session 5-bending workshop

7/29/2019 Session 5-Bending Workshop

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Thomas Telford Guide to EC2-2 C R Hendy/D A Smith

12/09/13

Worked Example (not in notes): Reinforced concrete beam

Find the ultimate sagging moment resistance of the beam in Figure 1 using the

simplified rectangular stress block with 7 No. 20 bars (As = 2199 mm2)

Reinforcement is B500B and C35/45 concrete

cc = 0.85 (bridges value)

s = 1.15

c = 1.5 cu3 = 0.0035

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cu3

s

x

d

b

As

x

z

Fc

Fs

fcd

x

zfAzFMyds

s ==

=

bdf

Afdz

cd

syd

21

cd

yds

fb

fAx

=

+

1

1

3cuss

yk

E

fd

x

dx>

Forfck

= 35 MPa, = 0.8 and = 1.0. d= 1500 50 10 = 1440 mm

Assuming reinforcement yields:

=

=

bdf

AfdfAM

cd

syd

yds2

1

Check that steel yields:

==cd

yds

fbfAx

mm

=d

x

=

+1

1

3cuss

yk

E

f

so reinforcement yields and resistance moment =kNm.

kNm


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12/09/13

Figure 1

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150

0m

m

7 No. 20

diameter bars

1000 mm

50 mm


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12/09/13

Worked Example 6.1-5: Prestressed concrete M beam

Figure 6.1-10 - Prestressed concrete M beam

Consider an M3 prestressed concrete beam with 160 mm deep insitu deck slab as

illustrated in Figure 6.1-10. Calculate the ultimate moment of resistance (i) using a

prestressing steel stress-strain relationship with a horizontal top branch, assuming the

following properties:Class C30/37 slab concrete, fck = 30 MPa

Class C40/50 beam concrete, fck= 40 MPa

Parabolic concrete stress distribution therefore cu2 = 0.0035 and c2 = 0.0020

Prestressing strands (using properties from EN 10138-3, Table 4):

21 No. 15 mm strands of type Y1670S7

Nominal diameter = 15.2 mm

Nominal cross sectional area = 139 mm2

Characteristic tensile strength, fpk = 1670 MPa

Characteristic value of maximum force 232101670139 3 == kN

Characteristic value of 0.1% proof force = 204 kN

14681670879.01670232

2041.0,

===kpf MPa

Ep = 195 GPa

s = 1.15

Stressing and losses:

Assume initial stressing to 75% fpkAllow 10% losses at transfer and a further 20% long term losses.

Prestrain = long term strand stress / Eptherefore prestrain =

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970 mm

1000 mm

160 mm

200 mm

60 mm

200 mm

80 mm

50 mm

160 mm

400 mm

160 mm

0 mm

330 mm

160 mm110 mm

60 mm

20 mm

300 mm 30 mm

80 mm


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The values for reinforcement stress-strain curve first need to be defined as:

5.127615.1

1468==pdf MPa

=

s

pkf

MPa

p

pd

E

f

(i) Consider horizontal top branch and a neutral axis depth, obtained by trial and error,

of 335 mm:

The strain profile is shown in Figure 6.1-11.

(a) Idealisation (b) Strains (c) Compressive stresses

Figure 6.1-11: Stress-strain profile for prestressed M beam

Therefore total strains at ULS in the four layers of strands including prestrain are:

=

=

=

=+=

4

3

2

10102.00046.00056.0

s

s

s

s

All strains are greater thanp

pd

Ef (=0.00655) therefore all stresses can be taken as fpd =

1276.5 MPa.

Thus total steel force, =sF kN.

The neutral axis is in the top flange of the beam therefore split the compression zone

into the following three sections and take account of the different concrete strengths:

(1) Rectangular part of stress block in top slab (top 143.6 mm)

(2) Parabolic part of stress block in top slab (bottom 16.4 mm)

(3) Parabolic part of stress block in top flange to neutral axis (175mm deep)Using geometry of the parabolic-rectangular stress-strain distribution it can be shown

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0.0020

0.0035

0.0018

0.0028

0.00450.0051

0.0056

930 mm

x =

335 mm

60 mm 110 mm

160 mm

330 mm

160 mm 1000 mm

250 mm

400 mm

335 mm

160 mm

185 mm 950 mm

fcd,slab

or fcd,beam

a1

a

2 a3

Fc1d

Fc2d

Fc3d


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that:

(1) Fc1 = 2440.7 kN with depth to centroid, a1 = 71.8 mm



Thus =cF kN

sF therefore section balances and neutral axis is at the correct level. (Note that

using the rectangular stress block would be much easier here).

Taking moments about the neutral axis level gives a resistance moment of:

=M

kNm.

Strictly, since the neutral axis is in the web, a new limiting strain diagram should be

used in compliance with 2-1-1/6.1(6) as discussed in section 6.1.2.2 of this guide.

This would require a strain limit of3

21002

= .c to be maintained at

49153

02160

22.

.

./h cuc == mm from the underside of the deck slab i.e. 68.6 mm

from the top. The analysis was conservatively repeated using a strain limit of3

2 1002

= .c at the top of the deck slab which gave a resistance moment of 2585

kNm.

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