Betonarme kirişsiz döşemelerde zımbalama dayanımı

Abstract

Döşemelerin doğrudan kolonlar tarafından taşındığı "kirişsiz döşeme" sistemlerinde kolon yöresinde oluşan asal çekme gerilmeleri oldukça yüksek çıkabilir ve betonun çekme dayanımını aşabilir. Bu sorun, betonarme kirişlerdeki eğik çekme sorununa çok benzemektedir. Araştırmada kolon döşeme bağlantısında meydana gelecek etkiler. incelenmiş ve bu etkilerin önlenmesi için iki tip zımbalama kayma donatısı denenmiştir. Konuya benzer kirişsiz döşemeler üzerinde literatür çalışması yapılmıştır. Zumbalama olayı, kırılma şekilleri, kirişsiz döşemelerde Teorik Taşınma Gücü ve Göçme Yükü için önerilen teorik, yarı teorik ve ampirik formüller incelenmiştir. Araştırmanın özünü betonarme kirişsiz döşeme de neyleri oluşturmuştur. Kolon civarındaki kayma göçmesi yükü ve eğilme göçmesi yükü teorik bir yöntemle araştırılmıştır. Deneyde kirişsiz döşemede şekil değiştirmeler mekanik ölçü aletleriyle ölçülmüştür. Her kirişsiz döşemenin betonuna ait özellikleri saptamak için çökme, birim ağırlık, 28 günlük küp ve silindir dayanımı tesbit edilmiştir. Silindir numunelerde betonun yanal deformasyonuda ölçülerek Poisson oranı bulunmuştur. 2 Betonarme kirişsiz döşeme deneyleri için 1600x1600 mm ve 120 mm yüksekliğinde 2 adet, 1600x1600 mm ve 140 mm yük sekliğinde 1 adet olmak üzere toplam 3 adet kirişsiz döşeme dökülmüştür. Döşemeler basit mesnetli oluşturulmuş ve yük kolon aksından uygulanmıştır. Her döşeme için 150x150 mm2 ve 400 mm yüksekliğinde kolonlar yapılmıştır. Kirişsiz döşemelerin donatıları A tipi, B tipi ve C tipi olmak üzere 3 tipte yapılmıştır. Her yük kademesinde her iki yönde plağın 1/4 noktalarında ki çökmeler deformasyon saatleri ile ölçülmüş, ve yük- çökme eğrileri elde edilmiştir. Her yük kademesinde meydana gelen çatlaklar yanına kademe numarası yazılarak plak üzerinde tesbit edilmiştir. Plak kırıldıktan sonra, plağın çatlama şeması çizilmiştir. Betonun çatlamadan önce homojen ve izotrop bir malzeme olduğu kabul edilerek yapılan hesapların uygunluğu, teorik olarak araştırılmıştır.

Concrete flat plates are subjected to large bending moment and shearing force at their connections with columns. This combined effect can cause failure by punching of the slab. A literature survey has been done, on similar concrete flat plates. In flexure, reinforced concrete slabs exhibits great deal of ductility, extensive deformations occur before their ultimate strength is reached. Design codes place increasing reliance on this ductile behaviour which enable slab systems to redistribute moments. Complete. redistribution of bending moments can generally be achieved in a slab system prior to failure provided that shear failure is prevented. It is therefore, important that in design, when considering punching, attention be given to both strength and ductility. Perhaps the greatest source of problems for slabs in practice, apart from excessive deflection, is shear. The so-called punching shear failure of slabs without shear reinforcement occours suddenly without warning so that, normal no remedial measures can be applied prior to failure Fig (1). In current practice, the corrections of a concrete flat plate and its supporting columns, are reinforced by Stirruos (Fig (2)) or by structural steel shear heads. Because stirrups are less expensive, they are used more frequently. However in relatively thin slabs (150 to 250 mm overall thickness) such reinforcement may be difficult to install. In slabs thinner than (250 mm), stirrups arranged as show in Fig (2). In this arrangement the stirrups are closed and enclose a longitudinal bar at each corner, stirrups will not have enough anchorage at the top and bottom of the shear reinforcement to develop yield strength. The zone above the column is congested with horizontal top and bottom reinforcing bars running in two perpendicular directions, with vertical bars of the column and with the stirrups (see. Picture 2\. A new type of shear reinforcement, in the form of a stud with an anchor head and a steel strip welded to its top and bottom. The advantages of the studs are : 1 - They are easy to install, even in thin slabs, 2 - They do not interfere with flexural reinforcement, 3- Anchorages at the top and bottom are sufficient to develop yield in the shear studs before failure, which occurs in a ductile fashion when this type of reinforcement is used. Fig (3). xn COLUMN BENOING CRACKS APPEARING AT LOW LOADING.^7^\ STIRRUPS DETAIL REACTION SECTION A-A REINFORCED ZONE FOR SHEAR TOP VIEW TOP ANCHOR PLATES STUDS FOR CLARITY, THE REMA1NOER OF SLAB REINFORCEMENT IS OMITTED Fig. 1- (left) Failure by punching of a concrete flat plate. Fig. 2- (above) Stirrups in slabs thinner than 10 in. (250 mm). ANCHOR AREA »10 TIMES STEM AREA h- A i -i <=r-p_zr:£ 2/3 0 HOLES FOR iS ATTACHMENT TO FORMWORK STEM OIA. 0 J rJ_ -2.0/Z >2.5 D SECTION A-A WELD BOTTOM ANCHOR STRIP Fig. 3- Stud shear reinforcement details. The essential part of the study includes the test of reinforced concrete flat plates. In tests, three reinforced concrete flat plates were casted. The specimens were simply supported along the periphery of a side length 1600 mm. Dimension of colomns and thicks were 150 by 150 ram and 120 mm, thick for a slab was 140 mm respectively. The slab was tested in a horizontal position the axial column force was applied. The design concrete strength was 20 N/mm2, steel strengths were 365 N/mm2 and 191 N/mm. The shear stud strength was 518 N/mm2. The slab was square 1600 by 1600 ram. X1X1 Three slabs in each group had three different types of reinforcement respectively (see. Pic.l, 2, 3). Deflections on the slab surfaces were measured by the mechanical extenscmeters. Slumps, densities, 28 days cube and cylinder strengths of concrete for each slab were determined. Lateral deformations of concrete on the cylinder specimens was measured Poisson's ratio was calculated. Calculations, based on the assumptions of homogeneous and izotropic material for concrete, have been checked against experimental results. Therefore, for a slab which was exposed to concentrated column load, Navier type solutions have been obtained considering the slab as a plate which have four edges free in solving this problem it was assumed that the effect of shear stresses on strains is negligible and also the normal stress distribution is linear. The load was expanded into a fourier series and with the general solution for the slab deflections, moments, shear forces, stress and slab deflections at the determined point have been calculated by computer program. The failure loads V,, V 2 and V ^ have been calculated corresponding to the11 failure modes below. The actual failure load is the smallest of the three values. 1. Flexural failure load. According to Johansen's yield line theory, the ultimate load for a slab supported and loaded as described above is given by vui = m Hjh,* 2n) c (1) According to Andra the ultimate load is given vul = 2nm (2) Where; b : the length of a supported side of the slab. c : the column side length m : ultimate flexural capacity per unit width of the slab given by. m = 0 pd^f ( 1-0,59 p-* ) y f * c (3) Where; p =/pxpy xiv p, p = reinforcement rations parallel to x-and y ^ directions respectively 1 d = - (dx+dy) 2 dx, dy = effective depths in the slab x-and y-directions f = the yield strength of flexural reinforcement The failure by punching shear can accur in one of the following ways. 2. Shear failure at a distance d/2 from the outhermost row of shear studs. Vu2 " 0vcUpd (4) U = 2[(b+d) + (h+d)] (4.) where: U : the perimeter of the critical section v ' : the nominal shear stress that can be resisted by the concrete without shear reinforcement. The value v can be calculated by c _..2. a-1, v ' =0,17 [1+ - (1 )]>/fTr (N/min) (5) 3^ 3 where; 3 : the ratio of longer side to shorter side of c the column a : the ratio of the distance between the column face and critical section to slab effective depth d. The values of a and 3 in Equation (5) must be equal to or greater than 1 and 2, respectively. Thus, at sections closer than d from the column face, use a= 1, similary, when the column aspect ratio is smaller than 2, use 3 = 2. c 3. Shear failure within the shear reinforced zone s : between the first row end the column face must not be smaller than d/4, and s and s the spacing s : between the rows must not exceed the following limits. IV When ; °'33/V~s V0 â °'5/I7~ (N/mm2) the limits for sQ â d/2 s < 3/4d When ; °'5/V â V0 S °'67/V"(N/mm2) s S 0,35 d o s â d/2 The upper limit for the strength when shear reinforcement is used is 0,67/F/ (N/mm2) For shear failure at a distance d/2 from the face of the column the ultimate shearing force is given by vu3 = 0 (vc+vs) (6) Where; V : the force resisted by concrete V : the force resisted by the shear reinforcement s The forces V and V are given by c s Vc = vcbod (7) V = A f d s v yv - (8) s Where; A : the area of shear reinforcement in one row v paralel to column periphery f : yield strength of shear studs s : spacing between rows of shear reinforcement elements, measured perpendicular to column face v : nominal shear stress resisted by concrete at ultimate, to be design of shear reinforcement The value v is given by v.V c(N/mm2) =f-+ °'08/V <9> XVI The values of the strength reduction factor 0 to be used in design are 0,90 Eq. (3) and 0,85 in Eq. (4) and (6) Unless the spacing, s between the rows of shear studs is increased away from the column, no other section needs to be checked within the shear-reinforced zone. The results of this investigation are : Shear reinforcement in the form of vertical bars mechanically anchored by heads at the top and by a steel strip at the bottom can be used to reinforce slabs against punching» Compared to conventional stirrups, this relatively new type of shear reinforcement offers more efficiency, cost saving in materials and simplicity of installment and control. The equations suqgested for design result in a smaller amounts of shear reinforcement because, as a result of good anchorage, a higher portion of the shear stress can be carried by concrete.