Beam Design
The design of beam is presented below steps wise: [ hope you have read details before, if not click here. ]
- Preliminary design of beam
The preliminary design of beam is used to find out the tentative dimension of the beam to initiate the design of the beam. We have already discussed about it before. Click here to reach there. The preliminary design is based on serviceability condition of deflection control criteria.
2. Load calculation
We have to calculate all the load that the beam have to support including its own self weight. Some of the loads that must of the beam encountered are:
- Partition load
It can be calculated by multiplying the unit weight of partition material with the plan dimension. The partition load is uniformly distributed load and calculated per meter.
- Self-weight of beam
Self-weight of beam = unit weight of beam * width * overall depth
The unit weight of reinforced concrete is taken as 25 KN/m2.
- Load from slab
- Live load
- Floor finish
- Self-weight of slab
The dead load is calculated from IS 875 part 1 and the live load is derived from IS 875 part 2.
3. Moment calculation
- By using codes
- Moment distribution method
- From design software
For the calculation of moment we have to go either manually or from design software. For manually, the easy method is using codes. The indian code IS 456:2000, table 12 and 13 gives bending moment and shear coefficient. But it has many restriction and limitation. For continuous beam under clause 22.5 states that “unless more exact estimates are made, for beam of uniform cross-section which support substantially uniformly distributed loads over three or more span which do not differ by more than 15 % of the longest”. For details you can have a view in code Is 456:2000 clause 22.
Here we do not discuss about the moment distribution method. You can also go through this for calculation of moment but it is so complex and tedious in calculation. So I recommend you to use design software like ETABS, SAP2000 or STAAD PRO.
Some of you may confuse that why to use design software just to calculate moment and shear thought these software also give reinforcement details. For the final year project, you are not allowed to use directly reinforcement detail for the design software, so. It is simple to calculate moment and shear from software rather than using tedious moment distribution method or codes with restrictions.
The above mentioned for continuous beam, for simple supported
Bending moment = wl2/8
And shear force = w l /2
The moment calculated from this formula can be redistributed for the economic design.
4. Check for singly or doubly reinforced beam
To know the either the beam is singly or doubly reinforced, we have to calculate limiting moment and compare from the moment calculated above.
Mu,lim = 0.36 fck b xu ( d – 0.42 xu)
We can get the value of xu from IS 456:2000 clause 38.1 page 70 if we know fy and effective depth.
If Mu,lim < Mu ; it is doubly reinforced beam otherwise single reinforced beam.
5. Calculation of tensile and compression reinforcement.
It is not necessary to calculate compression reinforcement if the beam is singly reinforced beam. Despite singly reinforced, nominal number of compression reinforcement should be provided ie. two number of bar in two corner to support the transverse reinforcement.
- For singly reinforced beam
Mu = 0.87 fy Ast [ d – fy Ast / fck b ]
Since all parameter are known, we can find out the Ast area of tensile reinforcement.
- For doubly reinforced beam
For tensile reinforcement;
Mu,lim = 0.87 fy Ast1 [ d – fy Ast1 / fck b ]
Mu – Mu,lim = 0.87 fy Ast2 [ d – d’]
Where d’ = effective clear cover
Total tensile reinforcement Ast = Ast1 + Ast2
For compression reinforcement;
Mu – Mu,lim = Asc (fsc – fcc) [ d – d’]
where, fsc and fcc can be calculated as,
strain at the level of centroid of compression steel is
Ԑ = 0.0035 [1- d’/xu,max]
Again from table ‘A’ of code sp 16
For the given strain calculate the stress which is called fsc .
And fcc = 0.446 fck
- Calculation of number of bar provided and check for min. and max. limits
As we have initially assumed the diameter of bar, we have to use that diameter to calculate the number of bar.
Area of a bar = pi * dia2 / 4
Then, number of bar = total calculated area / area of a bar
Check:
- the maximum area of reinforcement provided = 4% of Bd
- the minimum area of reinforcement required = 0.85 * bd/fy
6. Arrangement of bar with spacing check
The horizontal distance between parallel reinforcement bars or groups, near the tension face of the beam shall be greater than the value given in Table 15 of IS 456:2000.
The horizontal distance between two parallel main reinforcing bars shall usually be not less than the greatest of the following
- Diameter of larger bar
- 5mm more than nominal maximum size of coarse aggregate
The bar should be arranged considering the maximum and minimum spacing of longitudinal bars. When there are two or more rows of bars, the bars shall be vertical in line and the minimum vertical distance between the bars shall be 15 mm or the maximum size of aggregate or the maximum size of the bar, whichever is greater.
7. Check for deflection
The depth of beam initially obtained is through different assumption in preliminary design. After calculating area of reinforcement provided, it is easier to calculate exact modification value. (IS 456:2000 clause 23.2.1)
The calculated ratio of span to depth should be greater than initially obtained from assumed value.
8. Check for shear
First calculate nominal shear stress Ⴀv = Vu / bd
The method of calculation of shear force Vu is already mentioned.
From table 19 of code IS 456:2000, calculate design shear stress Ⴀc and from table 20, calculate the maximum shear stress Ⴀc,max .
If Ⴀv < Ⴀc , no need to calculate the shear reinforcement. But we have to provide nominal reinforcement.
If Ⴀv > Ⴀc,max we have to revise the dimension of beam
If Ⴀv < Ⴀc < Ⴀc,max, shear reinforcement are to be designed.
9. Calculation of transverse reinforcement with spacing
The shear reinforcement can be calculated from the code IS 456:2000 of clause 40.4 (page 73).
Near the support, there is only stirrups to resist the shear force.
Shear resisted by reinforcement = Total shear minus shear resisted by concrete.
Vus = 0.87 fy Asv d / sv
Where, Asv = area of stirrups [Note: for two-legged stirrups, area of bar should be multiplied by 2]
At the point of bar bent-up
There are bent-up bars to resist shear stress beside the traverse reinforcement. The contribution of bent-up bar can’t be taken more than 50% of total shear resisted by reinforcement.
The spacing of transverse reinforcement should not be greater than 300 mm or 0.75d .
ALSO VISIT
