Figure 4.3.1 Figure 4.3.1
A building usually comprises several, generally nonstandard, types of slabs (oneway, twoway etc.). In this case, each slab is analysed separately by using tables. The moments M at the edge (border) of two slabs could be taken as the average of moments M_{1} (left slab moment) και M_{2} (right slab moment).
The difference ΔΜ=M_{2}M_{1} is redistributed in all directions of slabs, affecting less the other support moments and even lesser the span moments. Special types of plates (inclined, of nonconstant thickness etc.) could be solved as equivalent rectangular slabs of constant thickness.
Application width of concentrated load on slab
Figure 4.3.2 Figure 4.3.2
The application width t of concentrated load P onto the middle level of the slab is expressed as: t = b_{0} + 2·(s+h_{s}/2) = b_{0} + 2·s + h_{s} , where s is the covering thickness. Thickness s should not be taken into consideration, when the concentrated load is applied directly onto the concrete's surface, e.g. in the case of brick masonry.
Distribution width of concentrated load
Figure 4.3.31: Oneway slab bearing concentrated load Figure 4.3.31: Oneway slab bearing concentrated load
In case of concentrated, linear or uniformly distributed load applied on oneway slab, unless more accurate analysis is performed, the calculated distribution width b_{m} is defined transversely to the direction of main reinforcement. Symbols used: b_{m,f} is the distribution width of span moment, b_{m,s} is the distribution width of the support bending moment (provided fixed support exists) and b _{m,v} (b_{m,vL}, b_{m,vR}) is the distribution width of shear force (at left and right support). These widths, generally, differ. Each stress resultant a_{w} at each position x of oneway slab LR, due to uniform load p_{d}=γ_{g}·g+γ_{q}·q, remains constant along the entire lengthl_{y} with units of moment or shear per m in direction y. Concentrated load P=γ_{g}·G+γ_{q}·Q causes additional stress resultant A_{P} (in units of moment or shear) corresponding to the distribution width BM, thus is equal to a_{P}=A_{P}/b_{m} per m in direction y . According to superposition principle, the total stress resultant applied on b_{m} wide strip is equal to a_{tot}=a_{w}+a_{P}, while on the rest width l_{y} is equal to a_{w}. The distribution width of the concentrated load is a function of the support types of the slab (fix, simple, free) and the distance x between the centre of mass and the supports. The following table gives the distribution width of partial uniform load for every type of support.
Distribution width of partial uniform load 
 b _{m} _{,} _{f} = t_{y} + 2.5·x·(1x/l) t _{x,max }= l, t _{y,max }= 0.8l  b _{m} _{,} _{vL} = t _{y} + 0.5·x t _{x,max }= l, t _{y,max }= 0.8l b _{m} _{,} _{vR} = t _{y} + 0.5·(lx) t _{x,max }= l, t _{y,max }= 0.8l   b_{m,f}=0.37+2.5x1.5x(10.375)=2.71 m  b_{m,vL}=0.37+0.5x1.5=1.12 m b_{m,vR}=0.37+0.5x2.5=1.62 m   b_{m}_{,}_{f}=0.37+2.5x2.0x(10.5)=2.87 m  b_{m}_{,}_{vL}=0.37+0.5x2.0=1.37 m b_{m,vR}=0.37+0.5x2.0=1.37 m  Simple with one end fixed 
 b _{m} _{,} _{f} = t _{y} + 1.5·x·(1x/l) t _{x,max }= l, t _{y,max }= 0.8l b _{m} _{,s} = t _{y} + 0.5 · x·(2x/l) t _{x,max }= l, t _{y,max }= 0.8l  b _{m,vL} = t _{y} + 0.3·x t _{x,max }= 0.2l, t _{y,max }= 0.4l b _{m,vR} = t _{y} + 0.4·(lx) t _{x,max }= 0.2l, t _{y,max }= 0.4l   b_{m,f}=0.37+1.5x1.5x(10.375)=1.78 m b_{m,s}=0.37+0.5x1.5x(20.375)=1.59 m  b_{m,vL}=0.37+0.3x1.5=0.82 m b_{m,vR}=0.37+0.4x2.5=1.37 m   b_{m,f}=0.37+1.5x2.0x(10.5)=1.87 m b_{m,s}=0.37+0.5x2.0x(20.5)=1.87 m  b_{m,vL}=0.37+0.3x2.0=0.97 m b_{m,vR}=0.37+0.4x2.0=1.17 m  
 b _{m} _{,} _{f} = t _{y} + x·(1x/l) t _{x,max} = l, t _{y,max} = 0.4l b _{m} _{,sL} = t _{y} + 0.5·x·(2x/l) b _{m} _{,sR} = t _{y} + 0.5·(lx)·(1+x/l) t _{x,max} = l, t _{y,max} = 0.4l  b _{m} _{,vL} = t _{y} + 0.3·x t _{x,max} = 0.2l, t _{y,max} = 0.4l b _{m} _{,vR} = t _{y} + 0.3·(lx) t _{x,max} = 0.2l, t _{y,max} = 0.4l   b_{m,f}=0.37+1.5x(10.375)=1.31 m b_{m,sL}=0.37+0.5x1.5x(20.375)=1.59 m b_{m,sR}=0.37+0.5x2.5x(1+0.375)=2.09 m  b_{m,vL}=0.37+0.3x1.5=0.82 m b_{m,vR}=0.37+0.3x(4.01.5)=1.12 m   b_{m,f}=0.37+2.0x(10.5)=1.37 m b_{m,s}=0.37+0.5x2.0x(20.5)=1.87 m b_{m,sR}=0.37+0.5x2.0x(1+0.5)=1.87 m  b_{m,vL}=0.37+0.3x2.0=0.97 m b_{m,vR}=0.37+0.3x(4.02.0)=0.97 m  
 b _{m} _{,s} = t _{y} + 1.5 · x t _{x,max }= l, t _{y,max }= 0.8l  b _{m} _{,} _{v} = t _{y} + 0.3·x t _{x,max} _{}= 0.2l, t _{y,max} _{}= 0.4l   b_{m,s}=0.37+1.5x1.5=2.62 m  b_{m,v}=0.37+0.3x1.5=0.82 m   b_{m,s}=0.37+1.5x2.0=3.37 m  b_{m,v}=0.37+0.3x2.0=0.97 m  Examples 1 and 2 are presented below. Example 1 (slab at the start of paragraph 3.3.1)
Figure 4.3.32 Figure 4.3.32
Dimensions of slab l_{x}=4.00 m, l_{y}=5.00 m, width h_{s}=170 mm. Subjected to covering load g_{e}=1.0 kN/m^{2}, live load q=5.0 kN/m^{2} and concentrated P (dead G=10.0 kN and live Q=10.0 kN), applied directly (s=0) onto the slab, with dimensions b _{ox}=0.50 m, b_{oy}=0.20 m and distance of center of mass from left support x=1.50 m. L, R are pin supports. Calculate bending moment and shear force diagrams. The analysis with uniform loads is the same as in the example of §3.3.2 (v_{w}=29.18 kN/m, m_{w}=29.18 kNm/m). This design load is uniform along the entire length l_{y} of the slab. The application width of the design concentrated load P is P=1.35·G+1.50·Q=1.35x10.0+1.50x10.0=28.5 kN t_{x}=b_{ox}+h_{s}=0.50+0.17=0.67 m t_{y}=b_{oy}+h_{s}=0.20+0.17=0.37 m The equivalent linear distributed load p in t_{x}^{}is^{} p=P/t_{x}=28.5kN/0.67m=42.5 kN/m The support shear forces are V_{L}=V_{x,L}=p·t_{x}·(l_{x}x)/l_{x}=42.54x0.67x2.5/4.0=17.8 kN VR=V_{x,R}=V_{x},Lp·t_{x}=17.8142.54x0.67=10.7 kN The zero shear force point is located at x'=V_{L}/p=17.81kN/(42.54kN/m)=0.419 m The maximum moment is (see § 3.3.3, sub note 10) The load widths are (previous table): t_{x,max}=l_{x}=4.0 m (>t_{x}=0.67 m) και t_{y,max}=0.8·l_{x}=0.8x4.0=3.20 m (>t_{y} =0.37 m) The additional design moment at a strip of width b_{m,f}=2.71 m is equal to m_{P}=M_{P}/b_{m,f}=24.48kNm/2.71m=9.03 kNm/m while the total design moment is equal to m_{tot}=m_{w}+m_{P}=29.18+9.03=38.21 kNm/m. The additional design shear force at a strip of width b_{m,vL}=1.12 m is equal to v_{P}=V_{L}/b_{m,vL}=17.81kN/1.12m=15.9 kN/m while the total design shear force is equal to v_{tot}=v_{w}+v_{P}=29.2+15.9=45.1 kN/m. In the previous slab, additional constant load g=8.0 kN/m is applied due to masonry from L to R (total load G=8.0kN/m·4.00m=32.0 kN) on width b_{oy}=0.20 m. Calculate bending moment and shear force diagrams. The analysis with uniform loads is the same as the previous example. The load strip dimensions are b_{ox}=4.00 m and b_{oy}=0.20 m.
Figure 4.3.33 Figure 4.3.33
The application width of the design concentrated load p is: p=γ_{g}·g=1.35x8.0kN/m=10.8 kN/m t_{x}=min(b_{ox}+h_{s}, l_{x})=min(4.00+0.17, 4.00)=4.00 m t_{y}=b_{oy}+h_{s}=0.20+0.17=0.37 m The support shear forces are V_{L}=V_{R}=p·l_{x}/2=10.8x4.00/2=21.6 kN M_{p}=p·l^{2}_{x}/8=10.8x4.00^{2}/8=21.6 kNm. The distribution load widths are (previous table) t_{x}=t_{x,max}=l_{x}=4.00 m, t_{y,max}=0.8·l_{x}=0.8x4.00=3.20 m(>t_{y}=0.37 m) b_{m,f}=t_{y}+2.5·x·(1x/l)=0.37+2.5x2.00x(10.50) b_{m,vL}=b_{m,vR}=t_{y}+0.5·x=0.37+0.5x2.00=1.37 m The additional design moment at a strip of width b_{m,f}=2.87 m is equal to m_{p}=M_{p}/b_{m,f}=21.60kNm/2.87m=7.5 kNm/m while the total design moment is equal to m_{tot}=m_{w}+m_{P}=29.2+7.5=36.7 kNm/m. The additional design shear force at a strip of width b_{m,vL}=1.37 m is equal to v_{P}=V_{L}/b_{m,v}=21.6kN/1.37m=15.8 kN/m while the total design shear force is equal to v_{tot}=v_{w}+v_{P}=29.2+15.8= =45.0 kN/m. 
The slab of the two previous examples is designed for bending with moment m_{w}=29.2 kNm/m (strip width 1.00 m). The respective reinforcement a _{s,w} (mm^{2}/m) is placed uniformly along full length l_{y}=6.00 m. 
Slab strip b_{m,f} should be designed for bending using strip width 1.00 m and moment m_{tot}, resulting to reinforcement a_{s,tot} (mm^{2}/m). The additional required reinforcement for strip b_{m,f} is equal to a_{s}=a_{s,tot}a_{s,w} (mm^{2}/m) and A_{s}=a_{s}·b_{m,f} (mm^{2}), is distributed on a strip of width b _{z}with t_{y}≤ b_{z} ≤b_{m,f}, symmetrically with respect to the load. 
In addition to longitudinal reinforcement, transverse reinforcement is also required both in upper and lower fiber of the strip, equal to at least 60% of a_{s} having length b_{m,f} + anchorage length. If a strip is selected, e.g. of width b_{z}=0.40 m, it can form a hidden beam shape, contributing to the support of upper transverse reinforcement (see volume C'). This means that the distribution area is reinforced as flange of web width b_{z}=0.40 m and slab width b_{m,f}. 
The slabs of the previous examples are designed for total shear v=45. 1 kN for strip width 1.00 m.
Figure 4.3.34 Figure 4.3.34
In the case of twoway slab, concentrated or linear loads are distributed all over the slab as equivalent uniform load. This load is added to other uniform loads of the slab. Example : The equivalent uniform linear load of masonries of the figure (e.g. the weight of masonry minus the weight of the door), along directions of 4.00 and 1.50 m are equal to 5.0 kN/m and 6.0 kN/m, respectively. Calculate the equivalent uniform load of the masonries distributed in the entire surface of the slab. 4.00m·5.0kN/m+1.50m·6.0kN/m=29.0 kN Equivalent uniform load of masonry: 29.0kN/(4.00m·5.00m)=1.45 kN/m^{2}
