ay) azole | Veh olay ly co = | _ Seismic EXAMS T_ _ Structural Department = = ) 4th year Civil =AIN SHAMS UNIVERSITY FACULTY OF ENGINEERING STRUCTURAL ENGINEERING DEPARTMENT 4" Year, Civil Engineering (Structural Division) Mid Term - Nov. 200 ae Question (2) For the given plan of a commercial building, located in El-Fayoum (zone 3), It is required to: 1- Calculate the equivalent static load at Y-direction. 2- Calculate the distribution of lateral load on each floor . 3- Calculate the factor of safety against overturning. 4- Calculate the percentage of lateral load to be carried by frame on axis (3). Gi t,=200mm & F.C. + Parition. = 3.0 kN/m? - Floor height = 3.0 m - No. of floors = 10 floors - Soil is dense sand (soil type B). - All columns are (300*700 mm) - All beams are (300*600 mm) >? % @ @ Shopping Area 6.0m Area Area 6.0mSolution 1- Simplified Modal Response Spectrum = W. FL= Sy(T) xa © According to Soil type and building location Table (8-2 El-Fayoum (Zone 3) One a = 0.15g Soil type (B) S = 135 Table (8-3 Tz = 0.05 Response spectrum curve Type (1) P. (3/5 code) To = 0.25 Cassia) sal gle Ube pill gbuwl) Ty= 1.20 T=¢, 4 Beams + Columns ———> R.C. Frames “PGS code)” C,= 0.075 Total Height of building (H) = 30m —> T,= 0.075 x 30°" = 0.96 sec. Check T,<(4Tc=1.0sec) O.K. T, < 2.0 sec. O.K. — Te< TS Ib yaecag” F4-8-13) : Te Sy (T= a, % 8 23 [%]> 0.20 ay R=5.00 ny = 1.00 Commercial building esa” Y= 1.208, (T,)= 0.15 gx 1.20 «1.35 x 25; [ 925 | 1.00 = 0.032 ¢ 0.20 a, y = 0.20 x 0.15 g x1.20 = 0.036g —» Take S, (T,) = 0.036g T,> 2T, = 0.50 see Frag sgy> % = 1.00 = DL.+ox LL. Ws =(tsy, + F.C. + Walls) +x L.L. Commercial building Table (8-7 a =0.50 P. (5/5 code) Wrotat= (D.L. + &x L.L.)x area x no. of floors + own weight of beams and columns W = (0.20 * 25 + 3.0 + 0.5 * 3.0) * (21 * 12) * 10 (shopping area) + (0.20 * 25 + 3.0 + 0.5 * 6.0) * (6 * 12) * 10 (storage area) b t H Ye no. of columns +0.3*0.7*30*25* 15 (columns) b (t-t,) Ye no. of floors + 0.3 * (0.60 - 9.20) #25 *(12* 5-427 *2) * 10 (beams) Wrotat = 37642.5 KN — Fy= Sa) xd ¥ = 0.036 gx 1.0 x 37128 F,= 1355.13 kN|2- Distribution of lateral force on each floor =o Floor] 4; No. i(m) F, (KN) H.= 2, Hj=30+27+24421 418-415 10 | 300 | 24639 +12+9+6+3=165m 9 | 270 | 221.75 R= (1355.13) qh Hy gs | 240 | 197.11 a aa a2 toe ania, E 8.21H; Gia nis Omer 83 5 | 150 | 123.19 4 | 120 | 9855 3 9.0 73.92 2 60 | 49.28 1 3.0 24.64 3- Total bending moment at base Myase=2F;. H, = 246.39 * 30 + 221.75 * 27 + 197.11 * 24.0 5.0 + 98.55 * 12.0 + 172.47 * 21.0 + 147.83 * 18.0 + 123.19 * + 73.92 * 9.0 + 49.28 * 6.0 + 24.64 * .3.0 Mase = 28457.73 KN.m4- Factor of safety against overturning Mooseus,= 28457.73 kN.m M = Moose ux. ‘overturning 1.40 E eemanea| | ——| =~28457.73 = 99326.95 kN. h =" 140. : Bo W Movertuming Resisting Moment = Wrotat xB |_B__| = 37642.5x42 = 225855 kNm Resisting Moment 225855 Factor Of Safety = GyerTuming Moment ~ 3032695 = !!-1> 15 -. Safe 5- Percentage of load carried by frame on axis (3) on 6.0m 7sm__| 7.5m__|, 6.0m The frame on axis (3) is located on the (C.R.) —> x,=zero % ofeach Frame =—Kni_ yy + hex (1x e") 2 Kr = Ky; (xi) % Frame on Axis (3) = Koxit eS 5xK = Kn ()AIN SHAMS UNIVERSITY FACULTY OF ENGINEERING STRUCTURAL ENGINEERING DEPARTMENT 4" Year, Civil Engineering (Structural Division) ic 003 Question (2) (40 % of maximum credit ) Figure 1 shows a structural plan of an industrial building . The structure has overall dimensions of 27.0 *22.5 mand contains no interior beams . A marginal beam on the outside perimeter of the building is provided .The building is supported on 16 circular columns of 400 mm diameter and 4.0 m height . An opening with dimensions of 2.0 x 2.0 m is provided as shown in the figure . If this building lies at Aswan and consists of basement and 10 floors with floor height and live load as shown in Fig (2) in addition to flooring & partitions = 2.0 kN/m’ for each floor . Foundation soil is loose sand. Due to earthquake loads it is required to: 1- Calculate the story shear at each floor level . 2- Draw the distribution of this shear on the height of the building . 3- Check the stability of the building against overturning . Assuming average floor thickness (ts) = 250 mmt 1 1 t = [SSS == D) ee ceed | eee eee le eee Figure (1) i LL. =|2.0 kN/m? LL. =|2.0 kN/m? LL. 52.0 KN/m? LL, 42.0 kN/m? LL. 52.0 kN/m? LL. 42.0 kN/m? LL. =2.0 kN/m? LL. 42.0 kN/m? LL. 5,0 kKN/m?™ LL. =| 5.0 KN/m? rete X40 m—l______— 8x3.0m UT STITT STOTT Figure (2)Question@ 1- Simplified Modal Response Spectrum = W. Fy= Sal) x ioe According to Soil type and building location Fig. (8-1 Table (8-2 = Aswan pf cody 20M 2) “pais coaay™ A= 01258 Weak soil —Zable@-D, soil type (D) P. (2/5 code) Soil type (D) = S = 1.80 Table (8-3. Ts = 0.10 Response spectrum curve Type (1) P. (3/5 code) To = 0.30 (as gil ja) gle Abe pal Gblidl) T= 1.20 T,=¢, Hu Flat Slab “PGi codey C,= 0.075 Total Height of building (H) = 35 m —> T,= 0.075 x35" = 1.08sec. Check T,<(4Tg=1.2sec) O.K. T, < 2.0 sec. O.K. —> << Eq. (8-13) Tp P. (3/5 code) Sq (T)=a, % 8 23 [¥]} > 0.20 a, R= 5.00 7 = 1.00 Eea Table (8-9 _ Ge Sais code) = 1.00 Sq (T,)= 0.125 1.00 «1.80 x 2% [ 9. = 0.0313 g 0.20 a, ¥, = 0.20 x 0.125g x1.00 = 0.025g < S, (7) OK. 930] «1.00 T, > 2Ty = 0.60 see SR cay” ® = 1.00 w,= D.L.+ 0x LL. Table (8-7 - P.(5/5codey” % = 9-25 Whoor=(D.L. + &x L.L.)x area D.L. =tsy, + F.C. + Walls = (0.25 x 25 + 2.0) =8.25 kN/n? FloorD&@ Waoor= (8.25 +0.25x5.0)x 536 = 5092 kN Floor@®—-@ Wacor=(8.25 +0.25x2.0)x 536 = 4690 kN W, Total =(5092 x2+ 4690x9) =52394 kN —> F,= S4(T)) xa ¥ = 0.0313 gx 1.0 xe I 1639.93 kN Fy2- Distribution of lateral force on each floor Fok x “iti oe Sw, Hy Moor! cm) [wkN)] wi Hy |B &N) [Q:GeN) [FH kN.m) 1 | 35.0 | 4690 | 164150 | 263.46 | 263.46 | 9221.25 sacl 10 | 32.0 | 4690 | 150080 | 240.88} 504.35 | 7708.21 ee 9 | 29.0 | 4690 | 136010 | 218.30| 722.64 | 6330.67 722.64 8 | 26.0 | 4690 | 121940 | 195.72| 918.36 | s08s.62 21836 7 23.0 | 4690 | 107870 173.13 | 1091.49] 3982.07 en 6 20.0 | 4690 | 93800 150.55 | 1242.05 3011.02 is0el 5_| 17.0 | 4690 | 79730 | 127.97| 1370.01] 2175.46 ans.a0 4 14.0 | 4690 | 65660 105.39 | 1475.40 1475.40 1558.20 3 | 11.0 | 4690} sisoo | 82.80 | 1558.20] 910.83 | j6is.a1 2 | 70 | soo2| 35644 | 57.21 [1615.41] 400.47 | gaol 1 3.0 5092 15276 24.52 | 1639.93 73.55 . Sw, H, Shear Diagram 021750 3- Check overturning Myseur.= =F xH; = 40377.55 kN.m — Moose uz, — 40377.55_ _ Movertuming= T4040 28841.11 kKN.m Resisting Moment =W,,, xB: = 52394 x 23:5. =589432.5KN.m Resisting Moment _ 589432.5 =20.4> 1.5 Factor Of Safety = Over Turning Moment ~ 28841.11 - SafeAIN SHAMS UNIVERSITY FACULTY OF ENGINEERING STRUCTURAL ENGINEERING DEPARTMENT 4" Year, Civil Engineering (Structural Division) Jan. 2008 Question (2) Figure 1 shows a structural plan of a 12-story office building with an overall dimensions . Of 16,0x28.0m. The slabs of ground and all typical floors are flat slabs. The floor slabs are supported on 9 columns and 6 shear walls. The dimensions of all columns in last floor are 400x3: 1e dimensions of all: shear walls are 300x3000.mm:and thé floor height is 3.75m.- HOEK es : : Assuring that the office building in’Figu is located in:Alexandria-on a medium ‘soil: The contribution of columns in resisting seismic load in Y-direction may be neglected. Duet wake loads i irection-only, it is required to: 1) Calculate the equivalent static loads at foundation level (Base shear-and torsion: moment): faite Z TA ae bes! 7 2) aos and draw the I ‘equivalent static load, the shear force diagram and bending moment diagram over the height of the building. 3) Check the giobal overturning and sliding of the building at foundation level.” 4) Neglecting the coupling of the shear walls, calculate the percentage of the base shear “V" carried by the shear wall (SW1) on axis “8”, Giv ts= 22.0 cm F.C. + Parition, = 3.0 KN/m? L.L. = 4.0 KN/m?Question (2) 1- Simplified Modal Response Spectrum F,= Sa(T) x2 Y According to Soil type and building location Fig. (8-1 Table (8-2. Alexandria iss” Zone (2) oslo” ay= 0.125 Table (8-1 Medium soil Toke” Soil type (C) Table (8-3 Tz = 0.20 R Type (2 P. (3/5 cod = esponse spectrum curve Type (2) code) To = 0.60 (ae gid ja! gle Aad! gblist!) Tp = 2.00 Soil type (C) S = 1.25 T=¢, we! Shear walls PGi codey” C,= 0.05 Total Height of building (H) = 45 m —> T,= 0.05 x 45°" = 0.87sec. Check T,<(4T.=2.4 sec) O.K. T, < 2.0 sec. O.K. — T. 0.20 a, 7, R=5.00 7 = 1.00 ol} gas Table (8-9), Y,= 1.00 P. (4/5 code)Sy (T,)= 0.125¢ «1.00 * 1.25 x 23, [ 0.60 | x1.00 0.87 = 0.0539 g 0.20 a, ¥, = 0.20 x 0.125g «1.00 = 0.025g < S, (T,) O.K. T,<2Ty = 1.20 see Ss coda” A = 0.85 w,= DL +axLL. W, =(ts% + F.C. + Walls) + Ox L.L. a Table (8-7 wal} Be Bei code) % = 9.25 w, =(0.22 x 25 + 3.0 ) + 0.25 x 4.0 =9.5 kN/m? slab shear walls columns Weloor = 9-5 X16 x28 + 0.33 x3.75 x25 x (6) +0.350.4 x3.75 x25 x (9) = 4880.375 KN Wrote = 4880.375 x 12 = 58564.5kN — Fy= Sg) xa ¥ = 0.0539 x 0.85 x S8POES F,,= 2683.13 kN| 2- Distribution of lateral force on each floor _ ee Hi : =gi eu ssa O59 OYisa & Hj = 45441.25+37.5+33.75433 +30+26.25+22.5+ 18.75 ‘ +15+11.25+7.5 $3.75 = 292.5 m 1 i = (2683.13) 597-5 Hi hea} ll E = 9.173 H; Neo | Him) | RN) | Q,N) | M,N) 12 45.00 412.79 412.79 1547.96 BT 41.25 378.39 791.18 4514.88 10 37.50 343,99 1135.17 8771.77 9 33.75 309.59 1444.76 14189.63 8 30.00 275.19 1719.96 20639.46 i 26.25 240.79 1960.75 27992.27 6 22.50 206.39 2167.14 36119.06 5 18.75 172.00 2339.14 44890.83 4 3 2 1 15.00 137.60 | 2476.74 54178.59 11.25 103.20 | 2579.93 63853.33 7.50 68.80 2648.73 73786.08 3.75 34.40 2683.13 83847.81 412.79 412.79 ; 378.39 7oi.is|_|12 1547.96 2 343.99 nosy 11 astaas 309,59 1444.76 Ww sm! a 17195 9 ais9.63/-19 240.19 1960.75 2 20639.46, 8 —— 2167.14 7 27992.27 7 2339.14, 6 Benen 6 2476.74 5 44890.83 5 2579.93 4 54178.59, fe 2648.73; s 63853.33, 2 2683.13 2 13786008 2 1 gsear81 : Load Diagram Shear Diagram Moment DiagramShearing force at base = |2683.13 kN Bending moment at base =| 83847.81 kKN.m 3- Check overturning Moseut.= 83847.81 kN.m Myce 83847.81 _ overtning= PQ = dG = 59891.29 KNam Resisting Moment =W,,,., x B x 58564.5 x “2 16. = 468516 kN.m Resisting Moment 468516 _ Factor Of Safety = Over Turning Moment ~ 59891.29 =78 >15 oa -. Safe Check ne oe Sliding Force =>. as = 2683-13 — 1916.5aN Resisting Force =H xW = 0.30 x58564.5= 17569.35 kN Factor Of Safety = Resisting Force, _ 1756935 92 >1.5 ; » Safe Sliding Force Coefficient of friction (1) 2535 ee e Check sliding lb als (4 4B yrs Lad gb Gaglladll GY alld yp = 0.3 Abad Gof ob Atladll 8 GAN 4g aslll Ge lal Uae» Check sliding4- Percentage of (V) carried by SW, 20 2 a 79 SS 3 y= P3039 = 0.675 mi! Xx > 1,xx; eee ene CR. > J; = 20,675 x0-+2x0.675x7+2%0.675 x21 — 6x0.675 ean ee L . 28 = Kgc > B =140m €=Xouy- Xcx =14.0-9.33 = 4.67m * e = e+0.05 L = 4.67 + 0.05 x 28.0 =6.07 m % of each Shear Wall = 0.675 2.33 jp UE ero) 1 0.675*6 20.675 (9.38 +2.33°+ 11.67) = Zero. off | 030° = 16.67 %AIN SHAMS UNIVERSITY FACULTY OF ENGINEERING STRUCTURAL ENGINEERING DEPARTMENT 4" Year, Civil Engineering (Structural Division) Jan. 2009 Question (2) (20% of maximum credit) Part (2-a): (14% of maximum credit] Figure (2-a) shows a typical plan of a residential building consisting of 7 typical floors. If the building is subjected to horizontal earthquake force (Base shear force = V) in the Y-direction. The earthquake force is resisted only by the two shear wails (W1 & W2). It is required to: A: Calculate the following in terms of "V": * 1- Distribute the base shear force (V) on each floor (assuming total uniform load including slab weight and floor cover weight on each floor = w, and Fiop =zero). 2- Calculate the lateral earthquake force on each wall taking effect of eccentricity into consideration (stiffness of wall 2 = twice the stiffness of wal) 1) 3- Calculate the maximum moment at base of each wall. B: Calculate the maximum base shear force (V) that the building can resist knowing that | a. maximum moment of resistance of wall W1 = 3500 kN.m b. maximum moment of resistance of wall W2 = 5000 kN.m ab o 410 m———v TS < x | a aro 4.0m——4.0 m— ———10.0m —__»Question (2) uestion ras 1- Distribution of lateral force on each floor re Ee ae awit Felon] Be) WA Ho yest 7 | 21.0 | 025%, SywiH, =H, au asa 5s oF 6 | 180] 021k = 5 | 15.0 | 0.18%, 2, Hy= 21418415412 + 94 643 = 84m 4 [120 | 014% - 1 3 | 90 | 01 R=-7 (&*HD 2 | 60 | 0078, F — 0.0119 (RH) 1130 | 0048, Moseut= = Bx; =0.25 F,x21+0.21 F,x18 +0.18 F,x 15+ 0.14 F,x 12 +0.11 F,x9 + 0.07 F,x6 +0.04 F,x3 = 14.94 F, 2- Center of rigidity and torsional moment =2K, (Given) $ ee > Kx x; : d il BO 2K pe SR gt hoo = Kyx0 +Kyx14 = 2K, x0+ Kyxdl4 M4. Ko+ Ky 2Ki+ Ky il =4.67m Gg iene aceeeoees yo 10-46 -1.0=7.0m 10 10 1.Q— ma : Xem- Xex =7.0- 4.67 = 2.33 m Xa e= e* = €+0.05L = 2.33 + 0.05 x16 =3.13 m3- Lateral load and moment carried by shear walls | Zl —_ 7.0 ——-+}-—— 7.0 hon yw =X xt + (1x3.13) ——__41 * 9.33 % Wi K+ Ke ( K,x9.33° +K5% 4.67" K K, x 9.33 9 =a xl + (13.13) = 56 % K,+2K, : K,*9.33°+2K,* 4.67" F, = 0.56 F, M, = 0.56%14.94 F, = 8.366 F, K Ky x 4.67 9, =o xt - (1x3.13) ——____3 $= ""*"___— Wi, = +K : K,X9.33° +K x 4.67" =—2Ki_x 1. (1x3,13) —__2K 1 *4.67 __ = 44 % K+ 2K, K,* 9.33 +2K, x 4.67 F, = 044K, M, = 0.4414.94 F, = 6.574 F, 4- Maximum base shear force M, = 3500 = 8.366 ,—» F, = 418.36 KN M,= 5000 = 6.574 F,—» F, = 760.57 kN —» Maximum base shear force F, = 760.57 kNExample (1) For the given elevated tank , located in Cairo.The tank has dimensions of (4.0 x 4.0 x 4.0m) with top horizontal beam (250x750mm). Thickness of all walls and floors are 250 mm, The tank is constructed on a weak soil. It is required to : 1- Calculate the base shear of the tank.(Case of Full Tank) 2- Check Stability against overturning. 3- Without any calculations sketch the shape of B.M.D. and N.F.D. on the frame supporting the water tank. 4- Sketch Reinforcement Details of Joint (A) Given that : —__—_—_——_ 2 All columns are (300 x300) All beams are (300x600) 40 | Joint (A) 40 ™. ao + TT -— Sec A-A | |Solution 1- Simplified Modal Response Spectrum FL= S(T) xB According to Soil type and building location Cairo (Zone 3) Table (8-2) 5 ap= 0.15g P. (2/5 code) Weak soil —T2le(B-D_» soit type (D) P. (2/5 code) Soil type (D) S = 1.80 Table (8-3) Ty = 0.10 Response spectrum curve Type (1) P. (3/5 code) To = 0.30 (Lee giall paul gle Ale pid) gb) Ty = 1.20 T=C, Hw" Beams + Columns > R.C. Frames PGs code” C,= 0.075 Total Height of building (H) = 16 m — + T,= 0.075 x 16°" = 0,60 sec. Check T,<(4T,=1.2sec) OK. T, < 2.0 sec. O.K. —> hs 7 P. (3/5 code) 2. Te Sa(T)=a, % 8 22 [3] >0.20a, R= 5.00 yn = 1.00ets Table), y = 1.20 P. (4/5 code) = 2.5 0.30 Sq (T))= 0.15 gx1.20 x1.80 x 25% [ 930 | x1.00 = 0.081g 0.20 a, ¥, = 0.20 x 0.15 g x 1.20 = 0.036g < Sy (T,) O.K. T, = 27 = 0.60 see Saggy & = 0.85 We DL.+ax LL. Sb sd Table (8-7 = OLLI Gis cody” = 10 Wy = (0.25 x0.75x25 )x(4x4) Top Hz. Beam + (0.25 x25 )x(4 x4) x4 Walls + 0.25 x25 )x(4 x4) Floor + (0.3 0.3x25) (12%4) Columns + (0.3x0.6%25)(4x4*2) Beams = 827 kN W, 1, =(4.0 4.0 x4.0) x10 Water = 640 KN Wrota) = DL. + ox LL. Wrotal = (827 + 1.0 640) =1467 kN —> Fy= Sy(T) xa = 0.081 gx 0.85 x ee F,= 101.003 KN2- Check Overturning Fy 101.003 kN 4 Mouse x, = 101.003 x14.0 =1414.042 kN.m M — Messe un. — 1414.042 overturning 1.40 1.40 mt =1010.03 kN.m Resisting Moment =W,.91 * B = 1467x 4p =2934 kN.m Resisting Moment 2934__9 904> 1.5 Factor Of Safety = Over Turning Moment ~ 1010.03 - Safe 3- Shape of B.M.D. and N.F.D. of Supporting Frame —— — © © ao w » TR wo © r = — B.M.D. N.F.D.4- Reinforcement Details of Joint (A) TT al Joint (A) YS was HL 4 JISI! ag oles! 5s tas posedt ales gles 9 gle BSI Gob geal abut gS UL) sal ddthe od Sl alt) Slog oe piry 2g) Bactie gd Base Gabi) Sdbos Jac pryAIN SHAMS UNIVERSITY , FACULTY OF ENGINEERING STRUCTURAL ENGINEERING DEPARTMENT, 4" Year Civil (Structural Division) Jan 2017 Time : 3:00 His. E Tee _ Reinforced Concrete Desigh (3) Bos Question (3) (25 % of maximum credit) Figure (3-A) shows a residential building (/, = 1.0) located in Cairo (Zone 3; ag= 0.15g), The building consists of 6 typical flat siab floors (ground and § floors) and rested on very dense sand (soil type B). The lateral loads are resisted by reinforced concrete cores. Nglecting the effect_of columns on the lateral_load resistance of the building and Considering the earthquake force in X-direction only, itis required to: a) Calculate the internal force due to earthquake (shear at base). b) Calculate the lateral force at each floor and draw its distribution. c) Calculate the lateral force on the cores above the ground floor only. d) If the cores were relocated as shown in Figure (3-B), which case has a better resistance fo fateral loads and why. (equations can be shown without calculation) Data: Slab thickness = 200mm, Live load = 2 kN/m? , floor cover and walls = 4 kN/m* Columns dimensions = 0.5 x 0.5m, Core thickness = 0.25 m 18m 3 K8 15m Tsm oe 3x85 180m ws 15m gm mam _,18™ 15m 3m. em om 15m & e ar . alfa . e El oer Ne = Corr 1 i . =) on . . . . . E . Leet) . 1.5m A a 9 6m__.3m tt mn Figure (3-A): Building dimensions Figure (3-8): Cores locations 5m 3A 14m > 15m 3m 1.5m 3mQuestion 3) 1- Simplified Modal Response Spectrum F,= S(T) x2 F According to Soil type and building location Cairo (Zone 3) ——® a,= 0.15 g (Given) Dense sand ——> Soil type (B) (Given) Table (8-3) Response spectrum curve Type(1) | P.(3/5 code) Soil type (B) Cane sid) pal gle Albee pall gllaall) | T=, He Shear walls Pi codey” C,= 0.05 Total Height of building (H)= 18m —> T,= 0.05 18° = 0.437sec. Check T,<(4Tc=1.0sec.) OK. T, < 2.0 sec. OK. — Tes <5 Th Gem” Eu (8-13) S, (T= 8 R = 5.00 1 = 1.00 Y= 1.00 (Given) 2.5 | Te 5, S He [Fn > 0.20 4,5, S = 135 Ty = 0.05 To = 0.25 T, = 1.20S,(T,)= 0.15 gx 1.00 x1.35 x 25, [ 225 | 1.00 = 0.0579g 0.20 a, Y, = 0.20 x 0.15 g 1.00 = 0.030g < S,(T,) T, < 2T, = 0.50 sec r= 0.85 P.(3/5 code)” —— w,= DL.+axL.L. We = (5%, + F.C.+ Walls) +x LL. Kas —Lable (8-7) H&S DER eodey” t= 9.25 W, =(0.20x 25 $4.0) 40.25 x 2.0 =9.5 kN/m™ o.w. of columns 0.W. Of cores eee errr eee Serene Wotoor= 95% 21% 14 + 0.5 x05%3x 25x 12 + 0.25% Ix 3x 25% 2 =3355.5 kN Wrotat = Wytoor * No. of floors Wrotal = 3355.5 x 6 =20133kN — B= Sy(T) xd x = 0.0579 g x 0.85% 2g83. F,,= 990.845 kN2- Distribution of lateral force on each floor he Rx ei : aw; A; _ bh. Ai ee sol o59 OF 2H 2 Ai i SW = 18+15+12+ 9+6+3 =63m iv i ee i (RxH;) F = 0.0159 (Fx H;) yoo" Hi (m)| Ff (kN) 6 | 18.0 | 283.1 283.1 5 | 15.0 | 235.9 235.9 4 | 120 | 188.7 188.7 3 | 90 | 1415 141.5 2 | 6.0 | 944 94.4 1 | 3.0 | 472 47.2 63.0 | 990.8 aaa Lateral load distribution 3- The lateral force on the cores CM.=CR. % of Core = Je xp +—e vi x(1x @min) zk = Ty Cin = 0.05 L = 0.05% 14.0 =0.70m | ye 60.25% 042 x1 502540.75 _p rs 6X 0.25 +2x*1,5* 0.25 y,., = 5.5 - 0.25=5.25m tor. % of Core = «1+ 4*525_ (1 x0.70) = 56.67 % ai 21x 5.25—> Lateral force on cores above ground = 47.2 x 0.5667 = 26.74 kN 6m__ 6m__ 6m 3m_6m Im Ty OTT ra . & . rd . el °F corer & § ” 4 stl e a . . skle . ° . 7 Je : my Core 1] Im 6m_|_3. I, Tey x5. % of Core =—» x F. eylaseas X Minin =Value t + > case(A) 2hey * 2 Iey (5.25) I. I, 5 % of Core =— x F, ey *5.25 X Mi jyin =Value t 4+——__ Case(B) 2key > 2 Icy (5.25) +21, (3.0) —» Case(B) is better as it causes less straining actions on cores