Shipbuilding and Repair Factory - 28

y q : beam end settlement caused by force Q = 1. Q: Beam end shear force caused by external force.

From (5-43) we have:

k

Q y 1 y 2 y 3 y 4 . P

(5-43)

(5-44)

2 y q

Car

Ray

big gap

beam

mattress

beam head

Figure (V - 39): Connection diagram of Beam - rail - beam end pad.

Pk Pk Pk Pk _ Q

+ Q

Figure (V - 40). Interaction diagram of beam segments connecting rails.

How to calculate Q:

- Let P ki = 1 use formula (5-35) to find y i

- Given Q = 1 placed at the end of the beam, find y q

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force

- Instead we find Q. With Q consider it as an external force to calculate moment, shear force and reaction.

Uneven settlement is determined by the following formula:

y max y min

2 l

In there:

1 

1000

1

2000

, (5-45)

- y max , y min - are the maximum and minimum settlement of the beam (see chapter IV);

- l - half the length of the beam. Specific calculation steps:

The beam bears a moving load, so when calculating, we let this load group move to each position and at each position we find the shear force at the beam end, then consider this force as an external force to calculate the foundation reaction p, moment M and shear force Q. With many positions, we will draw a diagram covering p, M, Q to calculate the steel reinforcement.

* The calculation steps of a load position are: 1- Calculate the wheel pressure P k ;

2- Calculate the unit settlement y i and y q ; 3- Calculate the shear force at the beam end;

4- Calculate p, M, Q (using Gorbunov-Poxatdov table) in the appendix of the document ''Foundation and Foundation''.

Horizontally, we cut out 1m and calculate according to plane deformation (also using the table above).

P

P

Figure (V - 41). Horizontal calculation diagram

of beam.

External force consists of 2 concentrated forces P (figure V-41), the value of force P is taken as 1/2 of the total ground reaction force on 1m length of the beam and taken at the position with the largest ground reaction force (when calculated vertically). However, it should be noted that the stress in the beam during operation is not large, so in most cases, the cross-sectional size of the beam is determined by the stress when lifting the beam (construction conditions). When lifting the beam, we often calculate it as the beam resting on 2 supports with the evenly distributed load of the self-weight multiplied by the dynamic coefficient taken as 1.5.

Some regulations when calculating beams on ballast foundation according to limit state.

1/- According to the first limit state, the basic formula lμ:

U ≤ m . R , (5-46)

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In which: U - Total calculated value of external force or stress that can destabilize or lose the strength of the structure corresponding to the most unfavorable load combination; m - Working condition coefficient; R- Corresponding bearing capacity of the material against destructive effects on the stability or strength of the structure (ie the limit value). The value m is taken as follows:

* When calculating longitudinally in the beam: Calculate for positive moment m = 1.3; Calculate for negative moment m = 0.9; Calculate for shear force m = 1.0.

* When calculating horizontally in the beam direction: Calculate by positive moment, take m = 1.0; Calculate for moment

negative take m = 0.85.

The modified stiffness of the beam is calculated according to:

B  . E . J td

, (5-47)

In which: E - Initial elastic modulus of concrete; J td - Moment of inertia of the modified cross section.

J td = J d + J ray ,

in which: J d - Converted moment of inertia of the entire cross-section of the reinforced concrete beam with respect to the common center of gravity (in the calculation, take the entire cross-section of the concrete plus the longitudinal reinforcement cross-section multiplied by the ratio of elastic modulus between steel and concrete); Jray - Moment of inertia of the rail; However, J of the rail is usually very small compared to J of the beam, so we ignore it;  - Coefficient considering the plasticity and creep of the concrete, Coefficient

can be taken from table (V - 12).

When calculating according to formula (5-46), an additional coefficient is added to take into account the level of the project.

The additional coefficient is denoted by M bs and is taken from table (5-13).

Table (V-12). Coefficient values ​​ .

Characteristics of concrete structures

reinforced concrete

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Project level

1.1

2/- According to the third limit state: All calculated stresses are multiplied by a system

3/- According to the second limit state:

tb

R H

(5 - 48)

 tb : average pressure transmitted by the structure to the ground.

= P'

, (5-49)

average b+ 2h

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In which: P ' - total ground reaction over 1 meter length of the beam taken at the position with the largest ground reaction; b - width of the beam bottom; h - thickness of the crushed stone layer, if the beam is placed on a compact sand foundation (not crushed stone) then h =

0; R H - standard strength of the foundation soil, calculated at a depth not exceeding one foundation width and taken according to formula (5-50).

1

d

2

dd

R H m  A ( B  2 h )  A ( d  h )  H D . C H, (5-50)

In which: m - working condition coefficient with: - dusty sand m = 0.6; - fine grained sand m = 0.8; - soil types

else m = 1.0; A 1

, A 2

, D - non-uniform coefficient depends on the standard internal friction angle of the soil H

Take according to table (5-14).

Table (V-14) Values ​​of coefficients A 1 , A 2 , D.

B: foundation bottom width tank; d - foundation depth; h

d , d

H - thickness and weight of the body

volume of the cushion layer; CH - unit cohesion value (kg/cm2 ) is taken as follows: gravel and large-grained sand is 0.01; fine-grained sand is 0.02; dusty sand is 0.04.

Example of calculating the force of a reinforced concrete beam placed on a compact sand foundation. Dimensions and load are as shown in the drawing, knowing that the wheel pressure Pk = 15 tons.

Figure (V-42) Diagram of a train carrier.

Solution : The beam consists of 2 longitudinal beams supporting 2 rails. When calculating, we consider them as separate beams, each beam supports 1 rail and works independently of each other.

So the calculation diagram is:

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1000 cm

P k 120 P k 360 P k 120 P k

Figure (V-42). Load diagram.

- wheel pressure P k = 15 tons (given)

- Calculate y i and y q

Park the car in an arbitrary position. For example, park the car 50 cm to the right of the beam. So the load diagram will be:

P k = 15,000kg P k = 15,000kg

350

120

360

120

-Q

+Q

According to (5-35)



y K . P . ,

2 k

In which: when we let P k = 1 to find y i and y q , we rewrite:

y   k. 

2 

In there:

k 4



4 EJ



b . l

2 d . c

b = 50 cm; l = 1000cm; d = 60 cm (distance between 2 cross beams); C = 4 kg/cm 3 (with compact sand).

Instead we have:

 50 . 1000

2 . 60

.4 

5 . 10 3

3

b . h 3

J  12

50 . 40 3

 12

32 . 10 5

 12

E  2 . 10 5

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Therefore:

4

3 . 4 . 2 . 10 5 . 32 . 10 5

5 . 10 3 . 12

k  0.0094 cm  1

Now use formula (5-35) to set up a table for y i and y q . Table (V-15) gives the results of calculating the unit magnitudes of y i and y q .

For x = 0 we have y q , and the x i will give y i .

3/- Calculate the shear force of the beam:

q

y i

Q 2 y . P k

Q ⎛0.02389  0.00553  0.00000214  0.0000054⎞

⎜

⎝

2(  0.0282)

⎟ .15000 

⎠

 0.0294167 .15000  7822.5 kG

0.0564

Answer: Q = 7822.5 kg.

IV. Structural slope calculation on pile foundation .

Beams on pipe pile foundations are calculated as single beams or multi-span continuous beams on rigid or elastic bearings. The load transmitted to the beam is considered perpendicular.

Some regulations when calculating beams on pile foundations according to limit state:

1/- With single beam :

* When calculating according to the first and third limit states:

- Working condition coefficient m in formula (5-46) is taken as 1 (when calculating cross-section).

- Converted moment of inertia:

J td

 J d  0.8J ray

J d : symbol as above;

0.8: coefficient considering the loose connection between rail and beam;

J ray : moment of inertia of the rail about the common centroid multiplied by the ratio of the elastic modulus between the rail and the beam.

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* When calculating according to the second limit state:

According to the condition: Δ≤ Δgh , (5-52)

in which: Δ- deformation of the structure under the effect of load; Δgh - limit deformation value of the structure.

Beam deformation is taken according to:

Δ g. h  Δ y. L d

l car

, (5-53)

In which: Δ y - the difference in deformation of the cushion layer at the wheel bearing the largest load and the wheel bearing the smallest load. According to the research results: Δ y = 33mm; L d - length of the slope beam span; l xe - length of the segmented vehicle.

2/- Continuous beam :

If pile step l ≤ 3S, calculate as beam on elastic foundation.

If l > 3S, then calculate as multi-span beam resting on elastic or rigid bearings according to the method

influence line

Here S is the elastic property of the beam determined by the following formula:

4B o

4

bK o

S = ,

(5-54)

In which: b - width of beam (m); Bo - converted stiffness of beam (kN.m 2 ), and is calculated as follows:

Bo = K p l  J td ( 5 - 55 )

E - elastic modulus of concrete (kH/cm 2 ); Jtd = J + 0.8Jray - calculated moment of beam and rail; Kpl - coefficient considering creep deformation and plastic deformation of beam concrete; Coefficient Ko is determined from experiment according to the following formula: Ko = P/ yF, (5 - 56)

in which: P - load placed on a pile during the load test (kN); y - settlement of the pile under the effect of that load, (m); F - area of ​​the beam resting on a pile, (m 2 ). When there is no test data, Ko can be calculated using the empirical formula

Ko =  .S.Lc / F, (5 - 57)

where S and Lc - cross-sectional circumference and length of pile, (m); - the proportional coefficient depends on the ground. When the distance between piles l = (4  6).d then = 5 MN/m 3 for clay, = 10 MN/m 3 for sandy soil, = 25 MN/m 3 for pedogenic soil.

V. Calculate other sub-details.

In addition to the ramp system, to serve the purpose of pulling the train up and down, people also arrange other parts such as pulley system and winch platform, and train towing vehicle. Therefore, in the calculation, it is necessary to conduct calculations on the durability and stability of those parts to ensure the entire system works safely and smoothly.

1-/ Check the stability of the train and vehicle system under the most adverse conditions, which is when the train is placed on a horizontal vehicle, being pulled up and subjected to the strongest wind pressure.

The test is conducted under the following conditions:

KM l M g (5 - 58)

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In which: K - safety factor, taken from 1.1 to 1.5 depending on the maximum allowable wind speed of the construction area; M l - total moment of the forces causing the vehicle-train system to overturn; M g - total moment of the anti-overturning forces of the vehicle-train system.

2-/ Calculate the pulling force acting on the vehicle (T). During operation, we need to know the maximum pulling force acting on the vehicle to allow us to decide on the choice of pulley cable and winch.

The tensile force T is determined by the following formula:

T = k.(T k + T y ) + P i + W t +T q (5 - 59)

In which: Tk - resistance caused by the wheel (or roller) when rolling on the rail Tk = ( Q + Q vehicle ).Cos . f 2 /R (5 - 60)

f 2 - coefficient of rolling friction, taken according to table (5 - 17).

Table (5 - 17): Rolling friction value f 2 (cm).

Numerical order

R - radius of wheel or roller;

- the angle of inclination of the slide relative to the horizontal; Q, Qxe - the weight of the train and vehicle;

T y - frictional force between the axle and the wheel bearing;

T y = ( Q + Q xe ).Cos .  .r/R, (5 - 61)

- coefficient of sliding friction, taken from table (5 - 18).

Table (5 - 18). Coefficient  ..

Numerical order

k - coefficient considering some other frictions, taken as (1.1 - 1.4); r - wheel axle radius;

W t - pounding, W t = pFcos  , (5 - 62)

p - wind pressure intensity, depending on wind level (Kg/m 2 ), with wind level 5, p = 20 kg/m 2 ;

F - windbreak area of ​​the hull; P i - resistance due to inertial force

P i = (Q + Q vehicle ). v/(gt) (5 - 63)

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