interakcijski diagrami nosilnosti ab precnih...

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UNIVERZA V LJUBLJANI FAKULTETA ZA GRADBENI ˇ STVO IN GEODEZIJO SEBASTJAN BRATINA MOJCA MARKOVI ˇ C IGOR PLANINC INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRE ˇ CNIH PREREZOV PRI DVOJNO EKSCENTRI ˇ CNI OSNI SILI LJUBLJANA, december 2006

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Page 1: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

UNIVERZA V LJUBLJANIFAKULTETA ZA GRADBENISTVO IN GEODEZIJO

SEBASTJAN BRATINA

MOJCA MARKOVIC

IGOR PLANINC

INTERAKCIJSKI DIAGRAMI NOSILNOSTIAB PRECNIH PREREZOV PRI

DVOJNO EKSCENTRICNI OSNI SILI

LJUBLJANA, december 2006

Page 2: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

Kazalo

1 Racunski primeri 31.1 Kvadratni precni prerez . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Pravokotni precni prerez . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 Interakcijski diagrami mejne nosilnosti pri dvojno ekscentricni osni sili 82.1 PRAVOKOTNI PREREZ, 4 palice (δz = 0.1, δy = 0.1), S500 . . . . . . . . . . . . 92.2 PRAVOKOTNI PREREZ, 4 palice (δz = 0.125, δy = 0.1), S500 . . . . . . . . . . . 152.3 PRAVOKOTNI PREREZ, 4 palice (δz = 0.1, δy = 0.125), S500 . . . . . . . . . . . 212.4 PRAVOKOTNI PREREZ, 4 palice (δz = 0.125, δy = 0.125), S500 . . . . . . . . . 272.5 PRAVOKOTNI PREREZ, 4 palice (δz = 0.15, δy = 0.125), S500 . . . . . . . . . . 332.6 PRAVOKOTNI PREREZ, 4 palice (δz = 0.125, δy = 0.15), S500 . . . . . . . . . . 392.7 PRAVOKOTNI PREREZ, 4 palice (δz = 0.15, δy = 0.15), S500 . . . . . . . . . . . 452.8 PRAVOKOTNI PREREZ, 8 palic (δz = 0.075, δy = 0.075), S500 . . . . . . . . . . 512.9 PRAVOKOTNI PREREZ, 8 palic (δz = 0.1, δy = 0.075), S500 . . . . . . . . . . . 572.10 PRAVOKOTNI PREREZ, 8 palic (δz = 0.075, δy = 0.1), S500 . . . . . . . . . . . 632.11 PRAVOKOTNI PREREZ, 8 palic (δz = 0.1, δy = 0.1), S500 . . . . . . . . . . . . . 692.12 PRAVOKOTNI PREREZ, 8 palic (δz = 0.125, δy = 0.1), S500 . . . . . . . . . . . 752.13 PRAVOKOTNI PREREZ, 8 palic (δz = 0.1, δy = 0.125), S500 . . . . . . . . . . . 812.14 PRAVOKOTNI PREREZ, 8 palic (δz = 0.125, δy = 0.125), S500 . . . . . . . . . . 872.15 PRAVOKOTNI PREREZ, 12 palic (δz = 0.05, δy = 0.05), S500 . . . . . . . . . . . 932.16 PRAVOKOTNI PREREZ, 12 palic (δz = 0.075, δy = 0.05), S500 . . . . . . . . . . 992.17 PRAVOKOTNI PREREZ, 12 palic (δz = 0.05, δy = 0.075), S500 . . . . . . . . . . 1052.18 PRAVOKOTNI PREREZ, 12 palic (δz = 0.075, δy = 0.075), S500 . . . . . . . . . 1112.19 PRAVOKOTNI PREREZ, 12 palic (δz = 0.1, δy = 0.075), S500 . . . . . . . . . . . 1172.20 PRAVOKOTNI PREREZ, 12 palic (δz = 0.075, δy = 0.1), S500 . . . . . . . . . . . 1232.21 PRAVOKOTNI PREREZ, 12 palic (δz = 0.1, δy = 0.1), S500 . . . . . . . . . . . . 129

2

Page 3: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

1 Racunski primeri

1.1 Kvadratni precni prerez

V prvem racunskem primeru bomo za kvadratni armiranobetonski precni prerez, ki je prikazan na sliki(1), s pomocjo izdelanih interakcijskih diagramov izracunali potrebno vzdolzno armaturo za danoobremenitev. Geometrijski podatki AB precnega prereza ter podatki o legi in razporeditvi armaturnihpalic so prikazani na sliki (1). Racunska obremenitev precnega prereza je:

Nxd =−500 kN,

Myd = 120 kNm,

Mzd = 90 kNm.

Slika 1: Kvadratni AB precni prerez, b/h = 40/40 cm.

Precni prerez je izdelan iz betona kvalitete C25/30, kar pomeni, da je njegova karakteristicna tlacnatrdnost fck = 2.5 kN/cm2, in armature kvalitete S500 s karakteristicno mejo tecenja fyk = 50.0 kN/cm2.Racunsko tlacno trdnost betona fcd izracunamo z izrazom:

fcd =fck

γc=

2.51.5

= 1.67 kN/cm2, (1)

kjer je γc parcialni varnostni faktor za beton. Racunsko trdnost armature pri meji tecenja fyd paizracunamo z izrazom:

fyd =fyk

γs=

50.01.15

= 43.48 kN/cm2, (2)

kjer je γs parcialni varnostni faktor za armaturo.

Normirana mejna obremenitev precnega prereza je enaka:

nd =Nxd

bhα fcd=

−50040 ·40 ·0.85 ·1.67

=−0.220, (3)

myd =Myd

bh2α fcd=

1200040 ·402 ·0.85 ·1.67

= 0.132. (4)

3

Page 4: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

Razmerje normiranih upogibnih momentov mzd/myd pa je:

mzd

myd=

Mzd

Myd

hb

=90120

4040

= 0.75. (5)

Parametra δz in δy, ki enolicno dolocata lego armature, izracunamo z izrazoma:

δz =az

h=

540

= 0.125, (6)

δy =ay

b=

540

= 0.125. (7)

Ko poznamo stevilo in lego armaturnih palic, kvaliteto armature ter razmerje normiranih upogibnihmomentov precnega prereza, na ustreznem normiranem interakcijskem diagramu poiscemo tockos koordinatama nd = −0.220 in myd = 0.132, ki doloca normirano obremenitev precnega prereza.Ob predpostavki, da je obremenitev precnega prereza (Nxd,Myd,Mzd) ravno enaka mejni nosilnostiprecnega prereza (NxRd,MyRd,MzRd) odcitamo reducirano potrebno stopnjo armiranja µ0. Ker zarazmerje normiranih upogibnih momentov mzd/myd = 0.75 interakcijski diagram v prirocniku ne ob-staja (razmerja 0.1→1.0 po koraku 0.1), izracunamo potrebno kolicino armature z uporabo dveh naj-blizjih normiranih interakcijskih diagramov ter dobljena rezultata linearno interpoliramo. Na ustrez-nih normiranih interakcijskih diagramih [(stran 90, slika (137)) in (stran 91, slika (138))] poiscemotocko, ki doloca normirano obremenitev precnega prereza, katere koordinate so nd = −0.220 inmyd = 0.132 ter odcitamo reducirano potrebno stopnjo armiranja µ0, ki znasa µ0(0.8) = 0.340 inµ0(0.7) = 0.302 (slika (2) in (3)).

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.7

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 2: Izbran interakcijski diagram.

4

Page 5: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.8

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 3: Izbran interakcijski diagram.

Nato izracunamo potrebno stopnjo armiranja za posamezno armaturno palico µ1, ki je:

µ1(0.8) =µ0

8α fcd

fyd=

0.3408

0.85 ·1.6743.48

= 1.39 ·10−3, (8)

µ1(0.7) =µ0

8α fcd

fyd=

0.3028

0.85 ·1.6743.48

= 1.23 ·10−3. (9)

Ploscina precnega prereza posamezne armaturne palice As,1, ki je potrebna, da podani AB precniprerez prevzame obremenitev (Nxd,Myd,Mzd), je glede na uporabljena interakcijska diagrama:

As,1(0.8) = µ1Ac = 1.39 ·10−3 ·40 ·40 = 2.224 cm2, (10)

As,1(0.7) = µ1Ac = 1.23 ·10−3 ·40 ·40 = 1.968 cm2. (11)

Dobljena rezultata As,1(0.8) = 2.224 cm2 in As,1(0.7) = 1.968 cm2 linearno interpoliramo. Potrebnaploscina precnega prereza posamezne armaturne palice As,1 je torej

As,1(0.75) =As,1(0.7)+As,1(0.8)

2=

2.224+1.9682

= 2.10 cm2. (12)

5

Page 6: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

1.2 Pravokotni precni prerez

V drugem racunskem primeru nas zanima, koliksna je potrebna vzdolzna armatura za pravokotniarmiranobetonski precni prerez pri dani obremenitvi. Geometrijski podatki o AB precnem prerezuter podatki o legi in razporeditvi armature so prikazani na sliki (4). Racunska obremenitev precnegaprereza je:

Nxd =−400 kN,Myd = 72 kNm,Mzd = 18 kNm.

Slika 4: Pravokotni AB precni prerez, b/h = 25/30 cm.

Pri dimenzioniranju precnega prereza uporabimo beton kvalitete C30/37 ( fcd = 2.0 kN/cm2) in ar-maturo kvalitete S500 ( fyd = 43.48 kN/cm2). Tudi sedaj s pomocjo izdelanih interakcijskih dia-gramov izracunamo potrebno kolicino armature.

Normirana mejna obremenitev je:

nd =Nxd

bhα fcd=

−40025 ·30 ·0.85 ·2.0

=−0.314, (13)

myd =Myd

bh2α fcd=

720025 ·302 ·0.85 ·2.0

= 0.188. (14)

Razmerje normiranih upogibnih momentov mzd/myd pa je:

mzd

myd=

Mzd

Myd

hb

=1872

3025

= 0.3. (15)

Lega armature je enolicno dolocena s parametroma δz in δy, ki sta:

δz =az

h=

3.830

= 0.1267≈ 0.125, (16)

δy =ay

b=

2.825

= 0.152≈ 0.150. (17)

6

Page 7: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

Na ustreznem normiranem interakcijskem diagramu (stran 40, slika (53)) poiscemo tocko, ki jo dolocanormirana obremenitev precnega prereza (nd = −0.314, myd = 0.188), ter nato odcitamo reduciranostopnjo armiranja µ0, ki je µ0 = 0.304 (glej sliko (5)). Na koncu izracunamo se stopnjo armiranja zaposamezno armaturno palico µ1, ki je enaka

µ1 =µ0

4α fcd

fyd=

0.3044

0.85 ·2.043.48

= 2.97 ·10−3. (18)

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.15

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.3

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 5: Izbran interakcijski diagram.

Precni prerez posamezne armaturne palice As,1, ki je potreben, da podani AB precni prerez prevzameprojektirano obremenitev (Nxd,Myd,Mzd), je:

As,1 = µ1Ac = 2.97 ·10−3 ·25 ·30 = 2.23 cm2. (19)

7

Page 8: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

2 Interakcijski diagrami mejne nosilnosti pri dvojno ekscentricniosni sili

8

Page 9: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

2.1 PRAVOKOTNI PREREZ, 4 palice (δz = 0.1, δy = 0.1), S500

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε

2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.2

0.30.4

0.50.6

µ0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.1

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 1: Pravokotni prerez: δz = 0.1, δy = 0.1, 4 palice.

9

Page 10: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.2

0.30.4

0.50.6

µ0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.2

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 2: Pravokotni prerez: δz = 0.1, δy = 0.1, 4 palice.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.2

0.30.4

0.50.6

µ0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.3

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 3: Pravokotni prerez: δz = 0.1, δy = 0.1, 4 palice.

10

Page 11: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.4

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 4: Pravokotni prerez: δz = 0.1, δy = 0.1, 4 palice.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.5

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 5: Pravokotni prerez: δz = 0.1, δy = 0.1, 4 palice.

11

Page 12: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.6

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 6: Pravokotni prerez: δz = 0.1, δy = 0.1, 4 palice.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.5

0.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.7

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 7: Pravokotni prerez: δz = 0.1, δy = 0.1, 4 palice.

12

Page 13: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.8

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 8: Pravokotni prerez: δz = 0.1, δy = 0.1, 4 palice.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.9

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 9: Pravokotni prerez: δz = 0.1, δy = 0.1, 4 palice.

13

Page 14: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 1

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 10: Pravokotni prerez: δz = 0.1, δy = 0.1, 4 palice.

14

Page 15: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

2.2 PRAVOKOTNI PREREZ, 4 palice (δz = 0.125, δy = 0.1), S500

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε

2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.2

0.30.4

0.50.6

µ0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.1

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 11: Pravokotni prerez: δz = 0.125, δy = 0.1, 4 palice.

15

Page 16: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.2

0.30.4

0.50.6

µ0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.2

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 12: Pravokotni prerez: δz = 0.125, δy = 0.1, 4 palice.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.3

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 13: Pravokotni prerez: δz = 0.125, δy = 0.1, 4 palice.

16

Page 17: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.4

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 14: Pravokotni prerez: δz = 0.125, δy = 0.1, 4 palice.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.5

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 15: Pravokotni prerez: δz = 0.125, δy = 0.1, 4 palice.

17

Page 18: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.6

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 16: Pravokotni prerez: δz = 0.125, δy = 0.1, 4 palice.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.7

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 17: Pravokotni prerez: δz = 0.125, δy = 0.1, 4 palice.

18

Page 19: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.8

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 18: Pravokotni prerez: δz = 0.125, δy = 0.1, 4 palice.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.9

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 19: Pravokotni prerez: δz = 0.125, δy = 0.1, 4 palice.

19

Page 20: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 1

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 20: Pravokotni prerez: δz = 0.125, δy = 0.1, 4 palice.

20

Page 21: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

2.3 PRAVOKOTNI PREREZ, 4 palice (δz = 0.1, δy = 0.125), S500

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε

2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.2

0.30.4

0.50.6

µ0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.1

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 21: Pravokotni prerez: δz = 0.1, δy = 0.125, 4 palice.

21

Page 22: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.2

0.30.4

0.50.6

µ0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.2

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 22: Pravokotni prerez: δz = 0.1, δy = 0.125, 4 palice.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.2

0.30.4

0.50.6

µ0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.3

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 23: Pravokotni prerez: δz = 0.1, δy = 0.125, 4 palice.

22

Page 23: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.4

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 24: Pravokotni prerez: δz = 0.1, δy = 0.125, 4 palice.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.5

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 25: Pravokotni prerez: δz = 0.1, δy = 0.125, 4 palice.

23

Page 24: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.6

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 26: Pravokotni prerez: δz = 0.1, δy = 0.125, 4 palice.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.7

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 27: Pravokotni prerez: δz = 0.1, δy = 0.125, 4 palice.

24

Page 25: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.8

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 28: Pravokotni prerez: δz = 0.1, δy = 0.125, 4 palice.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.9

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 29: Pravokotni prerez: δz = 0.1, δy = 0.125, 4 palice.

25

Page 26: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 1

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 30: Pravokotni prerez: δz = 0.1, δy = 0.125, 4 palice.

26

Page 27: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

2.4 PRAVOKOTNI PREREZ, 4 palice (δz = 0.125, δy = 0.125), S500

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε

2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.2

0.30.4

0.50.6

µ0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.1

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 31: Pravokotni prerez: δz = 0.125, δy = 0.125, 4 palice.

27

Page 28: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.2

0.30.4

0.50.6

µ0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.2

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 32: Pravokotni prerez: δz = 0.125, δy = 0.125, 4 palice.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.3

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 33: Pravokotni prerez: δz = 0.125, δy = 0.125, 4 palice.

28

Page 29: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.4

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 34: Pravokotni prerez: δz = 0.125, δy = 0.125, 4 palice.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.5

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 35: Pravokotni prerez: δz = 0.125, δy = 0.125, 4 palice.

29

Page 30: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.6

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 36: Pravokotni prerez: δz = 0.125, δy = 0.125, 4 palice.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.7

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 37: Pravokotni prerez: δz = 0.125, δy = 0.125, 4 palice.

30

Page 31: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.8

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 38: Pravokotni prerez: δz = 0.125, δy = 0.125, 4 palice.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.9

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 39: Pravokotni prerez: δz = 0.125, δy = 0.125, 4 palice.

31

Page 32: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 1

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 40: Pravokotni prerez: δz = 0.125, δy = 0.125, 4 palice.

32

Page 33: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

2.5 PRAVOKOTNI PREREZ, 4 palice (δz = 0.15, δy = 0.125), S500

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.2

0.30.4

0.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.15

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.1

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 41: Pravokotni prerez: δz = 0.15, δy = 0.125, 4 palice.

33

Page 34: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.15

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.2

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 42: Pravokotni prerez: δz = 0.15, δy = 0.125, 4 palice.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.15

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.3

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 43: Pravokotni prerez: δz = 0.15, δy = 0.125, 4 palice.

34

Page 35: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.15

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.4

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 44: Pravokotni prerez: δz = 0.15, δy = 0.125, 4 palice.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.15

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.5

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 45: Pravokotni prerez: δz = 0.15, δy = 0.125, 4 palice.

35

Page 36: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.15

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.6

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 46: Pravokotni prerez: δz = 0.15, δy = 0.125, 4 palice.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.15

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.7

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 47: Pravokotni prerez: δz = 0.15, δy = 0.125, 4 palice.

36

Page 37: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.15

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.8

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 48: Pravokotni prerez: δz = 0.15, δy = 0.125, 4 palice.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.15

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.9

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 49: Pravokotni prerez: δz = 0.15, δy = 0.125, 4 palice.

37

Page 38: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.15

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 1

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 50: Pravokotni prerez: δz = 0.15, δy = 0.125, 4 palice.

38

Page 39: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

2.6 PRAVOKOTNI PREREZ, 4 palice (δz = 0.125, δy = 0.15), S500

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.2

0.30.4

0.50.6

µ0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.15

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.1

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 51: Pravokotni prerez: δz = 0.125, δy = 0.15, 4 palice.

39

Page 40: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.5

0.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.15

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.2

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 52: Pravokotni prerez: δz = 0.125, δy = 0.15, 4 palice.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.15

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.3

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 53: Pravokotni prerez: δz = 0.125, δy = 0.15, 4 palice.

40

Page 41: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.15

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.4

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 54: Pravokotni prerez: δz = 0.125, δy = 0.15, 4 palice.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.15

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.5

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 55: Pravokotni prerez: δz = 0.125, δy = 0.15, 4 palice.

41

Page 42: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.15

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.6

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 56: Pravokotni prerez: δz = 0.125, δy = 0.15, 4 palice.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.15

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.7

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 57: Pravokotni prerez: δz = 0.125, δy = 0.15, 4 palice.

42

Page 43: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.15

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.8

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 58: Pravokotni prerez: δz = 0.125, δy = 0.15, 4 palice.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.15

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.9

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 59: Pravokotni prerez: δz = 0.125, δy = 0.15, 4 palice.

43

Page 44: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.15

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 1

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 60: Pravokotni prerez: δz = 0.125, δy = 0.15, 4 palice.

44

Page 45: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

2.7 PRAVOKOTNI PREREZ, 4 palice (δz = 0.15, δy = 0.15), S500

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.15

δy = a

y / b = 0.15

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.1

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 61: Pravokotni prerez: δz = 0.15, δy = 0.15, 4 palice.

45

Page 46: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.15

δy = a

y / b = 0.15

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.2

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 62: Pravokotni prerez: δz = 0.15, δy = 0.15, 4 palice.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.15

δy = a

y / b = 0.15

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.3

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 63: Pravokotni prerez: δz = 0.15, δy = 0.15, 4 palice.

46

Page 47: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.15

δy = a

y / b = 0.15

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.4

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 64: Pravokotni prerez: δz = 0.15, δy = 0.15, 4 palice.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.15

δy = a

y / b = 0.15

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.5

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 65: Pravokotni prerez: δz = 0.15, δy = 0.15, 4 palice.

47

Page 48: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.15

δy = a

y / b = 0.15

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.6

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 66: Pravokotni prerez: δz = 0.15, δy = 0.15, 4 palice.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.15

δy = a

y / b = 0.15

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.7

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 67: Pravokotni prerez: δz = 0.15, δy = 0.15, 4 palice.

48

Page 49: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.15

δy = a

y / b = 0.15

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.8

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 68: Pravokotni prerez: δz = 0.15, δy = 0.15, 4 palice.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.15

δy = a

y / b = 0.15

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.9

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 69: Pravokotni prerez: δz = 0.15, δy = 0.15, 4 palice.

49

Page 50: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0.22 0.24 0.26 0.28 0.3

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.15

δy = a

y / b = 0.15

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 1

µ1 = µ

0 / 4 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 70: Pravokotni prerez: δz = 0.15, δy = 0.15, 4 palice.

50

Page 51: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

2.8 PRAVOKOTNI PREREZ, 8 palic (δz = 0.075, δy = 0.075), S500

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε

2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.2

0.30.4

0.50.6

µ0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.1

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 71: Pravokotni prerez: δz = 0.075, δy = 0.075, 8 palic.

51

Page 52: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.2

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 72: Pravokotni prerez: δz = 0.075, δy = 0.075, 8 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.3

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 73: Pravokotni prerez: δz = 0.075, δy = 0.075, 8 palic.

52

Page 53: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.4

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 74: Pravokotni prerez: δz = 0.075, δy = 0.075, 8 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.5

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 75: Pravokotni prerez: δz = 0.075, δy = 0.075, 8 palic.

53

Page 54: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.6

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 76: Pravokotni prerez: δz = 0.075, δy = 0.075, 8 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.7

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 77: Pravokotni prerez: δz = 0.075, δy = 0.075, 8 palic.

54

Page 55: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.8

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 78: Pravokotni prerez: δz = 0.075, δy = 0.075, 8 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5 40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.9

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 79: Pravokotni prerez: δz = 0.075, δy = 0.075, 8 palic.

55

Page 56: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5 40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 1

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 80: Pravokotni prerez: δz = 0.075, δy = 0.075, 8 palic.

56

Page 57: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

2.9 PRAVOKOTNI PREREZ, 8 palic (δz = 0.1, δy = 0.075), S500

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.1

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 81: Pravokotni prerez: δz = 0.1, δy = 0.075, 8 palic.

57

Page 58: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.2

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 82: Pravokotni prerez: δz = 0.1, δy = 0.075, 8 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.3

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 83: Pravokotni prerez: δz = 0.1, δy = 0.075, 8 palic.

58

Page 59: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.4

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 84: Pravokotni prerez: δz = 0.1, δy = 0.075, 8 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.5

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 85: Pravokotni prerez: δz = 0.1, δy = 0.075, 8 palic.

59

Page 60: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.6

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 86: Pravokotni prerez: δz = 0.1, δy = 0.075, 8 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.7

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 87: Pravokotni prerez: δz = 0.1, δy = 0.075, 8 palic.

60

Page 61: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.8

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 88: Pravokotni prerez: δz = 0.1, δy = 0.075, 8 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.9

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 89: Pravokotni prerez: δz = 0.1, δy = 0.075, 8 palic.

61

Page 62: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0.22 0.24 0.26 0.28 0.3

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 1

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 90: Pravokotni prerez: δz = 0.1, δy = 0.075, 8 palic.

62

Page 63: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

2.10 PRAVOKOTNI PREREZ, 8 palic (δz = 0.075, δy = 0.1), S500

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε

2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.2

0.30.4

0.50.6

µ0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.1

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 91: Pravokotni prerez: δz = 0.075, δy = 0.1, 8 palic.

63

Page 64: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.2

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 92: Pravokotni prerez: δz = 0.075, δy = 0.1, 8 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.3

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 93: Pravokotni prerez: δz = 0.075, δy = 0.1, 8 palic.

64

Page 65: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.4

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 94: Pravokotni prerez: δz = 0.075, δy = 0.1, 8 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.5

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 95: Pravokotni prerez: δz = 0.075, δy = 0.1, 8 palic.

65

Page 66: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.6

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 96: Pravokotni prerez: δz = 0.075, δy = 0.1, 8 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.7

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 97: Pravokotni prerez: δz = 0.075, δy = 0.1, 8 palic.

66

Page 67: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5 40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.8

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 98: Pravokotni prerez: δz = 0.075, δy = 0.1, 8 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5 40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.9

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 99: Pravokotni prerez: δz = 0.075, δy = 0.1, 8 palic.

67

Page 68: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0.22 0.24 0.26 0.28 0.3

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 1

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 100: Pravokotni prerez: δz = 0.075, δy = 0.1, 8 palic.

68

Page 69: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

2.11 PRAVOKOTNI PREREZ, 8 palic (δz = 0.1, δy = 0.1), S500

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.1

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 101: Pravokotni prerez: δz = 0.1, δy = 0.1, 8 palic.

69

Page 70: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.2

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 102: Pravokotni prerez: δz = 0.1, δy = 0.1, 8 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.3

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 103: Pravokotni prerez: δz = 0.1, δy = 0.1, 8 palic.

70

Page 71: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.4

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 104: Pravokotni prerez: δz = 0.1, δy = 0.1, 8 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.5

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 105: Pravokotni prerez: δz = 0.1, δy = 0.1, 8 palic.

71

Page 72: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.6

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 106: Pravokotni prerez: δz = 0.1, δy = 0.1, 8 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.7

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 107: Pravokotni prerez: δz = 0.1, δy = 0.1, 8 palic.

72

Page 73: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.8

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 108: Pravokotni prerez: δz = 0.1, δy = 0.1, 8 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.9

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 109: Pravokotni prerez: δz = 0.1, δy = 0.1, 8 palic.

73

Page 74: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0.22 0.24 0.26 0.28 0.3

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 1

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 110: Pravokotni prerez: δz = 0.1, δy = 0.1, 8 palic.

74

Page 75: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

2.12 PRAVOKOTNI PREREZ, 8 palic (δz = 0.125, δy = 0.1), S500

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.1

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 111: Pravokotni prerez: δz = 0.125, δy = 0.1, 8 palic.

75

Page 76: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.2

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 112: Pravokotni prerez: δz = 0.125, δy = 0.1, 8 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.3

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 113: Pravokotni prerez: δz = 0.125, δy = 0.1, 8 palic.

76

Page 77: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.4

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 114: Pravokotni prerez: δz = 0.125, δy = 0.1, 8 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.5

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 115: Pravokotni prerez: δz = 0.125, δy = 0.1, 8 palic.

77

Page 78: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.6

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 116: Pravokotni prerez: δz = 0.125, δy = 0.1, 8 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.7

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 117: Pravokotni prerez: δz = 0.125, δy = 0.1, 8 palic.

78

Page 79: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.8

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 118: Pravokotni prerez: δz = 0.125, δy = 0.1, 8 palic.

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0.22 0.24 0.26 0.28 0.3

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.9

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 119: Pravokotni prerez: δz = 0.125, δy = 0.1, 8 palic.

79

Page 80: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0.22 0.24 0.26 0.28 0.3

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 1

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 120: Pravokotni prerez: δz = 0.125, δy = 0.1, 8 palic.

80

Page 81: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

2.13 PRAVOKOTNI PREREZ, 8 palic (δz = 0.1, δy = 0.125), S500

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.1

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 121: Pravokotni prerez: δz = 0.1, δy = 0.125, 8 palic.

81

Page 82: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.2

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 122: Pravokotni prerez: δz = 0.1, δy = 0.125, 8 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.3

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 123: Pravokotni prerez: δz = 0.1, δy = 0.125, 8 palic.

82

Page 83: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.4

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 124: Pravokotni prerez: δz = 0.1, δy = 0.125, 8 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.5

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 125: Pravokotni prerez: δz = 0.1, δy = 0.125, 8 palic.

83

Page 84: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.6

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 126: Pravokotni prerez: δz = 0.1, δy = 0.125, 8 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.7

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 127: Pravokotni prerez: δz = 0.1, δy = 0.125, 8 palic.

84

Page 85: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.8

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 128: Pravokotni prerez: δz = 0.1, δy = 0.125, 8 palic.

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0.22 0.24 0.26 0.28 0.3

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.9

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 129: Pravokotni prerez: δz = 0.1, δy = 0.125, 8 palic.

85

Page 86: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0.22 0.24 0.26 0.28 0.3

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 1

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 130: Pravokotni prerez: δz = 0.1, δy = 0.125, 8 palic.

86

Page 87: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

2.14 PRAVOKOTNI PREREZ, 8 palic (δz = 0.125, δy = 0.125), S500

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.1

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 131: Pravokotni prerez: δz = 0.125, δy = 0.125, 8 palic.

87

Page 88: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.2

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 132: Pravokotni prerez: δz = 0.125, δy = 0.125, 8 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.3

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 133: Pravokotni prerez: δz = 0.125, δy = 0.125, 8 palic.

88

Page 89: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.4

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 134: Pravokotni prerez: δz = 0.125, δy = 0.125, 8 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.5

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 135: Pravokotni prerez: δz = 0.125, δy = 0.125, 8 palic.

89

Page 90: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.6

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 136: Pravokotni prerez: δz = 0.125, δy = 0.125, 8 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.7

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 137: Pravokotni prerez: δz = 0.125, δy = 0.125, 8 palic.

90

Page 91: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.8

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 138: Pravokotni prerez: δz = 0.125, δy = 0.125, 8 palic.

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0.22 0.24 0.26 0.28 0.3

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.9

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 139: Pravokotni prerez: δz = 0.125, δy = 0.125, 8 palic.

91

Page 92: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0.22 0.24 0.26 0.28 0.3

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.125

δy = a

y / b = 0.125

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 1

µ1 = µ

0 / 8 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 140: Pravokotni prerez: δz = 0.125, δy = 0.125, 8 palic.

92

Page 93: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

2.15 PRAVOKOTNI PREREZ, 12 palic (δz = 0.05, δy = 0.05), S500

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε

2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.2

0.30.4

0.50.6

µ0=0.7

myd

n d

δz = a

z / h = 0.05

δy = a

y / b = 0.05

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.1

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 141: Pravokotni prerez: δz = 0.05, δy = 0.05, 12 palic.

93

Page 94: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.2

0.30.4

0.50.6

µ0=0.7

myd

n d

δz = a

z / h = 0.05

δy = a

y / b = 0.05

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.2

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 142: Pravokotni prerez: δz = 0.05, δy = 0.05, 12 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.05

δy = a

y / b = 0.05

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.3

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 143: Pravokotni prerez: δz = 0.05, δy = 0.05, 12 palic.

94

Page 95: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5 40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.05

δy = a

y / b = 0.05

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.4

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 144: Pravokotni prerez: δz = 0.05, δy = 0.05, 12 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5 40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.05

δy = a

y / b = 0.05

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.5

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 145: Pravokotni prerez: δz = 0.05, δy = 0.05, 12 palic.

95

Page 96: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5 40/−3.5

40/40

0.10.20.3

0.40.50.6

µ0=0.7

myd

n d

δz = a

z / h = 0.05

δy = a

y / b = 0.05

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.6

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 146: Pravokotni prerez: δz = 0.05, δy = 0.05, 12 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5 40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.05

δy = a

y / b = 0.05

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.7

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 147: Pravokotni prerez: δz = 0.05, δy = 0.05, 12 palic.

96

Page 97: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5 40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.05

δy = a

y / b = 0.05

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.8

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 148: Pravokotni prerez: δz = 0.05, δy = 0.05, 12 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5 40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.05

δy = a

y / b = 0.05

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.9

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 149: Pravokotni prerez: δz = 0.05, δy = 0.05, 12 palic.

97

Page 98: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5 40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.05

δy = a

y / b = 0.05

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 1

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 150: Pravokotni prerez: δz = 0.05, δy = 0.05, 12 palic.

98

Page 99: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

2.16 PRAVOKOTNI PREREZ, 12 palic (δz = 0.075, δy = 0.05), S500

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.3

0.40.50.6

µ0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.05

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.1

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 151: Pravokotni prerez: δz = 0.075, δy = 0.05, 12 palic.

99

Page 100: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.05

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.2

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 152: Pravokotni prerez: δz = 0.075, δy = 0.05, 12 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.05

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.3

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 153: Pravokotni prerez: δz = 0.075, δy = 0.05, 12 palic.

100

Page 101: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.05

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.4

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 154: Pravokotni prerez: δz = 0.075, δy = 0.05, 12 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.05

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.5

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 155: Pravokotni prerez: δz = 0.075, δy = 0.05, 12 palic.

101

Page 102: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.05

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.6

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 156: Pravokotni prerez: δz = 0.075, δy = 0.05, 12 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.05

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.7

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 157: Pravokotni prerez: δz = 0.075, δy = 0.05, 12 palic.

102

Page 103: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5 40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.05

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.8

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 158: Pravokotni prerez: δz = 0.075, δy = 0.05, 12 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5 40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.05

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.9

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 159: Pravokotni prerez: δz = 0.075, δy = 0.05, 12 palic.

103

Page 104: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5 40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.05

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 1

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 160: Pravokotni prerez: δz = 0.075, δy = 0.05, 12 palic.

104

Page 105: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

2.17 PRAVOKOTNI PREREZ, 12 palic (δz = 0.05, δy = 0.075), S500

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε

2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.2

0.30.4

0.50.6

µ0=0.7

myd

n d

δz = a

z / h = 0.05

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.1

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 161: Pravokotni prerez: δz = 0.05, δy = 0.075, 12 palic.

105

Page 106: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.2

0.30.4

0.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.05

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.2

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 162: Pravokotni prerez: δz = 0.05, δy = 0.075, 12 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.05

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.3

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 163: Pravokotni prerez: δz = 0.05, δy = 0.075, 12 palic.

106

Page 107: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5 40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.05

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.4

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 164: Pravokotni prerez: δz = 0.05, δy = 0.075, 12 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5 40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.05

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.5

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 165: Pravokotni prerez: δz = 0.05, δy = 0.075, 12 palic.

107

Page 108: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5 40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.05

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.6

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 166: Pravokotni prerez: δz = 0.05, δy = 0.075, 12 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5 40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.05

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.7

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 167: Pravokotni prerez: δz = 0.05, δy = 0.075, 12 palic.

108

Page 109: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5 40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.05

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.8

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 168: Pravokotni prerez: δz = 0.05, δy = 0.075, 12 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5 40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.05

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.9

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 169: Pravokotni prerez: δz = 0.05, δy = 0.075, 12 palic.

109

Page 110: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5 40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.05

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 1

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 170: Pravokotni prerez: δz = 0.05, δy = 0.075, 12 palic.

110

Page 111: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

2.18 PRAVOKOTNI PREREZ, 12 palic (δz = 0.075, δy = 0.075), S500

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.4

0.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.1

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 171: Pravokotni prerez: δz = 0.075, δy = 0.075, 12 palic.

111

Page 112: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.2

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 172: Pravokotni prerez: δz = 0.075, δy = 0.075, 12 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.3

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 173: Pravokotni prerez: δz = 0.075, δy = 0.075, 12 palic.

112

Page 113: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.4

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 174: Pravokotni prerez: δz = 0.075, δy = 0.075, 12 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.5

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 175: Pravokotni prerez: δz = 0.075, δy = 0.075, 12 palic.

113

Page 114: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.6

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 176: Pravokotni prerez: δz = 0.075, δy = 0.075, 12 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5 40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.7

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 177: Pravokotni prerez: δz = 0.075, δy = 0.075, 12 palic.

114

Page 115: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5 40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.8

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 178: Pravokotni prerez: δz = 0.075, δy = 0.075, 12 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5 40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.9

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 179: Pravokotni prerez: δz = 0.075, δy = 0.075, 12 palic.

115

Page 116: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5 40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 1

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 180: Pravokotni prerez: δz = 0.075, δy = 0.075, 12 palic.

116

Page 117: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

2.19 PRAVOKOTNI PREREZ, 12 palic (δz = 0.1, δy = 0.075), S500

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.1

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 181: Pravokotni prerez: δz = 0.1, δy = 0.075, 12 palic.

117

Page 118: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.2

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 182: Pravokotni prerez: δz = 0.1, δy = 0.075, 12 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.3

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 183: Pravokotni prerez: δz = 0.1, δy = 0.075, 12 palic.

118

Page 119: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.4

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 184: Pravokotni prerez: δz = 0.1, δy = 0.075, 12 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.5

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 185: Pravokotni prerez: δz = 0.1, δy = 0.075, 12 palic.

119

Page 120: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.6

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 186: Pravokotni prerez: δz = 0.1, δy = 0.075, 12 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.7

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 187: Pravokotni prerez: δz = 0.1, δy = 0.075, 12 palic.

120

Page 121: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.8

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 188: Pravokotni prerez: δz = 0.1, δy = 0.075, 12 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.9

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 189: Pravokotni prerez: δz = 0.1, δy = 0.075, 12 palic.

121

Page 122: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0.22 0.24 0.26 0.28 0.3

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5 40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.075

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 1

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 190: Pravokotni prerez: δz = 0.1, δy = 0.075, 12 palic.

122

Page 123: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

2.20 PRAVOKOTNI PREREZ, 12 palic (δz = 0.075, δy = 0.1), S500

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.1

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 191: Pravokotni prerez: δz = 0.075, δy = 0.1, 12 palic.

123

Page 124: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.2

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 192: Pravokotni prerez: δz = 0.075, δy = 0.1, 12 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.3

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 193: Pravokotni prerez: δz = 0.075, δy = 0.1, 12 palic.

124

Page 125: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.4

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 194: Pravokotni prerez: δz = 0.075, δy = 0.1, 12 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.5

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 195: Pravokotni prerez: δz = 0.075, δy = 0.1, 12 palic.

125

Page 126: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.6

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 196: Pravokotni prerez: δz = 0.075, δy = 0.1, 12 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5 40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.7

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 197: Pravokotni prerez: δz = 0.075, δy = 0.1, 12 palic.

126

Page 127: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5 40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.8

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 198: Pravokotni prerez: δz = 0.075, δy = 0.1, 12 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5 40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.9

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 199: Pravokotni prerez: δz = 0.075, δy = 0.1, 12 palic.

127

Page 128: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0.22 0.24 0.26 0.28 0.3

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5 40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.075

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 1

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 200: Pravokotni prerez: δz = 0.075, δy = 0.1, 12 palic.

128

Page 129: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

2.21 PRAVOKOTNI PREREZ, 12 palic (δz = 0.1, δy = 0.1), S500

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.1

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 201: Pravokotni prerez: δz = 0.1, δy = 0.1, 12 palic.

129

Page 130: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5

40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.2

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 202: Pravokotni prerez: δz = 0.1, δy = 0.1, 12 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.3

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 203: Pravokotni prerez: δz = 0.1, δy = 0.1, 12 palic.

130

Page 131: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.4

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 204: Pravokotni prerez: δz = 0.1, δy = 0.1, 12 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.5

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 205: Pravokotni prerez: δz = 0.1, δy = 0.1, 12 palic.

131

Page 132: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.6

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 206: Pravokotni prerez: δz = 0.1, δy = 0.1, 12 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.7

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 207: Pravokotni prerez: δz = 0.1, δy = 0.1, 12 palic.

132

Page 133: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5

40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.8

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 208: Pravokotni prerez: δz = 0.1, δy = 0.1, 12 palic.

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5 40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 0.9

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 209: Pravokotni prerez: δz = 0.1, δy = 0.1, 12 palic.

133

Page 134: INTERAKCIJSKI DIAGRAMI NOSILNOSTI AB PRECNIH …fgg-web.fgg.uni-lj.si/KMLK/Sebastjan/masivni...Razmerje normiranih upogibnih momentov m zd/m yd pa je: m zd m yd = M zd M yd h b = 90

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0.22 0.24 0.26 0.28 0.3

−2.0

−1.8

−1.6

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1.0

ε2/ε

1 = −2.0/−2.0°/

°° ε2/ε

1 =

−1/−2.75°/°°

εs/ε

1 =

0/−3.5°/°°

εs/ε

1 =

1/−3.5°/°°

2/−3.5

5/−3.5

10/−3.5

20/−3.5 40/−3.5 40/40

0.10.20.30.40.50.6µ

0=0.7

myd

n d

δz = a

z / h = 0.1

δy = a

y / b = 0.1

Normirana obremenitev:n

d = N

d / (A

c α f

cd)

myd

= Myd

/ (Ac h α f

cd)

mzd

Mzd

hm

yd M

yd b= = 1

µ1 = µ

0 / 12 ⋅ α f

cd / f

ydA

s,1 = µ

1 A

c

Slika 210: Pravokotni prerez: δz = 0.1, δy = 0.1, 12 palic.

134