Design Of Reinforced Concrete Structures (II) PDF

29d ago
1 Views
0 Downloads
1.74 MB
33 Pages
Transcription

Design of ReinforcedConcrete Structures (II)DiscussionEng. Mohammed R. Kuheil

ECIV 4316Design of Reinforced Concrete Structures (II)2017ReviewThe thickness of one-way ribbed slabsAfter finding the value of total load (Dead and live loads), the elements are designed. Based on themechanism of load transfer, the ribs are the first elements to take the load applied. The design of itis based on three requirements that must be fulfilled (deflection, shear, and flexure).To fulfill deflection requirement, the following table that shows minimum thicknesses for ribsand beams is used. When choosing this thickness, the deflection requirement is accomplished.CasesMin. requiredthicknessSimplyOne EndBoth ileverL/8L: is the span length in the direction of bending form center to center of support (In ribs, thesupport is beam).Hint: when the case is cantilever the length of span taken from the face of the support to the end ofspan.Loads1. Dead loadDead load in buildings is include own weight, covering materials, and equivalent partitionload and external walls.2. Live loadTo find the live load applied on a building must be refer to the general codes (IBC, UBC and ASCE 710).The value of live load varies according to the usage of the building, According ASCE 7-10, Ch. (4):For regular residential building 200 kg/m2 0.20 t/m2 .For dance halls and ballrooms 490 kg/m2Eng. Mohammed R. Kuheil1

ECIV 4316Design of Reinforced Concrete Structures (II)20173. Special loadsThis loads are include seismic forces, wind load, and other dynamic loads. Our concern in thiscourse are to find live and calculate the dead loads only and the special loads are used in advancedcourses.The dead load details:1. Own weightTotal own weight Block weight concrete weightBlock weight (the thickness of block in cm) kgExample: when the thickness of block is equal 20 cm the weight of block is equal 20 kg.Concrete weight the volume of concrete ΥcΥc the unit weight of reinforcement concrete𝐇𝐢𝐧𝐭: 1. the unit weight of plain conctete (with out reinforcement) 2.4 t/m32. the unit weight of reinforcement conctete 2.5 t/m3The volume of concrete the total volume of the representative sample the volume blockThe total volume of the representative sample (the length of block the width of rib) (block width) total thicknessthe total own weight per unit area Conrete weight (ton) Block weight (ton)t/m2Area of the representative sample2. Covering MaterialsMaterialThickness (cm)Unit Weight (t/m3)Sand101.7Mortar32.2Tile2.52.5Plaster22.2Eng. Mohammed R. Kuheil2

ECIV 4316Design of Reinforced Concrete Structures (II)2017The total weight of covering material (Thickness (h) unit weight)Covering Materials3. Equivalent partition load (EPL) (internal walls)1 m2 from the block wall 1 12.50 Blocks.0.4 0.2The weight of 1 m2 12.5 the weight of 1 blockThe weight of plaster t/m2 thickness (2 faces) unit weightThe total weight of EPL every 1m2 from the wall weight of plaster weight of blockThe total weight of EPL (ton) (total weight / m2 ) height of story total length of EPLThe total weight of EPL (t/m2 ) the total weight of EPL (ton)net area of slab𝐇𝐢𝐧𝐭: The net area of slab total area of slab all open areasEng. Mohammed R. Kuheil3

ECIV 4316Design of Reinforced Concrete Structures (II)20174. External wallsThe calculation of the external wall is the same of the internal wall but the external wall is carrydirectly on the exterior beams and the internal wall carried on the slab.1 m2 from the block wall 1 12.50 Blocks.0.4 0.2the weight of 1 m2 12.5 weight of 1 blockthe weight of plaster t/m2 thickness (2 faces) unit weightthe total weight of EPL every 1m2 from the wall weight of plaster weight of blockthe total weight of EPL (t/m′ ) (total weight m2 ) height of storyLoad combinations:Dead and live loads (DL LL):1.4 D1.2 D 1.6 LDead (D), live (L) and wind (W):1.2 D 1.0 L1.2 D 0.8 W1.2 D 1.6 W 1.6 L0.9 D 1.6 WDead (D), live (L) and Earthquake (E):1.2 D 1.0 L 1.0 E0.9 D 1.0 EEng. Mohammed R. Kuheil4

ECIV 4316Design of Reinforced Concrete Structures (II)2017Example:Calculate the factored load (dead and live loads) per unit area for residential building.Given:Total area of the floor 250 m2Total thickness of slab 30 cmTopping slab 8 cmThe width of rib 12 cmThe total length of EPL 60 m’The total length of exterior walls 65 m2The thickness of EPL 12 cmThe thickness of plaster 1.5 cm for all elements, but in the exterior face is equal 2 cmThe thickness of sand 13 cmThe mortar thickness 3 cmThe thickness of plaster 2 cmThe thickness of tile 2.5 cmEng. Mohammed R. Kuheil5

ECIV 4316Design of Reinforced Concrete Structures (II)2017Hint:The area of stair 13 m2Another open area 15 m2Solution:1. Dead load:Own weightVT (0.12 0.40) 0.25 0.30 0.039 m3VB (0.22 0.25 0.40) 0.022 m3VC (0.039 0.022) 0.017 m3WC 0.017 2.5 0.0425 tonWB 0.022 tonWT 0.0425 0.022 0.0645 tonwT (per unit area) WC WBArea of the representative samplethe weigth of total sample 0.0425 0.022 0.50 t/m20.52 0.25Equivalent partition loadthe total weight of EPL /m2 from the wall (12.5 0.012) (0.04 2.2) 0.238 tonthe total weight of EPL(ton) 0.238 3.00 60 42.84 tonthe total weight of EPL (t/m2 ) Eng. Mohammed R. Kuheil42.84 0.193 t/m2250 13 156

ECIV 4316Design of Reinforced Concrete Structures (II)2017𝐂𝐨𝐯𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐞𝐫𝐢𝐚𝐥𝐬The total weight of covering materials (Thickness (h) unit weight)The total weight of covering materials (0.13 1.7 0.03 2.2 0.02 2.2 0.025 2.5)The total weight of covering materials 0.394 t/m2The total service dead load own weight EPL covering materialsThe total service dead load 0.50 0.193 0.35 1.043 t/m2 .2. Live loadFrom the ASCE 7-10 Chapter 4 L.L. for regural residential building 200 kg/m2 0.20 t/m2Factored load max(1.4 DL or 1.2 DL 1.6 LL)U (t/m2 ) max(1.4 1.043 or 1.2 1.043 1.6 0.2)U (t/m2 ) max(1.46 or 1.57)U (t/m2 ) 1.57 t/m2 .Moment and Shear design:According the following example.Eng. Mohammed R. Kuheil7

ECIV 4316Design of Reinforced Concrete Structures (II)2017Example: For the one way ribbed slab shown in figure below, design any of the typical ribs andmain interior beam.Solution: From deflection control3.88 4 1.75The thickness of slab (hmin. ) max. (,,) 0.219 m 22 cm18 21 8Used Hollow block 25 40 17 cm.Topping slab thickness 22 17 5 cm.The topping slab is designed as a continuous beam supported by the ribs. Due to the large numberof supporting ribs, the maximum bending moment is taken as Mu Wu l2c /12Assume DL 0.85 t/m2 , LL 0.20 t/m2 f′c 200 kg/cm2 fy 4200 kg/cm2Wu 1.2 0.85 1.6 0.2 1.34 t/m2For a strip 1 m width Wu 1.34 t/m′Mu t 1.34 0.42 0.018 t. m123 Muϕb f′cEng. Mohammed R. Kuheil8

ECIV 4316t Design of Reinforced Concrete Structures (II)3 0.018 1050.9 100 2002017 2.06 cm 5 cmBut the t is not to be less than 1/12 the clear distance between ribs, nor less than 5.00 cmt max. (2.06,1 40, 5) 5 cm.12Area of shrinkage reinforcement As 0.0018 b h 0.0018 100 5 0.9 cm2 /mUse 4 ϕ 6 mm or ϕ 8 mm @ 50 cm in both direction.Now, we must be check for shear and bending moment (Using ROPOT structural analysis software)To find the load per meter length, take a strip (shown in the figure above).Factored load (1.2 0.85 1.6 0.20) 0.52 (width of strip) 0.70 t/m′The result from ROPOT structural analysis softwareS.F.D.B.M.D.Eng. Mohammed R. Kuheil9

ECIV 4316Design of Reinforced Concrete Structures (II)2017Shear design:Vu (max.) 1.65 ton.Now, calculate the capacity for rib for shear ϕVcResistance force(ϕVc ) must be greater than applied force Vu (max.)ϕVc ϕ 0.53 f′c b dd thickness of slab (h) cover stirrup 0.5 dbd 22 2 0.6 0.6 assume ϕ 12 mm reinforcing bars and ϕ 6 mm stirrupsd 18.80 cmϕVc 0.75 0.53 200 12 18.80 1.27 ton.1000Shear strength provided by rib concrete ϕVc may be taken 10 % greater than those for beams.It is permitted to increase shear strength using shear reinforcement or by widening the ends of ribs.1.1 ϕVc 1.1 1.27 1.4 ton.Use 4 ϕ 6 mm U-stirrups per meter run are to be used to carry the bottom flexural reinforcement.Since critical shear section can be taken at distance d from faces of support (beam).the previous of Vu (max.) (1.65 ton) form the center of the support.so we will take the distance 0.5 beam width d 0.5 75 19.40 56.9 cm.so the critical shear section will be at a distance 56.9 cm from the center of support (beam).Eng. Mohammed R. Kuheil10

ECIV 4316Design of Reinforced Concrete Structures (II)2017From the previous figure:Vu (critical shear section) 1.26 ton 1.1ϕVc (1.4 ton) OKThe rib shear resistance is adequate.Eng. Mohammed R. Kuheil11

ECIV 4316Design of Reinforced Concrete Structures (II)2017𝐖𝐡𝐞𝐧 𝐕𝐮 𝛟𝐕𝐜 ? ? ?.We have six choices:1. Increase the compressive strength of concrete f′c.2. Increase the width of rib.3. Increase the depth of the slab, which is uneconomic choice.4. Can be used stirrups as a shear reinforcement to resist the applied force.5. Change the direction of blocks at maximum shear area so that the width of rib is increased. thefigure below describes this solution.6. Enlarge the beam width.Eng. Mohammed R. Kuheil12

ECIV 4316Design of Reinforced Concrete Structures (II)20172. Moment designMu (max.) 1.59 t/m2 .###AMidSpan ABBMidSpan BCMidCSpanDCDMidSpan DEEMidFSpan o 89518Db (mm)1410101010101010121010d . eck ScOKOKOKOKOKOKOKOKOKOKOKρ 0.85 f′c2.353 105 Mu[1 1 ]fy0.9 bw d2 f ′ c0.80 f ′ c 140.80 200 14ρmin. max. (, ) max. (,) 0.0033fyfy42004200ρmax. 0.31875 0.85 f′c 0.31875 0.85 200 0.0129fy42000.85 (200)2.353 105 1.59 ρ [1 1 ] 0.011742000.9 12 18.80 200ρmin. ρ ρmax. OK ρused 0.0117d 22 2 0.6 0.6 18.80 cmEng. Mohammed R. Kuheil13

ECIV 4316Design of Reinforced Concrete Structures (II)2017As ρused b d 0.0117 12 18.80 2.64 cm2try ϕ 12 mm# of bars As2.642.64 2.332Area of one bar0.7854 1.220.7854 dbtry ϕ 14 mm# of bars As2.642.64 1.71Area of one bar0.7854 db 2 0.7854 1.42use 1ϕ12 mm 1ϕ14 mm.Eng. Mohammed R. Kuheil14

ECIV 4316Design of Reinforced Concrete Structures (II)Eng. Mohammed R. Kuheil201715

ECIV 4316Design of Reinforced Concrete Structures (II)2017Beam designDesign the continuous beam (B2) shown in the figure below.Use f ′ c 200 kg/cm2 and fy 4200 kg/cm2 .DL 1.00 t/m2 and LL 0.20 t/m2 .𝐇𝐢𝐧𝐭: the thickness of slab is equal 25 cm.Solution:Wu (t/m2 ) 1.2 1.00 1.6 0.20 1.52 t/m2 .Wu (t/m′) Wu (t/m2 ) Hatch Area 1.52 4 6.08 t/m′.The width of Hatch Area Eng. Mohammed R. Kuheil3.25 3.25 0.75 4 m2216

ECIV 4316Design of Reinforced Concrete Structures (II)2017By using Autodesk ROPOR structural analysis softwareS.F.D.B.M.D.Vu (max.) 19.00 ton. Vu (at a critical section "d from the face of column") 15.4 ton.Mu (max.) ( ve) 10.69 t/m2 .Mu (max.) ( ve) 19.00 t/m2 .Eng. Mohammed R. Kuheil17

ECIV 4316Design of Reinforced Concrete Structures (II)20172. Shear designHidden beamϕVc ϕ 0.53 f′c b dd 25 4 0.8 0.7 19.50 cm. assume db 14mm and dstirrup 8mmϕVc 0.75 0.53 200 75 19.50 8.22 ton1000Vu VC VsϕVs Vu ϕVC 15.40 8.22 9.6 ton.ϕ0.75Check for ductility Assume db ϕ14 mm & dstirrups ϕ8 mm .2.2 f′c bw d 2.2 200 75 19.50 45.5 ton Vs (9.6 ton).1000The dimensions of the cross section are adequate for ensuring a ductile mode of failure.Shear zones:Zone (A)Vu ϕV2No shear reinforcement is required, but it is recommended to use minimum area of shearreinforcementTrying two legged ϕ8 mm vertical stirrupsAv0.20 f ′ c bw 3.5 bw0.20 200 75 3.5 75( ) max. (,) (,) 0.0625cm2 /cmS min.fyfy42004200Smin.π2Av (min.) # of legges area of cross section for one leg 2 (4 0. 8 ) 16 cm0.06250.06250.0625Eng. Mohammed R. Kuheil18

ECIV 4316Design of Reinforced Concrete Structures (II)2017Hint:1. When Vs 1.1 f′c bw d the maximum stirrup spacing is not to exceed the smaller of d/2 or60 cm.dSmin. max. (2 , 60 cm)2. When 2.2 f′c bw d Vs 1.1 f′c bw d the maximum stirrup spacing is not toexceed the smaller of d/4 or 30 cm.dSmin. max. (4 , 30 cm)3. When Vs 2.2 f′c bw d must be enlarge the section dimensions.Vs 9.6 ton. f ′ c bw d 200 75 19.519.5 20.68 ton Vs Smin. max. (, 60 cm) 9.75 cm.10002dSused min. ( , 60cm, Smin. ) ( 9.75 cm, 60 cm, 16 cm) 10 cm.2NoteWhen the thickness of slab is small, (d) the spacing between shear stirrups are be very small andwe should be enlarge the depth of beam (drop beam).Zone (B)ϕVc Vu ϕV, minimum shear reinforcement is required.2Similar to zone AZone (C)Vu ϕVc , shear reinforcement is required.Vu VC VsϕAvVs14.4 1000( ) 0.176 cm2 /cmS Calculated. fy d 4200 19.5Eng. Mohammed R. Kuheil19

ECIV 4316Design of Reinforced Concrete Structures (II)2017AvAv( ) ( ) OKS Calculated.S min.Vs S Vu ϕVC 19.00 8.22 14.4 ton.ϕ0.75Av fy d 2 (0.5) 4200 19.5 8.53 cmVs9.6 1000dSused min. ( , 60cm, Smin. , S) ( 9.75 cm, 60 cm, 16 cm, 8.5) 8 cm (NOT practical).2Note:The spacing between the stirrups is small because the criteria of S d/2 in minimum zones and inmaximum shear zone (C) because the Vs is high.To avoid these problems in shear should be enlarge the depth of section (Drop beams) to overcomethe criteria S d/2.Eng. Mohammed R. Kuheil20

ECIV 4316Design of Reinforced Concrete Structures (II)2017Design of columnsClassification of columns according to support type:Columns are classified according to the type of support into two categories:1. Pin supported columnsIn this type of columns, the supports don’t resist moment and so do the columns. This type ofcolumns resists only axial load (Design 1).2. Moment resisting columnsThis type of columns resists both axial load and bending moment, and the design of this typeof columns depends on the interaction between both forces.The type of column is depends on the details of steel reinforcement at the joints of columns.Classification of Frames (Sway and Non-Sway Frames):Moment resisting frames are classified into two types according to lateral movement probability:1- Sway frames: at which the lateral movement is

ECIV 4316 Design of Reinforced Concrete Structures (II) 2017 Eng. Mohammed R. Kuheil 1 Review The thickness of one-way ribbed slabs After finding the value of total load (Dead and live loads), the elements are designed. Based on the mechanism of load transfer, the ribs are the first elements to take the load applied. ...