## 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)

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)

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

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

γfL concrete

1.0

1.0

1.15

1.15

γfL surfacing

1.2

1.2

1.75

1.75

γfL HA

1.2

1.0

1.5

1.25

γfL HB

1.1

Temperature Difference

γfL

0.8

1.0#

Beam C40/50    fcu = 50 N/mm2,  fci = 40 N/mm2
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. Section properties of Y5 beam were obtained from a Paper by Taylor, Clark and Banks (download .pdf extract from paper by ).
Note: Level 2 is at the nib level and not at the top of the beam.

Property

Beam Section

Composite Section

Area(mm2)

449.22×103

599.22×103

Centroid(mm)

456

623

2nd Moment of Area(mm4)

52.905×109

103.515×109

Modulus @ Level 1(mm3)

116.020×106

166.156×106

Modulus @ Level 2(mm3)

89.066×106

242.424×106

Modulus @ Level 3(mm3)

179.402×106

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 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 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 × φ
F = -200 × 10-6 × 34 × 1000 × 150 × 0.43 = -439 kN
Eccentricity acent = 502mm
Restraint moment Mcs = -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.

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

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

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 × 24.02 / 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.7.4.3.2

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

Comb 1
(HA)

Comb 1
(HB)

Comb 3

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

-8.94

-8.94

-8.94

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

-7.88

-7.17

-6.78

Reverse Temperature = γfL × -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/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 × 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 × 106 / 5011 × 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

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 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 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/mm2

Moment 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. 6.7.2.5

Allowing for 2% relaxation loss in steel after transfer,
concrete shrinkage εcs = 300 × 10-6
and concrete specific creep ct = 1.03 × 48 × 10-6 per N/mm2
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 = (εcs × Es × Aps ) = 300 × 10-6 × 196 × 32 × 139 = 262

cl. 6.7.2.5

Concrete creep = ( ct ×  fco × Es × Aps ) = 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 = 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/mm2

Level 2 f2,0.82P = 0.82 × - 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 × 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

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

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 =

-49

MFp5 = 168 × -0.265 =

-45

MFp4 = 675 × 0.375 =

253

MFp3 = 1382 × 0.475 =

656

MFp2 = 1746 × 0.525 =

917

MFp1 = 1060 × 0.575 =

610

MFf = 2400 × 0.49 =

1176

MFw = 3258 × 0.207 =

674

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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