What does superimposed dead weight mean

Top 4 types of steel bridges (with examples)

This article highlights the four top steel bridges. The types are: 1. Rolled steel girder bridges 2. Bridged girder bridges 3. Plate girder bridges 4. Truss girder bridges.

Type # 1. Rolled steel girder bridges:

This is the simplest steel bridge with RSJ as the girder and steel trough slab, filled with concrete or reinforced concrete slab as the bridge deck, as shown in Fig. 14.1.

These bridges have very small spans and are built over canals or small canals where cleaning is negligible and shallow foundations are possible to reduce foundation costs. Because the load-bearing capacity of these bridges is limited, these bridges are suitable for village streets where both the weight and the frequency of vehicle traffic are lower.

Type # 2. Coated Girder Bridges:

Coated girder bridges can cover comparatively larger spans than RSJ bridges, as their section modulus is increased by increasing the flange area through additional plates attached to the flanges by riveting or welding (Fig. 14.2).

Type # 3. Plate girder bridges:

If the span of the bridge exceeds the span of the clad beam bridges, plate girder bridges are used. In such bridges, the depth of the girder, taking into account bending and deflection, is such that rolled steel girders are not suitable, and therefore the girders with plates and angles are made either by riveting or by welding.

If the bridge is continuous, only two beams can be used on each side. In the case of deck-type bridges, however, any number of girders can be used, depending on economic considerations.

The section module required for the plate carrier in different sections, such as in the middle, a third section, a quarter section, etc., varies depending on the moment at these sections, and as such the flange plates can be shortened at a few moments at B. on the ends for simply supported straps.

The components of a plate carrier are as follows (Fig.14.4):

1. Web plate

2. Flange plates

3. Flange angle

4. Rivet or weld the flange angle to the flange plates and the web plate.

5. Vertical stiffeners attached to the multi-walled sheet at intervals along the length of the beam to prevent kinking of the multi-walled sheet.

6. Horizontal stiffeners that are attached to the multi-walled sheet in the depth direction, one or more, to prevent the multi-wall sheet from buckling.

7. Bearing stiffeners at the ends above the center of the bearing and at intermediate points under the point loads.

8. Track connecting plates for connecting the two track plates.

9. Flange connection plates for connecting the two flange plates.

10. Angle splice plates to connect the two flange angles.

11. Bearing plates at the ends that rest on the pillars / abutments.

The entire length of panels and angles for making the panel carrier that require splicing may not be available. The flange plates are usually spliced ​​near the ends for easy supported spans while the web plate is spliced ​​in or near the center.

To prevent the multi-skin sheet from buckling, vertical and horizontal stiffeners are provided using ms angles. At each end and also at concentrated heavy loads, bearing stiffeners are required for the transfer of loads. The bearing stiffeners are not curled and the packing plate is used between the web and the stiffener bracket, but intermediate bracket stiffeners are usually curled.

The construction of a plate carrier involves the following steps:

1. The calculation of BM and SF in different sections says a quarter, a third and a half.

2. Estimation of the required section modules at different sections.

3. Design of the mesh from shear considerations.

4. Construction of flange angles and flange plates in order to obtain the necessary cross-sectional modules at different sections.

5. Shortening of flange plates and flange angles, taking into account reduced values ​​of the required cross-sectional modules in the vicinity of the end sections.

6. Construction of rivets or weld seams that connect various elements such as flange angle with web plate and flange angle with flange plates.

7. Design of splices such as flange splice and rail splice.

8. Design of stiffeners.

9. Construction of the bearing plates.

Example 1:

A simply supported plate girder bridge with a span of 20 meters carries a dead load of 50 KN / m without the dead weight of the girder and a payload of 60 KN / m per girder. Design the panel support in the center of the span taking into account the impact deviation according to the IRC code.


Dead load = 50 KN / m.

Load with impact = 60 x 1, 269 = 76, 14 KN / m. Total superimposed load with impact without dead weight of the girder = 50 + 76, 14 = 126, 14 KN / m.

The weight of the plate carrier per meter of length is approximately given by WL / 300, where W is the total superimposed load per meter and L is the span in m.

. . . Weight of the plate carrier = WL / 300 = (126, 14 x 20) / 300 = 8, 41 KN / m

Design of the multi-skin sheet:

Assume the thickness of the multi-skin sheet, t w = 12 mm. The economic depth of a plate carrier is given by

Where M = maximum bending moment; f b = permissible bending stress; t w = Thickness of the multi-skin sheet.

Web width = 2000 mm.

Design of the flange plates:

Net Flange Area Required for the Stress Flange, A t = M / f b d = 6750 x 10 6/138 x 2000 = 24 456 mm 2 . If 4 No. 22 mm. Dia rivets are used to connect flange plates with flange angles and 4 rivets are used to connect flange angles to the web plate and 2 numbers. 500 mm x 16 mm. Flange plates and 2 no. To manufacture the plate carrier, flange angles of 200 mm x 100 mm x 15 mm are used. The following net flange areas are then available:

The details of the plate carrier are shown in Fig. 14.5.

Check the bending stress:

Check for shear stress:

Enter # 4. Truss bridges:

Truss or truss bridges have a top or top chord, bottom or bottom chord, and web links that are vertical and diagonal. With a simple truss bridge, the upper chord is compressed and the lower chord is tensioned.

The bridge links can only be diagonals as in the Warren Truss (Fig.14.6a) or a combination of verticals and diagonals as in the modified Warren Truss (Fig.14.6b) or Pratt Truss (Fig.14.6c & 14.6d) or Howe Truss ( Fig.14.6e) or Parker Truss (Fig.14.6g).

In the case of larger spans, the panels are again subdivided for structural reasons, e.g. B. for trusses with diamond bracing (Fig.14.6f), Pettit truss (Fig.14.6h) or K-truss (Fig.14.6i). The span for a single supported truss bridge is 100 to 150 meters.

The truss bridges can either be of the deck type or of the passage type (Fig. 14.7), that is, the bridge deck is near the upper chord in the first type and near the lower chord in the second type.

Needless to say, the parallel tendon trusses shown in Figs. 14.6a to 14.6c can either be of the deck type or of the continuous type, as in Figs. 14.7a and 14.7b, but trusses with a curved Lopps tendon, as shown in Fig. 14.6g to 14.6i are always of the passage type (Fig. 14.7c).

The bridge deck is on longitudinal girders that rest on transverse girders that transfer the loads to the trusses at each panel joint. Details of a truss bridge are shown in Fig. 14.8. Since the truss elements are not loaded except at the plate connections, the truss elements are only subject to direct tension, either tension or compression, and no bending moment or shear force occurs in the truss elements.

The plate joints where the members meet are assumed to be articulated and therefore no bending moment develops in the truss members even due to the deflection of the scaffolding.

Determination of forces in statically determined traverses:

The forces in the truss members are determined using the following methods if the trusses are statically determined:

1. Graphical method based on stressor force diagrams.

2. Method of Sections.

3. Method of dissolution.

The above methods are illustrated by way of an illustrative example.

Example 2

A simple equilateral triangular truss with a load of 30 kN on joint 2 of the trestle is shown in Fig. 14.9a. Calculate the forces in the truss members using the three methods mentioned above.

Graphic method:

The members are numbered 0 in the middle of the framework and A, B, C on the outside and are counted clockwise. Hence the reactions are AB and CA. The members are OB, OC and OA. Reaction AB = Reaction CA = 15 KN.

Since the loads and reactions are vertical, a force diagram is drawn on a suitable scale (Fig. 14.9b), which is also vertical. In this diagram, bc stands for W downwards, ca upwards for R 2 and up for R 1 . Since R 1 + R 2 = 30 KN, the force diagram also shows bc = ca + ab = 15 + 15 = 30 KN.

Now the force diagram is drawn. Looking at connection 1 of the frame, a line bo in the force diagram is drawn parallel to BO and a line ua is drawn in the force diagram parallel to AO. The triangle, oab, is the force triangle for connection 1 and ab, bo, oa, around the reaction R 1 and to scale the internal forces in BO or OA.

Similarly, in connection 2 W is the external load or force represented by bc in the force diagram. The lines ob and oc are drawn parallel to members OB and OC.

The triangle, bco, is the force triangle for connection 2 and bc, whether the reaction W and the internal forces should be scaled in OC or OB. The force triangle for joint 3, namely cao, is drawn similarly; ca, ao and oc represent the reaction R 2 and to scale the internal forces in the term AO and OC, respectively.

The values ​​of the internal forces in the elements are known from the force diagram shown above. The nature of the force, namely whether the force is tensile or compressive, can also be determined from the same force diagram.

In every force triangle diagram, the path of the forces emanating from the known force is traced in the same direction, and these directions are indicated in the frame diagram. For example, in the triangle of the force diagram above, it is known that ab (= reaction R 1 ) acts upwards.

In this way, the direction of force bo and oa is shown as in the force diagram and also displayed in the frame diagram. A force toward a joint in the frame diagram indicates a compressive force, and a force away from the joint is a pulling force.

Thus, in joint 1, the known force is ab = R 1, which acts upwards and follows this path, the directions of force for bo and oa in the force diagram and for the element BO and OA in the frame diagram. The direction of force BO is in the direction of the joint and is therefore a compressive force.

Similarly, the direction of force OA is remote from the joint and is therefore a tensile force. In the same way and based on the force whose direction is known, the directions of all forces are shown in the frame diagram and thus the nature of all forces is known.

Method of sections:

In this procedure, the element whose force is to be determined is cut by a line that also intersects some other elements of the frame. The start occurs at a point where only one force is unknown. The frame remains in equilibrium even through the section when external forces act in the section elements as in Fig. 14.10 in the same simple frame as in Fig. 14.9.

The forces can be determined by taking a moment over a suitable joint so that only a known and an unknown force are involved. For example, in Fig. 14.10b, a cut XX is made in the frame cutting element AO and BO.

Take moment over joint 2, f OA x

/ 2 x 6 = 15 x 3 or f OA = 8, 66 KN ie take the moment away from the joint via joint 3, f IF x
/ 2 x 6 = 15 x 3. . . f IF = 17.32 kN, ie in the direction of the joint, ie compressive force.

Similarly, the force f OC be recognized by a cut YY and take a moment around joint 1.

Therefore, the forces in the elements determined by the method of the sections are as follows:

f IF = f OC = 17, 32 KN (pressure), f OA = 8, 63 KN (pull)

Method of dissolution:

With this method, all forces and external loads on a connection in the horizontal and vertical direction are dissolved and equal to zero, since the connection is in equilibrium. The start takes place from the connection on which the external load acts and there are no more than two unknowns.

The same numerical example as shown in 15.9 is intended to illustrate this procedure as well. The force towards a joint is compressive and the force away from the joint is pull.

Taking into account joint 1 and resolving f IF in horizontal and vertical directions and equal to zero, f is IF sin 60 ° + 15 = 0 or f IF = (-) {[15 x2] / √3} = (-) 17.32 KN ie, compression and f IF cos 60 ° + f 0 = 0 or f 0 = (-) f IF cos 60 ° = (-) 17, 32 x 1/2 = (-) 8, 66 KN i.e. tension.

If one considers connection 3, then f OC cos 60 ° + f O 0 = 0 or f OC = (-) 8, 66 x 2 = (-) 17, 32 KN pressure.

The forces obtained in the dissolution process in the frame are: f IF = f OC = 17.32 KN pressure. f 0 = 8.66 kN tensile strength.

It can therefore be noted that the forces are the same to the extent that they are worked out by the method of sections and the method of dissolution. The values ​​determined with the graphical method differ slightly because they have to be sealed and such a measurement error occurs. For all practical purposes, however, these values ​​are acceptable and the design can be carried out without hesitation.

Determination of the forces in trusses with a redundant member :

Hence, some other methods are to be used to find out the forces in such trusses, two of which are discussed below:

1. Method based on the least labor principle.

2. Maxwell method.

Method based on the principle of least labor:

A consequence of Castigliano's theorem is that the work done in stressing a structure under a given load system is as little as possible, which is consistent with maintaining balance. Therefore, the difference coefficient of the work done in relation to one of the forces in the structure is zero. this is that "Principle of Least Labor", which is used to evaluate the forces in statically indeterminate trusses.

The strain energy stored in any element of length L and cross-sectional area A under a direct force P is given by

And the work done throughout the structure is:

When evaluating the forces in the cross member, proceed as follows:

1. Remove the superfluous member and calculate the forces that result in the remaining members of the truss (which is now statically determined) due to the external load. The forces in the limbs due to the above are F. 1, F 2, F 3 (we say).

2. Remove the external load and apply a unit pull to the superfluous component to determine the forces in the trusses.

3. If K 1, K 2, K 3 etc. are the forces in the limbs created by pulling the unit in the superfluous limb, and if the actual force in the superfluous limb of the truss due to the external load is T, then the total force in the limbs is T for that redundant link (since F = 0) and (F 1 + K 1 T), (F 2 + K 2 T), (F 3 + K 3 T) etc. for other members.

4. The total work done in the structure, including that in the redundant member, is:

5. The difference coefficient of the work done in relation to the force T in the superfluous term is therefore given by:

Maxwell's method:

This process is also based on the overall work done in emphasizing the structure. The fundamental difference of this method, however, is that no internal force T is induced in the redundant element, but that this force acts as an external load.

This means that in the previous method, which is based on the principle of least work, the strain energy of the superfluous member is also included in the total work, since the force T in the superfluous member is an internal one, but in the Maxwell method it is the force T. An external work and therefore does not contribute to the overall work done by the stress on the structure.

In the Maxwell Method, Castigliano's first theorem is used to evaluate the forces in the superfluous limb as follows:

1. Step 1 through Step 4 as in the previous method. However, in step 3, unit load and T are external loads along the redundant element.

2. The total work without the dismissed member will be:

According to Castigliano's first theorem, the differential coefficient of the total deformation energy in a structure with respect to any load indicates the deformation of the structure along the direction of the load.

Therefore ∂U / ∂T indicates the deformation of the redundant element in the direction T.

As a consequence of the force T in the superfluous link, the deformation of the link is also given by the following relation:

Where L 0 and A 0 Length and cross-sectional area of ​​the superfluous element.

A minus sign in equation 14.7 is used because the deformation in equation 14.6 gives the value of δ in the direction of T, but as a result of the tensile force T, the deformation in the element will be in the opposite direction.

The values ​​of T can be determined from Equation 14.8, since all other values ​​except T are known. If the value of T is known, the forces in all members of the truss can be determined, e.g. B. T in the redundant term and (F 1 + K 1 T), (F 2 + K 2 T), (F 3 + K 3) T) etc. in other members.

It can also be noted that although the carrier with a redundant element is analyzed by two different methods, the result is the same, as can be seen from equations 14.4 and 14.8.

Example 3:

A bridge structure with a redundant bar on the central plate and with 200 KN vertical and 100 KN horizontal loads acting on one of the top panel nodes is shown in Fig. 14.11. Find the forces in all members of the framework.

The framework is hinged to one support and is provided with roller bearings on the other support. To simplify the calculation, it can be assumed that the ratio of length to cross-sectional area is the same for all elements.

Least labor method solution:

1. The superfluous member BE is removed and the forces in all remaining members of the truss, which are now statically determined, are determined by one of the following methods:

(i) Graphical method using a stress or force diagram

(ii) Method of Sections

(iii) Dissolution Process.

This is listed in Table 14.1. Fig. 14.12a shows external stresses and reactions.

2. The external loads are removed, a unit train is applied in the redundant element (Fig. 14.12b) and the forces K 1, K 2, K 3 etc. found in different elements. This is also shown in Table 14.1.

Determination of the forces in trusses with two or more redundant members:

The procedure for determining the forces in truss with two or more redundant members is the same with some modifications due to the presence of more than one redundant member, and the principle of least labor can also be used in this way.

This is explained below:

1. Remove the redundant links so that the carrier becomes perfect and will not be distorted after removing the redundant links. The truss in Fig.14.13a has two redundant links BG and DG, which are removed as shown in Fig.14.13b. This latter beam is statically determined and the forces in the links with the external loads are determined. The forces in the limbs are e.g. F 1, F 2, F 3 etc.

2. Remove the external load and pull on the unit for the redundant link BG (Fig. 14.13c). If K 1, K 2, K 3 etc. are the forces in the links that arise from the pulling of the unit in the superfluous link BG, and if the actual force in the superfluous link BG due to the external load is T, then the total forces in the other part are members ( F. 1 + K 1 T), (F 2 + K 2 T) etc.

3. Next, put on a unit train in the redundant element DG (Fig. 14.13d), if K ' 1, K ' 2, K ' 3 etc. are the forces in the members due to a unit pull in the superfluous element DG and If the actual force in the superfluous member DG due to the external load is T ', then the forces in the other members are K' 1 T, K ' 2 T 'etc. due to the force T in the redundant term DG.

The actual forces in the other elements due to steps 1 through 3 are (F. 1 + K 1 T + K ' 1 T), (F 2 + K 2 T + K ' 2 T) etc.

5. The overall performance of the structure including that of the dismissed members will

All of the terms in Equations 14.13 and 14.14 are known except for T and T ', and as such, by solving these two simultaneous equations, the values ​​of T and T' can be calculated. Knowing the values ​​of T and T 'will determine the forces in other elements from step 4, i.e. (F 1 + K 1 T + K ' 1 T), (F 2 + K 2 T + K ' 2 T) etc as done in example 3.

Influence lines for truss bridges:

The bridge girders are exposed to moving loads, so that the forces in the truss links cannot be assessed if the lines of influence are not supported.

Therefore, it is important to draw the lines of influence for the forces in the various truss elements, and the maximum value for each truss element is thus determined after the moving loads have been placed for maximum effect. The moving loads from the carriageway come on each traverse on both sides of the carriageway only with panel connections.

The total load is shared equally by each truss. The influence line diagram for the upper and lower chords is drawn for the BM, while the influence lines are drawn for the diagonals and the vertical terms for the SF

The bridge types commonly used are shown in Fig. 14.6, and the lines of influence vary depending on the type of truss and the position of the component in the truss. However, the principle of drawing the line of influence is explained for a parallel chord Pratt framework by an illustrative example.

Example 4:

Draw the lines of influence for the force in the lower chord AB, the upper chord LK, the diagonals AL & LC and the vertical BL of the Pratt truss bridge shown in Fig. 14.14. Also calculate the maximum force in the diagonal AL and the lower chord AB when the IRC class AA is passed over the bridge by a single track. Plate length = 6 m and cross member height = 8 m.

Influence line for the force in the diagonal, AL:

Cut the lower tendon AB and the diagonal AL with a cutting line 1-1 (see Fig. 14.15a). Draw a vertical line BN from B to AL. If a unit load moves from one end of the bridge to the other, let the responses at A and G R 1 or R 2 . The left part of the cut truss is in equilibrium for each position of the load unit in the bridge deck.

Influence line for Bottom Chord AB:

Look at section line 1-1 as before.

Take moment at L, f FROM xh = R 1 a or f FROM = R 1 a / h = M 1 / h (voltage)

Therefore the line of influence for the force in the lower tendon AB is equal to 1 / h times the line of influence for M L, which is a triangle with an ordinate equal to x (L - x) / L, ie 5a / 6. Therefore, the ordinate is the line of influence for f FROM at L the same

as shown in Fig.14.15c.

Influence line for vertical BL:

When a unit load moves from A to B, the tension in the vertical member BL becomes from zero to one. Again, the voltage in BL decreases from one to zero as the unit load moves from B to C. After that, the voltage in BL is always zero when the unit load moves from C to G. Hence, the line of influence is for the vertical element. BL is a triangle with a maximum ordinate equal to one, as shown in Fig. 14.15d.

Influence line for diagonal LC:

Look at section line 3-3 and notice that the unit load moves from A to B. If the equilibrium of the right side of section line 3-3 is taken into account, it turns out that the force in the diagonal LC is close to the joint C is down because the external force, i.e. the reaction R 2that is to be balanced by the force in LC is directed upwards.

Hence the force in LC is compressive and its magnitude is given by f LC sin? = R 2 or f LC = R 2 / Sin? = R 2 cosec? (Compression)

Next, consider the balance of the truss to the left of section line 3-3 as the unit load moves from C to G. With the same reasoning as before, the force in LC near the junction L is directed downwards, since the reaction R 1 acts upwards. Hence the diagonal LC is in tension and the magnitude is given by: f LC sin θ = R 1 or f LC = R 1 cosec θ (tension)

The influence line for R 1 and R 2 are triangles with ordinates one and zero at A and G for R. 1 and with the ordinates zero and one at A and G for R, respectively 2 . Therefore the line of influence for LC cosec is 8 times as large as the line of influence for R 2 from A to B and compressing.

The line of influence for LC is cosec & thetas; times the influence line for R 1 from C to G and the tensile strength. The line of influence for LC between B and C is a line connecting the ordinates at B & C, which are 1/6 cosec θ (pressure) and 2/3 cosec θ (tension). The influence line for LC is shown in Fig. 14.5c.

Influence line for Top Chord LK:

Look at the truss to the left of section line 3-3. Let's take a moment around C, f LK xh = R 1 x 2a or f LK = 1 / hx 2 aR 1 (Compression). 2aR 1 however, is the moment of the freely supported framework in C. . . f LK = Mc / h (compression).

Maximum forces in members due to IRC Class AA shift

Truss length = 6a = 6 x 6 = 36 m

Truss height = h = 8m.

Total load per traverse = 35 tons

Loading length = 3.6 m.

Load intensity per meter of length = 9.62 tons.

Distribution factor due to 10 eccentricity of load = 1, 2 (say)

Efficiency factor = 10%.

Force in the diagonal AL:

Power in the lowest chord AB: