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Prestressed Concrete Beam Example to British Standards

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/m^{2} 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/m^{3}

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)

Note: The loading has been simplified to demonstrate the method of designing the beam (See BS 5400 Pt2, or DB 37/01 for full design loading)

Nominal Dead Loads :

slab = 24 × 0.15 × 1.0 = 3.6 kN/m

beam = say Y5 beam = 10.78 kN/m

surfacing = 24 × 0.1 × 1.0 = 2.4 kN/m

Nominal Live Load :

HA = 10 × 1.0 + 33.0 = 10 kN/m + 33kN

25 units HB = 25 × 10 / 4 per wheel = 62.5 kN per wheel

Load factors for serviceability and ultimate limit state from BS 5400 Part 2 Table 1:

SLS

ULS

Comb 1

Comb 3

Comb 1

Comb 3

Dead Load

γ_{fL} concrete

1.0

1.0

1.15

1.15

Superimposed Dead Load

γ_{fL} surfacing

1.2

1.2

1.75

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 C40/50 f_{cu} = 50 N/mm^{2}, f_{ci} = 40 N/mm^{2}

Slab C32/40 f_{cu} = 40 N/mm^{2}

BS 5400 Pt. 4

Section Properties

cl.7.4.1

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

Section properties of Y5 beam were obtained from a
Paper by Taylor, Clark and Banks

Note: Level 2 is at the nib level and not at the top of the beam.

__Property__

__Beam Section__

__Composite Section__

Area(mm^{2})

449.22×10^{3}

599.22×10^{3}

Centroid(mm)

456

623

2nd Moment of Area(mm^{4})

52.905×10^{9}

103.515×10^{9}

Modulus @ Level 1(mm^{3})

116.020×10^{6}

166.156×10^{6}

Modulus @ Level 2(mm^{3})

89.066×10^{6}

242.424×10^{6}

Modulus @ Level 3(mm^{3})

—

179.402×10^{6}

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 × 10^{-6} per °C.

From BS 5400 Pt4 Table 3 : E_{c} = 34 kN/mm^{2} for f_{cu} = 50N/mm^{2}

Hence restrained temperature stresses per °C = 34 × 10^{3} × 12 × 10^{-6} = 0.408 N/mm^{2}

a) __Positive temperature difference __

Force F to restrain temperature strain :

0.408 × 1000 × [ 150 × ( 3.0 + 5.25 ) ] × 10^{-3} +
0.408 × ( 300 × 250 × 1.5 + 750 × 200 × 1.25 ) × 10^{-3} =
504.9 + 122.4 = 627.3 kN

Moment M about centroid of section to restrain
curvature due to temperature strain :

0.408 × 1000 × [ 150 × ( 3.0 × 502 + 5.25 × 527 ) ] × 10^{-6} +
0.408 × ( 300 × 250 × 1.5 × 344 - 750 × 200 × 1.25 × 556 ) × 10^{-6}
= 261.5 - 26.7 = 234.8 kNm

b) __Reverse temperature difference__

Force F to restrain temperature strain :

- 0.408 × [ 1000 × 150 × ( 3.6 + 2.3 ) + 300 × 90 × ( 0.9 + 1.35 )
] × 10^{-3}

- 0.408 × 300 × ( 200 × 0.45 + 150 × 0.45 ) × 10^{-3}

- 0.408 × 750 × [ 50 × ( 0.9 + 0.15 ) + 240 × ( 1.2 + 2.6 ) ] × 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 × [ 150000 × ( 3.6 × 502 + 2.3 × 527 ) + 27000 × ( 0.9 ×
382 + 1.35 × 397 ) ] × 10^{-6}

- 0.408 × 300 × ( 200 × 0.45 × 270 - 150 × 0.45 × 283 ) × 10^{-6}

+ 0.408 × 750 × [ 50 × ( 0.9 × 358 + 0.15 × 366 ) + 240 × ( 1.2 ×
503 + 2.6 × 543 ) ] × 10^{-6}

= - 194.5 - 0.6 + 153.8 = - 41.3 kNm

BS 5400 Pt.4 cl.7.4.3.4

Use cl.6.7.2.4 Table 29 :

Total shrinkage of insitu concrete = 300 × 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 × 10^{-6} ,

hence the differential shrinkage is 200 × 10^{-6}

BS 5400 Pt.4 cl.7.4.3.5

Force to restrain differential shrinkage : F = - ε_{diff}
× E_{cf} × A_{cf} × φ

F = -200 × 10^{-6} × 34 × 1000 × 150 × 0.43 = -439 kN

Eccentricity a_{cent} = 502mm

Restraint moment M_{cs} = -439 × 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 × 3.6)+(1.0 × 10.78) = 14.4kN/m

Design serviceability moment = 14.4 × 24^{2} / 8 = **1037 kNm**

Combination 1 Loading

Super. & HA live load for SLS:

= [(1.2 × 2.4)+(1.2 × 10)]udl & [(1.2 × 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 × 62.5

= **2.9 kN/m** & 4 wheels @ 68.75 kN

Total load for ultimate limit state:

= [(1.15 × 3.6)+(1.15 × 10.78)+(1.75 × 2.4)+(1.5 × 10)]udl & [(1.5 × 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 × 24.0^{2} / 8 + 39.6 × 24 / 4

= **1310 kNm**

25 units HB Design SLS moment:

= 2.9 × 24.0^{2} / 8 + 982.3(from grillage analysis)

= **1191.1 kNm**

Design ultimate moment:

= 35.7 × 24.0^{2} / 8 + 49.5 × 24 / 4

= **2867 kNm**

Combination 3 Loading

Super. & HA live load for SLS:

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

= (2.88 + 10.0)udl & 33kel

= **12.9 kN/m & 33kN**

Total load for ultimate limit state:

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

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

= **33.2 kN/m & 41.3kN**

Design serviceability moment:

= 12.9 × 24.0^{2} / 8 + 33 × 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/mm^{2} max.

cl.6.3.2.4 b)

Tension = 1.0 N/mm^{2}

At serviceability limit state :

cl.7.4.3.2

Compression (1.25 × Table 22)

1.25 × 0.4f_{cu} = 25 N/mm^{2}

Tension = 0 N/mm^{2} (class 1) & 3.2 N/mm^{2} (class 2 - Table 24)

Stresses at Level 1 due to SLS loads (N/mm^{2}) :

Comb 1

(HA)

Comb 1

(HB)

Comb 3

Dead Load M / Z = (1037 × 10^{6}) / (116.020 × 10^{6})

-8.94

-8.94

-8.94

Super. & Live Load M / Z = M / (166.156 × 10^{6})

-7.88

-7.17

-6.78

Reverse Temperature = γ_{fL} × -1.69 = 0.8 × -1.69

—

—

-1.35

Differential shrinkage

__-0.60__

__-0.60__

__-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/mm^{2} (use allowable stress of 20 N/mm^{2})

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

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

P = 449.22 × 10^{3} × ( 20 × 116.02 - 89.066) / ( 116.02 + 89.066) × 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 × P_{u})

Area of tendon = 139mm^{2}

Nominal tensile strength = f_{pu} =1670 N/mm^{2}

Hence 32 tendons required.

Initial force Po = 32 × 174 = **5568 kN
**P = 0.9 × 5568 =

**5011 kN**

Substituting P = 5011 kN in (eqn. 2)

e <= Z_{level 2} / A + Z_{level 2} / P = (89.066 × 10^{6} / 449.22 × 10^{3}) + (89.066 × 10^{6} / 5011 × 10^{3})

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__

32

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} × (A_{ps} / A) × (1 + A × e^{2} / I) / E_{ci} ]

P = 0.99 × Po / [ 1 + 196 × ( 32 × 139 / 449220) × (1 + 449220 × 219^{2} / 52.905 × 10^{9}) / 31 ]

P = 0.91 Po = 0.91 × 5568 = **5067 kN**

Initial stresses due to prestress at end of transmission zone :

Level 1 : P / A × ( 1 + A × e / Z_{level 1} ) = 11.3 × ( 1 + 219 / 258 ) = 20.89 N/mm^{2}

*(20.89 N/mm ^{2} is slightly greater than the allowable of 20 N/mm^{2} so a number of tendons will need to be debonded near the ends of the beam)*.

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

Moment due to self weight of beam at mid span = 10.78 × 24^{2} / 8 = 776.2 kNm

Stress due to self weight of beam at mid span :

@ Level 1 = - 776.2 / 116.02 = - 6.69 N/mm^{2}

@ Level 2 = 776.2 / 89.066 = 8.71 N/mm^{2}

Initial stresses at mid span :

cl. 6.7.2.5

Allowing for 2% relaxation loss in steel after transfer,

concrete shrinkage ε_{cs} = 300 × 10^{-6}

and concrete specific creep c_{t} = 1.03 × 48 × 10^{-6} per N/mm^{2}

Loss of force after transfer due to :

cl. 6.7.2.2

Steel relaxation = 0.02 × 5568 = 111

cl. 6.7.2.4

Concrete shrinkage = (ε_{cs} × E_{s} × A_{ps} ) = 300 × 10^{-6} × 196 × 32 × 139 = 262

cl. 6.7.2.5

Concrete creep = ( c_{t} × f_{co} × E_{s} × A_{ps} ) = 1.03 × 48 × 10^{-6} × 12.76 × 196 × 32 × 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 × 20.89 = 17.08 N/mm^{2}

Level 2 f_{2,0.82P} = 0.82 × - 1.20 = - 0.98 N/mm^{2}

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/mm^{2} (> 0 hence O.K.)

Level 1, combination 3 : f = 17.08 - 17.67 = - 0.59 N/mm^{2} (> - 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 × 3.15) = 8.8 N/mm^{2} (< 25 O.K.)

Ultimate Capacity of Beam and Deck Slab

(Composite Section)

Ultimate Design Moment = γ_{f3} × M = 1.1 × 2867 = **3154 kNm**

cl. 6.3.3

Only
steel in the tension zone is to be considered :

Centroid of tendons in tension zone = (6×60 + 10×110 + 8×160 +
4×260) / 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 × 28 × 139 × 1670 × 10-3 = 5655
kN

Compressive
force in concrete flange :

F_{f }= 0.4 × 40 × 1000 × 150 × 10^{-3} = 2400 kN

Let X = depth to neutral axis.

Compressive force in concrete web :

F_{w} = 0.4 × 50 × [393 - (393 - 200) × (X - 150) / (671 ×
2)] × (X - 150) × 10^{-3 }

F_{w} = ( -2.876X^{2} + 8722.84X - 1243717) × 10^{-3}

Equating forces to obtain X :

5655 = 2400 + ( -2.876×^{2} + 8722.84X - 1243717) × 10^{-3}

X = 659 mm

Stress in tendon
after losses = f_{pe} = 4144 × 10^{3} / (32 × 139) =
932 N/mm^{2}

Prestrain ε_{pe} = f_{pe}
/ E_{s} = 932 / 200 × 10^{3} = 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} =
ε_{6} + ε_{pe} = -459 × 0.0035 / 659 + 0.0047 = 0.0022

ε_{pb5} =
ε_{5} + ε_{pe} =
-359 * 0.0035 / 659 + 0.0047 = 0.0028

ε_{pb4} =
ε_{4} + ε_{pe} =
281 * 0.0035 / 659 + 0.0047 = 0.0062

ε_{pb3} =
ε_{3} + ε_{pe} =
381 * 0.0035 / 659 + 0.0047 = 0.0067

ε_{pb2} =
ε_{2} + ε_{pe} =
431 * 0.0035 / 659 + 0.0047 = 0.0069

ε_{pb1} =
ε_{1} + ε_{pe} =
481 * 0.0035 / 659 + 0.0047 = 0.0072

f_{pb6} = 0.0022 × 200 × 10^{3} = 444 N/mm^{2}

f_{pb5} = 0.0028 × 200 × 10^{3} = 551 N/mm^{2}

f_{pb4} = 1162 + 290 × (0.0062 - 0.0058) / 0.0065 = 1178 N/mm^{2}

f_{pb3} = 1162 + 290 × (0.0067 - 0.0058) / 0.0065 = 1201 N/mm^{2}

f_{pb2} = 1162 + 290 × (0.0069 - 0.0058) / 0.0065 = 1213 N/mm^{2}

f_{pb1} = 1162 + 290 × (0.0072 - 0.0058) / 0.0065 = 1225 N/mm^{2}

Tensile force in tendons :

F_{p6} = 2 × 139 × 444 × 10^{-3}

= 124

F_{p5} = 2 × 139 × 551 × 10^{-3}

= 153

F_{p4} = 4 × 139 × 1178 × 10^{-3}

= 655

F_{p3} = 8 × 139 × 1201 × 10^{-3}

= 1336

F_{p2} = 10 × 139 × 1213 × 10^{-3}

= 1686

F_{p1} = 6 × 139 × 1225 × 10^{-3}

= 1022

F_{t} = ∑ F_{p1 to 6} = 4976 kN

Compressive force in concrete :

F_{f} = 0.4 × 40 × 1000 × 150 × 10^{-3} = 2400

F_{w} = 0.4 × 50 × 0.5 × (393 + 247) × (659 - 150) × 10^{-3} = 3258

F_{c} = F_{f} + F_{w} = 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 :

M_{Fp6} = 134 × -0.365 =

-49

M_{Fp5} = 168 × -0.265 =

-45

M_{Fp4} = 675 × 0.375 =

253

M_{Fp3} = 1382 × 0.475 =

656

M_{Fp2} = 1746 × 0.525 =

917

M_{Fp1} = 1060 × 0.575 =

610

M_{Ff} = 2400 × 0.49 =

1176

M_{Fw} = 3258 × 0.207 =

674

M_{u} = ∑ M_{Fp1 to 6} + M_{Ff} + M_{Fw} = **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.

cl. 6.3.4

The Shear Resistance of the beam needs to be determined in accordance with clause 6.3.4. and compared with the ultimate shear load at critical sections.