ankastre kolon hesabı
August 17, 2017 | Author: teknokolik | Category: N/A
Short Description
Download ankastre kolon hesabı...
Description
Ankastre Kolon Hesabı
2.1
Yan Duvar Kolonları Yük Analizi
Öz Ağırlık Kar
Öz Ağırlık Kar
Rüzgar
2.1.1
Rüzgar
Bölüm:2
Yükleme Durumu
VA (ton)
VB (ton)
H (ton)
Öz Ağırlık:
3.218
3.218
0
Kar:
3.713
3.713
0
Soldan Rüzgar:
-1.027
-1.368
-0.078
Sağdan Rüzgar:
-1.368
-1.027
-0.078
B
A soldan rüzgar sağdan rüzgar
2.1.2
Yükleme Kombinezonları
Yukarıdaki yükleme durumlarının yanında kolonda kendi ağırlığı, örtü malzemesi, yatay kirişler ve stabilite bağlantılarının öz ağırlıkları nedeniyle oluşan ek kuvvetler aşağıdaki gibi hesaplanmıştır: Birim Etki Ettiği Özellikler: Ağırlık: Alan/Uzunluk: Toplam Kuvvet: (kg) Kolon Öz Ağırlığı Oluklu Eternit Cam Yatay Kirişler Düşey Stabilite Bağları
I 380 (kabul) Çift kat Pimapen U120 (kabul) L 50.50.5
84 kg/m 40 kg/m2 28.55 kg/m2 13.4 kg/m 3.77 kg/m
6.8 m 5x3.3 =16.5 m2 5x0.8 =4 m2 5x4 =20 m 0.65+0.57 =1.25 m
84x6.8= 40x16.5= 28.55x4= 13.4x20= 3.8x1.25=
Σ= Z- Ekseni Öz ağırlık Yüklemesi: Kar Yüklemesi: Rüzgar Yüklemesi: Rüzgar yayılı yükü:
3.218+1.784 =5.002 ton 3.713 ton -1.027 ton
571.2 660 114.2 268 4.75 1618 kg 1.784 ton
X- Ekseni
-0.078 ton 0.8x0.088x5 =0.352 ton/m
Sap 2000 programı kullanılarak yukarıdaki yüklerin sisteme etkimesi sağlanmış ve aşağıdaki normal kuvvet, kesme kuvveti, moment diyagramları elde edilmiştir.
Skx= 2x6.8 =13.6 m Sky= 1x3.4 =3.4 m
34
1.Yükleme Durumu Öz Ağırlık + Kar
2.Yükleme Durumu Öz Ağırlık + Kar + Rüzgar
3.Yükleme Durumu Öz Ağırlık + Rüzgar
35
2.1.3
Kesit Hesabı
Seçilen Profil..........\\
I 320 2
F prof := 77.8cm 3
W x := 782cm
4
Ix := 12510cm I. Durum İçin:
λx :=
σ1 :=
h := 32cm
N1 := 8.71ton
N3 := 3.97ton
S kx := 13.6m
b := 131mm
N2 := 7.69ton
M 3 := 7.61ton ⋅ m
S ky := 3.4m
M 2 := 7.61ton ⋅ m
M 3 k := 1.77ton ⋅ m
s := 11.5mm t := 17.3mm
M 2 k := 1.77ton ⋅ m
ix := 12.7cm iy := 2.67cm
λx = 107.087
ix N1⋅ w
σ1 = 0.365
F prof
λ y :=
S ky
F prof σçem⋅ 1.15 w S ky
(TS 648-Madde 3.4) 2
4
σeb = 89.669
bulunur.
F bas = 28.133cm
I ybas = 324.1cm
12
ise,
if( σ1 ≤ σçem, "OK" , "NOT OK !" ) = "OK"
2
3
N2
λ ybas :=
2
cm
max( λ) = 127.341 w := 3.26
cm
h − 2⋅ t ⎞ 1 ⋅ ⎝ 2 ⎠ 3
σ bem :=
2
ton
Gerilme Kontrolü:
ton
F bas := b ⋅ t + s ⋅ ⎛⎜
σeb :=
σçem := 1.44
cm
λ y = 127.341
iy
II. Durum İçin: (Normal Kuvvet + Moment)
t⋅ b
kg
(Sadece Normal Kuvvet)
S kx
I ybas :=
σa := 2400
iybas :=
kg λx :=
2
cm
σ bem = 705.304
kg 2
cm
σeb σ bem
I ybas
iybas = 3.394cm
F bas S kx ix
λx = 107.087 ise,
= 0.127
σ bx :=
M2 Wx
w := 2.13 bulunur.
σ bx = 882.823
kg 2
cm
λ ybas = 100.173
iybas
⎛ M2 k ⎞ + 0.3⋅ ⎛ M2 k ⎞ ⎜ M ⎝ M2 ⎠ ⎝ 2⎠
c bi := 1.75 − 1.05⋅ ⎜
c bi := if( c bi ≤ 2.3, c bi , 2.3)
c bi = 1.576
⎡ ⎛ 3⋅ 107 kg ⎞ ⋅ c ⎡ ⎛ 107 kg ⎞ ⋅ c ⎤ ⎤ ⎢ ⎜ bi ⎢⎜ bi ⎥ ⎥ 2 2 2 ⎡2 σa⋅ ( λ ybas) ⎤⎥ cm ⎠ cm ⎠ ⎝ ⎝ ⎢ ⎢ ⎥⎥ ⎢ σ bx1 := if ≥ λ ybas , − ⋅ σa , 2 ⎢ ⎢ ⎢ ⎥ kg 3 7 σa ( λ ybas) ⎥⎦ ⎥⎥ 9⋅ 10 ⋅ c bi ⎣ ⎢ ⎢ ⎥ 2 cm ⎣ ⎣ ⎦ ⎦
36
84 kg
3
σ bx1 = 1.192 × 10
cm
S ky⋅
σ bx2 := if( σ bx1 ≤ 0.6⋅ σa , σ bx2 , 0)
⎡ 829⋅ 104 ⎢ 2 cm σex := ⎢ ⎢ (λ )2 x ⎣ kg
⎤ ⎥ ⎥ ⎥ ⎦
4
2
cm
σ bx2 :=
2
ton
⋅ 10 ⋅ c bi ⋅
h
1
3
σ bx2 = 3.105 × 10
1000
2
cm
F bas Enkesit Koşulları Uygun Olmadığından,
kg
3
σ bx2 = 3.105 × 10
2
cm
σ bx2 := 0 σex = 722.91
kg
kg
Alınır
c m := 0.85 Alınır.
2
cm
Kontroller: c m⋅ σ bx σeb σeb ⎡ σeb ⎥⎤ ≤ 0.15, , + σeb ⎞ σ bem σ bem ⎢ σ bem ⎛ ⎥ ⋅ max( σ bx1 , σ bx2) 1− ⎜ ⎢ ⎥ σex ⎠ ⎣ ⎝ ⎦
kontrol1 := if⎢
kontrol2 :=
σeb σa⋅ 0.6
kontrol1 = 0.127
σ bx
+
kontrol2 = 0.803
max( σ bx1 , σ bx2)
if[ ( kontrol1 < 1) ⋅ ( kontrol2 < 1) , "OK" , "NOT OK" ] = "OK" III. Durum İçin: (Normal Kuvvet + Moment)
σeb :=
N3
σeb = 46.292
F prof
σ bem :=
λ ybas :=
σçem⋅ 1.15 w
kg λx :=
2
cm
σ bem = 705.304
S ky
(TS 648-Madde 3.4) S kx
σeb
kg
σ bem
2
cm
λx = 107.087 ise,
ix
= 0.066
σ bx :=
M3 Wx
w := 2.13 bulunur.
σ bx = 882.823
kg 2
cm
λ ybas = 100.173
iybas
⎛ M3 k ⎞ + 0.3⋅ ⎛ M3 k ⎞ ⎜ M ⎝ M3 ⎠ ⎝ 3⎠
c bi := 1.75 − 1.05⋅ ⎜
c bi := if( c bi ≤ 2.3, c bi , 2.3)
c bi = 1.576
⎡ ⎛ 3⋅ 107 kg ⎞ ⋅ c ⎡ ⎛ 107 kg ⎞ ⋅ c ⎤ ⎤ ⎢ ⎜ bi ⎢⎜ bi ⎥ ⎥ 2 2 2 ⎤ ⎡ σa⋅ ( λ ybas) ⎥ 2 cm ⎠ cm ⎠ ⎝ ⎝ ⎢ ⎢ ⎥⎥ ⎢ σ bx1 := if ≥ λ ybas , − ⋅ σa , 2 ⎢ ⎢ kg ⎢ ⎥ 3 7 σa (λ ybas) ⎥⎦ ⎥⎥ ⋅ c bi 9⋅ 10 ⎣ ⎢ ⎢ ⎥ 2 cm ⎣ ⎣ ⎦ ⎦ 84 3
σ bx1 = 1.192 × 10
kg 2
cm
σ bx2 :=
ton 2
cm
S ky⋅
4
⋅ 10 ⋅ c bi h
⋅
1 1000
3
σ bx2 = 3.105 × 10
kg 2
cm
F bas
37
bas
σ bx2 := if( σ bx1 ≤ 0.6⋅ σa , σ bx2 , 0)
⎡ 829⋅ 104 kg ⎢ 2 cm σex := ⎢ ⎢ (λ )2 x ⎣
⎤ ⎥ ⎥ ⎥ ⎦
Enkesit Koşulları Uygun Olmadığından,
kg
3
σ bx2 = 3.105 × 10
2
cm
σ bx2 := 0 σex = 722.91
kg
Alınır
c m := 0.85 Alınır.
2
cm
Kontroller: σeb σeb c m⋅ σ bx ⎡ σeb ⎥⎤ ≤ 0.15, , + σ bem σ bem σeb ⎞ ⎢ σ bem ⎛ ⎥ ⋅ max( σ bx1 , σ bx2) ⎜1 − ⎢ ⎥ σex ⎠ ⎣ ⎝ ⎦
kontrol1 := if⎢
kontrol2 :=
σeb σa⋅ 0.6
+
kontrol1 = 0.066
σ bx
kontrol2 = 0.773
max( σ bx1 , σ bx2)
if[ ( kontrol1 < 1) ⋅ ( kontrol2 < 1) , "OK" , "NOT OK" ] = "OK"
Yukarıdaki hesaplamalardanda anlaşılacağı gibi I320 profilinin kullanılmasıyla tüm yükleme durumları için gerekli şartlar sağlanmaktadır. Kolon profili olarak I320 seçilmiştir.
0 mm
Kolon Üst Ucu Mesnet Detayı
r =8
100 mm
l =10 cm 42
2.1.4
8
A
8.71 ton
l
10 cm
s
100 mm
Mesnet Parçası Eğrilik Yarıçapı: r
A 8.68. l
r = 7.56 2
r
s
h1
r
h2
50 mm h 1
2
2
r
80 mm
alınmıştır
h 1 = 17.55 mm h 2 = 32.45 mm
38
Temas Gerilmesi Kontrolü: bi
2 . ( 50 mm 10 mm)
Ft
12 cm. 1.37 cm
bi = 120 mm 1.37 cm) . 1 cm
( 10 cm
2
Ft = 25.07 cm σ
A. Ft
σ = 0.347
1 ton
ton
if σ 1.1
2
cm
ton 2
, "OK" , "NOT OK !" = "OK"
cm
Kolon Nervürleri Hesabı: 1.A .
An
2 Ft
( 10 1.37) . 1 ton . cm
2
An . . 2 0.3 cm ( 15 cm
τk
2.1.5
An = 1.499 ton τ k = 0.174
2 . 0.3 cm)
ton 2
cm
if τ k 1.1
ton 2
, "OK" , "NOT OK !" = "OK"
cm
Ankastre Kolon Ayağı Veriler:
N M T
N
8.71 ton
T
2.32 ton 7.61 ton . m
M0
Z D
I320
b kolon
131 mm
h kolon
320 mm
e1
100 mm
t1
10 mm
tguse a2
650.270.10
M24
10 mm 5 mm
akaynak
M24
5 mm 1.1
σ em
1120
M28 M24
σ çem
2
cm 1.440
3
M24
M24
M24
p em
350
M24
80-80
ton 2
cm
3
5
kg
M24
5
10
150
75
650.350.30
2
cm
80-80 650
ton
τ em
550
ton 2
( BS 20)
m
39
•
•
Taban Levhasının Boyutlandırılması: B := b kolon + 2⋅ 100mm
A := 650mm
Seçilmistir
B = 331mm
B := 350mm
Seçilmistir
D ve Z Kuvvetlerinin Hesabı: A
e1 :=
2
e2 :=
D = 20.415ton
e1 + e2
Z :=
D A 4
p 0 = 358.953
8
⋅A
e2 = 243.75mm
M 0 − N⋅ e1 e1 + e2
⋅B
ton
Z = 12.054ton
if( p 0 ≤ p em , "OK" , "NOT OK !" ) = "OK"
2
m
Ankraj Bulonlarının Kontrolü:
(
)
d 1 := ⎡⎣ 5⋅ 0.01⋅ min( t1 , tguse) ⋅ 10cm d 1 :=
floor( d 1⋅ 1000)
Fz :=
2
m := 1
2
Fz = 3.346cm
4
σz :=
min( t1 , tguse) = 10mm
adet := 3
için,
π⋅ ( 0.86⋅ d 1)
⎤⎦ − 0.1cm
0.5
d 1 = 21mm
1000
d 1 := 24mm
•
3
Beton Basınç Gerilmelerinin Kontrolü: p 0 :=
•
e1 = 225mm
M 0 + N⋅ e1
D :=
•
− e0
Z
σz = 1089.41
adet ⋅ Fz
kg
if( σz ≤ σem , "OK" , "NOT OK !" ) = "OK"
2
cm
Taban Levhası (t) Kalınlığı: c :=
B − b kolon − t1
c = 10.45cm
2 3
M 1 = 1.778 × 10 kg⋅ cm − 0.5
tlevha := 2.45cm
⋅
M 1 := p 0⋅
c
2
2
⋅ 1cm
M 2 :=
p 0⋅ B 2
⋅
B 4
− c ⋅ 1cm
M 2 = 968.77kg⋅ cm
max( M 1 , M 2)
σçem
tlevha :=
ceil( tlevha⋅ 1000) 1000
tlevha = 29mm
40
•
Guse Levhasının h 1 Yüksekliği: P := h 1 :=
N 4
+
M0
P = 14.068ton
2⋅ h kolon
P a1⋅ τem
+ 2⋅ a1
h 1 :=
ceil( h 1⋅ 100)
h 1 = 27cm
100
2 h1 tlevha ⎞ B⋅ ( tlevha) ⎛ 2⋅ h 1⋅ tguse⋅ ⎜ + + 2 2 ⎠ 2 ⎝ eguse :=
eguse = 6.138cm
2⋅ h 1⋅ tguse + B⋅ tlevha
⎛ ⎝
Sx := B⋅ tlevha⋅ ⎜ eguse −
Ix := 2⋅
tguse⋅ ( h 1)
3
12
tlevha ⎞
2 2 h1 tlevha ⎞ ⎞ ⎛ ⎛ 4 4 + 2⋅ h 1⋅ tguse⋅ ⎜ + tlevha − eguse + tlevha⋅ B⋅ ⎜ eguse − Ix = 1.12 × 10 cm 2 ⎠ ⎝ 2 ⎠ ⎝
⎛ ⎝
h kolon ⎞
⎛ ⎝
h kolon ⎞
M 1 := Z⋅ ⎜ e1 − M 2 := D⋅ ⎜ e2 −
3
Sx = 475.842cm
⎠
2
M 1 = 78.35ton ⋅ cm
⎠
2
M 2 = 170.98ton ⋅ cm
⎠
2
max( M 1 , M 2) = 1.551 × 10 kg m 3
Q := if( Z < D , D , Z)
σ0 :=
max( M 1 , M 2) Ix
⋅ ( h 1 − tlevha − eguse) σ0 = 0.274
τ k :=
•
Q⋅ Sx
τ k = 0.868
Ix⋅ 2⋅ a2
Q = 20.415ton
ton
if( σ0 ≤ σçem, "OK" , "NOT OK ! " ) = "OK"
2
cm
ton
if( τ k ≤ τem , "OK" , "NOT OK ! " ) = "OK"
2
cm
T Yatay Kuvvetinin Aktarılması: a3 := 3mm
σ3 :=
T
2⋅ a3⋅ ( sx kolon − 2⋅ a3)
σ3 = 0.144
ton 2
cm
if( σ3 ≤ σçem, "OK" , "NOT OK ! " ) = "OK"
41
•
Kamada U100 Kullanılırsa: h kama := 10cm
b kama := 10cm
t kama := 5cm σ kama :=
a kama := 3mm
lkama := 10cm T
σ kama = 42.093
lkama⋅ ( h kama − t kama)
kg 2
cm
if( σ kama ≤ p em , "OK" , "NOT OK !" ) = "OK" I kama := ( h kama + 2⋅ a kama) ⋅
⎛ ⎝
M kama := T⋅ ⎜ h kama − F k := 2⋅ h kama⋅ a kama
σ kama :=
τ kama :=
M kama I kama T Fk
( b kama + 2⋅ a kama) 3 12
h kama − t kama ⎞
⎠
2
−
b kama⋅ ( h kama)
3
12
4
I kama = 218.731cm
4
M kama = 1.579 × 10 kg⋅ cm
2
F k = 6 cm
⎛ b kama − a ⎞ kama ⎝ 2 ⎠
⋅⎜
τ kama = 0.387
σ kama = 0.374
ton
ton 2
cm
if( σ kama ≤ σ kem , "OK" , "NOT OK ! " ) = "OK"
2
cm
if( τ kama ≤ τem , "OK" , "NOT OK ! " ) = "OK"
•
Ankraj Bulonlarında Kontrol Veriler: G
100 mm
D
76 mm
N
250 mm
d1
30 mm
F
15 mm
C
570 mm
p em = 550
τ em
40
ton 2
m
ton 2
m
Çekme Yoluyla Aktarılan Kuvvet:
⎡
Ze1 := ⎢ 2⋅ G − 2
2
π⋅ D
⎣
4
−
π⋅ ( d 1)
4
2
⎤ ⎥ ⋅ pe ⎦
3
Ze1 = 7.363 × 10 kg
42
•
Aderans Yoluyla Aktarılan Kuvvet: 5
Ze2 := ⎡⎣ π⋅ D⋅ N + π⋅ d 1⋅ ( C − N − F) ⎤⎦ ⋅ τem
Ze2 = 8.825 × 10 kg
5
Ze0 := Ze1 + Ze2
Ze0 = 8.899 × 10 kg
Z = 8.328ton if⎛⎜ Ze0 ≥
⎝
2.2
Z 3
, "OK" , "NOT OK !" ⎞ = "OK"
⎠
Kalkan Duvar Kolonları
En elverişsiz pozisyonda duran kolonun S4 kolonu olduğu tesbit edilmiş ve aşağıdaki gibi boyutlandırılmıştır:
S3 – S8
Kolonu İçin Hesap Veriler: h kolon := 734cm
h eternit := h kolon − 370cm
a1 := 2.8m
h cam := 80cm
a2 := 2.8m
Mmax
N2
p wind
h kolon
2.2.1
q wind := 80 (N)
⎛ a1 + a2 ⎞ ⎝ 2 ⎠
p wind := 0.8⋅ q wind⋅ ⎜
N2 :=
a1 + a2
Q ust :=
2
M maxm :=
p wind = 179.2
2
Q ust = 0.725ton
p wind⋅ ( h kolon) 8
2
2
m
q eternit := 40
q cam := 28.55
kg 2
m
kg 2
m
kg m
⋅ ( q cam⋅ h cam + q eternit⋅ h eternit) + p kolon⋅ h kolon
p wind⋅ h kolon
kg
Qalt := Q ust
N2 = 1.97ton Qalt = 0.725ton
3
M maxm = 1.207 × 10 kg⋅ m
43
Eksenel Basınçlı Eğilme Durumunda Hesap:
Seçilen Profil..........\\
I 260 2
F prof := 53.4cm 3
W x := 442cm
4
Ix := 5740cm
b := 113mm
N2 = 1.97ton
s := 9.4mm
M 2 := M maxm
t := 14.1mm
M 2 = 1.33ton ⋅ m
ix := 10.4cm iy := 2.32cm h := 26cm
h − 2⋅ t ⎞ 1 ⋅ ⎝ 2 ⎠ 3
I ybas :=
σeb :=
t⋅ b
σeb = 33.464
σ bem :=
λ ybas :=
σçem⋅ 1.15
S ky := S kx
S ky = 734cm
kg λx :=
2
cm
3
S ky
2
cm
S kx = 734cm
I ybas
iybas :=
σ bem = 1.008 × 10
w
ton
2
4
F prof
σçem := 1.44
(TS 648-Madde 3.4)
I ybas = 169.54cm
N2
2
cm
F bas = 19.565cm
3
12
kg
S kx := h kolon
II. Durum İçin: (Normal Kuvvet + Moment) F bas := b ⋅ t + s ⋅ ⎛⎜
σa := 2400
2
cm
S kx
σeb
kg
iybas = 2.944cm
F bas
σ bem
λx = 70.577
ix
= 0.033
σ bx :=
ise,
M2 Wx
w := 1.49 bulunur.
σ bx = 273.035
kg 2
cm
λ ybas = 249.341
iybas
c bi := 1.75
⎡ ⎛ 3⋅ 107 kg ⎞ ⋅ c ⎡ ⎛ 107 kg ⎞ ⋅ c ⎤ ⎤ ⎢ ⎜ bi ⎢⎜ bi ⎥ ⎥ 2 2 2 ⎡2 σa⋅ ( λ ybas) ⎥⎤ cm ⎠ cm ⎠ ⎝ ⎝ ⎢ ⎢ ⎥⎥ ⎢ σ bx1 := if ≥ λ ybas , − ⋅σ , ⎢ ⎢3 ⎥ a ⎢ (λ )2 ⎥ ⎥ 7 kg σa 9⋅ 10 ⋅ c bi ybas ⎦⎥ ⎢ ⎢ ⎥ ⎣ 2 cm ⎣ ⎣ ⎦ ⎦ 84 σ bx1 = 281.481
kg 2
cm
σ bx2 := if( σ bx1 ≤ 0.6⋅ σa , σ bx2 , 0)
σ bx2 :=
ton 2
cm
S ky⋅
4
⋅ 10 ⋅ c bi h
⋅
1
3
σ bx2 = 1.367 × 10
1000
kg 2
cm
F bas 3
σ bx2 = 1.367 × 10
kg 2
cm
Enkesit Koşulları Uygun Olmadığından, σ bx2 := 0
Alınır
44
⎡ 829⋅ 104 kg ⎢ 2 cm σex := ⎢ ⎢ (λ )2 x ⎣
⎤ ⎥ ⎥ ⎥ ⎦
3
σex = 1.664 × 10
kg 2
cm
c m := 0.85 Alınır.
Kontroller: σeb σeb c m⋅ σ bx ⎡ σeb ⎥⎤ ≤ 0.15, , + σ bem σ bem σeb ⎞ ⎢ σ bem ⎛ ⎥ 1− ⋅ max( σ bx1 , σ bx2) ⎜ ⎢ ⎥ σex ⎠ ⎣ ⎝ ⎦
kontrol1 := if⎢
kontrol2 :=
σeb σa⋅ 0.6
+
σ bx
max( σ bx1 , σ bx2)
kontrol1 = 0.033
kontrol2 = 0.993 if[ ( kontrol1 < 1) ⋅ ( kontrol2 < 1) , "OK" , "NOT OK" ] = "OK"
•
S3 – S8
Kolon Ayağı Veriler: N := 1.97ton H := 0.725ton h kolon := 260mm sx kolon := 22.3cm b kama := 50mm t beton := 30mm t kama := 8mm σçem := 1.44
τ kem := 1.1
p em := 55
ton 2
cm
ton 2
cm
kg 2
cm
45
(BS 20)
•
Taban Levhasının Boyutları: A := 340mm B := 600mm F ta :=
N
2
F ta = 64.987cm
pem
if( A ⋅ B < F ta , "NOT OK !" , "OK" ) = "OK"
2
•
Taban Levhasının t Kalınlığı: N
p 0 :=
3
p 0 = 8.761 × 10
A⋅B
kg 2
m
t1 := 10mm c1 := A − h kolon − 2⋅ t1 t1
c0 := c1 + M 1 := p 0⋅ M 2 :=
c0 = 65mm
2
( c0) 2
M 1 = 18.507kg
2
p 0⋅ A
⋅ ⎛⎜
A
⎝4
2
− c0 ⎞
t1 = 3.699mm
σçem
M 0 := max( M 1 , M 2)
M 2 = 29.786kg
⎠
M0
t1 := 2.45⋅
•
c1 = 60mm
t1 :=
ceil( t1⋅ 1000) 1000
M 0 = 29.786kg
t1 = 4 mm
Guse Levhasının Yüksekliği: t2 := 10mm a maxm := 0.7⋅ min( t1 , t2)
a minm := 3mm a maxm = 2.8mm
N
h 1 :=
•
4⋅ a1⋅ τ kem
+ 2⋅ a1
a1 := 3mm
h 1 = 20.924mm
h 1 :=
Seçilmiştir.
ceil( h 1⋅ 1000) 1000
h 1 = 21mm
Guse Levhasının Uç Kesitinde Gerilme: A ⋅ t 1⋅ y :=
Ix :=
t1 2
⎛ ⎝
t2 ⎞⎤ 2
⎥ ⎠⎦
y = 3.652mm
A ⋅ t1 + 2⋅ h 1⋅ t2 A ⋅ ( t1)
3
12
M := ( p 0⋅ A ) ⋅ σ0 :=
⎡ ⎣
+ 2⋅ ⎢h 1⋅ t2⋅ ⎜ t1 +
M Ix
⎛ ⎝
+ A ⋅ t 1⋅ ⎜ y −
( c 1) 2 2
⋅ ( h1 + t1 − y)
t1 ⎞ 2
⎠
2
+
t2⋅ ( h 1)
3
12
⎛ h1 + t − y ⎞ 1 ⎝ 2 ⎠
+ h 1⋅ t2⋅ ⎜
2
4
Ix = 3.796cm
M = 5.361kg m σ0 = 0.332
ton 2
cm
if( σ0 ≤ σçem, "OK" , "NOT OK !" ) = "OK"
46
•
Guse Levhasını Taban Levhasına bağlayan a
Kaynak Kordonlarında Gerilme:
2
a2 := 3mm Q := p 0⋅ A ⋅ c1
Q = 178.715kg
⎛ ⎝
Sx := A ⋅ t1⋅ ⎜ y − τ k :=
•
t1 ⎞ 2
Q⋅ Sx
⎠
3
Sx = 2.246cm τ k = 0.194
Ix⋅ 2⋅ a2
ton
if( τ k ≤ τ kem , "OK" , "NOT OK !" ) = "OK"
2
cm
a3 Dikişinde Gerilme Kontrolü: a3 := 3mm τ k :=
•
H
if( τ k ≤ τ kem , "OK" , "NOT OK !" ) = "OK"
2
cm
H
lkama :=
( b kama − t beton) ⋅ pem
ceil( lkama⋅ 1000) 1000
lkama = 60mm
Kama Kol Kalınlığının Tahkiki: pi :=
W := σ0 :=
•
ton
Kama Boyunun Bulunması: lkama :=
•
τ k = 0.056
2⋅ a3⋅ ( sx kolon − 2⋅ a3)
H
lkama⋅ ( b kama − t beton)
( t kama) 2
M := pi⋅
( b kama − t beton) 2 2
M = 0.121ton
2
W = 0.107cm
6 M
σ0 = 1.133
W
ton
if( σ0 ≤ σçem, "OK" , "NOT OK !" ) = "OK"
2
cm
a4 Dikişinde Gerilme Kontrolü: a4 := 3mm I k := ( lkama + 2⋅ a4) ⋅ F k := 2⋅ b kama⋅ a4 τ k := M := σ k :=
H
2
⎛ b kama + a ⎞ 4 Ik ⎝ 2 ⎠
M
⋅⎜
− lkama⋅
(b kama) 3 12
4
I k = 34.089cm
2
F k = 3 cm ton 2
cm
⋅ H⋅ ( b kama + t beton)
if⎡⎢⎛⎜ σ k ≥ 0.75
⎣⎝
12
τ k = 0.242
Fk 1
(b kama + 2⋅ a4) 3
M = 26.308kg m σ k = 0.238
ton 2
cm
⎞ ⋅ ⎛ τ ≥ 0.75 ton ⎞ ⋅ ⎡ ( σ ) 2 + ( τ ) 2 > 1.25 ton ⎤ , "NOT OK !" , " OK" ⎤ = " OK" k k k ⎥ 2 ⎜ 2 ⎢ 2⎥ cm ⎠ ⎝ cm ⎠ ⎣ cm ⎦ ⎦ ton
47
S4 Kolonu İçin Hesap
Veriler: h kolon := 776cm
h eternit := h kolon − 370cm
a1 := 2.8m
h cam := 80cm
a2 := 2.8m
Mmax
N2
p wind
h kolon
2.2.2
q wind := 80
⎛ a1 + a2 ⎞ ⎝ 2 ⎠
N2 :=
a1 + a2
Q ust :=
2
2
Q ust = 0.766ton
p wind⋅ ( h kolon) 8
2
q cam := 28.55
kg 2
m
kg 2
m
kg m
⋅ ( q cam⋅ h cam + q eternit⋅ h eternit) + p kolon⋅ h kolon
p wind⋅ h kolon
M maxm :=
p wind = 179.2
2
m
q eternit := 40
(N)
p wind := 0.8⋅ q wind⋅ ⎜
kg
Qalt := Q ust
N2 = 2.105ton Qalt = 0.766ton
3
M maxm = 1.349 × 10 kg⋅ m
48
Eksenel Basınçlı Eğilme Durumunda Hesap:
Seçilen Profil..........\\
I 280 2
F prof := 61.6cm 3
W x := 542cm
4
Ix := 7590cm
b := 119mm
N2 = 2.105ton
s := 10.1mm
M 2 := M maxm
t := 15.2mm
iy := 2.45cm h := 28cm
I ybas :=
σeb :=
t⋅ b
S kx = 776cm
S ky := S kx
S ky = 776cm
F prof
σ bem :=
λ ybas :=
2
4
σeb = 30.995
σçem⋅ 1.15
kg λx :=
2
cm
3
S ky
I ybas
iybas :=
σ bem = 1.036 × 10
w
2
cm
(TS 648-Madde 3.4)
I ybas = 213.453cm
N2
ton
F bas = 22.29cm
3
12
2
cm
S kx := h kolon
II. Durum İçin: (Normal Kuvvet + Moment) h − 2⋅ t ⎞ 1 ⋅ ⎝ 2 ⎠ 3
kg
M 2 = 1.487ton ⋅ m σçem := 1.44
ix := 11.4cm
F bas := b ⋅ t + s ⋅ ⎛⎜
σa := 2400
2
cm
S kx
σeb
kg
iybas = 3.095cm
F bas
σ bem
λx = 68.07
ix = 0.03
ise,
σ bx :=
M2 Wx
w := 1.45 bulunur.
σ bx = 248.87
kg 2
cm
λ ybas = 250.762
iybas
c bi := 1.75
⎡ ⎛ 3⋅ 107 kg ⎞ ⋅ c ⎡ ⎛ 107 kg ⎞ ⋅ c ⎤ ⎤ ⎢ ⎜ bi 2 2 ⎤ ⎢ ⎜⎝ cm2 ⎠ bi ⎥ ⎥ ⎡2 σ ⋅ ( λ ) cm ⎠ a ybas ⎝ ⎢ ⎥⎥ ⎥ ⋅ σa , ⎢ σ bx1 := if ≥ λ ybas , ⎢ − 2 ⎢ ⎥⎥ ⎢ ⎢3 ⎥ 7 kg σa ( λ ybas) 9⋅ 10 ⋅ c bi ⎣ ⎦⎥ ⎢ ⎢ ⎥ 2 cm ⎣ ⎣ ⎦ ⎦ 84 σ bx1 = 278.302
kg 2
cm
σ bx2 := if( σ bx1 ≤ 0.6⋅ σa , σ bx2 , 0)
σ bx2 :=
ton 2
cm
S ky⋅
4
⋅ 10 ⋅ c bi h
⋅
1
3
σ bx2 = 1.368 × 10
1000
kg 2
cm
F bas 3
σ bx2 = 1.368 × 10
kg 2
cm
Enkesit Koşulları Uygun Olmadığından, σ bx2 := 0
Alınır
49
⎡ 829⋅ 104 kg ⎢ 2 cm σex := ⎢ ⎢ (λ )2 x ⎣
⎤ ⎥ ⎥ ⎥ ⎦
3
σex = 1.789 × 10
kg 2
cm
c m := 0.85 Alınır.
Kontroller: σeb σeb c m⋅ σ bx ⎡ σeb ⎥⎤ ≤ 0.15, , + σ bem σ bem σeb ⎞ ⎢ σ bem ⎛ ⎥ 1− ⋅ max( σ bx1 , σ bx2) ⎜ ⎢ ⎥ σex ⎠ ⎣ ⎝ ⎦
kontrol1 := if⎢
kontrol2 :=
σeb σa⋅ 0.6
+
σ bx
max( σ bx1 , σ bx2)
kontrol1 = 0.03
kontrol2 = 0.916 if[ ( kontrol1 < 1) ⋅ ( kontrol2 < 1) , "OK" , "NOT OK" ] = "OK"
•
S4
Kolon Ayağı Veriler:
N := 2.105ton H := 0.766ton h kolon := 280mm sx kolon := 24cm b kama := 50mm t beton := 30mm t kama := 8mm σçem := 1.44
τ kem := 1.1
p em := 55
ton 2
cm
ton 2
cm
kg 2
cm
50
(BS 20)
•
Taban Levhasının Boyutları: A := 340mm B := 600mm F ta :=
N
2
F ta = 69.441cm
pem
if( A ⋅ B < F ta , "NOT OK !" , "OK" ) = "OK"
2
•
Taban Levhasının t Kalınlığı: N
p 0 :=
3
p 0 = 9.361 × 10
A⋅B
kg 2
m
t1 := 10mm c1 := A − h kolon − 2⋅ t1 t1
c0 := c1 + M 1 := p 0⋅ M 2 :=
c0 = 45mm
2
( c0) 2
M 1 = 9.478kg
2
p 0⋅ A
⋅ ⎛⎜
A
⎝4
2
− c0 ⎞
t1 = 5.408mm
σçem
M 0 := max( M 1 , M 2)
M 2 = 63.654kg
⎠
M0
t1 := 2.45⋅
•
c1 = 40mm
t1 :=
ceil( t1⋅ 1000) 1000
M 0 = 63.654kg
t1 = 6 mm
Guse Levhasının Yüksekliği: t2 := 10mm a maxm := 0.7⋅ min( t1 , t2)
a minm := 3mm a maxm = 4.2mm
N
h 1 :=
•
4⋅ a1⋅ τ kem
+ 2⋅ a1
a1 := 3mm
h 1 = 21.947mm
h 1 :=
Seçilmiştir.
ceil( h 1⋅ 1000) 1000
h 1 = 22mm
Guse Levhasının Uç Kesitinde Gerilme: A ⋅ t 1⋅ y :=
Ix :=
t1 2
⎛ ⎝
t2 ⎞⎤ 2
⎥ ⎠⎦
y = 4.419mm
A ⋅ t1 + 2⋅ h 1⋅ t2 A ⋅ ( t1)
3
12
M := ( p 0⋅ A ) ⋅ σ0 :=
⎡ ⎣
+ 2⋅ ⎢h 1⋅ t2⋅ ⎜ t1 +
M Ix
⎛ ⎝
+ A ⋅ t 1⋅ ⎜ y −
( c 1) 2 2
⋅ ( h1 + t1 − y)
t1 ⎞ 2
⎠
2
+
t2⋅ ( h 1)
3
12
⎛ h1 + t − y ⎞ 1 ⎝ 2 ⎠
+ h 1⋅ t2⋅ ⎜
2
4
Ix = 5.392cm
M = 2.546kg m σ0 = 0.123
ton 2
cm
if( σ0 ≤ σçem, "OK" , "NOT OK !" ) = "OK"
51
•
Guse Levhasını Taban Levhasına bağlayan a
2
Kaynak Kordonlarında Gerilme:
a2 := 3mm Q := p 0⋅ A ⋅ c1
Q = 127.308kg
⎛ ⎝
Sx := A ⋅ t1⋅ ⎜ y − τ k :=
•
t1 ⎞ 2
Q⋅ Sx
⎠
3
Sx = 2.895cm τ k = 0.126
Ix⋅ 2⋅ a2
ton
if( τ k ≤ τ kem , "OK" , "NOT OK !" ) = "OK"
2
cm
a3 Dikişinde Gerilme Kontrolü: a3 := 3mm τ k :=
•
H
if( τ k ≤ τ kem , "OK" , "NOT OK !" ) = "OK"
2
cm
H
lkama :=
(b kama − t beton)⋅ pem
ceil( lkama⋅ 1000) 1000
lkama = 64mm
Kama Kol Kalınlığının Tahkiki: pi :=
W := σ0 :=
•
ton
Kama Boyunun Bulunması: lkama :=
•
τ k = 0.055
2⋅ a3⋅ ( sx kolon − 2⋅ a3)
H
l kama⋅ ( b kama − t beton)
(t kama)2
M := pi⋅
(b kama − t beton)2 2
M = 0.12ton
2
W = 0.107cm
6 M
σ0 = 1.122
W
ton
if( σ0 ≤ σçem, "OK" , "NOT OK !" ) = "OK"
2
cm
a4 Dikişinde Gerilme Kontrolü: a4 := 3mm I k := ( lkama + 2⋅ a4) ⋅
M := σ k :=
H
2
⎛ b kama + a ⎞ 4 Ik ⎝ 2 ⎠ ⋅⎜
if⎡⎢⎛⎜ σ k ≥ 0.75
⎣⎝
ton 2
cm
⋅ H⋅ ( b kama + t beton)
M
12
4
I k = 35.776cm
F k = 3 cm
τ k = 0.255
Fk 1
12
− lkama⋅
(b kama)3
2
F k := 2⋅ b kama⋅ a4 τ k :=
(b kama + 2⋅ a4)3
M = 27.796kg m σ k = 0.24
ton 2
cm
⎞ ⋅ ⎛ τ ≥ 0.75 ton ⎞ ⋅ ⎡ ( σ ) 2 + ( τ ) 2 > 1.25 ton ⎤ , "NOT OK !" , " OK" ⎤ = " OK" k k ⎜ k ⎥ 2 ⎢ 2⎥ cm ⎠ ⎝ cm ⎠ ⎣ cm ⎦ ⎦ ton
2
52
Kolonu İçin Hesap
N2
p wind
h kolon
Veriler: h kolon := 818cm
h eternit := h kolon − 370cm
a1 := 2.8m
h cam := 80cm
a2 := 3m
Mmax
S5
2.2.3
q wind q eternit
(N)
⎛ a1 + a2 ⎞ ⎝ 2 ⎠
p wind := 0.8⋅ q wind⋅ ⎜
N2 :=
a1 + a2
Q ust :=
2
2
Q ust = 0.837ton
p wind⋅ ( h kolon) 8
2
kg 2
m 40
q cam
28.55
kg 2
m
kg 2
m
kg m
⋅ ( q cam⋅ h cam + q eternit⋅ h eternit) + p kolon⋅ h kolon
p wind⋅ h kolon
M maxm :=
p wind = 185.6
80
Qalt := Q ust
N2 = 2.319ton Qalt = 0.837ton
3
M maxm = 1.552 × 10 kg⋅ m
53
Eksenel Basınçlı Eğilme Durumunda Hesap:
Seçilen Profil..........\\
I 300 2
F prof := 69.1cm 3
W x := 653cm
4
Ix := 9800cm
b := 125mm
N2 = 2.319ton
s := 10.8mm
M 2 := M maxm
t := 16.5mm
iy := 2.56cm h := 30cm
I ybas :=
σeb :=
t⋅ b
S kx = 818cm
S ky := S kx
S ky = 818cm
F prof
σ bem :=
λ ybas :=
2
4
σeb = 30.45
σçem⋅ 1.15
kg λx :=
2
cm
3
S ky
I ybas
iybas :=
σ bem = 1.029 × 10
w
2
cm
(TS 648-Madde 3.4)
I ybas = 268.555cm
N2
ton
F bas = 25.431cm
3
12
2
cm
S kx := h kolon
II. Durum İçin: (Normal Kuvvet + Moment) h − 2⋅ t ⎞ 1 ⋅ ⎝ 2 ⎠ 3
kg
M 2 = 1.711ton ⋅ m σçem := 1.44
ix := 11.9cm
F bas := b ⋅ t + s ⋅ ⎛⎜
σa := 2400
2
cm
S kx
σeb
kg
iybas = 3.25cm
F bas
σ bem
λx = 68.739
ix = 0.03
σ bx :=
ise,
M2 Wx
w := 1.46 bulunur.
σ bx = 237.729
kg 2
cm
λ ybas = 251.721
iybas
c bi := 1.75
⎡ ⎛ 3⋅ 107 kg ⎞ ⋅ c ⎡ ⎛ 107 kg ⎞ ⋅ c ⎤ ⎤ ⎢ ⎜ bi 2 2 ⎤ ⎢⎢ ⎜⎝ cm2 ⎠ bi ⎥⎥ ⎥⎥ ⎡2 σ ⋅ ( λ ) cm ⎠ a ybas ⎝ ⎢ ⎥ ⋅ σa , ≥ λ ybas , ⎢ − σ bx1 := if ⎢ ⎢3 ⎥ ⎢ (λ )2 ⎥ ⎥ 7 kg σa ⋅ c bi ybas 9⋅ 10 ⎦⎥ ⎢ ⎢ ⎥ ⎣ 2 cm ⎣ ⎣ ⎦ ⎦ 84 σ bx1 = 276.185
kg 2
cm
σ bx2 := if( σ bx1 ≤ 0.6⋅ σa , σ bx2 , 0)
σ bx2 :=
ton 2
cm
S ky⋅
4
⋅ 10 ⋅ c bi h
⋅
1
3
σ bx2 = 1.382 × 10
1000
kg 2
cm
F bas 3
σ bx2 = 1.382 × 10
kg 2
cm
Enkesit Koşulları Uygun Olmadığından, σ bx2 := 0
Alınır
54
⎡ 829⋅ 104 kg ⎢ 2 cm σex := ⎢ ⎢ (λ )2 x ⎣
⎤ ⎥ ⎥ ⎥ ⎦
3
σex = 1.754 × 10
kg 2
cm
c m := 0.85 Alınır.
Kontroller: σeb σeb c m⋅ σ bx ⎡ σeb ⎥⎤ ≤ 0.15, , + σ bem σ bem σeb ⎞ ⎢ σ bem ⎛ ⎥ 1− ⋅ max( σ bx1 , σ bx2) ⎜ ⎢ ⎥ σex ⎠ ⎣ ⎝ ⎦
kontrol1 := if⎢
kontrol2 :=
σeb σa⋅ 0.6
+
σ bx
max( σ bx1 , σ bx2)
kontrol1 = 0.03
kontrol2 = 0.882 if[ ( kontrol1 < 1) ⋅ ( kontrol2 < 1) , "OK" , "NOT OK" ] = "OK"
•
S5
Kolon Ayağı Veriler: N := 2.319ton H := 0.837ton h kolon := 300mm sx kolon := 25.7cm b kama := 50mm t beton := 30mm t kama := 8mm σçem := 1.44 τ kem := 1.1
p em := 55
ton 2
cm
ton 2
cm
kg 2
cm
55
(BS 20)
•
Taban Levhasının Boyutları: A := 340mm B := 600mm F ta :=
N
2
F ta = 76.5cm
pem
if( A ⋅ B < F ta , "NOT OK !" , "OK" ) = "OK"
2
•
Taban Levhasının t Kalınlığı: N
p 0 :=
4
p 0 = 1.031 × 10
A⋅B
kg 2
m
t1 := 10mm c1 := A − h kolon − 2⋅ t1 t1
c0 := c1 + M 1 := p 0⋅ M 2 :=
c0 = 25mm
2
( c0) 2
M 1 = 3.223kg
2
p 0⋅ A
⋅ ⎛⎜
A
⎝4
2
− c0 ⎞
t1 = 6.952mm
σçem
M 0 := max( M 1 , M 2)
M 2 = 105.188kg
⎠
M0
t1 := 2.45⋅
•
c1 = 20mm
t1 :=
ceil( t1⋅ 1000) 1000
M 0 = 105.188kg
t1 = 7 mm
Guse Levhasının Yüksekliği: t2 := 10mm a maxm := 0.7⋅ min( t1 , t2)
a minm := 3mm a maxm = 4.9mm
N
h 1 :=
•
4⋅ a1⋅ τ kem
+ 2⋅ a1
a1 := 3mm
h 1 = 23.568mm
h 1 :=
Seçilmiştir.
ceil( h 1⋅ 1000) 1000
h 1 = 24mm
Guse Levhasının Uç Kesitinde Gerilme: A ⋅ t 1⋅ y :=
Ix :=
t1 2
⎛ ⎝
t2 ⎞⎤ 2
⎥ ⎠⎦
y = 4.927mm
A ⋅ t1 + 2⋅ h 1⋅ t2 A ⋅ ( t1)
3
12
M := ( p 0⋅ A ) ⋅ σ0 :=
⎡ ⎣
+ 2⋅ ⎢h 1⋅ t2⋅ ⎜ t1 +
M Ix
⎛ ⎝
+ A ⋅ t 1⋅ ⎜ y −
( c 1) 2 2
⋅ ( h1 + t1 − y)
t1 ⎞ 2
⎠
2
+
t2⋅ ( h 1)
3
12
⎛ h1 + t − y ⎞ 1 ⎝ 2 ⎠
+ h 1⋅ t2⋅ ⎜
2
4
Ix = 7.362cm
M = 0.701kg m σ0 = 0.027
ton 2
cm
if( σ0 ≤ σçem, "OK" , "NOT OK !" ) = "OK"
56
•
Guse Levhasını Taban Levhasına bağlayan a
Kaynak Kordonlarında Gerilme:
2
a2 := 3mm Q := p 0⋅ A ⋅ c1
Q = 70.125kg
⎛ ⎝
Sx := A ⋅ t1⋅ ⎜ y − τ k :=
•
t1 ⎞ 2
Q⋅ Sx
⎠
3
Sx = 3.395cm τ k = 0.059
Ix⋅ 2⋅ a2
ton
if( τ k ≤ τ kem , "OK" , "NOT OK !" ) = "OK"
2
cm
a3 Dikişinde Gerilme Kontrolü: a3 := 3mm τ k :=
•
H
2⋅ a3⋅ ( sx kolon − 2⋅ a3)
if( τ k ≤ τ kem , "OK" , "NOT OK !" ) = "OK"
2
cm
H
lkama :=
( b kama − t beton) ⋅ pem
ceil( lkama⋅ 1000) 1000
lkama = 70mm
Kama Kol Kalınlığının Tahkiki: pi :=
W := σ0 :=
•
ton
Kama Boyunun Bulunması: lkama :=
•
τ k = 0.056
H
l kama⋅ ( b kama − t beton)
( t kama)2
M := pi⋅
(b kama − t beton)2 2
M = 0.12ton
2
W = 0.107cm
6 M
σ0 = 1.121
W
ton
if( σ0 ≤ σçem, "OK" , "NOT OK !" ) = "OK"
2
cm
a4 Dikişinde Gerilme Kontrolü: a4 := 3mm I k := ( lkama + 2⋅ a4) ⋅
M := σ k :=
H
2
⎛ b kama + a ⎞ 4 Ik ⎝ 2 ⎠ ⋅⎜
if⎡⎢⎛⎜ σ k ≥ 0.75
⎣⎝
12
4
I k = 38.307cm
ton 2
cm
⋅ H⋅ ( b kama + t beton)
M
( b kama) 3
F k = 3 cm
τ k = 0.279
Fk 1
12
− lkama⋅
2
F k := 2⋅ b kama⋅ a4 τ k :=
(b kama + 2⋅ a4) 3
M = 30.373kg m σ k = 0.245
ton 2
cm
⎞ ⋅ ⎛ τ ≥ 0.75 ton ⎞ ⋅ ⎡ ( σ ) 2 + ( τ ) 2 > 1.25 ton ⎤ , "NOT OK !" , " OK" ⎤ = " OK" k k ⎜ k ⎥ 2 ⎢ 2⎥ cm ⎠ ⎝ cm ⎠ ⎣ cm ⎦ ⎦ ton
2
57
S7
Kolonu İçin Hesap Veriler: h kolon := 776cm
h eternit := h kolon − 370cm
a1 := 2.8m
h cam := 80cm
a2 := 3m
Mmax
N2
p wind
h kolon
2.2.4
q wind q eternit
(N)
⎛ a1 + a2 ⎞ ⎝ 2 ⎠
p wind := 0.8⋅ q wind⋅ ⎜
N2 :=
a1 + a2
Q ust :=
2
2
Q ust = 0.794ton
p wind⋅ ( h kolon) 8
2
kg 2
m 40
q cam
28.55
kg 2
m
kg 2
m
kg m
⋅ ( q cam⋅ h cam + q eternit⋅ h eternit) + p kolon⋅ h kolon
p wind⋅ h kolon
M maxm :=
p wind = 185.6
80
Qalt := Q ust
N2 = 2.18ton Qalt = 0.794ton
3
M maxm = 1.397 × 10 kg⋅ m
58
Eksenel Basınçlı Eğilme Durumunda Hesap:
Seçilen Profil..........\\
I 280 2
F prof := 61.1cm 3
W x := 542cm
4
Ix := 7590cm
b := 119mm
N2 = 2.18ton
s := 10.1mm
M 2 := M maxm
t := 15.2mm
M 2 = 1.54ton ⋅ m
ix := 11.1cm iy := 2.45cm h := 28cm
h − 2⋅ t ⎞ 1 ⋅ ⎝ 2 ⎠ 3
I ybas :=
σeb :=
t⋅ b
F prof
σ bem :=
λ ybas :=
σçem⋅ 1.15
S ky := S kx
S ky = 776cm
kg λx :=
2
cm
3
S ky
2
cm
S kx = 776cm
I ybas
iybas :=
σ bem = 1.022 × 10
w
ton
2
4
σeb = 32.364
σçem := 1.44
(TS 648-Madde 3.4)
I ybas = 213.453cm
N2
2
cm
F bas = 22.29cm
3
12
kg
S kx := h kolon
II. Durum İçin: (Normal Kuvvet + Moment) F bas := b ⋅ t + s ⋅ ⎛⎜
σa := 2400
2
cm
S kx
σeb
kg
iybas = 3.095cm
F bas
σ bem
λx = 69.91
ix
= 0.032
ise,
σ bx :=
M2 Wx
w := 1.47 bulunur.
σ bx = 257.758
kg 2
cm
λ ybas = 250.762
iybas
c bi := 1.75
⎡ ⎛ 3⋅ 107 kg ⎞ ⋅ c ⎡ ⎛ 107 kg ⎞ ⋅ c ⎤ ⎤ ⎢ ⎜ bi 2 2 ⎤ ⎢⎢ ⎜⎝ cm2 ⎠ bi ⎥⎥ ⎥⎥ ⎡2 σ ⋅ ( λ ) cm ⎠ a ybas ⎝ ⎢ ⎥ ⋅ σa , ≥ λ ybas , ⎢ − σ bx1 := if ⎢ ⎢3 ⎥ ⎢ (λ )2 ⎥ ⎥ 7 kg σa 9⋅ 10 ⋅ c bi ybas ⎦⎥ ⎢ ⎢ ⎥ ⎣ 2 cm ⎣ ⎣ ⎦ ⎦ 84 σ bx1 = 278.302
kg 2
cm
σ bx2 := if( σ bx1 ≤ 0.6⋅ σa , σ bx2 , 0)
σ bx2 :=
ton 2
cm
S ky⋅
4
⋅ 10 ⋅ c bi h
⋅
1
3
σ bx2 = 1.368 × 10
1000
kg 2
cm
F bas 3
σ bx2 = 1.368 × 10
kg 2
cm
Enkesit Koşulları Uygun Olmadığından, σ bx2 := 0
Alınır
59
⎡ 829⋅ 104 kg ⎢ 2 cm σex := ⎢ ⎢ (λ )2 x ⎣
⎤ ⎥ ⎥ ⎥ ⎦
3
σex = 1.696 × 10
kg 2
cm
c m := 0.85 Alınır.
Kontroller: σeb σeb c m⋅ σ bx ⎡ σeb ⎥⎤ ≤ 0.15, , + σ bem σ bem σeb ⎞ ⎢ σ bem ⎛ ⎥ ⋅ max( σ bx1 , σ bx2) 1− ⎜ ⎢ ⎥ σex ⎠ ⎣ ⎝ ⎦
kontrol1 := if⎢
kontrol2 :=
σeb σa⋅ 0.6
+
σ bx
max( σ bx1 , σ bx2)
kontrol1 = 0.032
kontrol2 = 0.949 if[ ( kontrol1 < 1) ⋅ ( kontrol2 < 1) , "OK" , "NOT OK" ] = "OK"
•
S7
Kolon Ayağı Veriler: N := 2.18ton H := 0.794ton h kolon := 280mm sx kolon := 24cm b kama := 50mm t beton := 30mm t kama := 8mm σçem := 1.44 τ kem := 1.1
p em := 55
ton 2
cm
ton 2
cm
kg 2
cm
60
(BS 20)
•
Taban Levhasının Boyutları: A := 340mm B := 600mm F ta :=
N
2
F ta = 71.915cm
pem
if( A ⋅ B < F ta , "NOT OK !" , "OK" ) = "OK"
2
•
Taban Levhasının t Kalınlığı: N
p 0 :=
3
p 0 = 9.694 × 10
A⋅B
kg 2
m
t1 := 10mm c1 := A − h kolon − 2⋅ t1 t1
c0 := c1 + M 1 := p 0⋅ M 2 :=
c0 = 45mm
2
( c0) 2
M 1 = 9.816kg
2
p 0⋅ A
⋅ ⎛⎜
A
⎝4
2
− c0 ⎞
t1 = 5.504mm
σçem
M 0 := max( M 1 , M 2)
M 2 = 65.922kg
⎠
M0
t1 := 2.45⋅
•
c1 = 40mm
t1 :=
ceil( t1⋅ 1000) 1000
M 0 = 65.922kg
t1 = 6 mm
Guse Levhasının Yüksekliği: t2 := 10mm a maxm := 0.7⋅ min( t1 , t2)
a minm := 3mm a maxm = 4.2mm
N
h 1 :=
•
4⋅ a1⋅ τ kem
+ 2⋅ a1
a1 := 3mm
h 1 = 22.515mm
h 1 :=
Seçilmiştir.
ceil( h 1⋅ 1000) 1000
h 1 = 23mm
Guse Levhasının Uç Kesitinde Gerilme: A ⋅ t 1⋅ y :=
Ix :=
t1 2
⎛ ⎝
t2 ⎞⎤ 2
⎥ ⎠⎦
y = 4.472mm
A ⋅ t1 + 2⋅ h 1⋅ t2 A ⋅ ( t1)
3
12
M := ( p 0⋅ A ) ⋅ σ0 :=
⎡ ⎣
+ 2⋅ ⎢h 1⋅ t2⋅ ⎜ t1 +
M Ix
⎛ ⎝
+ A ⋅ t 1⋅ ⎜ y −
( c 1) 2 2
⋅ ( h1 + t1 − y)
t1 ⎞ 2
⎠
2
+
t2⋅ ( h 1)
3
12
⎛ h1 + t − y ⎞ 1 ⎝ 2 ⎠
+ h 1⋅ t2⋅ ⎜
2
4
Ix = 5.972cm
M = 2.637kg m σ0 = 0.119
ton 2
cm
if( σ0 ≤ σçem, "OK" , "NOT OK !" ) = "OK"
61
•
Guse Levhasını Taban Levhasına bağlayan a
Kaynak Kordonlarında Gerilme:
2
a2 := 3mm Q := p 0⋅ A ⋅ c1
Q = 131.844kg
⎛ ⎝
Sx := A ⋅ t1⋅ ⎜ y − τ k :=
•
t1 ⎞ 2
Q⋅ Sx
⎠
3
Sx = 3.003cm τ k = 0.122
Ix⋅ 2⋅ a2
ton
if( τ k ≤ τ kem , "OK" , "NOT OK !" ) = "OK"
2
cm
a3 Dikişinde Gerilme Kontrolü: a3 := 3mm τ k :=
•
H
if( τ k ≤ τ kem , "OK" , "NOT OK !" ) = "OK"
2
cm
H
lkama :=
(b kama − t beton)⋅ pem
ceil( lkama⋅ 1000) 1000
lkama = 66mm
Kama Kol Kalınlığının Tahkiki: pi :=
W := σ0 :=
•
ton
Kama Boyunun Bulunması: lkama :=
•
τ k = 0.057
2⋅ a3⋅ ( sx kolon − 2⋅ a3)
H
l kama⋅ ( b kama − t beton)
(t kama) 2
M := pi⋅
( b kama − t beton) 2 2
M = 0.12ton
2
W = 0.107cm
6 M
σ0 = 1.128
W
ton
if( σ0 ≤ σçem, "OK" , "NOT OK !" ) = "OK"
2
cm
a4 Dikişinde Gerilme Kontrolü: a4 := 3mm I k := ( lkama + 2⋅ a4) ⋅
M := σ k :=
H
2
⎛ b kama + a ⎞ 4 Ik ⎝ 2 ⎠ ⋅⎜
if⎡⎢⎛⎜ σ k ≥ 0.75
⎣⎝
12
4
I k = 36.62cm
ton 2
cm
⋅ H⋅ ( b kama + t beton)
M
(b kama)3
F k = 3 cm
τ k = 0.265
Fk 1
12
− lkama⋅
2
F k := 2⋅ b kama⋅ a4 τ k :=
( b kama + 2⋅ a4)3
M = 28.812kg m σ k = 0.243
ton 2
cm
⎞ ⋅ ⎛ τ ≥ 0.75 ton ⎞ ⋅ ⎡ ( σ ) 2 + ( τ ) 2 > 1.25 ton ⎤ , "NOT OK !" , " OK" ⎤ = " OK" k k ⎜ k ⎥ 2 ⎢ 2⎥ cm ⎠ ⎝ cm ⎠ ⎣ cm ⎦ ⎦ ton
2
62
63
View more...
Comments
