# HOW TO DESIGN OF REINFORCED BRICK SLAB (R.B. Slab) With Example

Today I am going to discuss **DESIGN OF REINFORCED BRICK SLAB (R.B. Slab)** with example. As you know that last time I publish a post about Load Factor Method. Some of my viewers comment us to discuss some hints about DESIGN OF REINFORCED BRICK SLAB.

# HOW TO DESIGN OF REINFORCED BRICK SLAB (R.B. Slab) With Example

Sometimes bricks are used along with concrete in the construction of a slab. Such a slab is called reinforced brick slab (R. B. Slab) this slab is weak as compared with R.C.C slab. Bu it is cheap. This is used in unimportant construction works. In this slab reinforcement is used at certain fixed distance. Because this distance depends on the size of bricks. The length of brick is 22.5cm (9 inch), 20 cm (9 inches) and 11 cm (4.5 inches). Such slabs are used for small spans only.

The capacity of heat resistance in an R.B. slab is more that in an R.C.C. slab. That is thermal conductivity in R.B. slab is less than that of an R.C.C. slab. The bricks used in an R.B. slab should be of first class. In spite of the use of good bricks, the R.B. slab it can absorb more humidity. This implies that the steel should not have direct contact with the bricks. In this case humidity will reach the steel and the steel will be rusted. In addition to this the bond will be weakened. There should not be vertical joints at the end of a wall, but instead half of the brick should be placed on the wall at joint.

A network of steel is made before putting bricks. The bricks should be made wet properly before using them so that they should not absorb water from the concrete. The bricks should be placed such that the distance between them is at least 4 cm (1.5 inches). Then concrete is filled between them. The ration of concrete mostly adopted in R.B. slab is (1:2:3) and in its preparation, the minimum water-cement ration is taken as 0.5, so that its work-ability is better. And that the concrete should fill then joints as well.

The method of designing of an R.B. slab is the same as that of an R.C.C. slab. The weight of an R.B. slab is taken 1950 Kg per cu.m (120lb/cu.ft). This stress should never exceed 1 kg/cm2. This stress should never exceed 3 kg/cm2.

**The values of other stressed are as follows:**Permissible compressive stress in R.B. = fcc=30Kg/cm2 (425/In2)

Permissible tensile stress in steel = fst = 1400 Kg/cm2.

Modular ratio = m= 40

Safe diagonal stress = fst = 1400 Kg/cm2 (142 lb/In2)

Local Bond stress = fb (l) = 10 kg/cm2 (142 lb/In2)

Average Bond stress = fb = 6 kg/cm2 85 lb/In2)

Lever arm factor = la =0.84

Constant for moment of resistance (Mr) = k= 5.8

Example Of REINFORCED BRICK (R.B.) Slab. Clear span between two brick walls, 20 cm thick, is 3 meter. It is proposed to cover it with R.B. slab. Design this slab for a live load of 150 kg/square meter and 50 kg/ square meter as floor finishes.

Fcb= 30 Kg/square meter, fst = 1400 Kg/square meter, m = 40.Given data:

Clear span = l1 =3 m

Thickness of walls = t = 20 cm

Live load = w1 = 150 kg/m2

Load of floor finish = w2 = 50 kg/m2

Bending stress = fcb = 30 kg/cm2

Tensile stress = fst = 1400 kg /cm2

Modular ratio = m = 40

Lever arm = la = 0.86 d1

Constant for moment of resistance = k = 5.8Solution:

Effective span = l + l1 +t = 3+0.20 = 3.20m

Thickness of slab = Span/30 = 3 /30 * 100 = 10cm

Effective span = l = l1 + d = 3+ 0.10 = 3.10m

The less value will accept, i.e., 3.10 m.

Self load of slab load = w3 = 1 *1 10/100 * 1920 = 192kg/m2

Total load = W = w1 +w2+w3=150+50+192 =392kg/m2

Maximum bending moment = B.M. = Wl2/8 = 392*(3.1)2/8 *100 =47089 kg.m

Effective Depth of slab =

d1 = √(B.M.)/kb =√47089/(5.8*100) = 47089 kg.cm

Overall depth = d=d1+c.c = 9.01 +1.8 = 10.81 cm ≈11cm

Actual effective depth =d1 = 11.00-1.8 =9.2cm

Area of steel for main bars

=Ast=(B.M.)/(fst*la)

=Ast=47089/(1400*0.86*9.2) =4.25cm2

Spacing of 12 mm diameter bars

= (Area of One bar)/Ast*100

= (1.13*100)/4.25=26.59cm = 26.5cm c/c

Area of steel for distribution bars

= Ast=0.15/100*100*11=1.65cm2

Spacing of 8 mm diameter bars

= (Area of one bar)/Ast*100

= 0.50/As1.65t*100=30.3c/c=26.5 cm c/c

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