Precast Concrete Bridge Beam Design

Prestressed Concrete Beam Design to BS 5400 Part 4

Problem:

Design a simply supported prestressed concrete Y beam which carries a 150mm thick concrete slab and 100mm of surfacing, together with a nominal live load udl of 10.0 kN/m² and kel of 33kN/m . The span of the beam is 24.0m centre to centre of bearings and the beams are spaced at 1.0m intervals.

γ_conc. = 24kN/mm³

25 units of HB to be considered at SLS for load combination 1 only (BS 5400 Pt4 Cl. 4.2.2)

Loading per beam (at 1.0m c/c)

Nominal Dead Loads :

slab = 24 x 0.15 x 1.0

= 3.6 kN/m

beam = say Y5 beam

= 10.78 kN/m

surfacing = 24 x 0.1 x 1.0

= 2.4 kN/m

Nominal Live Load :

HA = 10 x 1.0 + 33.0

= 10 kN/m + 33kN

25 units HB = 25 x 10 / 4 per wheel

= 62.5 kN per wheel

Load factors for serviceability and ultimate limit state from BS 5400 Part 2 (or BD 37/01) Table 1:

SLS

ULS

Comb.1

Comb.3

Comb.1

Comb.3

Dead Load

γ_fL concrete

1.0

1.15

Superimposed Dead Load

γ_fL surfacing

1.2

1.75

Live Load

γ_fL HA

1.2

1.0

1.5

1.25

γ_fL HB

1.1

Temperature Difference

γ_fL

0.8

1.0

Concrete Grades
Beam f_cu = 50 N/mm², f_ci = 40 N/mm²
Slab f_cu = 40 N/mm²

BS 5400 Pt. 4

Section Properties

cl.7.4.1

Modular ratio effect for different concrete strengths between beam and slab may be ignored.

Property

Beam Section

Composite Section

Area(mm²)

449.22x10³

599.22x10³

Centroid(mm)

456

623

2nd Moment of Area(mm⁴)

52.905x10⁹

103.515x10⁹

Modulus @ Level 1(mm³)

116.020x10⁶

166.156x10⁶

Modulus @ Level 2(mm³)

89.066x10⁶

242.424x10⁶

Modulus @ Level 3(mm³)

179.402x10⁶

Temperature Difference Effects

Apply temperature differences given in BS 5400 Pt2 Fig.9 (Group 4)to a simplified beam section.
Cl. 5.4.6 - Coefficient of thermal expansion = 12 x 10^-6 per 慢.
From BS 5400 Pt4 Table 3 : E_c = 34 kN/mm² for f_cu = 50N/mm²
Hence restrained temperature stresses per 蚓 = 34 x 10³ x 12 x 10^-6 = 0.408 N/mm²

a) Positive temperature difference

Force F to restrain temperature strain :
0.408 x 1000 x [ 150 x ( 3.0 + 5.25 ) ] x 10^-3 +
0.408 x ( 300 x 250 x 1.5 + 750 x 200 x 1.25 ) x 10^-3 = 504.9 + 122.4 = 627.3 kN

Moment M about centroid of section to restrain curvature due to temperature strain :
0.408 x 1000 x [ 150 x ( 3.0 x 502 + 5.25 x 527 ) ] x 10^-6 +
0.408 x ( 300 x 250 x 1.5 x 344 - 750 x 200 x 1.25 x 556 ) x 10^-6 = 261.5 - 26.7 = 234.8 kNm

b) Reverse temperature difference

Force F to restrain temperature strain :
- 0.408 x [ 1000 x 150 x ( 3.6 + 2.3 ) + 300 x 90 x ( 0.9 + 1.35 ) ] x 10^-3
- 0.408 x 300 x ( 200 x 0.45 + 150 x 0.45 ) x 10^-3
- 0.408 x 750 x [ 50 x ( 0.9 + 0.15 ) + 240 x ( 1.2 + 2.6 ) ] x 10^-3 = - 385.9 - 19.3 - 295.1 = - 700.3 kN

Moment M about centroid of section to restrain curvature due to temperature strain :
- 0.408 x [ 150000 x ( 3.6 x 502 + 2.3 x 527 ) + 27000 x ( 0.9 x 382 + 1.35 x 397 ) ] x 10^-6
- 0.408 x 300 x ( 200 x 0.45 x 270 - 150 x 0.45 x 283 ) x 10^-6
+ 0.408 x 750 x [ 50 x ( 0.9 x 358 + 0.15 x 366 ) + 240 x ( 1.2 x 503 + 2.6 x 543 ) ] x 10^-6
= - 194.5 - 0.6 + 153.8 = - 41.3 kNm

Differential Shrinkage Effects

BS 5400 Pt.4 cl.7.4.3.4

Use cl.6.7.2.4 Table 29 :
Total shrinkage of insitu concrete = 300 x 10^-6
Assume that 2/3 of the total shrinkage of the precast concrete takes place before the deck slab is cast and that the residual shrinkage is 100 x 10^-6 ,
hence the differential shrinkage is 200 x 10^-6

BS 5400 Pt.4 cl.7.4.3.5

Force to restrain differential shrinkage : F = - ε_diff x E_cf x A_cf x φ
F = -200 x 10^-6 x 34 x 1000 x 150 x 0.43 = -439 kN
Eccentricity a_cent = 502mm
Restraint moment M_cs = -439 x 0.502 = -220.4 kNm

Self weight of beam and weight of deck slab is supported by the beam. When the deck slab concrete has cured then any further loading (superimposed and live loads) is supported by the composite section of the beam and slab.

Dead Loading (beam and slab)

Total load for serviceability limit state =

(1.0 x 3.6)+(1.0 x 10.78) = 14.4kN/m

Design serviceability moment =

14.4 x 24² / 8 = 1037 kNm

Combination 1 Loading

Super. & HA live load for SLS =

[(1.2 x 2.4)+(1.2 x 10)]udl & [(1.2 x 33)]kel

(2.88 + 12.0)udl & 39.6kel

14.9 kN/m & 39.6kN

Super. & HB live load for SLS =

2.88 & 4 wheels @ 1.1 x 62.5

2.9 kN/m & 4 wheels @ 68.75 kN

Total load for ultimate limit state =

[(1.15 x 3.6)+(1.15 x 10.78)+(1.75 x 2.4)+(1.5 x 10)]udl & [(1.5 x 33)]kel

(4.14 + 12.40 + 4.20 + 15.0)udl & 49.5kel

35.7 kN/m & 49.5kN

HA Design serviceability moment =

14.9 x 24.0² / 8 + 39.6 x 24 / 4

1310 kNm

25 units HB Design SLS moment =

2.9 x 24.0² / 8 + 982.3(from grillage analysis)

1191.1 kNm

Design ultimate moment =

35.7 x 24.0² / 8 + 49.5 x 24 / 4

2867 kNm

Combination 3 Loading

Super. & HA live load for SLS =

[(1.2 x 2.4)+(1.0 x 10)]udl & [(1.0 x 33)]kel

(2.88 + 10.0)udl & 33kel

12.9 kN/m & 33kN

Total load for ultimate limit state =

[(1.15 x 3.6)+(1.15 x 10.78)+(1.75 x 2.4)+(1.25 x 10)]udl & [(1.25 x 33)]kel

(4.14 + 12.40 + 4.20 + 12.5)udl & 41.3kel

33.2 kN/m & 41.3kN

Design serviceability moment =

12.9 x 24.0² / 8 + 33 x 24 / 4

1127 kNm

Allowable stresses in precast concrete

At transfer :

cl.6.3.2.2 b)

Compression ( Table 23 )
0.5f_ci (<=0.4f_cu) = 20 N/mm² max.

cl.6.3.2.4 b)

Tension = 1.0 N/mm²

At serviceability limit state :

cl.7.4.3.2

Compression (1.25 x Table 22)
1.25 x 0.4f_cu = 25 N/mm²

Tension = 0 N/mm² (class 1) & 3.2 N/mm² (class 2 - Table 24)

Stresses at Level 1 due to SLS loads (N/mm²) :

Comb.1
(HA)

Comb.1
(HB)

Comb.3

Dead Load M / Z = (1037 x 10⁶) / (116.020 x 10⁶)

- 8.94

Super. & Live Load M / Z = M / (166.156 x 10⁶)

- 7.88

- 7.17

- 6.78

Reverse Temperature = γ_fL x -1.69 = 0.8 x -1.69

-1.35

Differential shrinkage

-0.60

Total Stress at Level 1 =

-17.42

-16.71

-17.67*

Hence Combination 3 is critical

Prestressing Force and Eccentricity

Using straight, fully bonded tendons (constant force and eccentricity).
Allow for 20% loss of prestress after transfer.
Initial prestress at Level 1 to satisfy class 2 requirement for SLS (Comb. 3).

Stress at transfer = ( 17.67 - 3.2 ) / 0.8 = 18.1 N/mm² (use allowable stress of 20 N/mm²)

The critical section at transfer occurs at the end of the transmission zone. The moment due to the self weight at this section is near zero and initial stress conditions are:

P/A + Pe/Z_{level 1} = 20

.....................(eqn. 1)

P/A - Pe/Z_{level 2} >= - 1.0

.....................(eqn. 2)

(eqn. 1) x Z_{level 1} + (eqn. 2) x Z_{level 2} gives :

P >= A x (20 x Z_{level 1} - 1.0 x Z_{level 2}) / (Z_{level 1} + Z_{level 2})

P = 449.22 x 10³ x ( 20 x 116.02 - 89.066) / ( 116.02 + 89.066) x 10^-3 = 4888 kN

Allow 10% for loss of force before and during transfer, then the initial force P_o = 4888 / 0.9 = 5431kN

Using 15.2mm class 2 relaxation standard strand at maximum initial force of 174kN (0.75 x P_u)
Area of tendon = 139mm²
Nominal tensile strength = f_pu =1670 N/mm²
Hence 32 tendons required.
Initial force Po = 32 x 174 = 5568 kN
P = 0.9 x 5568 = 5011 kN

Substituting P = 5011 kN in (eqn. 2)

e <= Z_{level 2} / A + Z_{level 2} / P = (89.066 x 10⁶ / 449.22 x 10³) + (89.066 x 10⁶ / 5011 x 10³)

e = 198 + 18 = 216 mm

Arrange 32 tendons symmetrically about the Y-Y axis to achieve an eccentricity of about 216mm.

Taking moments about bottom of beam :

2 @

1000 =

2000

2 @

900 =

1800

4 @

260 =

1040

8 @

160 =

1280

10 @

110 =

1100

6 @

60 =

360

7580

e = 456 - 7580 / 32 = 456 - 237 = 219mm

Allowing for 1% relaxation loss in steel before transfer and elastic deformation of concrete at transfer :

cl. 6.7.2.3

P = 0.99 P_o / [ 1 + E_s x (A_ps / A) x (1 + A x e² / I) / E_ci ]

P = 0.99 x Po / [ 1 + 196 x ( 32 x 139 / 449220) x (1 + 449220 x 219² / 52.905 x 10⁹) / 31 ]

P = 0.91 Po = 0.91 x 5568 = 5067 kN

Initial stresses due to prestress at end of transmission zone :

Level 1 : P / A x ( 1 + A x e / Z_{level 1} ) = 11.3 x ( 1 + 219 / 258 ) = 20.89 N/mm²

Level 2 : P / A x ( 1 - A x e / Z_{level 2} ) = 11.3 x ( 1 - 219 / 198 ) = - 1.20 N/mm²

Moment due to self weight of beam at mid span = 10.78 x 24² / 8 = 776.2 kNm

Stress due to self weight of beam at mid span :
@ Level 1 = - 776.2 / 116.02 = - 6.69 N/mm²
@ Level 2 = 776.2 / 89.066 = 8.71 N/mm²

Initial stresses at mid span :

cl. 6.7.2.5

Allowing for 2% relaxation loss in steel after transfer, concrete shrinkage ε_cs = 300 x 10^-6
and concrete specific creep c_t = 1.03 x 48 x 10^-6 per N/mm²
Loss of force after transfer due to :

cl. 6.7.2.2

Steel relaxation = 0.02 x 5568 = 111

cl. 6.7.2.4

Concrete shrinkage = (ε_cs x E_s x A_ps ) = 300 x 10^-6 x 196 x 32 x 139 = 262

cl. 6.7.2.5

Concrete creep = ( c_t x f_co x E_s x A_ps ) = 1.03 x 48 x 10^-6 x 12.76 x 196 x 32 x 139 = 550

Total Loss = 111 + 262 + 550 = 923 kN

Final force after all loss of prestress = P_e = 5067 - 923 = 4144 kN (P_e/P = 0.82)

Final stresses due to prestress after all loss of prestress at :

Level 1 f_1,0.82P = 0.82 x 20.89 = 17.08 N/mm²

Level 2 f_2,0.82P = 0.82 x - 1.20 = - 0.98 N/mm²

Combined stresses in final condition for worst effects of design loads, differential shrinkage and temperature difference :

Level 1, combination 1 HB : f = 17.08 - 16.71 = 0.37 N/mm² (> 0 hence O.K.)

Level 1, combination 3 : f = 17.08 - 17.67 = - 0.59 N/mm² (> - 3.2 hence O.K.)

Level 2, combination 1 : f = - 0.98 + 1037 / 89.066 + 1310 / 242.424 + 1.64 = 17.71 (< 25 O.K.)

Level 3. combination 3 : f = (1127 / 179.402) + (0.8 x 3.15) = 8.8 N/mm² (< 25 O.K.)

Ultimate Capacity of Beam and Deck Slab
(Composite Section)

Ultimate Design Moment = γ_f3 x M = 1.1 x 2867 = 3154 kNm

cl. 6.3.3

Only steel in the tension zone is to be considered :
Centroid of tendons in tension zone = (6x60 + 10x110 + 8x160 + 4x260) / 28 = 135mm
Effective depth from Level 3 = 1200 - 135 = 1065mm

Assume that the maximum design stress is developed in the tendons, then :
Tensile force in tendons Fp = 0.87 x 28 x 139 x 1670 x 10-3 = 5655 kN

Compressive force in concrete flange :
F_f= 0.4 x 40 x 1000 x 150 x 10^-3 = 2400 kN

Let X = depth to neutral axis.
Compressive force in concrete web :
F_w = 0.4 x 50 x [393 - (393 - 200) x (X - 150) / (671 x 2)] x (X - 150) x 10^-3
F_w = ( -2.876X² + 8722.84X - 1243717) x 10^-3
Equating forces to obtain X :
5655 = 2400 + ( -2.876X² + 8722.84X - 1243717) x 10^-3
X = 659 mm

Stress in tendon after losses = f_pe = 4144 x 10³ / (32 x 139) = 932 N/mm²
Prestrain ε_pe = f_pe / E_s = 932 / 200 x 10³ = 0.0047

Determine depth to neutral axis by an iterative strain compatibility analysis
Try X = 659 mm as an initial estimate
Width of web at this depth = 247mm

ε_pb6 = ε₆ + ε_pe = -459 x 0.0035 / 659 + 0.0047 = 0.0022

ε_pb5 = ε₅ + ε_pe = -359 * 0.0035 / 659 + 0.0047 = 0.0028

ε_pb4 = ε₄ + ε_pe = 281 * 0.0035 / 659 + 0.0047 = 0.0062

ε_pb3 = ε₃ + ε_pe = 381 * 0.0035 / 659 + 0.0047 = 0.0067

ε_pb2 = ε₂ + ε_pe = 431 * 0.0035 / 659 + 0.0047 = 0.0069

ε_pb1 = ε₁ + ε_pe = 481 * 0.0035 / 659 + 0.0047 = 0.0072

f_pb6 = 0.0022 x 200 x 10³ = 444 N/mm²

f_pb5 = 0.0028 x 200 x 10³ = 551 N/mm²

f_pb4 = 1162 + 290 x (0.0062 - 0.0058) / 0.0065 = 1178 N/mm²

f_pb3 = 1162 + 290 x (0.0067 - 0.0058) / 0.0065 = 1201 N/mm²

f_pb2 = 1162 + 290 x (0.0069 - 0.0058) / 0.0065 = 1213 N/mm²

f_pb1 = 1162 + 290 x (0.0072 - 0.0058) / 0.0065 = 1225 N/mm²

Tensile force in tendons :

F_p6 = 2 x 139 x 444 x 10^-3

= 124

F_p5 = 2 x 139 x 551 x 10^-3

= 153

F_p4 = 4 x 139 x 1178 x 10^-3

= 655

F_p3 = 8 x 139 x 1201 x 10^-3

= 1336

F_p2 = 10 x 139 x 1213 x 10^-3

= 1686

F_p1 = 6 x 139 x 1225 x 10^-3

= 1022

4976 kN

Compressive force in concrete :

F_f = 0.4 x 40 x 1000 x 150 x 10^-3

= 2400

F_w = 0.4 x 50 x 0.5 x (393 + 247) x (659 - 150) x 10^-3

= 3258

5658 kN

Fc > Ft therefore reduce depth to neutral axis and repeat the calculations.
Using a depth of 565mm will achieve equilibrium.
The following forces are obtained :

F_p6 = 134

F_f = 2400

F_p5 = 168

F_w = 2765

F_p4 = 675

F_c = 5165

F_p3 = 1382

F_p2 = 1746

F_p1 = 1060

F_t = 5165

Taking Moments about the neutral axis :

F_p6 = 134 x -0.365 =

-49

F_p5 = 168 x -0.265 =

-45

F_p4 = 675 x 0.375 =

253

F_p3 = 1382 x 0.475 =

656

F_p2 = 1746 x 0.525 =

917

F_p1 = 1060 x 0.575 =

610

F_f = 2400 x 0.49 =

1176

F_w = 3258 x 0.207 =

674

M_u =

4192 kNm > 3154 kNm hence O.K.

cl. 6.3.3.1

Mu / M = 4192 / 3154 = 1.33 ( > 1.15 ) hence strain in outermost tendon O.K.