As per our last article How to Structural Design a Building/House Step by Step Part-1 & 2. In this article, we design slab load, slab depth, slab steel design. Now in this article, we calculated a slab beam design.
Slab Beam Design
Below steps slab beam design.
1. Beam Thickness & Depth Design
2. Find Ast
3. Slab beam Steel DesignÂ
Beam Thickness & Depth Design
Mu lim = 0.138 . fck . b d2 | SP. 16, P-10, Table-c
d = 2 b
Find Ast Min Steel Design
Ast = (pt / 100) x b x d | Sp. 16, P – 229
Pt= 50 (fck/fy)x( 1-√(1-(4.6xMu/Fck b.d.d)))
Beam Ring Design
Vus = Vu – Ï„c . b . d
Vu = w l /2
Ï„c | IS 456-2000, P-73, Table-19
Sv = (0.87 x fy x Asv x d) / Vus
Check for Maxminum SpacingÂ
1. 0.75 d
2. 300 mm
3. Sv = (0.87 x fy x Asv) / (0.4 x d) | IS : 456 – 2000, P 47
Beam Thickness & Depth Design- Example
Total Weight = As per Slab total Weight Mu X Area = 11.82 kn.m x 3.0 x 3.0 = 89.438 kn.m.
Beam load per running. m = 89.438 kn.m. / (3.0 + 3.0 + 3.0 + 3.0 ) = 14.031 Kn.m.
Along Lx Beam weight = 7.453 x 3.0 = 22.36Â kn.m.
Along Ly Beam weight = 7.453Â x 3.0 = 22.36 kn.m.
Along Lx Find Breadth & Depth
Mu lim = 0.138 . fck . b d2
22.36 x 106Â = 0.138 x 20 x b x (2b)2
Breadth = 138.9 mm ≅ 230 mm
Depth = 2 x 230 mm = 460 mm
Along Ly Find Breadth & Depth
Mu lim = 0.138 . fck . b d2
22.36 x 106Â = 0.138 x 20 x b x (2b)2
Breadth = 138.9 mm ≅ 230 mm
Depth = 2 x 230 mm = 460 mm
Find Ast Min Steel Design
Along Lx
Ast = (pt / 100) x b x d | Sp. 16, P – 229
Pt= 50 (fck/fy) x ( 1-√(1-(4.6 x Mu / Fck b.d.d)))
Pt= 50 (20 / 415 ) x ( 1-√(1-(4.6 x 22.36 x 106/ 20 x 230 x 4602 )))
Pt = 0.131 %
Ast = ( 0.131 / 100) x 230 x 460
Ast = 138.6 mm2
Select = 12 mm dia
ast = ( π / 4 ) x 122 = 113.04 mm2
No Bars. = 138.6/ 113.04 = 1.22 ≅ 2 Nos.Â
2 Nos – 12 mm DiaÂ
Along Ly
Ast = (pt / 100) x b x d | Sp. 16, P – 229
Pt= 50 (fck/fy) x ( 1-√(1-(4.6 x Mu / Fck b.d.d)))
Pt= 50 (20 / 415 ) x ( 1-√(1-(4.6 x 22.36 x 106/ 20 x 230 x 4602 )))
Pt = 0.131 %
Ast = ( 0.131 / 100) x 230 x 460
Ast = 138.6 mm2
Select = 12 mm dia
ast = ( π / 4 ) x 122 = 113.04 mm2
No Bars. = 138.6/ 113.04 = 1.22 ≅ 2 Nos.Â
2 Nos – 12 mm DiaÂ
Beam Ring Design
Lx Beam Self Wight = (b x d x l) x Density of RCC = ( 0.230 x 0.460 x 3.0) x 25 = 7.94 Kn.m.
Ly Beam Self Wight = (b x d x l) x Density of RCC = ( 0.230 x 0.460 x 3.0) x 25 = 7.94 Kn.m.
Lx Total Weight = 22.36 kn.m. + 7.94 kn.m. = 30.3 kn/m.
Ly Total Weight = 22.36 kn.m. + 7.94 kn.m. = 30.3 kn/m.
Along lx Vu =Â ( w . l)/2 = ( 30.3 x 3) / 2 = 45 kn
Along ly Vu =Â ( w . l)/2 = ( 30.3 x 3) / 2 = 45 kn
Along with lx pt = 0.131%Â
Along with ly pt = 0.131%Â
Along with lx Ï„c = 0.28 N/mm2 | IS 456-2000, P 73, Table -19
Along with ly Ï„c = 0.28 N/mm2 | IS 456-2000, P 73, Table -19
Find Vus
Vus = Vu – (Ï„c x b x d)
Along with lx
Vus = 45 – (0.28 x 230 x 460) = 15826 N
Along with ly
Vus = 45 – (0.28 x 230 x 460) = 15826 N
Beam Ring Design
Consider 2 – legged 8 mm ∅ Vertical stirrups of Fe – 415 steel
Asv = 2 x (Ï€ /4)Â x d2 = 2 x (3.14/4) x 82 = 100.51 mm2
Sv -1Â
Along lx, ly = (0.87 x fy x Asv x d) / Vus = ( 0.87 x 415 x 100.51 x 460) / 15826 = 1054 mm.
Sv -2
Along lx, ly = 0.75 x d = 0.75 x 460 = 345 mm
Sv -3
Along lx, ly = 300 mm | IS 456-2000, P-47
Sv -4
Along lx, ly = (0.87 x fy x Asv) / 0.4 b = (0.87 x 415 x 100.51) / 0.4 x 230 = 238 mm
Minmum sapcing of vartical strips = 238 mm
Slab Beam Design
FAQ: Slab Beam Design
What is slab beam design?
Slab beam design involves determining the appropriate dimensions (thickness and depth) and reinforcement (steel bars) required for beams that support concrete slabs in building structures.
Why is slab beam design important?
Slab beams are crucial for distributing loads from slabs to columns and walls efficiently, ensuring structural integrity and safety of the building.
What factors influence slab beam design?
Factors include the type of loads (dead, live, wind), span length, building usage, and local building codes and regulations.
What are the steps involved in slab beam design?
Steps typically include determining beam dimensions based on bending moments, calculating required steel reinforcement (Ast), and designing shear reinforcement (stirrups).
How is the depth of a slab beam calculated?
The depth (d) of a slab beam is calculated based on bending moment considerations and is influenced by factors such as span length and load distribution.
How do you calculate the steel reinforcement (Ast) for slab beams?
Ast is calculated using the formula (pt / 100) x b x d, where pt is the percentage of tensile reinforcement required based on the bending moment and material properties.
What types of reinforcement are used in slab beams?
Slab beams typically use longitudinal reinforcement (main bars) and shear reinforcement (stirrups or ties) to resist bending moments and shear forces.
How are shear forces handled in slab beam design?
Shear forces are resisted using stirrups or ties, which are spaced at calculated intervals along the beam length to prevent failure due to shear stress.
What are the design considerations for slab beam connections?
Connections between slab beams and columns or walls must be designed to transfer forces effectively and accommodate movement and settlement.
How does slab beam design comply with building codes and standards?
Design calculations and detailing must adhere to local building codes and standards (e.g., IS 456-2000), ensuring structural safety and durability.
