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An-Najah National University Engineering Collage Civil Engineering Department

An-Najah National University Engineering Collage Civil Engineering Department. Graduation project: Static Design of “Midad paper factory " Supervised by: Dr. Riyad Abdl- karim Prepared By :Imad Rasheed. Chapter One: Introduction. Project description .

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An-Najah National University Engineering Collage Civil Engineering Department

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  1. An-Najah National UniversityEngineering CollageCivil Engineering Department Graduation project: Static Design of “Midad paper factory" Supervised by: Dr. Riyad Abdl- karim Prepared By :Imad Rasheed

  2. Chapter One: Introduction

  3. Project description • “Midad" factory is to be constructed in Bethlehem cumulative area is 1404 m2 area. It consists of two basement and one floor , the two basements will be a storages and the first floor used as storage and equipment's work the table below show the area of each story:

  4. Mechanical properties • Materials: • Concrete: For beams and slabs: Fc = 280 Kg/cm2 ,ρ= 2500 Kg/m3 For Columns: F’c = 300 Kg/cm2 • Reinforcing Steel: Fy =4200 Kg/cm2 • Soil Bearing capacity = 2.5 Kg/cm2

  5. Chapter twoPreliminary design

  6. 2.Preliminary design *one way ribbed slab (in x-direction) : hmin = Ln\18.5 = 5.4/18.5 =29.2 cm Use H=32cm Use concrete blocks of 40*25*24cm

  7. 2.Preliminary design Two way flat plate: ,(550\33))=max(18.33 ,17)(H min=max((550\30 Use H=20cm

  8. Preliminary design 1- Beams • for 6m beam span h min. = (600/21= 29 cm) Preliminary Dimensions : 32*30cm • For 18m beam span H=Ln \16=18\16=112.5cm Preliminary Dimensions=100*40cm

  9. Preliminary design 2-Column : *for C1 column group assume dimensions=60*60cm *for C2 group assume dimensions=100*80cm

  10. Chapter Three:3D MODELING CHECKS

  11. Modeling Checks • 1- check Compatibility

  12. Modeling Checks • 2- check Equilibrium result from SAP DL=22633 KN , LL=10525 KN as shown in figure below

  13. Modeling Checks • 2- check Equilibrium Manual calculation DL=22368 KN , LL=10692 KN The percentage of live load error=(1.6%)<(5%) ok The percentage of dead load error=(1.2%)<(5%) ok

  14. Modeling Checks • 3- stress-strain relationships Manual check: Take beam (E1-E4) that length=18m

  15. Modeling Checks Depth=1 m length=18 m width=0.4 m own weight=1*0.4*25=10 KN\m Ultimate load on slab=11.134 KN\ m2 Ultimate load on beam=10*1.2+(11.134*6) =78.804 KN\m Mu =Wu*L2\8=78.804*182\8=3191.56 KN.m

  16. Modeling Checks Sap results:

  17. Modeling Checks Mu=1121.4+1857.3=2978.7 KN.m The percent of error=(2978.7- 3191.5)\3191.5 = 6.6% < 10% ok *After there checks the designer can confident with the result of sap

  18. Modeling ChecksCheck stability

  19. Chapter Four:Static design

  20. Static design *After checking the structure by SAP, failure in the edge and interior beams due to torsion and shear was found as shown in Figure below.

  21. Torsion and depth problem This problem will be solved by increasing the width of beams B1 & B2 to (40cm) and B4 to (60cm) and increase the depth to 60cm intead of 30cm.

  22. Beams after modification

  23. Static design • Slab • Beams • Columns • Footings • Basement walls design • Shear walls design

  24. 1.one way ribbed Slab A-Bending resistance B-Check shear for slab

  25. A-Bending resistance Slab loads: Own weight=(0.24*2*0.2*12)+[((0.08*0.52)+(0.12*0.24))*25] =1.152+1.76=2.912KN\m.rib S.D.L=0.015*23=0.345KN\m2 Wd=0.345+(2.912\0.52)=5.945KN\m2 WL=2.5KN\m2 Wu=1.2*5.945+1.6*2.5=11.134KN\m2 Wu=11.134*0.52=5.72KN\m

  26. A-Bending resistance Max.negative moment= WuLn2\10= 17.62 KN.m Max.positive moment= WuLn2\14= 12.59 KN.m As=2ø12(top) As=2ø12(bottom steel) As(shrinkage) = 0.0018 Ag As shrinkage=0.0018*1000*80=144mm2\m Max.spacing=min(5h or 30cm)=(40 or 30)cm As(shrinkage)=1ø8\30cm

  27. B-Check shear for slab Check of shear Max.Vu=1.15*Wu*Ln\2=1.15*5.72*5.55\2=18.25 KN øVc=(1.1*0.75*(1\6)*120*220*√28)\1000=19.21 KN Vu< øVc(no need for stirrup)

  28. Reinforcement detail in the one way ribbed slab

  29. Two way solid slab design After check the thickness for wide beam and punching shear the thickness change to 32 cm .Then the The designer take the area of steel as follow:

  30. C. Check deflection (immediate deflection): Deflection limitation=L/180 Deflection limitation=L/180=600\180=3.33cm

  31. C. Check deflection For beam=18m,deflection limitation=1800\180=10cm

  32. long term deflection: Deflection limitation=L/240 Deflection limitation=L/240=600\240=2.5 cm For beam=18m,deflection limitation=1800\240=7.5 cm

  33. long term deflection:

  34. Design of beams

  35. Design of beams Take beam (C1 – C4) .A-Slab loads: slab (o.w)=(0.24*2*0.2*12)+[((0.08*0.52)+(0.12*0.24))*25] =1.152+1.76=2.912 KN\m.rib S.D.L=0.015*23=0.345 KN\m2 Wd=0.345+(2.912\0.52)=5.945 KN\m2 Wu=1.2*5.945+1.6*2.5=11.134 KN\m2

  36. Design of beams: B- Beam design: H=60 cm b=40 cm Own weight=0.6*0.4*25=6 KN\m Wu=6*11.134+1.4*(6) =75.204 KN\m Max.positive moment= Wu*Ln2\14=75.204 *5.552\14=165.5 KN.m Max.negative moment= Wu*Ln2\10=75.204 *5.552\10= 231.6 KN.m

  37. Design of beams As=5ø18(top) in internal support As=4ø18(bottom) Max.Vu=1.15*Wu*Ln\2=1.15*75.204 *5.55\2=240 KN øVc=0.75*(1\6)*400*540*√28=142.87 KN Vu at d= Vu-Wu*d=240-(75.204 *0.54)= 199.4 KN Vn=199.4\0.75=266 KN VS=Vn-Vc=266-190.5=75.5 KN vs<2vc 75.5<(2*190.5) Smax=min(d\2,600mm)=min(270,600mm) take S=25cm

  38. Design of beams Use S=250mm Av\sAt\s:

  39. Column design After model the structure in sap the assumption of primarily column dimensions are not economical then take the ultimate axial force in column and take dimensions: Pu=2922KN For 1% steel reinforcement: ΦPn=Ag(0.438Fc+0.0052fy) Ag=44*44cm take Dimension = 45cm*45cm

  40. column design A= = 2.16 B= =

  41. column design For braced frames:. K=0.75

  42. column design ns = ≥ 1 Pc = EI = Ec = 4700 = 25742960 KN/m2 Ig = 0.45(0.45)3/12 =0.0034 m4 Bd = Bd = = 0.45 EI = = 24145.12KN.m2 Pc = = 30614 KN. Cm = 1.0 (single curvature)

  43. column design ns = =1.14>1 Mc = ns×M Mc=1.14*83.227=95.36KN.m  = = 0.82 At X-axis Mu\bh2=95.36*10002\4503=1.04 At y-axis Pu\bh=2922*1000\4502=14.4MPa From the figure of interaction diagram

  44. column design

  45. column design  = 1% =min=0.01 As = 0.01*Ag = 0.01*450*450 = 2025mm2 Results from SAP = 2025 mm2

  46. column design

  47. 4.5- design of footing Design of group F1: *Required area of the footing Af req = = =8.51 m2. B= 3 m, L= 3 m Ultimate pressure under the footing: qu = = = 327.44 KN\m2 qu=0.327 N\mm2

  48. 4.5- design of footing *Check for punching shear: Ultimate shear force: Vu = qult[ B*L – (c1+d)2 ] = 0.327[ 30002 – (450 + 530)2] = 2.62*106 N Ø Vc = 0.75*1\3 bo d bo = 4(c+d) = 4(450+530) = 3920 mm Ø vc=0.75* 1\3* 3920*530= 2.84*106 N> vu So d=530 mm and H=600 mm Mu = = = 266.15KN.m

  49. 4.5- design of footing p p=0.00277 > min = 0.0018( min for shrinkage) A st = p*B*d = 0.00277*1000*530 = 1468 mm2\m. Use 8 Ø 16\m in both directions. Shrinkage steel: A shrinkage =0.0018BH A shrinkage = (0.0018*1000mm*600) =1080mm2/m.  use 6 Ø 16 \m in both directions.

  50. 4.5- design of footing

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