DOUBly Reinforced Beam Design (Simplified)
Where the design moment is greater than the ultimate moment of resistance i.e. M>Mu the concrete will have insufficient strength in compression to generate this moment and maintain an under-reinforced mode of failure. An alternative to increasing the strength/area of concrete in compresion is to use a Doubly-Reinforced section.
Here we must design the compression reinforcement to be able to withstand the moment in excess of Mu. This ensures that the compressive strength of the concrete is not exceeded and an under-reinforced mode of failure will be seen.
Considering the case of a doubly reinforced rectangular beam
The area of compression reinforcement is thus calculated from:
where d’ is the depth to the compression steel from the top surface.
We must now increase the area of tensile reinforcement to maintain compatibility by an equal amount.
and K’ = 0.156
In the above it has been assumed that the compression steel has yielded i.e. steel stress = 0.87 fy, but this will only apply if
if d’/x>0.43, the compression steel will be at a stress less than yield, in which case this stress can be found from the elasto-plastic stress strain curve shown in the introductory notes.(fig2.2 BS8110)
Example 4. Design of Bending Reinforcement for a Singly Reinforced Beam
A simply supported rectangular beam of 9m span carries a characteristic dead (gk ) load (inc. Self wt. of beam ), and imposed (qk ) loads of 6 kN/m and 8 kN/m respectively.
The beam dimensions are breadth b, 225mm and beam height h, 400mm. Assuming fcu =30N/mm2 and fy =460N/mm2 calculate the area of reinforcement required.
Since Mu<M we can NOT design as a singly reinforced beam, we must design as doubly reinforced .
Assuming the compression steel to be 20mm diameter bars
From this a choice is made which is influenced be our previouis assumptions if they appear reasonable at this point:
943mm2 3 Nr. 20mm Dia.
Other combinations of bars resulting in a reinforcement provision greater than the area of steel required Or change our basic assumptions.
Say PROVIDE 3T20 ( A’s prov. = 943mm2 )
Hence PROVIDE 4T25 (As prov = 1960 mm2 )
As with singly reinforced beams an alternative method of determining the area of steel reinforcement required makes use of the Design Charts provided in Part 3 of the BS. These charts are available in a variety of combinations for various concrete grades, steel grade = 460N/mm2 and d’/d ratios.
1. Check Mu<M
2. Calculate d’/d
3. Select appropriate chart based on concrete grade and d’/d ratio
4. Calculate M/bd2
5. Plot M/bd2 ratio on chart and read off corresponding 100A’s/bd2 and 100As/bd2 (Assume x/d=0.5)
6. Calculate A’s and As
For comparison purposes we can check the above example using Chart No. 2 from BS 8110 Pt. 3:
from the Chart 7 for using the fcu =30N/mm2 , fy =460N and d’/d = 0.15 with x/d = 0.5 as our limitting value.
therefore As = 1918mm2
Say Provide 4T25 ( As prov. = 1960mm2 )
therefore A’s = 744mm2
Say Provide 3T20 ( A’s prov. = 943mm2 )
Same as before? NOT QUITE
by: Dr. David W. Begg
University of Portsmouth