Reinforced Concrete Slabs 02
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Reinforced Concrete Slabs 02 Presentation Transcript

  • 1. Reinforced Concrete Slabs • TYPES OF SLABS • In reinforced concrete construction, slabs are used to provide flat, useful surfaces. A reinforced concrete slab is a broad, flat plate, usually horizontal, with top and bottom surfaces parallel or nearly so. It may be supported by reinforced concrete beams (and is usually cast monolithically with such beams), by masonry or reinforced concrete walls, by structural steel members, directly by columns, or continuously by the ground. • Slabs may be supported on two opposite sides only, as shown in Fig. 1 a, in which case the structural action of the slab is essentially one-way, the loads being carried by the slab in the direction perpendicular to the supporting beams. • There may be beams on all four sides, as shown in Fig. 1 b, so that two-way slab action is obtained. Concrete slabs in some cases may be carried directly by columns, as shown in Fig. 1 d, without the use of beams or girders. Such slabs are described as flat plates and are commonly used where spans are not large and loads not particularly heavy. Flat slab construction, shown in Fig. 1 e, is also beamless but incorporates a thickened slab region in the vicinity of the column and often employs flared column tops. Both are devices to reduce stresses due to shear and negative bending around the columns. They are referred to as drop panels and column capitals, respectively. Closely related to the flat plate slab is the two-way joist, also known as a grid or waffle slab, shown in Fig. 1 f . To reduce the dead load of solid-slab construction, voids ar formed in a rectilinear pattern through use of metal or fiberglass form inserts. A two way ribbed construction results. Usually inserts are omitted near the columns, so a solid slab is formed to resist moments and shears better in these areas FIGURE 1 Types of structural slabs
  • 2. DESIGN OF ONE-WAY SLABS • The structural action of a one-way slab may be visualized in terms of the deformed shape of the loaded surface. Figure 2 shows a rectangular slab, simply supported along its two opposite long edges and free of any support along the two opposite short edges. If a uniformly distributed load is applied to the surface, the deflected shape will be as shown by the solid lines. Curvatures, and consequently bending moments, are the same in all strips s spanning in the short direction between supported edges, whereas there is no curvature, hence no bending moment, in the long strips I parallel to the supported edges. The surface approximately cylindrical. • For purposes of analysis and design, a unit strip of such a slab cut out at right angles to the supporting beams, as shown in Fig. 2, may be considered as a rectangular beam of unit width, with a depth h equal to the thickness of the slab and a span la equal to the distance between supported edges. This strip can then be analyzed by the methods that were used for rectangular beams, the bending moment being computed for the strip of unit width. The load per unit area on the slab becomes the load per unit length on the slab strip. The loads recommended by ASCE for different usage of slabs are shown in table 1. Since all of the load on the slab must be transmitted to the two supporting beams, it follows that all of the reinforcement should be placed at right angles to these beams, with the exception of any bars that may be placed in the other direction to control shrinkage and temperature cracking. A one-way slab, thus, consists of a set of rectangular beams side by side. • This simplified analysis, which assumes Poisson's ratio to be zero, is slightly conservative. Actually, flexural compression in the concreteFig. 2
  • 3. • in the direction of la will result in lateral expansion in the direction of lb unless the compressed concrete is restrained. In a one-way slab, this lateral expansion is resisted by adjacent slab strips, which tend to expand also. The result is a slight strengthening and stiffening in the span direction, but this effect is small and can be disregarded. • Factored moments and shears in one-way slabs can be found either by elastic analysis or through the use ofthe same coefficients as used for beams . If the slab rests freely on its supports, the span length may be taken equal to the clear span plus the depth of the slab but need not exceed the distance between centers of supports, according to ACI Code 8.9.1. In general, center-to-center distances should be used in continuous slab analysis, but a reduction is allowed in negative moments to account for support width as discussed in Chapter 12. For slabs with clear spans not more than 10 ft that are built integrally with their supports, ACI Code 8.9.4 permits analysis as a continuous slab on knife edge supports with spans equal to the clear spans and the width of the beams otherwise neglected. If moment and shear coefficients are used, computations should be based on clear spans. • ACI Code 9.5.2 specifies the minimum thickness in Table 2 for nonprestressed slabs of normal weight concrete (wc = 145 pcf) using Grade 60 reinforcement, provided that the slab is not supporting or attached to construction that is likely to be damaged by large deflections. Lesser thicknesses may be used if calculation of deflections indicates no adverse effects. For concretes having unit weight wc in the range from 90 to 115 pcf, the tabulated values should be multiplied by 1.65 - 0.005wc' but not less than 1.09. For reinforcement having a yield stress other than 60,000 psi, the tabulated values should be multiplied by 0.4 +1/100,000. Slab deflections may be calculated, if required, by the same methods as for beams . The total slab thickness h is usually rounded to the next higher Table. 2 Minimum thickness h of non prestressed one-way slabs
  • 4. ¼ in. for slabs up to 6 in. thickness, and to the next higher ½ in. for thicker slabs. • Shear will seldom control the design of one-way slabs, particularly if low tensile reinforcement ratios are used. It will be found that the shear capacity of the concrete Vc will, almost without exception, be well above the required shear strength Vn at factored loads. • The concrete protection below the reinforcement should follow the requirements of ACI Code 7.7.1, calling for ¾ in. below the bottom of the steel . In a typical slab, 1 in. below the center of the steel may be assumed. • The lateral spacing of the bars, except those used only to control shrinkage and temperature cracks (see Section 13.3), should not exceed 3 times the thickness h or 18 in., whichever is less, according to ACI Code 7.6.5. Generally, bar size should be selected so that the actual spacing is not less than about 1.5 times the slab thickness, to avoid excessive cost for bar fabrication and handling. Also, to reduce cost, straight bars are usually used for slab reinforcement, cut off where permitted are as described for beams . • Since concrete is weak in tension, these temperature and shrinkage stresses are likely to result in cracking. Cracks of this nature are not detrimental, provided their size is limited to what are known as hairline cracks. This can be achieved by placing reinforcement in the slab to counteract contraction and distribute the cracks uniformly. In one-way slabs, the reinforcement provided for resisting the bending moments has the desired effect of reducing shrinkage and distributing cracks. However, as contraction takes place equally in all
  • 5. • directions, it is necessary to provide special reinforcement for shrinkage and temperature contraction in the direction perpendicular to the main reinforcement. This added steel is known as temperature or shrinkage reinforcement, or distribution steel. • Reinforcement for shrinkage and temperature stresses normal to the principal reinforcement should be provided in a structural slab in which the principal reinforcement extends in one direction only. ACI Code 7.12.2 specifies the minimum ratios of reinforcement area to gross concrete area (i.e., based on the total depth of the slab) shown in Table 13.2, but in no case may such reinforcing bars be placed farther apart than 5 times the slab thickness or more than 18 in. In no case is the reinforcement ratio to be less than 0.0014. The steel required by the ACI Code for shrinkage and temperature crack control also represents the minimum permissible reinforcement in the span direction of oneway slabs; the usual minimums for flexural steel do not apply. Table. 3 Minimum ratios of temperature and shrinkage reinforcement in slabs based on gross concrete area
  • 6. DESIGN LIMITATIONS ACCORDING TO THE ACI CODE The following limitations are specified by the ACI Code. 1. A typical imaginary strip 1ft (or 1m) wide is assumed. 2. The minimum thickness of one-way slabs using grade 60 steel according to the ACI Code, for solid slabs and for beams or ribbed one-way slabs should be equal to the following: For simply supported spans: solid slabs, h = Ll20 (ribbed slabs, h = L/16). For one-end continuous spans: solid slabs, h = Ll24 (ribbed slabs, h = Ll18.5). For both-end continuous spans: solid slabs, h = Ll28 (ribbed slabs, h = Ll21). For cantilever spans: solid slabs, h = LItO (ribbed slabs, h = Ll8). For fy other than 60 ksi, these values shall be multiplied by 0.4 + 0.01 fy, where fy is in ksi. This minimum thickness should be used unless computation of deflection indicates a lesser thickness can be used without adverse effects. 3. Deflection is to be checked when the slab supports are attached to construction likely to be damaged by large deflections. Deflection limits are set by the ACI Code, Table 9.5b. 4. It is preferable to choose slab depth to the nearest ¼ in. (or to mm). 5. Shear should be checked, although it does not usually control. 6. Concrete cover in slabs shall not be less than ¾ in. (20 mm) at surfaces not exposed to weather or ground. In this case, d = h - (3/4 in.) - (half-bar diameter). 7. In structural slabs of uniform thickness, the minimum amount of reinforcement in the direction of the span shall not be less than that required for shrinkage and temperature reinforcement (ACI Code, Section 7.12). . 8. The principal reinforcement shall be spaced not farther apart than three times the slab thickness nor more than 18 in. (ACI Code, Section 7.6.5). 9. Straight-bar systems may be used in both tops and bottoms of continuous slabs. An alternative bar system of straight and bent (trussed) bars placed alternately may also be used. 10. In addition to main reinforcement, steel bars at right angles to the main must be provided. This additional steel is called secondary, distribution, shrinkage, or temperature reinforcement
  • 7. One-way slab design. • Example. 1. A reinforced concrete slab is built integrally with its supports and consists of two equal spans, each with a clear span of 15 ft. The service live load is 100 psf,v and 4000 psi concrete is specified for use with steel with a yield stress equal to 60,000 psi. Requiremaent 2. Design the slab, following the provisions of the ACI Code. Solution 3. Thickness of Slab. h= L/28 Table 3 h= 15x12/28 = 6.43 in Say 6.5 in 4. Factored Loads Dead Load = 1.2x( 6.5x150/12) = 97 psf Live Load = 1.6x 100 = 160 “ Total = 257 “ 5.For this case, factored moments at critical sections may be found using the ACI moment coefficients(see Table 12.1): 6. Max Steel ratio Notes
  • 8. 7. Determine As for max negative BM (6.43 lb ft) ( Assume a=1 and find As) ( Check a for As determined ) As = 0.27 sq in at mid support As = 0.17 sq in at mid span As = 0.1 sq in at ext support 8. Temperature and shrinkage Reinforcement As = 0.0018 X 12 X 6.50 = 0.14 in2 9. Shear Force . The factored shear force at a distance d from the face of the interior support is Ok 10. Check spacing of reinforcement etc Notes
  • 9. Two -Way Slab • When the slab is supported on all four sides and the length is less than twice the width. The slab will deflect in two directions. and the loads on the slab arc transferred to all four supports. This slab is referred to as a two-way slab. The bending moments and deflections in such slabs are less than those in one-way slabs: thus. the same slab can carry more load when supported on four sides. The load in this case is carried in two directions, and the bending moment in each direction is much less than the bending moment in the slab if the load were carried in one direction only. Typical two-way slabs is shown in Fig. 1 • To visualize its flexural performance, it is convenient to think of it as consisting of two sets of parallel strips, in each of the two directions, intersecting each other. Evidently, part of the load is carried by one set and transmitted to one pair of edge supports, and the remainder by the other. Figure 1 shows the two center strips of a rectangular plate with short span la and long span lb' If the uniform load is q per square foot of slab, each of the two strips acts approximately as a simple beam, uniformly loaded by its share of q. Because these imaginary strips actually are part of the same monolithic slab, their deflections at the intersection point must be the same. Equating the center deflections of the short and long strips gives Fig. 1
  • 10. where qais the share of the load q carried in the short direction and qb is the share of the load q carried in the long direction. Consequently, One sees that the larger share of the load is carried in the short direction, the ratio of the two portions of the total load being inversely proportional to the fourth power of the ratio of the spans. • This result is approximate because the actual behavior of a slab is more complex than that of the two intersecting strips. For instance for a simply supportId square slab, qa = qb = q/2 . If only bending were present, the maximum moment in each strip would be • The exact theory of bending of elastic plates shows that actually the maximum moment in such a square slab is only O.048ql2, so that in this case the twisting moments relieve the bending moments by about 25 percent. Inelastic redistribution also modifies these bending moments further. A complicated situation is therefore obtained. Various design method are suggested to resolve this situation.
  • 11. Choice of Concrete Slabs • Various types of floor systems can be used for general buildings. such as residential. office. And institutional buildings. The choice of an adequate and economic floor system depends on the type of building. architectural layout. aesthetic features. and the span length between columns. In general. the superimposed live load on buildings varies between 80 and 150 psf. A general guide for the economical use of Boor systems can be summarized as follows. • Flat plates are most suitable for spans of 20 to 25 ft and live loads between 60 and 100 psf. The advantages of adopting flat plates include low-cost formwork, exposed fl al ceilings. and fast construction. Flat plates have low shear capacity and relatively low stiffness. which may cause noticeable deflection. Flat plates are widely used in buildings either as reinforced or prestressed concrete slabs. • Flat slabs are most suitable for spans of 20 to 30 ft and for live loads of 80 to 150 psf. They need more fonnwork than flat plates. especially for column capitals. In most cases. only drop panels without column capitals are used. • Waffle slabs are suitable for spans of 30 to 48 ft and live loads of 80 to 150 psf They carry heavier loads than flat plates and have attractive exposed ceilings. Formwork. including the use of pans. is quite expensive.
  • 12. • Slabs on beams are suitable for spans between 20 and 30 ft and live loads of 60 to 120 psf . The beams increase the stiffness of the slabs. producing relatively low deflection. Additional form wor k for the beams is needed. • One-way slabs on beams are most suitable for spans of 10 to 20 ft and a live load of 60 to 100 psf. They can be used for larger spans with relatively higher cost and higher slab deflection . Additional formwork for the beams is needed. • A one-way joist floor system is most suitable for spans of 20 to 30 ft and live loads of 80 to 120 psf. Because of the deep ribs. the concrete and steel quantities are relatively low, but expensive formwork is expected. The exposed ceiling of the slabs may look attractive. • Two-way slabs a re extremely complex and statically indeterminate. Many attempts, analytical and empirical, have been made to determine the division of the moments an d shears between the two spans and the distribution of these along the principal axes of the slab. Many elastic analyses of .two-way slabs have been proposed. All have shortcomings. They neglect Poisson's ratio, torsion, changes in stiffness, ultimate capacity, edge restraint, variation of moments and shears along span, and others. • The direct design method. DDM (ACI Code. Section 13.6). is an approximate procedure for the analysis and design of two-way slabs. It is limited to slab systems subjected to uniformly distributed loads and supported on equally or nearly equally spaced columns. The method uses a set of coefficients to determine the design moments at critical sections. Two-way slab systems that do not meet the limitations of the ACI Code. Section 13.6.1, must be analyzed by more accurate procedures. Design of Two-way Slabs
  • 13. • The equivalent frame method. EFM (ACI Code. Section 13.7). is one in which a three dimensional building is divided into a series of two-dimensional equivalent frames by cutting the building along lines midway between columns. The resulting frames are considered separately in the longitudinal and transverse directions of the building and treated 1100r by floor. as shown in Fig. • In elastic analysis, a concrete slab may be treated as an elastic plate. The flexure, shear, and deflection may be calculated by the fourth differential equation relating load to deflection for thin plates with small displacements, as presented by Timoshenko . Finite difference as well as finite element solutions have been proposed to analyze In the finite element method, the slab is divided into a mesh of triangles or quadrilaterals. The displacement functions of the nodes (intersecting mesh points) are usually established, and the stiffness matrices are developed for computer analysis slabs and plates. • For plastic analysis, three methods are available. The yield line method was developed by Johansen to determine the limit state of the slab by considering the yield lines that occur in the slab as a collapse mechanism. The strip method was developed by Hillerborg . The slab is divided into strips, and the load on the slab is distributed in two orthogonal directions. The strips are analyzed as simple beams. The third method is optimal analysis. There has been considerable research into optimal solutions. Rozvany and others presented methods for minimizing reinforcement based on plastic analysis. Optimal solutions are complex in analysis and produce complex patterns of reinforcement. • Nonlinear analysis simulates the true load deformation characteristics of a reinforced concrete slab when the finite element method takes into consideration the nonlinearity of the stress strain-relationship of the individual elements [11,12]. In this case, the solution becomes complex unless simplified empirical relationships are assumed.
  • 14. Coefficient Method • This method has been used extensively since 1963 for slabs supported at the edges by walls, steel beams, or monolithic concrete beams having a total depth not less than about 3 times the slab thickness. While it was not a part of the 1977 or later ACI Codes, its continued use is permissible under the ACI 318-08 code provision (13.5.1) that a slab system may be designed by any procedure satisfying conditions of equilibrium and geometric compatibility, if it is shown that the design strength at every section is at least equal to the requires strength, and that serviceability requirements are met. • The method makes use of tables of moment coefficients for a variety of conditions. These coefficients are based on elastic analysis but also account for inelastic redistribution. In consequence, the design moment in either direction is smaller by an appropriate amount than the elastic maximum moment in that direction. • Table 1-4 gives coefficients for different conditions for bending moments and shear force. These also specify formula to determine design parameters. The method since falls with in framework of ACI 318-08; all other specification of code do govern relevant design parameters. • Span length of members not built integrally with supports shall be considered as the clear span plus the depth of the member, but need not exceed distance between centers of supports. ACI 8.9.1 • Deflection of two-way slab should not exceed the limits specified in table 9.5(b)
  • 15. TABLE 9.5(b) — MAXIMUM PERMISSIBLE COMPUTED DEFLECTIONS • For slabs with beams spanning between the supports on all sides, the minimum thickness, h, shall be as follows: For αfm greater than 0.2 but not greater than 2.0, h shall not be less than but not less than 5 in. For αfm greater than 2.0, h shall not be less than but not less than 3.5 in. . Max spacing of reinforcement should not exceed twice thickness of slab or that given by temperature and shrinkage . . The twisting moments are of consequence only at exterior corners of a two-way slab system, where they tend to crack the slab at the bottom along the panel diagonal, and at the top perpendicular to the panel diagonal. Special reinforcement should be provided at exterior corners in both the bottom and top of the slab, for a distance in each direction from the corner equal to one-fifth the longer span of the corner panel. The spacing and size should be same as for positive bending moment.
  • 16. Example 1. Design the roof slab, beam and column of house given in figure 1. Concrete compressive strength (fc′) = 3 ksi. Steel yield strength (fy) = 40 ksi. Load on slab: 4″ thick mud. 2″ thick brick tile. Live Load = 40 psf. Requirement 2. Design of slab for room and veranda. Solution Veranda Slab 3 Ratio of spans lb/la = 24.75/8 = 3.09 > 2 “one way slab” 4 Assume slab thickness of slab = 5 in Clear span = 8 + 5/12 = 8.42 ft ACI 8.9.1 or c/c of support = 8 + 13/24 + .5 = 9.0625 ft Take l = 8.42 ft Min slab thickness = l/20 x (0.4+fy/100000)x12 ACI = 4.04 in 5 in is ok 5 d = 5 – 0.75 – 3/16 = 4 in 6 Factored loads dead load Slab = 0.15x5/12 = 0.0625 ksf Mud = 0.12x4/12 = 0.04 ksf Choka = 0.12x2/12= 0.02 ksf total =0.1225 ksf Live load = 0.04 ksf Wu = 1.2x 0.1225 + 1.6x 0.04 = 0.211 ksf Mu= (0.211x8.42x8.42/8)x12 = 22.44 in -kip/ft
  • 17. 7 Asmin = 0.002bhf (for fy 40 ksi, ACI 10.5.4) = 0.002 × 12 × 5 = 0.12 in2 a = Asminfy/ (0.85fc′b) = 0.12 × 40/ (0.85 × 3 × 12) = 0.156 in ΦMn(min) = ΦAsminfy (d – a/2) = 0.9 × 0.12 × 40 × (4 – 0.156/2) = 16.94 in-k < Mu 8 Determine As for applied Mu = 22.44 in-kip Mu/0.9 = As x fy x d – sq(Asxfy)/(1.7xfcxb) As = 0.16 sq in using #4 bars Ab= 0.2 sq in Spacing S= 0.2 x 12 /0.16 = 15 in c/c using #3 bar S=0.11x12/0.16 = 7.5 in or 6 in c/c is ok. use #3 @ 6 in c/c 9 Temperature and Shrinkage Steel Ast = 0.002 x b x hf = 0.002 x 12 x 5 = 0.12 sq in S =0.11 x 12 /0.12 = 11 in c/c 10 Maximum spacing for main steel in one way slab according to ACI is minimum of: 3hf =3 × 5 =15 in or 18 in Therefore 6″ spacing is O.K.
  • 18. 11 Maximum spacing for shrinkage steel in one way slab according to ACI 7.12.2 is minimum of: 5hf =5 × 5 =25in or 18in Therefore 9 in spacing is O.K. Design of Room Slab 12 Span ratio = 16/12 = 1.33 < 2 “ two-way slab” Min h = Perimeter/180 = 2 x ( 16 + 12 ) x 12 /180 = 3.73 in not less than 5 in Aci hf = 5 in is ok. 13 Wu= 0.211 ksf 14 Tables to be consulted for design m = short span/long span = 12/16 = 0.75 Since slab is continuous on long edge only; case 6 would apply Coefficient for this slab Ca,neg = 0.088 Cb,neg = 0 Ca,dl = 0.o48 Cb,dl = 0.012 Ca,ll = 0.055 Cb,ll = 0.016 15 Calculate Bending moment
  • 19. 16 Total design Moments Max negative BM in Short direction(12 ft) = 32.04 in k “ “ long “ (16 ft) = 0 in k max positve BM in short Direction (12 ft ) =18.36 in-k “ “ long “ (16 ft ) = 8.544 in-k 17 Design of Slab As min= 0.002x12x5 =0.12 sq in Mu(min) = 0.9x0.12x40{(4 – 0.12x40/(1.7x3x12)} = 19.94 in-k Neg ative BM in Short span = # 3 @ 4.5 in c/c Positive BM in short span = # 3 @ 9.0 in c/c Negative BM in long span = nil Positve BM in long span = # 3 @ 11 in c/c not to be greater than 2xhf = 2x5=10 in as per ACI 13.3.2 may reduce to 9 in spacing for simplicity. 18 Min Reinforcement is also provided at discont edges against partial fixations
  • 20. Reinforcement Plan
  • 21. Example 1 The floor system shown in Fig. consists of solid slabs and beams in two directions supported on 20-in. square columns. Design a typical interior slab pannel to carry a live load of 100 psf appart from self weight. Use f`c = 3 ksi and fy = 60 ksi. 2 Solution Since slab is supported on beams, we determine its thickness. To apply the equation of ACI we need to find out Ib and Is etc. Effective X-Sec of T beam is Shown in fig. The moment of inertia of the slab in the long direction assuming slab thickness of 7 in The moment of inertia of the slab in the short direction
  • 22. Also, hmin = 3.5 in. Therefore, h = 6.27 in. controls. A slab thickness of 6.5 in. or 7.0 in. may be adopted. Note that in most practical cases, Eq. 17.2 controls. Eq. 9.12 -------- Eq. 9.13 ---------
  • 23. 3. Span ratio = 20/24 = 0.83 two-way slab thickness of slab = 7 in 4. Factored Load Wu = 1.2x(7/12x.15) + 1.6x0.1 = 0.265 k/sq ft 5. Coefficient from Tables ( Case 2 ) Ca neg = 0.062 Cb neg =0.029 Ca dl = 0.053 Cb dl = 0.012 Ca ll = 0.039 Cb ll =0.018 6. Bending Moments Ma neg = 0.062x0.265x20x20x12 = 78.86 in-k Ma pos = (0.053x0.105+0.039x0.16)x20x20x12 = 56.66 in-k Mb neg =0.029x0.265x24x24x12 = 53.12 in-k Mb pos =(0.012x0.105+0.018x0.16)x24x24 = 28.62 in-k 7. Design of slab in short dir for neg moment As= 0.228 sq in #4 @ 10 in c/c “ pos “ As= 0.161 “ #4 @ 14 “ In long dir for neg moment As= 0.166 “ #4 @ 14 “ “ pos “ As= 0.088 “ #4 @ 14 “ Max spacing = 2xhf = 14 in or temp & shrinkage rft Temp= 0.0018xbxhf =0.151 sq in #4 @ 15 in c/c Max spacing of 14 in c/c governs
  • 24. TABLE. 1 COEFFICIENTS FOR NEGATIVE MOMENTS IN SLABS *A cross-hatched edge indicates that the slab continues across or is fixed at the support; an unmarked edge indicates a support at which torsional resistance is negligible ie simple support * * where w = total uniform Factored dead plus live load
  • 25. TABLE.2 COEFFICIENTS FOR DEAD LOAD POSITIVE MOMENTS IN SLABS *A cross-hatched edge indicates that the slab continues across or is fixed at the support; an unmarked edge indicates a support at which torsional resistance is negligible ie simple support. * * where w = total uniform Factored dead plus live load
  • 26. TABLE.3 COEFFICIENTS FOR LIVE LOAD POSITIVE MOMENTS IN SLABS *A cross-hatched edge indicates that the slab continues across or is fixed at the support; an unmarked edge indicates a support at which torsional resistance is negligible ie simple support. * * where w = total uniform Factored dead plus live load
  • 27. TABLE.4 RATIO OF LOAD lV IN A AND B DIRECTIONS FOR SHEAR IN SLAB AND LOAD ON SUPPORTS *A cross-hatched edge indicates that the slab continues across or is fixed at the support; an unmarked edge indicates a support at which torsional resistance is negligible ie simple support.