# Distribution Factor for a Typical Interior Beam

Document Sample

```					                      NU2000-200 ft Single Span Example

Given:
Use as many depression points as needed
Span Data                                       for design. Max. 12 strands depressed at any
Overall Girder Length =200 FT                 one point.
Design Span = 199 FT
Girder is simply supported                    Reinforcing Steel (Non-Prestressed)
f y = 60 KSI
E s = 29,000 KSI
Bridge Cross Section Data
Number of girder lines = 6                   Welded Wire Reinforcement
Girder Spacing = 9.00 FT                          Use standard reinforcement sizes and
Roadway Width= 51.00 FT                      spacing. Use 75 ksi steel for shear
Overall Width = 53.50 FT                     reinforcement
Deck Thickness
H = 70 % (Relative Humidity)
Note: the deck is integrally precast
with the girder
Actual     = 7.5 IN
Structural = 7.0 IN
Haunch thickness = 1 IN

Girder Type
NU 2000, with 150 mm (5.91 in.) web
Location: Interior

See table below

Girder Concrete
fc  (f’ci/0.8) KSI

fci = to be determined
w c = normal weight aggregate concrete

Prestressing Steel
Type: 0.7-IN Diameter 270 KSI Low-
Relaxation Seven-Wire Strand
Area = 0.294 in2
Pull: 75%

1
DF         0.689      moment
DF         0.884      shear                                        Table of Moments, K-FT

Location                           0   critical   0.1 L      0.2 L     0.3 L         0.4 L     0.5 L     0.6 L     0.7 L     0.8 L    0.9 L    critical    L
ft                                 0    6.94      19.9       39.8      59.7          79.6      99.5      119.4     139.3     159.2    179.1    192.06     199

Acting on Non-comp. Section
Girder, klf               0.941    0    627.1      1676.9   2608.5    3912.8        4471.7    4658.1    4471.7    3912.8    2981.2    1676.9    627.1     0.0
Deck , klf                0.816    0    543.6     1453.5    2584.0    3391.5        3875.9    4037.4    3875.9    3391.5    2584.0    1453.5    543.6     0.0
haunch, klf               0.049    0     32.3      86.5      153.8     201.8         230.6     240.3     230.6     201.8     153.8     86.5      32.3     0.0

Subtotal:                          0   1203.0     3216.9    5346.2    7506.0        8578.3    8935.8    8578.3    7506.0    5718.9    3216.9   1203.0     0
Acting on Compostie section
Barrier, KLF                0.09   0     60.0      160.4     285.1     374.2         427.7     445.5     427.7     374.2     285.1     160.4     60.0     0
Future wearing Surface     0.213   0    142.0      379.6     674.8     885.7        1012.2    1054.4    1012.2     885.7     674.8     379.6    142.0     0
Live Load + Impact                 0    698.9     1878.2    3367.8    4414.9        5040.8    5191.5    5040.8    4414.9    3367.8    1878.2    698.9     0

SUBTOTAL-Service I                 0    900.9     2418.2     4327.7    5674.8        6480.7    6691.4    6480.7    5674.8    4327.7   2418.2    900.9     0
SUBTOTAL-Service III               0    761.1     2042.5     3654.2    4791.8        5472.5    5653.1    5472.5    4791.8    3654.2   2042.5    761.1
Total-Strength I                   0   3014.9     8077.8    13945.1   18904.9       21597.2   22393.3   21597.2   18904.9   14410.9   8077.8   3014.9     0
Total-Strength IV                  0   1804.6     4825.3     8019.4   11259.1       12867.5   13403.6   12867.5   11259.1    8578.3   4825.3   1804.6     0

Moments at release, K-FT
Location                           0   transfer   0.1 L      0.2 L     0.3 L         0.4 L     0.5 L     0.6 L     0.7 L     0.8 L    0.9 L    transfer    L
ft                                 0     1.50      20         40        60            80        100       120       140       160      180      198.50    200

Girder, KLF                0.941   0    140.1     1693.8    3011.2    3952.2        4516.8    4705.0    4516.8    3952.2    3011.2    1693.8    140.1     0.0

1
1.1 Section Properties

1.1.1 Integral Precast Girder/Deck:

Properties of NU2000:
A = 903.8 IN2
I =    790,592 IN4
h =    78.7 IN
bW =   5.9 IN
yb =   35.7 IN
yt =   43.0 IN

1.1.2 Composite Section

108"
Effective Width
68.04"
Trasformed Width

Deck C.G.
83.20"
Girder C.G.
35.70"

Figure 1.1.2-1 Cross Section of Single Girder with Composite Deck

1
Needs revision!!!!!

Table 1.1.2-1 Composite Section Properties

Component
Area          yb         A yb     A(yb- ybc)2      Io           Ic
Girder              903.80       35.70     32,265.66   282,684.28 790,592.00 1,073,276.28
Haunch              32.39        79.20     2,565.50    21,586.30      2.70       21,589.00
Effective
508.07       83.20     42,271.26   451,627.27    2,074.61   453,701.88
Deck
Total              1,444.26                77,102.43                            1,548,567.16

Applicable Limit States

Strength I : Qu  1.25 DC  1.5DW  1.75 ( LL  IM )

Service I : Qu  1.0( DC  DW )  1.0( LL  IM )

Service III : Qu  1.0( DC  DW )  0.8( LL  IM )

2
3
Shear Force, K
Location                            0      critical   0.1 L   0.2 L    0.3 L       0.4 L   0.5 L   0.6 L    0.7 L    0.8 L    0.9 L    critical     L
ft                                  0         6.94    19.9    39.8     59.7        79.6    99.5    119.4    139.3    159.2    179.1      192.06    199

Acting on Non-comp. Section
Girder                    0.941    93.6       87.1    74.9    56.2     37.5        18.7    0.0     -18.7    -37.5    -56.2    -74.9      -87.1    -93.6
Deck                      0.816    81.2       75.5    64.9    48.7     32.5        16.2    0.0     -16.2    -32.5    -48.7    -64.9      -75.5    -81.2
haunch                    0.049     4.8        4.5     3.9     2.9      1.9         1.0    0.0      -1.0     -1.9     -2.9     -3.9       -4.5     -4.8

Subtotal:                          179.6     167.1    143.7   107.8    71.8        35.9    0.0     -35.9    -71.8    -107.8   -143.7    -167.1    -179.6
Acting on Compostie section
Barrier                     0.09     9.0       8.3      7.2     5.4     3.6         1.8     0.0     -1.8     -3.6      -5.4     -7.2      -8.3      -9.0
Future wearing Surface     0.213    21.2      19.7     17.0    12.7     8.5         4.2     0.0     -4.2     -8.5     -12.7    -17.0     -19.7     -21.2
Live Load + Impact                 138.9     132.1    119.4   101.2     84          68     53.2     -68      -84     -101.2   -119.4     132.1    -138.9

SUBTOTAL-Service I                 169.0     160.1    143.5   119.3     96.1        74.0   53.2     -74.0    -96.1   -119.3   -143.5     104.1    -169.0
SUBTOTAL-Service III               141.3     133.7    119.6    99.0     79.3        60.4   42.6     -60.4    -79.3    -99.0   -119.6      77.6    -141.3
Total-Strength I                   510.6     480.0    423.0   337.6    254.0       172.5   93.1    -172.5   -254.0   -337.6   -423.0     -17.7    -510.6
Total-Strength IV                  314.6     292.7    251.7   188.8    125.9        62.9    0.0     -62.9   -125.9   -188.8   -251.7    -292.7    -314.6

Table 4-2 Shear Forces for a Typical Interior Girder

4
5- Total number and arrangement of strands

The total number and arrangement of strands was estimated using service III and NCHRP 18-07
(LRFD 2005) detailed loss method. For this iterative process to be done quickly, the Loss spreadsheet
is used. The total number of bottom strands can be found = 82-0.6 in. strands: 46 straight strands at the
bottom flange and 36 harped strands. To avoid excessive uplift force on the prestressing bed, a
maximum of 12 strands will be harped at any point. Thus, the 36 strands will be harped in three groups.
The drape point locations will be assumed at 0.2L, 0.3L, and 0.4 L as shown in figures 5-1 and 5-2.
Further design checks may require altering this arrangement. In addition, 4 straight top strands will be
used, mild reinforcement support and for control of top cracking at prestress release. For the top
strands, the tension will be specified at 10 ksi.. Effect of the top strands on the total prestress force will
be ignored.

0.2L
0.3L
0.4L
Figure 5-1 Harped strands profile

End Section (0.0L)                     Section at (0.5L)
Figure 5-2 Pretensioning Scheme

All strands eccentricities are shown in Table ???? below:

Table 5-1 Pretensioning eccentricity at different sections

5
critical
Strand type          0L       0.035 L      0.1 L      0.2 L       0.3 L       0.4 L        0.5 L

12 draped strands       32.2       27.18       17.85        3.5         3.5         3.5         3.5
12 draped strands       50.2       45.34       36.30      22.40         8.5         8.5         8.5
12 draped strands       68.2       63.41       54.53      40.85       27.18        13.5        13.5
46 straight strands     4.09        4.09        4.09       4.09        4.09        4.09        4.09

82 Bottom strands      24.33      22.19       18.20      12.06       8.03         6.03        6.03

6- Strength I at Mid-Span
The reinforcement bars in the deck slab and the top flange strands are ignored in these calculations.
The I-beam top flange is assumed to be a rectangle of equal area to the actual flange area. The depth
can found to be 3.93 in for the actual width of 48.2 in. of the NU I girder top flange shape. Strain-
compatibility approach was used to calculate the section strength. See spreadsheet #??? For the results.

The last cycle of the iterative analysis will be redone by long hand below to explain the spreadsheet
analysis and to check its results: (See attached Excel Spread Sheet Strength I)

Straight strands              As1=46×0.217=9.982 in.2           y1 = 81.60 in. from top of slab
Harped Strands                As2=36×0.217=7.812 in.2           y2 = 76.20 in. from top of slab

The value of the compression block depth, from the spreadsheet, a = 11.07 in. This indicates that the
compression block is within the top flange of the beam: 11.07-7 (slab) – 1 (haunch) = 3.07.

Average beta formula Source PCI BDM and LRFD

Deck ß1 = 0.85-0.05×(f’c-4) = 0.85-0.05×(5-4) = 0.8
Girder ß1 = 0.85-0.05×(f’c-4) = 0.85-0.05×(10-4) = 0.65
Average ß1 = (Ac1×f’c1×ß1+ Ac2×f’c2×ß2)/ (Ac1×f’c1 + Ac2×f’c2) = 0.76
c = a/ ß1 = 14.578 in.

The compression force in the deck slab and moment due to this force at the top of the deck slab
Fc1 = 0.85×5×(9×12×7)               = 3213.00 kips
Mc1 = 3213.00 ×7/2                  = 11245.50 k.in.

Similarly, the compression force in the haunch and moment due to this force at the top of the deck slab
are 204.85 kips and 1536.38 kip-in.; the compression force in the beam top flange and moment due to
this force at the top of the deck slab are 1259.17 kips, and 12,008.27 kip-in.

The spreadsheet is able to treat each strand layer individually. For convenience in the hand calculation
check, we will assume two groups of strands, the straight strands and the harped strands, clustered at
two points.

6
 82.61        172
Straight strands strain: 1  0.003            1            0.020
 c          28500
Using the “Power formula” PCI BDM Section ?????
                                     
27613
fps  ε ps 887                                   270

        
1  112.4ε ps      
7.36 1 / 7.36 

fps1 = 263.21 ksi
Fs1 = 263.21×9.982=2627.36 kips
Moment at top face Ms1 = 2627.36 ×82.61 = 217,046.39 kip-in.

Tension force in the harped strands and moment due to this force at the top of the deck slab
 78.2  172
Straight strands strain: 1  0.003        1       0.019
 c       28500
Using” Power formula” fps2 = 262.37 ksi
Fs1 = 262.37 ×7.182=2,049.63 kips
Ms1 = 2,049.63 ×78.2 = 160,281.4 k.in

Sum of forces = 2,049.63 +2627.36 -204.85 -4543.8682-3213.00 = 0.0
Since strain in the extreme bottom stand layer is larger than 0.005, the resistance factor  =1
Sum of M =1× (160,281.4 +217,046.39 -12,008.27 -1536.38 -
11245.50)/12=29378.3 k.ft > Mu = 22660.6 k.ft
Thus the provided strength is about 30% larger than the required strength. This check will not need to
be revisited unless the final number of strands has been extremely reduced. This also tells us that the
assumed 10 ksi final strength may be too high. We will find out when we do Strength IV check. Also,
we have an option to try to use a lower deck strength at this time.

7
Flexural Strength Using Strain Compatibity and Mast's Variable   0.75 to 1.0
W1
W2                                                                      cu                      0.003
W3
c=                       14.578
T 2 U pper
T 2 L ow er
1                                           a                        11.073
2                   T2
Sum of
d si
3                                                                 forces                     0.00

A si
Units in kips and inches                                                                                 4                                                                      ANSWER:
                      1.00
 Mn kip-in          352539.1
W4                                                                     kip*ft                29378.3

Concrete Layers                     f'c                Width, W     Thick., T     Depth, dc         1           Tupper      Tlower       Revised T Beta1calcuation
1              5.000        108.000         7.000       3.500             0.800         0.000        7.000      7.000            3024      3780
2              5.000         48.200         1.000       7.500             0.800         7.000        8.000      1.000           192.8       241
3             10.000         48.200         3.930       9.537             0.650         8.000      11.930       3.073    962.8911331 1481.37097
4             10.000           5.910      74.770       11.930             0.650       11.930       86.700       0.000               0         0
5                                                      86.700             0.850       86.700       86.700       0.000               0         0
6                                                      86.700             0.850       86.700       86.700       0.000               0         0
7                                                      86.700             0.850       86.700       86.700       0.000               0         0
Total:                                   86.7                                                                       4179.691133 5502.37097

Steel Layers                        Area Asi           Grade                              Depth dsi Es
Effective Prest.                    Q                  fpy            R          K                  so                          Total  s
Grade 60 Bars                   1                                    60                 0                29000                  0             60        100             1.096          0.0000            0.0000               -0.0030
2                                    60                 0                29000                  0             60        100             1.096          0.0000            0.0000               -0.0030
3                                    60                 0                29000                  0             60        100             1.096          0.0000            0.0000               -0.0030
4                                    60                 0                29000                  0             60        100             1.096          0.0000            0.0000               -0.0030
5                                    60                 0                29000                  0             60        100             1.096          0.0000            0.0000               -0.0030
6                                    60                 0                29000                  0             60        100             1.096          0.0000            0.0000               -0.0030
7                                    60                 0                29000                  0             60        100             1.096          0.0000            0.0000               -0.0030
8                                    60                 0                29000                  0             60        100             1.096          0.0000            0.0000               -0.0030
9                                    60                 0                29000                  0             60        100             1.096          0.0000            0.0000               -0.0030
10                                    60                 0                29000                  0             60        100             1.096          0.0000            0.0000               -0.0030
11                                    60                 0                29000                  0             60        100             1.096          0.0000            0.0000               -0.0030
12                                    60                 0                29000                  0             60        100             1.096          0.0000            0.0000               -0.0030
13                                    60                 0                29000                  0             60        100             1.096          0.0000            0.0000               -0.0030
Grade 70 Plate                  1                  0                 70                 0                29000                  0             70        100              1.06          0.0000            0.0000               -0.0030
Gr. 150 Rods                    1                  0                150                 0                29000              0.016         127.50       3.75              1.04          0.0000            0.0000               -0.0030
Gr 270                          1              9.982                270               172     82.610     28500              0.031           243        7.36             1.043          0.0060            0.0140                0.0200
Gr 270                          2              7.812                270               172     78.200     28500              0.031           243        7.36             1.043          0.0060            0.0131                0.0191
Gr 270                          3                                   270               172                28500              0.031           243        7.36             1.043          0.0060            0.0000                0.0030
Gr 270                          4                                   270               172                28500              0.031           243        7.36             1.043          0.0060            0.0000                0.0030
5                                   270                20                28500              0.031           243        7.36             1.043          0.0007            0.0000               -0.0023
6                                   270               172                28500              0.031           243        7.36             1.043          0.0060            0.0000                0.0030
7                                   270               172                28500              0.031           243        7.36             1.043          0.0060            0.0000                0.0030
8                                   270               172                28500              0.031           243        7.36             1.043          0.0060            0.0000                0.0030
9                                   270             172.3                28500              0.031           243        7.36             1.043          0.0060            0.0000                0.0030
10                                   270             172.3                28500              0.031           243        7.36             1.043          0.0060            0.0000                0.0030
11                                   270             172.3                28500              0.031           243        7.36             1.043          0.0060            0.0000                0.0030
12                                   270             172.3                28500              0.031           243        7.36             1.043          0.0060            0.0000                0.0030
13                                   270             172.3                28500              0.031           243        7.36             1.043          0.0060            0.0000                0.0030
Sum of M                                                                                                                                  MAXIMUM            :                     0.0140

Figure 6-1 Strength I calculation at Mid-Span Section

8
7. The Concrete Strength at Release

Because of the extreme length of this beam, it will be required to set the lifting points some
distance away from the beams ends to help resist buckling of the top flange. We will assume that
a careful lifting analysis has resulted in identifying the lifting points at 20 ft from each end.

NDOR Policy January-February 2001 PCI paper by Tadros et al., “Strength Design of
Pretensioned Flexural Concrete Members at Prestress Transfer” the strength design for prestress
transfer can be approached in a manner similar to that for non-prestressed reinforced concrete.
The member can be treated as a “reinforced concreted column subjected to moment combined
with axial compression force equal to the force in the prestressing steel just before prestress
transfer.” Therefore, we can solve for the neutral axis location”c” and f’ci by using the
equilibrium equations.

j=9
j=8
j=7
j=6

Asi

y si                         j=5
j=4
j=3
j=2
Tj
j=1
dcj

bj

9
STRENGTH DESIGN PROGRAM FOR PRESTRESS TRANSFER

Directions:
1 Input       Output
2 Sign convention: compressive stress
in concrete and tensile stress in steel
are considered positive. Prestress force
is always positive. Moment in the same
sense as prestress force moment is positive.
3 Units in kips and inches
4 Make sure the compression fibre.
dc & ys based on compressed concrete edge
5 Functions are used below

Output    f'c    Top Steel Stress
7.270     -26.37

DATA FOR CONTROLING
MExt                    m                  ?          p          cu
0                           0.85          0.75    1.15       0.003

CONCRETE LAYERS
Width, b               Thick., T Sum of T Compr. T Depth, dc   Ac                  Fc        Mc
1                            38.4           5.3      5.3     5.3     2.65   203.52            1257.652   3332.777
2                           22.15           5.5    10.8      5.5     8.05 121.825             752.8176   6060.182
3                             5.9        63.59    74.39    63.59   42.595 375.181             2318.431   98753.57
4                           27.05         1.75    76.14     1.75   75.265 47.3375             292.5221   22016.67
5                            48.2         2.56     78.7 0.240667 76.26033 11.60016            71.68318   5466.583
6                                                      0       0          0      0                   0          0
7                                                      0       0          0      0                   0          0
8                                                      0       0          0      0                   0          0
9                                                      0       0          0      0                   0          0
10                                                      0       0          0      0                   0          0
78.7                                                4693.106   135629.8

STEEL LAYERS
As   Initial Prest fpi      Depth ys      Es         s       Stress  f
s   wi         Fs         M   s
Grade         f py     K      Q       R
Gr 270           1    9.982                  202.5        4.09     28500    -0.00289   -82.35416           1   2339.036   9566.655           270     243    1.04   0.031   7.36
2    2.604                  202.5        32.2     28500   -0.002132   -60.75526           1   666.4267   21458.94           270     243    1.04   0.031   7.36
3    2.604                  202.5        50.2     28500   -0.001646   -46.92311           1   702.4456   35262.77           270     243    1.04   0.031   7.36
4    2.604                  202.5        68.2     28500   -0.001161   -33.09079           1    738.465   50363.31           270     243    1.04   0.031   7.36
5    0.868                  202.5       76.95     28500   -0.000925   -26.36674           0   246.6277     18978            270     243    1.04   0.031   7.36
6                                                 28500           0            0          1          0          0           270     243    1.04   0.031   7.36
7                                                 28500           0            0          1          0          0           270     243    1.04   0.031   7.36
8                                                 28500           0            0          1          0          0           270     243    1.04   0.031   7.36
9                                                 28500           0            0          1          0          0           270     243    1.04   0.031   7.36
10                                                 28500           0            0          1          0          0           270     243    1.04   0.031   7.36
Grade 70 Plate    1                                                 29000           0            0          1          0          0            60      70       1       0     10
Gr. 150 Rods      1                                                 29000           0            0          1          0          0           150    127.5    1.4   0.016   4.99
Grade 60 Bars    1                                                  29000           0            0          1          0          0            60      60     1.4       0     25
2                                                 29000           0            0          1          0          0            60      60     1.4       0     25
3                                                 29000           0            0          1          0          0            60      60     1.4       0     25
4                                                 29000           0            0          1          0          0            60      60     1.4       0     10
5                                                 29000           0            0          1          0          0            60      60     1.4       0     10
6                                                 29000           0            0          1          0          0            60      60     1.4       0     10
             c                     a        F         M                                4693.001   135629.7
0.6865         111.2609864          76.38 0.1050751   0.105079

The equations that were used in the analysis are as follows:
 y 
1. Strain calculation: Δε si  0.003  1  si 
    c 
2. Force equilibrium:
        p                            y  a w                                                                     1
 F  0.85f c  A cj   A si  f si   f si  0.85f c w i   0 ysi  a w i
                                                                                                                   0
j        i                                            si      i

3. Moment equlibrim:
         p                         m                                                                            y si  a w i  1
 M  0.85f c  A cj y cj   A si  f si   f si  0.85f c w i y si   M  0
                                                                                                                  
j          i                                                                                                                        y si  a w i  0
                       
                       
                       
4. Power formula to get strands stress:               Q         1 Q        
fsi  ε siE s
      ε E R  
1/ R

      si s   
    1         

      kfpy   
             

10
By solving the two equilibrium equations, the two variables “c” and f’ci can be obtained. The
analysis can be done by trying “c” and f’ci and check equilibrium equations.

The following is the last trial of the analysis.
c= 111.2609864 in., f’ci = 7.270 ksi
ß = 0.85-0.05×(f’c-4) = 0.85-0.05×(7.270 -4) = 0.6865
a = c * ß = 76.38 in.
Divide NU2000 into 5 layers.
Width              Thickness
38.4         5.3
22.15         5.5
5.9       63.59
27.05        1.75
48.2        2.56

1. concrete layer1
Fc1 = 0.85×10×( 38.4×5.3)= 1257.652 k
Mc1 = 1257.652 ×5.3/2 = 3332.777 k. in

2. concrete layer2
Fc2 = 0.85×10×( 22.15×5.5) = 752.8176 k
Mc2 = 752.8176 ×(5.5/2+5.3) = 6060.182 k. in

3. concrete layer 3
Fc3 = 0.85×10×( 5.9×63.59) = 2318.431k
Mc3 = 2318.431×42.595 = 98753.57k. in

4. concrete layer 4
Fc4 = 0.85×10×( 27.05×1.75) = 292.5221k
Mc4 = 292.5221 ×75.265 = 22016.67 k. in

5. concrete layer 5
Fc5 = 0.85×10×( 48.2×2.56) = 71.68318 k
Mc5 = 71.68318 ×76.26033 = 5466.583 k. in
Add them together, F = 4693.106 k, M = 135629.8 k.in

    y si 
Use Δε si  0.003  1         
    c 
Top steel
    4.09 
Δε s1  0.003  1         0.00289
 111 .26 

11
                                                                                           
                                                                                           
                         
f s1  ε s1E s Q 
1 Q           0.00289  285000.031

1  0.031                   
  -82.3542ksi
                    1/ R 

1 / 7.36

         ε E R                                      0.00289  28500  7.36 
                          
  si s                                                                           
        1
                                            1                 
          kfpy   
                         
          
      1.04  270                  

       
if the effective concrete includes strands inside, the exact force of concrete should be 0.85f’c (Ac-
As)=0.85 f’cAc-0.85 f’cAs.
0.85f’cAs was considered when we calculate force of strands in order to avoid the calculation of
centroid of the concrete area.
Fint erior  Fexternal
γp
Fc  Fs           A ps fpi
φ
γp
 0.85 fc ( A c  A s )   A s Δfsi  φ A ps fpi
Rearrange the items in the equation, we get
      γp                   y  a w  1
F  fc  A cj   A si  Δfsi  φ fsi  0.85fc w i   0 y si  a w i  0
                      
j             i                             si          i

Hence, Fs=9.982(-82.3542+0.75/1.15*202.5+0.85*7.27=2339.036k.
Ms = 339.036*4.09=9566.655 k. in
The calculations of the other strands are the same. So the total force and moment of strands is:
Fs  4693.001k
M  135629.7k.in
s

F  F  0 k
c         s

M  M  0 k.in
c         s

12
8. Service III Using Transformed Section Properties and NCHRP 18-07

8.1.1 Moments
Mg    =4,658 k.ft         = 55,897 k.in.
Md    =4278 k.ft          = 51,334 k.in
MSIDL =2450 k.ft          = 29,404 k.in
ML.L =5192 k.ft           = 62,298 k.in

8.1.2 Material properties.
Modulus of elasticity
E c  33000( w )1.5 f c'
E ci  33000(0.15)1.5 7.5 = 5,250 ksi
E c  33,000(0.15)1.5 10.0 = 6,062 ksi
E cd  33,000(0.15)1.5 5.0 = 4,074 ksi
8.2 Shrinkage Strain

8.2.1 Precast

First period (from release to Final).
S bif  0.00048 k cp k s k h k st k t
k cp  1 One day steam curing
1064  94(3.2)
ks                      1.038
735
k h  2.00  0.0143 (70 )  1.0
5
kf             0.588
1  7.5
k sh  (1)(1.038 )(1.0)( 0.588 )  0.61

k td 
20000  1           0.998
61  4(7.5)  20000  1
S bif  292 x10 6

8.2.2 Second period (from release to deck placement).
S bid  0.00048 k cp k s k h k st k t
k cp  1 One day steam curing
1064  94(3.2)
ks                      1.038
735
k h  2.00  0.0143 (70 )  1.0
5
k st           0.588
1  7.5
k sh  (1)(1.038 )(1.0)( 0.588 )  0.61

13
k td 
60  1         0.656
61  4(7.5)  60  1
S bid  192 x10 6
S bdf  S bif  S bid
S bdf  100 x10 6

8.2.3 Deck Slab
S di-I  0.00048 k cp k s k h k st k t
k cp  1 for 7 days moisture curing
1064  94(7.0 / 2)
ks                        1.0
735
k h  2.00  0.0143 (70 )  1.0
5
k st         0.833
1 5
k td 
20000  60          0.998 at final
61  4(7.5)  20000  60 
S ddf  0.00048 (1.0)(1.0)(1.0)( 0.833 )( 0.998 )  399 x10 6

8.3 Creep Coefficient
8.3.1 Precast
First period (from release to final).
bif  1.9k st k s k h k la k t
1064  94(3.2)
ks                         1.038
735
5
kf             0.588
1  7.5
k hc  1.56  0.008 (70 )  1.0



k la  t la0.118  1.0

k td 
20000  1
 0.998
61  4(7.5)  20000  1
bif    1.9(1.038 )( 0.588 )(1.0)(1.0)( 0.998 )  1.157

8.3.2 Second period (release to Deck placement).

bid  1.9k st k s k h k la k t
1064  94(3.2)
ks                         1.038
735
5
kf             0.588
1  7.5
k hc  1.56  0.008 (70 )  1.0

14


        
k la  t la0.118  10.118  1.0

k td 
60  1         0.656
61  4(7.5)  60  1
bid  1.9(1.038 )( 0.588 )(1.0)(1.0)( 0.656 )  0.761

8.3.3 From deck placement to final

bdf  1.9k st k s k h k la k t
1064  94(3.2)
ks                         1.038
735
5
kf            0.588
1  7.5
k hc  1.56  0.008 (70 )  1.0


            
k la  t la0.118  60 0.118  0.617

k td 
20000    60 
 0.998
61  4(7.5)  2000  60 
bdf    1.9(1.038 )( 0.588 )(1.0)( 0.617 )( 0.998 )  0.714

8.3.4 Deck Slab
ddf  1.9k f k s k h k la k t
1064  94(3.5)
ks                        1.0
735
5
kf                1.0
1  0.8(5)
k hc  1.56  0.008 (70 )  1.0


        
k la  t la0.118  10.118  1.0

k td 
20000  1        0.998
61  4(5)  2000  1
bid  1.9(1.0)(1.0)(1.0)(1.0)( 0.998 )  1.877

Stage                                           Shrinkage   Creep
Strain      coefficients
I-beam    Release to final                                     292x10-6    1.157
Release to deck placement                            192x10-6    0.761
Deck placement to final                              100x10-6    0.396
Deck Slab Deck placement to final                              399x10-6    1.788

8.4 Section Properties

8.4.1 Modular ratios

15
E s 28 ,500
Initial steel modular ration = n i                5.429
Ei   5,250
E s 28,500
Final steel modular ration = n              4.701
E     6,062
E     28 ,500
Deck concrete modular ration = n d  s             6.996
Ed     4,074

8.4.2 Transformed Section Properties
Transformed Section Properties at Release ( non-composite )

Ati  Ag  (n i  1)Ap  982.6in 2

Ybi 
 AY  33.32in
A
I ti  854 ,412 in 4
Transformed Section Properties at stage II (composite)
A t  A g  (n i  1)Ap  1,510.12in 2

Yb 
 AY  51.32in   2

A
I t  1,666 ,755 in 4

8.5 Losses at the first period

8.5.1 Elastic Shortening due to the initial pretension force and girder self weight.
ES i  f c ni
 1       ep  M g * ep
2

f c  Pi                       5.023 ksi
 A ti I ti        I ti
              
ES i  (5.023 ) * (5.429 )  27 .3ksi
8.5.2 Shrinkage between release and deck place
SH  K id E s  bid
1
K id                                      0.712
         A       ep 
2

1  n i P 1  2 1  C u 

         Ag  r 
             

6
SH  (0.712 ) * (28,500 ) * (192 x10 )  3.896 ksi
Creep between release and deck place
CR  k id n i  id f cgp
CR  (0.712 ) * (5.429 ) * (5.023 ) * (0.761 )  14 .776
Relaxation between release and deck place
CR  1.2ksi

16
Elastic due to deck weight
Using the transformed section
ES  f c n
M d * e tr
fc              1.681
I tr
ES  (1.681) * (4.701)  7.9ksi

Elastic due to superimposed DL (on composite section)
Using the composite transformed section
ES  f c n
M SIDL * e Com _ Tr
fc                     0.799
I ComTr
ES  (0.799) * (4.701)  3.756ksi

From deck placement and SIDL to final time

Time dependent losses between deck placement and final
SH  KE s  SH
1
K df                                  0.718
       A   ep 2

1  n i P 1  2 1  Cu 

       Ag  r 
                


SH  (0.718 ) * (28,500 ) * (100 x10 6 )  2.046 ksi

Creep of beam bet.deck place and final, initial loads
CR  k df n i  bdf f cir
CR  (0.718 ) * (5.429 ) * (5.023 ) * (0.396 )  7.75ksi
Creep of beam due to deck and SIDL
CR  k df n i ddf f cir
CR  8ksi

Relaxation Losses
CR  1.2ksi

Shrinkage of deck

E p  ddf A d E cd  1 e pc e d   
f pSS                                    K df 1  0.7 bdf   0.6ksi
E c 1  0.7 df   A c
       Ic     

Elastic due to LL

17
M LL * e Com _ Tr
fc                       8.0ksi
I Com _ Tr
Strand Stress at service
f s  202 .5  27 .3  3.9  14 .776  1.2  7.9  3.8  2.1  7.8  8  1.2  0.6  8  172 .0ksi

Service III

fc 
 1      ep yb     Mg * yb
Pi                 
 A ti    I ti       I ti
                 
 1           e p _ net _ i y b _ ner _ i   
 (long - term prestress force loss(initi al to deck placemnt)                                           
 A net _ i           I net _ i             
                                           
M d * y tr

I tr
M SIDL * y Com _ Tr

I Com Tr
 1            e p _ net _ Tr y b _ net _ Tr     
 (long - term prestress force loss( deck placemnt t o final)                                                
 A net _ Tr             I net _ Tr              
                                                
M LL * y Com _ Tr

I Com _ Tr
 0.555ksi  0.19 f c'  0.601ksi

18
9. Strength IV Limit State (Non-composite section only)

9.1.1 This limit states is used to check the capacity of precast section only. The excel strain-
compatibility program was used to design the section. The capacity of the precast section
without any mild reinforcement in compression zone was estimated at:
M n  12314 .0 KIP  FT against the demand of :

M u  13403 .6 KIP  FT

The mild reinforcement was provided in the top flange to obtain the adequate moment capacity.
8 #9 was used to satisfy that limit state. This reinforcement is extended to the 0.4 L. The 6#9 are
extended to the 0.3 L.
Section at point 0.4L with 6 #9 in the top flange has capacity of:
M n  12922 .0 KIP  FT > M u  12867 .5 KIP  FT ,                    OK

Section at 0.3L without any top reinforcement has capacity:
M n  11774 .0 KIP  FT > M u  11259 .51 KIP  FT ,                   OK

For brevity, hand calculation only for last iteration is shown hereafter (midspan).

f c' =10,000 psi                          1 =0.65

c = 55.34 IN
a = c1 = 0.65(55.34) = 35.97 IN

dp      f pe 
 ps  0.003
     1  
 E 

 c      p

We can combine prestressing strand into the clusters:
1. 46 straight strands
d p  78 .7  4.09  74 .61 IN

 74 .61   172 .0 
 ps  0.003        1                   3
  7.187 x10 IN
 55 .34   28500 
2. 12 draped strands- First Cluster
d p  78 .7  3.5  75 .2 IN

 75 .2      172 .0 
 ps  0.003        1                     3
  7.219 x10 IN
 55 .34   28500 

19
3. 12 draped strands- Second Cluster
d p  78 .7  8.5  70 .2 IN

 70 .2      172 .0        
 ps  0.003           1                            3
  6.948 x10 IN
 55 .34   28500           
4. 12 draped strands–Third Cluster
d p  78 .7  13 .5  65 .2 IN

 65 .2      172 .0 
 ps  0.003           1                     3
  6.678 x10 IN
 55 .34   28500 
5. 4 partially tensioned straight strands in the top flange
d p  2.0 IN

 55 .34  2   10 
 ps  0.003               
3
  2.541 x10 IN
 55 .34   28500 

Strain in compression steel find from ratio:
d= 2.0 IN

 55 .34  2 
 s'  0.003                           3
  2.892 x10 IN
 55 .34 

f s   s E s  f y  60 KSI

f s  2.892x103 (29000)  83.87KSI  f y  60 KSI

Thus, take f s  60 KSI

Find stress of prestressing strands using the “Power formula”:

                                  
             27,613               
f ps     ps 887                          1 
 270 KSI

      
1  112.4 ps 
7.36

7.36 

1. 46 Straight strands

20
                                                    
 7.187 x10 887                                                  199.80 KSI
3                 27,613
f ps                                                               1 

           
1  112.4(7.187 x10 3 )    
7.36
   7.36 

2. 12 Draped strands- First Cluster

                                                     
 7.219x10 887                                                   200.54KSI
3                27,613
f ps                                                               1 

            
1  112.4(7.219x10 3 )
7.36
           7.36 

3. 12 Draped strands- Second Cluster

                                                     
 6.948x10 887                                                   194.14 KSI
3                27,613
f ps                                                               1 

            
1  112.4(6.948x10 3 )
7.36
           7.36 

4. 12 Draped strands –Third Cluster

                                                     
 6.678x10 887                                                   187.48KSI
3                27,613
f ps                                                               1 

            
1  112.4(6.678x10 3 )
7.36
           7.36 

5. 4 Partially tensioned straight strands

                                                 
 2.541x10 887                                               72.41KSI
3                27,613
f ps                                                           1 

            
1  112.4(2.541x10 3 )
7.36
        
7.36 


Assume the compression flange is divided into separate sections .
The rectangular shape was used to substitute the trapezoidal form of the top flange. The width of
the rectangular shape “b” is as follows:
b = (48.2 + 5.9)/2=27.05 IN
Compression forces:
C1  0.85 f c' Ac  0.85 (48 .2)2.56 (10 )  1048 .8 KIP

C 2  0.85 f c' Ac  0.85 (27 .05 )1.77 (10 )  407 .0 KIP

C 3  0.85 f c' Ac  0.85 (5.9)(35 .97  2.56  1.77 )(10 )  1586 .8KIP

C4  APS ( f ps  0.85 f c' )  4(0.217)(72.41  0.85(10))  55.5KIP

21
C 5  AS ( f s  0.85 f c' )  9(0.79 )( 60 .0  0.85 (10 ))  366 .2 KIP

Tension forces:
T1  A ps f ps  46 (0.217 )199 .8  1994 .4 KIP

T2  A ps f ps  12 (0.217 )200 .54  522 .2 KIP

T3  A ps f ps  12 (0.217 )194 .14  505 .5 KIP

T3  A ps f ps  12 (0.217 )187 .48  488 .2 KIP

To get the moment capacity of the section we can take sum of the moments around any point.
Let’s use bottom of compression block
4                                                                                    3
M n   A psi f psi y psi  A top ps ( f ps  0.85 f c' ) y ps  As ( f s  0.85 f c' ) y s   0.85 f c' Acj y j =
i 1                                                                                 j 1

=1,994.4(38.64) + 522.2(39.23) + 505.5(34.23) + 488.3(29.23) + 1,048.8(34.69)
+ 407.0(32.53) + 1,586.8(15.82) + (55.5+366.2)33.97 =
=77,063.6 + 20,485.9 +17,303.3 + 14,273.0 + 36,382.9 + 13,239.7 + 25,103.2 + 14,325 =
218,176.7 KIP*IN = 18,181.4 KIP*FT

Resistance factor  shall be takes according to 5.5.4.1.
For flexure and tension of prestressed concrete it is equal to 1.
For unified design method the next formula of  calculation is going to be used:
0.75    (1.75  250  extreme) / 3  1.00

Proposed for LRFD by Seguirant et al. (PCI Journal 2004)
 78 .7  2 
 extreme  0.003            1  1.158 x10 3 IN
 55 .34      
  (1.75  250 (1.158 x10 3 ) / 3  0.68           Thus take   0.75
M n  0.75 (18,181 .4)  13,636 .0 KIP FT
From the Table 4-1 the factored moment for Strength IV is equal to:
M u  13,403 .6 kip-ft.

22
The Excel program treats each layer of strands individually and provides more accurate result.
Refer to Figure 9-2 and the attached Excel Program for Strength IV, cell M17.(see next page)

Its output for M n  13,505 .7 KIP  IN that is greater than M u  13,403 .6 KIP  IN

OK

23
Flexural Strength Using Strain Compatibity and Mast's Variable 0.75 to 1.0
W1
W2
u                      0.003
W3
c=                      55.343
T2Upper
T2Lower
1                                           a                       35.973
2               T2
Sum of
dsi
3                                                            forces                    0.00

Asi
Units in kips and inches                                                                     4                                                                 ANSWER:
                     0.75
Mnkipin         162404.8
W4                                                                kip*ft              13533.7

Concrete Layers                f'c                Width, W     Thick., T    Depth, dc  1            Tupper      Tlower       Revised T Beta1calcuation                                                    Area
1            10.000         48.200         2.560      1.280       0.650         0.000        2.560      2.560         802.048   1233.92                                           123.392
2            10.000         27.050         1.770      3.445       0.650         2.560        4.330      1.770      311.21025    478.785                                            47.879
3            10.000           5.900      74.370      20.152       0.650         4.330      78.700      31.643    1213.518563 1866.95164                                           186.695

Total:                                   78.700                                                                   2330.213887 3584.94444

Steel Layers                   Area Asi           Grade                          Depth dsi Es
Effective Prest.                 Q              fpy            R          K                 eso              De           Total es      Stress
Grade 60 Bars              1               7.11              60              0.0       2.000  29000            0            60        100              1.096                0    0.002891856 -0.002891856        -60

Gr 270                     1              3.472             270           172      -2.000    28500         0.031         243.00       7.36             1.043 0.006035088         0.003108414   0.002926674 83.40713
Gr 270                     2              3.472             270           172      -4.000    28500         0.031            243       7.36             1.043      0.0060              -0.0032         0.0028    80.32
Gr 270                     3               2.17             270           172      -6.000    28500         0.031            243       7.36             1.043      0.0060              -0.0033         0.0027    77.23
Gr 270                     4              0.868             270           172      -8.000    28500         0.031            243       7.36             1.043      0.0060              -0.0034         0.0026    74.14
5              0.868             270            10       2.000    28500         0.031            243       7.36             1.043      0.0004              -0.0029        -0.0025   -72.41
6              2.604             270           172      -3.500    28500         0.031            243       7.36             1.043      0.0060              -0.0032         0.0028    81.09
7              2.604             270           172      -8.500    28500         0.031            243       7.36             1.043      0.0060              -0.0035         0.0026    73.37
8              2.604             270           172     -13.500    28500         0.031            243       7.36             1.043      0.0060              -0.0037         0.0023    65.64

24
25
10. Check Minimum Reinforcement:

M n  the lesser of 1.2 M cr or 1.33 M u                                             (LRFD 5.7.3.3.2)
Compute Mcr using Std Specs Article 9.18.2.1
M cr   f r  f pe  S c  M d / nc S c S b  1                      (Std Specs 9.18.2.1)
where
f r  0.24 f c = 0.24 10.00 KSI = 0.759 KSI                                              (LRFD 5.4.2.6)
f pe = compressive stress in concrete due to effective prestress forces only (after …

losses) at extreme fiber of section where tensile stress is caused by externally applied

Prestressing force is P  82(0.217)172.0  3060.6 KIP

f pe = P  Pe y nc  3060 .6  3060 .6(35 .7  6.03) 35 .7 = 7.487 KSI
A I nc         903 .8         790 ,592
M d / nc = 8,935.8 K-FT (Refer to Table 4-1)
S c = Sbcg = composite section modulus for the tension face
S b = non-composite section modulus for the tension face
 27,372 IN3    
M cr  0.759 KSI  7.487 KSI  27,372IN3   8,935K  FT(12) 
 22145 IN 3
 1

               
= 225,709.5 – 25,307.7 = 251,017.2 KIP-IN = 20,918.1 KIP - FT

1.2 Mcr = 1.2 (20,918.1) = 25,101.7 K-FT                GOVERNS since 1.2 Mcr < 1.33 Mu

1.33 Mu = 1.33 (22,660.6) = 30,138.6 K-FT

Moment capacity of the composite section at midspan:

M n  29 ,728 .5 KIP  FT

(see attached Excel Program Strength I midspan, cell M17)

M n = 29,728.5 K-FT > 1.2Mc r = 17,195 K-FT                     0.K

The requirement of minimum reinforcement has been met

26
11. Shear Design

11.1. Transverse Shear Reinforcement
In this example, the girder will be designed for vertical shear at the
critical section for shear. In a full design, other sections along the
length of the girder would have to be designed as well.

11.1.1 Critical Section for Shear

Shear design using the Sectional Design Model is an iterative process that begins by
assuming a value for .

Assume an initial value for the inclination of the compression field, θ, of 22º.

Critical section for shear is greater of:                                          (LRFD 5.8.3.2)
• 0.5 dv cotθ or dv

dv = Effective shear depth
= Distance between resultants of tensile and compressive forces
The depth of the compression block, a, was computed in determining the moment
capacity of the section (see Section 3.1.4.5). You can also use the output of Excel
program “Strength I, Strain-Compatibility Approach AASHTO LRFD @ critical
section (cell M8)

d v  d e   hg  h f  c.g. @ critical sec tion  = (78.7 + 7.0 + 1 -22.19) - (10.05/2)
a                                         a
2                                         2
= 59.49 IN
But dv need not be taken less than the greater of:                                 (LRFD 5.8.2.7)
0.9 de = (0.9) (78.7 + 7 + 1 – 22.19) = (0.9) (64.51) = 58.06 IN
0.72 h = (0.72) (78.7 + 7 + 1) = 62.42 IN                  GOVERNS
Therefore, use dv = 62.42 IN

Critical section for shear is greater of:
• 0.5 dv cot θ = 0.5 (62.42) cot(22º) = 77.24 IN or         GOVERNS            (LRFD 5.8.3.2)
• dv = 62.42 IN
Note: dv will govern for θ > 26.6º.
Therefore the critical section for shear is:
0.50 FT + 77.24 IN / 12 = 6.94 FT from centerline of support.
At the critical section for shear, Vu = 485.0 KIP

11.1.2 Component of Shear Resistance from Prestress, Vp
Pf = 82 (0.217)172.0 = 3060.6 KIP

27
Angle of center of gravity of strand profile with respect to horizontal, :
 = tan-1 [(eCL - eend) / (dist. to depression point)]
= tan-1 [((24.33 IN – 18.20 IN) / 12) / (0.1(200)) FT] = 1.463o
Vp = Pf sin (1.463o) = (3060.6 KIP) [sin (1.463o)] = 78.2 KIP

11.1.3. Governing Equations for Shear
Vu  Vr = Vn                                                                (LRFD 5.8.2.1-2)
 = 0.90 for shear                                                         (LRFD 5.5.4.2.1)
Vn = Vc + Vs + Vp                                                             (LRFD 5.8.3.3-1)
Compute maximum shear capacity of section:

Vn max = 0.25fc b v dv  Vp                                                      (LRFD 5.8.3.3-2)
Vn max = (0.25) (10.00 KSI) (5.9 IN) (62.42 IN) + 78.2 KIP = 998.9 KIP
Vn max = (0.90) (985.1) = 998.9 KIP > Vu = 485.0 KIP           O.K.

11.1.4 Concrete Contribution to Shear Resistance, Vc
Vc  0.0316        fc bv dv                                                      (LRFD 5.8.3.3-3)
To use this equation, the quantity  must be determined. This quantity is a factor that
represents the efficiency of shear transfer by concrete. Note that 0.0316 fc KSI  =


1.0 f c PSI  ,
            so a  value of 2 would provide a concrete contribution similar to the
familiar simplified value of Vc  2 fc PSI  bd.


v                          v
To obtain , the quantities          and  are needed, where    is a relative shear stress

fc                          
fc
and  is the inclination of the compression field.
Vu  V p 485.0 KIP  0.9 78.2 KIP
v              
0.9 5.9 IN 62.42 IN 
= 1.25 KSI                    (LRFD 5.8.3.4.2-1)
 bv d v
v
1.25 KSI
= 10 .0 KSI = 0.125

fc

Begin iterations using the previously assumed value for .

Trial 1: Assume  = 22 (previously assumed and used to determine location of critical
section for shear).
Mu
 0.5 N u  0.5(Vu  V p )cot  A ps f po
dv
x 
2 E s As  E p A ps 
 0.002                  (LRFD 5.8.3.4.2-2)

Mu = 3050.9 K-FT = 36,610.8 K-IN (see Summary of Dead and Live Load Effects)
Moment in this formula is not supposed to be taken less than Vu dv (LRFD C5.8.3.3)

28
Vu d v  485.0(62.42)  30,273.7  36,610 K-IN
Nu = 0 - no applied axial loads
fpo = 0.7 (270) = 189.0 KSI                                               (LRFD C5.8.3.4.2)
Aps = area of prestressing steel on flexural tension side of the member, i.e.,
the straight strands
Aps = 46(0.217) = 9.982 IN2

 0.5 485 .0  78 .2  cot22   9.982 189 .0
36 ,610 .8
 x  62 .42                                                           0.002
2  28,500 9.982 
586 .5  503 .4  1886 .6  796 .7
x                             
2284 ,487 
= - 0.0014
568 ,974
Because x is negative, use Eq. 5.8.3.4.2-3:
Mu
 0.5 Nu  0.5Vu cot  A ps fpo
dv
x 
                  
2 E c A c  E s A s E p A ps
(LRFD 5.8.3.4.2-3)

Ac = Area of concrete on flexural tension side
= Area of girder below h/2 = (78.7+7+1)/2 = 43.35 IN
= (38.4)(5.3) + (5.5)(38.4+5.9)/2 + (43.35-5.3-5.9)(5.9) = 517.4 IN2
586 .5  503 .4  1886 .6       796 .7
x                                 
26062 517 .4   284 ,487  6,841,932 = -1.16x10
-4

v
From Table 5.8.3.4.2-1, with x = -1.16x10-4 &          = 0.125, find  = 21.5 and  = 3.03.

fc
That value is very close to the assumed value, so convergence has been achieved.

With these values, the concrete contribution, Vc, can now be computed.
VC  0.03163.03 10.0 KSI 5.9 IN 62.42 IN  = 111.5 KIP

11.1.5 Required Shear Reinforcement, Vs
Required Vs = Vu /  - Vc – Vp = 485.0 / 0.9 – 111.5 – 78.2 = 349.2 KIP
Assuming vertical stirrups,
A v f y d v cot 
Vs                                                                       (LRFD C5.8.3.3-1)
s
Compute Av on an IN2/FT basis (s = 12 IN):
12 Vs
Av 
fy dv cot 

Av 
12 IN 349.2 KIP 
75 KSI 62 .42 IN  cot21 .5 = 0.352 IN /FT
2

29
Check minimum transverse reinforcement:
bv s
A v  0.0316         
fc                                                             (LRFD 5.8.2.5)
fy

Av  0.0316          10 .00 KSI
5.90 IN 12 IN 
75 KSI        = 0.094 IN2/FT < 0.352 IN2/FT   O.K.

Check maximum stirrup spacing:                                              (LRFD 5.8.2.7-2)
Vu = 485.0 KIP > 0.1 f’c bv dv = (0.1) (5.9) (10.00) (62.42) = 368.3 KIP
Therefore, maximum WWR spacing is 12 IN.
For D18 WWR max spacing is:
s = 0.18*2*12/0.352 = 12.27 IN
Use D18 WWR @ 12 IN
Av,prov’d = 0.18(2)12/12 = 0.36 IN2/FT
The design of the other sections is in the attached Spread Sheet.

11.2 Longitudinal Reinforcement Requirement
In this example, the longitudinal reinforcement requirement will be
checked at the critical section for shear. The Specifications require that
this requirement must be satisfied at each section of the girder.
Therefore, in a full design, other sections along the length of the girder
would also have to be checked.

11.2.1 Required Longitudinal Force
Required Longitudinal Force:
M               N    V                     
Treqd =  u  0.5 u   u  0.5 Vs  Vp  cot
                                                         (LRFD 5.8.3.5-1)
 dv                              

However, at the inside edge of bearing at the simply-supported ends,
 Vu               
Treqd = 
         0.5 Vs  Vp  cot
                                                    (LRFD 5.8.3.5)
                 
where:
Vs = shear resistance provided by transverse reinforcement, not to exceed Vu / .
A v f y d v cot 
=                            (Use final values from shear design above)     (LRFD C5.8.3.3-1)
s
0.36 IN 75 KSI 62.42 IN  cot21.5
2

=                           12.0 IN                      = 356.5 KIP

 485                        
Treqd =        0.5 356 .5  78 .2  cot 21 .5 = (538.9 – 178.3 – 78.2) cot (21.5)
 0.9                        
= (282.5) (2.54) = 717.0 KIP

30
11.2.2 Available Longitudinal Force
The force to resist Treqd must be supplied by the reinforcement on the flexural tension
side of the member. In this case, the available reinforcement consists of the straight
strands. The available force that can be provided by these strands at the critical
section for shear must be determined considering the lack of full development due to
the proximity to the end of the girder.
The location at which T must be provided is where the failure crack assumed for this
analysis, which radiates from inside face of the support, crosses the centroid of the
straight strands. The angle  determined above during shear design at this location is
used here. The inside face of the support is 12 IN from the end of the girder.

Figure 11.1: Assumed Failure Crack and Location Where
Crack Crosses Straight Strands

The total effective prestress force for the straight strands is:
Pes = Aps fpe = 9.982 IN2 (172.0 KSI) = 1716.9 KIP
The distance from the bottom of the girder to the centroid of these strands is:
dg = c.g. straight strands = 4.09 IN

Measured from the end of the girder, the crack crosses the centroid of the straight
strands at:
x =  b  dg cot  = 12 IN + 4.09 IN (cot 21.5°) = 22.38 IN
This location is within the transfer length t , so the available stress is less than the
effective prestress force for the straight strands. The available prestress force, Tavail,
at x is therefore computed assuming a linear variation in stress from the end of the
girder to the transfer length. The transfer length, t , is 60 db or 36 IN.        (LRFD 5.11.4.1.)
Tavail = Pes x  t  = 1716.9 KIP (22.38 IN / 36 IN) = 1067.34 KIP
Since Tavail = 1067.34 KIP > Treqd = 717.0 KIP, the straight strands are adequate to
resist the required longitudinal force at this location and no additional reinforcement
is required.
would have been added to provide the remainder of the required force.

31
12. The design moment diagram M n and the factored moment Mu at 10th
points
30000

25000
 Mn
Bending Moment (K-ft)

20000

15000

Mu
10000

5000

0
0   0.1   0.2   0.3    0.4    0.5    0.6     0.7    0.8   0.9   1
Girder Sectionns X/L

32
13. Interface Shear Reinforcement
The girder will be designed for interface shear at the initial critical
section for shear.
The width of the shear interface is equal to the width of the top flange
of the girder, which is 48.20 IN. Therefore, bv = 48.20 IN.
Assume that the top surface of the girder was intentionally roughened
to an amplitude of 0.25 IN and cleaned prior to placement of the deck
concrete. The requirement for intentional roughening of the top of the
girder should be indicated on the plans.
Compute the factored horizontal shear, Vh:
Vh = Vu / de                                                             (LRFD C5.8.4.1-1)
The definition for de given for this equation is the same as dv.
Therefore use dv as computed above.
Vh = Vu / dv = 485 / 62.42 = 7.770 KIPS/IN
Since Vh   Vn and  = 0.9,
Vn reqd = Vh /  = 7.770 / 0.9 = 8.633 KIPS/IN
Check limits on Vn:
Vn  0.2 fcAcv or 0.8 Acv : 0.8 Acv is controlling the design
Acv = area of concrete engaged in shear transfer
= b v  = (48.2 IN) (1.0 IN) = 48.2 IN2
Use  = 1.0 IN to compute Vh on a per inch basis.
Vn reqd = 7.770 KIPS/IN  0.2 fcAcv = (0.8) (48.2) = 38.56 KIPS / IN      O.K.
Compute the nominal interface shear resistance, Vn:

Vn  c A cv   A vf f y  Pc                                                  (LRFD 5.8.4.1-1)
where:
c = 0.100 KSI and  = 1.000 for an intentionally roughened surface             (LRFD 5.8.4.2)
Avf = area of shear reinforcement crossing the shear plane
Pc = permanent net compressive force normal to the shear plane
= 0
Solve for the required Avf:
Vn reqd  c A cv
A vf 
f y

7.77  0.100 48 .2 
A vf 
1.00 60        = 0.049 IN2/IN or 0.59 IN2/FT

Minimum steel requirement:

33
s
A vf  0.05 b v                                                        (LRFD 5.8.4.1-4)
fy

Avf = (0.05) (48.2 IN) (12 IN) / 60 KSI = 0.482 IN2 / FT
Use 2 # 5 @ 12 IN (Av provd = 0.62 IN2/FT Say OK – Note that this limit depends directly on
the width of the interface – more steel required for a wider interface)

34
14. Design and detail the end zone reinforcement

14.1 Anchorage Zone Reinforcement
Article 5.10.10.1 requires that the factored bursting resistance of a pretensioned
anchorage zone be at least 4.0% of the total prestressing force. This resistance is
provided by vertical reinforcement close to the ends of pretensioned girders.
The factored bursting resistance is given by:
Pr = fs As                                                          (LRFD 5.10.10.1-1)
where:
Pr = (0.04) Po = (0.04) [(0.75) (270 KSI) (17.794IN2)] = 144.13 KIP
Note: The total jacking force prior to any losses is used as the total prestressing
force Po in this calculation:
fs is the working stress in the reinforcement, not to exceed 20 KSI
Solving for the required area of reinforcement, As:
P 144 .13 KIP
As  r                           2
fs    20 KSI  = 7.20IN
Therefore, at least 6.00 IN2 of vertical reinforcement must be placed within h/5 = 78.7
IN / 5 = 15.7 IN from the end of the member. Stirrups placed for vertical or interface
shear can also be used to satisfy this requirement since this reinforcement is only
required to resist forces at release.
Take # 7 stirrups @ 3 inches spacing
provided
As         6(2)0.6  7.20 IN 2
Note: The first spacing from the end of the girder to the first stirrup is 3 inches as
well.

14.2 Confinement Reinforcement
In accordance with Article 5.10.10.2, confinement reinforcement not less than # 3
bars at a spacing of not more than 6.0 IN shall be placed within 1.5 d (say 1.5 h =
9.00 FT) from the end of the girder. These bars shall be shaped to enclose the
strands.

35
15 and 16 - Mid-Span Camber at release and erection and live load deflection

Curvature Values at mid-span
fpi A ps  einitial
Due to Initial prestress                    φ                                      = 2.24E-05
E ciIinitial
 Mgdr
Due to member weight                        φ                                      =-1.23E-05
E ciIinitial
Δfps A ps  e final
Due to loss (from initial to erection)      φ                                      =-2.23E-06
EciIfinal
 M deck
E c I final
 Mdl
E cIfinal composite
Δfps 2 A ps  e final composite
Due to prestress loss (erection to final)   φ                                      =-1.39E-06
E cIfinal compositel
 M ll
Due to Live load                                                                  =-1.21E-05
E c Ifinal com posite

36
Table 5-1 Approximate Camber and Deflection Calculation
Elastic Curvature           Elastic                           Initial         Erection 
It is valid for prestressing with straight, one-poind draped
Midspan              End                                                                                     and two-point draped strands
Initial prestress                    2.24E-05           1.11E-05           1.55E+01            1.55E+01        2.70E+01                                                      L2                L2
Member weight                       -1.23E-05          0.00E+00           -7.33E+00           -7.33E+00       -1.28E+01                       due to prestress    Δ  φc         (φ e  φ c )
Loss (initial to erection)          -2.23E-06          -1.11E-06          -1.55E+00          ------           -2.35E+00                                                      8                 6
Pe
Loss (erection to final)            -1.39E-06          -9.85E-07           -8.96E-01         ------            ------                         a = distance from end to drape point
Live load                           -1.21E-05          0.00E+00          ------              ------            ------
5ML2
Total                                                                                               8.18             11.89                    due to distributed load           Δ
48 EI

Table 5-2 Accurate Camber and Deflection Calculation
Curvature                                                           Elastic              Initial         Erection 
End               0.1L                0.2L               0.3L             0.4L            Midspan
Initial prestress                    1.11E-05           1.50E-05            1.86E-05           2.11E-05         2.24E-05           2.24E-05          1.48E+01             1.48E+01              2.57E+01
Member weight                       0.00E+00           -4.65E-06           -7.09E-06          -1.05E-05        -1.18E-05          -1.23E-05         -7.20E+00            -7.20E+00             -1.25E+01
Loss (initial to erection)          -1.11E-06          -1.50E-06           -1.92E-06          -2.14E-06        -2.26E-06          -2.23E-06         -1.49E+00           ------                 -2.27E+00
Dead load on precast                0.00E+00           -3.75E-06           -6.55E-06          -8.48E-06        -9.61E-06          -1.00E-05         -5.94E+00           ------                ------
Dead load on composite              0.00E+00           -1.48E-06           -2.58E-06          -3.34E-06        -3.79E-06          -3.95E-06         -2.34E+00           ------                ------
Loss (erection to final)            -9.85E-07          -1.24E-06           -1.44E-06          -1.46E-06        -1.44E-06          -1.39E-06         -1.01E+00           ------                ------
Live load                          0.00E+00            -4.56E-06           -8.03E-06          -1.04E-05        -1.18E-05          -1.21E-05         -7.24E+00           ------                ------
Total                                                                                                                                                                          7.56                  10.90

The formula for deflection at midspan as a function of curvatures at (span/10) points, for simple spans and for interior spans.
It may be used for end spans without much loss of accuracy. Symmetrical moment diagram about midspan is assumed) .
1                                 7
Δ  0 .01L2 ( Φ 0  Φ 1  2Φ 2  3 Φ 3  4Φ 4  Φ 5 )
6                                 3

37

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