Precast/Prestressed Girder Help

Tutorial 1: Continuous 3-Span 72 Inch Bulb Tee Girder Design (U.S. Units) (LFD)

Concrete Properties

Art.8.7.1

E c = w c 1.5 ( 33 ) ( f ` c ) = 150 1.5 ( 33 ) 6000 = 4696000 p s i = 4696.0 k s i
E c t = 150 1.5 ( 33 ) 3500 = 35866000 p s i = 3586.6 k s i

Girder Line Data

T-Trib=Tributary width of beam 1

Tributary Width of Beams

T-Trib=Tributary width of beam 1

T T r i b 1 = 4.25 + 8.5 ÷ 2 = 8.50 f t

Continuity analysis

W1= Magnitude of total superimposed dead load

W 1 = ( 30.5 p s f ) ( 51 f t ) = 1555.5 p l f = 1.556 k l f

Model Used by Precast/Prestressed Girderfor SDL Continuity Analysis

Results of independent frame analysis program:

  Independent Program Precast/Prestressed Girder
M (Span 1, 58.5ft) 1597.0 k-ft 1597.0 k-ft
V (Span 1, 58.5ft) -18.2k -18.2k
Recation 1 -72.8k -72.8k
Reaction 2 -200.2k -200.2k

Moving Load Analysis

Position of Truck to Cause Maximum Positive Moment at Span 1: 46.9 ft.

Position of Lane Load to Cause Mmax(-) at Span 1: 117.00 ft.

Results of independent frame analysis program:

  Independent Program Precast/Prestressed Girder
M (Span 1, 58.5ft) 1597.0 k-ft 1597.0 k-ft
V (Span 1, 58.5ft) -18.2k -18.2k
Recation 1 -72.8k -72.8k
Reaction 2 -200.2k -200.2k

Position of Lane Load to Cause Vmax at Span 1:117.00 ft.

Position of Lane Load to Cause Mmax(-) at Span 1: 117.00 ft.

Position of Lane Load to Cause Vmax at Span 1: 117.00 ft.

Note: Concentrated load is slightly to the left of the support.

Results of independent frame analysis program:

  Independent Program Precast/Prestressed Girder
Mmax(+) @46.90 ft 1827.7 k-ft 1830.1 k-ft
Mmax(-) @ Support 2 1757.7 k-ft 1757.8 k-ft
Vmax @ Support 2 90.2 k 90.2 k

Effective Width

Effective width Art 8.10.1.1

  1. L / 4 = 115.0 f t / 4 = 28.75 f t = 345.0 i n
  2. Beam spacing= 8.50ft= 102.0in(controls)
  3. *Web width + (12)(Flange thickness + Haunch thickness + Topping thickness)=6in+(12)(2 in+ 0.50in+ 8.00in)=132.0in
  4. *Precast flange width + (12)(Topping thickness)=42in+(12)(8in)=138.0in
Note: *See Theory section for details.

Effective width = 102.0 in

Impact Factor

Eq.3-1 I = 50 L + 125

Impact (moment):

1 + I = 1 + 50 115.0 + 125 = 1.208

Distribution Factor

Wheel Load Distribution Factor

table 3.23.1 S 5.5 = 8.5 5.5 = 1.5455

Precast/Prestressed Girder performs a moving live load analysis using a full truck and lane load (two wheel lines). Therefore, for consistency, the distribution factor must be divided by two.

DF: 1.5455/2=0.773

Section Properties

Self-weight (Precast)=(A)(Density Concrete)

767 i n 2 144 i n 2 / f t 2 × ( 150 p c f ) = 799.0 p l f

Self-weight (Composite)

799.0 p l f + 42 × 8.5 + 60 × 8 144 i n 2 / f t 2 × ( 150 p c f ) = 1670.0 p l f

Transformed widths of slab parts:

Transformed topping width = (0.7638)(42 in) = 32.078in

Transformed gap width = (0. 7638)(60 in) = 45.825 in

Element Width Thick A Yb Ayb I
Topping 32.087 8.5 272.663 76.25 20791 1642
Gap 45.825 8.5 366.603 76.50 28045 1955
Beam     767.000 36.60 28072 545894
Total     1406   76908 549491

H t o f c . g . = A y b A = 76908 1406 = 54.70 i n

Moment of Inertia:

Σ I + Σ ( y b c y b ) 2 A = 549491 + ( 272.663 ) ( 54.70 76.25 ) 2 2 + ( 366.603 ) ( 54.70 76.25 ) 2 2 + ( 767.0 ) ( 54.70 36.6 ) 2 2 = 1101619 i n 4

Tributary Fraction

DL-Comp Trib Frac = Fraction of total moments and shears of SDL continuity analysis

DL-Comp Trib Frac = T r i b u t a r y W i d t h B r i d g e W i d t h = 8.5 51.00 = 0.167

Tributary Fraction

DL-Comp Trib Frac = Fraction of total moments and shears of SDL continuity analysis

DL-Comp Trib Frac = T r i b u t a r y W i d t h B r i d g e W i d t h = 8.5 51.00 = 0.167

Miscellaneous

Kern point:

U p p e r = k u = y b + S b A = 36.60 + 545894 i n 4 36.60 i n 767.0 i n 4 = 56.05 i n
L o w e r = k 1 = y b + S t A = 36.60 + 545894 i n 4 ( 72 i n 36.60 i n ) 767.0 i n 4 = 16.49 i n

Prestressed Steel

Depressed at 0.4L:

0.4 × 116.0 f t = 46.4 f t f r o m b e a m e n d
46.4 f t 0.5 f t = 45.9 f t f r o m sup p o r t

Total strand area = (38)(0.153 in2) = 5.814 in2

Initial pull = (5.814 in2)(0.75)(270 ksi) = 1178.1 k

Trans len= Transfer length Art. 9.20.2.4

= 50 × ( s t r a n d d i a m e t e r s ) = 50 × 0.500 i n = 25.00 i n = 2.083 f t

Dev. len = Development length of strands at ultimate Eq.9-32

= ( f s u 2 3 f s e ) D = f s u = 264.8 k s i
= f s e = 0.75 × 270 k s i × ( 1 0.215 ) = 159.0 k s i

Dev. len

( 264.8 2 3 ( 159.0 ) ) ( 0.5 ) = 79.41 i n = 6.62 f t