example 9.2– part vi pci bridge design manual bulb “t” (bt-72) three spans, composite deck...

EXAMPLE 9.2– Part VIPCI Bridge Design ManualEXAMPLE 9.2– Part VI

PCI Bridge Design Manual

BULB “T” (BT-72)

THREE SPANS, COMPOSITE DECK

LRFD SPECIFICATIONS

Materials copyrighted by Precast/Prestressed Concrete Institute, 2011. All rights reserved. Unauthorized duplication of the material or presentation prohibited.

ALTERNATE SHEAR DESIGNALTERNATE SHEAR DESIGN

• In Version 4 of the LRFD Specifications, AASHTO restored the use of Vci and Vcw from the old Standard Specifications.

• This was the result of NCHRP Report 549.– Concluded that the sectional method (modified compression

field) was more accurate.– However, even the most accurate method has a COV of 25%– While conservative, this method is not unreasonable for

design.– Modified to cover both prestressed and non prestressed

members.

Flexural shear – Vci and Web Shear - Vcw

The flexural shear capacity, Vci, is the shear force required to cause a flexural crack to “bend over” and become a shear crack. (This is conservative because the cracks bend over long before the strut will crush!).

The web shear capacity, Vcw is the shear force that causes a diagonal crack in the web, usually near the support.


Flexural shear - Vci

It is much easier to understand Vci if you look at the derivation.

To find Vci, it is first necessary to determine the shear in the section when the beam cracks. A prestressed beam will form a flexural crack when the moment at a section reaches Mcr. The shear at the section which exists at the time of cracking is called Vcr. Note that the beam cracks due to moment, not shear. Vcr is the shear in the section associated with Mcr.



Experiments have shown that if the shear at the section increases by 0.02(fc’)bvdv, the flexural crack will grow to a shear crack.

The flexural shear strength can be written as:

ci cr c v vV V 0.02 f ' b d

Note that this is the same increment as in the old Standard Specification, just changed to KSI units!



The term Vcr is the shear in the section when the section cracks at Mcr. It is assumed that the shear and the moment increase proportionally:

cr u

cr u

ucr cr

u

uci cr c v v

u

V V

M M

or

V V (M )

M

The equation becomes:

VV (M ) 0.02 f ' b d

M



It was assumed that the shear and moment increase proportionally. However, the dead load doesn’t increase proportionally, so subtract it out of the proportionality portion of the equation.

The equation becomes:

ici cre c v v d

max

d

max

VV (M ) 0.02 f ' b d V

M

where:

V is the shear due to unfactored dead load.

M is the maximum moment at the section due to all imposed factored loads

(all but non-composit

cre

i max

e dead load).

M is the cracking moment adjusted for dead load

V is the factored shear at the section corresponding M .



Mcre must be adjusted to reflect the fact that the dead load effect has been accounted for:

Where Sc = composite section modulusfr = modulus of rupture = 0.2√fc’ (5.4.2.6)fcpe = compression stress at the flexural tension fiber due to effective prestressing forceMdnc = moment applied to the non-composite sectionSnc = non-composite section modulus.

dnccre c r cpe

nc

MM S f f

S

5.8.3.4.3-2



There is a limit placed on Vci :

For a composite section, the Commentary of 5.8.3.4.3 permits:

Vi = Vu – Vd and Mmax = Mu – Md

Where Vu and Mu are the FACTORED total shear and moment while Vd and Md are the UNFACTORED dead load shear and moment.

ici cre c v v d

max

ci c w

VV (M ) 0.02 f ' b d V

M

V 0.6 f ' b d

5.8.3.4.3-1


Web shear - Vcw

In a beam, there are shear stresses from flexure. The maximum shear stress occurs at the neutral axis. For most beams, there is no normal stress at the neutral axis. However, in a prestressed beam there is a normal stress from the P/A term in the stress equation. In a composite beam, the neutral axis of the composite beam is not the same as in the non-composite beam. At the neutral axis of the composite section, there will also be normal stresses from bending, caused by the prestressing and the dead load applied to the non-composite section.


Web shear - Vcw

The normal stress fpc is:

ng.prestressi the ofty eccentrici the is e

loads. dead composite-non for moment the is M

beams. composite-non for 0, Use sections. composite-non and

composite the of axis neutral the between distance the is y

inertia. of moment composite-non the is I

area. composite-non the is A

force. ngprestressi effective the is P

:where

I

yM

I

eyP

A

Pf

dnc

c

nc

nc

eff

nc

cdnc

nc

ceff

nc

effpc


Web shear - Vcw

Vcw can be calculated using the shearing stress equation

v = (VcwQ)/(It),

where v is the shear stress which causes a maximum principal tensile stress of 4(fc’)½ psi when the normal stress is fpc:


Web shear - Vcw

In place of the calculation of principal stress, the following approximate equation may be used:

cw c pc v v pV (0.06 f ' 0.3f )b d V

Vp is the vertical component of prestressing force from harped strands.

fpc is the axial stress in centroid of the cross section due to effective prestressing force

5.8.3.4.3-3


0

cot cot sin

cot 1 45

cot 1.0 3 1.8'

v y vs

ci cw

pcci cw

c

A f dV

s

IF V V

fIF V V

f

For shear stirrups, the theta angle is needed:

5.8.3.4.3-4

Remember, is the angle between the longitudinal axis of the beam and the stirrups. For normal, upright stirrups, = 90o.

ALTERNATE SHEAR DESIGN

Check Vci and Vcw at 0.2L

Using the values of shear and moment calculated in the table in the Bridge Manual:

Mu = 1.25(873+1335-24)+1.5(-42)+1.75(1044) = 4494 k-ft

Vu = 1.25(28.8+44.0+5.0)+1.5(10.0)+1.75(113.8) = 311.4 k


Next, the UNFACTORED non-composite shear and moment are needed

Mdnc = (873+1335-24) = 2184 k-ft

Vd = (28.8+44.0+5.0) = 77.8 kips


Article 5.8.3.4.3 states that Mmax is the FACTORED moment from all superimposed loads.

Vi is the FACTORED shear at the section associated with Mmax. It is NOT necessarily the maximum shear due to superimposed loads at the section!


As previously stated, the Commentary of 5.8.3.4.3 states that, for composite sections it is permissible to use:

Vi = Vu – Vd

Mmax = Mu – Md

i

max

V 311.4k 77.8k 233.6k

M 4494k ft 2184k ft 2310k ft


Now find the cracking moment.

From previous calculations:

P = 1071 k (after all losses)e = 25.8 inches (at 0.2L = 24 ft)

cpeb

cpe 2 3

P Pef

A S

1071k 25.8in1071kf 3.25ksi

767in 14915in


33

0.2 0.2 7 0.529

2184 (12)20545 0.529 3.25

14595

40750 3396

dnccre c r cpe

nc

r c

cre

cre

MM S f f

S

f f ksi ksi

k ftM in ksi ksi

in

M k in k ft

Note that Sc = Sbc and Snc = Sb.Mdnc must be in K – IN!


ici cre c v v d

max

ci

ci c v v

ci

VV (M ) 0.02 f ' b d V

M

233.6kV 3396k ft 0.02 7ksi 6in 71.82in 77.8k

2310k ft

V 443k 0.06 f ' b d 68.4k

V 443k

Note that Vi and Vd are in KIPS.Mmax and Mcre must have the same units.


Web shear - Vcw

The normal stress fpc is:

2 4 4

1071 25.8 57.2 36.6 2184 12 57.2 36.61071

767 545894 5458941.34

eff eff c dnc cpc

nc nc nc

pc

pc

P P ey M yf

A I I

k in in k ft inkf

in in inf ksi


cw c pc v v p

cw

cw

V (0.06 f ' 0.3f )b d V

V 0.06 7ksi 0.3 1.34ksi 6in 71.82in 35.2k

V 276.8k

Web shear - Vcw

Vcw controls!


2

cot 1.0 3 1.8'

1.34cot 1.0 3 2.5 1.8

7

cot 1.8

cot 0.6 60 71.82 1.8193.4

24

pcci cw

c

v y vs

fV V

f

ksi

ksi

A f d in ksi inV k

s in

Assuming #5 @ 24 inches and =90o:


0.9 276.8 193.9 423 311.4

u c s

u

V V V

k k k k V

Finally:

OK!


example 9.2– part vi pci bridge design manual bulb “t” (bt-72) three spans, composite deck...

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