Prelmnary Desgn of Moment-Resstng Frames Preprnt Aamer Haque Abstract A smple method s developed for prelmnary desgn of moment-resstng frames. Preprnt submtted to Elsever August 27, 2017
1. Introducton 1.2. Frame Parameters and Geometry 1.1. Background Informaton The lateral load resstng system of a buldng may consst of a frame, truss, shear wall, or some combnaton thereof. Moment-resstng frames consst of beams and columns wth rgd connectons. The lateral loads are ressted prmarly by flexural deformatons of the beams and columns. Ths system s often employed n multstory buldngs of moderate heght.e. up to 30 stores). Informaton on lateral load systems s found n Councl on Tall Buldngs and Habtat [3], Smth and Coull [6], Taranath [7]. We shall focus our dscusson on steel buldngs. Furthermore, we assume frames are constructed as regular grds of prsmatc members. Irregular frames are not consdered. Informaton on prelmnary desgn of moment-resstng frames s seldom found n desgn texts. Iyengar [5] descrbes a method of prelmnary desgn based on approxmate 1st order analyss of lateral loads. The same method s explaned n more detal n Smth and Coull [6]. We extend the method to nclude 2nd order P- effects. The approxmate 2nd order analyss procedure of Haque [4] s used as the bass of the desgn procedure. The method has the followng assumptons and features: Lateral deflecton s caused by flexural deformaton of the frame members. The mdponts of the beams and columns are the ponts of contra-flexure. For frst story columns, the pont of contra-flexure s located at 2/3 the heght of the columns. The 2nd order P- effect s ncluded n the desgn procedure. The maxmum permssble 1st order story drft s computed. The desgner chooses the fracton of 1st order story drft produced by column rotatons and beam rotatons. The columns and beams are desgned based on ther contrbuton to 1st order story drft. The ndvdual columns are checked for stablty usng Euler bucklng crtera. To smplfy the desgn, we use the same column secton and beam secton for an entre story. A 10-story buldng s used as an example for the desgn procedure. It must be emphaszed that the desgn method s for prelmnary desgn only. The desgn should be refned snce the prelmnary desgn may be under-conservatve. Furthermore, desgn optmzaton procedures can be appled to produce a more effcent desgn. Accurate nonlnear structural analyss software must be used to verfy that the fnal desgn s n complance wth the approprate buldng codes. Fgure 1: Frame geometry Member Story Lne Length I Connects Beam j L j I b,j Column lne j to j +1 Column k h I c,k Beam lne 1 to Table 1: Frame member parameters In order to properly defne a frame, the geometry and secton propertes must be defned. The frame has N s stores, N b bays, and N c columns between floors. Notce that N c = N b + 1. Let h be the heght of story and L j be the wdth of bay j. The moment of nerta s the most mportant secton property. I,j b s the moment of nerta of the beam on top of story n bay j. Smlarly, I,k c s the moment of nerta of the column on story along column lne k. We assume that Young s modulus E s constant throughout the frame. Fgure 1 llustrates the frame geometry. The top of a story s a beam lne of the same ndex as the story ndex. Table 1 summarzes the beam and column nformaton of fgure 1. 2
1.3. Loads and Forces Fgure 2 shows the lateral and gravty loads appled to the frame. Lateral loads are assumed to act at the jonts on one sde of the frame. Gravty loads may occur anywhere long the beams. Such loads are not dsplayed n fgure 2. Instead, we lump these loads to the jonts snce we are not nterested n beam deflectons. Computaton of lateral and gravty loads s descrbed n ASCE [2]. Also, the approprate load factors are appled n accordance wth the desgn code beng used. The free body dagram n fgure 3 cuts the frame through the ponts of contra-flexure of story. Thus there are no moments n the columns at the cut locatons. Moment equlbrum s acheved by axal loads n the columns due to the lateral loads. The total shear n the story s computed usng horzontal equlbrum: N c V = V,k = k=1 N s m= w m 1.1) Equaton 1.1) does not provde a way of computng the shear V,k n the columns. Further assumptons are requred to compute the column shears. The total axal load n the column located at story and column lne k s: P t),k = P,k +P w),k 1.2) where P,k s the total gravty load and P w),k s the axal load due to the lateral loads. The gravty loads n a column are computed as: Fgure 2: Frame loads N s P,k = p m,k 1.3) m= Computaton of the axal loads assocated wth lateral loads s more complex and wll now be descrbed. We shall use the cantlever method of analyss. Assume that all columns on a story have the same cross-sectonal area. Then we locate the neutral axs x along the horzontal cut n fgure 3) n the followng way: { 0 f k = 1 x k = k m=2 L k 1 f k > 1 x = 1 N c N c k=1 The dstances of columns from the neutral axs are computed usng: r k = x k x, k = 1,...,N c Notce that r 1 = x. The heghts of the lateral forces above the cut are: y ) m = h 2 + m q=+1 h q, x k m =,...,N s Fgure 3: Free body dagram for story 3
Preprnt Fgure 4: Free body dagram for story Fgure 4 shows the forces nvolved for moment equlbrum of the lateral forces. The appled moment s: M ) ext = N s m= w m y ) m The column forces are assumed to be a lnear functon of dstance from the neutral axs. We choose the frst column forcep w),1 as a reference. The column forces can be computng usng: ) P w),k = rk Pw),1 1.4) The nternal moment s computed as: M ) nt = M ) nt = M ) nt N c r 1 P w),k r k k=1 N c P w),1 k=1 = Pw),1 r 1 rk r 1 N c rk 2 k=1 Settng M ) nt = M) ext and solvng for P w),1 : P w),1 = r Ns 1 m= w my m ) Nc k=1 r2 k ) r k 1.5) The remanng column forces are computed usng equaton 1.4). 4
2. Approxmate 2nd Order Lateral Deflecton Preprnt Table 2 summarzes the smplfed procedure to approxmate 2nd order lateral deflecton of a moment-resstng frame. These equatons are derved by Haque [4] and wll be used to develop the desgn procedure. Quantty Equaton Label Column gravty loads Frame gravty loads P,k = N s m= p m,k 2.1) P = N c k=1 P,k 2.2) Addtonal gravty loads P = N gc) j=1 P,j 2.3) Gravty loads for stablty ˆP = P + P 2.4) Shears Column parameters Beam parameters Story drft parameters V = N s m= w m 2.5) C = N c I c,k k=1 h 2.6) B = N b I b,j j=1 L j 2.7) α 1 = 4 3, β 1 = 12 43 Fxed base = 1 α 1 = 4, β 1 = 3 2 Pnned base = 1 2.8) α = 1, β = 1 = 2,...,N s 1st order story drfts 1) [ ] = Vh2 α C + β B 2.9) 1st order deflectons u 1) Story bucklng loads Amplfcaton factors γ = 1 1 ˆP /P ) cr = m=1 1) m 2.10) P ) cr = V h / 1) 2.11) 2.12) 2nd order story drfts 2) = γ 1) 2.13) 2nd order deflectons u 2) = m=1 2) m 2.14) Table 2: Summary of deflecton computatons 5
3. Desgn Procedure 3.1. Story Drft Lmt The story drft lmt s defned usng equaton 2.13): 2) = γ 1) h b 3.1) where b s a constant. A common choce for the drft lmt s: b = 500. We wsh to remove the amplfcaton factor γ from our formulaton to lmt 1). We solve equaton 2.12) for the frst order story drft 1) : γ = 1 γ ˆP P cr ) 1 P ) cr 1 1 ˆP /P ) cr = 1 ˆP P ) cr = 1 1 γ = 1) V h = 1) = 1 1 ) γ 1ˆP 1 1 ) γ 1ˆP 1 1 ) V h γ ˆP Ths expresson s substtuted nto equaton 3.1): γ 1) γ 1) V h ˆP Usng ths result n 1) = h b = h b γ = 1+ ˆP bv = h /γb) gves: 1) = h b+ ˆP /V 3.2) 3.2. Story Drft Parttonng We partton the 1st order drft 1) nto drft c due to column rotatons and drft b due to beam rotatons. Ths s accomplshed usng equaton 2.9): 1) = c + b 3.3) c = V h 2 ) α 3.4) C b = V h 2 ) β 3.5) We choose the fracton η of 1) caused by column rotatons. The drft parttonng s wrtten as: B c = η 1) 3.6) b = 1 η) 1) 3.7) The drft fracton η s a desred value chosen durng the ntal desgn process. The actual drft parttons are determned from equatons 3.3) to 3.5) after the desgn s determned. 3.3. Desgn Columns The requred moment of nerta for the columns s determned from equatons 3.4) and 3.6). Solvng for the column moment of nerta: ) η 1) N c k=1 C I c,k h = V h 2 = V h 2 = V h 2 α C α η 1) α η 1) The sum of the column moment of nertas on story s: ) N c I,k c = V h 3 α 3.8) k=1 η 1) If we choose the columns on the story to be dentcal, then we set I,k c = Ic for k = 1,...,N c. ) I c = V h 3 α 3.9) η 1) N c Rolled or bult-up shapes are chosen wth moment of nerta whch exceeds ths value. Elastc bucklng of the columns must also be nvestgated. We consder the Euler bucklng load of a pn-pn column snce ths procedure s conservatve for fxed columns. The largest column axal force on a story s: P max ) = max P t) k=1,...,n,k 3.10) c where P t),k s defned by equaton 1.2). We must check that the moment of nerta I clg for the radus of least gyraton satsfes: I clg µp) maxh 2 π 2 E where the parameter µ 1 s a factor of safety. ) ) 6
3.4. Desgn Beams The beams are desgned n the same manner as the columns. The requred moment of nerta for the beams s determned from equatons 3.5) and 3.7). Solvng for the beam moment of nerta: 1 η) 1) = V h 2 β B [ B = V h 2 ) β 1 η) 1) The sum of the beam moment of nertas on story s: [ ] N b I,j b = V h 2 β 3.11) L j=1 j 1 η) 1) ] Preprnt If we choose the beams on the story to be dentcal, then we set I b,j = Ib for j = 1,...,N b. I b = V h 2 [ β ] N b 1 η) 1) j=1 1 L j 1 3.12) 7
3.5. Summary Table 3 summarzes the desgn of a frame wth dentcal columns and dentcal beams on a story. Preprnt Quantty Column gravty loads Frame gravty loads Equaton P,k = N s m= p m,k P = N c k=1 P,k Addtonal gravty loads P = N gc) j=1 P,j Gravty loads for stablty ˆP = P + P { 0 f k = 1 x k = k m=2 L k 1 f k > 1 x = 1 N c Nc k=1 x k Column wnd loads r k = x k x y ) m = h 2 + m q=+1 h q P w),1 = Ns r1 m= wmy) m Nc k=1 r2 k ) P w),k = Pw) rk,1 r 1 Total column loads P t),k = P,k +P w),k Shears 1st order story drfts 1) V = N s m= w m = h b+ˆp /V Story drft parameters Column desgn Beam desgn Column stablty check α 1 = 4 3, β 1 = 12 43 Fxed base = 1 α 1 = 4, β 1 = 3 2 Pnned base = 1 α = 1, β = 1 I c = Vh3 I b = Vh2 [ α η 1) N c β 1 η) 1) P ) max = max k P t),k I clg µp) max h2 π 2 E ) = 2,...,N s ] [ Nb j=1 1 L j ] 1 Table 3: Summary of desgn procedure 8
4. Desgn Example 4.1. Geometry and Loads A 10 story buldng s provded as an example for the desgn procedure. The buldng floor plan s 100 ft x 60 ft. Each story s 12 feet n heght. The buldng has moment-resstng frames along ts permeter. Interor columns carry only gravty loads. The framng plan for a floor s shown n fgure 5. We only desgn the frame along the short drecton.e. 60 ft long). Also, we are not be concerned wth any archtectural requrements or detals. Fgure 5 colors the trbutary areas for the corner columns n cyan and the other permeter columns n magenta. The area shaded n yellow s the addtonal trbutary area for gravty loads whch affect the frame stablty. The red lne on the left sde represents the trbutary wdth for wnd loads. The floor s a one-way slab system. Beams carry the floor loads to grders whch are connected to the columns. Notce that for the frame that we desgn, the beams do not carry floor loads. The floor loads are transferred to the columns from the grders. The frame elevaton of fgure 6 shows that the gravty and wnd loads are appled to the frame as pont loads. These loads are computed n the tables below. Fgure 5: Frame floor plan Load Gravty Wnd Value 150 psf 50 psf Table 4: Basc gravty and wnd loads Pont Load Corner column gravty Other column gravty Top story wnd Other story wnd Value p/2 = 150 psf 10 ft 10 ft = 15 kp p = 150 psf 20 ft 10 ft = 30 kp w/2 = 50 psf 50 ft 6 ft = 15 kp w = 50 psf 50 ft 12 ft = 30 kp Table 5: Frame loads Fgure 6: Frame elevaton 9
j = 1 j = 2 j = 3 j = 4 = 1 150 300 300 150 = 2 135 270 270 135 = 3 120 240 240 120 = 4 105 210 210 105 = 5 90 180 180 90 = 6 75 150 150 75 = 7 60 120 120 60 = 8 45 90 90 45 = 9 30 60 60 30 = 10 15 30 30 15 j = 1 j = 2 j = 3 j = 4 = 1-94.35 218.55 381.45 394.35 = 2-60.75 204.75 335.25 330.75 = 3-32.55 189.15 290.85 272.55 = 4-9.75 171.75 248.25 219.75 = 5 7.65 152.55 207.45 172.35 = 6 19.65 131.55 168.45 130.35 = 7 26.25 108.75 131.25 90.75 = 8 27.45 84.15 95.85 62.55 = 9 23.25 57.75 62.25 36.75 = 10 13.65 29.55 30.45 16.35 Table 6: Column gravty loads P,j n kp Table 8: Total column loads P t),j n kp j = 1 j = 2 j = 3 j = 4 = 1-244.35-81.45 81.45 244.35 = 2-195.75-62.25 62.25 195.75 = 3-152.55-50.85 50.85 152.55 = 4-114.64-38.25 38.25 114.64 = 5-82.35-27.45 27.45 82.35 = 6-55.35-18.45 18.45 55.35 = 7-33.75-11.25 11.25 33.75 = 8-17.55-5.85 5.85 17.55 = 9-6.75-2.25 2.25 6.75 = 10-1.35-0.45 0.45 1.35 Table 7: Column wnd loads P w),j n kp V max k P,k max k P w),k max k P t),k 1 285 300 244.35 394.35 4500 2 255 270 195.75 330.65 4050 3 225 240 152.55 272.55 3600 4 195 210 114.75 219.75 3150 5 165 180 82.35 172.35 2700 6 135 150 55.35 130.35 2250 7 103 120 33.75 93.75 1800 8 75 90 17.55 62.55 1350 9 45 60 6.75 36.75 900 10 15 30 1.35 16.35 450 Table 9: Accumulated loads n kp ˆP 10
4.2. Desgn The desgn procedure s appled to our example frame wth a drft parttonη = 0.5. The desgn of the frame s performed every two stores. We choose an AISC secton wth moment of nerta slghtly larger than requred. Tables 10 and 11 lst the requred and chosen secton propertes. Requred I c Secton I c Secton 1 5840 6000 W14x398 2 3920 6000 W14x398 3 3459 3840 W14x238 4 2999 3840 W14x238 5 2539 2660 W14x211 6 2079 2660 W14x211 7 1618 1710 W14x145 8 1158 1710 W14x145 9 698 722 W14x68 10 237 722 W14x68 Table 12 lsts the requred and avalable column strength. The AISC strength gven n AISC [1] s smaller than the Euler bucklng strength because other modes of falure are also consdered. Our chosen columns satsfy the column strength requrements. All values are rounded to the nearest kp. max k P t),k AISC φp n Euler load P el Secton 1 394 4850 29952 W14x398 2 331 4850 29952 W14x398 3 273 3440 1988 W14x238 4 220 3440 1988 W14x238 5 172 2550 1422 W14x211 6 130 2550 1422 W14x211 7 94 1750 9345 W14x145 8 63 1750 9345 W14x145 9 37 701 1670 W14x68 10 16 701 1670 W14x68 Table 12: Column strength n kp Table 10: Column desgn; unts for I c s n4 Requred I b Secton I b Secton 1 2717 9600 W24x279 2 8710 9600 W24x279 3 7687 8490 W24x250 4 6664 8490 W24x250 5 5640 6260 W24x192 6 4617 6260 W24x192 7 3596 4020 W24x131 8 2573 4020 W24x131 9 1550 1830 W24x68 10 527 1830 W24x68 Table 11: Beam desgn; unts for I b s n4 4.3. Story Drft and Dsplacement The 2nd order story drfts and dsplacements for the desgn are gven n table 13. All values are rounded to the nearest hundredth of an nch. 2) u 2) 1 0.18 0.18 2 0.22 0.40 3 0.26 0.66 4 0.22 0.89 5 0.26 1.15 6 0.22 1.37 7 0.26 1.63 8 0.18 1.82 9 0.26 2.08 10 0.09 2.17 Table 13: Story drft and dsplacement; unts are n 11
References [1] AISC. AISC Steel Constructon Manual. Amercan Insttute of Steel Constructon, fourteenth edton, 2011. [2] ASCE. Mnmum Desgn Loads for Buldngs and Other Structures. Amercan Socety of Cvl Engneers, seventh edton, 2010. [3] Councl on Tall Buldngs and Urban Habtat. Structural Systems for Tall Buldngs. Tall Buldngs and Urban Envronment Seres. McGraw-Hll, New York, 1995. [4] Aamer Haque. Approxmate lateral deflecton of momentresstng frames. 2017. [5] S. Hal Iyengar. Prelmnary desgn and optmzaton of steel buldng systems. In Bruce G. Johnson, edtor, Structural Desgn of Tall Steel Buldngs, volume 2 of Plannng and Desgn of Tall Buldngs. Amercan Socety of Cvl Engneers, 1972. [6] B.S. Smth and A. Coull. Tall Buldng Structures, Analyss and Desgn. Wley, New York, 1991. [7] B.S. Taranath. Structural Analyss and Desgn of Tall Buldngs. McGraw-Hll, New York, 1988. 12