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1.2.9 Racunanje vzdolzne normalne napetosti in dolocanje jedra ...

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60 1 Upogib z osno silo<br />

PREGLEDNICA 1.5 (nadaljevanje): Upogibnice <strong>in</strong> notranje sile za nekatere nosilce<br />

h) Upogibnica enkrat statično nedoločenega nosilca obteženega z momentom M<br />

Pomik <strong>in</strong> zasuk v polju 1:<br />

w 1 =<br />

M x2 (<br />

a 3<br />

4 L 3 + 3 a 2 b − 2 b 3 + (a − 2 L)a x ) ,<br />

E I y<br />

ω y1 = −<br />

M x (<br />

2(a 3<br />

4 L 3 + 3 a 2 b − 2 b 3 ) + 3 (a − 2 L)a x<br />

E I y<br />

Pomik <strong>in</strong> zasuk v polju 2:<br />

w 2 =<br />

M a (<br />

2 a L 3<br />

4 L 3 − 4 L 3 x − 3 L(a − 2 L)x 2 + (a − 2 L)x 3 ) ,<br />

E I y<br />

ω y2 =<br />

M a<br />

4 L 3 E I y<br />

(<br />

4 L 3 + 6 L(a − 2 L)x − 3 (a − 2 L)x 2 ) .<br />

Pomik <strong>in</strong> zasuk pri x = a ter zasuk pri x = L:<br />

w(a) =<br />

M a2 b<br />

4 L 3 (a 2 − 2 b 2 ),<br />

E I y<br />

ω y (a) =<br />

M a<br />

4 L 3 (a 3 + 4 b 3 ), ω y (L) = − M a (2 b − a).<br />

E I y 4 L E I y<br />

Pri a mej = 0.582925 L je |w 1,ekst | = |w 2,ekst |.<br />

Ekstremni pomik, če je a > a mej :<br />

x ekst = 2 L(2 b2 − 2 a b − a 2 )<br />

, w ekst =<br />

3 a(a − 2 L)<br />

M<br />

27 a 2 (a + 2 b) 2 E I y<br />

(a 2 + 2 a b − 2 b 2 ) 3 .<br />

Ekstremni pomik, če je a < a mej :<br />

√<br />

( √ ) 2 b − a<br />

x ekst = L 1 −<br />

, w ekst = − M a 3(2 b − a) 3<br />

3(2 b + a)<br />

18 L E I y 2 b + a .

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