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SIN 325: Concrete Design Lecture Notes R145,00
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Class notes

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 planni...

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  • May 30, 2021
  • 182
  • 2019/2020
  • Class notes
  • Prof e kearsley
  • 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 -




⑦
{ MEO : 12 FA t
6 FB
=
12 Wx 6


FB =
12W
-

2 FA -




②
°


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 -




④

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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