UNIVERZITET U TUZLI RGGF PREDMET: BETONSKE KONSTRUKCIJE I
TREĆI PROGRAMSKI ZADATAK: ZADATAK: DIMENZIONIRANJE NA SMICANJE I TORZIJU. POKRIVANJE DIJAGRAMA SILA ZATEZANJA I SILA SMICANJA. GRANIČNO STANJE UPOTREBLJIVOSTI.
Uradio: M!i" Ma#$r T%&a' ().)*(+
SADR,AJ 1. An Anvel velope ope dijag dijagram rama a unutarn unutarnjih jih sila( sila(ULS ULS))
(.(. Di-o%i/i!a 0o-a1a (.). A02$&oa di!a3ra#a #o#$0a4a -a2i!a0!a (.5. A02$&oa di!a3ra#a 4ra0-2$r%a&0i6 -i&a (.7. A02$&oa di!a3ra#a 0or#a&0i6 -i&a (.+. A02$&oa di!a3ra#a #o#$0a4a 4or%i!$ 2. Dimenz Dimenzion ionira iranje nje grede grede na na savij savijanj anjee
).(. Di#$0%io0ira0!$ 3r$d$ 8&!$94$0! A; ).). Di#$0%io0ira0!$ 3r$d$ o&! ).5. Di#$0%io0ira0!$ 3r$d$ r$-!$8 5<5 =&i!$2o od -4>a? ).7. Ko04ro&a 4$@i94a ar#a4r$ 3. Dimenzion Dimenzioniranj iranjee grede grede na popre poprečne čne sile i torziju torziju
5.(. Di#$0%io0ira0!$ 3r$d$ 8&!$94$0! A; 5.). Di#$0%io0ira0!$ 3r$d$ r$-!$/i#a 0a 5# i +'+# od 8&!$94$0!a 5.5. Di#$0%io0ira0!$ 3r$d$ o&! 5.7. Di#$0%io0ira0!$ 3r$d$ r$-!$8 5<5 =&i!$2o od -4>a? . Dime Dimenz nzio ioni nira ranj njee stu stu!a !a
7.(. Di#$0%io0ira0!$ -4>a ra2/ o-$ )<) 7.). Di#$0%io0ira0!$ -4>a ra2/ o-$ (<( ". #o$riv #o$rivanj anjee dijagra dijagrama ma zate% zate%u&i u&ih h sila ' račun račun . ontrola ontrola graničnog graničnog stanja stanja upotre upotre!ljiv !ljivosti osti gred gredee (SLS) (SLS) *. #o$rivanj #o$rivanjee armaturom armaturom dijagra dijagrama ma sila zatezan zatezanja ja i dijagrama dijagrama sila sila smi+anja smi+anja ,. #lan armature armature $ompletnog $ompletnog rama i spe+i-i$ spe+i-i$a+ija a+ija armature armature
1. Anvelope dijagrama unutarnjih sila
(.(. Di-o%i/i!a 0o-a1a
(.). A02$&oa di!a3ra#a #o#$0a4a -a2i!a0!a
(.5. A02$&oa di!a3ra#a 4ra0-2$r%a&0i6 -i&a
(.7. A02$&oa di!a3ra#a 0or#a&0i6 -i&a
(.+. A02$&oa di!a3ra#a #o#$0a4a 4or%i!$
2. Dimenzioniranje grede na savijanje
).(. Di#$0%io0ira0!$ 3r$d$ 8&!$94$0! A;
aterijal/
BetonC 30 / 37 : f ck =30 MPa ;f cd =
30 =20 MPa 1,5
Armatura B 500: f yk =500 MPa ; f yd =
500 = 435 MPa 1,15
Statič$i uti+aji/
M SD=−2086,25 kNm
N SD =−207,59 kN
U o2o# -&1a! o!a- !$ %a4$304' r$-!$8 radi#o 8ao ra2o3ao0i
RD , lim ¿ ∙ b w ∙ h2 ∙ f cd RD , lim ¿=μ ¿ M ¿
z lim ¿ ∙ f + yd
RD, lim ¿=740,88 kNm M ¿
RD, lim ¿ ∆ M = M SD ,S − M ¿
N ∆ M − SD =29,94 + 50,72− 4,77 =75,89 cm2 ( d −d 2 ) ∙ f yd f yd M RD, lim ¿ ¿ A S 1=¿
A S 1 =80,05 cm2 ; A S 2=49,26 cm2 ⇒ A S,! =129,31 cm2
∆ M =1389,48 kNm
U-2a!a0!$# o4r$>0$ ar#a4r$ do>i!a -$ i%ra%i4o 2$&i8 ro/$0a4 ar#ira0!a. Ta8o$r a8o %#$#o o>%ir i%ra%i4o 2$&i8$ i-4o2r$#$0$ 4i/a!$ or$10i6 -i&a i 4or%i!$ o2o# r$-!$8' %a8&!1!$#o da 4r$>a -2o!i4i 2$"$ di#$0%i!$ or$10o3 r$-!$8a 8o!$ "$ >i4i o3od0i!$ %a di#$0%io0ira0!$ S&i!$di o2$"a0!$ r$-!$8a:
RD, lim ¿ ∙ b w ∙ d 2 ∙ f cd RD , lim ¿= μ¿ M ¿
z lim ¿ ∙ f + yd
RD, lim ¿=1134 kNm M ¿
RD , lim ¿ ∆ M = M SD − M ¿
∆ M =1006,22 kNm
N ∆ M − SD =42,77 + 33,06 − 4,77=71,06 cm2 ( d −d 2 ) ∙ f yd f yd M RD , lim ¿
¿ A S 1 =¿
A S 1 =73,63 cm2 ; A S 2=34,36 cm2 ⇒ A S,! =107,99 cm2 < A S ,ma" =176 cm2 2,45 < 4
#$%o&eno : A S 1 15 ϕ 25 ; A S 2 7 ϕ 25
).). Di#$0%io0ira0!$ 3r$d$ o&! aterijal/
BetonC 30 / 37 : f ck =30 MPa ;f cd =
30 =20 MPa 1,5
Armatura B 500: f yk =500 MPa ; f yd =
500 = 435 MPa 1,15
Statič$i uti+aji/
M SD=1055,22 kNm
N SD =−207,59 kN
S&1a! (. Pro2!$ra da &i N.O. -i!$1$ &a09
μSD =
1055,22 + 207,59 ∙ 0,4386 2
1,2 ∙ 0,75 ∙ 20
=0,085 ⇒ ' $=10 ( ; ' c =2 ( ; " =0,167 ∙ 75 =12,53 cm
)to zna*+da N ! $+&e*e -.o*u+ -re$&ek rad+mo kao -ra%ou/aon++ ¿ &edno$truko arm+ran0 M N 1146,27 ∙ 100 − SD =37,47 − 4,77=32,7 cm2 A S 1 = SD ,S = z ∙ f yd 0,938 ∙ 75 ∙ 43,48 f yd
#$%o&eno : A S 1 7 ϕ 25 A S ,! =34,36 cm2
).5. Di#$0%io0ira0!$ 3r$d$ r$-!$8 5<5 =&i!$2o od -4>a? aterijal/
BetonC 30 / 37 : f ck =30 MPa ;f cd =
30 =20 MPa 1,5
Armatura B 500: f yk =500 MPa ; f yd =
500 = 435 MPa 1,15
Statič$i uti+aji/
M SD=−1341,55 kNm
N SD =−207,59 kN
RD, lim ¿ ∙ b w ∙ d 2 ∙ f cd RD , lim ¿= μ¿ M ¿
z lim ¿ ∙ f + yd
RD, lim ¿=1134 kNm M ¿
RD , lim ¿ ∆ M = M SD − M ¿
∆ M =261,52 kNm
N ∆ M − SD =42,77 + 8,59− 4,77= 46,59 cm2 ( d −d 2 ) ∙ f yd f yd M RD ,lim ¿
¿
A S 1=¿
A S 1 =49,09 cm2 ; A S 2=9,82 cm2 ⇒ A S ,! =58,91 cm2 < A S,ma" =176 cm2 ('57 < 4 #$%o&eno : A S 1 10 ϕ 25 ; A S 2 2 ϕ 25
).7. Ko04ro&a 4$@i94a ar#a4r$
d 1 2=
∑ a $ ∙ d + = 6 ∙ 4,91 ∙ 4,25+2 ∙ 4,91∙ 9,25 =5,5 cm 1
A $ 1
39,28
2ado%o.&a%a -ret-o$ta%.&enate3+4taarmature 0
3. Dimenzioniranje grede na poprečne sile i torziju
5.(. Di#$0%io0ira0!$ 3r$d$ 8&!$94$0! A; Statič$i uti+aji/
5 SD =860,48 kN SD=143,1 kNm
5 RD 1 =[ 6 RD ∙ k ∙ ( 1.2 + 40 ∙ 7. ) + 0.15 ∙ 8 c- ] ∙ bw ∙ d Z>o3 i-4o2r$#$0o3 d!$&o2a0!a #o#$04a 4or%i!$ ⇒
5 RD 1 =0
⇒ o4r$>0a or$10a ar#a4ra
da r$%#$ /!$&o80 -i&.
5 RD 2 =
9 ∙ f cd ∙ z ∙ b w 0,55 ∙ 20 ∙ 67,5 ∙ 40 = =1485 kN cot : + tan : 2
5 SD =0,56 > 0,2 5 RD 2
⇒ $w,ma" ¿ 5* /#
5 RD 2 >5 SD
$w , %¿
A $w ∙ f yd ∙ m ∙ z 5 D
m= 4 ϕ 10
$ w , % ¿ 11,03 cm
⇒
Djelovanje torzije:
RD 1=
2 ∙ 9 1 ∙ f cd ∙t k ∙ A k cot : + tan :
< SD
A 40 ∙ 80 =13,33 cm t w = w = u w 2 ∙ 40 + 2 ∙ 80 1 3
= n= ∙ ( b+ ∙t +3 )
t f =
A f = 80 ∙ 15 =6,31 cm u f 2 ∙ 15 +∙ 2 ∙ 80
1 3
= w = ∙ [ 40 ∙ 13,333 ∙ 2+ ( 80−2 ∙ 13,33 ) ∙ 13,333 ∙ 2 ]
1 3
= f = ∙ [ 15 ∙ 6,313 ∙ 2+ ( 80−2 ∙ 6,31 ) ∙ 6,313 ∙ 2 ]
= w =147389,54 cm4
= f =13798,1 cm4
#d+o -o&ed+n+h d+&e.o%au no$+%o$t+natorz+&u &e : R>BR! 91,44 ;?@AN A 8,56 A kw =2653,47 cm2 1
RD 1=
2∙ 9
A kf =640,37 cm2
∙ f cd ∙ ( t kf ∙ A kf + t kw ∙ A kw ) 2 ∙ 0,385 ∙ 20 ∙ ( 6,31 ∙ 640,37 + 13,33∙ 2653,47) −3 = ∙ 10 cot : + tan : 2
RD 1=303,47 kNm
Po%r4+na -re$&eka -o-re*ne armature :
$ w ,t ¿ (5'7 /# $w ¿
$ w ,% ∙ $ w, t =6,14 cm $ w ,% + $w , t
A $w, =
SD,w ∙ $w , ϕ 10 2 ∙ A kw ∙ f yd ∙ cot : ⇒
Usvojena poprečna armatura za prva 3 metra raspona:
R>BR! Rϕ 10 / 5 cm
?@AN)A Rϕ 8 / 5 cm
Po%r4+nauzdu3ne armature za moment torz+&e : A $. =
SD ∙u k 2 ∙ A k ∙ f yd
ukw =2 ∙ ( 80 −13,33 )+ 2 ∙ ( 40 −13,33 ) A $. =
ukw =186,68 cm
130,79 ∙ 186,68 =10,58 cm2 2 ∙ 2653,47 ∙ 435
#$%o&eno : 8 ϕ 14
#z -ret-o$ta%ku -r+t+$nut+hd+&a/ona.a: = 45 +$ra*unat+9 =0,385 $.+&ed+dokaz no$+%o$t+ d = h−5 =75 cm
%'+ /#
1
9 ∙ f cd ∙ z ∙ b w 0,385 ∙ 20 ∙ 67,5 ∙ 40 = =1039,5 kN 5 RD 2 1 = cot : + tan : 2
Dokazno$+%o$t+ -r+t+$nuto/rebra :
1,03 1 ma.o -rekora*en&e mo3emo
2
5 RD 2 1 > 5 SD
2
[ ][ ] SD
RD 1
+
5 SD
5 RD 2 1
<1
zanemar+t+
5.). Di#$0%io0ira0!$ 3r$d$ r$-!$/i#a 0a 5# i +'+# od 8&!$94$0!a
< 0a 5 #:
5 SD +*7'7( 8N $w , %¿
A $w ∙ f yd∙ m ∙ z 5 D
m= 2 ϕ 10 ⇒
$ w , % ¿ 9,19 cm
$ w ,t ¿ (5'7 /# $w ¿
$ w ,% ∙ $ w, t =5,4 cm $ w ,% + $w , t
U-2o!$0o: Rϕ 10 /5 cm
< 0a +'+ #:
5 SD ))7'(H 8N A $w ∙ f yd∙ m ∙ z 5 D
$w , %¿
m=2 ϕ 10 ⇒
$ w , % ¿ 20,6 cm
$ w ,t ¿ (5'7 /# $w ¿
$ w ,% ∙ $ w, t =8,08 cm $ w ,% + $w , t
U-2o!$0o: Rϕ 10 / 7,5 cm
5.5. Di#$0%io0ira0!$ 3r$d$ o&! Statič$i uti+aji/
5 SD =0 kN SD=23,85 kNm Potreban proračun armature samo na torziju. Poprečna armatura:
A $w, =
SD,w ∙ $w , ϕ 10 2 ∙ A kw ∙ f yd ∙ cot : ⇒
u $ w,ma" = k =¿ 23,33 cm 8 U-2o!$0o: Rϕ 10 / 20 cm
Podužna armatura:
A $. =
SD ∙u k (' cm2 2 ∙ A k ∙ f yd
$ w ,t ¿ '75 /#
#$%o&eno : 6 ϕ 12
5.7. Di#$0%io0ira0!$ 3r$d$ r$-!$8 5<5 =&i!$2o od -4>a? Statič$i uti+aji/
5 SD =734,33 kN SD=0 kNm
5 RD 1 =[ 6 RD ∙ k ∙ ( 1.2 + 40 ∙ 7. ) + 0.15 ∙ 8 c- ] ∙ bw ∙ d
[
5 RD 1 =
0,034 ∙ 1 ∙ ( 1,2 + 40 ∙ 0,015 ) + 0,15 ∙
k =1,6− d =0,85 k =1
]
207,59 4400
∙ 40 ∙ 75
Prora*un -otrebnearmature od ut+ca&a -o-re*ne$+.e : $w , %¿
A $w ∙ f yd ∙ m ∙ z H'+ /# 5 D
U-2o!$0o: Rϕ 10 / 7,5 cm
. Dimenzioniranje stu!a
aterijal/
BetonC 30 / 37 : f ck =30 MPa ;f cd =
30 =20 MPa 1,5
Armatura B 500: f yk =500 MPa ; f yd = Statič$i uti+aji/
500 = 435 MPa 1,15
7.= 0,015
5 RD 1 =204,75 kN
5 SD < 5 RD 3=5 RD 1+ 5 D
M SD=−988,07 kNm
N SD =−3763,88 kN
SD=0 kNm
l *
= β ⋅ l col = *' ⋅ +'* = 5'+ m
7.(.
Di#$0%io0ira0!$ -4>a ra2/ o-$ )<) l *
=
λ
i
=
5+* (7'75
e*( = e*'#i0
M Sd
e*)
=
$ *(
≤ $ *)
ν u
=
N Sd
=
=
N Sd f cd ⋅ Ac
(+
λ crit =
ν u
= )7')+ h
)*
=
+* )*
DHH*G 5GF5'HH
=
) )***
*'D7
cm
= )F')+ cm
− 5F5'HH
(+
=
= )'+
= *'D7
= (+'7G
λ λ/ri4 → o4r$>0a !$ 4$ori!a ). r$da
Prora10 #!$roda20o3 $8-/$04ri/i4$4a o4$r$"$0!a $ 4o4 : $ 4o4
= $* + $a + $)
< doda40i $8-/$04ri/i4$4 0a o-0o2 ro#!$0&!i2$ ra-od!$&$ #o#$04a: ee
= *'F ⋅ e*) + *'7 ⋅ e*( = *'F ⋅ )F')+ + *'7 ⋅ )'+ = (F'G+cm
< 0$@$&!$0i $8-/$04ri/i4$4: $a
= ν( ⋅
ν (
=
&* )
( (** ⋅ l col
=
( (** ⋅ +'*
=
( ))5'F
<
( )**
⇒
ν (
=
( )**
(
=
ea
)**
⋅
5+* )
= *'H+ cm
< $8-/$04ri/i4$4 r$#a 4$ori!i ). r$da:
= K ( ⋅
e)
l *)
( ⋅ (* r
za 5+ > λ > (+ K (
λ
= ( 2!" # !,$% = !,&'2%
( = ) ⋅ K ) ⋅ ε d r *'D ⋅ d ε d =
=
K )
C d ES
=
757'H )*****
N d − N Sd N d − N >a&
= *'**)(
≤(
=do94$0o !$ da -$ %#$ K ) ( ' -ro40o# o4r$>0a i4$ra/i!a? d = 5+ cm
( = ) ⋅ ⋅*'**)( = *'*(* ( m r *'D ⋅ *'7+ e)
= *'7F)+
etot
5'+ ) (*
⋅ *'*(* = *'**F( m = *'F(cm
= e* + ea + e) = (F'G+ + *'HG+ + *'F( = (H')5+ cm
M SD= 0,182 ∙ 3768,88=685,03 kNm
μSD =
M SD 2
b w ∙ h ∙ f cd
=0,343 9 SD =
N SD bw ∙ h ∙ f cd
N SD =−3763,88 kN
=−0,941
A S 1 + A S 2=C∙ b w ∙ d ∙
Sa d+&a/rama +nterakc+&e ⇒ C =0,5
A S! =46 cm2 < A S,ma"= 80 cm2
2
f cd = 46 cm2 f yd
A S! =46 cm > A S,m+n=12,98 cm
2
2,3 < 4
#$%o&eno : 10 ϕ 25
= β ⋅ l col = *' ⋅ +'* = 5'+ m
l *
7.).
Di#$0%io0ira0!$ -4>a ra2/ o-$ (<(
l *
=
λ
i
*( =
e
ν u
=
= e
5+* (('+7
*) =
N Sd f cd ⋅ Ac
(+
λ crit =
ν u
= 5*'55
*
=
=
− 5F5'HH ) )***
(+ *'D7
= *'D7
= (+'7G
λ λ/ri4 → o4r$>0a !$ 4$ori!a ). r$da
Prora10 #!$roda20o3 $8-/$04ri/i4$4a o4$r$"$0!a $ 4o4 : $ 4o4
= $* + $a + $)
< 0$@$&!$0i $8-/$04ri/i4$4: $a
= ν( ⋅
ν (
=
ea
=
&* )
( (** ⋅ l col ( )**
⋅
5+* )
=
( (** ⋅ +'*
=
( ))5'F
= *'H+ cm
< $8-/$04ri/i4$4 r$#a 4$ori!i ). r$da:
<
( )**
⇒
ν (
=
( )**
= K ( ⋅
e)
l *)
( ⋅ (* r
za 5+ > λ > (+
K (
λ
= ( 2!" # !,$% = !,$''%
( = ) ⋅ K ) ⋅ ε d r *'D ⋅ d ε d =
=
K )
C d ES
=
757'H )*****
N d − N Sd N d − N >a&
= *'**)(
≤(
=do94$0o !$ da -$ %#$ K ) ( ' -ro40o# o4r$>0a i4$ra/i!a? d = 5+ cm
( = ) ⋅ *'**)( = *'*(5H ( m r *'D ⋅ *'5+ e)
= *'FF+
etot
5'+ ) (*
⋅ *'*(5H = *'*()D m = ('5cm
= e* + ea + e) = * + *'HG+ + ('5 = )'(G+ cm
M SD= 0,0218 ∙ 3768,88= 82,05 kNm
μSD =
N SD =−3763,88 kN
M SD N SD = = =−0,941 9 0,051 SD bw ∙ h ∙ f cd b w ∙ h2 ∙ f cd
A S 1 + A S 2= ∙ b w ∙ d ∙
Sa d+&a/rama +nterakc+&e ⇒ C =0,12
f cd =11,04 cm2 Po-4o!$"a ar#a4ra: f yd
=%ado2o&!a2a?
Potrebna -o-re*na armatura :
4 ϕ 25
A St%
2 ('5 cm
ϕ 5
*')+ ϕ ')+ ##
u$%o&eno R ϕ H Razmak %+.+ca : $ =12 ϕ 5** ## #$%o&eno : Rϕ 8 / 25 cm
". #o$rivanje dijagrama zate%u&ih sila
Porivanje dija)rama u polju
-
*ila zatezanja:
M 1146,27 −207,59 =1490,6 kN Sd= Sd − N Sd = z 0,675 +orizontalni poma linije sile zatezanja (odreivanje večine pomaa al": z a. = ∙ ( cot: −cot ) 2
a. =
675 ∙ cot 45=33,75 cm 2
- usvojeno a. = 3% cm
*ila oju preuzima jedna ipa : (/s0,prov = &,10cm2"
Rd= f yd ∙ A $ 1, -ro%=43,48 ∙ 4,91= 213,43 kN
/nerisanje rajeva ipi armature ϕ f yd
. b= ∙ 4
f bd
= 25 ∙ 434,8 = 905,83 mm 4
. b,net = a ∙. b ∙
3,0
AS ,reE <. A S,-ro% b,m+n
inimalna dužina sidrenja
. b,m+n =0,3 ∙ . b < 10 ϕ < 100 mm . b,m+n =0,3 ∙ . b=0,3 ∙ 905,83=271,75 mm ; 10 ϕ =250 mm; 100 mm . b,net =860,2 mm<.b ,m+n=271,75 mm
Dužina sidrenja nad osloncem
-
/nerisanje rajeva ipi armature ϕ f yd
. b= ∙ 4
f bd
=
25 434,8 905,83 = =1294,04 mm ∙ 4 3,0 0,7
. b,net = a ∙. b ∙
AS ,reE <. A S,-ro% b,m+n
. b,net =1230 mm
-
Dužina sidrenja u uljetenju
. b,net =1250 mm
.
ontrola graničnog stanja upotre!ljivosti grede (SLS)
ombinacije optere4enja za )ranično stanje upotrebljivosti (*5*"
*6/578 8P699;9
8P699;9 7<9 P0
Č$-4a 8o#>i0a/i!a M 1$-4o
= ∑ ! M G' ! + ('( ⋅ M '( + ∑i>( )'i ⋅ M
M *e$to= 460,2 +0,50 ∙ 289,3 =0,605 MNm K2a%i<-4a&0a 8o#>i0a/i!a M 82a%i<-4a&0o =
∑ M +∑ 3' !
!
i ≥(
)'i ⋅ M O'i
M k%az+−$ta.no =4 60,2+ 0,30 ∙ 289,3 =0 , 547 MNm
B+** − − > f yd E s
=
f yk
γ s
= 75+ MPa
= )** GPa
C 5* P 5G − − > f cd
=
f ck
γ c
= )* MPa
= )'D MPaL f ctk '*.*+ = ) MPa E cm = 5('D GPa ) A s( = 57'5Fcm α s = F')G f ctm
N r
A + ( cm2 )
e + ( cm)
A + ∙e +
A + ∙e +2
= +
A c 1=15 ∙ 100 A c 2=35 ∙ 60 A $ 1=5,8 ∙ 41,
( H**
'+
(5 +**
(*( )+*
55 +*
) **
7'+
()5 +**
+ H )+*
(+ 7(' ∅
)(+'77
A $ 2=5,8 ∙ 2,23
7'+
( *+*')H
∅
('+
7')+
F
I id$a&.
)('
( (()'*+
(+5 5(('+
(7 5+H
∅
7 '*(
=∑
$ id$a&.
( (+ 7+'
A i ⋅ $i
∑A
7 ('
= (+55(('D+ = 5)'GHcm 7FGG'*(
i
) = ∑ I i + ∑ A i ⋅ $ i ) − $ id$a&. ⋅ ∑ A i = *'*GF(( # 7
I id$a&.
=
Qdo&$
6 − $ id$a&.
*'*F((
=
*'H − *'5)H
= *'(F() # 5
r,m= ctm>?dole=2,1>!,0'02 ≡!,&'$ 7m Da >i -$ do>i&a d$or#a/i!a ar#a4ri -&!$d #o#$04a Mr'# %a -4adi! I' o4r$>0o !$ odr$di4i #o#$04 o4ora r$-!$8a 0i2o 4$@i94a ar#a4r$. Qar#a4r$
=
I id$a& 6 − $ id$a&
− *'*+
=
*'*F(( *'H − *'5)H − *'*+
= *'(D(F # 5
Nao0 >$4o0 0i2o 4$@i94a ar#a4r$: R /'-r(
=
M r' #
= *'7FG = )'77 MNP# )
Qar#a4r$
*'(D(F)
D$or#a/i!a >$4o0a 0o2i 4$@i94a ar#a4r$ od d$!-42o# #o#$04a o42ara0!a r-&i0$: T -r'(
=
R /'-r( E/
=
)'77
5(D**
= *'****GF+ = *'*GF+S *
Na8o0 94o -$ o42or$ r-&i0$ -&i8a d$or#a/i!a -$ #i!$0!a. N$4ra&0a o- ro&a%i 8ro% 4$@i90i/ id$a&0o3 or$10o3 r$-!$8a.
V II
U ⋅ A = - >
V II
6 C = *'(+#
⋅ − ( +
(+
) ⋅ > ⋅ d U- ⋅ A-
⇒ ri#!$0!!
F')G ⋅ 57'5F ⋅ (* −7 = ⋅ − ( + (')
-$ i%ra%i %a 3r$d WWTWW r$-!$8a:
(+
) ⋅ (') ⋅ *'G)+ F')G ⋅ 57'5F ⋅ (* − 7
= *'(+7(#
V II
− 8 ) +
=
8 )
+ > X ⋅ ( ) ⋅ d ⋅ A $ + 6 C ⋅ 8 ( )
)
> X
+ > X ⋅ V II ) V II z II = d − 6 5 ⋅ > X ⋅ V II + 8 ( ⋅ ) − C V II 8 ( ⋅ 6 C ⋅ 5 − ) ⋅
6 C
#zu%oGen&eodno$a : ❑$= > $ / >$ ; A e =❑$ ∙ A $ ; k 1=hf ∙ ( b−b w ) ;k 2= A e + k 1 " == =0,1518 m 2 == = 0,6877 m Nao0 ar#a4ri 0a8o0 o42ara0!a r-&i0a !$: M r'#
=
R -r)
Z II ⋅ A -(
=
*'7FG *'FHGG ⋅ 57'5F ⋅ (* −7
= (DG'F5MNP# )
D$or#a/i!$ ar#a4ri 0a8o0 o42ara0!a r-&i0a !$: T -r)
=
R -r) E-
=
(D'F5 )*****
= *'***DHH(
D$or#a/i!$ ri r$&a-8 i% -4adi!a I II
❑r= $r 2−$ r 1= 0,9881 −0,0765=0,9116 D$or#a/i!$ ar#a4ri %a 1i-4i; -4adi! II D$or#a/i!a ar#a4ri -&!$d M82a%i<-4a&0o : R -)'O-
=
M 82a%i−-4a&0o
ε -)'O-
=
R -)
Z II ⋅ A -
E-
=
=
*'+7 *'FH ⋅ 57'5F ⋅ (* − 7
)5('7D )*****
= )5('7DMNP# )
= *'**((+
D$or#a/i!a ar#a4ri -&!$d M1$-4o: R -)'6
=
M 1$-4o Z II ⋅ A -
=
*'F*+ *'FH ⋅ 57'5F ⋅ (* − 7
= )+F'*7MNP# )
T -)'6
=
)+F'*7 )*****
= *'**()H
❑$ 2=❑$ 2 h=❑$ 2, E$ +❑$ 2, M ❑$ 2, M =1,28−1,157 =0,123 ( T -)'OT -)
=
('(+ (')H
= *'D*5D *
dio M82a%i<-4a&0o ε-): T -)'∆M T -)
=
*'()5 (')H
= *'*DF *
dio ∆M ε-):
❑$m=❑$ 2−❑t ∙ ❑r =1,28 −0,25 ∙ 0,9039 ∙ 0,9116 − 0,096 ∙ 0,25 ∙ 0,9116 ❑$m=0,0010 52
Sr$d0!$ ra-4o!a0!$ r-&i0a -r#
S rm=50 + 0,25 ∙ k 1 ∙ k 2 ∙ H /❑r A c,eff =2,5 ∙ ( h −d ) ∙ bw =550 cm2
Y r
=
57'5F ++*
= *'*F)
S rm=50 + 0,25 ∙ 0,8 ∙ 0,5 ∙
25 = 90,32 mm 0,062
k =1,7 ∙ 0,0010 52 ∙ 90,32 =0,161 I 0,16 mm Do8a% ro3i>a 3r$d$
@ranični pro)ib: lim ¿=
@ 250
=
960 =5,6 cm 250
9¿ 9 tot = k ∙ @2 ∙
1
r tot
k =∙ ❑$m ∙ S rm /r =0,3 mm − zanorma.ne AB kon$trukc+&e
k =
5 ∙ ( 1−0,1 ∙ J ) 48
Kri2i0a %a 0ao0-8o -4a0!$ I:
Msd
=
B I
= E c ⋅ I ideel
B I
=
*'7D+
ϑ r (
H*D*'7D
= *'****D)F
( m
Kri2i0a %a 0ao0-8o -4a0!$ II: ϑ r )
=
Msd B II
=
*'G7D+ )H*'57
= *'**)FG
(
B II
m
= E s ⋅ A s ⋅ z II ⋅ =d − x II ?
M 8 $r= cr = 197,63 MPa z ∙ A $ 1
Ko$i/i!$04 r$ra-od!$&$:
( )
8 $r 2 =1−1 ∙ 0,5 ∙ K =1− J 1 ∙ J 2 ∙ 8 $ 1
rm
(
197,63 251,04
)=
0,618
=K ∙ + ( 1−K ) ∙= 0,618 ∙ 0,00267 + ( 1−0,618 ) ∙ 0,0000 926=0,00168
Pro3i> -r$di0i ra-o0a: f = k ⋅ l eff ⋅ ϑ m )
k = *'(*7 ⋅ =( − *'( ⋅
M A − M B M C
? = *'(*7 ⋅ =( − *'( ⋅
k = *'*FD f = *'*FD ⋅ (7 ) ⋅ *'**(FH f = *'*)) m = )') cm < f dop
= +'F cm
D()' + F)+'5 7F*')
?
1
m
