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/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.

γconc. = 24kN/m3

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

4.2.2 Serviceability limit state. For the limitations given in 4.1.1.1 a) for reinforced concrete, load combination 1 only sh ould be considered. Where type HB loading is to be taken into account, only 25 units should be considered. The type HA wheel load need not be considered except for cantilever slabs and the top flanges in beam-and-slab, voided slab and box-beam construction

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:

Wednesday, November 4, 2015 11:05 AM

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Concrete Grades

Beam C40/50 fcu = 50 N/mm2, fci = 40 N/mm

2

Slab C32/40 fcu = 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.

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 : Ec = 34 kN/mm2

for fcu = 50N/mm2

Hence restrained temperature stresses per °C = 34 × 103

× 12 × 10-6

= 0.408 N/mm2

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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

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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 × 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 = - εdiff × Ecf × Acf × φF = -200 × 10-6 × 34 × 1000 × 150 × 0.43 = -439 kNEccentricity acent = 502mmRestraint moment Mcs = -439 × 0.502 = -220.4 kNm

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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 × 242

/ 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 × 24.02

/ 8 + 39.6 × 24 / 4

= 1310 kNm

25 units HB Design SLS moment:

= 2.9 × 24.02

/ 8 + 982.3(from grillage analysis)

= 1191.1 kNm

Design ultimate moment:

= 35.7 × 24.02

/ 8 + 49.5 × 24 / 4

= 2867 kNm

Combination 3 LoadingSuper. & 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:

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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 & 41.3kNDesign serviceability moment:= 12.9 × 24.02 / 8 + 33 × 24 / 4= 1127 kNm

Allowable stresses in precast concrete

At transfer :

cl.6.3.2.2 b)

Compression ( Table 23 )

0.5fci (≤ 0.4fcu) = 20 N/mm2

max.

cl.6.3.2.4 b)

Tension = 1.0 N/mm2

At serviceability limit state :

cl.7.4.3.2Compression (1.25 × Table 22)

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At serviceability limit state :

cl.7.4.3.2Compression (1.25 × Table 22)1.25 × 0.4fcu = 25 N/mm2

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

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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/Zlevel 1 = 20 ..................... (eqn. 1)P/A - Pe/Zlevel 2 >= - 1.0 ..................... (eqn. 2)

(eqn. 1) × Zlevel 1 + (eqn. 2) × Zlevel 2 gives :P >= A × (20 × Zlevel 1 - 1.0 × Zlevel 2) / (Zlevel 1 + Zlevel 2)

P = 449.22 × 103 × ( 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 Po = 4888 / 0.9 = 5431kN

Using 15.2mm class 2 relaxation standard strand at maximum initial force of 174kN (0.75 ×

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Allow 10% for loss of force before and during transfer,

then the initial force Po = 4888 / 0.9 = 5431kN

Using 15.2mm class 2 relaxation standard strand at maximum initial force of 174kN (0.75 ×

Pu)

Area of tendon = 139mm2

Nominal tensile strength = fpu =1670 N/mm2

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 <= Zlevel 2 / A + Zlevel 2 / P = (89.066 × 106

/ 449.22 × 103) + (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.

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 Po / [ 1 + Es × (Aps / A) × (1 + A × e2 / I) / Eci ]

P = 0.99 × Po / [ 1 + 196 × ( 32 × 139 / 449220) × (1 + 449220 × 2192 / 52.905 × 109) / 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 / Zlevel 1 ) = 11.3 × ( 1 + 219 / 258 ) = 20.89 N/mm2

(20.89 N/mm2

is slightly greater than the allowable of 20 N/mm2

so a number of tendons will

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Level 1 : P / A × ( 1 + A × e / Zlevel 1 ) = 11.3 × ( 1 + 219 / 258 ) = 20.89 N/mm2

(20.89 N/mm2

is slightly greater than the allowable of 20 N/mm2

so a number of tendons will

need to be debonded near the ends of the beam).

Level 2 : P / A × ( 1 - A × e / Zlevel 2 ) = 11.3 × ( 1 - 219 / 198 ) = - 1.20 N/mm2Moment due to self weight of beam at mid span = 10.78 × 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. 6.7.2.5Allowing for 2% relaxation loss in steel after transfer,concrete shrinkage εcs = 300 × 10-6and concrete specific creep ct = 1.03 × 48 × 10-6 per N/mm2Loss of force after transfer due to :cl. 6.7.2.2Steel relaxation = 0.02 × 5568 = 111cl. 6.7.2.4Concrete shrinkage = (εcs × Es × Aps ) = 300 × 10-6 × 196 × 32 × 139 = 262cl. 6.7.2.5Concrete creep = ( ct × fco × Es × Aps ) = 1.03 × 48 × 10-6 × 12.76 × 196 × 32 × 139 = 550Total Loss = 111 + 262 + 550 = 923 kNFinal force after all loss of prestress = Pe = 5067 - 923 = 4144 kN (Pe/P = 0.82)Final stresses due to prestress after all loss of prestress at :Level 1 f1,0.82P = 0.82 × 20.89 = 17.08 N/mm2Level 2 f2,0.82P = 0.82 × - 1.20 = - 0.98 N/mm2Combined 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 × 3.15) = 8.8 N/mm2 (< 25 O.K.)

Ultimate Capacity of Beam and Deck Slab

(Composite Section)

Ultimate Design Moment = γf3 × M = 1.1 × 2867 = 3154 kNm

cl. 6.3.3

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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 :

Ff = 0.4 × 40 × 1000 × 150 × 10-3

= 2400 kN

Let X = depth to neutral axis.

Compressive force in concrete web :

Fw = 0.4 × 50 × [393 - (393 - 200) × (X - 150) / (671 × 2)] × (X - 150) × 10-3

Fw = ( -2.876X2

+ 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 = fpe = 4144 × 103

/ (32 × 139) = 932 N/mm2

Prestrain εpe = fpe / Es = 932 / 200 × 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

εpb6 = ε6 + εpe = -459 × 0.0035 / 659 + 0.0047 = 0.0022

εpb5 = ε5 + εpe = -359 * 0.0035 / 659 + 0.0047 = 0.0028

εpb4 = ε4 + εpe = 281 * 0.0035 / 659 + 0.0047 = 0.0062

εpb3 = ε3 + εpe = 381 * 0.0035 / 659 + 0.0047 = 0.0067

εpb2 = ε2 + εpe = 431 * 0.0035 / 659 + 0.0047 = 0.0069

εpb1 = ε1 + εpe = 481 * 0.0035 / 659 + 0.0047 = 0.0072

fpb6 = 0.0022 × 200 × 103

= 444 N/mm2

fpb5 = 0.0028 × 200 × 103

= 551 N/mm2

fpb4 = 1162 + 290 × (0.0062 - 0.0058) / 0.0065 = 1178 N/mm2

fpb3 = 1162 + 290 × (0.0067 - 0.0058) / 0.0065 = 1201 N/mm2

fpb2 = 1162 + 290 × (0.0069 - 0.0058) / 0.0065 = 1213 N/mm2

fpb1 = 1162 + 290 × (0.0072 - 0.0058) / 0.0065 = 1225 N/mm2

Tensile force in tendons :

Fp6 = 2 × 139 × 444 × 10-3

= 124

Fp5 = 2 × 139 × 551 × 10-3

= 153

Fp4 = 4 × 139 × 1178 × 10-3

= 655

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Fp4 = 4 × 139 × 1178 × 10-3

= 655

Fp3 = 8 × 139 × 1201 × 10-3

= 1336

Fp2 = 10 × 139 × 1213 × 10-3

= 1686

Fp1 = 6 × 139 × 1225 × 10-3

= 1022

Ft = ∑ Fp1 to 6 = 4976 kN

Compressive force in concrete :

Ff = 0.4 × 40 × 1000 × 150 × 10-3

= 2400

Fw = 0.4 × 50 × 0.5 × (393 + 247) × (659 - 150) × 10-3

= 3258

Fc = Ff + Fw = 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 :

Fp6 = 134

Ff = 2400

Fp5 = 168

Fw = 2765

Fp4 = 675

Fc = 5165

Fp3 = 1382

Fp2 = 1746

Fp1 = 1060

Ft = 5165

Taking Moments about the neutral axis :

MFp6 = 134 × -0.365 = -48.91

MFp5 = 168 × -0.265 = -44.52

MFp4 = 675 × 0.375 = 253.125

MFp3 = 1382 × 0.475 = 656.45

MFp2 = 1746 × 0.525 = 916.65

MFp1 = 1060 × 0.575 = 609.5

MFf = 2400 × 0.49 = 1176

MFw = 3258 × 0.207 = 674.406

Mu = ∑ MFp1 to 6 + MFf + MFw = 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.

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