Project   Laurencekirk WWTW   Page   1/4Job   33052   Updated   12/03/2014Version   1.1   Reviewed   17/03/2014Designer   Miro Stefanko   Reviewer

1 Continuous beam design (BS8110)

All units are[kN, mm]

A typical floor plan of a small building structure is shown in Figure 1.1.2. Designcontinuous beams 3A/D and B1/5 assuming the slab supports an imposed loadof 4kN m2 and finishes of 1.5kN m2. The overall sizes of the beams and slab areindicated on the drawing. The columns are 400×400mm. The characteristic strengthof the concrete is 35N mm2 and of the steel reinforcement is 500N mm2. The coverto all reinforcement may be assumed to be 30 mm.

1.1 Loading

Figure 1.1.1: Beam cross-section at center line 3.

Figure 1.1.2: Floor plan layout.

Dead load, gk, is the sum of 

weight of slab = 0.15× 3.75× 24 = 13.5weight of downstand = 0.3× 0.4× 24 = 2.88finishes = 1.5× 3.75 = 5.625

22.0kN m−1

Imposed load, q k  = 4× 3.75 = 15kN m1

Design uniformly distributed load, = (1.4gk + 1.6q k) = (1.4× 22 + 1.6× 15) =

 

Project   Laurencekirk WWTW   Page   2/4Job   33052   Updated   12/03/2014Version   1.1   Reviewed   17/03/2014Designer   Miro Stefanko   Reviewer

54.8kN m1

Design load per span, F  = ω × span = 54.8× 8.5 = 465.8kN F  = 465.8kN 

1.1.1 Design moment and shear forces

From clause 3.4.3 of BS 8110, as gk  > q k, the loading on the beam is substantiallyuniformly distributed and the spans are of equal length, the coefficients in Table 3.19can be used to calculate the design ultimate moments and shear forces. The resultsare shown in the table below. It should be noted however that these values areconservative estimates of the true in-span design moments and shear forces since thecoefficients in Table 3.19 are based on simple supports at the ends of the beam. Inreality, beam 3A/D is part of a monolithic frame and significant restraint momentswill occur at end supports.

Figure 1.1.3: Bending moment and shear forces

1.2 Steel reinforcement

1.2.1 Middle of 3A/B (and middle of 3C/D)

Assume diameter of main steel, φ  = 25mm, diameter of links, φ0 = 8mm and nominalcover,  c = 30mm. Hence

Eff ective depth,

d =  h − φ/2 − φ0− c = 550 − 25/2 − 8 − 30 = 499mm

d = 499mmThe eff ective width of beam is the lesser of 

(a) actual flange width = 3750 mm(b) web width + bz/5, where bz  is the distance between points of zero moments

which for a continuous beam may be taken as 0.7 times the distance between centresof supports. Hence

lz  = 0.7× 8500 = 5950mm(critical)

lz  = 5950mmb = 300 + 5950/5 = 1490mm

b = 1490mm

K  =  M 

f cubd2 =

  356.3× 106

35× 1490× 4992 = 0.0274   K  = 0.0274

 

Project   Laurencekirk WWTW   Page   3/4Job   33052   Updated   12/03/2014Version   1.1   Reviewed   17/03/2014Designer   Miro Stefanko   Reviewer

z  = d(0.5 +p 

(0.25 − K/0.9)) ≤ 0.95d = 0.95× 499 = 474mm) z limited to 0.95d= 474mm

x = (d − z )/0.45 = (499 − 474)/0.45 = 56mm < flangethickness   x = 56mm

Area of steel reinforcement,

As =  M 

0.87f yz  =

  356.3× 106

0.87× 500(0.95× 499) = 1728mm2

Provide 4H25 (As = 1960mm2).   As = 1960mm2

Figure 1.2.1: Flexural reinforcement middle of 3A/B and 3C/D.

1.2.2 At support 3B (and 3C)

Assume the main steel consists of two layers of 25 mm diameter bars, diameter of links,  φ0 = 8mm and nominal cover, c = 30mm. Hence Eff ective depth,

d =  h − φ− φ0− c = 550 − 25 − 8 − 30 = 487mm

d = 487mmSince the beam is in hogging, b = 300 mm

M u = 0.156f cubd2 = 0.156× 35× 300× 4872 × 10−6 = 388.5kN m M  u = 388.5kN m

Since  M u   < M (= 435.5kN m), compression reinforcement is required. Assume di-ameter of compression steel, φ = 25mm, diameter of links,  φ0 = 8mm, and cover toreinforcement, c = 30mm. Hence eff ective depth of compression steel d0 is

d0 = c + φ0 + φ/2 = 30 + 8 + 25/2 = 51mm d0 = 51mm

z  = d(0.5 +p 

(0.25 − K 0/0.9)) = 487(0.5 +p 

(0.25 − 0.156/0.9)) = 378mm z  = 378mm

Depth to neutral axis,

x = (d − z )/0.45 = (487 − 378)/0.45 = 242mm

d’/x = 51/242 = 0.21<0.37. Therefore, the compression steel has yielded, i.e.f 0s = 0.87f y  and

Area of compression steel,

A0

s =  M  − M u

0.87f y(d − d0) =

  (435.5 − 388.5)106

0.87× 500(487 − 51) = 248mm2

 

Project   Laurencekirk WWTW   Page   4/4Job   33052   Updated   12/03/2014Version   1.1   Reviewed   17/03/2014Designer   Miro Stefanko   Reviewer

Provide 2H25 (A0

s = 982mm2).   A0

s = 982mm2

Area of tension steel,

As =  M u0.87f yz 

 + A0

s =  388.5× 106

0.87× 500× 378 = 2610mm2

Provide 6H25 (As = 2950mm2).   As = 2950mm2

Figure 1.2.2: Flexural reinforcement at support 3B and 3C.

1.2.3 Middle of 3B/C

From above, eff ective depth, d = 499mm and eff ective width of beam, b = 1490mm.Hence, As   is

As =  M 

0.87f yz  =

  277.2× 106

0.87× 500(0.95× 499) = 1344mm2

Provide 3H25 (As = 1470mm2).   As = 1470mm2

Figure 1.2.3 shows a sketch of the bending reinforcement for spans 3A/B and 3B/C.The curtailment lengths indicated on the sketch are in accordance with the simplifiedrules for beams given in clause 3.12.10.2 of BS 8110.

Figure 1.2.3: Bending reinforcement for spans 3A/B and 3B/C.