Ex 7.2 - Simply Supported (Flange Beam)

Simply Supported (Flange Beam) Figure below shows part of the first floor plan of a reinforced concrete office building. During construction, slabs and beams are cast together. The overall thickness of slab is 125mm and the dimensions of the beams are as given in the diagram. The finishes, ceiling and services form a characteristic permanent action of 1.5kN/m 2. The characteristic variable action is 3.0kN/m2. Three meter high brickwall weighing 2.6kN/m2 is placed over the entire span of all beams. The construction materials consist of Grade C25 concrete and Grade 500 steel reinforcement. For durability consideration a nominal cover of 30mm is required. Based on the information provided: a) Calculate the design action carried by beam 2/B-C b) Sketch the bending moment and shear force diagrams of beam 2/B-C c) Design the beam for ultimate and serviceability limit states.

1

2 2500 2500

A

3500 3500

3

2000 2000

250 x 500

250 x 500

250 x 600

250 x 500

250 x 600

C

5500 6000

B

250 x 500

1.0 SPECIFICATION Slab thickness Slab Characteristic actions: ----Finishes etc. ----Variable, qk ----Unit weight brick-wall Materials: ----Characteristic strength of concrete, fck ----Characteristic strength of steel, fyk ----Characteristic strength of link, fyk ----Unit weight of reinforced concrete ----Nominal cover Assumed: ----Øbar1 ----Øbar2 ----Ølink

= 100mm = 1.5kN/m = 3.0kN/m = 2.6kN/m2 = 25N/mm2 = 500N/mm2 = 500N/mm2 = 25kN/m3 = 30mm = 25mm = 12mm = 6mm

2.0 BEAM SIZE beff hf h

bw

bw = 250mm hf = 100mm h = 500mm b1 = 1625mm b2 = 1125mm lo= 6000mm beff,i = 0.2bi + 0.1lo

0.2lo

= 0.2b1 + 0.1lo = 915mm

0.2lo = 1200mm beff,1 = 915mm, beff,2 = 825mm b = b1 + b2 + bw

= 2950mm

Effective flange width, beff =

= 915 + 825 + 250 = 1990mm < b

3.0 LOADING AND ANALYSIS Loads on slab: Self-weight

= 25 x 0.1 = 2.5kN/m2 = 1.5kN/m2 = 4.0kN/m2 = 3.0kN/m2

Finishes etc. Characteristic permanent action, gk Characteristic variable action, qk Loads on beam 2/B-C

2500

6000

2000

3500

w1

w2

Distribution of load from slabs to beam are as follows: Panel B-C/2-3: Ly/Lx

= 6/2.5 = 2.4 > 2.0 (One-way slab) Shear coefficient, = 0.50

Panel B-C/1-2: Ly/Lx

= 6/3.5 = 1.71 < 2.0 (Two-way slab) Shear coefficient, = 0.57

Characteristic permanent load wo = 0.25 (0.50-0.10) x 25 w1 = nLx= 0.57 x 4.00 x 3.50 w2 = nLx = 0.50 x 4.00 x 2.50 w3 = 2.6 x 3.0 (Brickwall) Gk

= 2.50kN/m = 7.98kN/m = 5.00kN/m = 7.8kN/m = 23.28kN/m

Qk

= 5.99kN/m = 3.75kN/m = 9.74kN/m

Characteristic variable load w1 = nLx= 0.57 x 3.00 x 3.50 w2 = nLx = 0.50 x 3.00 x 2.50

Total design load on beam 2/B-C wd=1.35Gk + 1.5Qk

= 46.03kN/m

46.03kN/m

6m

Shear Force, V = wdL/2 V = 138.1kN Bending Moment, M = wdL2/8 M = 207.1kNm

4.0 MAIN REINFORCEMENT Effective depth: d = h – Cnom – Ølink – Øbar/2 MEd = 207.1kNm Mf = 0.567fckbhf(d-0.5hf)

MEd < Mf

= 451.5mm

= 0.567 x 25 x 1990 x 100(451.5-50) = 1133kNm

Neutral axis within the flange

No compression reinforcement

K = M/bd2fck

= 207.1x106/(1990 x 451.12 x 25) = 0.20

z = d[0.5+ z = 0.98d Area of tension reinforcement As = MEd/0.87fykz

= 207.1x106/(0.87 x 25 x 0.98 x 451.5) = 1110mm2

Use: 3H 25 (1473mm2)

Min. and max. reinforcement area, As,min = 0.26(fctm/fyk)bd

As,max = 0.04Ac

= 0.26(2.56/500)bd = 0.0013bd (use 0.0013bd) = 0.0013 x 250 x 451.5 = 151mm2 = 0.04 x 250 x 500 = 5000mm2

5.0 SHEAR REINFORCEMENT VEd = 138.1kN Concrete strut capacity VRd,max = 0.36bwdfck(1-fck/250)/(cotθ+tanθ)

= 318kN (θ=22, cotθ=2.5) = 457kN (θ=45, cotθ=1.0)

VEd < VRd,max cotθ=2.5 VEd < VRd,max cotθ=1.0 Use θ=22, cotθ=2.5 Shear links Asw/s=VEd/(0.78fykdcot )

= 138.1 x 103/(0.78x500x452x2.5) = 0.317

Try link H6, Asw=57mm2 Spacing, s=57/0.317=178 mm Use H6-150 Max. spacing, S=0.75d

= 0.7 x 452 = 339mm

Minimum links:Asw/s = (0.08fck1/2bw)/fyk

= (0.08x251/2x250)/500 = 0.200

Try link H6, Asw = 57mm2 Spacing, s = 57/0.200 = 283mm < S Use H6-275 Shear resistance of minimum links Vmin = (Asw/s)(0.78dfykcot )

= (57/275)(0.78 x 452 x 500 x 2.5) = 90kN

138kN 90kN x = (138-90)/46 x = 1.05m

90kN 138kN

H6-150

H6-275

H6-150

1.05m

3.89m

1.05m

Transverse steel in the flange The longitudinal shear stresses are the greatest over a distance x measured from the point of zero moment, x = 0.5(L/2)

= 6000/4 = 1500mm

The change in moment over distance x from zero moment M = (wL/2)(L/4) – (wL/4)(L/8) = 3wL2/32

The change in longitudinal force, Fd = [ M/(d-0.5hf] x [(b-bw)/2b]

= (3 x 46 x 6.02)/32 = 155.35kNm

= [155.35 x 103/(452 - 50)] x = [(1990 - 250)/(2 x1990)] = 169kN

Longitudinal shear stress ved = Fd/(hf x)

= 169x103/(100 x 1500) = 1.13N/mm2

ved > 0.27fctk = 0.27 x 1.80 = 0.48N/mm2 Transverse shear reinforcement is required Concrete strut capacity in the flange ved,max = 0.4fck(1-fck/250)/( cotθ+tanθ)

= 3.59N/mm2 (θ=26.5, cotθ=2.0) = 4.50N/mm2 (θ=45, cotθ=1.0)

ved < vRd,max cotθ=2.5 ved < vRd,max cotθ=1.0 Use θ=26.5, cotθ=2.0 Transverse shear reinforcement Asf/sf = vedhf/0.87fykcotθ

= 1.13 x 100 / (0.87 x 500 x 2.0) = 0.13

Try H10, Asf = 79mm2 Spacing, sf = 79/0.13 = 608mm Minimum transverse steel area, As,min = 0.26(fctm/fyk)bhf

= 0.26(2.56/500)bhf = 0.0013 bhf (use 0.0013 bhf) = 0.0013 x 1000 x 100 = 133mm2/m Use H10 – 500 (157mm2/m)

Additional longitudinal reinforcement Additional tensile force, Ftd = 0.5Vedcotθ

= 0.5 x 138 x 2.48 = 171kN

Med,max/z

= 207.1x106/428.9 = 483kN

As,req = Ftd/0.87fyk

= 171x103/(0.87 x 500) = 393mm2 Use 1H25 (491mm2)

6.0 DEFLECTION Percentage of required tension reinforcement, =As,req/bd

Reference reinforcement ratio, =(fyk)1/2x10-3

Percentage of required compression reinforcement =As’,req/bd

= 1110/(250X452) = 0.010 = (25)1/2x10-3 = 0.005

= 0/(250x587) = 0.000

Factor for structural system, K = 1.0 use equation (2)

= 1.0(11+3.81) = 14.8 Modification factor for flange width, b/bw > 3

= 0.80

Modification for span less than 7m

= 1.0

Modification factor for steel area provided

=As,prov/As,req =1473/1110 =1.33 1.5

Therefore allowable span-effective depth ratio

=(l/d)allowable =14.81 x 0.80 x 1 x 1.33 =15.7

Actual span-effective depth

=(l/d)actual =6000/451.5 =13.3 (l/d)allowable OK

7.0 CRACKING Limiting crack width, w max

= 0.3mm

Steel stress,

= 285N/mm2 Max. allowable bar spacing

=100mm

Bar spacing, S

= [250-2(30)-2(6)-20]/2 = 76.5mm 100mm (OK)

S

S

8.0 DETAILING A

X

B

6000 2H12 - 3 H8-250 3H25 - 1

2H12 - 3

X

X-X