensile Base Plate Rods may be placed outside the flanges or between the flanges The width of plate resisting bending can be established by distributing the force on a 45 angle from the rod, assuming one way bending Bending will be resisted by the width b eff for each rod. Can not overlap or must use reduced b eff b eff 4 in. 4 in Example 3 The W1x96 column of Example 1 is also called upon to resist an LRFD net uplift force of 50 kips. The ¾ in. F1554 anchor rods will be placed in a 4 in. square as shown. t = in. b w eff 4 tw = = ( ) = 3.45 in. 50 T u = = 1.5 kips < 14.4 kips 4 b eff s+ b = = 7.45 < h= 9.74 in. eff 4 in. 4 in

27 Required moment M u Nominal moment Required thickness Example = Tul = = 1.6 in.-kips b t 3.45t M F t 4 4 t p eff p p n = y = 36 = 31.1 p n φ M = = tp b eff 4 in. = = in. 0.9( 31.1) Use a 1.0 in. plate 4 in Example 3 Determine the required weld to connect the column to the base plate The maximum weld load is the force per rod divided by the effective width per rod Tu 1.5 ru = = = 3.6 kips/in. b 3.45 eff The minimum weld size is based on the thinner part being connected, the column web, t w = in. (Table J.4 min fillet weld ¼ in.)

28 Example 3 Weld strength with an E70 electrode and using the 50% strength increase for load perpendicular to the weld The increase for directionality 1.5 R = 0.6F sin θ A n EXX we ( 1 4) kips/in. = = 11.1 kips/in. φ R = = 8.33 > 3.6 kips/in. n J-4, J-5 We have more than enough weld strength so use the minimum fillet weld on each side of the web 8.55 Example 3 The column web stress is the force per inch divided by the area of the web per inch 3.6 f 13. ksi 0.9F 0.9( 36) 3.4 ksi uw = = < y = = J To complete the design, the rods in the concrete must be evaluated

29 Example 3 ASD If the W1x96 column of Example 1 is called upon to resist an ASD net uplift force of 34 kips, the solution is similar for the 4 in. square pattern t = in. w b eff 4 tw = = ( ) = 3.45 in. 34 T a = = 8.5 kips < 9.60 kips 4 b eff 4 in. 4 in Example 3 ASD Required moment M a = Tal = = 14.7 in.-kips Nominal moment b t 3.45t M F t 4 4 eff p p n = y = 36 = 31.1 p Required thickness M t t n Ω= 31.1 p 1.67 = 18.6 p = 14.7 b eff 4 in t p = = in. Use a 1.0 in. plate in. 8.9

30 Example 3 ASD Determine the required weld to connect the column to the base plate The maximum weld load is the force per rod divided by the effective width Ta 8.5 ra = = =.46 kips/in. b 3.45 eff The minimum weld size is based on the thinner part being connected, the column web, t w = in. (Table J.4 min fillet weld ¼ in.) 8.59 Example 3 ASD Weld strength with an E70 electrode and using the 50% strength increase for load perpendicular to the weld The increase for directionality 1.5 R = 0.6F sin θ A n EXX we ( 1 ) J-4, J-5 = = 11.1 kips/in. 4 Rn 11.1 = = 6.65 kips/in. >.46 kips/in. Ω 1.67 We have more than enough weld strength so use the minimum fillet weld on each side of the web

31 Example 3 ASD The column web stress is the force per inch divided by the area of the web per inch (.46) Fy 36 f aw = = 8.95 ksi < = = 1.6 ksi J To complete the design, the rods in the concrete must be evaluated 8.61 Base plates are designed to resist moment as either small moment or large moment base plates For small moment base plates, the moment is resisted fully by the bearing stress on the concrete For large moment base plates, the moment must rely on a contribution from tension in the anchor rods

32 For both small and large moment base plates, the compressive bearing stress will be assumed to have a uniform distribution The distinction between small and large moment base plates will be established by determining a critical eccentricity If this critical eccentricity is exceeded that must be treated as a large moment base plate 8.63 Small moment base plate model e P r M r P r = M e = P r r T = 0 ɛ qy q T = 0 ɛ qy q Y/ Y/ N Y/ Y/ N

33 Small moment base plate model e P r q For equilibrium, the line of action of the load, P r, and the bearing force, qy, must coincide so that e=ε and Pr = fpby = qy If e exceeds ɛ max the applied load, P r, can not be resisted by qy alone. N Y ε= as Y decreases, ɛ increases. Y reaches its T = 0 ɛ qy smallest value when q reaches its maximum. Y/ Y/ Thus Pr N Ymin = where qmax = fp(max) B qmax 8.65 Small moment base plate model T = 0 e ɛ P r qy Y/ Y/ N q Since ɛ reaches its maximum when Y is at its minimum N Ymin ε max = N Pr = qmax Therefore, N P e r critical =ε max = q max Thus, if e e critical there is no tension in the rod and the base plate will be treated as a small moment base plate

34 Example 4 Design a base plate on a concrete pedestal for a W1x96 to resist the loading shown P r PL = 160 M r PD = 100 M M L D kips Fy = 36 ksi kips f c = 4 ksi = 400 in.-kips For LRFD = 50 in.-kips Pu = 1.( 100) + 1.6( 160) = 376 kips M u = 1.( 50) + 1.6( 400) From Example = 940 in.-kips Try B = 16.5 in., N = 18.5 in Example 4 Determine the eccentricity of load M u 940 e = = =.5 in. P 376 u Determine the maximum bearing stress f A =φ ( 0.85 f ) = 0.65( 0.85( 4.0) )( 1.0) =.1 ksi A up(max) c c 1 Determine the maximum force per inch qmax = fup(max) B= = 36.5 kips/in

35 Example 4 Determine the critical eccentricity P u N ecritical = = = 4.10 in. > e=.5 in. q 36.5 max Since e < e critical this base plate qualifies as a small moment base plate Determine the length of bearing Y from N Y e =ε=, Y = N e= = 13.5 in Example 4 Determine the actual bearing stress Pu 376 q = = = 7.9 < qmax = 36.5 kips/in. Y 13.5 f ksi.1 up = = < fup(max) = ksi 16.5 To determine the plate thickness, first determine m and compare to Y N 0.95d m= = = 3. in. < Y =

36 Example 4 Since Y > m, we can treat determination of the base plate thickness as we did for concentrically loaded base plate Determine n and compare to m B 0.80b f n = = = 3.37 in. Note that these are the same cantilever dimensions we had in Example, since we have the same column and the same base plate. Thus n controls 8.71 Example 4 Base plate thickness is determined similarly to Example using l = n and substituting Y for N t p ( 376) Pu = l = 3.37 = 1.09 in. φ FBY b y Use a PL1¼ in. x 16½ in. x 1.0 ft 6½ in. base plate on a concrete pedestal of the same dimensions

37 Example 4 If Y < m, the moment on the cantilever of length m is not for a uniformly loaded cantilever If we set the required moment for this N e m Y P r q case equal to a simple cantilever moment of equivilant length l and solve for l Y qleq Mr = qym = leq = my Y we can compare this leq to m and n and determine the plate thickness for the longest cantilever eq eq 8.73 Example 4 ASD Design a base plate on a concrete pedestal for a W1x96 to resist the loading shown P r PL = 160 M r PD = 100 M M kips kips = 400 in.-kips = 50 in.-kips L D From Example Fy = 36 ksi f = 4 ksi c For ASD Pa = = 60 kips M a = = 650 in.-kips Try B = 16.5 in., N = 18.5 in

38 Example 4 ASD Determine the eccentricity of load M a 650 e = = =.5 in. P 60 a Determine the maximum bearing stress ( 0.85 f ) ( 0.85( 4.0 c A )) f = = ( 1.0 ) = 1.47 ksi ap(max) Ω A 1.31 Determine the maximum force per inch qmax = fap(max) B= = 4.3 kips/in Example 4 ASD Determine the critical eccentricity P a N ecritical = = = 3.90 in. > e=.5 in. q 4.3 max Since e < e critical this base plate qualifies as a small moment base plate Determine the length of bearing Y from N Y e =ε=, Y = N e= = 13.5 in

39 Example 4 ASD Determine the actual bearing stress Pa 60 q= = = 19.3 < qmax = 4.3 kips/in. Y 13.5 f ap = 19.3 = 1.17 ksi < f 1.47 ap(max) = ksi 16.5 To determine the plate thickness, first determine m and compare to Y N 0.95d m= = = 3. in. < Y = Example 4 ASD Since Y > m, we can treat determination of the base plate thickness as we did for concentrically loaded base plate Determine n and compare to m B 0.80b f n = = = 3.37 in. Note that these are the same cantilever dimensions we had in Example and n controls

40 Example 4 ASD Base plate thickness is determined as it was in Example except that Y is substituted for N t p 1.67 ( 60) ( 36)( 16.5)( 13.5) ΩbPa = l = 3.37 = 1.11 in. FBY y Use a PL1¼ in. x 16½ in. x 1.0 ft 6½ in. base plate on a concrete pedestal of the same dimensions 8.79 Large moment base plate model f f e P r Tension in the anchor rod is a distinguishing feature of large moment baseplates B x M r P r = B x T ɛ q max Y q max T ɛ q max Y q max f+n/-y/ Y/ N f+n/-y/ Y/ N

41 Large moment base plate model f e P r Vertical force equilibrium requires that T = q Y P max r x Summation of moments about B must equal zero B T ɛ q max Y f+n/-y/ Y/ N q max N Y q Y f P e f max + r + = 0 Solving for Y gives N N Pr e Y = f + ± f + q ( + f ) max 8.81 Large moment base plate model f e P r A real solution will only exist for B T x ɛ q max Y f+n/-y/ Y/ N q max if N N Pr e Y = f + ± f + q N Pr e f + q ( + f ) max ( + f ) Therefore, if this inequality is not satisfied, a larger base plate is required max

42 Example 5 Design a base plate on a concrete pedestal for a W1x96 to resist the loading shown P r PL = 160 kips Fy = 36 ksi M r PD = 100 kips f c = 4 ksi M L = 1, 500 in.-kips For LRFD M D = 1, 000 in.-kips Pu = 1.( 100) + 1.6( 160) = 376 kips M u = 1. 1, ,500 = 3, 600 in.-kips Try B = 19.0 in., N = 19.0 in Example 5 Determine the eccentricity of load M u 3, 600 e = = = 9.57 in. P 376 u Determine the maximum bearing stress f A =φ ( 0.85 f ) = 0.65( 0.85( 4.0) )( 1.0) =.1 ksi A up(max) c c 1 Determine the maximum force per inch qmax = fup(max) B= = 4.0 kips/in

43 Example 5 Determine the critical eccentricity P u N ecritical = = = 5.0 in. < e= 9.57 in. q 4.0 max Since e > e critical this base plate must be treated as a large moment base plate Check to be sure that these dimensions will result in a solution for Y 8.85 Example 5 Confirm that the base plate dimensions are sufficiently large f e = 9.57 in. Assumed 1½ in. x B ɛ q max Y T q max P r Assume a 1 ½ in. edge distance, therefore N 19.0 f = 1.5 = 1.5 = 8.0 in. ( + f ) ( + ) f+n/-y/ Y/ Thus, a larger base plate is required N 8.86 N Pr e f + q max <

44 Example 5 A larger base plate is required, try a.0 in. x.0 in. plate q = f B= = max up(max) kips/in. P u N ecritical = = = 7.13 in. < e= 9.57 in. q 48.6 max No change So this is still a large moment base plate. But note that e and e critical are now closer. It is possible that you might increase the plate dimensions sufficiently to make it a small moment base plate so this check is always required Example 5 Confirm that the new base plate dimensions are sufficiently large f e = 9.57 in. P r With 1½ in. N P ( e+ f ) x B T ɛ q max Y f+n/-y/ Y/ q max N.0 f = 1.5 = 1.5 = 9.5 in. f + ( + ) N 8.88 max So the base plate is large enough for a solution r q 8.44

45 Example 5 Determine the bearing length Y N N Pr e Y = f + ± f + q ( + f ) = ± = 0.5 ± 11. = 9.3 in. max Determine the total rod force, T u Tu = qmaxy Pu = = 76.0 kips 8.89 Example 5 Determine the actual bearing stress Pu + Tu q = = = 48.6 kips/in. = qmax = 48.6 kips/in. Y fup = =.1 ksi = fup(max) =.1 ksi.0 To determine the plate thickness, first determine m and compare to Y N 0.95d m= = = 5.0 in. < Y = 9.3 in

46 Example 5 Since Y > m, we can treat determination of the base plate thickness as we did for concentrically loaded base plate Determine n and compare to m B 0.80b f n = = = 6.1 in. Since n > m, as far as compression is concerned, n controls 8.91 Example 5 For the compression side, the base plate thickness is determined similarly to what we have already done, using l = n and Y for N and now the force (P u + T u ) t p ( ) Pu = l = 6.1 =.6 in. φ FBY b y

47 Example 5 For the tension side, the required moment strength must be less than the available moment strength l Required moment strength M u = Tl Here l was conservatively assumed to the flange center line. u Remember, rod force, T u, is the full tension force required, not that for a single rod Example 5 For the tension side, the required moment strength must be less than the available moment strength l Available moment strength for the full width of the base plate, B Bt p bmn bf y φ =φ 4 This assumes that there are a sufficient number of rods that the full width of the plate is engaged

48 Example 5 Solving for the minimum plate thickness based on the tension side l Therefore so φ φ M b n u p bfy u t p Bt 4 M b y Tl Tl u φ F B 8.95 Example 5 For the tension side, determine the cantilever length and the required moment strength l t f N d l = = = 3.60 in. M u = T l = = 74 in.-kips u

49 Example 5 For the tension side, the plate thickness is t p Tl u = = = 1.3 in. φ FB b y l Thus, the compression side controls Use a PL½ in. x in. x 1.0 ft 10 in. base plate on a concrete pedestal of the same dimensions 8.97 The uniform stress distribution is consistent with the procedures adopted by ACI For base plates that are required to resist both an axial load and a moment, it is permissible to use a triangular or trapezoidal stress distribution

50 Alternate stress distribution, small moment q 1 e P r M e = P r r P q + q 1 c = N ( + 1) ( q + q ) N q q N ε= 3 1 N/ N ɛ P c q ε=e 8.99 Alternate stress distribution, small moment N/ N e ɛ P r P c M e = P q r r P q N c = ε= e ( ) ( q ) N q N N ε= = 3 6 Within the Kern or middle third

51 Alternate stress distribution, the real limit N/ f N e ɛ P r P c M e = P q r r q N f c = + P N f ε= 3 3 ε=e Within the Kern or middle third Alternate stress distribution, large moment T c f e Y P r M e = P N/ ɛ q P c N r r P q Y c = N Y ε= 3 q T P P = = Y P c c r r

52 Alternate stress distribution The design approach is similar to that for uniform stress distribution Determine the maximum bearing stress Select a plate size Determine plate thickness based on the moment in the plate due to the stress distribution you have Alternate stress distribution This alternate stress distribution will often require a slightly thicker base plate It may also require slightly smaller anchor rods Generally there is no advantage, either for economy or safety, in using this slightly more complicated approach

53 We have addressed Summary Concentrically loaded base plates experiencing a compressive load Concentrically loaded base plates resisting a tension load Small moment base plates Large moment base plates OSHA minimum requirements An alternate triangular bearing stress distribution Conclusion For the 8 lessons of Night School 10 we have considered beams and columns with shapes/conditions that are just a bit different than most of our experience We have provided guidance so that the designer may feel more comfortable extending their designs to these and other shapes that are less familiar

54 Thank You One East Wacker Drive Chicago, IL Individual Webinar Registrants CEU/PDH Certificates Within business days You will receive an on how to report attendance from: Be on the lookout: Check your spam filter! Check your junk folder! Completely fill out online form. Don t forget to check the boxes next to each attendee s name! 8.54

55 Individual Webinar Registrants CEU/PDH Certificates Within business days Reporting site (URL will be provided in the forthcoming ). Username: Same as AISC website username. Password: Same as AISC website password. 8-Session Registrants CEU/PDH Certificates One certificate will be issued at the conclusion of all 8 sessions. 8.55

56 Access to the quiz: Information for accessing the quiz will be ed to you by Thursday. It will contain a link to access the quiz. COMES FROM [email protected] Quiz and Attendance records: Posted Wednesday mornings. - click on Current Course Details. Reasons for quiz: 8-Session Registrants Quizzes EEU must take all quizzes and final to receive EEU CEUs/PDHS If you watch a recorded session you must take quiz for CEUs/PDHs. REINFORCEMENT Reinforce what you learned tonight. Get more out of the course. NOTE: If you attend the live presentation, you do not have to take the quizzes to receive CEUs/PDHs. 8-Session Registrants Recording Access to the recording: Information for accessing the recording will be ed to you by this Thursday. The recording will be available for two weeks. For 8-session registrants only. COMES FROM [email protected]. CEUs/PDHS If you watch a recorded session you must take AND PASS the quiz for CEUs/PDHs. 8.56