Week 9 PCIBDM 9.1 Shear

7/29/2019 Week 9 PCIBDM 9.1 Shear

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Example 9.1

PCI Bridge Design Manual

Noncomposite Adjacent Box Girder

Simple SpanStandard Specifications


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Connection of Beams


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Shear

Key


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9.1.2 Material Properties (1)

Concrete

f'ci = 4,000 psi = 4.0 ksi

f'c = 5,000 psi = 5.0 ksi

wc = 150 pcf = 0.150 kcf

The AASHTO Standard Specifications require a

pretensioned member to be designed at 5000 psi (6000 psi

at the discretion of the engineer).


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9.1.2 Material Properties (2)

Strand

0.5-in., seven wire low relaxation;

fpu = 270 ksi; fpi = 0.75fpu = 202.5 ksi

Es = 28,500 ksiNonprestressed Steel

fsy = 60 ksi

Es = 29,000 ksi


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9.1.3 Cross-Section Properties (1)

A = 813 in.2 I = 168,367 in.4

yb = 19.29 in. Sb = 8,728 in.3

yt = 19.71 in. St = 8,542 in.3


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9.1.3 Cross-Section Properties (2)

At release

Eci = 33,000 (0.150)1.5 (4.0)1/2

= 3,834 ksi

At 28 days

Ec = 33,000 (0.150)1.5 (5.0)1/2

= 4,287 ksi

Modulus of elasticity is needed for various calculations:


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9.1.4.1.1 Dead Loads per Girder

Beam self weight = 0.847 kip/ft

Diaphragm weight = 0.73 kip/diaphragm at pts)

Barrier weight = 0.086 kip/ft

Wearing surface = 0.134 kip/ft


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Shear and Moment at h/2

Compute Vci: Vci=0.6(fc)1/2bd+Vd+(ViMcr)/Mmax

Vd = 50.0 kips = shear due to dead loads

VLL+I = 32.7 kips = shear due to live loads

Md = 82.8 ft-kips = moment due to dead loads

MLL+I = 53.2 ft-kips = moment due to live load

Vu = 136.0 kips = factored shear force


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Strand Pattern-Midspan (Fig 9.1.5.4-1)

Whats with the 2 at the top?? Discussed later!

Pattern is symmetrical must be or beam bends sideways!


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Strand Pattern-Beam End (Fig. 9.1.8.2-1)Debond 7 strands in bottom row for 5 ft.

PATTERN STAYS SYMMETRICAL


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DEBONDEDSTRAND

Strand is

covered

with teflonto prevent

bonding.


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9.1.11 Shear Design

Members should be designed such thatVu < f(Vc + Vs)

Critical section is h/2 from face of support

Shear forces at this section are given in

Table 9.1.4.2.4-1

In reinforced concrete, ACI and AASHTO use d

from face of support, but in prestressed concrete,d often varies due to debonding and/or harping.

Thus, h/2 is used.


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9.1.11 Shear Design

Compute Vci: Vci=0.6(fc)1/2bd+Vd+(ViMcr)/Mmax


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9.1.11 Shear Design

Mu = 223.1 ft-kips factored moment

Vmu = 136.0 kips factored shear forceoccurring with the factored moment

Mmax = 140.3 ft-kips = Mu - Md

Vi = 86.0 kips = Vmu - Vd

fpe = Pse/A + (Pseec)/Sbe = eccentricity at h/2 = 13.96 in.

Pse = 24(0.153)(171.6) = 630.1 kips

fpe = 1.783 ksi


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fd = Md/Sb = (82.8(12))/8,728 = 0.114 ksi

Mcr= (6(fc)1/2+fpe-fd)Sb =

(((6(5000)1/2)/1000)+1.783-0.114) (8,728/12) =

1,522.5 ft-kips

9.1.11 Shear Design


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9.1.11 Shear Design (4)

d = 39.0-5.33 = 33.67 in. > 0.8(h) = 31.20 in.,

therefore d = 33.67 in.

Vci = 0.6(5000)1/2(10)(33.67)+50.0+((86.0(1,522.5))

1000 140.3= 997.5 kips

Is Vci

> minimum Vci

?

minimum Vci= 1.7(fc)1/2bd

= (1.7(5000)1/2)/1000)(10)(33.67) = 40.5 kips

< Vci = 997.5 kips OK


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9.1.11 Shear Design

Compute Vcw

Vcw= (3.5(fc)1/2+0.3fpc)bd+Vpfpc = Pse/A = 630.1/813 = 0.775 ksi

Vp = 0 for straight strands

Vcw = (((3.5(5000)1/2

)/1000)+0.3(0.775))(10)(33.67)+0 = 161.6 kips


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Vc = 161.6 kips (lesser of Vci and Vcw)

Vu = 136.0 < fVc = 0.9(161.6) = 145.4 kipsSince Vc>Vu the concrete resists the total

shear and shear steel is not required to assist in

resisting Vu. Therefore provide min. reinforcement.


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Graph of Vu and concrete shear strengths

0

100

200

300

0 10 20 30 40Length (ft.)

Sheark

phi Vcw

phi Vci

Vu


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Minimum shear reinforcement

Av-min= (50bs)/fy = minimum area of webreinforcement

Set s = 12 in.

Av-min = 50(10)(12)60,000

= 0.10 in.2 (per foot)

Maximum spacing = 0.75h = 0.75(39) = 29.25 in.

use 24.0 in.Use #4 at 24 in. spacing in each web

(Av = 0.20 in.2/ft)

Week 9 PCIBDM 9.1 Shear

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