Round Bars Stock Item Set

Pyörötankojen varastonimikesarja

Let us drawn up on the basis of the strength a logical set of round bar stock items. This can be obtained by combining the Fibonacci and ratio numbers by a factor of 1.25. When the diameter of the steel bar increases by a factor of 1.25 (general machine construction fatigue factor), the cross-sectional areas follow ​​the Fibonacci numbers.

Fibonacci numbers        ( 1 ) -  1   -   2  -   3  -  5  -   8 -    13   -   21  -   34   -  55

Ratio numbers (D cm)    (1)- 1.2(5) - 1.6 - 2 -  2.5 - 3,2  -   4   -    5  -    6,3  -   8     

A round bar  cm²          0,785   1       2   3,15  4,91  8,04 12,57  19,63  33,18   51

                                                            1.     2.      3.      4.      5.      (6.).

The proportionality limit is five steps, where the accuracy stays.

The Diameter series for machine construction

10 - 12 - 16 - 20 - 25 - 32 - 40 - 50 - 65 - 80 - 100.... mm

An Example:

Round bar diameter 8 cm  =>  A = (3,14... x D²) / 4

                                                 A = (3,14... x 8²) 4

                                                 A = 50,27 cm²

Fibonacci numbers round bar diameter 55, has a cross sectional area of ​​50.27 cm². Steel cross-section diameters have a + tolerance, in which case the value of 50.25 cm² may be increased by the value of 51 or lowered by value of 50 cm². The calculation of this does not matter, when the steel rod cross-section is increased by a factor of 1.25 (in machine-building the general fatigue factor), then the cross-section surface areas follow the Fibonacci numbers.

The Golden Ratio

The ratio number increases two steps, increases the diameter by a factor of 1.6(18).

             Step 1         Step 2

Esim.   D 2.0 cm -  D 3.15 cm          3.15/2 = 1.6(18)

            D 4,0 cm -  D 6.3 cm            6.3 / 4 = 1.6(18)

            D 6.3 cm -  D 10.0 cm          10 /6.3 = 1.6(18)

The cross-sectional surface area increases by a factor of 1.6(18)²

                                Step 1            Step 2

          D2.0 cm   -    3.15 cm²   -   33.18 cm²              8,04 / 3.15  = 1.6(18)²

           D4.0 cm   -  12.57 cm²   -   33.18 cm²           33,18 / 12.57 = 1.6(18)²

The EP-calculation is not calculation of decimal places.

Static Values

D cm           1    -  1.2(5)  -  1.5(1.6) -  2  -   2.5  -   3.2  -   4  -     5  -   6.3  -   8  -  10

Ix  cm⁴     0.0491    0.12     0.32     0.785   1.92   5.15   12.56  30.7   77.3    201   490.1

Wx cm³    0.0981   0.192     0.40    0.785   1.53   3.21    6.28   12.3   24,5    50.3  98.2

                  (+2)      (+1.6)    (+1.25)   (1)    (-1.25) (-1.6) (-2)   (-2.5)  (-3.15)  (-4)   (-5)

                                   Wx / Ix                                           Ix / Wx    

a)  Free fall of a body 4,91 m /1 sec

b)  The gravity force 9.81 m/s²

c)  The specific gravity of steel 0.785

d)  Calculation factors 1.25 and 1.12

e)  The full angle 2 x pi radians (6.28)

f)   Fibonacci numbers 1 - 1 - 2

                                    1 + 1 = 2

                D10    0,0491  Ix cm⁴     -      0,0981 Wx cm³ etc.

                0.0491 cm⁴  + 0.0491 cm⁴  = 0.0982  cm⁴

                 D10 0.0981 Wx cm³   -  D12(.5) 0.192  Wx cm³

The shown set covers the cross-sections on the basis of strength. The selected set features will be presented in the strength calculation.

17.3.2015*20:28 (1016 - 699)
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