Home
Add Document
Sign In
Register
07 - Blok Dijagram
Home
07 - Blok Dijagram
GRF, betonFull description...
Author:
Igor
18 downloads
256 Views
455KB Size
Report
DOWNLOAD .PDF
Recommend Documents
Smithov dijagram
Smithov dijagram
.
Fe C Dijagram
fgvb
06 Klasni dijagram
Full description
Fe-Fe3C dijagram
Fe-Fe3C dijagram sa objasnjenjima i linijom
dijagram zdruzene sadnje
Bodeov dijagram - Zadaci(1).pdf
Bodeov dijagram -
UML Dijagram klasa
Full description
Moody Jev Dijagram
Moodyev dijagram
08. DTP Dijagram Toka Podataka
Dijagram Toka PodatakaFull description
Blok Ukur Dan Blok Sudut
metroFull description
Blok Ukur Dan Blok Sudut
Blok Ukur Dan Blok SudutDeskripsi lengkap
Blok Ukur dan Blok Sudut
Blok Ukur Dan Blok Sudut
Blok Ukur Dan Blok SudutFull description
blok 1.4
pbl
DIAGRAM BLOK
Deskripsi lengkap
07 -
Rotational kinematics and mechanics - freshman physicsDescripción completa
07
7Descrição completa
07
diseñoDescripción completa
07
PerforadoraDescrição completa
07
Descrição completa
07
07
Full description
Diagram Blok
DIAGRAM BLOKFull description
1
Preseci Pre seci nepravilno nepravil no g o blika pritisnute zone betona b(y)
T
x
Gb
M u d
σb = α×f B
εb
Abp
y d x 8 . 0 x y
σb(y) y
T
x
Dbu
NEUTRALNA LINIJA
h
b
z
x h
Aa1 1
1
a
Z au au
a
εa1
b1
∑ M
a1
= 0 :
Dbu × z b
= M au ⇒
s
1
a
εb
(y)
y d
x
y
σb εb(y) y
y = x
Dbu
= ∫ σ b ( y ) b( y ) dy y =0
σb =
f B 4
× (4 − ε b ) × ε b σb = f B
2
0 ≤ ε b
≤ 2 ‰ 2 ‰ ≤ ε b ≤ 3.5 ‰
σb(y) Dbu
3
x
Gb
σb = α×f B
εb
b(y)
x 8 . 0
y d
Abp
y
x 2 . 0 y = x
Dbu
= ∫ σb ( y ) b( y ) dy y =0
Dbu
= σb × Abp = α × f B × Abp
T
x
Dbu
4
Dimenzion isanje trape trapeznog znog p reseka
Gb
Abp
Dbu NEUTRALNA LINIJA
b
h
T
x
x 8 . x 0
M u d
σb = α×f B
εb
2
(0.8x)
b
z
x h
Aa1 1
1
a
1
a
b1
∑ M = 0 : D ∑ N = 0 : a1
Z au au a
εa1
× z b = M au = M u + N u × (y b1 − a1 ) ⇒ Dbu − N u Dbu − Z au = N u ⇒ Aa1 = σv
bu
s
5
Prora n t a l o m a k ru r u n o g p re r e s ek ek a u n m o m e nt Aap
ε b
a
x 8 . 0 x
Abp M u N u
r r 2 2 = = D a D
σq Dbu
x 8 . 0
x 1
ϕa ϕb
a
σb = α×f B
a NEUTRALNA LINIJA SISTEMNA LINIJA
x h
r
r a
a
a
εa,max.
Aaz
∑ N = 0 : ∑ M = 0 :
Dbu
σv
+ Dau − Z au = N u ⇒
Dbu × a1 + Dau × a2 + Z au × a3
s
= M u
Z au au
Dau 2
a
3
a
Prora n t a l o m a k ru r u n o g p re r e s ek ek a u n m o m e nt ε b σb = α×f B Abp
x 8 . x 0
ϕb
r 2 = D
Dbu
NEUTRALNA LINIJA
r
Dbu
x 8 . 0
6
ϕb
r − 0 .8 x = arc cos r
2 sin ϕb 2 = α × f B × Abp = 0 .95 × f B × r × ϕb − 2
Prora n t a l o m a k ru r u n o g p re r e s ek ek a u n m o m e nt 5 . 4
URØ8/20 1 4 0 = 5 a D
12RØ19 A a = 34.02 cm 2 ( 1 2 RØ R Ø1 9 ) D a = 50 – 2×4.5 = 41 c m
5 . 4
aa
=
Aa π × Da
=
34.02 π × 0 .41
cm 2 = 26 .41 m
7
8
Prora n t a l o m a k ru r u n o g p re r e s ek ek a u n m o m e nt Aap εb σq a x
ϕa
a
r 2 =
x – r
x
N.L.
a
D
Dau
x - Z au au h
r
r a
a
a
Aaz
Dau = Aap × σ q
εa,max.
= aa × ϕa × Da × σq
Z au = Aaz × σv = aa × (π − ϕa ) × Da × σv
σv
ϕa
r − x = arc cos r a
×
Report "07 - Blok Dijagram"
Your name
Email
Reason
-Select Reason-
Pornographic
Defamatory
Illegal/Unlawful
Spam
Other Terms Of Service Violation
File a copyright complaint
Description
×
Sign In
Email
Password
Remember me
Forgot password?
Sign In
Our partners will collect data and use cookies for ad personalization and measurement.
Learn how we and our ad partner Google, collect and use data
.
Agree & close