Static Values from Deflection 1

Staattiset arvot taipumasta

The idea of the this calculation is there that sometimes the  cross-section has no static values. The simple example is the hollow tube section in a place where is not possible take thickness dimension or weight. That we know is the static values of the HEB100 section. They are easy to remember, thanks to number five. Once we know this, we know quite a lot things of the hollow section

5 x 90 = 450

Profile      1:1000      σ_t            Ix           Wx          Iy        Wy                  Area

Code           cm      kN/cm²     cm⁴        cm³         cm⁴        cm³        kg/m    A cm²

HEB100      518         2,2         450          90         167       33,5       20,4      26

Bending case 1

Example

Square tube 100 x 100, length 480 cm is enclosed by welding at the both ends and there is no possibility to see the tube wall thickness. The tube length of 470 cm, bends 0.47 cm of the 1 kN concentrated load in the middle of the girder. The deflection/length ratio is 1:1000.

The basic load of the calculation is 1 kN. Later the load of 1 kN may differ. At first need to find out the location where the deflection/length ratio is 1:1000. This requires the sufficient tube length to determine the deflection/length ratio, otherwise the deflection is determined e.g. 1:2000.

Profile      L1000      σ_t           Ix          Wx          Iy         Wy                    Area

Nimike       cm      kN/cm²     cm⁴          cm³           cm⁴        cm³      kg/m    A cm²

HEB100      518         2,2         450          90         167       33,5       20,4      26

The length 470 cm, related to the measure of 518 cm, in relation to factor of 1.1.

518 / 1.1 = 471 cm

2.2 kN /cm² x 1.1 x 1.03(4) = 2.5 kN/cm²        (2.45 kN/cm² by calculation program)

                                                                         Universal friction = 1.03

The manufacturing tolerance of the tube gives greater deviation than calculation. The tube has shorter support length 1:1000 than HEB 100 cross-section. This indicates the square tube has lower second moment of area than 100 HEB cross-section. This because deflection is relative to the bending stress increase.

Profile      L1000      σ_t           Ix          Wx          Iy         Wy                    Area

Nimike       cm      kN/cm²     cm⁴          cm³           cm⁴        cm³      kg/m    A cm²

HEB100      518         2,2         450          90         167       33,5       20,4      26

Putkipalkki

Tubular girder 100 x 6,3

=> Ix = 450 cm⁴ / (1,25x1,034) = 348 cm⁴.

Taulukoitu putkipalkin 100 x 6,3 neliömomenttii Ix =  341cm⁴.

The tabled second moment of area (Tube 100 x 6.3) has the value Ix = 341cm4.

Laskelma poikkeaa 2 %. Taulukoituna on taattava arvo, jolloin laskenta voi olla tarkka.

Calculation differs 2%. The tabulated value is guaranteed, when the calculated can be precise.

Of this can continue to bending resistance

Wx = 341 cm⁴/ 5 cm = 68,2 cm³ 

(5 cm is the distance from the neutral axis to edge of the profile).

This example illustrates how the shape, load and the loading case have the same meaning through the proportionality. Determination of strength, through the proportionality does not require any formulas. Pascal's triangle and the ratio queu of calculation are sufficient.

                                          -------------------------------------------

Check

Deflection of the Concentrated Force 1 kN

F = 1 kN

f = F x L³ / (48 x I x E)

f = 1 x 470³ / (48 x 341 x 20600)

f = 0.31 cm  (0.30 cm)

Deflection of the Girder Weight

F = Girder weight 0.85 kN

f = taipuma / deflection

f = 5 x F x L³ / (384 x E x  I)

f = 5 x 0.85 x 470³ / 384 x 20600 x  341

f = 0.16 cm  (0.16 cm)

0.31 cm + 0.16 cm = 0.47 cm

We do not need formulas, we need the understanding of the nature and its mechanism.

26.2.2012 13:14 (46950 - 493)
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