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SIN 325: Concrete Design Lecture Notes

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These are some of the SIN 325 handwritten notes I made during my time at the University of Pretoria. These notes cover examples, explanations and additional research. They are written on either the given lecture slides or in a separate notebook. Have a look at the bundle deals if you are planning on buying more than one set of notes, and please do not hesitate to contact me over the app. Happy studying students!

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Uploaded on
May 30, 2021
Number of pages
182
Written in
2019/2020
Type
Class notes
Professor(s)
Prof e kearsley
Contains
1,2,4,6

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JKEARNEY
NOTES




SIN 325

,Chapter
-
1 :
-
Properties Load
of Reinforced Concrete .
(
pg
7)




{Compression
A




lreinfoqe.me#I- 1¥ N#%
µ µ #
Tension .




Section strain
Et




whenwehavecracks.i s
:÷÷÷÷÷t÷i¥.¥÷÷÷:÷÷÷¥¥ :
the steel .




whenwehavemas.si#ecrack:fsYotcoenhe8wpfaFfE?
FIFI F Easy:#age
*



concrete starts
to crush .




Concrete :
( 8)
T pg



!
f ↳o£
EE



SA cube
use
strength ( 28 water
150mm cubes)
day
cured
strength
-




USA use
cylinders ( 150mm diameter ,
300mm


high)
Euro code use both Cuckoo has a
of
cylindercube
strength 40M Pa
of
or a
strength 50 MPa .

, -ypicagth (
Pg 'd
fcy Cylinder
-




(
{tubeegth

#
Houses
20125 .

for .
-




( 25/30 structures .



Strength
( 30/37




tus is:/:÷÷:÷÷:÷÷:¥÷÷
Cassius >
Class
cylinder kubeato.se)
cube
i.e
{ff}
Coolio →
E- value 220+0.2 fcu strength



( 60/75 .
( 70/85

C.
80/95
60/105 fcu


fcu Cube
strength
-




Fck
cylinder strength ( OR
fcy)
-

, cylinder
cube




tension /
strength using
* Note : Use cube when

compression steel
.




Use
cylinder strength
when
working
with shear .

,Steel (
pg " )
mild steel ( 250hPa ) R
SA use
fy
-

-




( Y
high yield
steel
fy 450hPa )
-




Euro code 2
fy 500hPa
-
use

O




Ki
Estee ,
=
200 G- Pa

fy - -




;



l
E

Ey
Bardia : ( Q)

8MM
steel
10mm High yield
12mm
f
16mm
20mm 10412 @ 200
+200mm
spacing
25mm
32 MM
I tzmm diameter .




40mm 10 bars

We in the tables ( last page)
get areas
* can

,Compositettction
concrete steel

strength in tension Poor Good

compressive strength Good will
buckle
fair Good
shear
strength
corrodes
Durability Good

Fire resistance . Good Poor .




Use concrete in compression & steel
in tension & shear Cover steel
concrete
.




in to it
protect .

,Chapter :


Li¥g cpgzos
Limit state is condition at which a
a

structure becomes unfit for its intended
USC .




Ultimate Limit State ( ULS) :
collapse .




Servis Limit state :(SLS)
ability
deflection ,
cracking durability
,
.




Limits use
ps t factors .




Von material
safety factor
-



.




safety factor
tf load
-


.




D.es#adeffect- Design resistance .




If Q n f
tfkm
where Qin -



nominal load
fr characteristic material
strength
-



.




Materfeto ( pg 3 formula
book )
ULS concrete : 1.5 steel : 1.15
SLS concrete : i. O steel : i. O

, Design strength fume
-
-




( Table 4)
Pat#fetyfactofoato( tf )
ULS 1.2dL ti - 6hL ( DL =
Gu &
LL Qu )
-

-




OR 0.9dL → for load reversal .




SLS : t.IDLtl.CL

value of action
Design Vfx characteristic
- -




tie action .




( permanent & Variable Action)

Dead Load & Live Load .




Loads
own
weight a
Imposed
I I
sometimes
Always .




1. ZDLTI . 6hL .
I ZDLTI .
6hL


IT IT TET
I # IT
Max
hogging .





⇐ XX •





Bending
Max

Bending
Max

,Note : we have
hargetxampke.x.DE 25k NIM
,
LL
-

-




IOKNIM
a
D E
few load cases



\
Afpc
that will produce f f f f f f f Hogging
on:iI¥imwEm moment
for FA FB
design .
Fc
CASI:
207 KNM Bumpy
o• ↳
loading
Full




#
.







116,4kNM




WULF 1.2×25+1.6×10
46 KNIM =




{ Fy : Fat FBTFE 46W -





{ MEO : 12 FA t
6 FB
=
12 Wx 6


FB =
12W
-

2 FA -





°


Hfc) =
FA -


wat FB Gc -


a>


MGC)
-
-



SVGc) doc =
FA -


x -




wyd
t FBGC -


6> + A

MCO)
-

-

O ÷ A


§,=M£ :EIf#=FAxzI w.gs#tFBGc-z65tB
-

o

, 3


ZI
"


FBxG
EIV =
FAX t + But C
wz 2g
-




EI (o) v
-
-


O o : c
-
-

o
EI VC6)
RAx6£ ¥634 G B O
= - =




% B =
9W -



GRA -30


EI VCR)
FARE -wz{y FB
# B
=
t t 12
x



E .
O =
24 FA -


72W t 3. FB t B -






③ in ④
0=24 FA 72W t 3 FB t 9W GFA
-
-




=
12 FA -



27W


I. RA =
103,5 kN

In ③ B =
9146 ) -



6/103 5) ,
=
207

M ( 6) 103,5 (6) 207 KNM
4621672
= - = -




Find zero moment : M (x ) =
0=103 5K -


46×2
,


2-
#




f. 3C
4,5M
=




Findmaxlminmomentoo UGO -

0=103,5 -


46k


2,25M where the maximum
sagging
o : so -


moment is .




M ( 2,25) =

103,512,25 ) t 4612,255L =

116,4k NM .




2



we now finished
* Note
hogging
: .
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I am a civil engineering graduate from the University of Pretoria with an honours degree in Engineering and Technology Management. As I prepared for exams, I created "super summaries" and some handwritten notes of each course. I make handwritten notes of every lecture I attend and hope that they can be used to improve your understanding of the course material. If there are any issues or questions regarding the notes please contact me over the app. Happy studying !

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