Appendix 1: Shrinkage stressesA contiguous composite bridge is located over a waterwayand consists of a series of Y8 precast prestressed concretebeams at 2m centres and with a 220mm deep in situ con-crete slab. Youngs modulus for the Y-beam concrete is50N/mm2 and for the in situ slab it is 35N/mm2. Determinethe stresses induced in the section due to shrinkage of thetop slab. (Figure 41 and Table 7 refer.)

1. Calculate properties of sectionModular ratio 50/35 1.429. Therefore eective width ofslab 2000/1.429 1400mm.Ix (slab) 140 223=12 124 227 cm4Distance of neutral axis from top 607 471/8927 68 cm.Ix (comp) 124 227 3080 68 112

118:86 105 5847 76:1 22 682

273 105 cm4

2. Calculate restrained shrinkagestresses

F 50 1400 220 200 106 3080 kNM 3080 0:68 0:11 1756 kNmRestrained shrinkage stress f0 3080 103=308 000

10N=mm2

3. Calculate balancing stresses

Direct stress f10 3080 103=892 700 3:45N/mm2Bending stresses My=I , Balancing stresses:f21 3:45=1:429 1756106 680=273109=1:429

2:41 3:06 5:47N/mm2

f22 2:41 1756 106 680 220=273 109 1:429 2:41 2:07 4:48N/mm2

f23 4:48 1:429 6:40N/mm2

f24 3080 103=892 700 1756 106 940=273 109 3:45 6:04 2:59N/mm2

It is clear that there is a substantial level of tension in thetop slab which cannot only cause cracking but also resultsin a considerable shear force at the slabbeam interfacewhich has to be resisted by shear links projecting fromthe beam.

Appendix 2: Primary temperaturestresses (BD 37/88)Determine the stresses induced by both the positive andreverse temperature dierences for the concrete box girderbridge shown in Figure 42 (A 940 000mm2,I 102 534 106 mm4, depth to NA 409mm,T 12 106, E 34 kN/mm2).1. Calculate critical depths oftemperature distributionFrom BD 37/88 Figure 9 this is a Group 4 section, there-fore:

h1 0:3h 0:3 1000 300 > 150; thus h1 150mmh2 0:3h 0:3 1000 300 > 250; thus h2 250mmh3 0:3h 0:3 1000 300 > 170; thus h3 170mm

+

10

+

Restrainedshrinkage force

Balancing forcesand stresses

Final stresses

5.47

4.486.40

2.59

4.53

5.52

3.6

2.59

Figure 41 Final stress distribution

Section A: cm2 y: cm Ay

Slab 3080 11 33 880

Y8 beam 5847 98.1 573 591

8927 607471

Table 7 Section properties

409

5911000

250 250

220

2201000

200070 surfacing

h1h2

h3

T1

T2

T3

Figure 42 Box girder dimensions and temperature distribution

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2. Calculate temperature distributionBasic values are given in Figure 9 of BD 37/01 which aremodied for depth of section and surface thickness by inter-polating from Table 24 of BD 37/01.

T1 17:8 17:8 13:520=50 16:18CT1 4:0 4:0 3:020=50 3:608CT1 2:1 2:5 2:120=50 2:268C3. Calculate restraint forces at criticalpointsThis is accomplished by dividing the depth into convenientelements corresponding to changes in the distributiondiagram and/or changes in the section (see Figure 3.2 ofBD 37/01):

F EcTTiAiF1 34 000 12 106 16:1 3:6 2000 150=1000

765 kNF2 34 000 12 106 3:6 2000 150=1000

441 kNF3 34 000 12 106 3:6 2:6=2 2000

220 150=1000 177 kNF4 34 000 12 106 2:6=2 2 250 70

250=1000 48 kNF5 34 000 12 106 2:26=2 1000 170=1000

78 kNTotal F 1509 kN (tensile)4. Calculate restraint moment about theneutral axis

M 765409 50 441409 75 177409 185 48409 270 78591 170 2=3=1000

M 431 kNm (hogging)

5. Calculate restraint stressesf EcTTif01 34 000 12 106 16:1 6:56N/mm2

f02 34 000 12 106 3:6 1:47N/mm2

f03 34 000 12 106 2:6 1:06N/mm2

f04 34 000 12 106 0 0:00N/mm2

f05 34 000 12 106 0 0:00N/mm2

f06 34 000 12 106 2:26 0:92N/mm2

6. Calculate balancing stressesDirect stress f10 1509 103=940 000 1:61N/mm2

Bending stresses f2i My=I :

f21 431 106

102 534 106 409 1:71N/mm2

f22 431 106

102 534 106 259 1:08N/mm2

f23 431 106

102 534 106 180 0:75N/mm2

f24 431 106

102 534 106 9 0:06N/mm2

f25 431 106

102 534 106 421 1:76N/mm2

f26 431 106

102 534 106 591 2:47N/mm2

7. Calculate final stressesThe nal stress distribution is shown in Figure 44. Similarcalculations for the cooling (reverse) situation are shownin Figure 45. Table 8 gives a summary of stresses.

F1F3

F2F4

F52.26

2.6

3.6

16.1Top slab 220 h1 = 150

h1 = 250

h3 = 170

430

180 409NA

Figure 43 Element forces

Restrainedstresses

Stresses dueto relaxing

force

Stresses dueto relaxingmoment

Final self-equilibrating

stresses

0.92 2.47 1.78

6.56

1.06

1.61150

250

430

170

1.71 3.241.14

1.67+

Figure 44 Final stress distribution (positive)

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Appendix 3: wind loads (BD 37/88)Calculate the worst transverse wind loads on the structureshown in Figure 46. Assume that v 28m/s; span 33m;H 10m.S1 K1 1:0: From Table 2, S2 1:54(i) Unloaded deck:

vt 28 1 1 1:54 43:13m/sq 43:132 0:613=103 1:14 kN/m2

From Table 4, d d2 1 1:94 2:94mFromTable 5, d2 1:94m, thus b=d2 9:52=2:94 3:24,

and Figure 5, CD 1:4.A1 2:94 33 97:02m2

Thus Pt 1:14 97:02 1:4 154:84 kN(ii) Loaded deck:

vt 35m/s (maximum allowed in the code)q 352 0:613 103 0:75 kN/m2

d2 2:94m > dL 2:5mFrom Table 5, d d2 thus b=d2 9:52=2:94 3:24, and

from Figure 5, CD 1:4.From Table 4,

d d3 dL slab thickness depth of steel beams 2:5 0:22 1:4 4:12m

Pt 0:75 1:4 4:12 33 142:76 kNThus design force greater of (i) and (ii)154.84 kN.

Restrainedstresses

Stresses dueto relaxing

force

Stresses dueto relaxingmoment

Final self-equilibrating

stresses

2.57 0.83 2.00

200

200

200

200

200

3.827 1.38 0.56 1.89

1.89

1.11

+

+

Figure 45 Final stress distribution (negative)

Restraintstresses

Balancingdirect stress

Balancingbending stress

Finalstresses

1 6.56 1.61 1.71 3.24 (C)2 1.47 1.61 1.08 1.14 (T)3 1.06 1.61 0.75 1.3 (T)4 0 1.61 0.06 1.67 (T)

5 0 1.61 1.76 0.15 (C)6 0.92 1.61 2.47 1.78 (C)

Table 8 Summary of stresses

9520

Closed parapet

220

1940

1400

1000

Figure 46 Steel beam and reinforced concrete deck

Note: BD 37/88 has been superseded by BD 37/01.

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