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RCD WSD – Working Stress Design Notes, Study notes of Reinforced Concrete Design

Explains the principles of Working Stress Design (WSD) for reinforced concrete Includes stress limits, service load design, and factor of safety concepts With sample computations for beams and columns under WSD Ideal for mastering classical design approaches

Typology: Study notes

2023/2024

Available from 06/06/2025

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AX\ALLY
LOADED
RC
COLUMNS
ACI
/NSCP
Code
Requremems
)For
moin
lorgittia
Reinorcig
ars
0.08
(a)
o.o1
(b)
Pa
min.
n0.
OF
ES
six
(o)
For
circular
arangentert
FOUr
(4)
For
rectglar
ariargemt
three
(3)
For
triangular
arrBrgement
(c)
min.
clear
distance
betueen
jogitinal
bars
(Sc)
For
lateral Ties
concrete
cover
:
40
mn(
minimum
)
(a)
minimum
size
1.
Pms
L40
mm
(b)
maximum
size
Po
(c)
maximum
spaing
oF
barr
(eTC)
megnitude
oF
usoe
column
AxYl
Lad
Stmg
(a)
Tied
Column
JDmm0
12
rmg
Ieast
column
dimension
Pu
0.80
p o
8sft
(Ag-As)
+Asf91
(b)
Spiral
Column
p
O-05
;tso.1h
(perpendicular)
0.85fé(
A9-
As)
+
AsfY
b
pure
axial
capacity
Pu
0.52
o-esf
(Ag
AS)
+
Asfy
Pu
85
o85f
(Ag-As)
*Agy
|
smY)l
ecentricity
Pu
0-w375
0.85f
(Ag-As)
+AfYJ
i
circular
buj
hed
cpacing
:
1-
4.5
%
Cco
nomical
ucually
more
eiclen
(0)
steel
iF
So
50,
dect
case
p
ma
raho
(odd
or
e
ven
)
by
2 ,
by
+(quale)
honer
tnan
beaM
S
bar
oF
Cegiegation
(
honeycom)
9)
Spiralsodd
or
Gven)
(a)
min
size
IOmm
(b)
max
s1ze
lon
(c)
ciear
pocing
oF
pilas,
min
2Gnn
fdagg
t
sp
(a)
mnimum
spra
teel
rano
Prmin
0.45
(e)
min
spoBùng
(theo.
compar
) s)
4ASPDe
-
psp)
AS
Ag
2
De
fsmin
Area
oF
Suel
Gros
Area
oF
Column
gross
stcel
raio
AC
Area
oF
Core
,0.C
De
=
Core
Diameter
De
=
Dg
-
2(cc)
Asp
Cross- sectional
Arca
of
spral
ar
pf3
pf4
pf5
pf8
pf9
pfa

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AX\ALLY LOADED RC COLUMNS

ACI (^) /NSCP (^) Code (^) Requremems )For (^) moin (^) lorgittia (^) Reinorcig ars

(a) o.o1 0.

(b)

Pa

min. n0. OF ES six (^) (o) For (^) circular (^) arangentert FOUr (4) (^) For (^) rectglar (^) ariargemt three (^) (3) For (^) triangular (^) arrBrgement (c) min. (^) clear (^) distance betueen (^) jogitinal bars

(Sc)

For lateral Ties

concrete cover : 40 mn( minimum )

(a) minimum size

  1. Pms

L40 mm

(b) maximum size

Po

(c) (^) maximum spaing oF (^) barr (^) (eTC)

megnitude oF usoe column AxYl Lad Stmg

(a) Tied Column

JDmm

12 rmg

Ieast column dimension

Pu 0.80 p o8sft (Ag-As) +Asf

(b) Spiral Column

p O-05 ;tso.1h (perpendicular)

0.85fé( A9- As) + AsfY

b pure axial capacity

Pu 0.52 o-esf (Ag AS) + Asfy

Pu 85 o85f (Ag-As) *Agy |

smY)l ecentricity

Pu 0-w375 0.85f (Ag-As)+AfYJ

i circular buj hed cpacing :

1- 4.5 % Cco nomical

ucually more

eiclen (0)

steel

iF So 50, dect case pma

raho

(odd or even )

by 2 , by +(quale)

honer tnan beaM S bar oF Cegiegation (honeycom)

9) Spiralsodd or Gven)

(a) min size IOmm

(b) max s1ze lon

(c) ciear pocing oF pilas,

min (^) 2Gnn (^) fdagg t (^) sp

(a) mnimum spra teel rano

Prmin 0.

(e) min spoBùng (theo. compar ) s) 4ASPDe - psp)

AS Ag

2 De fsmin ’ Area oF Suel Gros Area oF Column gross stcel raio

AC Area oF Core

,0.C

De = Core Diameter

De = Dg - 2(cc)

Asp Cross- sectional Arca of spral ar

PROBLEM 1. Design the Q P 9o KN. Ute fC

lomn laterol tier

pg Ag

AS

AS

445

Pg A

A : 02 Ag

445

S 48 OLT

syuare hcd colum to camYy seryicC axial load

  1. MPa, fy 414 MP, T 219GMP•, Po > 2

leat dim.

Pudes 2D +-6L

:. S 400 mm

2000 xI Ag

Pudes : 2000 KN

Puder Pucap Pu

Pu

= 12(J00)^ +).o/800)

s Ag sAg

0-80 o-85t^ (Ag-^ As)^ +^ Afy^ 0-b 6-80(0-b5)0-8s (21)(A9 - 0-02A9) + (0-02A9) (414)|

AS 0-02Ag

0.02 (445)(445) Ac =3900.5^ mm^2

3900-

T (25)

n25 8.

minimum spacing oF Lateral Ties

l(25)

use |2 -25mm p

raurd up by 5 (design , exan)

S 440. 450mm Say 445 mm

4

S- 2CC 20Lt nme

okay!

445- 2(4o)- 2(1o)

Sc 8).0Mm

Should be ymmeti^ cal^ (by^ 4)^4 peride

1-5(25) >^ 37.5^ /

40

Detas

+(25)

4 45

445

Lyuse

C-

Po lotN 25mm d ma

445 x445 mm |2-25mn man Bar

use l0mn LateralTies

1@ SOmm, rest 40Dmn

194,00|. 4744 mm 2 A

R cOLUMN^ VBUECTED^ To^ AXAL^

BENDING MOMENT

Mp ML

ML

PE

Frame A (FA)

P ME PE

"Eceyicitby, e

F(ame 1 (F-1)

Benavior OF Column

f balanced

MOF-A

rension cotiecl Compresion contraled

mF-A

mudes

Mudes @x-•xis

Pu des

MO F-

ME F-

" Mudes @y-oxis

Col. int oF

grids A 1

col. @int oF

gids A

Pudes >(>F-A t Pp4) 102d combination

Column

M =Pe

Bending Moment

AbDUt y- xis

Kung ano mag-gogoven sa

Po

Bendne amen t

About Xaxis

DL+ LL + EL per rame Y

A-

DF-A (^) pop-t

cno0sSe tne max or greatest bton Fame A Reme 1 (bigget value)

gogamitin sa load com bination

N.A.

Actual e < ebal

Balanced Condition (fs *f9) Compression Controls

Tension Contro)S

PIastic Centroid

fe 20. mpa

fy= 44 Mpa

Bi 0.

4- 28mnd

4- 28mm

350

similar o centric

(rectangular, regular shape)

PROBLEM 1. Determin the nomnal axial capcity , Pn ,

@balanced cordition.

T- Ce -Cs t Pn

Pn =^ 0-8sft^ ab

Pn

AS -

yF# >0^ (@^ FOrce^ Diagram)

d-d'

Pa epal

(ExIo) tbal +

G00(drc)

a

e< eal

G00(530-c)

a

bal 452.^448 mm

e

Asfy - 0-95feab^ -^ Asfy^ +^ Pn^ o

C 313.ol mm cl

(@ Force magiam)

co.

: 0"85^ (20)^ A(3s0)^ y^

D-

530- 7O

eb•l

2a. So8mm A

784T

0 85{

Force Dig am

As fy

T ASfy

As fr

ASfY

)- cs (d-d') - ce (a-)

Pn

Ce C1 0.85fcab

Cr :C

Ccor Ci

fi

ASfy or A[fS

ASOT f9)

and ecenttricity, e,

(@ center of a

As (ps orf9)

balanced eebal

As fY As >AS

aoo(c -d)

Goo(D-0)

  • Asfy (d-d')^ -^ 08sfiab(d-)^ o

Comp. steel yiedr

  • (A847)(414)(50-0)^ -^ 085(20-*)D()(530^ -

PROBLEM 3. Same / poolem 2 t Coctua So0mm

AS : 24-03Ol mm 2 AS =

fy

p

  1. ol mm 2 20¬ MPa 414 MPa

ebal 452. 44B nn D)

Pn -Cs - Ce +T

Pn - Asfy -0-85ft qb +

Compreucion Contros

Pn = loi9%86.|4 + 5234.5125C

Afe

Asfy

a(0-)

C = 302. 122 m c

Pn - z409.01( **) - o8s(o1) (0.85c) (30) +2443-0|I

Pn 2230- I|9 kN

Bc

Pn (tac t) -cs ( drd') - Co (d-)

(30 t

d530mn

B (530-)

E(530-)

d' 70mn

fn (eaet

n o00mm

6co(

1477800 (530 rc)

) -(820-7o)^ - (Dc{^50 -^ )^ -o

+)-A[fy (d-d') - o.8sfiab (d-) o

INTERACTION DIAGRAM

PROBLEM 1. using^ the^ iteraction^ diagram^ ,^

determine tne^ value^ oF^ Pn^ Fr^ te^ dort

tied column^ Shoan^ pr:

(a) ecemcity^ along^ yraxis (0) eccenicity^ along^ yaxis

fc 20. MPa fy =^ 43.^70 MPa d =534.5 mm

b 3 5Oom

(o)

(o)

02.5 mn

Pg

X 0.792 A

Po

O. o.

490

As Ag

AS

200 (

495

475

X

(1949- 25)(2)

SEES

(356) (900)

0-

(847.25)(2 ider) (3s0)(o)

200 mn

interpolYto

e44.2s

X 0-

X 2-83 MPa

X -0-

0-90- 0.

Ag

0-05 PO

3D (uco)

20- 3

X

Pn

AQ

o 48

35D(0)

X

plot g(let)^ (ngn)

he value on the

X axjs

X-0. o-55- 0-4gksi

Pn 1 484.503 kN

X = 3.45 NPa p

20-7 MPa 3 kui

Po - 2U34. 206 KN

SLENDER OR^ LONG^ COLUMNS

Conditons @wich^ sleerness^ cFrects^ may^

be neylectad^ (NCP^ 2o5)

por columns^ braced^

agsint sideay

Klu

  • por^ columns^ rot^ braced^ 8ganst

Where

KLu 2 22

Por

K erFective^ lengtn^ Factor

KLu3 eFFeCtive length

Lu unsupported^ jengtn^ oF^ the^

column

Magni ied Moment i(Mo)

Mo M

1

34 4 12

: radus^ oF^ gyration

" Case^ I.^ Frames Cm

rectargular ciraul ar

(ET) eFF -

Pu 0.75 Per

+ Magnipled mament

Cm 0.b0^ 0.^ o

(Kiu)*

pin fo pin

Mi

1.2D Mp Mu

Larger OF^ the^ nd^ moments

Moment magnification Factor

0.40E, Ig

pn to ixed

()

Braced AgainstSderway

Pu ultmate Facorod Loat

Per Citiçal Buckling Load

K 1.

and

cidesoay

K 0.

r0.30h

efrecive value OF EI

k 05

klu

Single Curvatue m - (egative)

Me M2ns

Double cufvoure

" case I. Frames not Braced Aqaist SidenaN

(a) Ss

() Ss=

(o) Ss

m2ns (DL) larger erd Fatorecd

moment (o sway)

(positive)

m2s (E^ tW)^ larer^ end^ Factored

perpercicular to tre

axir of bendirg

t s M2s

2Pu Ao Vule

Pu

O.75fer

yPU 0 15 Pcr

Le CTcs

Vu Stoey cnear

Ao 1t order

rlative dertection

2På = Um of all ctual FoTes of •ll e

columns in tne ctorey aralye

2Per = um o all ciical loac oF tne

Columns ln tne storey analyzd

nc 518. 083 KN-m

D (9%)

S D 2949

4040

0.15Pcr

Pu

Cm

1

Pr: 81 49. 90oKNC

(Ku)e

(ET)erf

Po 1900 kN

fy 400 MP•

:1 (fixed)

2

dergn,

(EJ )err: 29.727,293.^58 xlo

o.40 (470Bo)

0.40EcIg

j9% 2(90)

A0-

Mu

1-2mD

AIGNMENTCHART

(EJ)err

W-(single curvoture)

  • |.2(1900) t.o(Wb0)

" )+-2(90 1-6(00)

(pinned)

pd

Cm 1

pd

Pu 4D4 kN

Pu -2Pp +1.oPL

theCalculate magnpiecdmonent

1.PROBLEMhe column ishna memeroFa Frame bracec gaint y.

z(4)

: joint rigidity

theoreical,

JAC kroN MORE LAND