sum query question Calculate the cumulative thickness based on the layered thickness

select , sum(

from a1 t1, a1 t2

where <=

group by

This one will definitely work.

Less than the more troublesome point, value(sum(, 0) is written as db2, if sum() is null, then 0Different SQL dialects should be converted appropriately.

select , value(sum(, 0)from tab t1

right join tab t2

on >

group by

If you are only calculating the actual depth of a layer, select sum(layer thickness) from a1 where

layerid: "The current layid is up.

If it's the actual depth of each layer, I think it's easier to implement with a cursor.

1。Define a cursor.

2。Control the cursor by looping so that the cursor goes down at once, i.e., next.

3。When the cursor moves to the nth record, take out the layerid of the current row, and then calculate the actual depth of the current layer: select sum(layer thickness) from a1 where

layerid: "Current layid.

4。Release the cursor.

Good luck.

Grades: Mainstay, contracted results, north-south, east-west.

1 s 3s 0 140 0 double none.

2 e ) 3nt -1 50 0 north and south.

3 s 2c (doubled) +1 270 ? Stuff 4 s 4h +2 680 ?There are doubles.

5 e 6c ±0 ？920 North and South.

6 W 2NT (double) +2 ? 1090 stuff 7 e 5d +1 ? 620 pairs have.

8 s 2s (doubled) 0 460 ? Double no 9 n 4s (double and double) -3 ? 1000 stuff 10 n 1nt +1 120 ? There are doubles.

11 w 3h +2 ？200 pairs of none.

12 e 1nt (double and double) +1 ? 760 North-South 13 n 7s 0 2210 ? There are doubles.

14 n 5h +1 ？Double none.

15 s 3nt(doubled) -4 1100 north and south 16 w 3c +2 ? 150 things.

Yes, it is the sum of the actual thickness of the insulation material of each layer. The insulation thickness expressed by the design value is the thickness that is actually needed, because the thickness of the material may not be the same as the design thickness, and the material is always accumulated in actual operation, and the total thickness must meet the thickness required by the design.

Thickness of each layer of backfill: (mm).

Compaction tools virtual paving thickness wood rammer, iron rammer "200 frog type hang, fire rammer 200-500 roller 200-300

Vibratory roller 400

Looking at the context, it should be the thickness of the virtual paving.

Notice the difference:

The parameters of the sum function are number1, number2,..This means that exce; When selecting a parameter for sum, a value is selected for each parameter, and if its parameter is an array, each value of this array is taken as a separate piece of data and then accumulated.

The third parameter in your current offset is the array, which can return an array of 1 10 thicknesses specified by the array row(a1:a10), but since sum only takes one number of these arrays, the first number of these 10 arrays is taken every time, and then the 400 is mined after the sum operation is completed.

The first parameter of the sumif function is different, it is range, that is, its first argument itself is an array, when the offset also returns 10 arrays, since you are using the array formula input, it can substitute each array returned by offset into parameter 1 separately, so that 10 different results are obtained, and then the minimum value is returned by min.

1. The reason for taking: the distribution of additional stress is not linear, if the division is too far away, it will produce a large error, so the thickness can not be too thick, in fact, this is also an empirical value, the largest upper limit is this, specifically in practice, due to the objective reasons of groundwater level and soil distribution, most of them are still less than.

2. Advantages: The physical concept is clear, and the calculation process is not difficult.

Disadvantages: The relationship between the calculated results and the measured values: medium and medium-grade, the calculated values are approximately equal to the measured values; Weak foundation, the calculated value is less than the measured value; Solid foundation, the calculated value is much larger than the measured value.

In addition, because the thickness of the foundation should be specified, the calculation workload is large and the process is cumbersome.

Hand hit,! ）

Steps: 1, according to the relevant requirements and soil properties for foundation stratification;

2. Calculate the self-weight stress of the layer, and take the effective gravity below the groundwater level for calculation;

3. Calculate the additional stress of each layering point, and find the average value of the additional stress at the interface of each layer;

4. The sum of the average self-weight stress and the average additional stress of each layer is the total stress of the layer after compression;

5. Determine the thickness of the compression layer;

6. Calculate the compression amount of each layer;

7. Calculate the average final settlement of the foundation.

The settlement of each layer is calculated first, and then the total settlement is accumulated.

A brief description of the branch and his calculation method ended up being a bit of a step down.

I don't know.

The code recommends a method for calculating the final settlement of a foundation.

The thickness of the formation is calculated according to the layering of each wire.

The thickness calculation formula: d l (sin cos sin cos sin).

where d: the true thickness of the formation (m).

l: wire slant distance (m).

True dip angle of rock formation (°).

Topographic slope angle (°.)

The sharp angle (°) between the profile traverse and the stratigraphic strike line

Note: When the slope direction is opposite to the rock strata inclination, the plus sign is used in the formula; When the aspect is the same as the tendency of the rock formation, the formula.

is calculated with a minus sign. ）

For example, a section of wire in a measured profile is recorded as follows:

Conductor 2 3, slope distance 25m, direction 35°, slope +10°

0 5m: Sandstone.

5 23m: Dolomitic limestone, occurring at 200° 60° (at 20m).

23 25m: Mudstone.

Calculate the thickness of the dolomitic limestone layer:

l 23 5 18m, 60°, 10°, 110° direction 35°

Substitute the formula d 18 (sin60° cos10° sin75° cos60° sin10°).

Understandable. According to the actual situation, the result can be calculated.