Graduation project 2 Structural design of alhuda building by hane daghlas , raed Abdullah Project description: This building consists of eight floors with roof the Total area 4482 m2 Floor Name Function Area(m2) Height (m) Ground Multipurpose store 469.68 5 First Car parking 469.68 2.5 Second Apartments 446.5 3 Third Apartments 446.5 3 Fourth Apartments 446.5 3 Fifth Apartments 446.5 3 Sixth Apartments 446.5 3 Seventh Apartments 446.5 3 eighth Apartments 452.8 3 Roof Apartments 410.7 3 Total area 4482 m2 Objective: In this project the concepts of structural engineering that we learned over all last years are applied. Many courses have been studied in the programing of Civil Engineering Department at An - Najah National University, Different types of analysis and design are done in this project in order to choose the most safe and economical type and method for the design. After completing the design phase in proper way we get feasible solutions of projects. In any structural system there are many alternatives that can be used. The successful one takes into consideration strength, serviceability, safety and economy aspects. Site and geology: The building is located in Nablus.. The soil in the site has bearing capacity of about 3 kg /cm2   Methodology: First, a preliminary analysis and design using ID and 2D models are made, and then the Structural elements are modeled as 3D dimensional elements (Beams, columns and footing) using (ETAPS program), after using 3D model to represent the structure, it was analyzed considering gravity, and compare the results with hand calculations to verify that 3D model results are ok. This is in brief, an overview of the project methodology, of which a full discussion is presented deeply later in this report. Codes and standard: The codes used in this project to design the members and the elements are the following: ACI318-14: building code requirements for structural concrete provides minimum requirements necessary to provide public health and safety for the design and construction of structural concrete buildings. ASCE7-10: is standard not code used for determine the value of minimum design loads for building. Materials: Concrete: Compressive strength of concrete (fc,=28MPa) for reinforced concrete slabs, beams, columns, shear walls, basement walls, and foundations and unit weight of reinforced concrete=25KN/m3 Steel: Steel reinforcement has a yielding strength of 420 MPa and modulus of elasticity (Es) of 200 GPa . loads: Dead load Superimposed load Live load Dead load: loads that relatively don’t change over time, such as the weight of all permanent components of a building including wall ,beams , columns, flooring material ,etc. Superimposed load: weight of plastering, filler, tiles, mortar and partition. Super imposed dead loads calculations: 12cm fill under tiles unit weight=16kN/m³ 3cm mortar under tiles, unit weight=23kN/m³ 3cm tiles, unit weight=25kN/m³ Super imposed dead loads = (thickness for each one square meter of the layer) *(density of this layer) S.I.D= (0.12*16)+(0.03*23)+(0.03*25)=3.36KN/m2 Live load: weight of everything superimposed on, or temporarily to a structure (people, machinery and equipment, furniture, appliances, etc.) Live load on floors and roofs =2 KN/m2 Live load on garage =2 KN/m2 load combinations load combinations Primary load 1.4D D 1.2D+1.6L D AND L Table shows load combination from ACI code for gravity Computer programs ETABS: This program is used to analyze and design the structural elements. AutoCAD: This program is used to draw structural details SAP: this program is used to analyze and design the structural elements.   Preliminary Design: The structure consists of several elements that work together to resist all the forces and loads that exposed to the building. Each element passes the load to another one until it reaches the soil and spreads through it .Therefore, the safe structural system is the one that can provide safe path for load. In this chapter, preliminary design for some of the elements will be discussed Hand calculations and the structural analysis and design software (ETAPS) will be used. Design Requirements: Strength Requirements • ASCE 7-CODE was used for live load determination. • Analysis and design are according to American Concrete Institute ACI-318-08. . • The nominal strength of members is reduced by reduction factor (Ø) • Design strength > required strength. • Ø Nominal strength > Ultimate strength. • The nominal strength Reduction factors Ø are: (1) Axial load and bending: 0.9 (2) Shear and torsion: 0.75 (3) Tied column: 0.65 The used load combinations are: Wu= 1.4DL Wu=1.2DL+1.6LL Where: DL: Dead Load. LL: Live load. Preliminary design calculation: Slab: Slabs are structural elements with a small thickness comparable to their dimensions in the other two directions. Slabs are usually used in floor and roof construction. Type of slab depends on the way it supported by. It may be supported by edge beams or walls, or they may be supported directly by columns (flat slab). The choice of type of slab for a particular floor depends on many factors. Economy of construction is obviously an important consideration, but this is a qualitative argument until specific cases are discussed, and it is a geographical variable. The design loads, required spans, serviceability requirements, and strength requirements are all important. One way solid slab will be used in this project For critical panel: L = 14.78 m / B= 4 m Check thickness of one way solid slab By taking a strip in the load direction with 1m width . First span is one end continues with 4 m long  Second span is two end continues with 4.2 m  Third span is two end continues with 4.4 m  Fourth span is one end continues with 4.37 m  Check slab for shear Own weight = 1 x 0.25 x 25 = 6.25 KN/ Wu= 1.2 (3.5 +6.25) + 1.6 (2 ) =16.5 KN/ Vu = Wu (0.5 Ln – d) Vu = 62.37 KN Vc = 134.72 > Vu  OK beams: The preliminary dimension of beams will be find by the following table   h=L/18.5 h=9.66/18.5=0.52 m. Try h = 550 mm Columns The tributary area for critical column equal 28 as shown in the following figure: Then the load on the column by using factored load combination Pu=1.2D+1.6L For live L=2 KN/ in all floors PL= tributary area*L*No. Of floors =28*3*9=756 KN For dead PD = Dslab + Dbeam Dslab = no. of floor × (superimposed dead load x tributary area + thickness of slab x tributary area x 25) Dslab = 9 x (3.5x28+ 0.25 x 28 x 25) = 2363 KN Dbeam = depth × width × length Dbeam = 0.5 x 0.5 x (2.1+2.237+4.83) x 25 = 57.3 KN PD = 2363 + 57.3 = 2420.3 KN Pu = (1.2 × 2420.3)+ (1.6 × 756) = 4114 KN To calculate area for column, will use the following equation: Pu = ϕ × λ × (0.85 f'c ×Ag × (1- ρ) + fy × ρ ×Ag) Assume ρ = 0.03 4114 x 10^3 = 0.65 x 0.8 (0.85 x 24 x 0.97 Ag + 420 x 0.03 x Ag) 4114000=16.83 Ag ⇛ Ag = 244445 mm2 Try column 700 x 400 mm2 Shear wall : Shear wall is 25.4 m long and 4 m height Minimum thickness (h) = Use h = 0.25 m Summary of preliminary design: Type Thickness (mm) Dimension (mm) Slab 250 --------- Column -------- 800 x 400 Beam -------- 550 x 350 Shear wall 250, 300 --------- Modeling , analysis and design: input data of the models: material: Material that used in this project is reinforced concrete fc = 28MPa for slab, beams, columns and shear wall   section on ETABS: After defining the material used in elements, all sections of these elements will define on ETAPS and that includes columns, beams, slabs and shear wall.   modifiers on ETAPS: Section Modifier Column 0.7 Beam 0.35 slab 0.25 Shear wall 0.7 2 Load combinations for seismic design on ETAPS Load Cases on ETAPS Load type Case Live load in each floor 2KN/m2 Superimposed load in each floor 3.5KN/m2 Load Cases on ETAPS: slab Check Wide beam shear check: = The strength reduction factor for shear and equal 0.75. VC =concrete shear capacity in KN. fc = concrete compressive strength in MPa . bw= section web width in mm . by taking 30mm as cover in the solid slab , total thickness = 250mm , and the effective depth = 220 mm For ETABS max V23 from 1.2D+1.6L = 23 KN For ETABS max V13 from 1.2D+1.6L = 23KN ØVC = = 134.2 KN It’s ok. Check slab for deflection: Check for compatibility We used the model shapes, and periods as an indication for compatibility check of the structure. Since all parts of the structure (slabs, beams and columns) move together and the period is reasonable that the compatibility check its ok. As shown in figure that the period =0.846s which is less than 1 s (its ok) Equilibrium check: Hand calculation Live load = (4482* 2) = 8964KN SD load = (4482 *3.5) = 15687 KN %Errorlive = = 0.454% %ErrorSD = = 0.5291% Stress strain check: Mhand = = 10.218 KN.m M=5.3433KN.m METABS = +5.9321 =10.67KN %error = =4.42% Seismic Analysis seismicity of site and structure To determine the earthquake load and seismic design category, some factors should be fined and calculated. These factors affected by type of soil and natural of structure and location of building, Risk category: It's classification for building according to the importance of building and the risk of failure structure. importance factor The importance factor will find according to risk category Site classification   According to the type of soil of site, the site of structure will be classified for A, B, C, D, E. And the classification of soil depends on the properties of soil on the depth of 30 m. According to the table 20.3-1 site classification in page 204, ASCE7, available data about the site soil is q, this value approximately not exactly equal to twice ofSu. q = 300 = 6265 psf Seismicity of the site: The Z factor will be used to calculate S1 and Ss Ss: it is acceleration will effect on the soil of site B within a short period of time and will be as a percent of g. S1: it is acceleration will effect on the soil of site B within 1 second and will be as a percent of g. Ss = 2.5 × Z = 2.5 × 0.20 = 0.50 S1 = 1.25 × Z = 1.25 × 0.20 = 0.25 SMS: acceleration will effect on the soil of the project within a short period of time and will be as a percent of g. SM1: it is acceleration will effect on the soil of the project within a 1 second and will be as a percent of g. SMS = Fa × Ss = 1.2 × 0.50 = 0.60 SM1 = Fv × S1 = 1.55 × 0.25 = 0.3875 Then SDS, SD1 will be calculated due to SMS, SM1 SDS: acceleration will effect on the structure within a short period of time and will be as a percent of g (design acceleration). SD1: acceleration will effect on the structure within a 1 second will be as a percent of g (design acceleration). SDS = 2/3SMS = 0.40 SD1 = 2/3SM1 = 0.2583 Seismic force resisting system: Special reinforced concrete moment frames. From ASCE 7 10 Table 12.2 1: R =8 ʊO =3 Cd =5.5 Ta = Ct *hn x = 0.0466*31.5 0.9 =1.039 s Dead load = 57542, 62 KN Live load = 8923 KN = 0.25 * 8923 = 2230.75 KN S.D Load = 15604 KN Weight of building = 75377.37KN Cs = SDS/ (R/Ie) = 0.4/ (8*1) =0.05 CS MAX = SD1/ T(R/Ie) = 0.2583/ 1.039(8/1) =0.03107 controlled CS MIN = 0.044(SDS* Ie) = 0.044*0.4*1 =0.0178 more then 0.01 V =CS * W = 0.03107 *75377.3 = 2341.9 KN   Seismic checks : ETABS check codes Beam design: shear reinforcement for beams: From ETAPS: Vu = 181.49 KN Assume cover = 60 mm = 151.24.33 KN Vs = Vu/0.75 – = 90.73 Max = 617.34 KN ok section size is adequate Min = 87.65/ (420*500) = 0.4173 ) min = max ) min = max (0.273, 0.291) Use 0.4409 Assume stirrup with diameter (10mm) two leg, Av =157 mm Spacing = = 356.08mm Smax = min (d/2, 600) = 500/2 = 250 mm Use 1Ǿ 10 /250 mm Longitudinal reinforcement for beams: To check longitudinal steel in this span Ꝭ=) Mu =265 KN.m bw = 350 mm d =490 mm fc = 28 MPa fy =420 MPa Ꝭ =0.00905 ≤ Ꝭmin =0.0033 Use Ꝭ = 0.00905 AS = 0.00905 * 350 *490 AS = 1552.39 mm2   Figure 27 : Rienforcment in beams At bottom Ꝭ =0.00413≤ Ꝭmin =0.0033 Use Ꝭ = 0.00413 AS = 0.00413* 350 *490 AS = 708 mm Torsion reinforcement for beams: columns Slab design Design slab for flexure :   Ꝭ=) In y direction bottom Mu = 28 KN.m Ꝭ =0.00187 As = 0.00187 *1000 *200 =375 mm2 /m In y direction top Mu = 23 KN.m Ꝭ =0.00153 As = 0.0018 *1000 *200 = 360mm2 /m In y direction bottom Mu =8 KN.m Ꝭ =0.000531 As = 0.0018 *1000 *200 =360 mm2 /m In y direction top Mu = vary small (KN.m ) Ꝭ =0.0018 As = 0.0018 *1000 *200 = 360 mm2 /m   Shear wall With according to section 14.5.3 ACI318-11, minimum thickness of bearing wall is the larger of 10cm and 1/25 the supported height or length, whichever is shorter. For wall in stories 2-10 The controlled value is: h = = 10 cm. Shear wall in these stories has h = 25 cm which more than 10. Then, it’s ok.     For G. story , h for wall = 5 m ., the minimum thickness is 20 cm. h for G story wall = 25 cm > 20 cm - Ok Footing design Figure 37 : Reinforcment detailing for single footing Design wall footing Figure 38 : Wall footing on Auto-CAD image1.jpeg image2.png image2.jpg image4.png image3.jpg image6.png image4.jpg image5.jpg image9.png image6.jpg image11.png image7.jpg image8.jpg image14.png image9.emf We define load combination for long term and we assume that 30% of live load sustained forever ∆𝑙𝑡=ձ∞∗ ሺ ∆𝐷+∆𝐿 ሻ + ձ𝑡𝑖𝑚𝑒∗∆𝑙 𝑠𝑢𝑠𝑡𝑎𝑖𝑛𝑒𝑑 ձ ∞ =2 ձ time = 2 ∆ LT = 2SD +2D +1.3L Figure 1: slab deflection Table 7 : deflection values for panel Allowable deflection = 𝐿 240 = 5.59 /240 = 40 mm Deflection from ETABS = deflection at center – (sum of deflection at points1, 2, 3, 4)/4 Deflection from ETABS = 26.069 – (3.168+2.841+6.315+4.1)/4 = -21.963 mm Deflection from ETABS ≤ allowable deflection (its ok) U Z Reading in(mm) U Z1 -4.1 U Z2 -6.315 U Z3 -2.841 U Z4 -3.168 U Z-max -26.069 image10.PNG image17.png image12.PNG image19.png image13.PNG image15.PNG image22.png image16.PNG image17.emf Seismicity of the site expressed by many parameters, the major one is peak ground acceleration which expressed by Z factor as the following: Figure 25 : Seismic Hazard Map and Seismic Zone Factor for Palestine The project is proposed to build in Nablus, which classified as zone 2B with Z=0.20. image18.emf Then; Ss, S1 and the site class will be used to determine the factors Fa and Fv. Figure 4.5: site coefficient for short period, Fa. As shown in the figure above Fa = 1.2 Figure 4.6: Site coefficient for long period, Fv. The factor of Fv will be calculated by interpolation: Fv = 1.6+1.5 2 = 1.55 SMS, SM1 will be calculated from Fa, Fv, Ss, S1 depending on the site c lass image19.emf Seismic design category: According to risk category and SDS and SD1 the seismic design category will be determined . Figure4.7: Seismic Design Category Based on Short Period Response Acceleration Parameter. Figure 4.8: Seismic Design Category Based on long Period Response Acceleration Parameter. As shown in figures above the design category will be D. image20.emf K = 1.298 Force to floor (KN ) Cvx Wi hi k Weight (KN) Height (m) Floor 461.588 0.1971 663815.93 7537.73 31.5 10 405.382 0.1731 582947.17 7537.73 28.5 9 350.68 0.14982 504579.65 7537.73 25.5 8 298.123 0.1273 428917.223 7537.73 22.5 7 247.53 0.1057 356209.5 7537.73 19.5 6 199.295 0.0851 286770.67 7537.73 16.5 5 153.62 0.0656 221010.95 7537.73 13.5 4 110.91 0.04736 159493.97 7537.73 10.5 3 70.662 0.0306 103055.244 7537.73 7.5 2 42.3181 0.01807 60884.06 7537.73 5 1 2340.1 SUM =0.9997 SUM = 3367684.367 image21.emf In this section will be making check on story drift (horizontal displacement of storey) Table 11: Allowable Story Drift, Δa. Values of displacement for story10 from ETAPS : Table 12 : Diaphragm center of mass displacement for story 10 image23.png image30.png image31.png image24.png image25.png image26.emf Longitudinal reinforcement in columns : Area of cross section for column = 320000 Area of steel = 0.01 * 320000 mm 2 = 12 bar Ǿ 20 Figure 1 : rebar percentage in columns Design stirrups for column Spacing for the stirrups is the smallest of: (use Ǿ 12) 1. 16 * d 20 = 32 cm 2. 48 diameter for stirrups = 57 cm 3. Short dimension = 40 cm Then we use 32 cm Stirrups Ǿ 12 / 30 cm image35.png image27.emf Shear wall design using ETABS will be check in this section with accordance to requirements of ch.14 and 11 in ACI318-11 code. Type of forces act on wall will be as shown in the following figure: Figure 1 : In-plane and out-of-plane forces And, the dimension of the wall will be as the following: Figure 2 : Defining shear wall height, length, and depth image37.png image28.emf . W2 has the larger tributary area = 18.44 m2 Wall 2 dimensions: Lw = 2 m h = 25 cm hw = 5 m d with accordance to section 11.9.4 ACI318-11 = 0.8 Lw. then, d= 1.6 m.  Design for out of plane shear which be perpendicular on wall plan treated as design shear in slab with  Vc = 1 6   ξ fc ′  h  d. Then for wall 2, max value for out of plane shear will be as in the following figure: Figure 1 : Walls layout. image29.emf Figure 1 : Out of plane shear for wall 2. Then,  Vc= 1 6   ξ fc ′  bw  d   Vc= 1 6  0.75  ξ 28  2000  160=211.6 KN≥ 0.8869 KN .. OK image30.emf  With according to section 11.9.6 ACI318-11, Vc value for out of plane shear will be determined by using equation 11-27, and equation 11-28 rejected because term (Mu/Vu – lw /2) is negative, then 11-27 will be applied as the following:  Vc=  0.27  ξ fc ′  h  d+ 𝑁𝑢  d 4  𝑙𝑤 Because of first combination will give max in plane shear, Nu for this combination will be used = 9743 KN  Vc=0.75  0.27  ξ 28  250  1600+ 9743000  1600 4  2000 =2377 KN Figure 1 : In plane shear for wall 2 Then, Vu=66.79 KN<  Vc=2377 KN then,it ′ s ok  According to section 11.9.9.2 ACI318-11, Ratio of horizontal shear reinforcement area to gross concrete area of vertical section, t, shall not be less than 0.0025. Then, area of horizontal steel = 0.0025  0.25  5  10^6 = 3125 𝑚𝑚 2 . image31.emf And Spacing of horizontal shear reinforcement shall not exceed the smallest of: 1- lw /5 = 0.4 m = 400 mm 2- 3h = 3  0.25 = 0.75 cm = 750 mm 3- 450 mm Then, S max = 40cm for horizontal steel. Use S = 15 cm for vertical steel  For vertical steel, ratio of vertical shear reinforcement area to gross concrete area of horizontal section, L, shall not be less than the larger of: 1- L=0.0025+0.5 ൬ 2.5− ℎ𝑤 𝑙𝑤 ൰ (t−0.0025) ( Error! No text of specified style in document..1) And, 0.0025 Then, L=0.0025. Then, area of vertical steel = 0.0025  250  5000 = 3125 𝑚𝑚 2 . And Spacing of vertical shear reinforcement shall not exceed the smallest of 1- lw /3 = 0.66 m 2- 3h = 0.75 m 3- 450 mm Then, Smax = 45cm for vertical steel. image32.emf Then, for vertical steel two layers will be placed with As for each layer = 1536 𝑚𝑚 2 along d for the 2 m long wall. which mean Two layers of bars 8mm in horizontal direction 1∅8/65mm will be used. Table Error! No text of specified style in document.-1: ETABS design result for wall-2. Figure 1 : Values of reinforcement from ETABS image33.emf Design the foundation in two different cases for a three story reinforced concrete building. Case 1: if the allowable bearing capacity of the soil is 300 KN/ m 2 Design the single, combined and wall footing. 5.5.1 Design single footing : q = P / A q = 300 KN /m 2 P = 1833 KN The area equal 6.11 m 2 Let b = 3 L = 2.2m P u = 2282 Kn Detriment d by shear equation q u = 362.22 KN /m 2 3000(1100-d) 0.36222 = 0.75/6*(24^0.5)*3000 *d 1837d = 1195260-1068.66d d = 41.44cm h = 50cm Checks the punishing shear b o =4.04m V= 0.75 *(24^0.5)*4000 *300 /3 V = 1469.6 KN 2282– 362 (0.8+0.41)(0.4+0.41)= 2003.26 KN It’s ok Ꝭ= 0.85𝑓𝑐 𝑓𝑦 (1−√(1− 2.61𝑀𝑢 𝑏𝑑 2 𝑓𝑐 ) M11 =325.24KN.m 0.00541139 As = 2218.6mm 2 /m 8 Ǿ 20 / m in x direction image34.png image36.png image38.png image39.emf w = 4704/2 =2352 KN/m Determine B 300= 2352 /(b*2) B= 4 m q u = 2945 /(4*2) =368.125 KN.m 2 Determint d 0.368125 (850-d)4000 = (0.75 *24 0.5 *4000*d )/6 2449.4d =1251625-1472.5d d =320mm h = 400 mm by SAP m11 = 265.41 kn.m Ꝭ= 0.85𝑓𝑐 𝑓𝑦 (1−√(1− 2.61𝑀𝑢 𝑏𝑑 2 𝑓𝑐 ) Ꝭ=0.00741 As =2371.42 8 Ǿ 20 / m in x direction M22 = 590kn.m Ꝭ=0.0188 As =5189 10Ǿ 26/m As min = 4 Ǿ 14 /m image40.png image41.png image42.png image43.emf Design wall footing 1. w = 757.28 KN/m Determine B 300= 757.28 /(b*1) B= 2.6 m q u = 931 /(2.6) =358.3KN/m 2 Determine d 0.3583 (1175-d)1000= (0.75 *24 0.5 *2600*d )/6 1592d =421002.5-358.3d d =230mm h = 300 mm by SAP m11 = 195 kn.m Ꝭ= 0.85𝑓𝑐 𝑓𝑦 (1−√(1− 2.61𝑀𝑢 𝑏𝑑 2 𝑓𝑐 ) Ꝭ=0.0109 As =2524.38 8 Ǿ 20 / m in x direction M22 = 920kn.m Ꝭ=0.004833 As =1111.7 6Ǿ 16/m As min = 3 Ǿ 14 /m image44.png /docProps/thumbnail.jpeg