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DOUBLY REINFORCED BEAM DESIGN

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

concrete design

The area of compression reinforcement is thus calculated from:

compression reinforcement

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.

maintain-compatibility

where

beam-design

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.

reinforced-beam

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.

concrete-beam-design

Since  Mu<M we can NOT design as a singly reinforced beam, we must design as doubly reinforced .

Compression Reinforcement

Assuming the compression steel to be 20mm diameter bars

Beam-concrete-design

Bar Number of Bars
Size 1 2 3 4 5 6 7 8 9 10
6 28.3 56.6 84.9 113 142 170 198 226 255 283
8 50.3 101 151 201 252 302 352 402 453 503
10 78.5 157 236 314 393 471 550 628 707 785
12 113 226 339 452 566 679 792 905 1020 1130
16 201 402 603 804 1010 1210 1410 1610 1810 2010
20 314 628 943 1260 1570 1890 2200 2510 2830 3140
25 491 982 1470 1960 2450 2950 3440 3930 4420 4910
32 804 1610 2410 3220 4020 4830 5630 6430 7240 8040
40 1260 2510 3770 5030 6280 7540 8800 10100 11300 12600

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 )

Tension Reinforcement

Hence PROVIDE 4T25 (As prov = 1960 mm2 )

Design Charts??

concrete-beam

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.

Procedure!

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:

momen-design

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.

concrete-design-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

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