ex 7.2 - simply supported (flange beam)

8/19/2019 Ex 7.2 - Simply Supported (Flange Beam)

1/9

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 is125mm and the dimensions of the beams are as given in the diagram. The finishes, ceilingand services form a characteristic permanent action of 1.5kN/m 2. The characteristic variableaction is 3.0kN/m 2. Three meter high brickwall weighing 2.6kN/m 2 is placed over the entirespan of all beams. The construction materials consist of Grade C25 concrete and Grade 500steel 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.

2500

2 0 0 0

5 5 0 0

1 2 3

A

B

C

3500

25003500

2 5 0

x 6 0 0

2 5 0

x 6 0 0

2 5 0

x 5 0 0

250 x 500

250 x 500

250 x 500


2/9

1.0 SPECIFICATION

Slab thicknessSlab Characteristic actions:----Finishes etc.----Variable, q k ----Unit weight brick-wallMaterials:----Characteristic strength of concrete, f ck ----Characteristic strength of steel, f yk----Characteristic strength of link, f yk ----Unit weight of reinforced concrete----Nominal coverAssumed:----Øbar1 ----Øbar2 ----Ølink

= 100mm

= 1.5kN/m= 3.0kN/m= 2.6kN/m 2

= 25N/mm 2 = 500N/mm 2 = 500N/mm 2 = 25kN/m 3 = 30mm

= 25mm= 12mm= 6mm

2.0 BEAM SIZE

bw = 250mmh f = 100mmh = 500mmb1 = 1625mmb2 = 1125mmlo= 6000mm

beff,i = 0.2b i + 0.1lo 0.2l o

0.2lo = 1200mmbeff,1 = 915mm, b eff,2 = 825mm

b = b1 + b2 + bw

= 0.2b 1 + 0.1lo = 915mm

= 2950mm

b eff

h

bw

h f


3/9

Effective flange width,beff = = 915 + 825 + 250= 1990mm < b3.0 LOADING AND ANALYSIS

Loads on slab:Self-weight

Finishes etc.Characteristic permanent action, g k Characteristic variable action, q k

Loads on beam 2/B-C

Distribution of load from slabs to beam are asfollows:

Panel B-C/2-3: L y/Lx

Panel B-C/1-2: L y/Lx

= 25 x 0.1= 2.5kN/m 2 = 1.5kN/m 2 = 4.0kN/m 2 = 3.0kN/m 2

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

= 6/3.5

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

w1 w2

3500 2500

2 0 0 0

6 0 0 0


4/9

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

Gk

Characteristic variable loadw1 = nLx= 0.57 x 3.00 x 3.50w2 = nLx = 0.50 x 3.00 x 2.50

Q k

Total design load on beam 2/B-Cwd=1.35 Gk + 1.5Q k

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

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

= 46.03kN/m

Shear Force,V = wdL/2V = 138.1kN

Bending Moment,M = wdL2/8M = 207.1kNm

6m

46.03kN/m


5/9

4.0 MAIN REINFORCEMENT

Effective depth:d = h – Cnom – Ølink – Øbar /2-MEd = 207.1kNmMf = 0.567f ckbh f (d-0.5h f )

MEd < Mf Neutral axis within the flange

K = M/bd 2f ck

z = d[0.5+ z = 0.98dArea of tension reinforcementAs = MEd/0.87f ykz

Use: 3H 25(1473mm 2)

Min. and max. reinforcement area,

As,min = 0.26(f ctm /f yk)bd

As,max = 0.04Ac

= 451.5mm

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

No compression reinforcement

= 207.1x10 6/(1990 x 451.1 2 x 25)= 0.20

= 207.1x10 6/(0.87 x 25 x 0.98 x 451.5)= 1110mm 2

= 0.26(2.56/500)bd= 0.0013bd (use 0.0013bd)= 0.0013 x 250 x 451.5= 151mm 2

= 0.04 x 250 x 500= 5000mm 2


6/9

5.0 SHEAR REINFORCEMENT

VEd = 138.1kN

Concrete strut capacityVRd,max = 0.36b wdf ck(1-f ck/250)/(cotθ+tanθ)

VEd < VRd,max cotθ=2.5 VEd < VRd,max cotθ=1.0

Use θ=22, cotθ=2.5

Shear linksAsw/s=VEd/(0.78f ykdcot )

Try link H6, Asw=57mm 2 Spacing, s=57/0.317=178 mm

Use H6-150

Max. spacing, S=0.75d

Minimum links:-Asw/s = (0.08f ck1/2 bw)/f yk

Try link H6, Asw = 57mm 2 Spacing, s = 57/0.200 = 283mm < S

Use H6-275

Shear resistance of minimum linksVmin = (Asw/s)(0.78df ykcot )

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

= 138.1 x 10 3/(0.78x500x452x2.5)= 0.317

= 0.7 x 452= 339mm

= (0.08x25 1/2 x250)/500= 0.200

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

= 90kN

138kN90kN

138kN90kN

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

H6-150 H6-275 H6-150

1.05m 3.89m 1.05m


7/9

Transverse steel in the flangeThe longitudinal shear stresses are the greatest over adistance x measured from the point of zero moment,

x = 0.5(L/2)

The change in moment over distance x from zeromoment

M = (wL/2)(L/4) – (wL/4)(L/8) = 3wL2/32

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

Longitudinal shear stressved = Fd/(h f x)

ved > 0.27f ctk = 0.27 x 1.80 = 0.48N/mm 2

Transverse shear reinforcement is required

Concrete strut capacity in the flangeved,max = 0.4f ck(1-f ck/250)/( cotθ+tanθ )

ved < vRd,max cotθ=2.5 ved < vRd,max cotθ=1.0

Use θ =26.5 , cotθ =2.0

Transverse shear reinforcementAsf /s f = ved h f /0.87f ykcotθ

Try H10, Asf = 79mm 2 Spacing, s f = 79/0.13 = 608mm

Minimum transverse steel area,As,min = 0.26(f ctm /f yk)bh f

Use H10 – 500

(157mm2

/m)

= 6000/4= 1500mm

= (3 x 46 x 6.0 2)/32= 155.35kNm

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

= 169x10 3/(100 x 1500)= 1.13N/mm 2

= 3.59N/mm 2 (θ=26.5, cotθ=2.0) = 4.50N/mm 2 (θ=45, cotθ=1.0)

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

= 0.13

= 0.26(2.56/500)bh f = 0.0013 bh f (use 0.0013 bh f )= 0.0013 x 1000 x 100= 133mm 2/m


8/9

Additional longitudinal reinforcementAdditional tensile force,

Ftd = 0.5Vedcot θ

Med,max /z

As,req = Ftd /0.87f yk

Use 1H25(491mm 2)

= 0.5 x 138 x 2.48= 171kN

= 207.1x10 6/428.9= 483kN

= 171x10 3/(0.87 x 500)= 393mm 2

6.0 DEFLECTION

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

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

Percentage of required compression reinforcement

=As’,req /bd

Factor for structural system,K = 1.0

use equation (2)

Modification factor for flange width, b/b w > 3

Modification for span less than 7m

Modification factor for steel area provided

Therefore allowable span-effective depth ratio

= 1110/(250X452)= 0.010

= (25)1/2 x10-3 = 0.005

= 0/(250x587)= 0.000

= 1.0(11+3.81)= 14.8

= 0.80

= 1.0

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

=(l/d) allowable

=14.81 x 0.80 x 1 x 1.33=15.7


9/9

Actual span-effective depth =(l/d) actual

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

7.0 CRACKING

Limiting crack width, w max

Steel stress,

Max. allowable bar spacing

Bar spacing, S

= 0.3mm

= 285N/mm 2

=100mm

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

8.0 DETAILING

X - X

S S

6000

H8-250

3H25 - 1

2H12 - 3

A B

2H12 - 3

X

X

ex 7.2 - simply supported (flange beam)

Documents