PSI - Issue 17

Ivan Baláž et al. / Procedia Structural Integrity 17 (2019) 734 – 741 Ivan Baláž, Yvona Koleková, Lýdia Moroczová / Structural Integrity Procedia 00 (2019) 000 – 000

738

5

F

F

C

F C

F

0.58906

0.12851

1.62999

3.01461

=

+

 w

 

if

(17)

cr a .

E

E w

E

F

F

C

F C

F

0.88638

0.02988

3.01461

3.80198

=

+

 w

 

if

(18)

cr a .

E

E w

E

E cr a F F = .

 E w F C  3.80198

if

(19)

4. Comparisons and verifying of the results

The following examples enable to compare the values of the critical force F cr obtained from diagrams, F cr.a obtained from approximate formulae and the exact value F cr.IQ calculated by computer program IQ 100 (2010). The comparisons are given in table 1. All critical forces in table 1 are calculated for the following input values: E = 210 GPa, I = 349.2 cm 4 , ℓ = 5 m. The value of the spring stiffness C w was chosen to obtain the maximum difference between curves and straight lines in the diagrams.

Table 1. Comparisons of the diagram values F cr , approximate values F cr.a and exact values F cr.IQ . n C w ( kN/m ) F cr ( kN ) F cr.a ( kN ) Difference between F cr.a and F cr ( % ) F cr.IQ ( kN ) 2 58.66 186.73 182.374 -2.73 187.50 3 99.732 231.603 223.707 -3.324 231.40 4 134.931 260.553 256.282 -1.841 261.09 5 74.505 202.652 200.398 -1.447 203.34 6 99.732 231.603 228.435 -1.282 231.40 7 134.931 260.553 257.235 -0.918 259.62

Table 1 shows that the maximum difference between diagram values F cr and approximate values F cr.a calculated from the above formulae is -3,324%. The F cr.a values are rather smaller than F cr being on the safe side. Differences between F cr and exact F cr.IQ values are negligible. It was verified that equation (2) with Φ ∞ valid for infinitive number of fields n = ∞ may be used for calculation of the approximate valve of the critical force of the compression member on the elastic foundation. In such case the foundation modulus k [kN/m 2 ] should by calculated as follows:

nl k n C w ( 1) − =

(20)

5. Critical force and buckling length of simply supported member on elastic foundation

The following tree cases are investigated: (i) member with free ends, (ii) member with hinged ends and (iii) member with fixed ends. Relative critical force and relative resistance of foundation are as follows:

EI F F L cr cr 2 2  =

EI k kL 4 =

,

(21)

where k [kN/m 2 ] is the resistance of the foundation (foundation modulus).

5.1. Member with fixed ends (case A in figure 5)

Exact value of the relative critical force may be obtained from the buckling condition: