CACULATING AND DESIGN FOR a RAILWAY BRIDGE USING FINITE ELEMENT METHOD

CACULATING AND DESIGN FOR A RAILWAY BRIDGE USING FINITE ELEMENT METHOD The Van Tran, Trong Nghia Hoang, Anh Tuan Do Hung Yen University of Technology and Education ABSTRACT: Finite el

CACULATING AND DESIGN FOR A RAILWAY BRIDGE USING FINITE

ELEMENT METHOD The Van Tran, Trong Nghia Hoang, Anh Tuan Do

Hung Yen University of Technology and Education

ABSTRACT:

Finite element method is a popular and

efficient method for mumerical solving the different

technical problems as stress analysis and strain

analysis in mechanical structures of the car

elements, high building structures and bridge

bars, etc In this study, a structure of the railway

bridge is modelling for calculating stress, strain and displacements of bars The calculated results obtained base on finite element method is compared with that of Matlab It shows that the built railway bridge model is statified with the real model

Keywords: railway bridge, finite element method, stress, strain, displacement

1 INTRODUCTION

A railway bridge is a structure designed to

carry freight and passenger trains across an

obstacle in the landscape These bridges

represent complex feats of engineering and

design, and often require the cooperation of a

team of engineers and builders While many

railway bridges are designed to cross bodies of

water, others span valleys, canyons, or other

obstacles that once prevented rail travel within the

area A railway bridge often has a major impact

on travel, allowing for shorter trips and faster

freight delivery, as the train no longer needs to

take a longer route around the obstacle As the

popularity of train travel declines, railway bridges

are often preserved or reconfigured for other

uses, such as hiking or bike trails

As rail travel is replaced by other forms of

travel, rail bridges continue to play an important

role in society Many are celebrated for their

beauty or structure, while others are adopted by

historic preservation groups In the US, "rails to

trails" programs are particularly popular As part of these programs, communities transform old railway paths and bridges into scenic trails for recreation and hiking

Compared to road bridges, railway bridges are different, because the trains that pass bridges bring about different requirements When trains pass a bridge, the traffic loads are higher, which means that the relation between dead load and live load is a different one as compared to road bridges Translated into the language of the engineer, higher forces move relatively fast over a structure, having implications on the design of the bridge itself as well as for the protection system

of the bridge

The maximum deflection of a railway bridge is dependent on speed of the train, span length, mass, stiffness and damping of the structure, axle loads of the train Up to now railway bridges have been designed only due to a static analysis

2 MODEL OF STRUCTURE

The basic railway bridge consists of a simple

beam or girder, and is designed to cross short

spans, such as a small creek The addition of

triangular trusses allowed for longer, stronger

railway bridges Railway engineers also took

advantage of the natural strength of the arch to

design bridges with an arch-shaped support Suspension bridges rely on high-tension cables for support, which allows them to span even greater distances than earlier bridge designs The most advanced units featured things like double-decker construction, allowing railcars to share the same bridge as vehicles or pedestrians

Because A railway bridge must be equipped to

handle the extreme loads of a train and its cargo,

as well as the additional forces generated by the

speed of the train However, this topic we assume

there is one train stop on the train, since the railway bridge is applied constant load

The railway bridge is modeled as following figure:

Figure 1 Some types of railway bridges

(a)

(b)

Figure 2 Some example about applying load on the railway bridge

3 THEORETICAL CALCULATING MODEL BY

FINITE ELEMENT METHOD

A railway bridge must be equipped to handle

the extreme loads of a train and its cargo, as well

as the additional forces generated by the speed of

the train These bridges should also be capable of

withstanding extreme wind and weather In this

paper, the train stopped on the bridge and the

applied load of train is static load (Figure 3(a))

3.1 Node numbering scheme

Figure 3(b) shows a node numbering scheme

characteristic matrix depends on the node numbering scheme and the number of degrees of freedom considered per node If the bandwidth can minimize, the storage requirements as well as solution time can also be minimized The

bandwidth (B) is defined:

 1 

B  D  f (1)

Where D is the maximum largest difference in

the node numbers occurring for all elements of

the assemblage, and f is the number of degrees

of freedom at each node The previous equation indicates that D has to be minimized in order to minimize the bandwidth Thus, a shorter

Node Element

bandwidth can be obtained simply by numbering

the nodes across the shortest dimension of the

body

Table 1 Node index of the element

Node i Node j

(a)

(b)

Figure 3 The model calculation (a) and Node numbering scheme (b) 3.2 Determine element stiffness matrix

The review a one general element:

cos v sin cos sin

v

i

i

u

u u       

  (2)

sin v cos -sin cos

v

i

i

u

  (3)

In matrix form:

cos sin





    

For the two nodes of a bar element:

cos sin sin cos

co

0

s sin sin co





(5)

The nodal forces are transformed in the same way:

cos sin sin cos

co

0

s sin sin cos 0





(6)

where f' and f are the force in the local and global

coordinate system, respectively

In the local coordinate system, the displacements

can be obtained:

'

'

'

1 0 1 0

'

'

1 0 1 0

'

i

j

v EA

u

v



(7)

Figure 4 Coordinate system for a node

Table 2 Node in local and global systems

coordin

ate

Element stiffness matrix:

2

2

EA

K

(8)

where C=cos, S=sin and  is the angle

between two elements

Table 3 Angle of elements in global coordinate system

elements Angle C2 S2 CS

e1, e5, e9 600 1

4

3 4

3 4

e2, e4, e6, e8, e10 00 1 0 0

e3, e7, e11 1200 1

4

3 4

3 4



Element stiffness matrix for elements e1, e5, e9:

1,5,9

1 3 1 3

4 4 4 4

3 3 3 3

4 4 4 4

1 3 1 3

4 4 4 4

3 3 3 3

4 4 4 4

EA

K

L



(9)

Element stiffness matrix for elements e2, e4, e6,

e8, e10:

2,4,6,8,10

EA K

L





(10)

Element stiffness matrix for elements e3, e7, e11:

3,7,11

1 3 1 3

4 4 4 4

3 3 3 3

4 4 4 4

1 3 1 3

4 4 4 4

3 3 3 3

4 4 4 4

EA K

L



(11)

Apply load and boundary conditions:

210.10 , 280.10



3.3 Calculating process for stress, strain and displacement of bars by Matab software

The block diagram of program is constructed

as following:

Reading input data:

materials, geometric structure, meshing control, load, connecting elements, bound conditions

Calculating element stiffness matrix [k]e = node and calculating element load vector[f]e = node

Defining global stiffness matrix [K] and load vector[F]

Determining displacement vector of nodes by solving equations [K][u] = [F]

Setting bound conditions

Calculating stress, strain, reaction force,…

Printing results:

Displacement, stress, strain, reaction force

Figure 5 The block diagram of Matlab program

- The first: input data of the structure as elastic

module (E), length of bars (l 1 ,l 2 ,l 3 ,…), load

(P1,P2,…), bound conditions is setup

- The second: element stiffness matrix and

element load vector is calculated

- The third: global stiffness matrix and global load

vector are defined from element stiffness matrix

and element load vector based on connecting

algorithm

- The fourth: bound conditions are setup as Eq

(12)

- The fifth: displacements of nodes determined

- Last one: The stresses, strains and reaction

forces are determined from displacement

4 NUMERICAL RESULTS

A railway bridge is assembled from steel, the

same cross-section of steel bars with each other

and equal 3250 mm2 The train stopped on the

bridge, which have to apply the load of the train

(as Figure 3(a)) The displacements at each

nodes stresses and strains in each bar, reactive

load is determined

By solving above simultaneous equation (Eq

(7)), the displacements can be obtained After

that, the stress in each bar is calculated as

following:

,

,

0 0

1 1

0 0

i

i

j

i i j j

u v E

u L

v

        

 

 

 

 

 

 

(13)

From stresses in each bar, the strain is calculated

as:

i

i

i

E



From the global FE equation, the reaction forces

is calculated

,

,

.

i

j

u

u



 

 



 



(15)

The displacements at each nodes stresses and strains in each bar, reactive load is presented in Table 4 and Table 5 It shows that the results are obtained from Matlab program is approximately with that of the results obtained by directly solving equations

Table 4 Deplacement and load of nodes on elements

Nodes

Deplacement of nodes

(mm)

Load of nodes

( )N

Matlab Hand Matlab Hand

1

0 0 0 7.805

0 0 233333 233326.547

2 2.937 2.937 0 0 -3.336 -3.337 0 0

3 0.711 0.711 0 0 -6.263 -6.263 -210000 -210000

4 1.516 1.516 0 0 -6.892 -6.892 0 0

5 2.2026 2.20281 0 0 -6.659 -6.660 -280000 -280000

6 -0.047 -0.047 0 0 -3.555 -3.556 0 0

7 2.984 2.985 0 0

0 0 256670 256659.968

Table 5 Strain and stress of the elements

Elements

Strain of elements

3

10 ( mm)



Stress of elements

2

(N mm/ )

Matlab Hand Matlab Hand

1 -0.3948 -0.3948 -82.8995 -82.9016

2 0.1974 0.1974 41.4467 41.4508

3 0.3948 0.3948 82.8995 82.9020

4 -0.3947 -0.3948 -82.8934 -82.9016

5 -0.0395 -0.0395 -8.29 -8.2903

6 0.4145 0.4145 87.038 87.0465

7 0.0395 0.0395 8.29 8.2901

8 -0.4342 -0.4343 -91.1827 -91.1915

9 0.4342 0.4343 91.1895 91.1917

10 0.2171 0.2171 45.5914 45.5963

11 -0.4342 -0.4343 -91.1895 -91.1919

6 CONCLUSIONS

This paper has proposed a highly practical

model for calculating and design railway bridge

using finite element method This method can be

used to solve more 1–D and 2-D problems The

algorithm of program is designed exactly Matlab program runs stability and gets good result the same calculating result The program is highly flexible

REFERENCES

[1] V C Nguyễn, V H Trần, T B Mạc, Phân

tích thiết kế cơ khí, Nhà xuất bản Khoa học

và Kỹ thuật (2016)

[2] X L Nguyễn, Phương pháp phần tử hữu

hạn, NXB GTVT (2007)

[3] I T Trần, N K Ngô, Phương pháp phần tử

hữu hạn, NXB Hà Nội (2007)

[4] H S Govinda Rao, Finite element method and classical methods, New age international publishers (2007)

[5] S S Rao, The Finite Element Method in Engineering, 4nd ed Elsevier Butterworth– Heinemann, USA (2005)

TÍNH TOÁN THIẾT KẾ CẦU ĐƯỜNG SẮT XE LỬA ỨNG DỤNG

PHƯƠNG PHÁP PHẦN TỬ HỮU HẠN

TÓM TẮT:

Phương pháp phần tử hữu hạn (PTHH) là một

phương pháp rất phổ biến và hữu hiệu cho lời giải

số các bài toán kỹ thuật khác nhau như phân tích

trạng thái ứng suất, biến dạng trong các kết cấu

cơ khí, các chi tiết trong ô tô, khung nhà cao tầng,

dầm cầu, Trong nghiên cứu này, kết cấu cầu xe

lửa đường sắt là được mô hình hóa cho việc tích

toán các trạng thái ứng suất, biến dạng và chuyển

vị trong các thanh Kết quả thu được từ việc tính toán dựa trên lý thuyết phần tử hữu hạn được so sánh với kết quả từ phần mềm Matlab Từ đó cho thấy, mô hình đã xây dựng phù hợp với thiết kế thực tế cho các kết cấu cầu đường sắt

Từ khóa: cầu đường sắt, phương pháp phần tử hữu hạn, ứng suất, biến dạng, chuyển vị