3.2 Reinforced Concrete Slabs Slabs are divided into suspended slabs. Suspended slabs may be divided into two groups:

Transcription

1 Sabah Shawkat Cabinet of Structural Engineering Reinforced Concrete Slabs Slabs are divided into suspended slabs. Suspended slabs may be divided into two groups: (1) slabs supported on edges of beams and walls () slabs supported directly on columns without beams and known as flat slabs. Supported slabs may be one-way slabs (slabs supported on two sides and with main reinforcement in one direction only) and two-way slabs (slabs supported on four sides and reinforced in two directions). In one-way slabs the main reinforcement is provided along the shorter span. In order to distribute the load, a distribution steel is necessary and it is placed on the longer side. Oneway slabs generally consist of a series of shallow beams of unit width and depth equal to the slab thickness, placed side by side. Such simple slabs can be supported on brick walls and can be supported on reinforced concrete beams in which case laced bars are used to connect slabs to beams. Figure 3.-1: One way slab, two-way slab, ribbed slab, flat slab, solid flat slab with drop panel, waffle slab In R.C. Building construction, every floor generally has a beam/slab arrangement and consists of fixed or continuous one-way slabs supported by main and secondary beams.

2 Sabah Shawkat Cabinet of Structural Engineering 017 Figure 3.-: Solid flat slab, solid flat slab with drop panels The usual arrangement of a slab and beam floor consists of slabs supported on crossbeams or secondary beams parallel to the longer side and with main reinforcement parallel to the shorter side. The secondary beams in turn are supported on main beams or girders extending from column to column. Part of the reinforcement in the continuous is bent up over the support, or straight bars with bond lengths are placed over the support to give negative bending moments. Figure 3.-3: Types of the reinforced concrete slab systems

3 Sabah Shawkat Cabinet of Structural Engineering Flat Slabs Flat plate is defined as a two-way slab of uniform thickness supported by any combination of columns, without any beams, drop panels, and column capitals. Flat plates are most economical for spans from,5 to 7,5m, for relatively light loads, as experienced in apartments or similar building. -A flat slab is a reinforced concrete slab supported directly on and built monolithically with the columns, the flat slab is divided into middle strips and column strips. The size of each strip is defined using specific rules. The slab may be in uniform thickness supported on simple columns. These flat slabs may be designed as continuous frames. However, they are normally designed using an empirical method governed by specified coefficients for bending moments and other requirements which include the following: 1. There should be not less than three rectangular bays in both longitudinal and transverse directions.. The length of the adjacent bays should not vary by more than 10 %. Figure : Post punching behaviour of slab- critical section The general layout of the reinforcement is based on the both bending moments (in spans) and bending moments in addition to direct loads (on columns).

4 Sabah Shawkat Cabinet of Structural Engineering 017 Figure 3..1-: Combined punching shear and transfer of moments Figure

5 Sabah Shawkat Cabinet of Structural Engineering 017 Figure 3..1-

6 Sabah Shawkat Cabinet of Structural Engineering Analysis and Design of Flat Plate To obtain the load effects on the elements of the floor system and its supporting members using an elastic analysis, the structure may be considered as a series of equivalent plane frames, each consisting of vertical members columns, horizontal members - slab. Such plane frames must be taken both longitudinally (in x-direction) and transversely (in y direction) in the building, to assure load transfer in both directions. For gravity load effects, these equivalent plane frames can be further simplified into continuous beams or partial frames consisting of each floor may be analysed separately together with the columns immediately above and below, the columns being assumed fixed at their far ends. Such a procedure is described in the Equivalent Frame Method. When frame geometry and loadings meet certain limitations, the positive and negative factored moments at critical sections of the slab may be calculated using moment coefficients, termed Direct Design Method. These two methods differ essentially in the manner of determining the longitudinal distribution of bending moments in the horizontal member between the negative and positive moment sections. However, the procedure for the lateral distribution of the moments is the same for both design methods. Figure : Steel shear heads, steel plats joined by welding

7 Sabah Shawkat Cabinet of Structural Engineering 017 Since the outer portions of horizontal members (slab) are less stiff than the part along the support lines, the lateral distribution of the moment along the width of the member is not Uniform. The procedure generally adopted is to divide the slab into column strips (along the column lines) and middle strips and then apportion the moment between these strips and the distribution of the moment within the width of each strip being assumed uniform. Figure : Moments and frames

8 Sabah Shawkat Cabinet of Structural Engineering 017 Figure: Example: 3.-1 Design and calculation of Flat Plate Geometric Shapes Slab thickness h d 300mm The geometry of the building floor plans: l 1 7.7m l k.3m l 3.6m l y 7.7m Construction height of object: k v.850m Dimensions columns: b s 00mm h s b s The peripheral dimensions of the beam: h o 0.5m b o 0.30m Figure: Load calculation Load per area Reinforced concrete slab thickness of 300 mm kn q do h d q do kn m 3 m

9 Sabah Shawkat Cabinet of Structural Engineering 017 floor layer: q 1d kn 3 1. q 1d. m Live load (apartments): kn m v d v d kn 3 kn.0 m 1.5 Total load on 1 m of slab: Force load Peripheral masonry thickness of 00 mm YTONG: Total load acting on the console: F 1d F 1 F 1d kN Investigation replacement frame in the X-axis Frame 1: Calculation model m q d q do q 1d v d q d kn F 1 kn 10 k v l y m 3 00mm1.35 m F kN load calculation Figure: 3..1-

10 Sabah Shawkat Cabinet of Structural Engineering 017 Load width in a direction perpendicular to the x: zs x l y Load in the x-direction: q dx q d zs x q dx kn m Calculation of internal forces Moment on a console: M k l k F 1d l k q dx M k 65.07kN m Moment of inertia: Transverse replacement frame: 3 l y h d I p I p 0.017m 1 Central girders replacement of frame: I st column: I p 3 h s I s m 1 I s 1 b s Bending stiffness: Transverse replacement frame: I p kn K p 1000 K p l 1 rad m.5kn m rad Central girders replacement frame: I st l 1000 kn K st K st rad m.813kn m rad Column K s I s 1000 k v kn K s rad m 1.87kN m rad

11 Sabah Shawkat Cabinet of Structural Engineering 017 Figure: M M M M M 1010' 1 M 10'10 1 M M 97 1 Primary moments in node 9: M 97o 1 0kN m M 910o l 1 M 911o 0kN m Primary moments in node 10: 1 M 109o q dx 1 q dx 1 l 1 M 108o 0kN m M 101o 0kN m 1 M 1010ò q dx 1 l M 10'10o M 1010ò Given M 97 kn m M 97o K s 3 9rad M 911 kn m M 911o K s 9rad M 910 kn m M 910o K p 9rad 10rad M 108 kn m M 108o K s 3 10rad M 109 kn m M 109o K p 10rad 9rad M 101 kn m M 101o K s 10rad M 1010' kn m M 1010ò K st 10rad M 10'10 kn m M 10'10o K st 10rad Equilibrium conditions: Node 9 M k M 97 kn m Node 10: M 910 kn m M 911 kn m 0kN m

12 Sabah Shawkat Cabinet of Structural Engineering 017 M 109 kn m M 101 kn m M 1010' kn m M 108 kn m 0kN v Find M 97 M 910 M 911 M 109 M 101 M 1010' M M 10'10 m The calculated moments of individual members of equilibrium conditions: M 910 v ( 1.0) kn m M kn m M 10'10 v ( 90) kn m M 10' kN m M 1010' v ( 5.0) kn m M 1010' kn m M 911 v (.0) kn m M kN m M 109 v ( 3.0) kn m M kN m The computation of shear forces in the individual members: l 1 V 910o q dx V 109o V 910o M 910 M 109 V 910 V 910o V kN M 910 M 109 V 109 V 109o V kn l 1 V 1010ò q dx V 1010' V 1010' V 1010ò 56.8kN l 1 l 1 v Maximum moment between 9-10 Mmax l 1 a V 910 a 3.99m V 910 V 109 M max a V 910 a M 910 q dx M max 371.3kN m Maximum moment between Mstr l l 1 M str V 1010' M 1010' q dx M str.31kn m

13 Sabah Shawkat Cabinet of Structural Engineering 017 Figure: Transformation moments for the part columned strip and between the columns M a M 910 M b M 109 M a mkN M b mkN M c' M max 1.5 M c M str 1.5 M c' 6.78kN m M c 5.539kN m Moments over support: p 0.75 kn m M a 1 p M a kn m M 1b pm b M 1b 09.3kN m M b M b 136.7kN m pm a M 1a M 1a M a M b 1 p Positively support moments: m 0.60 M 3c M c' m M 3c kN m M c M c' 1 m M c kN m

14 Sabah Shawkat Cabinet of Structural Engineering 017 Dimensioning of the reinforcement: Figure: Material characteristic of concrete f ckcyl and steel fyk f yd 375MPa f cd 1MPa The top reinforcement for moments: effective height: d h d 3cm width, which act the moment b l y b 3.85m Column strip M 1a: kn m M 1a 0.518MNm M 1a f cd 1MPa b 3.85m d 0.7m M 1a bd f cd b bdf cd df cd 100cm 59.5cm MN cm

15 Sabah Shawkat Cabinet of Structural Engineering 017 Among the columned strip M a : kn m M a 0.17MNm M a f cd 1MPa b 3.85m d 0.7m M a bd f cd bdf cd 100cm 17.95cm MN cm Column strip M 1b : M 1b 09.3kN m M 1b 0.09MNm f cd 1MPa b 3.85m d 0.7m M 1b bd f cd bdf cd 100cm.857cm MN cm Among the columned strip M b : M b 136.7kN m M b 0.136MN m f cd 1MPa b 3.85m d 0.7m M b bd f cd bdf cd 100cm 1.83cm MN cm The lower reinforcement for moments:

16 Sabah Shawkat Cabinet of Structural Engineering 017 Column strip M 3c : M 3c kN m M 3c 0.78MN m b 3.85m d 0.7m f cd 1MPa M 3c bd f cd bdf cd 100cm 31.06cm MN cm Among the columned strip M c : M c kN m d 0.7m M c 0.185MNm f cd 1MPa b 3.85m M c bd f cd bdf cd 100cm cm MN Investigation replacement frame in y Frame Calculation Model

17 Sabah Shawkat Cabinet of Structural Engineering 017 Load calculation Figure: q d q d l 1 l q d kn m Calculation internal forces Support part: 1 M a 1 l y M a 83.6 q d Among the supports: 1 M c' q d 16 l y M c' 36.73kN m a magnification between support: M c M c' 1.5 M c 53.1kN m kn m Transformation moments for the part columned strip and among columned support t of Ma M a1 pm a M a kN m p 0.75 M a M a M a 1 p kn m Between the support of M c M c1 mm c M c1 7.07kN m m 0.6

18 Sabah Shawkat Cabinet of Structural Engineering 017 M c M c kN m M c 1 m Dimensioning of reinforcement Upper reinforcement of moment: Effective depth: d h d 3cm Column strip M 1a : The width on which acting the moment: b l 1 l b.85m Column strip M 1a : M 1a 0.518MNm d 0.7m f cd 1MPa b.85m M 1a bd f cd bdf cd 100cm 61.06cm MN Between the column strip M a : M a 0.17MNm d 0.7m f cd 1MPa b.85m M a bd f cd bdf cd 100cm 18.65cm MN

19 Sabah Shawkat Cabinet of Structural Engineering 017 Column strip M 1a : M c1 0.7MNm d 0.7m f cd 1MPa b.85m M c1 bd f cd bdf cd 100cm 9.99cm MN Between column strip M a : M c 0.181MNm d 0.7m f cd 1MPa b.85m M c bd f cd bdf cd 100cm 19.77cm MN Investigation extreme frame replacement Calculation Model:

20 Sabah Shawkat Cabinet of Structural Engineering 017 Calculation of load: Figure: From the slab: q 3do l 1 q d l k q 3do kn m Peripheral masonry thickness of 00 mm YTONG: F 1 F 1 kn m kn 10 k v m 3 00mm1. Total load replacement frame: q kd q 3do F 1 q kd kn m Calculation of internal forces Moment of the end strip:

21 Sabah Shawkat Cabinet of Structural Engineering 017 Support bending moment: 1 M ka q kd 1 l y M ka Between the column bending moment: kn m 1 M kc q kd 16 l y M kc kN m Transformation moments for the part columned bands and among columned columned strip width: l 1 b p3 l k b p3 6.15m Moments over support: M ka l k M exta 1 M exta b p3 M inta M ka M exta M ka pm inta M k3a 1 p kn m M inta M k3a kN m M ka kn m Between the column moments: M kc l k M extc 1 M extc kN m b p3 M intc M kc M extc M kc mm intc M kc kN m M intc M k3c 10.36kN m M k3c 1 m Design the reinforcement to the reinforced concrete slab The top reinforcement for moments:

22 Sabah Shawkat Cabinet of Structural Engineering 017 effective height: d d 3cm Column extreme strip M exta : width which act moment b l k b.3m Column extreme strip M exta : see diagram B3-B3.3 M exta 0.65 MNm d 0.m f cd 1MPa b.3m M exta bd f cd bdf cd 100cm cm MN Column strip inside M ka : width, which acts moment, see diagram B3-B3.3 b l 1 b 1.95m M ka 0.56 MNm d 0.m f cd 1MPa b 1.95m M ka bd f cd bdf cd 100cm 33.07cm MN cm Among the columned strip M k3a : width, which acts moment

23 Sabah Shawkat Cabinet of Structural Engineering 017 b l 1 b 1.95m M k3a MNm d 0.m f cd 1MPa b 1.95m M k3a bd f cd bdf cd 100cm 9.61cm MN cm The lower reinforcement for moments: Column extreme strip M extc : width, which acts moment, see diagram B3-B3.3 b l k b.3m M extc 0.198MNm d 0.m f cd 1MPa b.3m M extc bd f cd bdf cd 100cm.893cm MN cm Column strip inside M kc : width, which acts moment b l 1 b 1.95m 0.153MNm d 0.m f cd 1MPa b 1.95m M kc

24 Sabah Shawkat Cabinet of Structural Engineering 017 M kc bd f cd bdf cd 100cm 19.09cm MN cm Among the columned strip M k3c : width, which acts moment b l 1 b 1.95m See diagram B3-B3.3 M k3c 0.10MNm d 0.m f cd 1MPa b 1.95m M k3c bd f cd bdf cd 100cm 1.136cm MN cm

25 Sabah Shawkat Cabinet of Structural Engineering 017 Example 3.-: In the example we are considering reinforced concrete slab flat, floor slab thickness is hd = 0.3m, Column diameter (round column) d =0.50 m, the maximum force applied one column at Nd= 1800 kn. d 0.5m h d 0.3m b 1m N d 1800kN Material characteristics: f cd 17MPa f ctm 1.MPa f yd 375MPa Figure:.3. 1 Coefficient of shear strength 1 stw 1 18mm A s1 n 5 na s m stw m bh d in both directions On 1m plate 1 stmin f ctm stmin f yd b 1 s n 1.0 f 1.5 b 1 h d h 1. h 1. 3 m g g 1.7 s n h f

26 Sabah Shawkat Cabinet of Structural Engineering 017 Carrying capacity of the concrete section q bu 0.h d g b f ctm q bu 6.86 m 1 Assess the resistance of the concrete section Maximum force per columns V cd N d V cd 1800 kn kn Basic critical perimeter u cr.51 m Shear force on the critical perimeter q d V cd q d 716. m 1 u cr kn Shear resistance of concrete q bu 6.86 m 1 kn q d 716. m 1 kn q d q bu We suggest shear reinforcement q d q bu 55.7 m 1 q bu kn Incorrect design, head to be designed so that they apply condition: q d q bu correct proposal Proposal of hidden head Maximum critical perimeter with hidden head U crmax 1.9u cr U crmax.78 m q da V cd q da m 1 kn q bu 6.86 m 1 U crmax kn q da q bu 55.7 m 1 q bu kn

27 Sabah Shawkat Cabinet of Structural Engineering 017 If we want to make a proposal without head, subject to the following parameters: d 1.0m f ctm0 1.0MPa Carrying capacity of the concrete section u cr0 d h d u cr0.08 m q bu0 0.h d g b f ctm0 u cr0 q bu kn V cd 1800 kn q bu0 Proposal visible head Geometry head 50.9 kn q bu0 V cd 5deg sin( ) 0.71 cos ( ) 0.71 h h q d 0.6m d h.0m V cd U cr U cr d h U cr 6.8 m q d 86.8 m 1 q bu 6.86 m 1 kn kn q d q bu We suggest shear reinforcement q d q bu Figure:.3. Proposal shear reinforcement - reinforced by bins q q q d bu su q su na ss ss s f yd f yd 190MPa ss 1 s 1 A ss 1m q su q d q bu q su 3.6 m 1 n 1 kn n is the number of bins reinforced, Ass area of reinforcement to a bin q Given su na ss ss s f yd m 1 A ss Find A ss A ss m

28 Sabah Shawkat Cabinet of Structural Engineering 017 n 1 5 number of bars in one bin / m 'ss = 0.5m 1 8mm diameter of one profile A sssku n 1 A sssku m Assessment of the punching according to EC design value of shear resistance of plate without shear reinforcement (per unit length of critical perimeter) v Rd1 Rd k shear resistance d A sssku A ss Rd 0.3 MN h d 0.3m k 1.6 h d m k 1.3 m b t 1m average width tension section f yk 10 MN kn min1 0.6b t min m h d f yk m min min h d b t min min1 min m min max min min concrete area A c h d b t A c 0.3 m The maximum degree of reinforcement max 0.0 A c max 0.01 m The average degree of reinforcement min max

29 Sabah Shawkat Cabinet of Structural Engineering 017 v Rd1 Rd k v Rd m 1 kn h d The maximum design value of shear resistance of plate with shear reinforcement (per unit length of critical perimeter) v Rd 1.6v Rd1 v Rd 71.5 m 1 kn Design value of shear resistance of plate with shear reinforcement (per unit length of critical perimeter) A s f yd sin ( ) i v Rd3 v Rd1 u Column diameter P s 0.5m Diameter of critical perimeter P u 1.5 P u 1. m h d P s Critical perimeter u P u u. m A cw P u P s A cw 1.3 m concrete shear area Assumption degree of shear reinforcement w A s wa cw A s 0 m f yd 360 MN m A s f yd sin( ) v Rd3 v Rd1 u v Rd m 1 kn carrying capacity

30 Sabah Shawkat Cabinet of Structural Engineering 017 The load effects V sd 1800kN Computing shear force 1.15 internal columns Figure:.3. 3 v sd V sd v sd 70.6 m 1 u as being applicable condition kn v sd v Rd3 vsd vrd3 incorrect design, design head Geometry head l h 0.9m h h 0.6m d crit 3.11m Critical perimeter with head u crit d crit u crit 9.77 m Concrete shear area d crit P s A cwh A cwh 7. m The expected level of reinforcement by shear reinforcement A sh wa cwh A sh 0.01 m A sh f yd sin( ) v Rd3 v Rd1 u crit V sd v sd v sd m 1 u crit v Rd m 1 kn v sd v Rd3 kn Slab with shear reinforcement to a void punching.

31 Sabah Shawkat Cabinet of Structural Engineering 017 w A sw sin ( ) i A x Space inside the critical perimeter less the contact surface A x d crit P s A x 7. m For dimensioning elements requiring shear reinforcement V Rd 0.5 f cd b w 0.9 d 1 cotg( ) f ck 5MPa f cd 13.3MPa f ck MPa if Smallest section width in the range of effective height b w 1.0m Height of the floor slab h d 0.3m cot ( ) 0 V Rd cot ( ) V Rd kn f cd b w h d Maximum distance of stirrups s max 0.3h d s max 0.09 m s max if s max 0.m s max 0.m s max 0.09 m V sd 1800 kn 3 V Rd kn Maximum diameter of reinforcement stirrups with a smooth surface s 0.01m Sectional area of shear reinforcement in the length range

32 Sabah Shawkat Cabinet of Structural Engineering 017 A sw s A sw m w A sw sin( ) w A x wmin wtab wmin 0.6 wtab Necessary degree of reinforcement EC: w if w wmin wmin w w Minimum design values of moments on columns in contact with the plate at the eccentric load x 0.15 Internal Column, top moment y 0.15 Internal Column, top moment V sd 1800 kn acting shear force m sdx m sdx 5 x V sd kn m sdy y V sd m sdy 5 kn Figure:.3.

33 Sabah Shawkat Cabinet of Structural Engineering 017 Example 3.-3: In the middle columns of dimensions as x bs from adjacent reinforced flat slab of thickness hs at a critical cross-section carries a full load slab, shear force Vcd = 00 kn, shear force from accidental load Vcd = 35 kn and the bending moment Mcd = 0 knm (moment transmitted from slab to reinforced column). Material characteristics: Figure: f ckcub 0MPa f ckcyl 0.8f ckcub f ckcyl 16MPa f cd 0.85 f ckcyl 1.5 f cd 9.067MPa 3 f ckcyl f ctm 1. MPa 10MPa f ctm 1.915MPa f yk f yk 35MPa f yd f yd 300MPa 1.15 where fctk is the characteristic tensile strength of concrete (5-percent fractile), fctm is the mean tensile strength and fck is the characteristic compressive strength of concrete measured on cylinders. The depth of reinforced concrete slabs h s 0.m Dimension columns: a s 0.0m b s 0.0m

34 Sabah Shawkat Cabinet of Structural Engineering 017 Bending moment and shearing forces: V cd1 00kN V cd 35kN M cd 0kNm U c1 h s a s U c1 0.6m U c h s b s U c 0.6m U cr U c1 U c U cr.m q dmax V cd1 q dmax m 1 kn U cr M kontr 0.V cd h s M kontr 13mkN If Mkontr less than the Mcd, should be respected Mcd 3 U c1 U c U c1 I cr I cr 0.1 m 3 6 Figure:.3.3- n 1 n 0. 3 U c 1 U c1 dmax V cd M cd n0.5 U c1 dmax 15.08m 1 kn U cr I cr Calculation of Qbu

35 Sabah Shawkat Cabinet of Structural Engineering 017 d d 18mm A s1 d A s m n 6 na s1 d m c 1 a s \ c b s stx d stx c h s h s stmin f ctm 1 stmin f yd sty stm d sty c h s h s stm stx sty s 1 50 b stm stmin s 1.1 h 1.m h s h 1.67 m 3 q bu f ctm h s 0. s h n q bur 0.f ctm m m q bu m 1 kn q bur m kn h s The reliability condition q dmax m 1 kn q bu 6.911m 1 kn q bur m kn q bueur m kn The cross-section without shear reinforcement does not comply I suggest shear reinforcement in the form of welded of mesh s 8mm A ss1 s A ss m ss 1 f yd ss A ss1 s ss ss m q dmax 0.5q bu

36 Sabah Shawkat Cabinet of Structural Engineering 017 Perimeter displaced the critical cross section c h s U crp c 1 h s ss ss U crp m q dmaxp V cd1 U crp q dmaxp m 1 kn It is less than qbu, that is, the cross section satisfies without shear reinforcement. Alternative we suggest shear reinforcement consisting of a flexible conduit at an angle =60.deg. q bu 6.911m 1 kn q dmax m 1 kn 60deg q dmax 0.5q bu U cr A sb A sb m sin( ) sf yd The proposal oh 1mm A soh oh A soh m A sb A soh

37 Sabah Shawkat Cabinet of Structural Engineering 017 Figure: Internal column of 500 x 500 h d 5cm f ctm 1.MPa b s 50cm h s 50cm f yd 375MPa P 1 856kN step 1: h s h d u cr1 b s h d Q bu1 0.h d f ctm u cr1 Q bu1 378kN P P 1 0.5Q bu1 P 667 kn A sb P A sb m 5mm 0.86f yd A s1 A s m A sb n n A s1 V 1 0.h d f ctm u cr1 V 1 378kN P 667 kn Step h s 3 h d u cr b s 3h d u cr 5 m Q bu 0.h d f ctm u cr Q bu 630kN

38 Sabah Shawkat Cabinet of Structural Engineering 017 P P P 1 0.5Q bu P 51kN A sb A sb m 0.86f yd A sb n n 3. 5mm A s1 A s m A s1 V 0.h d f ctm u cr V 630kN u cr P u cr.937 m 0.h d f ctm u cr m Column of 00x 500 extreme h d 5cm f ctm 1.MPa b s 0cm h s 50cm f yd 375MPa Step 1: u cr1 b s 0.5h d h s h d Q bu1 0.h d f ctm u cr1 Q bu1 6.8kN 0.5Q bu kn When applied to the plate even bending moment, then we take 0.5 qbu P 1 577kN P P 1 0.5Q bu1 P 63.6 kn For P we calculate the required shear reinforcement. A sb A s1 P A sb m 0mm 0.86f yd A s m A sb n n A s1 V 1 0.h d f ctm u cr1 V 1 6.8kN P 63.6 kn V 1 P does not comply

39 Sabah Shawkat Cabinet of Structural Engineering 017 Step b s 1.5 h d u cr h s 3h d u cr.8m Q bu 0.h d f ctm u cr Q bu 35.8 kn We expand the circumference in order to prevent the creation of a new shear crack P P P 1 0.5Q bu P 00.6kN A sb A sb m 0.86f yd 0mm A s1 A s m A sb n n V 0.h d f ctm u cr V 35.8 kn P A s kn V P does not comply Step 3: Figure.3.3-: Shear reinforcement at slab-column connection b s.5 h d u cr3 h s 5h d u cr3 3.8 m Q bu3 0.h d f ctm u cr3 Q bu kn P P 1 0.5Q bu3 P 337.6kN

40 Sabah Shawkat Cabinet of Structural Engineering 017 A sb P A sb m 0.86f yd 5mm A s1 A s m A sb n n.13 A s1 V 3 0.h d f ctm u cr3 V kN P kn V 3 P OK V3 is greater than P, thus the determination of the reinforcement to avoid the punching in reinforced concrete slab flat over the column is o