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Chapter 3 LOADS ON BRIDGES contents 3.1 Loads on bridges 3.2 Dead loads on highway and railway bridges Fig(3-1) 3.3 Live loads on bridges 3.4 Impact loads 3.5 Centrifugal force Fig(3-4) 3.6 Temperature effect ((5-5),(6-5)) Fig(3-4)

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## Chapter 3

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**Chapter 3**LOADS ON BRIDGES**contents**• 3.1 Loads on bridges • 3.2 Dead loads on highway and railway bridges Fig(3-1) • 3.3 Live loads on bridges • 3.4 Impact loads • 3.5 Centrifugal force Fig(3-4) • 3.6 Temperature effect ((5-5),(6-5)) Fig(3-4) • 3.7 Wind pressure ((5-9),(6-9)) Fig(3-5) , Fig (3-6)**3.8 Braking force Fig (3-7)**• 3.9 Lateral shock effect (6-7) Fig (3-8) • 3.10 Frictional resistance of bearings ((5-10),(6-10)) • 3.11 Settlement of supports ((5-11),(6-11)) Fig (3-9) • 3.12 Forces due to erection ((5-14),(6-14))**3.1 Loads on bridges**Loads acting on bridges are divided into:- 1. Primary loads. 2. Secondary loads. A load is considered primary or secondary according to the part of the bridge which shall be designed. Wind loads are secondary loads in designing the main girders and primary loads in designing the wind bracings.**3.1.1 Primary loads on highway and railway bridges**1-Dead loads. 2-Live loads. 3-Impact loads (dynamic effect). 4-Centrifugal forces. 3.1.2 Secondary loads on highway and railway bridges 1-Wind pressure or earthquake. 2-Braking force. 3-Lateral shock effect. 4-Temperature effect.**5-Frictional resistance at movable bearing.**6-Forces due to settlement of supports. 7-Effect of shrinkage and creep of concrete. 8-Forces due to erection. 3.2 Dead loads on highway and railway bridges Fig(3-1) It consists of the weight of steel structure and the bridge floor. The weight of the floor is found from the dimensions and the unit weight of the different materials. Weight of an open timber floor for a single track railway bridge = (250+350)=600 kg/m’ (9.3.4(p149)). back**The weight of the steel structure is first approximately**determined from similar existing bridges or from empirical formula. * Approximate weight of the steel structure for single track Railway Bridge with open timber floor (standard grade steel) - For through bridge W = 0.75 + 0.50 L t \ m’ - For deck bridge W = 0.50 + 0.50 L t \ m’.**For deck bridge without stringers and cross girders**W = 0.25 + 0.50 L t \ m’ Where, W = weight of total steel in (t) for one meter of bridge, L = effective span (leff) of bridge in meters, (for continuous bridge, leff = (0.70 – 0.80) L = the distance between two sequence points with zero total moment. We have to increase the above steel weight by 90 % for double track and 80 % more for triple track * For a ballasted floor the weight of steel is 20 – 40 % greater.*** For roadway bridge with heavy traffic the weight of steel**structure is approximately for 1 m2 of roadway and side walks as follows:- * Outside side walks W1 = 200 + 4 L + 0.03 L2 kg/ m2 for roadway W2 = 100 + 3 L kg/ m2 for side walks W = W1 + W2 kg/ m2 of bridge*** Inside side walks**W = 200 + 4 L + 0.03 L2 for roadway W2 = 100 + 3 L for side walks W = W1 + W2 kg/ m2 of bridge L = effective span of bridge in meter**Rail**1.5 m Rail Sleeper 100 cm Sleeper h = L/ 10 B = Bridge Width Railway Through Bridge W2 W1 W2 bs 0.5 0.5 bs B a b' Roadway Open Through Bridge Side Walk Outside W1 bs bs B a b' Roadway Open Through Bridge Side Walk Inside home Fig(3-1) back**3.3Live loads on bridges**Railway Bridge (p6-1)Fig(3-2) The type of trains is different for different countries according to the importance of lines. In Egypt we shall consider one type of three train types (D, H, and L), train type “D” is the heaviest train is used in Egypt. Train type “D” consists of a two locomotives and two tenders followed on one side only by an unlimited number of wagons Locomotive +Tender + Locomotive + Tender + unlimited number of wagons**Where total weight of one Locomotive = 100 ton, and its**length = 10.50 m(p6-3). While total weight of one Tender = 80 ton, and its length = 8.40 m. If two tracks are loaded at the same time, only 90 % of loads are used. In case of three tracks only 80 % of loads are used while in case of four tracks we use 75 % of loads are used(p6-2).**Locomative = 100 t**Tender = 80 t Locomative = 100 t Tender = 80 t Wagon = 80 t + Any Number of Wagons 12.5 12.5 12.5 12.5 12.5 12.5 10 10 10 10 10 10 10 10 10 10 10 10 6.25 6.25 6.25 6.25 1.5 1.75 1.75 1.5 1.5 1.8 1.8 1.8 1.5 1.5 1.75 1.75 1.5 1.5 1.8 1.8 1.8 1.5 1.5 1.75 5.5 1.75 1.5 2 2 2 2 [100 t] 10.5 m [80 t] 8.4 m [100 t] 10.5 m [80 t] 8.4 m [80 t] 12 m [180 t] 18.9 m [180 t] 18.9 m 12 m 9.52 t/m' 6.666 t/m' Loads of Train Type 'D' home Fig(3-2)**Roadway Bridges Fig(3-3)**Within the kerb to kerb width of the roadway, the standard vehicles are assumed to travel parallel to the length of bridge, and to occupy any positions, which produce the maximum stress. For the standard vehicle, all the axles of a unit of vehicle are considered as acting simultaneously in a position causing maximum stresses. The vehicle in adjacent lanes is taken as headed in the direction producing maximum stresses. The maximum bending moment and maximum shear force on the plate girders are found by longitudinal location of loading. For main roadway bridge, the L.L shall be that type of vehicular rolling load and/ or distributed load representing it.**a.The main girder.**As well as the floor system shall be designed for truck concentrated axle loads the standard type shown together with distributed load of 500 kg/m2 cover the main lane of three meter width lane(p5-1), the second lane of three meter width each shall be covered with one truck moving in the same direction and parallel to the axes of the bridge. The remaining parts of the floor are covered with a uniform load of 300 kg/ m2 (p5-3). Also, side walks shall be covered by the same distributed load. The impact will be considered for the loads on the main lane only. b.The elements of side walks. Fig(3-4) It shall be designed for 500 kg/ m2, and then we check**also for a vertical concentrated load of (5 t, acting**without uniform loads) in the position giving maximum stresses(5-2-2). The handrail, shall be designed for line distributed load not less than 150 kg/m' at top level of the handrail(5-2-6).**300 kg/m (Without Impact)**2 10 t 10 t 10 t 500 kg/m 500 kg/m 2 2 10 t 10 t 10 t (With Impact) (With Impact) 0.6 0.2 5 t 5 t 5 t 300 kg/m (Without Impact) 300 kg/m (Without Impact) 2 2 5 t 5 t 5 t 0.6 0.2 300 kg/m (Without Impact) 2 Loads on Traffic Lanes of Roadway Bridges home Fig(3-3)**Foot Bridges**It shall be designed for uniform load of 500 kg/ m2, without impact(5-2-4). 3.4 Impact loads Impact is the dynamic effect on the bridge due to the moving loads. If we measure the deflection at a certain point of the bridge for slowly moving train (static L.L), and for rapidly moving train (static L.L + Impact), the increase of deflection in the later case is due to impact. back**The chief factors causing Impact are:-**1.Roughness and unevenness of the track of a railway bridge or of the roadway surface of a roadway bridge. The smoother of the surface the smaller will be the impact. In Railway Bridge the joints of rails increase the impact. It is recommended to use long rails on bridges or to weld the joints. 2.Irregular and eccentric wheels are defective springs. · The proportion is called Impact coefficient This coefficient depends on the loaded length and on the type of**· This coefficient depends on the loaded length and on**the type of the bridge. · In case of bigger loaded length we have smaller Impact coefficient. ·Rigid parts are more affected by impact than elastic parts. · For main truss member, the impact decreases as the loaded length increases, since the time necessary to cover a greater length is more and the load is applied less suddenly. · Impact formula for Railway Bridges (6-3) is:-**where, 0.25 I 0.75**Where I is the factor by which the live load is to be multiplied to give the addition due to dynamic effect · L = loaded length in meters of track or the sum of loaded lengths of double or multiple tracks producing maximum stresses in members. EXAMPLE Span of bridge = 50 m. Distances between X.G. = 5.0 m**1. Single Track Bridge**For stringer use I = 0.75 For X.G. use I = 0.704 For M.G. use I = 0.325 2. Double Track Bridge For stringer use I = 0.75 For X.G. use I = 0.546**For M.G. use I = 0.25**Impact formula for Roadway Bridges (5-2-3) is:- , L = loaded length · Where I is the impact coefficient (for the main lane only) due to vertical concentrated loads (60 t) and uniform distributed live load (500 kg/ m2). · L = loaded length in meters of traffic main lane producing maximum stresses in members. back**3.5Centrifugal force Fig(3-4)**For bridges in curves, the stresses due to the centrifugal action and the super elevation of the track must be considered in designing the members. A vertical load w moving in a curve of radius R and a speed V. For Railway bridges, (6-4) W = axle load in tons**(R = 400 – 600 m)**(V = 70 – 90 km/ hr) C is a horizontal force acting at the center of gravity of masses 2.00 above the top of rail (6.4.1) . It produces an increase of the vertical reaction on the outside rail and a decrease of the vertical reaction on the inside rail. For Roadway bridges, (5-4) R = radius of curve in m C = centrifugal force in tons every 50 m back**3.6 Temperature effect ((5-5),(6-5)) Fig(3-4)**When steel structure is not free to expand or contrast under variation of temperature, the stresses due a variation of 30 C. From local main must be considered. The coefficient of expansion for steel and concrete is 0.00001. If we consider unequal variation of temperature, in some structures which are not affected by equal changes, we allowed only for 15 C. In two hinged arches and suspension bridges the equal change of temperature has an effect on the internal forces. In continuous bridges the equal change of temperature has no effect because the girders are free to expand, but the unequal change has an effect. The horizontal displacement at point b in the main system**10 = F1tL**110 = S1tL get 10 10 + X111 = 0**· Temperature has large effect.**· The modulus of elasticity E = 2100 t/cm2 for steel. E = 1000 t/cm2 for cast iron. E = 210 t/cm2 for concrete**bs**bs B Curb Side walk Stringer X.G. Bracket b' Loads on Side Walk C Outside Rail 2 Inside Rail W Centrifugal Force X1= 1t X1= 1t L Temperature Force home back Fig(3-4)**3.7 Wind pressure ((5-9),(6-9)) Fig(3-5) , Fig (3-6)**For bridges we consider either the case of unloaded bridge with a wind pressure of 200 kg/m2 or the case of loaded bridge with a wind pressure 100 kg/m2 on exposed surfaces of bridge and train. The effective height of a train in railway bridges is 3.50 m from the rail level, and that for crowds or road vehicles is 3.00 m. The train is considered as having on single vertical plane only.**Wind Loads**200 kg/m 2 Unloaded 3.50 m 100 kg/m 2 Loaded B = Bridge Width 3.50 m 100 kg/m 2 Loaded B = Bridge Width home Fig(3-5)**Wind Loads**Unloaded 2 200 kg/m 2 100 kg/m 3 Loaded 2 100 kg/m 2 200 kg/m Unloaded 2 100 kg/m 2 100 kg/m 3 Loaded home back Fig(3-6)**3.8 Braking force Fig (3-7)**In railway bridges (6-6) we have to consider the stresses resulting from the application of brakes to the live load while passing on the bridge. The braking force is equal to 1/7 of the maximum Live Load, without impact, supported by one track only. In case of several tracks, the braking force on the second track is equal to 1/14 maximum L.L (of the second track). The braking force has a great effect on the design of the towers and also on the abutments and piers supporting the fixed bearing of bridges (hinged bearing). In roadway bridges (5-6) the braking forces ( 90 t) = 0.25 Loads on main lane (L) (= [(L - 6) 30.50 + 60]0.25) back**3.9 Lateral shock effect (6-7) Fig (3-8)**In railway bridges a single force 6t (without impact) is taken normal to the track at top rail level and in position giving maximum stresses. The stresses due to the lateral shock of locomotive wheel are considered in the design of:- 1-Stringer. 2-Stringer bracing. 3-Wind bracing. 4-End X-frame. 5-The bearings. 6-Rail connections.**7-The piers, the foundation.**(If there is My (due to lateral shock) use B.F.I.B, for stringer, to support My.) For railway bridges on a curve, only the greater of the centrifugal force or the lateral shock must be considered.**Braking force**B B B/ 2 B/ 2 B B X.G. X.G. Width of bridge Width of bridge M My max = B*a = M X.G. X.G. B B B/ 2 B/ 2 B B M.G. M.G. M.G. M.G. S X.G. X.G. B B B/ 2 B/ 2 B B stringer S stringer stringer X.G. X.G. B/ 2 B/ 2 B B B B B/ 2 neglect braking force neglect braking force B B/ 2 B/ 2 B B X.G. width of bridge 12.5 12.5 12.5 12.5 12.5 12.5 10 10 10 10 10 10 10 10 6.25 6.25 M.G. 3 1.8 1.8 1.8 3 1.75 1.75 3 1.8 1.8 1.8 2 2 2 2 L=Span of bridge Max. L.L. = Max. Sum of wheel loads on the bridge span B = {[Max. L.L./ 7] } / No. of X.G. home Figure (3-7)**Lateral Shock**6 t 6 t X.G. 6 t 6 t M.G. M.G. M.G. M.G. S 6 t 6 t X.G. X.G. S stringer stringer stringer X.G. X.G. 6 t X.G. 6 t 6 t M.G. M.G. S X.G. S stringer stringer stringer stringer X.G. home back Fig(3-8)**3.10Frictional resistance of bearings ((5-10),(6-10))**Forces due to friction at the expansion bearing under dead load only must be considered and the coefficient is:- F = RD.L. For roller bearing with one or two rollers = 0.03 For roller bearing with three or more rollers = 0.05 For sliding of steel on hard copper = 0.15 For sliding of steel on cast iron or cast steel = 0.25 back**3.11 Settlement of supports ((5-11),(6-11)) Fig (3-9)**Stresses due to unequal settlement of continuous structures supported on piers or abutments shall be added for all members. (Fig. ) (Maximum allowable settlement is 2.50 cm.) Settlement may be lead to the continuous structure to be simple structure; hence the internal forces will be increased.**Settlement of Supports**y1 y2 y3 L L 12 23 Figure (3-9) home back**3.12 Forces due to erection ((5-14),(6-14))**1-Erection by cantilever method. Additional stressed will be exists due to erection by cantilever method, so it must be considered during the design of bridge, also the allowable stresses are increased by percentage of 25 %, (0.58Fy 1.25 = 0.73 Fy) (2.5P8). If the erection of the bridge is done by the cantilever method, the biggest possible forces in the members during the erection must be considered in the design of these members. A higher working stress may be used (or 0.73 y) than for the complete bridge.**2-Erection by floating method.**It is used in cases of simple beam bridges. Where a loaded ship carries the structure up to the site of erection, then the loads are removed slowly till the structure has in its required erection level. back

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