Slope calculation with crushed stone foundation structure.

1

. Q 0

. Q

(5-32)

3 2 k 0 4 k 3 EJ 0

C  1 . M

4 2 EJk 2 0

Instead of (5-32) we have:

0

Y y A

 0 . B

4k2​​

 . M . C

 4K​

. Q . D

, (5-33)

C

In there:

0 x K x H

0 x 0 x

0

A x chkx .cos kx

B  1 ( chkx sin kx  Shkx cos kx )

x2

C 1 chkx sin kx

x2

D  1 ( chkx sin kx  Shkx cos kx )

x4

(5-34)

Solving further we find the formula to calculate Y, M, Q.

2. Calculate and test the rail.

The rails are located on small and evenly spaced sleepers, so they can be calculated as continuous beams placed on elastic bearings. However, because the distance between the sleepers is quite small and the contact surface between them and the rails is large, they can be calculated as beams on elastic foundations. Through calculations, it can be seen that the results of the two methods are not different, while the calculation by the second method is much simpler. Applying the formulas for calculating infinitely long beams, we have:

From formula (5 - 31) can be rewritten as follows:

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Y = - k . P k . 

2

M = 1

4. k

. P k . 

( 5 - 35 )

Q = - 1 . P . 

2k​

In the wild: 





= e -kx .( sin kx  cos kx )

= e -kx .( sinkx - coskx ) ( 5 - 36 )

= e -kx . coskx .

x - is the distance from the point of force application to the point under consideration.

4



4. EJ

K = ; 1 / Cm (5 - 37)

In which: - The anti-bending coefficient of the sleeper foundation is determined by the following formula:

= C. ab/l, (kg/cm2 ) ; a,b - are the width and length of the sleeper, when the sleeper supports two rails, b is taken as half the length; l - is the distance between the sleepers; E,J - is the bending stiffness of the rail; C - Cushion coefficient, depending on the type of foundation, look up in table (5 - 7).

Table (5 - 7). Allowable stress and cushion coefficient of some types of foundation.

Material Type

Table (V-8). Rail properties according to  C 8161 - 63, 7174 - 65 of the Soviet Union.

Features

The specific calculation sequence is as follows:

-) Calculate the pressure due to the wheel P k ;

-) Select a rail and look up all its geometric properties;

-) Calculate the deflection (settlement) of rail Y and bending moment M;

-) Check the stress in the rail according to the formula M/W  .

For example : Choose a rail that can withstand wheel pressure Pk = 20 T, placed on a gravel sleeper base. The sleepers have the following dimensions ahb = 25x20x260 cm, placed 50 cm apart, cushion coefficient C = 6 Kg/cm3, the load-bearing diagram is as shown below:

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Pk Pk Pk Pk

1m 4m 1m

Figure (V-36).

Solution:

1-) Choose P38 rail with the following characteristics:

x

J x = 1,223 cm4, E = 2,100,000 Kg/cm2, W x

2-) Calculate the working coefficient of the rail: 

(above) = 181.95 cm 3 , W

(below) = 180.29 cm3.

= Cab/ 2l = ( 6x25 x 260 )/ (2x50) = 390 kg/cm2;

3-) Calculate the K value:

390

4

4x2100000x1223

- 1

K = 4

4. E . J =

= 0.01396 cm

4-) Calculate settlement Y and bending moment M:

From the formulas

Y = - k

2. 

P k .  ;

M = 1

4.k

. P k.  .

we look up the table of functions and equivalent to functions  1,  2 in the textbook ''Foundation and Foundation'', the calculation results are given in table (5 - 9).

Note:

*) If any force is more than 3 m from the cross-section under consideration, its effect on the point is not considered.

already;

*) To ensure accuracy, relatively thick points (cross-sections) should be taken;

*) The origin of coordinates is always taken at the point where the force is applied.

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Table (5 - 9). Calculation results of Y and M.

Cross section

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Pk=20 T 100 cm Pk = 20 T 200 cm

I II III IV V VI VII VIII IX X XI XII

42908.

286138.97 286138.97

M chart (kg.cm)

-51561

0.0404

-0.501 -0.4599 Y-slope graph

Figure (V-37). (cm)

kg/ cm2 .

.

5-) Check rail stress: M/Wx = 286,138.97/180.29 = 1,587.1 kg/cm 2 2.850 Conclusion, the rail we chose is too durable, so to save money we need to re-select rail P33, then recalculate.

3. Calculate the parrot.

Sleepers are structures with short dimensions and relatively large stiffness, so it is impossible to use formulas in calculating infinitely long beams, but must use short beam calculations. Applying the methods stated in the subject "Foundation and Foundation" we will find the internal force diagram and deformation diagram. From there we proceed to check the durability and deformation conditions of the foundation and sleepers.

4. Check other conditions.

a) Check local candle stress on the beam:

 cb ; (5 - 38) In which: P - Force acting on the sleeper;

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- Area of ​​the pad under the rail;

 cb - Allowable local compressive stress of concrete for sleepers depends on concrete grade.

b) Check the stress on the ballast layer surface:

 .ab)  bl , (5 - 39)

In which : P - force acting on the sleeper;

ab - are the dimensions of the sleepers.

- Coefficient considering uneven settlement of sleepers, it depends on the structure of the floor

buffer through coefficient C and is looked up in table (5 - 10).

Table (5 - 10): Coefficient lookup table  .

Floor coefficient

C buffer (kg/cm 3 )

 bl - Allowable stress of ballast layer, taken according to table (5 - 7).

c) Check the thickness of the ballast layer:

The thickness of the balustrade layer is selected according to the following two conditions:

-)First condition:

53.87

h = 1.25 N ,

5 - 40 

  

N = 

In which: - allowable stress of the ground under the crushed stone layer, kg/cm 2 );

- actual stress at the bottom elevation of the crushed stone layer, (kG/cm 2 ).

The value of h can be taken from table (5 - 11).

Table (5 - 11): Table for h values ​​(cm).

N

l

-)Second condition: h = 1.25 53.87. a ,

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In which: l - distance between two sleeper centers; a - width of sleeper. The value h will be chosen as the larger of the two values ​​above.

d) Checking foundation stress: We assume that the pressure is transmitted at an angle of 30  , then the average stress at the bottom of the foundation will be:

*) For 2-rail sleepers:

 tb

h 

 a + 2htg30

2P

o  b  2 htg 30 o 

 

 ; (5-41)

*)For sleepers placed on one rail:  tb 

h 

a + 2htg30

2P

o  b  2 htg 30 o 

 

 . (5-42)

in which the symbols are as above, - volume of the cushion layer.

If  , then it is necessary to increase the thickness of the cushion layer or increase the size of the sleeper.

Note: When arranging the sleepers, we should keep the distance between them (l) so that the load-bearing areas transmitted by the sleepers to the ground surface through the cushion layer do not intersect (Figure V-38).

P

l

P

h

a

a

30 0 30 0 30 0 30 0

Figure (V-38).

III. Slope calculation with beam structure on crushed stone foundation.

Reinforced concrete beams bear mobile loads. According to the working properties of the structure, it belongs to the type of space beams on elastic foundation. However, if the beam width is small, it can be considered as half-space and can be calculated according to plane deformation and use one of the methods to calculate beams on elastic foundation.

presented in the documents ''Foundation and Foundation''.

The calculated length of the beam is the distance between 2 settlement joints (each beam section is usually 8  12m). When calculating, we can ignore the influence of the pressure of the beam end acting on the foundation.

for the settlement of the beam under consideration (ie the influence of the side load). Although doing so makes the calculated ground reaction smaller than the actual reaction, at the same time we do not consider the influence of the cushion plate.

beam ends (which reduce beam settlement), these two effects can cancel each other out (Figure V-39).

On the contrary, the influence of the rail cannot be ignored. Since the rail crosses the settlement joint, it can be assumed that the deflection of the rail at the two adjacent ends of the girder is equal. However,

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