Betonarme Kopru Hesabi

PLAK MOMENTLERİNİN HESABI Yük: H 15 − S 20 F T =

1)

F AKS

=

2

120 2

= 38cm

u0

v0

= 20cm

= 60kN

Zati Zati Yük için için Plak Plak Mome Moment ntle lerin rinin in Hesa Hesabı bı:: hmeyil

η 1

=

c

1120

2

2

= × 0!015 =

310 4625

× 0!015 = 84cm

× 6"5 = 466 m

η 2

=

3#0 4625

× 6"5 = 556 m η ort = 556 + 6"5 = 623m

$o%lam Plak &'ırlı'ı: (*eme: &s+alt: ,a%: Me-il .!:

0 2 × 25 × 185 × 430

= 3# kN

01 × 18 × 185 × 430 = 1332kN 004 × 22 × 185 × 430 = 652kN 0063 × 22 × 185 × 430 = 84#kN

/ $o%lam $o%lam  6531 k

M

a

1

b

M

2

2

 nol abaktan : a 185 = = 04625 b 430 b 430 = 216 ρ = = a 185

M 1 = 00485

ρ =

M 2 = 00080

M a = ( M 1 + M 2 × η ) × G = ( 00485 + 02 × 00080) × 6531 = 328kNm  m M b = ( M 2 + M 1 × η ) × G = ( 00080 + 02 × 00485) × 6531 = 116kNm  m

ürekli Plakta abit Yükten (ola-ı &çıklık e Mesnet Momentleri: = 08 × 328 = 263kNm  m M a  mes = −05 × 328 = −164 kNm  m M b  aç = 08 × 116 = 0"3kNm  m M b  mes = −05 × 116 = −058kNm  m M a  aç

Merke7i am-on $ekerle'i $ekerle'i 9çin Plak Momentleri: Yük: H 15 − S 12 F T =

F AKS 2

=

u0

120 2

= 38cm

v0

= 20cm

= 60kN

u = u 0 + 2 × ( hk + h plak  2 = 38 + 2 × ( 201 + 10 ) = "82cm v = v0 + 2 × ( hk + h plak  2 = 20 + 2 × ( 201 + 10) = 802cm ρ =

a 185 = = 04625 b 430

&bak 8a ;
δ

= 04 =

u a u a

=

=

"82 185

"82 185

v

= 0521

= 0521 =

b v b

=

802 430

=

802 430

= 01"6

= 01"6 ⇒ M 1 = 0138

&bak #a ;
δ

= 05 =

u a

"82 185

= 0521

=

= 04625 için enter%olas-on -a%ılırsa

&bak 8b ;
δ

= 04 =

u

&bak #b ;
δ

= 05 =

u

δ

=

a

a

=

"82

=

"82

185

185

= 0521 = = 0521 =

= 04625 için enter%olas-on -a%ılırsa

v b

=

802 430

= 01"6 ⇒ M 1 = 0136

M 1 = 013# v b v b

=

802

=

802

430

430

= 01"6 ⇒ M 2 = 0062 = 01"6 ⇒ M 2 = 00"6

M 2 = 0082

M a q = ( M 1 + M 2 × η ) × F T = ( 013# + 0 2 × 0082 ) × 60 = "21kNm  m M b q = ( M 2 + M 1 × η ) × F T = ( 0082 + 02 × 013# ) × 60 = 65# kNm  m

$ekerlek Yükü için &çıklık e Mesnet Momentleri: = 08 × "21 = #3#kNm  m M a  mes = −05 × "21 = −461kNm  m M b  aç = 08 × 65# = 526 kNm  m M b  mes = −05 × 65# = −32"kNm  m M a  aç

$a*ıma>ü?üne >re betonarme @esabın
= 14 × 263 + 16 × 13 × #3# = 1"02# kNm  m M a mes = 14 × ( − 164 ) + 16 × 13 × ( − 461) = −118"kNm  m M b  aç = 14 × 0"3 + 16 × 13 × 526 = 1225kNm  m M b  mes = 14( − 058) + 16 × 13 × ( − 32") = −#66kNm  m M a  aç

PLAK DONATISININ HESABI Mal7eme 220B20

ısa
A s =

6 Md = 1"02 × 10 = 604 mm 2  m k z ×  yd × d 0"#1 × 1"1 × 1#0

eçilen
C7n
Md = 1225kNm  m 6 Md 1225 × 10 k m = = = 04"3 → k z = 0"#8 bw × d 2 1000 × 158 2

A s =

6 Md = 132 × 10 = 416 mm 2  m k z ×  yd × d 0"#8 × 1"1 × 158

eçilen
KONSOL MOMENTİNİN HESABI

"1

= 210 − 145

2

= 13#5m

" 2 = 210 2 = 105 m

"3

= 1"

3 = 064 m

" 4 = 210 − 015 2 = 2025 m "T = 210 − 06 = 150 m

Zati -ük için konsol momenti G1 = γ × A1 = 25 × ( 040 × 145) = 145kN G2 = γ × A2 = 25 × ( 025 × 210 ) = 13125kN ( 060 − 02) G3 = γ × A3 = 25 × × 1"0 = "5kN 2 G4 = γ × A4 = 25 × ( 02 × 015)

= 0#5kN

= −[145 × 13#5 + 13125 × 105 + "5 × 064 + 0#5 × 2025] = −4131#5kNm  m

M k#

% = 08 × $ T + 115 = 08 × " + 115 = 08 × ( 210 − 060) + 115 = 235m M k

=−

M ko!sol

F T

× $

%

=−

60 × 15 235

× 11 = −42128kNm

= 14 × A −4132) + 16 × 13 × A −42128) = −1455kNm  m

Mesnet (onatısının Hesabı ısa
= −118" kNm  m

6 Md 118" × 10 k m = = = 042 → k z = 0"80 bw × d 2 1000 × 1#0 2

A s =

6 Md = 146# × 10 = 3#4 mm 2  m k z ×  yd × d 0"80 × 1"1 × 1#0

1!e 4 ana kiri*te A s  mev = φ 12  36 + φ 12  110 5 = 314 + 10## = 13"1mm 2  m > A s  #er = 3#4

Dk erek -ok 2! e 3 ana kiri*te A s mev = φ 12  36 + φ 12  36 = 628 mm 2  m > A s  #er = 3#4

Dk erek -ok

C7n
−#66 kNm  m

6 Md #66 × 10 k m = = = 02# → k z = 0"88 bw × d 2 1000 × 1#0 2

6 Md = #66 × 10 = 23# mm 2  m k z ×  yd × d 0"88 × 1"1 × 1#0 A s mev = φ 12  40 + φ 12  40 = 565 mm 2  m > A s  #er = 23#

A s =

Dk erek -ok

onsol
= −1455kNm  m 6 Md 1455 × 10 k m = = = 045 → k z = 0"#" bw × d 2 1000 × 5#0 2

Md

1455 × 10 6 Md = = 1366mm 2  m A s = k z ×  yd × d 0"#" × 1"1 × 620 A s mev

= φ 12  36 = 314mm 2  m

A s  ek = A s − A s  mev = 1366 − 314 = 1052 mm 2  m

eçilen ek
φ 12  10 5

(a'ıtma
φ 12  20

ANA KİRİŞ HESABI %rü birim a'ırlı'ının @esabı &s+alt ka%lama EEEEEEEEEEEEEE:

"25 × 1 × 01 × 20 = 185kN  m

,a%EEEEEEEEEEEEEEEEEE!: "25 × 1 × 004 × 22 = 814kN  m (*emeEEEEEEEEEEEEEEEE!!!: "25 × 1 × 02 × 25 = 4625kN  m Me-il beton EEEEEEEEEEEEE!:

 

"25 × 1 ×

006"5 2

× 22 = #08kN  m

 02 + 06  × 1"0 × 25 = #03#5kN  m    2  

$retarlarEE: 2 (145 × 04) + ( 045 × 015) + 

&na kiri*lerEEEEEEEEEE!!!: ( 2 × 06 × 140 + 2 × 06 × 1) × 25 = #2kN  m Dnleme kiri*lerEEEEEEEEEE!: ( 3 × 185 × 06 × 04 × 25) 44 = #5#kN  m orklklar EEEEEEEEEEEEEEEEEEEEEE!: = 4kN  m / $o%lam # = 233"15kN  m &na kiri*lerin atalet momentlerinin @esabı (ı* ana kiri* atalet momenti

A = 6 × 16 + "25 × 2 +

( 2 + 6) 2

× 1" = 1"05dm 2

  2 1 4   6 × 16 × 16  2 + 1" × 2 × A16 − ) + 1" × 4 × ×  16 − 2 −  + "25 × 2 × 15   2 2  3    = 1105dm y 0 = 1"05

 6 × 16 3   2 3 × "25 2    ( ) + 6 × 16 × ( 8 − 1105 ) 2  + + × × − − "  25 2 16 11  05 1   12  12      3   4 3 × 1" 4 × 1"  +  2 × 1" + 2 × 1" × (16 − 1105 − 1) 2  +  + × (16 − 1105 − 2) 2  = 3"8"45dm 4   2  12   12 

& "1 = 

9ç ana kiri* atalet momenti

A = 6 × 12 + 2 × 2 × "25

y 0 =

= 10"dm 2 ( 6 × 12 × 12  2 + 2 × 2 × "25 × 11)

= 8#4dm

10"

 6 × 12 3   2 3 × "25 2    + 6 × 12 × ( 2#4 ) 2  + × + × × − − = 1505dm 4 2 "  25 2 ( 12 8  #4 1 )     12   12 

& " 2 = 

Köprü açılık ortası mom!tlr"!"! #sa$ı %

$ 1

=

"340 × "40

M # =

1880

# × l 2 8

=

= 46"

233"15 × 1880 2 8

$ 2 = $ 3 =

( "40 − 425 ) × 46"

= 102#"4# kNm

"40

= 2564

M HS

= 120 × 46" + 120 × 2564 + 30 × 2564 = "4#4kNm

M # =

q × l 2 8

=

435 × 1880 2 8

= 1"116 kNm

(inamik katsa-ısının @esabı ϕ = 1 +

15 1880 + 3#

= 126" ≤ 13

Yük eksantriklik katsa-ısının @esabı

& $ 1 = 3"8"45dm 4 & $ 2 = 1505dm 4 & 2 = & 3 = & & 1 = & 4 =

3"8" 45 & = 265 & 1505

  & i ∑  ∆i = 1 + × e × ρ i  2  ∑ & i × ρ i   

∑ &

i

= ( 2 × 265 + 2 × 1) × & = #3&

∑ & × ρ i

i

2

= ( 265 × 36#52 ) + (1×12252 ) + (( − 1225) 2 ×1 + ( 265 × ( − 36#5) 2 = #458&

1! ana kiri* için ρ 1 = 36#5 m

#30 & ∆1 =  × e × 36#5  1 +  = 1 + 0360 ⋅ e  #458 &  2! ana kiri* için ρ 1 = 25m

 #30 &  ∆ 2 = 1 + × e × 1225  = 1 + 0112 ⋅ e  #458 &  Dleri*si7 kam-on *erit sa-ısının blnması e kam-on -üklerinelen a?ıklık ortası momentlerinin ıtılması : 1 ,erit
e1 = 4625 − 15 = 3125 m & M i = M HS × i × i × ∆ i & i

∑

     K i Folsu!

K 11

= 1×

265 #3

× (1 + 0360 × 3125) = 0##14

2 ,erit
e 2 = 4625 − 3 = 1625m K 12

= 2 × 265 × (1 + 0360 × 1625) = 1151 #3

3 ,erit
e3 K 13

= 3×

265 #3

= 4625 − 45 = 0125m

× (1 + 0360 × 0125) = 1138

K 11 = 0##14 K 12 = 1151

  K = K = 1151  maG 12 K 13 = 1138 × 0" = 10242 M i

= M HS × K i × ϕ = "4#4 × 1151× 126" = 1383#"kN ⋅ m = M 12

 × (1 + 0112 × 3125) = 0184"  #3  1 K 22 = 2 × × (1 + 0112 × 1625) = 03238  K maG = K 23 = 03#51 #3  1 × (1 + 0112 × 0125) × 0" = 03#51 K 23 = 3 × #3  K 21 = 1 ×

1

M i

= M HS × K i × ϕ = "4#4 × 03#51 × 126" = 450"6kN ⋅ m = M 23

El&r"'s"( trt)&ar sa*ısı!ı! $)l)!ması % 1 $retar
145 = 535 m 2 265 K 11F = 1 × × (1 + 0360 × 535) = 10622  K F #3  maG 265 F K 12 = 2 × × (1 + 0360 × 0) = 0#261   #3

e1′ = 4625 +

= K 11F = 10622

M 11F = M qtr × K 11 = 1"116 × 10622 = 20305kN ⋅ m 2 $retar
e2′ = 0

× (1 + 0112 × 535) = 021"1  K F = K F = 02#3" #3  maG 22 265 F  = × × + × = (1 0112 0) 02#3" K 22 2  #3 M 22F = M q × K 22 = 1"116 × 02#3" = 523#kN ⋅ m

F = 1× K 21

265

tr

+at" *ükt! ,ola*ı a!a k"r"'lr, ol)'a! açıklık ortası mom!tlr" % e # = 0 265 ⋅ & × 1 × 102#"4# = 3#315"kNm M 1 ( # ) = #3 ⋅ & =

M 2 ( # )

& #3 ⋅ &

× 1 × 102#"4# = 14082kNm

So!)ç ms!t mom!tlr"!"! ta'ıma -ü.ü! -ör soprpo("s*o!) %

∑ M = 14 × M i

i#

+ 16 × ( M iHS + M iq tr )

∑ M ∑ M

= 14 × 3#315" + 16 × (1383#" + 20305) = ##631#kN ⋅ m 2 = 14 × 14082 + 16 × ( 450"6 + 523# ) = 2##681kN ⋅ m

1

A!a k"r"'lr, ol)'a! maks"m)m ms!t raks"*o!) %

$ 1 = 1 ' #

=

' HS

' tr

$ 2

=

(1880 − 425) 1880

= 0##4

233"15 × 1880

= 21"2"5kN 2 = 120 × 1 + 120 × 0##4 + 30 × 054# = 22"2"kNm

=

435 × 1880 2

= 40#"kN

Ms!t raks"*o!ları!ı! a!a k"r"'lr ,a-ıtılması %

$ 3

=

(1880 − 2 × 425) 1880

= 054#

' i

= ' × K i = 21"2"5 × 265 × 1 = #"608kN

' 1 ( # ) ' 2 ( # )

= 21"2"5 ×

#3 1

#3

× 1 = 30041kN

= ' × K i × ϕ ' 1 ( HS ) = 22" 2" × 1151 × 126" = 334"0 kN ' 2 ( HS ) = 22"2" × 03#5 × 126" = 10"15kN ' i = ' × K i ' 1 ( q tr ) = 40#" × 10622 = 4333kN ' 2 ( q tr ) = 40#" × 02#34 = 1115kN So!)ç ms!t raks"*o!ları!ı! ta'ıma -ü.ü! -ör soprpo("s*o!) % ∑' i = 14 × ' i# + 16 × (' iHS + ' iqtr ) ' i

∑ ' = 14 × #"608 + 16 × ( 334" + 4333) = 1#1"68kN ∑ ' = 14 × 30041 + 16 × (10"15 + 1115) = 613054kN 1

2

A!a k"r"' Do!atı Hsa$ı % K!ar a!a k"r"' ,o!atı #sa$ı %

Mal7eme B20420 Md

<  160  15  145 ?m .  1"0/60/"253425 ?m a = h  kablü -a%ılır

= 10240# kNm

My = 085 ×  cd × b × h  × d − 05h  My = 085 × 133 × 3425 × 20 × (145 − 05 × 20 ) My = 1006#10 kN ⋅ cm ≥ Md = ##681#kN ⋅ cm My ≥ Md ol<'nen @esabı -a%ıla?ak M ##681# × 10 4 k m = = = 115# ⇒ k z = 0"24 bw × d 2 3425 × 1450 2 A s

=

##681# × 10

4

0"24 × 1450 × 365

= 15#5"mm 2

A s min = 08

 ctd 104 × b × d = 08 × × 600 × 1450 = 1"84 mm 2 365  yd

eçilen
I
olarak seçilir

esme
= 1#1"68kN ' cr = 065 ×  ctd × bw × d = 065 × 0104 × 60 × 145 = 568kN ' c = 08 × ' cr = 08 × 568 = 454kN

= ' d − ' c = 1#1"68 − 454 = 126568kN Ki+t tesirli etri-e kllanılırsa A φ 14 ) ' s

A sw = 2 × 154 = 308mm 2 ' × s 126568 × 10 3 × s A sw = s ⇒ 308 = ⇒ s = 12435mm 365 × 1450  yd × d d s < = #25mm s = 124 mm 2

s < bw = 600 mm

s > 100 mm

İç A!a K"r"' ,o!atı #sa$ı

s = 120 mm seçil
Mal7eme B20420

<  120  10  110 ?m .  "25/60/"25245 ?m a = h  kablü -a%ılır

Md = 2##681kN ⋅ m

My = 085 ×  cd × b × h  × d − 05h  My = 085 × 133 × 245 × 20 × (100 − 05 × 10) My = 4"8550kN ⋅ cm > Md = 2##681kN ⋅ cm 4 2##681 × 10 Md = = 113 → k z = 0"3# k m = bw × d 2 2450 × 1100 2 Md 2##681 × 10 4 A s = = = 8120mm 2  m k z ×  yd × d 0"3# × 365 × 1100

 ctd × b × d = 08 × 104 × 600 × 1100 = 1505mm 2 365  yd eçilen
A s min = 08

I
olarak seçilir

esme
' s

= 613054 kN ' cr = 065 ×  ctd × bw × d = 065 × 0104 × 60 × 110 = 4056kN ' c = 08 × ' cr = 08 × 4056 = 325kN = ' d − ' c = 613054 − 325 = 288054kN

Ki+t tesirli etri-e kllanılırsa A φ 14 ) A sw = 2 × 154 = 308mm 2 ' s × s 288054 × 10 3 × s ⇒ 308 = ⇒ s = 3"0mm A sw = 365 × 1100  yd × d d s < = 500mm s = 3"0 mm 2

s < bw = 600 mm

s > 100 mm

s = 200 mm seçil
E!lm k"r"' #sa$ı %

1 =  h 80 4  %K !*++,%-G 4F de! ⇒ µ = 22828 bw 40 2  = = b 180 " 

h 

=

20

& e!leme

=

& iç  a!!kir !

b⋅h

µ

3

=

16 × 8 3 22828

= 403#2dm 4 = 00404m 4

= 01505 m 4 λ = 245m

$ekerlek -ükten
"1 = 0 "2 = 18m M 0A " ) = β = 4

% 4 ⋅ β

"3

= 3m " 4 = 48m

"5

= 12m

⋅ e −β " ( ?os β " − sin β " )

48 ⋅ % ⋅ . a!iç K A 48 ⋅ % ⋅ . a!iç b ⋅ K → K a!!kir ! = ⇒ K = = 3 ⇒ λ ⋅ b 4 ⋅ % ⋅ . % -3 - ⋅ λ ⋅ b

"6

= 3m

β = 4

b ⋅ 48 ⋅ % ⋅ . A! +ç 12 ⋅ . A! +ç b ⋅ K 4 =4 = 3 4 ⋅ % ⋅ . % 4 ⋅ % ⋅ . % ⋅ - ⋅ λ ⋅ b λ ⋅ -3 ⋅ . % β = 4

12 × . A! +ç

λ × -

3

× . %

F = 60kN

1 =

"

M ( "1 )

0⇒

=4

M ( " )

=

12 × 01505 245 × 1880 3 × 00404

=

F

= 022"3#(1  m )

× e − β " ( ?os β " − sin β " )

4 × β

60

× e 0 ( ?os 0 − sin 0) = 653"65kN ⋅ m

4 × 022"3#

" 2 = 18m ⇒ M ( " 2 )

=

60 4 × 0 22"3#

"3 = 3m ⇒

× e −0 22"3#×18 ( ?os( 022"3# × 18) − sin( 022"3# × 18) ) = 12413# kN ⋅ m

M ( "3 )

=

60 4 × 022"3#

× e −0 22"3#×3 ( ?os( 022"3# × 3) − sin ( 022"3# × 3) ) = 45144kN ⋅ m

"4 = 48m ⇒ M ( "4 )

=

60 4 × 022"3#

× e −0 22"3#×48 ( ?os( 022"3# × 48) − sin ( 022"3# × 48) ) = 624#5kN ⋅ m

" 5 = 12 m ⇒ M ( "5 )

=

60 4 × 022"3#

" 6 = 3m ⇒

∑ M

.S

× e −0 22"3#×1 2 ( ?os( 022"3# × 12 ) − sin ( 022"3# × 12 ) ) = 486314kN ⋅ m

M ( "6 )

=

60 4 × 022"3#

× e −0 22"3#×3 ( ?os( 022"3# × 3) − sin ( 022"3# × 3) ) = 45144kN ⋅ m

= 653"65 + 12413# + 45144 + 624#5 + 48631 + 45144 = 15"3"##kN ⋅ m

Zati -ükten
"1 = 0 " 2 = 25m

"3

= 25m F 1 = 08 × 04 × 225 × 25 = 1025kN

F 2

1 =

"

M ( "1 )

0⇒

=

2535 4 × 022"3#

2  185 2  λ  6531   = ⋅   = ×  = #55kN b ⋅ λ  2  43 × 185  4   F = F 1 + 2 × F 2 = 1025 + 2 × #55 = 2535kN

G

× e 0 ( ?os 0 − sin 0) = 168"41kN ⋅ m

" 2 = 245m ⇒ M ( " 2 )

=

2535 4 × 022"3#

× e −0 22"3#×2 45 ( ?os( 022"3# × 245) − sin ( 022"3# × 245) ) = 30181kN ⋅ m

∑ M

o A # )

∑ M

o

= 168"41 + 2 × 30181 = 22"303kN ⋅ m

= 14 × M o# + 16 × ϕ × M o HS = 14 × 22"303 + 16 × 126" × 15"3"## = 355#4kN ⋅ m

Dnleme
<  80  4  #6 ?m

6 Md 355#4 × 10 k m = = = 153" → k z = 0"2# 2 bw × d 2 400 × #60 Md 355#4 × 10 64 A s = = = 13834mm 2  m k z ×  yd × d 0"2# × 365 × #60  104 A s min = 08 ctd × b × d = 08 × × 400 × #60 = "33mm 2 365  yd

eçilen 
d 2

φ 12

25

= 380mm

s > 100 mm

KENAR A/AK HESABI

s = 200 mm

φ 14

20

seçil
∆h =

08m

h = 15 + 45 + 5 25 + 160 = 1285m t = 45 + 15 = 6m b = 2  + c = 2 × 1 45 + "25 = 1215m

= ' a# + / × ' aHS + 2 × ' a tr = 21"2"5 + 3 × 22"2" + 2 × 40#" = 2#3311kN F 0 = 003 × ' A = 003 × 2#3311 = 82 kN Arlol mesnet sürtünme keri) ' A

G1

= 1215 × 3 × 1135 × 18 = #446#4kN

G2 G3

G4

= 1215 × 15 × "35 × 25 = 426010kN = 1215 × 65 × 15 × 25 = 2"615#kN = 1215 × 15 × 45 × 18 = 14#626kN

∑G

i

= #446#4 + 426010 + 2"615# + 14#626 = 161446# kN

∑' = ∑ G i

i

+ ' A =161446# + 2#3311 = 188###8kN

F 0T = 05 × 188###8 = "4388" kN

enar a-a'a etki-en akti+ e %asi+ to%rak itkilerinin blnması e a1 = λ a × γ t × ∆h = 033 × 18 × 08 = 480 kN  m 2 e a1 = λ a × γ t × A h + ∆h) = 033 × 18 × A1285 + 08) = 81"0 kN  m 2

= ea1 × h × d = 480 × 1285 × 1215 = #4"5kN h × d 1285 × 1215 = A81" − 48) × = 6018#2kN % a 2 = ( ea 2 − ea1 ) × % a1

2

2

e p = λ p × γ t × t = 3 × 18 × 6 = 324 kN  m 2

K!ar a*ak ta#k"klr" % D&r"lm ta#k"k" % kuvvet V(A) G1 G2 G3 G4 !

#1

Şiddeti (kN) 2733,11 7446,46 4260,1 2961,57 1476,26 11809,8

moment kolu (m) 2,25 4,50 2,25 3 0,75 6"3

M(kNm) 6149,50 33509,07 9585,225 8884,71 1107,2 23619,6

30687,56

to!l#m

82855,305

749,5

12,85"2

4815,54

% p = e p ×

t × d 2

= 324 × 6 × 1215 = 1180" 8kN 2

#2

%$6018,72 82

12,85"3 11,25

25780,184 922,5

6850,22

to!l#m

31518,224

L noktasına >re
∑ M ∑ M

1ire!e!

82855305

=

31518224

1evire!

= 2628 > 15 ol<'nüenli

Ka*ma m!"*t" ta#k"k" % #1 #2

%$&iddeti 749,5 6018,72 82

to!l#m

6850,22

! $%t to!l#m

11809,8 9438,89 21248,69

s K =

∑' ∑'

1ire!e!

=

1evire!

212486" 685022

= 3101 > 15 ol<'nüenli

+m"! -r"lm ta#k"k" % A = 6 × 1215

kuvvet V(A) G2 G4 $% #1 #2

= #2"m

&iddeti 2733,11 4260,1 1476,26 82 749,5 6018,72

2

moment kolu 0,75 0,75 2,25 11,25 12,85"2 12,85"3

2 =

62

× 1215 6

M(kNm) 2049,84 3195,08 3321,59 922,5 4815,54 25780,19

= #2"m 3

! G1

11809,8 7446,8

'6"3 '1,5 to!l#m

'23619,6 '11170,1 5295,03

σ 1

=

∑' + ∑ M = 188###8 + 52"503 = 33158kN  m

σ 2

=

∑' + ∑ M = 188###8 − 52"503 = 18633kN  m

σ 0

σ − σ   33158 − 18633  = 258"5kN  m 2 < σ = 400kN  m 3 = σ 2 +   1 2  = 18633 +   z em 2    2 

σ 1

=

A

A

2

2

#2"

#2"

#2"

2

#2"

2

#2"

∑' + ∑ M = 188###8 + 52"503 = 33158kN  m A

2

#2"

2

< 13 × σ z em = 13 × 400 = 520kN  m 3

$emel bo-tları 7emin emni-et >erilmeleri tara+ınn<r!

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