ŽILINSKÁ UNIVERZITA V ŽILINE. Fakulta stavebná BRIDGES. Examples. Jozef GOCÁL

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1 ŽILINSKÁ UNIVERZITA V ŽILINE Fakulta stavebná BRIDGES Examples Joz GOCÁL Ž I L I N A

2 ŽILINSKÁ UNIVERZITA V ŽILINE, Stavebná fakulta BRIDGES, Examples Podľa prednášok a návodu na cvičenia preložil / According to lectures and instructions for training translated: Ing. Joz Gocál, PhD. Za jazykovú (a vecnú) správnosť zodpovedá autor. Text neprešiel jazykovou úpravou. The author is responsible for the language (and objective) correctness. The text was not subject to language correction. Vydala Žilinská univerzita v Žiline, Žilina 2008 Issued by University of Žilina, Žilina 2008 Translation: GOCÁL Joz, 2008 Tlač / Printed by ISBN Vydané s podporou Európskeho sociálneho fondu, projekt SOP ĽZ 2005/NP1-007 Issued with support of European Social Foundation, project SOP ĽZ 2005/NP1-007

3 CONTENTS 1 PROJECT SETTINGS LOADS AND INTERNAL FORCES PERMANENT LOADS Part 1 (carried by steel beams) Part 2 (carried by steel and concrete composite beams) VARIABLE TRAFFIC LOADS CROSS-SECTION DESIGN FOR ULTIMATE LIMIT STATE COMBINATION OF LOADS FOR THE ULTIMATE LIMIT STATES CHARACTERISTICS OF CROSS-SECTION ASSESSMENT OF THE CROSS-SECTION BENDING RESISTANCE IN THE MID-SPAN ASSESSMENT OF SHEAR RESISTANCE OF THE CROSS-SECTION AT THE SUPPORT CHECK OF CROSS-SECTION FOR SERVICEABILITY LIMIT STATE CHARACTERISTICS OF COMPOSITE CROSS SECTION FOR SHORT-TERM LOAD EFFECTS CHARACTERISTICS OF COMPOSITE CROSS SECTION FOR LONG-TERM LOAD EFFECTS CLASSIFICATION OF EFFECTIVE CROSS-SECTION CALCULATION OF NORMAL STRESSES Stresses caused by part 1 of permanent loads (carried by steel beam) Stresses caused by part 2 of permanent loads (carried by composite beam) Variable traffic load Stresses caused by shrinkage of concrete CROSS-SECTION CHECK DESIGN OF SHEAR CONNECTION REFERENCES... 30

4 t f = DESIGN OF STEEL AND CONCRETE COMPOSITE ROAD BRIDGE 1 PROJECT SETTINGS The objective is to design steel and concrete composite road bridge transferring the road C9.5 and two footpaths of width 1.5 m over a river. The main steel beams, made of steel S355, with span 40 m and relative spacing 3.4 m are simply supported on concrete gravity abutments. The steel beams are connected with reinforced concrete slab of depth 250 mm, made of concrete C 30/37, by means of stud connectors 18 mm with ultimate strength f u = 340 MPa. The bridge will be built without temporary staging. The cross section of the bridge is presented in Fig w = 2300 f w = 9500 c w = 2300 f CAST-IN-PLACE FOOTPATH aerated concrete C35/45 PARAPET PRECAST CORDON RH-85 P 20x300 P 14x2000 P 40x400 GUARD-RAIL C 9.5/80 CARRIAGE WAY 85mm INSULATION 5mm CONCRETE SLAB C30/37 MIN. 250 mm 2,5% 2,5% 120 a c= 1700 a = 3400 a = 3400 a = 3400 a c= w = s t = 250 t = 90 s c Fig. 1 Cross section of the bridge GUARD-RAIL UPE 220 CAST-IN-PLACE FOOTPATH aerated concrete C35/45 PARAPET 4.0% PRECAST CORDON RH LOADS AND INTERNAL FORCES 2.1 Permanent loads The permanent loads are represented by self weight of the superstructure as well as the weight of all permanently embedded parts of bridge. With regards to un-propped construction of the bridge, the permanent loads have to be divided into two parts. The part 1 corresponds to the first construction phase, when the concrete slab is not stiffened yet and therore all the loads have to be carried by steel beams only. The second part of permanent loads corresponds to the construction phase after the concrete slab stiffening, when the steel and concrete composite structure is acting. According to EN 1990, Annex 2, the partial safety factor should be considered for all permanent loads by value G,sup = However, the national annex to STN EN 1990, Annex 2 differs two values of partial safety factor G,sup for the self weight of bridge elements made in special workshops it is dined by value 1.25, and for the self weight of bridge elements made in building site it is dined by value 1.35.

5 Part 1 (carried by steel beams) Loads: Characteristic values G Design values - concrete slab: tsws kn m kn m -1 - steel beams: n A kn m kn m -1 a - bracing: g n 1 a L kn m kn m -1 br The 1st part of permanent loads falling to one steel beam: G1, k g 1,k = n kn m-1 g 1,d = G 1,k = kn m -1 G 1,d = kn m -1 1, d kn m-1 Gn 4 Characteristic values of internal forces and moments in decisive cross sections: g1, k 1, k 2 2 M 1 8 g L knm V 1 2 g L kn g1, k 1, k Design values of internal forces and moments in decisive cross sections: g1, Ed 1, d 2 2 M 1 8 g L knm V 1 2 g L kn g1, Ed 1, d Part 2 (carried by steel and concrete composite beams) Loads: Characteristic values G Design values - carriage way: tc w c kn m kn m -1 - insulation: ti w s kn m kn m -1 - footpaths: 2t w kn m kn m -1 f f - parapets: (estimation) = 0.60 kn m kn m -1 - guard-rails: (estimation) = 0.80 kn m kn m -1 The 2nd part of permanent loads falling to one composite beam: g 2,k = G2, k kn m -1 g 2,d = n 4 G 2,k = kn m -1 G 2,d = kn m -1 2, d kn m-1 Gn 4 Characteristic values of internal forces and moments in decisive cross sections: g2, k 2, k 2 2 M 1 8 g L knm V 1 2 g L kn g2, k 2, k Design values of internal forces and moments in decisive cross sections: g2, Ed 2, d 2 2 M 1 8 g L knm V 1 2 g L kn g2, Ed 2, d

6 6 2.2 Variable traffic loads Variable loads are represented mainly by vertical fects of road traffic load, which should be generally considered by means of four loading models, dined in STN EN The main loading system is represented by Load model 1 (LM1), consisting of concentrated tandem system (TS) and uniformly distributed load (UDL system) situated in carriage way divided into several notional lanes. The LM1 expresses majority of passenger and freight service and should be used for verification of global and local fects of traffic as well. The load model 2 (LM2) is characterised by single axle vehicle and it may be determining mainly for short structural members with length 3 7 m. It is intended mainly for verification of local fects of traffic. The load model 3 (LM3) is intended for modelling special vehicles (e.g. for industrial traffic), moving pass the special permitted road tracks reserved for heavy loads. This model should be used for verification of global and local fects of traffic as well. The load model 4 (LM4) expresses the load caused by moving mass of people and it should by used for general verification of a structure. It may be simply proved that the LM2 and LM4 have less unfavourable fects on the superstructure strain than the LM1. The LM3 will not be considered for simplification. Consequently, only fects of the LM1 will be taken into account. These fects should be combined with horizontal fects of traffic load (braking forces) and pedestrian load of footpaths as well. While the braking forces will be neglected, the fect of pedestrian load of footpaths will be taken into account by reduced combination value. Transversal arrangement of LM1 should be considered using the influence line of transversal distribution obtained by method of rigid transversal bracing (Fig. 2). F=1 e a ed F=1 eb ec a b c d aa ab ad ac = a b c d aa,1 ab,1 ac,1 ad,1 + M=F e a a b c d aa,2 ab,2 ac,2 ad,2 Fig. 2 Influence line of transversal distribution for the outer beam a

7 7 Taking into account the bridge symmetry, only beams a and b need to be verified. Coordinates of the influence line for the outer beam a are given by equation: ya a ya a ai ai,1 ai,2 e d d i e d i 2 n 2 I yi I yi ei ei ia ia ia aa ab ac ac a a a a I e I 1 e 1 1.5a 1.5a a 0.5a a 0.5a a 0.5a a 1.5a a 0.5a 2 d ia ai 1.0 Similarly, coordinates of the influence line for the inner beam b are given by equation: yb b yb b bi bi,1 bi,2 e d d i e d i 2 n 2 I yi I yi ei ei ia ia ia ba bb bc bd a a a a I e I 1 e 1 0.5a 1.5a a 0.5a a 0.5a a 0.5a a 1.5a a 0.5a 2 d ia ai 1.0 Arrangement of the Load Model 1 together with the pedestrian load of footpaths in transversal direction is presented in Fig. 3. All the loads are considered in the most unfavourable positions with regard to the particular influence line of transversal distribution. The lightening fects are not taken into account, however, the tandem systems are always considered as complete. Classification factors Qi and qi may be generally considered within an interval The values applied here are considered according to national annex to STN EN

8 q q2 2 Q3Q 3/2 Q3Q 3/2 q q3 3 q qr r = = = = = = = = = = = = A f1 = A 1 = A 2 = A 3 = A r = A f2 = A f = A = A 2 = A 3 = Width of carriage way = F R F q f Q1Q 1/2 Q1Q 1/2 Q2Q 2/2 Q2Q 2/2 q1 q1 q f a b c d - a + b + 1 FIRST NOTIONAL LANE 2 SECOND NOTIONAL LANE 3 THIRD NOTIONAL LANE R F REMAINING AREA FOOTHPATH Fig. 3 Arrangement of LM1 with pedestrian load and coordinates of the influence lines The outer beam a is affected by following loads: - from tandem system (TS): F Q 2 Q 2 Q kn TS Q Q Q from UDL system: p q A q A q A UDL q1 1 1 q2 2 2 q kn m -1 - from pedestrian load of footpaths: p f q f A f kn m -1

9 The inner beam b is affected by following loads: - from tandem system (TS): F Q 2 Q 2 Q kn TS Q Q Q from UDL system: p q A q A q A q A UDL q1 1 1 q2 2 2 q3 3 3 qr r r kn m -1 - from pedestrian load of footpaths: p f q f Af 1 Af kn m -1 Characteristic values of internal forces and moments in decisive cross sections: The influence lines for calculation of internal forces and moments in longitudinal direction are presented in Fig. 4. All the loads from transversal distribution are considered in the most unfavourable positions with regard to the particular influence line. With regards to the higher values of all particular loads, only the outer beam a needs to be further evaluated. 2 F TS 2 F TS p f p UDL p f p UDL 19.4 L = 40.0 m L = 40.0 m M =9.7 =9.7 A =200.0 =1.00 =0.97 A =20.0 V Fig. 4 Influence lines for bending moment and shear force MLM1, k MTS, k MUDL, k FTS 1 pudl A = knm M f, k p f A knm VLM 1, k VTS, k VUDL, k FTS 1 2 pudl A = kn Vf, k p f A kn According to STN EN 1990, Annex 2, the load model LM1 consisting of TS system and UDL system and the reduced pedestrian load of footpaths make together the load group gr1a, which should be considered as one multi-part variable load. The resultant characteristic values of internal forces and moments corresponding to the load group gr1a are: M M M M = knm gr1 a, k TS, k UDL, k f, k V V V = kn gr1 a, k LM1, k f, k

10 h=2310 h a =2060 t c =250 t f2 =40 d=2000 t f1 = z a1 =835 z a2 = CROSS-SECTION DESIGN FOR ULTIMATE LIMIT STATE According to STN EN , the fects of creep and shrinkage of concrete, as well as the fects of sequence of construction and temperature fects may be neglected in analysis for verifications of ultimate limit states other than fatigue, providing that all cross-sections of the composite member are in class 1 or 2. Then, the verification consists in comparison of the design bending moment and the design shear force, respectively, caused by the most unfavourable combination of actions, with the design bending resistance of composite crosssection and shear resistance of the steel beam web, respectively. 3.1 Combination of loads for the ultimate limit states Design value of bending moment affecting the outer beam a at the mid-span is: M M M M M M Ed Gj gk, j Q, gr1a gr1 a, k g1, Ed g2, Ed Q, gr1a gr1 a, k = = knm Design value of shear force affecting the outer beam a at the support is: V V V V V V Ed Gj gk, j Q, gr1a gr1 a, k g1, Ed g2, Ed Q, gr1a gr1 a, k = = kn 3.2 Characteristics of cross-section Effective width of the concrete slab at the mid-span of beam a is dined by an expression b f = b 0 + b ei. Estimating distance of the outer queues of studs b 0 = 180 mm, the partial fective widths are: b e,1 = b e,2 = L e / 8 = / 8 = mm > b 1 = b 2 = mm b e,1 = b e,2 = mm. Consequently, the total fective width of concrete slab is b f = 3400 mm, regardless of the distance of outer queues of studs b 0. b =1610 b =180 b = b e1 b e2 b =300 f1 Steel beam: A a = m 2 C gw C ga t =14 w I ya = m 4 Concrete slab: A c = m 2 a =1700 c b =400 f2 a=3400 I yc = m 4 Fig. 5 Cross section of the composite beam

11 z 11 Material parameters of steel: (according to STN EN ) Class of steel: S355 Characteristic yield strength: f yk = 355 MPa Partial safety factor: M0 =1.0, M1 =1.1 Design yield strength: f yd = f yk / M = 355 MPa Modulus of elasticity: E = MPa Material parameters of concrete: (according to STN EN ) Class of concrete: C30/37 Characteristic cylinder compression strength: f ck = 30 MPa Partial safety factor: C = 1.5 Design cylinder compression strength: f cd = f ck / C = 20 MPa Modulus of elasticity: E cm = MPa 3.3 Assessment of the cross-section bending resistance in the mid-span Location of plastic neutral axis and classification of composite cross section Basically, there are three possible locations of the neutral axis. Either it lies in concrete slab, or in upper flange of steel beam, or it passes through the web of the steel beam. Its location results from the equilibrium condition of normal forces in the cross section at ultimate limit state. bf tf2 d tf1 tc b f1 Cga b f2 za + N a f yd 0.85 f cd - N c Fig. 6 Tenseness at ultimate limit state for neutral axis lying in concrete slab If A c 0.85 f cd A a f yd, the neutral axis lies in concrete slab (Fig. 6) and its location results from the following equilibrium condition N c N a bf z 0.85 fcd Aa f yd z b f A a f yd 0.85 f Te whole steel part is in tension, and so the classification of cross section is not needed. The plastic bending resistance results from the moment equilibrium condition in the cross section M pl, Rd Na za 0.5z Aa fyd za 0.5z. cd

12 at z 12 bf b f f cd za tf2 d tf1 tc Cgc Cgac Cga Cgat b f2 zac N at + f yd - N c - N ac f yd Fig. 7 Tenseness at ultimate limit state for neutral axis lying in steel beam If A c 0.85 f cd < A a f yd, the neutral axis is situated in the steel beam (Fig. 7) and the equilibrium condition of normal forces at ultimate limit state is dined by equation N N N c ac a A 0.85 f A f A f A A f c cd ac yd at yd a ac yd A 0.85 f 2 A f A f A c cd ac yd a yd ac A f A 0.85 f a yd c cd When the area of steel compression part A ac b f1 t f1, the neutral axis lies in the upper flange, in other case it passes through the web of steel beam. The location of neutral axis is then given as follows A t for A b t z A b t t t for A b t ac c ac f 1 f 1 bf 1 ac f 1 f 1 c f 1 ac f 1 f 1 tw In the case of neutral axis laying in the upper flange of steel beam the classification of cross section is not needed. If the neutral axis passes through the web, classification of the compressed part of the web according to STN EN is necessary. The steel compression flange that is restrained from buckling by fective attachment to the concrete flange by shear connectors may be assumed to be in class 1, if the spacing of connectors is in accordance with the shear connection detailing given in STN EN Providing that the compression part of steel web is classified into class 1 or 2, the plastic bending resistance results from the moment equilibrium condition in the composite cross section 2 f M N z t N z t N z t N z t pl, Rd at at c ac ac c a a c ac ac c M A z t A z t f pl, Rd a a c ac ac c yd yd

13 13 Applying the approach mentioned above, for the cross section presented in Fig. 5 it is valid A 0.85 f MN < A f MN c cd a yd The area of steel compression part is A ac Aa f yd Ac 0.85 fcd = m 2 2 f yd neutral axis lies in steel beam Aac m 2 < b f1 t f1 = = m 2 neutral axis lies in upper flange The location of plastic neutral axis is Aac z tc m b 0.3 z ac f 1 z tc m 2 2 Since the neutral axis lies in the upper flange of steel beam, therore the cross section may be assumed to be in class 1. Check of the plastic bending resistance of composite cross section in the mid-span: M A z t A z t f M pl, Rd a a c ac ac c yd pl, Rd M pl, Rd knm > M Ed knm Assessment of shear resistance of the cross-section at the support The shear resistance of the cross section is given by shear resistance of the web: Vpl, Rd Aw fyd kn > VEd kn 3 The web of steel beam is satisfactory. However, the fect of shear buckling should be checked yet according to STN EN The shear buckling coficient k is (for the web without longitudinal stiffeners and with rigid transverse stiffeners at distances a = 10 m): 2 2 k hw a The web slenderness parameter may be taken as: hw tw w > 1, k Contribution from the web to shear buckling resistance is w w

14 h=2310 h a =2060 t c =250 t f2 =40 d=2000 t f1 =20 z a4 =835 z a3 =1225 r r = 1350 a r c z e4 z e3 =z e2 14 Neglecting the contribution of flanges f, the reduction factor for shear buckling is V w f The design shear resistance (with shear buckling fect) is V b, Rd Vb, Rd 3 V f yw hw tw kn M kn > VEd kn 4 CHECK OF CROSS-SECTION FOR SERVICEABILITY LIMIT STATE In the case of composite members with cross-section classified in class 1 or 2, it should be further proved that the nominal stresses in the limiting cross sections resulting from the characteristic load combinations (partial safety factors Gi = Qi =1.0) do not exceed the design strength of materials, i.e. the beam stays in elastic conditions. The calculation of stresses should take into account (apart from other fects) the fect of shear lag, creep and shrinkage of concrete and sequence of construction. 4.1 Characteristics of composite cross section for short-term load fects Effective width of the concrete slab at the mid-span of beam a is the same as in ultimate limit states. Calculation of the elastic resistance to bending is based on the fective crosssection, in which the concrete part is transformed to equivalent steel part by means of the modular ratio for short-term loading n 0 = E a / E cm. 2 b = 3400 f 1 3 C gc b =300 f1 ze1 C g Y C ga t =14 w 4 b =400 f2 Fig. 8 Effective cross section of the composite beam 1(2) upper (bottom) edge of concrete slab, 3(4) upper (bottom) edge of steel beam C ga (C gc ) [C g ] centre of gravity of steel beam (concrete slab) [fective cross-section]

15 15 Characteristics of concrete slab: A c = = m 2 1 I yc = m4 12 Characteristics of steel beam: A a = = m z a4 = m 0.05 z a3 = = m 1 I ya = m 4 Characteristics of fective cross section of composite beam: n 0 = E a / E cm = 210 / 33 = A = A a + A c / n 0 = / = m 2 Aa r r c = m A r a = Ac n0 r m A I y, = I ya + I yc / n 0 + A a r a 2 + A c / n 0 r c 2 = I ya + I yc / n 0 + A r a r c I y, = / = m Characteristics of composite cross section for long-term load fects Concrete is a material underlying to creep, which may be simply dined as an increase of dormation in time under persistent long-term loading. This fect may be mathematically described as a degradation of modulus of elasticity, which in composite girder results in gradual migration of normal stresses from the concrete slab to the steel beam. The fects of creep may be taken into account by using modular ratios n L for calculation of the fective cross-section characteristics. The modular ratios depending on the type of loading are given by expression n n L 0 1 L t where n 0 is the modular ratio for short-term loading, t is the creep coficient (t,t 0 ) depending on the age (t) of concrete at the moment considered and the age (t 0 ) at loading, and L is the creep multiplier depending on the type of loading, which may be taken as 1.1 for permanent loads, 0.55 for primary and secondary fects of shrinkage and 1.5 for pre-stressing by imposed dormations. According to EN , Annex B, the creep coficient (t,t 0 ) may be calculated from expression: ( t, t ) ( t t ), 0 0 c 0 where 0 is the theoretical creep coficient given by expression ( f ) ( t ). 0 RH cm 0

16 16 Coficient RH expresses an fect of relative humidity of environment RH (%) on the theoretical creep coficient and is given by expression RH 1 RH for fcm 35 MPa 0,1 3 h0 1 RH for fcm 35 MPa ,1 h0 where h 0 (mm) is equivalent depth of slab h 0 2A c u the perimeter of member that is in contact with atmosphere., A c is area of the concrete slab and u is Coficient (f cm ) expresses an fect of concrete strength on the theoretical creep coficient and is given by expression ( f ) cm 16,8 f cm where f cm = f ck + 8 is the mean compression strength of concrete at the age 28 days. Coficient (t 0 ) expresses an fect of the age of concrete at loading on the theoretical creep coficient and is given by expression ( t ) 1 0 0,20 0,1 t0. Coficient c (t-t 0 ) describes development of creep in time and may by estimated by using following expression t t0 H tt0 c ( t t0 ) 0,3 Coficient H depends on relative humidity RH (%) and equivalent size of member h 0 and is given by expression H 18 1,5 1 (0, 012 RH ) h for fcm ,5 1 (0, 012 RH ) h for fcm 35 Coficients 1, 2, 3 taking into account the concrete strength are dined as follows fcm fcm fcm. Creep coficient for second part of permanent loads: On the assumption that the second part of permanent loads will start to affect the composite girder two months after concreting (t 0 = 60 days), the bridge will be introduced into service four months after concreting (t in = 120 days) and the planned service life of the bridge is 100 years (t fin = days), then at the relative humidity RH = 80% it will be valid:

17 17 h 0 2A c m u f cm = = 38 MPa > 35 MPa H 1,5 1 (0,012 80) < , c ( t t 0 ) for t = 120 days 0, c ( t t 0 ) for t = days 1 ( t0 ) ,20 0, ,8 ( f cm ) = RH , = The creep coficient at the time of introducing into service (t = 120 days) is: (120,60) = t The creep coficient at the end of service life (t = days) is: (36 525,60) = t Creep coficient for fect of shrinkage of concrete: Shrinkage of concrete is a long-term process, independent on acting load, which is characteristic by gradual decreasing volume of concrete. Since the steel beam restrains to free dormation of the concrete slab, there occur additional strains in the composite girder. According to STN EN , the creep coficient due to shrinkage should be calculated on the assumption that the shrinkage will start from the first day after concreting (t 0 = 1 day). 0, c ( t t 0 ) for t = 120 days 0, c ( t t 0 ) for t = days 1 ( t0 ) ,20 0, =

18 18 The creep coficient at the time of introducing into service (t = 120 days) is: (120,60) = t The creep coficient at the end of service life (t = days): (36 525,60) = t Modular ratios for long-term load fects: - for the second part of permanent loads at the time t = 120 days: n n L 0 L t - for the second part of permanent loads at the time t = days: n n L 0 L t - for the shrinkage fects at the time t = 120 days: n n L 0 L t - for the shrinkage fects at the time t = days: n n L 0 L t Characteristics of fective cross-section of composite beam for long-term load fects: The fective cross-section characteristics corresponding to calculated modular ratios n L are summarised in Tab. 1. Tab. 1 Effective cross-section characteristics Load 2nd part of permanent loads Shrinkage Age of bridge [days] (t,t 0 ) [-] L [-] n 0 = E a /E cm [-] n L = n 0 (1+ L (t,t 0 )) [-] A a [ 10 m 2 ] A c /n [ 10 m 2 ] A =A a +A c /n [ 10 m 2 ] r c = A a a /A [m] r a = A c a /(n A ) [m] I a [ 10 m 4 ] I c /n [ 10 m 4 ] A a c a a [ 10 m 4 ] I [ 10 m 4 ] z e 1 [m] z e 2 [m] z e 3 [m] z e 4 [m]

19 e4 d= Classification of fective cross-section It is sufficient to verify classification of the fective cross-section only for the case with maximum distance of the upper edge of steel beam from the elastic neutral axis z e3. The limiting value of the width to thickness ratio for class 3 is dined STN EN in dependence on the stress ratio 42 when 1: c / t 0, 67 0,33 when 1: c / t 62 (1 ) ( ) C g t =14 w Y - <f 3 y + z ze3 = 4 3 Fig. 9 Stress distribution in the cross-section From the classification viewpoint, the most unfavourable location of elastic neutral axis is dined for fects of shrinkage at the end of service life (t = 100 years) by the value z e3 = m. It means that the maximum stresses will be at the bottom edge of steel beam (Fig. 9) and the stress ratio may be expressed 4 ze , z e3 and the limiting proportion is Providing that the fective depth of fillet weld connecting the flanges and web of steel beam is a = 7 mm, the slenderness ratio of the web is w = c t w < the cross-section may be classified as 14 class 3

20 e3 e2 a4 a Calculation of normal stresses Stresses caused by part 1 of permanent loads (carried by steel beam) M MPa g1, k g1,3 za3 I ya C ga + 4 Fig. 10 Stress distribution in the steel cross-section M g1, k g1,4 za4 I ya MPa Stresses caused by part 2 of permanent loads (carried by composite beam) Stresses at the time of loading (t 0 = 60 days) M g2, k g2,1 ze1(60) I y, (60) n0 M g2, k g2,2 ze2(60) I y, (60) n MPa 0.95 MPa M g2, k g2,3 ze3(60) I y, (60) M g2, k g2,4 ze4(60) I y, (60) MPa MPa ze4 z =z ze1 C g M rc N + Fig. 11 Stress distribution in composite cross-section 4

21 Stresses at the time of introducing bridge into service (t = 120 days) Gradual migration of normal stresses from the concrete slab to the steel beam due to creep of concrete may be considered by means of relaxation method, according to which the calculation is divided to two phases. In the first phase it is assumed that after applying a long-term load (at the time t 0 ) the composite beam is fixated, in consequence of which the increase of dormation due to creep of concrete is zero. The creep of concrete causes that stresses in concrete slab gradually decrease this phenomenon is denoted as relaxation. The relaxed value of stresses at the time of evaluation t may be calculated from expression tt, n ( t) ( t ) 1 ( t ) 1 t, t ( t ) t, t , 0 L t t nl where (t,t 0 ) is denoted as relaxation factor. The initial stresses in the concrete slab (t 0 ) may be replaced by corresponding normal force N and bending moment M (see Fig. 11): N Ac 1 2 Ic 1 2 M Ic 2 t 2 t c c Due to relaxation, these internal forces will decrease to N N t, t I 0 M M t, t I 0 Then, the reactions in fictitious support are NII N NI N 1 t, t M II M M I M 1 t, t 0 0 In the second phase, after unlocking the fixation from the first phase, these reactions affect the composite cross-section, which is strained by eccentric compression and bending. The final stresses in the composite cross section are given by superposition of both steps of the relaxation method. In our case, the fect of stresses g2,1 and g2,2 in the concrete slab at the time t 0 = 60 days may be replaced by normal force N and bending moment M N kn M knm 0.25 The relaxation factor at the time of introducing bridge into service (t = 120 days) is equal to n , , n L(120)

22 22 The stresses in concrete and corresponding internal forces decrease to following values NI MI N 120, kn M 120, knm g2,1, I g2,1 120, MPa g2,2, I g2,2 120, MPa The reactions from fictitious support affecting the composite cross-section are NII N 1 120, kn MII M 1 120, knm The moment due to eccentricity of the normal force N II is M II NII rc knm Corresponding stresses in the composite cross-section are as follows 1 N M M z MPa II II II g2,1, II e1 nl A I y, 1 N M M z MPa II II II g2,2, II e2 nl A I y, N M M z MPa II II II g2,3, II e3 A Iy, N M M z MPa II II II g2,4, II e4 A Iy, Resultant stresses at the time of introducing bridge into service (t = 120 days) due to 2nd part of permanent loads including the creep fects are = 1.71 MPa g2,1 g2,1, I g2,1, II = 1.03 MPa g2,2 g2,2, I g2,2, II = MPa g2,3 g2,3(60) g2,2, II = MPa g2,4 g2,4(60) g2,4, II

23 Stresses at the end of service life (t = days) The relaxation factor at the end of service life (t = days) is equal to n , , n L(36 525) The stresses in concrete and corresponding internal forces decrease to following values NI MI N , kn M , knm g2,1, I g2, , MPa g2,2, I g2, , MPa The reactions from fictitious support affecting the composite cross-section are NII N , kn MII M , knm M II NII rc knm Corresponding stresses in the composite cross-section are as follows 1 N M M z MPa II II II g2,1, II e1 nl A I y, 1 N M M z MPa II II II g2,2, II e2 nl A I y, N M M z MPa II II II g2,3, II e3 A Iy, N M M z MPa II II II g2,4, II e4 A Iy, Resultant stresses at the end of service life (t = days) due to 2nd part of permanent loads including the creep fects are = 1.56 MPa g2,1 g2,1, I g2,1, II = 1.03 MPa g2,2 g2,2, I g2,2, II = MPa g2,3 g2,3(60) g2,2, II = 47,48 MPa g2,4 g2,4(60) g2,4, II

24 e3 e Variable traffic load Normal stresses caused by the load group gr1a (LM1 + pedestrian load of footpaths) are independent on the time of evaluation, because variable load represents a short-time fect that is not affected by creep. M gr1, a k gr1 a,1 ze1 I y, n MPa M gr1, a k gr1 a,2 ze2 I y, n MPa M gr1, a k gr1 a,3 ze3 I y, M gr1 a, k gr1 a,4 ze4 I y, MPa MPa Stresses caused by shrinkage of concrete Analogous to the creep fect, the fect of shrinkage of concrete may be calculated again in two steps. In the first step, it is assumed that the steel beam, restraining to free dormation of the concrete slab due to shrinkage, is absolutely rigid, in consequence of which there arise primary normal tensile stresses cs, uniformly distributed past the depth of slab (Fig. 12). These stresses may be replaced by normal force N cs = cs A c. In the second step, the reaction from fictitious support equal to this normal force affects the composite cross-section as an eccentric compression force. The final stresses in the composite cross section are given again by superposition of stresses obtained in the both steps. L cs tc cs + N cs ze4 z =z ze1 C g rc Fig. 12 Primary stresses in the concrete slab due to shrinkage According to STN EN , the total strain due to the shrinkage of concrete cs is given by summation of two parts the strain due to draining shrinkage cd and the strain due to autogenous shrinkage ca cs = cd + ca.

25 25 Development of the strain due to draining shrinkage cd in time is dined by expression t t, t k,0 cd ds s h cd where k h cd,0 is the final value of strain due to the drying shrinkage ( cd, ). The factor k h (see 2A Tab. 2) depends on equivalent depth of slab h c 0 [mm], where A c is area of the concrete u slab and u is the perimeter of member that is in contact with atmosphere. Tab. 2 Values of k h h 0 k h Coficient ds (t,t s ) describes development of drying shrinkage in time and may by calculated by using following expression ds tt, s 3 t t 0.04 h s t t s 0 where t is an age of concrete [days] at the time of evaluation and t s is an age of concrete at begin of the drying shrinkage, which should be generally assumed to be one day according to STN EN The basic strain due to drying shrinkage may be obtained as follows fcm ds2 fcm0 6 cd, ds1 e 10 RH RH 3 RH where f cm = f ck + 8 [MPa] is the mean compression strength of concrete at the age 28 days and f cm0 = 10 MPa. The coficients ds1, ds2 depend on the type of cement and are given by values: for cement class S: ds1 = 3, ds2 = 0.13 for cement class N ds1 = 4, ds2 = 0.12 for cement class R ds1 = 6, ds3 = 0.11

26 26 Development of the strain due to autogenous shrinkage ca in time is dined by expression t t ca as ca where coficient as (t) describing development of autogenous shrinkage in time may by obtained from expression as t e s t and the final value of strain due to the autogenous shrinkage is given by expression ca f ck In our case, on the assumption that the shrinkage will start to affect the composite girder from the first day after concreting (t s = 1 day) and the cement class N will be applied, then at the relative humidity RH = 80% it will be valid: 3 RH cd, e ds ds tt, s tt, s cd cd ca as e = as e = ca ca for t = 120 days for t = days The total strains due to shrinkage are: at the time t = 120 days: cs = = at the time t = days: cs = =

27 Stresses at the time of introducing bridge into service (t = 120 days) The primary normal stress in the concrete slab due to shrinkage is dined as follows: Ecm n cs cs t cs t Ecm n MPa Corresponding normal force is L t L N cs = cs A c = = kn The moment due to eccentricity of the normal force N cs is Mcs Ncs rc knm The final stresses in the composite cross section due to shrinkage are 1 N M z cs cs cs1 cs e1 nl A I y, 1 N M z cs cs cs2 cs e2 nl A I y, 0.22 MPa 0.50 MPa N M z MPa cs cs cs3 e3 A Iy, N M z MPa cs cs cs4 e4 A Iy, Stresses at the end of service life (t = days) The primary normal stress in the concrete slab due to shrinkage and corresponding internal forces affecting the composite cross-section are n cs cs t Ecm MPa n L N cs = cs A c = = kn Mcs Ncs rc knm The final stresses in the composite cross section due to shrinkage are 1 N M z cs cs cs1 cs e1 nl A I y, 1 N M z cs cs cs2 cs e2 nl A I y, 0.47 MPa 0.84 MPa N M z MPa cs cs cs3 e3 A Iy, N M z MPa cs cs cs4 e4 A Iy,

28 Cross-section check The final check of cross-section is presented in Tab. 3, where the values of normal stresses due to partial load fects are summarized. The variable loads are represented only by the load group gr1a and so no combination rules needs to be applied. Tab. 3 Resultant normal stresses Loads At the time of introducing into service (t = 120 days) Stresses due to partial loads [MPa] At the end of service life (t = days) Part 1 of permanent loads Part 2 of permanent loads Shrinkage Variable load - "gr1a" Resultant stresses [MPa] Design strength [MPa] DESIGN OF SHEAR CONNECTION Shear connection (together with transverse reinforcement) should be provided to transmit the longitudinal shear force between the concrete and the structural steel element, ignoring the fect of natural bond between the two. At present, headed stud shear connectors welded to the upper steel flange are mostly used. We propose the headed studs with diameter 20 mm and overall nominal height 100 mm, made of steel with ultimate limit strength f u = 340 MPa. The studs will be arranged in two lines. Characteristic resistance of the stud is dined as the smaller value from: Rk u P 0.8 f d kn 2 2 ck cm PRk 0.29 d f E kn where h 0.2 sc 1 d for 3 h sc / d 4 1 for h sc / d > 4 (in our case h sc / d = 100 / 20 > 4) It means that characteristic resistance of the stud is kn. Then the design value is Prd Prk V kn The shear connectors should transmit the longitudinal force in the slab between the cross section with zero slip and the nearest cross section in which the maximum slip would occur in case of absence of shear connection (critical cross sections in Fig. 13). The necessary number of studs results from equilibrium condition

29 29 n P rd = N c where N c represents the maximum longitudinal force on the steel-concrete interface between the critical cross sections (mid-span and support), which is equal to normal resistance of the concrete slab. Nc x x x Nc L Fig. 13 Slip of the concrete slab in case of absence of shear connection (x critical cross sections) N c = A c f cd = = kn Then, the necessary theoretical number of studs (in the half of span) is: Nc n P rd We propose 250 studs in two lines. Theoretical distances between the couples of studs are: L e 160 mm n We propose following arrangement of studs in the half of span (Fig. 14): mm mm (124 gaps or 125 couples of studs) e =140 e = L/2 = Fig. 14 Arrangement of studs on the beam

30 30 REFERENCES [1] STN EN 1990: 2004 Eurocode: Basis of structural design [2] STN EN 1990/A1 Annex2 Eurocode. Basis of structural design: Application for bridges [3] STN EN 1990/A1/NA Eurocode. Basis of structural design: Application for bridges. National annex. [4] STN EN : 2006 Eurocode 1: Actions on structures: Part 2: Traffic loads on bridges [5] STN EN /NA Eurocode 1: Actions on structures: Part 2: Traffic loads on bridges. National annex [6] STN EN : 2006 Eurocode 2: Design of concrete structures Part 1-1: General rules and rules for buildings. [7] STN EN : 2006 Eurocode 3: Design of steel structures Part 1-1: General rules and rules for buildings [8] STN EN : 2008 Eurocode 3: Design of steel structures- Part 1-5: Plated structural elements [9] STN EN : 2008 Eurocode 4: Design of composite steel and concrete structures Part 2: General rules and rules for bridges. [10] Bujňák, J. Hric, M.: Steel structures and bridges (Some problems of design and assessment) University of Žilina, In Slovak.