GUIDELINES FOR BEAM DESIGN | REQUIREMENTS FOR BEAM DESIGN | LCETED

  

Usual widths of the beam in India: 230, 300, 350, 400, 450 mm (230 and 300 mm are common).

Usual depths of beam in India: 300 (12”), 350 (14”), 450(18”), 500(20”), 600(24”), 750 mm.

Timber planks available in India are: 4’’(100), 6”(150), 9” (230), 12” (300), 14” (350 mm)

Depth of beam = 85 mm/m of span; or assume B= 300 mm, then 𝐷/𝐵 ≥ 0.3

Recommended depth/width = 1.5 - 2.0 (or width = 1/2 - 2/3 depth).

 

Approx. area of main steel,

𝐴𝑠𝑡 = 𝑀𝑢/0.8𝑑.𝑓𝑦

(𝑀𝑢 = 0.138𝑓𝑐𝑘𝑏𝑑²)

 

Xu,lim	Mu,lim	Pt,lim	𝑅 = 𝑀𝑢,𝑙𝑖𝑚 /𝑏𝑑2
Fe 250, 𝑥𝑢,𝑚𝑎𝑥 = 0.53𝑑	𝑀𝑢,𝑙𝑖𝑚 = 0.148𝑓𝑐𝑘𝑏𝑑2	Pt,lim = 1.75%	2.99
Fe 415, 𝑥𝑢,𝑚𝑎𝑥 = 0.48𝑑	𝑀𝑢,𝑙𝑖𝑚 = 0.138𝑓𝑐𝑘𝑏𝑑2	Pt,lim = 0.95%	2.77
Fe 500, 𝑥𝑢,𝑚𝑎𝑥 = 0.46𝑑	𝑀𝑢,𝑙𝑖𝑚 = 0.133𝑓𝑐𝑘𝑏𝑑2	Pt,lim = 0.75%	2.67

 

 

If an area of steel is known, then 𝑀𝑅 = 0.71𝐴_𝑠𝑡𝑓_𝑦.𝑑

It is economical to select singly reinforced sections with 𝑝_𝑡 = 0.75 − 0.8 P_{𝑡,𝑙𝑖𝑚}

P_{𝑡,𝑙𝑖𝑚} means max. % of steel for balanced section (roughly 1.0 % 𝑜𝑓 𝑏𝐷)

For the preliminary design of doubly reinforced beam, Assume 𝑝𝑡 = 1.2 − 1.5%

Assume 𝐴𝑠𝑐 = (0.5 − 0.65)n

Minimum % of tension steel = 0.3% 𝑜𝑓 𝑏𝐷 (for singly, doubly and flanged beam)

Minimum % of compression steel = 0.2% 𝑜𝑓 𝑏𝐷

As per IS 456, min. 40% tension steel to be provided as comp. steel.

𝑝𝑡 𝑚𝑎𝑥. = 2.5% 𝑜𝑓 𝑏𝐷 (for both tension and compression steel, too high as per IS 13920).

Generally, we provide 1.2-1.5% of 𝐴_𝑠𝑡 and 𝐴_𝑠𝑐 = (1/2 – 2/3) 𝐴_𝑠𝑡

Minimum dia. of anchor or holding bars = 10 mm. these bars are considered as compression steel if the percentage is > 0.2%.

The size of secondary beams should be smaller by 50 mm than the main beams to accommodate bottom bars in secondary beams.

Maximum no. of bars in one layer = 6#.

Common bar sizes for beams = 12, 16, 20, 22, 25 and 32 mm (some are not easily available)

Prefer a combination of two adjacent bar sizes (e.g, 16 and 12 mm; 20 and 16 mm)

Minimum dia. of the bar for main steel = 12 mm, for anchor bar = 10 mm.

Max. no. of bars in one layer in beams with widths 230, 300 and 450 mm are 3, 3 and 4.

For diff. sizes of bars in one layer, provide larger dia. near faces and smaller dia. Towards

The horizontal distance between bars, greater of:

1. Dia. of larger bar

2. Aggregate size + 5 mm

The vertical distance between bars, greater of:

1. Dia. of larger bar

2. 15 mm

3. 2/3 of max. agg. size

Checking of beam width = 𝑁 × 𝜙 + 𝑁 + 1 × 25

Suppose we have 3 # 20𝜙 in a beam 230 mm wide; = 3 × 20 + 3 + 1 × 25 = 160 𝑚𝑚 < 230 𝑚𝑚 𝑂𝐾

Check for spacing of bars; let 𝜙=16 mm, b=230 mm, No of bars=5;

230 − 25 × 2 − 8 × 2 − 5 × 16 4 = 21 𝑚𝑚 < 25 𝑚𝑚(𝑛𝑜𝑡 𝑎𝑐𝑐𝑒𝑝𝑡𝑎𝑏𝑙𝑒)

In any building, beams are generally doubly reinforced.

For doubly reinforced beam, 𝑀𝑢/𝑏𝑑2 = 3 − 5 (𝑡𝑎𝑘𝑒 4)

𝑀_𝑢 to be calculated from code coefficients.

For continuous beam 𝐿/𝐷 = 10 − 15 (𝑡𝑎𝑘𝑒 12). For S.S beam, take 15 (for singly)and take 12 (for doubly)

If actual 𝑙/𝑑 < basic 𝑙/𝑑 (ok)

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