Table of Contents
Design of Secondary Beam:
The structural member which is primarily subjected to flexure or bending are known as beam. Beam supports load and their own self weight by internal moment and shear. On the basis of end connections to structural member, there are two types of beam: 1. Primary beam and 2. Secondary beam. Primary beam means that beam which is supported by two column and secondary beam means that beam which is supported on beam on one or both end. Here the design of secondary beam is explained with an example.
Secondary beams are those beam which are connected with primary beam for transferring load of structure to the primary beam. They are provided for supporting the slab by reducing it thickness. The use of secondary beam drastically reduces the dead load of the building which becomes advantageous for seismic resistance. Normally, secondary beams are provided in such a way that it divides the longer span. But sometimes it is provided dividing the shorter span to eliminate different problem during construction and design. For design simplicity, secondary beam is treated as simply supported beam and this assumption gives satisfactory result.
Numerical:
Design example of secondary beam :
Design a secondary beam of the given institutional building plan which is aligned in y-axis. Grade of concrete = M30 and grade of steel = 500D. Slab thickness = 130 mm and No partition wall. Live load = 3 KN/m3 and floor finish = 1.5 KN/m3. Assume other data if necessary.
There is presence of secondary beam along y-axis dividing the slab in equal half [shown in fig of SAP2000 analysis]
Pink bold line= primary beam
Light green bold line = secondary beam
For design of beam along y-axis, let’s take longest span beam (10.02m c/c).
- Preliminary design of beam
The detail about the preliminary design is discussed before. If you want to go there, then click here.
The preliminary design is based on the serviceability condition of deflection criteria. From IS 456:2000 clause 23.2.1,
L/d ≤ αβγδλ
Assume d = l/17 [ for simple-supported beam and Fe 500D ]
Where l is effective span which can be calculated from clause 22.2 of IS 456:2000.
So, d = 10.02/17 = 0.585 m
Assume diameter of bar = 20 mm and clear cover = 25 mm
For secondary beam, we can decrease depth by some value [based on experiences]
Adopt overall depth of secondary beam equal to primary beam = 600 mm
So effective depth = overall depth – clear cover – half of diameter of bar
= 600 – 25 – 20/2 = 565 mm
And width = 400 mm
2. Load calculation
- Self-weight of beam
Size of beam = 250mm * 600 mm
Unit weight of concrete = 25 kN/m3
So, self-weight of beam = 25 * 0.4 * 0.6 = 6 kN/m
- Load from slab
Thickness of slab = 130 mm [From detailed design of slab, to learn more click here.]
Live load = 3 kN/m3
Floor finish = 1.5 kN/m3
So, weight of slab = unit weight * thikness of slab
= 25 * 0.13
= 3.25 kN/m2
Width of the slab loaded on beam = half the effective distance in left and half in right [since one way slab]
= 3.82/2 + 3.82/2
= 3.82 m
So, total load from slab = (live load + floor finish + slab weight) * loaded width
= (3 + 1.5 + 3.25) * 3.82
= 29.6 KN/m
Total factored load = 1.5 * (self-weight of beam + total load from slab)
= 1.5 * (6 + 29.6)
= 53.4 KN/m
3. Moment calculation
Since the beam does not meet the criteria for code. So we have either to use moment distribution method or structural design software. Here I use SAP2000, but you can also use other design analysis software.
Since the secondary beam is assumed as simply supported, there is negligible moment at the support .
from SAP2000 , as in figure above,
Maximum bending moment = 69.16 KN-m
Maximum shear force = 113.14 KN
4. Check for singly or doubly reinforced beam
For this, we have to calculate limiting moment
Mu,lim = 0.133 fck b d2 [ for fe 500]
= 0.133 * 30 * 400 * 5652 [fck = 30 N/mm2]
= 509.48 KN-m
So singly reinforced beam
5. Calculation of tensile and compression reinforcement.
For singly reinforced beam
Mu = 0.87 fy Ast [ d – fy Ast / fck b ]
69.16 * 106 = 0.87 * 500 * Ast [565 – 500 * Ast/ 30 * 400]
Ast = 287.49 mm2
6. Calculation of number of bar provided and check for min. and max. limits
The assumed diameter of bar = 20 mm
Area of a bar = 3.14 * 202/4
= 314 mm2
Checking limits:
- the maximum area of reinforcement provided = 4% of bD = 9600 mm2
- the minimum area of reinforcement required = 0.85 * bd/fy = 384.2 mm2
satisfying the limit so,
provide two bottom bar of diameter 20 mm throughout
provide two top bar of dimeter 20 mm throughout
7. Arrangement of bar with spacing check
From code IS 456:2000 table 15, the clear distance between the bars should not be greater than 150 mm for Fe 500 with no moment redistribution.
Provide 3/3 number of 20mm bar on both top and bottom layer of beam at midspan.
Checking spacing = (width – 2 * clear cover – nos * diameter of bar ) / ( nos of bar -1)
= (400 – 2 * 50 – 3 * 20) / 2 = 120mm ok
8. Check for deflection
For deflection control criteria, using IS 456:2000 clause 23.2.1
For modification factor for tension reinforcement,
fs = 0.58 fy * area required / area provided
= 0.58 * 500 * 287.5/942.47
= 88
Percentage of steel = 3*314.16 *100 / 400*600 = 0.39 %
From fig.4, modification factor for tension reinforcement = 2
From fig. 5, for compression reinforcement 0.39% , modification factor = 1.1
From fig. 6 modification value = 1
Allowable l/d ratio = basic value * modification value
= 26 *10/10.02 * 2 * 1.1 *1 = 57.08 >> Assumed (17)
9. Check for shear
First calculate nominal shear stress Ⴀv = Vu / bd
= 113.14 * 1000 / 400 * 565
= 0.5 N/mm2
From table 19 of code IS 456:2000, calculate design shear stress Ⴀc
For steel percent 0.39% and M30 grade of concrete,
Ⴀc = 0.37 +(0.50-0.37)*(0.39-0.25)/(0.5-0.25)
= 0.44 N/mm2
and from table 20, the maximum shear stress Ⴀc,max = 3.5 N/mm2
If Ⴀv < Ⴀc < Ⴀc,max, shear reinforcement are to be designed.
10. Calculation of transverse reinforcement with spacing
The shear reinforcement can be calculated from the code IS 456:2000 of clause 40.4 (page 73).
Shear resisted by reinforcement = Total shear minus shear resisted by concrete.
= (0.5 – 0.44) * 400 * 565
= 13.56 KN
Now,
Assume diameter of stirrups = 8 mm
Since two-legged stirrups so,
Area = 2 * 3.14 * 82/ 4
= 100.5 mm2
So,
Vus = 0.87 fy Asv d / sv
13560 = 0.87 * 500 * 100.5 * 565 / sv
Sv = 1821.56 mm
The spacing of transverse reinforcement should not be greater than 300 mm or 0.75d .
0.75 d = 0.75 * 565 = 423.75
So provide stirrups of 8 mm diameter @ 300 mm c/c throughout.
From the whole design procedure, it shows that the beam of size 400 mm * 600 mm gives very uneconomical result. And for economic design you can go for next trial with smaller dimension and check out all procedure.
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