Nosné Konštrukcie drevo

Nosné Konštrukcie I.

Drevo

Marek Zahradník

3. ročník / 3. krúžok

2010 / 2011

Zadanie č.3: Drevený strop

A = 1; drevo S I, snehová oblasť III – 1,5kN/m2, obchodný priestor

vložiť nákres vrstiev stropu/strechy + pôdorys + tabuľku

1. Výpočet zaťaženia na jednotlivé prvky konštrukcie

q0celkové = (qstále + qpremenné) . ZŠ + q0

nosníka

q0nosníka = A0 . ρ . 1,35

približné rozmery nosníka pre ln = 5,25m

h = <1/15ln; 1/20ln> -> h0 = 0,34999999999999998

b/h = 5/7 -> b0 = 0,25

q0celkové =(2062,9349999999999 + 3750) . 1,222 + 472,50000000000006 = 7575,9065699999992 N/m

2. Návrh prierezu dreveného nosníka

Dimenzačná podmienka pre ohyb Wy,PL ≥ Msd0 / fm,d

Msd0 = 1/8 . q0celkové . ln2

Msd0 = 1/8 . 7575,9065699999992 . 5,74500000000000012 = 31255,37321756428 Nm

fm,d = kmod . fm,k / γM

fm,d = 0,8 . 16000000 / 1,3 = 9846153,846153846

kmod = modif. činiteľ dĺžky trvania zaťaženia = 0,8

fm,k =pevnosť materiálu – pri tomto dreve = 16000000 Pa

γM = súčiniteľ materiálu = 1,3

Wy ≥ 31255,37321756428 / 9846153,846153846

Wy = Iy / z = 1/12 bh3 / (h/2) = 1/6 bh2

b/h = 5/7 -> b = 5/7 h

Wy = 5/42 h3 ≥ Msd0 / fm,d

h ≥ 3√(42 . Msd0 / 5 / fm,d) = 3√(42 . 31255,37321756428 / 5 / 9846153,846153846)

h = 0,29999999999999999 m

b = 5/7 h = 0,20999999999999999 m

3. Posúdenie podľa 1. medzného stavu = namáhanie ohybom

q0nosníka = A . ρ . 1,35 = 0,20999999999999999 . 0,29999999999999999 . 4000 . 1,35 =

340,20000000000005 N/m

q0celkové =(2062,9349999999999 + 3750) . 1,222 +340,20000000000005 = 7443,606569999999 N/m

Msd = 1/8 . qcelkové . ln2 = 1/8 . 7443,606569999999 . 5,74500000000000012 = 30709,552616626781 Nm

q = 7443,606569999999N/mBh = 0N

A = 21381,759872324998NBv = 21381,759872324998N5,7450000000000001m

fm,d ≥ Msd0 / Wy

Wy = 5/42 h3 = 3,2142857142857142E-3 m3

Msd0 / Wy = 30709,552616626781 / 3,2142857142857142E-3 = 9554083,0362838879 < fm,d = 9846153,846153846 -> vyhovuje

Reakcie

ΣMb = 0

-q.5,74500000000000012/2 + A.5,7450000000000001 = 0

A = 7443,606569999999.5,74500000000000012 / (2.5,7450000000000001) = 21381,759872324998 N

ΣFiy = 0

A + Bv – q. 5,7450000000000001 = 0

Bv = - 21381,759872324998 + 7443,606569999999. 5,7450000000000001 = 21381,759872324998 N

ΣFix = 0

Bh = 0 N

4. Posúdenie podľa 1. medzného stavu - šmyk

τ – šmykové napätie vznikajúce v priereze od sily Vd

Vd – max. priečna sila v priereze; Vd = ½ q . ln = ½ 7443,606569999999 . 5,7450000000000001 = 21381,759872324998 N

Sy(c) – statický moment k bodu c; Sy(c) = bh/4 . h/2 = 0,20999999999999999 . 0,299999999999999992 / 8 = 0,0023625 m3

Iy – moment zotrvačnosti v priereze k osi y; Iy = 1/12 b.h3 = 1/12 . 0,20999999999999999 . 0,299999999999999993 = 4,7249999999999994E-4 m4

fv,k – charakteristická pevnosť v šmyku – pre toto drevo fv,k = 1800000 Pa

Dimenzačná podmienka pre šmyk: τ ≤ fv,d

τ = Vd . Sy(c) / Iy / b = 21381,759872324998 . 0,0023625 / 4,7249999999999994E-4 / 0,20999999999999999 = 509089,5207696429 Pa

fv,d = kmod . fv,k / γM = 0,8 . 1800000 /1,3 = 1107692,3076923077 Pa

0N

21381,759872324998 N

- 21381,759872324998 N

30709,552616626781 Nm

τ < fv,d -> profil 0,20999999999999999 x 0,29999999999999999 m vyhovuje pre 1. medzný stav

y c

h = 0,29999999999999999 m

τ MAX

z

b = 0,20999999999999999 m

5. Posúdenie podľa 2. medzného stavu – použiteľnosť: profil 0,20999999999999999 x 0,29999999999999999 m

uNET, INST ≤ uLIM

uNET, INST = u1, INST . (1+kdef,1) + u2, INST . (1+kdef,2)

kdef – súčiniteľ zväčšenia deformácie v čase dotvarovaním; kdef,1 = 0,59999999999999998; kdef,2

=0,5

ED – modul pružnosti dreva; ED = 1010 Pa

qKnosníka = A . ρ = 0,20999999999999999 . 0,29999999999999999 . 4000 = 252 N/m

u1, INST = 5/384 . (qKstále . ZŠ + qK

nosníka) . ln4 / (ED . Iy )

u1, INST = 5/384 . (1528,0999999999999 . 1,222 + 252) . 5,74500000000000014 / (1010 .

4,7249999999999994E-4) = 6,3620542102861702E-3 m

u2, INST = 5/384 . (qKpremenné . ZŠ) . ln4 / (ED . Iy )

u2, INST = 5/384 . (2500 . 1,222) . 5,74500000000000014 / (1010 . 4,7249999999999994E-4) =

9,9273033786756466E-3 m

uNET, INST = 6,3620542102861702E-3 . (1+0,59999999999999998) + 9,9273033786756466E-3 . (1+0,5)

= 2,5070241804471344E-2 m

uLIM = 1/200 ln = 1/200 . 5,7450000000000001 = 2,8725000000000001E-2 m

uNET, INST < uLIM -> profil 0,20999999999999999 x 0,29999999999999999 m vyhovuje pre 2. medzný stav


Drevo

Marek Zahradník


2010 / 2011

Zadanie č.3: Drevený strop oprava 1





nosníka



h = <1/15ln; 1/20ln> -> h0 =

b/h = 5/7 -> b0 =

q0celkové =(2062,9349999999999 + 3750) . 1,222 + 0 = 7103,4065699999992 N/m




Msd0 = 1/8 . 7103,4065699999992 . 5,74500000000000012 = 29306,013928501779 Nm


fm,d = 0,8 . 16000000 / 1,3 = 9846153,846153846




Wy ≥ 29306,013928501779 / 9846153,846153846

Wy = Iy / z = 1/12 bh3 / (h/2) = 1/6 bh2

b/h = 5/7 -> b = 5/7 h

Wy = 5/42 h3 ≥ Msd0 / fm,d

h ≥ 3√(42 . Msd0 / 5 / fm,d) = 3√(42 . 29306,013928501779 / 5 / 9846153,846153846)

h = 0 m

b = 5/7 h = 0 m


q0nosníka = A . ρ . 1,35 = 0 . 0 . 4000 . 1,35 = 0 N/m

q0celkové =(2062,9349999999999 + 3750) . 1,222 +0 = 7103,4065699999992 N/m


fm,d ≥ Msd0 / Wy

q = 7103,4065699999992N/mBh = 0N

A = 20404,535372324997NBv = 20404,535372324997N5,7450000000000001m

Wy = 5/42 h3 = 0 m3

Msd0 / Wy = 29306,013928501779 / 0 = 0 < fm,d = 9846153,846153846 -> vyhovuje

Reakcie

ΣMb = 0

-q.5,74500000000000012/2 + A.5,7450000000000001 = 0

A = 7103,4065699999992.5,74500000000000012 / (2.5,7450000000000001) = 20404,535372324997 N

ΣFiy = 0

A + Bv – q. 5,7450000000000001 = 0

Bv = - 20404,535372324997 + 7103,4065699999992. 5,7450000000000001 = 20404,535372324997 N

ΣFix = 0

Bh = 0 N




Sy(c) – statický moment k bodu c; Sy(c) = bh/4 . h/2 = 0 . 02 / 8 = 0 m3

Iy – moment zotrvačnosti v priereze k osi y; Iy = 1/12 b.h3 = 1/12 . 0 . 03 = 0 m4



τ = Vd . Sy(c) / Iy / b = 20404,535372324997 . 0 / 0 / 0 = 0 Pa

fv,d = kmod . fv,k / γM = 0,8 . 1800000 /1,3 = 1107692,3076923077 Pa

τ < fv,d -> profil 0 x 0 m vyhovuje pre 1. medzný stav

0N

20404,535372324997 N

- 20404,535372324997 N

29306,013928501779 Nm

y c

h = 0 m

τ MAX

z

b = 0 m

5. Posúdenie podľa 2. medzného stavu – použiteľnosť: profil 0 x 0 m

uNET, INST ≤ uLIM



=0,5


qKnosníka = A . ρ = 0 . 0 . 4000 = 0 N/m



u1, INST = 5/384 . (1528,0999999999999 . 1,222 + 0) . 5,74500000000000014 / (1010 . 0) = 0 m


u2, INST = 5/384 . (2500 . 1,222) . 5,74500000000000014 / (1010 . 0) = 0 m

uNET, INST = 0 . (1+0,59999999999999998) + 0 . (1+0,5) = 0 m

uLIM = 1/200 ln = 1/200 . 5,7450000000000001 = 2,8725000000000001E-2 m

uNET, INST < uLIM -> profil 0 x 0 m vyhovuje pre 2. medzný stav


Drevo

Marek Zahradník


2010 / 2011






nosníka



h = <1/15ln; 1/20ln> -> h0 =

b/h = 5/7 -> b0 =

q0celkové =(2062,9349999999999 + 3750) . 1,222 + 0 = 7103,4065699999992 N/m




Msd0 = 1/8 . 7103,4065699999992 . 5,74500000000000012 = 29306,013928501779 Nm


fm,d = 0,8 . 16000000 / 1,3 = 9846153,846153846




Wy ≥ 29306,013928501779 / 9846153,846153846

Wy = Iy / z = 1/12 bh3 / (h/2) = 1/6 bh2

b/h = 5/7 -> b = 5/7 h

Wy = 5/42 h3 ≥ Msd0 / fm,d

h ≥ 3√(42 . Msd0 / 5 / fm,d) = 3√(42 . 29306,013928501779 / 5 / 9846153,846153846)

h = 0 m

b = 5/7 h = 0 m


q0nosníka = A . ρ . 1,35 = 0 . 0 . 4000 . 1,35 = 0 N/m

q0celkové =(2062,9349999999999 + 3750) . 1,222 +0 = 7103,4065699999992 N/m


fm,d ≥ Msd0 / Wy

q = 7103,4065699999992N/mBh = 0N

A = 20404,535372324997NBv = 20404,535372324997N5,7450000000000001m

Wy = 5/42 h3 = 0 m3


Reakcie

ΣMb = 0

-q.5,74500000000000012/2 + A.5,7450000000000001 = 0

A = 7103,4065699999992.5,74500000000000012 / (2.5,7450000000000001) = 20404,535372324997 N

ΣFiy = 0

A + Bv – q. 5,7450000000000001 = 0

Bv = - 20404,535372324997 + 7103,4065699999992. 5,7450000000000001 = 20404,535372324997 N

ΣFix = 0

Bh = 0 N








τ = Vd . Sy(c) / Iy / b = 20404,535372324997 . 0 / 0 / 0 = 0 Pa

fv,d = kmod . fv,k / γM = 0,8 . 1800000 /1,3 = 1107692,3076923077 Pa


0N

20404,535372324997 N

- 20404,535372324997 N

29306,013928501779 Nm

y c

h = 0 m

τ MAX

z

b = 0 m


uNET, INST ≤ uLIM



=0,5


qKnosníka = A . ρ = 0 . 0 . 4000 = 0 N/m



u1, INST = 5/384 . (1528,0999999999999 . 1,222 + 0) . 5,74500000000000014 / (1010 . 0) = 0 m


u2, INST = 5/384 . (2500 . 1,222) . 5,74500000000000014 / (1010 . 0) = 0 m

uNET, INST = 0 . (1+0,59999999999999998) + 0 . (1+0,5) = 0 m

uLIM = 1/200 ln = 1/200 . 5,7450000000000001 = 2,8725000000000001E-2 m



Drevo

Marek Zahradník


2010 / 2011






nosníka



h = <1/15ln; 1/20ln> -> h0 =

b/h = 5/7 -> b0 =

q0celkové =(2062,9349999999999 + 3750) . 1,222 + 0 = 7103,4065699999992 N/m




Msd0 = 1/8 . 7103,4065699999992 . 5,74500000000000012 = 29306,013928501779 Nm


fm,d = 0,8 . 16000000 / 1,3 = 9846153,846153846




Wy ≥ 29306,013928501779 / 9846153,846153846

Wy = Iy / z = 1/12 bh3 / (h/2) = 1/6 bh2

b/h = 5/7 -> b = 5/7 h

Wy = 5/42 h3 ≥ Msd0 / fm,d

h ≥ 3√(42 . Msd0 / 5 / fm,d) = 3√(42 . 29306,013928501779 / 5 / 9846153,846153846)

h = 0 m

b = 5/7 h = 0 m


q0nosníka = A . ρ . 1,35 = 0 . 0 . 4000 . 1,35 = 0 N/m

q0celkové =(2062,9349999999999 + 3750) . 1,222 +0 = 7103,4065699999992 N/m


fm,d ≥ Msd0 / Wy

q = 7103,4065699999992N/mBh = 0N

A = 20404,535372324997NBv = 20404,535372324997N5,7450000000000001m

Wy = 5/42 h3 = 0 m3


Reakcie

ΣMb = 0

-q.5,74500000000000012/2 + A.5,7450000000000001 = 0

A = 7103,4065699999992.5,74500000000000012 / (2.5,7450000000000001) = 20404,535372324997 N

ΣFiy = 0

A + Bv – q. 5,7450000000000001 = 0

Bv = - 20404,535372324997 + 7103,4065699999992. 5,7450000000000001 = 20404,535372324997 N

ΣFix = 0

Bh = 0 N








τ = Vd . Sy(c) / Iy / b = 20404,535372324997 . 0 / 0 / 0 = 0 Pa

fv,d = kmod . fv,k / γM = 0,8 . 1800000 /1,3 = 1107692,3076923077 Pa


0N

20404,535372324997 N

- 20404,535372324997 N

29306,013928501779 Nm

y c

h = 0 m

τ MAX

z

b = 0 m


uNET, INST ≤ uLIM



=0,5


qKnosníka = A . ρ = 0 . 0 . 4000 = 0 N/m



u1, INST = 5/384 . (1528,0999999999999 . 1,222 + 0) . 5,74500000000000014 / (1010 . 0) = 0 m


u2, INST = 5/384 . (2500 . 1,222) . 5,74500000000000014 / (1010 . 0) = 0 m

uNET, INST = 0 . (1+0,59999999999999998) + 0 . (1+0,5) = 0 m

uLIM = 1/200 ln = 1/200 . 5,7450000000000001 = 2,8725000000000001E-2 m



Drevo

Marek Zahradník


2010 / 2011

Zadanie č.:





nosníka



h = <1/15ln; 1/20ln> -> h0 =

b/h = 5/7 -> b0 =

q0celkové =( + ) . + = N/m




Msd0 = 1/8 . . 2 = Nm


fm,d = 0,8 . / =


fm,k =pevnosť materiálu – pri tomto dreve = Pa


Wy ≥ /

Wy = Iy / z = 1/12 bh3 / (h/2) = 1/6 bh2

b/h = 5/7 -> b = 5/7 h

Wy = 5/42 h3 ≥ Msd0 / fm,d

h ≥ 3√(42 . Msd0 / 5 / fm,d) = 3√(42 . / 5 / )

h = m

b = 5/7 h = m


q0nosníka = A . ρ . 1,35 = . . 4000 . 1,35 = N/m

q0celkové =( + ) . + = N/m

Msd = 1/8 . qcelkové . ln2 = 1/8 . . 2 = Nm

fm,d ≥ Msd0 / Wy

Wy = 5/42 h3 = m3

q = N/mBh = 0N

A = NBv = Nm

Msd0 / Wy = / = < fm,d = -> vyhovuje

Reakcie

ΣMb = 0

-q.2/2 + A. = 0

A = .2 / (2.) = N

ΣFiy = 0

A + Bv – q. = 0

Bv = - + . = N

ΣFix = 0

Bh = 0 N



Vd – max. priečna sila v priereze; Vd = ½ q . ln = ½ . = N

Sy(c) – statický moment k bodu c; Sy(c) = bh/4 . h/2 = . 2 / 8 = m3

Iy – moment zotrvačnosti v priereze k osi y; Iy = 1/12 b.h3 = 1/12 . . 3 = m4

fv,k – charakteristická pevnosť v šmyku – pre toto drevo fv,k = Pa


τ = Vd . Sy(c) / Iy / b = . / / = Pa

fv,d = kmod . fv,k / γM = 0,8 . /1,3 = Pa

τ < fv,d -> profil x m vyhovuje pre 1. medzný stav

0N

N

- N

Nm

y c

h = m

τ MAX

z

b = m

5. Posúdenie podľa 2. medzného stavu – použiteľnosť: profil x m

uNET, INST ≤ uLIM


kdef – súčiniteľ zväčšenia deformácie v čase dotvarovaním; kdef,1 = ; kdef,2 =


qKnosníka = A . ρ = . . 4000 = N/m



u1, INST = 5/384 . ( . + ) . 4 / (1010 . ) = m


u2, INST = 5/384 . ( . ) . 4 / (1010 . ) = m

uNET, INST = . (1+) + . (1+) = m

uLIM = 1/200 ln = 1/200 . = m

uNET, INST < uLIM -> profil x m vyhovuje pre 2. medzný stav


Drevo

Marek Zahradník


2010 / 2011

Zadanie č.4: Drevená strecha





nosníka



h = <1/15ln; 1/20ln> -> h0 = 0,34999999999999998

b/h = 5/7 -> b0 = 0,25

q0celkové =(1651,7249999999999 + 882) . 1,081 + 472,50000000000006 = 3211,456725 N/m




Msd0 = 1/8 . 3211,456725 . 4,70000000000000022 = 8867,6348819062514 Nm


fm,d = 0,8 . 16000000 / 1,3 = 9846153,846153846




Wy ≥ 8867,6348819062514 / 9846153,846153846

Wy = Iy / z = 1/12 bh3 / (h/2) = 1/6 bh2

b/h = 5/7 -> b = 5/7 h

Wy = 5/42 h3 ≥ Msd0 / fm,d

h ≥ 3√(42 . Msd0 / 5 / fm,d) = 3√(42 . 8867,6348819062514 / 5 / 9846153,846153846)

h = 0,20000000000000001 m

b = 5/7 h = 0,14000000000000001 m


q0nosníka = A . ρ . 1,35 = 0,14000000000000001 . 0,20000000000000001 . 4000 . 1,35 =

151,20000000000002 N/m

q0celkové =(1651,7249999999999 + 882) . 1,081 +151,20000000000002 = 2890,1567249999998 N/m


q = 2890,1567249999998N/mBh = 0N

A = 6791,8683037499995NBv = 6791,8683037499995N4,7000000000000002m

fm,d ≥ Msd0 / Wy

Wy = 5/42 h3 = 9,523809523809526E-4 m3

Msd0 / Wy = 7980,4452569062505 / 9,523809523809526E-4 = 8379467,5197515609 < fm,d = 9846153,846153846 -> vyhovuje

Reakcie

ΣMb = 0

-q.4,70000000000000022/2 + A.4,7000000000000002 = 0

A = 2890,1567249999998.4,70000000000000022 / (2.4,7000000000000002) = 6791,8683037499995 N

ΣFiy = 0

A + Bv – q. 4,7000000000000002 = 0

Bv = - 6791,8683037499995 + 2890,1567249999998. 4,7000000000000002 = 6791,8683037499995 N

ΣFix = 0

Bh = 0 N




Sy(c) – statický moment k bodu c; Sy(c) = bh/4 . h/2 = 0,14000000000000001 . 0,200000000000000012 / 8 = 7,0000000000000021E-4 m3

Iy – moment zotrvačnosti v priereze k osi y; Iy = 1/12 b.h3 = 1/12 . 0,14000000000000001 . 0,200000000000000013 = 9,3333333333333357E-5 m4



τ = Vd . Sy(c) / Iy / b = 6791,8683037499995 . 7,0000000000000021E-4 / 9,3333333333333357E-5 / 0,14000000000000001 = 363850,08770089276 Pa

fv,d = kmod . fv,k / γM = 0,8 . 1800000 /1,3 = 1107692,3076923077 Pa

0N

6791,8683037499995 N

- 6791,8683037499995 N

7980,4452569062505 Nm

τ < fv,d -> profil 0,14000000000000001 x 0,20000000000000001 m vyhovuje pre 1. medzný stav

y c

h = 0,20000000000000001 m

τ MAX

z

b = 0,14000000000000001 m

5. Posúdenie podľa 2. medzného stavu – použiteľnosť: profil 0,14000000000000001 x 0,20000000000000001 m

uNET, INST ≤ uLIM



=0,25


qKnosníka = A . ρ = 0,14000000000000001 . 0,20000000000000001 . 4000 =

112,00000000000001 N/m



u1, INST = 5/384 . (1223,4999999999998 . 1,081 + 112,00000000000001) .

4,70000000000000024 / (1010 . 9,3333333333333357E-5) = 9,7661934172481851E-3 m


u2, INST = 5/384 . (588 . 1,081) . 4,70000000000000024 / (1010 . 9,3333333333333357E-5) =

5,0895454055078122E-3 m

uNET, INST = 9,7661934172481851E-3 . (1+0,59999999999999998) + 5,0895454055078122E-3 .

(1+0,25) = 2,1987841224481861E-2 m

uLIM = 1/200 ln = 1/200 . 4,7000000000000002 = 0,0235 m

uNET, INST < uLIM -> profil 0,14000000000000001 x 0,20000000000000001 m vyhovuje pre 2. medzný stav

Documents

Nosné Konštrukcie drevo