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 d ^{2}** | 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=Â 50 (fck/fy)x( 1-âˆš(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 d ^{2}**

22.36 x 10^{6}Â = 0.138 x 20 x b x (2b)^{2}

**Breadth = 138.9 mm â‰… 230 mm**

**Depth =** 2 x 230 mm** = 460 mm**

### Along Ly Find Breadth & Depth

**Mu lim = 0.138 . fck . b d ^{2}**

22.36 x 10^{6}Â = 0.138 x 20 x b x (2b)^{2}

**Breadth = 138.9 mm â‰… 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=Â 50 (fck/fy) x ( 1-âˆš(1-(4.6 x Mu / Fck b.d.d)))**

Pt=Â 50 (20 / 415 ) x ( 1-âˆš(1-(4.6 x 22.36 x 10^{6}/ 20 x 230 x 460^{2} )))

**Pt = 0.131 %**

Ast = ( 0.131 / 100) x 230 x 460

**Ast = 138.6 mm ^{2}**

**Select = 12 mm dia**

**ast** = ( Ï€ / 4 ) x 12^{2}** = 113.04 mm ^{2}**

No Bars. = 138.6/ 113.04 = 1.22 â‰… **2 Nos.Â **

**2 Nos – 12 mm DiaÂ **

**Along Ly**

**Ast = (pt / 100) x b x d** | Sp. 16, P – 229

**Pt=Â 50 (fck/fy) x ( 1-âˆš(1-(4.6 x Mu / Fck b.d.d)))**

Pt=Â 50 (20 / 415 ) x ( 1-âˆš(1-(4.6 x 22.36 x 10^{6}/ 20 x 230 x 460^{2} )))

**Pt = 0.131 %**

Ast = ( 0.131 / 100) x 230 x 460

**Ast = 138.6 mm ^{2}**

**Select = 12 mm dia**

**ast** = ( Ï€ / 4 ) x 12^{2}** = 113.04 mm ^{2}**

No Bars. = 138.6/ 113.04 = 1.22 â‰… **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/mm ^{2}** | IS 456-2000, P 73, Table -19

**Along with ly Ï„c = 0.28 N/mm ^{2}** | IS 456-2000, P 73, Table -19

**Find Vus**

Vus = Vu – (Ï„cÂ 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 âˆ… Vertical stirrups of Fe – 415Â steel

**Asv =** 2 x (Ï€ /4)Â x d^{2} = 2 x (3.14/4) x 8^{2} = 100.51 mm^{2}

**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.