EXAMPLE 9.2 – Part IV PCI Bridge Design Manual BULB “T” (BT-72) THREE SPANS, COMPOSITE DECK...

EXAMPLE 9.2 – Part IVPCI Bridge Design Manual

EXAMPLE 9.2 – Part IVPCI 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.

UNFACTORED SHEARS AND MOMENTS LIVE LOAD ENVELOPE INCLUDES IM FACTOR

UNFACTORED SHEARS AND MOMENTS LIVE LOAD ENVELOPE INCLUDES IM FACTOR

Midspan Values, symmetrical

STRENGTH LIMIT STATEPositive Moment Zones

STRENGTH LIMIT STATEPositive Moment Zones

Since the dead loads and live loads produce stresses of the same sign, use maximum load factors.

Find Mu at midspan.

Mu = 1.25 DC + 1.5 DW + 1.75 (LL+IM)

= 1.25(1391 + 2127 + 73) + 1.5 (128) + 1.75 (2115)

= 8382 k-ft.

This is the applied FACTORED load at midspan of a center span interior beam.

Consider the section at strength limit state. Assume the stress block is entirely within the deck

slab (a< 7.5”). If this is true, treat as a rectangular section.

STRENGTH LIMIT STATE

Positive Moment Zones

Because the stress block is assumed to be in the flange (slab), the properties of the SLAB

concrete are used. Therefore, there is no need to “transform” the slab concrete to beam

concrete. The actual effective width of 144” is used, not the transformed width.



If the stress block had fallen into the precast beam:

The section would be treated as a “T” beam.

The section would be assumed to be made of “beam” concrete, so the beam concrete properties

would be used. For this reason, the transformed slab and haunch widths would also be used.



Equilibrium:

Compression = Tension

0.85 fc’ b a = Aps fps



ps pup

c ps ps

c ps pup

ps pu

puc ps

p

cf f k

d

f ba A f

a c

cf b c A f k

d

A fc

ff b kA

d

1

1

1

1

0.85 '

0.85 ' 1

0.85 '

The value of fps can be found from:

Then:

(Eq’n 5.7.3.1.1-1)



ps pu

puc ps

p

A fc

ff b kA

d

10.85 '

c = depth of neutral axis

b = width of compression block (flange in this case)

Aps = area of TENSILE prestressing steel

dp = depth to centroid of tensile prestressing steel

k = a constant for the prestressing steel

k = 0.28 for low relaxation steel

Reminder: When a < ts , the stress block is in the slab. Use the effective width of the slab NOT the

transformed width and use 1 of the slab concrete.



If there is mild (nonprestressed) tensile steel, As, and mild compression steel, As’, and

both yield (strength of fy ), the equation for c becomes:

ps pu s y s y

puc w ps

p

A f A f A fc

ff b kA

d

1

' '

0.85 '

c s y s y ps pup

cf b c A f A f A f k

d

1.85 ' ' ' 1

Rectangular section assumed. (Eq’n 5.7.3.1.1-4)



Reminder of previously calculated values:

Aps = 44 strand(0.153 in2) = 6.72 in2

fpu = 270 ksi

dp = beam depth+haunch+slab-ybs

= 72” + 0.5” + 7.5” – 5.82” = 74.18”

ybs = distance from bottom of beam to centroid of prestressing steel.

b = 144”

fc’ = 4.0 ksi

1 = 0.85 (for 4 ksi concrete)

k = 0.28 for low relaxation strand



1

2

2

1

0.85 '

6.732 270

2700.85(4 )(0.85)(144") 0.28(6.732 )

74.18"4.30"

0.85(4.30") 3.65" 7.5

ps pu

puc ps

p

A fc

ff b kA

d

in ksi

ksiksi in

c

a c in

OK – Stress Block in Flange – Rect. Section



Find the approximate stress in the prestressing steel:

1

4.30(270 ) 1 0.28

74.18

265.6

ps pup

ps

ps

cf f k

d

f ksi

f ksi

This equation gives the approximate stress in the prestressing steel. A more accurate value can be

found from strain compatibility (see the PCI Design Handbook)



From Equilibrium:

2

* ( / 2)

* ( / 2)

2

3.65"(6.732 )(265.6 ) 74.18"

2

129372 10780

n ps ps p

n

n

Moment Compression d a

Tension d a

aM A f d

M in ksi

M k in k ft



22)('85.0

2'''

22

1f

fwcsys

sysppspsn

hahbbf

adfA

adfA

adfAM

The complete moment equation, assuming prestressed and nonprestressed tensile steel, compression steel and a flanged

section (T beam) is given by Eq’n 5.7.3.2.2-1 :

ds is the depth from the compression fiber to the tensile mild steel and ds’ is the depth from the compression fiber to

the compression mild steel, bw is web width and hf is flange thickness.

In this design example: As = As’= 0 and b = bw , so the equation simplifies to Mn = Apsfps(dp – a/2)



1.0

8382

1.0(10780 )

u r n

u

n

M M M

for precast

M k ft

k ft M

OK for Strength Limit State in (+) Moment Zones IF the section is tension

controlled.



STRENGTH LIMIT STATENegative Moment Zones

STRENGTH LIMIT STATENegative Moment Zones

Since the dead loads and live loads produce stresses of the same sign, use maximum load factors.

Also, only the loads carried as by the continuous span cause negative moment. Thus, the only DC

is the barrier:

Mu = 1.25 DC + 1.5 DW + 1.75 (LL+IM)

= 1.25(-197) + 1.5 (-345) + 1.75 (-2328)

= -4837 k-ft.

This is the applied FACTORED load over the pier of a center span interior beam.

In the negative moment area, the bottom of the bottom flange is the compression area (use BT72

concrete properties). The tensile steel is placed in the deck.


Negative Moment Zones

Consider the section at strength limit state. Assume the stress block is entirely within the bottom

flange (a< 6”). If this is true, treat as a rectangular section.



The general equation for equilibrium is:

ps pu s y s y

puc ps

p

A f A f A fc

ff b kA

d

1

' '

0.85 '

c s y s y ps pup

cf b c A f A f A f k

d

1.85 ' ' ' 1



a = As fy / 0.85 fc’ b

This is the general equation with Ap = As’ = 0 and

a= 1c.



n ps ps p s y s

fs y s c w f

a aM A f d A f d

ha aA f d f b b h

1

2 2

' ' ' 0.85 '( )2 2 2

Once again, the complete moment equation, assuming prestressed and nonprestressed tensile

steel, compression steel and a flanged section (T beam) is given by Eq’n 5.7.3.2.2-1:

In the negative moment case, Ap = As‘= 0 and b = bw , so the equation simplifies to:

Mn = Asfs(ds – a/2)



Mn = As fy (ds – a/2)

In this equation, As and a are unknown.



Although As and “a” are unknown, it is possible to estimate “a”. Since a is usually an

order of magnitude smaller than d, even a gross error in “a” leads to a small error in the

moment arm,

ds -a/2, and a reasonably accurate estimate of As.



,min

4837

48375374

0.9

u

un imum

M k ft

M k ftM k ft

= 0.9 for reinforced concrete in flexure IF tension controlled.

The bottom flange is 6” high. We want to keep the assumption of a rectangular section and it

is better to overestimate “a” as this will underestimate the moment arm.

Assume a = 6”



n s y s

s

s

aM A f d

k ft A ksi

A in

2

2

6"5374 (12) (60 ) 76.25"

2

14.67

If the steel centroid is in the center of the slab:

ds = 72” + 0.5” + 7.5”/2 = 76.25”



The bars should be placed within the lesser of:

The effective flange width = 144”

1/10 of the average length of adjacent spans=

[(119+120)/2](12)/10 = 143”

Note that if the 1/10 span controls, additional, nonstructural bar is required in the areas outside of 1/10 span width

but within the effective width. The amount of bar required is 0.4% of the area outside of the width determined by

1/10 span.

In this case, the two are practically equal. It would be impractical to supply additional steel in the extra 1/2 inch on

side with the required amount of steel, which would be 0.03 in2

.



The bars are be placed within the effective width of 144”. The bridge design handbook suggests:

#5 @ 12” Top mat = 9 bars X 0.31 in2

= 2.79 in2

#4 @ 12” Bottom mat = 9 bars X 0.22 in2

= 1.99 in2

#7 Split between top and bottom mat

= 18 bars x 0.6 in2

= 10.8 in2

Total = 15.6 in2

#7 placed between each #4 on bottom mat and between each #5 on top mat. Spacing is 8 in c/c.

Now ds = 75.6 in.



As =15.6 in2

:

2

0.85 '

15.6 60

0.85 7 26"

6.04 6

s y

c

A fa

f b

in ksi

ksi

in in

This is close enough to 6 inch that the section can be assumed a rectangle.



2

2

6"0.9(15.6 )(60 ) 75.6"

2

61158 5096

4837

n s y

n

n

n u

aM A f d

M in ksi

M in k ft k

M k ft M

OK, IF the section is tension controlled.

Check Maximum Moment:



Tension controlled, compression controlled and transition sections are defined by the strain

in the extreme tension steel at Mn. Strain in the extreme tensile steel is defined in the

diagram.

This definition applies to prestressed and non-prestressed steel.


TENSION CONTROLLED SECTIONTENSION CONTROLLED SECTION

• A section is TENSION CONTROLLED if the extreme steel strain > 0.005.

• This applies to prestressed and non-prestressed steel.

• If tension controlled– = 1 for prestressed– = 0.9 for non-prestressed– is interpolated for partially prestressed.

COMPRESSION CONTROLLED SECTION

COMPRESSION CONTROLLED SECTION

• A section is COMPRESSION CONTROLLED if the extreme steel strain < balanced.

• For non-prestressed steel, the limit is fy/Es

• For prestressed steel, the limit is 0.002.• If compression controlled, = 0.75 for

members with ties or spirals.

TRANSITION SECTIONTRANSITION SECTION

• A transition section is has an extreme tensile steel strain between tension and compression controlled.

• For transition sections, is interpolated based in extreme tensile steel strain.

0.6

0.65

0.7

0.75

0.8

0.85

0.9

0.95

1

1.05

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007

Extreme Steel Strain

Ph

i F

act

or

CompressionControlled Transition

Tension Controlled

Prestressed:Strain = 0.004Phi = 0.92

Prestressed

Reinforced

Definition of for prestressed steel and GR 60 non-prestressed steel.

SECTION FACTORS

Check for Tension Control:

The tension control limit can be found as:

So a section is tension controlled if:

0.003 0.005 0.003

0.375

0.375

t

t

t

d c

c

d

c

d

SECTION FACTORS

In the prestressed girder (positive moment zone), it was found that:

c = 4.30 inches dt = 72+7.5-2 = 77.5 in

c/ dt = 4.30/77.5 = 0.055 < 0.375

Tension controlled

= 1.0

SECTION FACTORS

In the reinforced (negative moment) section

c = a/1 = 6.04/.7 = 8.60”

from the calculation of Mn

Assume 2.5 inches clear cover to the #5 top mat steel:

dt = 72 + 7.5 – 2.5 – (5/8)/2=76.7 in

c/dt = 8.60 / 76.7 = 0.112 < 0.375

Tension Controlled

SECTION FACTORS

Article 5.7.3.3.2 requires:

Mn =Mr > lesser of 1.2 Mcr or 1.33Mu

Mcr = Cracking Moment

1.33Mu = 1.33 (8381 k-ft) = 11150 k-ft

MINIMUM STEEL POSITIVE MOMENT SECTION


bccr r pb bc d / nc

b

r c

pe pe cpb

b b

b

bc

d / nc

SM f f S M

S

f . f ' Modulus of rupture

P P ef Stress at precast tensile fiber

A S

S Section Modulus to tensile fiber noncomp.

S Section Modulus to tensile fiber comp.

M noncomp. DL mome

1

0 37

nt

Equation 5.7.3.3.2-1


bccr r pb bc d nc

b

r

b

pb

d nc g s

bc

cr

SM f f S M

S

f ksi ksi

S in

f ksi

M M M k ft k in

S in

M

/

3

/

3

1

0.37 7 0.980

14915

1072 1072(30.78)3.610

767 149151391 2127 3518 42216

20545

205450.980 3.610 20545 42216 1

14915

783

cr

k in k ft

M k ft

66 6530

1.2 7837


Since 1.2Mcr = 7837 k-ft < 1.33Mu = 11147 k-ft

1.2Mcr controls.

Mn = 10649 k-ft > 1.2Mcr = 7837 k-ft OK


MINIMUM STEEL NEGATIVE MOMENT SECTION

MINIMUM STEEL NEGATIVE MOMENT SECTION

bccr r pb bc d nc

b

r

d nc

bc

pb

cr

cr u

n r cr

SM f f S M

S

f ksi ksi

M

S in

f

M ksi in k in k in

M k ft M k ft k ft

M M k ft M OK

/

/

3

3

1

0.37 4 0.740

0

57307

0

0.740 57307 42407 3534

1.2 4241 1.33 1.33 4837 6433

5115 1.2

DEVELOPMENT LENGTH OF PRESTRESSING STEEL

DEVELOPMENT LENGTH OF PRESTRESSING STEEL

d ps pe b

d

d

f f d

ksi ksi in

in ft

2

3

21.6 265.6 159.2 0.5

3

127 10.6 .

The term = 1.6. (5.11.4.2)

The prestressing steel must be embedded 10.6 ft from the point of maximum stress. The point of

maximum stress is at midspan and the embedment is 59.5 ft. OK

Eq’n 5.11.4.2-1

CONTROL OF CRACKING BY DISTRIBUTED REINFORCING IN THE SLAB


7002e

cs s

s df

According to Article 5.7.3.4 the spacing of the mild steel reinforcement in the layer closest

to the tension face shall satisfy equation 5.7.3.4-1.

e = Exposure factor = 1.00 for Class 1 exposure 0.75 for Class 2 exposure condition

fs = Tensile stress in steel reinforcement at the service limit state, ksi

dc= Thickness of concrete cover measured from extreme tension fiber to center of the flexural reinforcement located closest therto

βs=

h= Overall height of the section, in

10.7( )

c

c

d

h d


To find fs, the cracked moment of inertia is needed:

2 2

290005.7

5072

5.7 15.6 89.2

s

c

s

E ksin

E ksi

nA in in


2

2

22 3

120 6 2 10 4.5 6 89.2

2

1 4.520 6 3 2 10 4.5 6

2 3

6 89.2 75.62

3 254.2 7441 0

23

i ii

Ax A x

in in in in in x in x

inin in in in in in

xin x in in

in x in x in

x in

Use bottom as reference point:


3 2

3 2

3 22

4

120 6 20 6 23 3

121 1

2 10 4.5 10 4.5 23 7.536 2

16 23 89.2 75.6 23

3

330350

cr

cr

I in in in in in in

in in in in in in

in in in in in

I in


Mserv = -197k-ft - 345 k-ft - 2328 k-ft

= -2870 k-ft = - 34400 k-in

If 2.5 in cover and Class 1 ( = 1)

dc = 2.94 in = 2.5 + (7/16)

ss

k in in inf ksi

in4

34400 75.6 235.7 31.3

330350

2.941 1.054

0.7 80 2.94

700(1.0)2 2.94 15.3

1.054 31.3

s

in

in in

s in inksi


So maximum spacing =

15. 3 inches for Class 1 ( = 1)

10.0 inches for Class 2 ( = 0.75)

Actual spacing is 8 inches c/c.

OK


EXAMPLE 9.2 – Part IV PCI Bridge Design Manual BULB “T” (BT-72) THREE SPANS, COMPOSITE DECK...

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