Project Laurencekirk Laurencekirk WWTW 33052 Job Version 1.1 Miro Stefanko Stefanko Designer
1
Page 1/4 Updated 12/03/2014 Reviewed 17/03/2014 Reviewer
Con Contin tinuous uous beam beam desi design gn (BS8 (BS811 110) 0)
A typica typicall floor plan of a small small building building structure structure is shown shown in Figure Figure 1.1.2. 1.1.2. Design Design All units are continuous beams 3A/D and B1/5 assuming the slab supports an imposed load [kN, mm] of 4kN m2 and finishes of 1. 1.5kN m2 . Th Thee overa overall ll sizes sizes of the beams beams and and slab slab are indicated indicated on the drawing. The columns are 400×400mm 400mm.. The characteristi characteristicc strength 2 of the concrete is 35N 35N mm and of the steel reinforcement is 500N 500N mm2 . The cove coverr to all reinforcement may be assumed to be 30 mm.
1.1
Load Loadin ing g
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. 0.15 × 3.75 × 24 weight of downstand = 0. 0.3 × 0.4 × 24 finishes = 1. 1.5 × 3.75
= = =
13.5 2.88 5.625 22. 22.0kN m
−1
Imposed load, q k = 4 × 3.75 = 15kN 15kN m1 Design uniforml uniformly y distributed distributed load, = (1. (1.4gk + 1. 1 .6q k ) = (1. (1.4 × 22 + 1. 1.6 × 15) =
Project Laurencekirk WWTW 33052 Job Version 1.1 Miro Stefanko Designer 54.8kN m1 Design load per span, F = ω
1.1.1
Page 2/4 Updated 12/03/2014 Reviewed 17/03/2014 Reviewer
× span =
54.8 × 8.5 = 465.8kN
F = 465.8kN
Design moment and shear forces
From clause 3.4.3 of BS 8110, as gk > q k , the loading on the beam is substantially uniformly distributed and the spans are of equal length, the coefficients in Table 3.19 can be used to calculate the design ultimate moments and shear forces. The results are shown in the table below. It should be noted however that these values are conservative estimates of the true in-span design moments and shear forces since the coefficients in Table 3.19 are based on simple supports at the ends of the beam. In reality, beam 3A/D is part of a monolithic frame and significant restraint moments will occur at end supports.
Figure 1.1.3: Bending moment and shear forces
1.2 1.2.1
Steel reinforcement Middle of 3A/B (and middle of 3C/D)
Assume diameter of main steel, φ = 25mm, diameter of links, φ = 8mm and nominal cover, c = 30mm. Hence Eff ective depth, 0
0
d = h − φ/2 − φ
− c =
550 − 25/2 − 8 − 30 = 499mm d = 499mm
The eff ective width of beam is the lesser of (a) actual flange width = 3750 mm (b) web width + b z /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 centres of supports. Hence lz = 0.7 × 8500 = 5950mm(critical) lz = 5950mm b = 300 + 5950/5 = 1490mm b = 1490mm 6
K =
M 356.3 × 10 = = 0.0274 f cu bd2 35 × 1490 × 4992
K = 0.0274
Project Laurencekirk WWTW 33052 Job Version 1.1 Miro Stefanko Designer z = d(0.5 +
Page 3/4 Updated 12/03/2014 Reviewed 17/03/2014 Reviewer
p
(0.25 − K/0.9)) ≤ 0.95d = 0.95 × 499 = 474mm)
x = (d − z )/0.45 = (499 − 474)/0.45 = 56mm < flangethickness
z limited to 0.95d = 474mm x = 56mm
Area of steel reinforcement, As =
M 356.3 × 106 = = 1728mm2 0.87f y z 0.87 × 500(0.95 × 499)
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, φ = 8mm and nominal cover, c = 30mm. Hence Eff ective depth, 0
0
d = h − φ − φ
− c =
550 − 25 − 8 − 30 = 487mm
Since the beam is in hogging, b = 300 mm
d = 487mm
M u = 0.156f cubd2 = 0.156 × 35 × 300 × 4872 × 10
−6
= 388.5kN m
Mu = 388.5kN m
Since M u < M ( = 435.5kN m), compression reinforcement is required. Assume diameter of compression steel, φ = 25mm, diameter of links, φ = 8mm, and cover to reinforcement, c = 30mm. Hence eff ective depth of compression steel d is 0
0
0
0
d = c + φ + φ/2 = 30 + 8 + 25/2 = 51mm z = d(0.5 +
p
(0.25 − K /0.9)) = 487(0.5 + 0
p
(0.25 − 0.156/0.9)) = 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 s = 0.87f y and 0
Area of compression steel, M − M u (435.5 − 388.5)106 As = = = 248mm2 0.87f y (d − d ) 0.87 × 500(487 − 51) 0
0
d = 51mm 0
z = 378mm
Project Laurencekirk WWTW 33052 Job Version 1.1 Miro Stefanko Designer
Page 4/4 Updated 12/03/2014 Reviewed 17/03/2014 Reviewer
Provide 2H25 (As = 982mm2 ). 0
As = 982mm2 0
Area of tension steel, M u 388.5 × 106 + As = = 2610mm2 As = 0.87f y z 0.87 × 500 × 378 0
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 M 277.2 × 106 As = = = 1344mm2 0.87f y z 0.87 × 500(0.95 × 499)
Provide 3H25 (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 simplified rules 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.
As = 1470mm2