9-mid term revision part _1

-Midterm Revision part (1)

Reinforced Concrete

Third Civil Year

)9(

3rd Civivil year Reinfo

orced Con

p

Revisi

Midt

ncrete - R

page No(-

sion pa

term 2

Revision

(- 1 -)

art (1)

2013

n part (1)

)-Notes NNo (9)

3rd Civil year Reinforced Concrete - Revision part (1)-Notes No (9)

page No(- 2 -)

)Interaction diagram ( : 1- or ) ( 2- . . ) ( 3-(e/t) or (e/b) ( ( ) ( -4

. )i.D ( :

Mu =Mur + Madd. Mu : design moment (from i.D) Mur : ultimate moment of resistance Madd. : Additional moment due to buckling

et = e load + buckling et : from (i.D) e load :

buckling :


page No(- 3 -)

Problem (1) Given:

-Pu=150 t & He= 5.0m & braced col.

fcu=250 kg/cm2 & fy=3600 kg/cm2

Required:

Find its ultimate moment resistance about minor axes?

Solution :

.

As= 12 18= 30.54 cm2

& As= *b*t 30.54= *35*60 get =0.0145

= *fcu * 10^-5 0.0145 = *250*10^-5

get =5.81 6.0

/ :

-fy=3600 kg/cm2 - =0.90 (shaker page 428)

-uniform steel distribution From chart using =6.0 & . . = ^ =0.286 get ( =0.095 ) about minor axes.


page No(- 4 -)

About minor:

- =0.095 = ^ ^ get Mu =12.80t.m Check column buckling about minor axes:

- = = .. =14.28 ( b 15 braced)short column Myadd. =0.0

)Mur( : Mur =Mu- Myadd. =12.80-0.0=12.80 t.m

)About major axes: (

( =0.095 = ^ ^ ) get Mu =29.93t.m Short column Mxadd. =0.0 Mur =Mu- Mxadd. =29.93-0.0=29.93t.m

Pr

A

ge

-Fr

ge

3rd Civ

roblem

As= 1

& As

= *

et =8

-fy=3- =0

-uniforrom cha

et (

=

ivil year

m (2)

10 22

s= *b

*fcu *

8.44

3600 kg0.90 rm steeart usin

==0.08

Reinfo

2= 38.0

b*t

* 10^-5

9.0

kg/cm2 el distring

=0.08 )

=

orced Con

pa

S

0 cm2

38.0 =

5 0.0

0

ribution=9.0 &

)

^ ^

ncrete - R

page No(-

Solutio

= *30

021=

:

n & . ab

Revision

(- 5 -)

on

0*60

*250*

/

. =bout m

get M

n part (1)

ge

*10^-5

(shake

^ major a

Mu =21

)-Notes N

et =

5

er page

=0.5

axes.

1.60 t.m

No (9)

=0.021

e 428)

55

m

e

et for

P

A-

A

A

3rd Civ

et (total e

= e loadr short

et = e

Problem

- Colum

As= 1

As=

= *

ivil year

eccentricity

d + but colum

load =0

m (3)

- 2

umn is

12 19

*b*t

*fcu * 1

Reinfo

y) =

uckling mn : bu0.086 m

s short:

= 34.0

34.0

10^-5

orced Con

pa

= .uckling =

m =8.6

S- 1 :

t:

02 cm2

02 =

0.016

ncrete - R

page No(-

=0.08=0.0

60 cm

Solutio 100

2

*35*6

16= *

Revision

(- 6 -)

086 m

on

60

*225*1

n part (1)

get

*10^-5

)-Notes N

et =0

get

No (9)

0.016

=7.

2


page No(- 7 -)

/ :


-uniform steel distribution From chart using =7.20 & . . = ^ =0.211 get ( =0.12 ) about minor axes. =0.12 = ^ ^

get Mu =19.85 t.m

et (total eccentricity) = = . =0.199 m et = e load + buckling for short column : buckling =0.0

et = e load =0.199 m =19.90 cm B- He = 8.0m about minor axes:

- = = .. =22.86 (10 < b 23 UN braced) long col. b = = . . = 0.091m et = e load + buckling

0.199= e load +0.091 e load =0.108 m =10.80 cm

Pr

1-

-fr

- A

* f

Pu

3rd Civ

roblem

- For M

Colum

from col

Ac = 4

for tied

u = 0.35

=0.35

= 292

ivil year

m (4)

Mu=0.0

mn subj

lumn c

40* 60

d colum

35 fcu . A

5 * 250

2 ton

Reinfo

0 &

bjected

cross se

= 240

mn :

Ac +0.

50 * 24

orced Con

p

S

No

d to (Pu

ection:

00 cm2

.67 As.

00 + 0

ncrete - R

page No(-

Solutio

buckl

Pu)only

2 - A

. fy

0.67 *3

Revision

(- 8 -)

on

ling (

y.

As= 12

34.02*

292

n part (1)

Madd

2 19

*3600=

)-Notes N

dd=0.0

= 34.0

= 2920

No (9)

0)

02cm2

056 kg

mi

2

kg

-

inor


page No(- 9 -)

2-Required (pu & Mu) about minor (un braced):

-ey=6.0cm -H0=3.0m - Case (1) & Case (1) out plane

Check column buckling

About minor: = . for un braced column (Case(1)top ; Case (1)bottom) get K from table (6-10) =1.20

= . .. =9.0 ( b 10 un braced) short col. b = 0.0 ey (total eccentricity) = ey load + bbuckling =0.06m

= .. = 0.15 As= 12 19 = 34.02cm2

& As= *b*t 34.02= *40*60 get =0.0141

= *fcu * 10^-5 0.0141 = *250*10^-5

get =5.67 6.0

/ :


-uniform steel distribution

From chart using =6.0 = 0.15 get


page No(- 10 -)

( =0.06 & . . =0.40 ) =0.06 = ^ ^ get Mu =14.40t.m

)Mur( : Mur =Mu- Myadd. =14.40 - 0.0=14.40 t.m

. . =0.40= ^ get Pu=240 t

Pr

Gi- U-seRe1-

2-

3rd Civ

roblem

iven: Un bracec. (35

Required largesmome

- largebuckli

As=

& As

= *

ivil year

m (5)

ced col5*60) d : st eccenent aboest ecceling len

10 19

s= *b

*fcu *

Reinfo

lumn - As=

ntricityout min

centricitnthg a

9= 28.

b*t

* 10^-5

orced Con

pa

= 10

y : if shinor ( 15ity : ifand subj

S

.35 cm

28.35

5 0.0

ncrete - R

age No(-

19

hort col15 & 10if sebject to

Solutio

m2

5 = *3

01 =

Revision

- 11 -)

lumn s10) t.m elendr

o ultimon

*30*60

*250*

n part (1)

subject t at top colum

mate load

0 g

*10^-5

)-Notes N

to initp and bmn witad 150 t

get =

5 get

No (9)

itial bottom ?th 7.0m ton?

=0.01

t =6.

? 0m

6.3


page No(- 12 -)

Case (1) : short column subject to initial moment about minor ( 15 & 10) t.m at top and bottom :

for short column : Madd.=0.0 Mi= My2 = 15.0 t.m Mu = Mi+ Madd. =15.0 +0.0=15.0 t.m

/ :

-fy=3600 kg/cm2 - =0.90 (shaker page 428) -uniform steel distribution

from (i.D) using = . ^ ^ =0.08 & =6.3 get . . =0.36 = ^ Pu =189 t. et (total eccentricity) = = . =0.079 m et = e load + buckling for short column : buckling =0.0

et = e load =0.079 m =7.90 cm )i.D ( )et/b( :

from (i.D) using = . ^ ^ =0.08 & =6.3 get et/b = 0.22 et=0.22*35=7.70 cm

.


page No(- 13 -)

Case (2) : selendr column with 7.0m buckling lenthg and subject to ultimate load 150 ton:

from (i.D) using ^ = 0.28 & =6.3 get =0.11 = ^ ^ = Mu = 20.21 t.m et (total eccentricity) = = . =0.134 m Or from (i.D) get (et/b) =0.36 et= 0.36 *35 =12.60cm for slender column :

= = .. =20.0(10< t 23 un braced)long col. b = = . . = 0.07m et = e load + buckling 0.134 = e load +0.07 e load = 0.064 m = 6.4 cm


page No(- 14 -)

Problem (6):

for the shown column given that:

-fcu=250 kg/cm2 & fy=3600 kg/cm2

-in plane: column is cantilever & H0=4.0m

-out plane: column is (fixed - fixed) & H0=2.50m

find its ultimate load resistance?

Solution

As= 10 19= 28.35 cm2

& As= *b*t 28.35 = *30*60 get =0.01

= *fcu * 10^-5 0.01 = *250*10^-5 get =6.3

2- Check buckling:

: )un braced( . In plane: H0=4.0m Fixed (case 1) bottom. K: UN braced column K from table (6-10) =2.20

Free (case 4) top. ((

= . = . .. =14.6(10< t 23 un braced)long col.


page No(- 15 -)

t = = . . = 0.064m Mx add. = Pu* t = 0. 064Pu

Out plane:

H0=2.50m

Fixed (case 1) bottom. K: UN braced column K from table (6-10) =1.20

Fixed (case 1) top.

= . = . .. =10.0( b 10 un braced) short col.

UN braced column

-Mx = Mxinitial + Mxadd.= 0.0 + 0. 064Pu = 0. 064Pu t.m

-get: e (eccentricity)= = . = 0.064 m -get: = . . = 0.106


page No(- 16 -)

/ :



from chart using ( =6.30 & =0.106)

get . . =0.45 = ^ Pu =202.50 t.


page No(- 17 -)

Problem (7): Design an un braced column with (30*75)cm cross section ; with (fixed - hinged)end conditions ;the clear height is 5.50 m ;to carry ultimate load 211 ton; the top and bottom eccentricities are 10 cm and 5 cm at the same side about the major axes. fcu = 250 kg/cm2 & fy=3600 kg/cm2

Solution Given: - Un braced column -sec. (30 *75) - fixed - hinged -Single curvature - H0 =5.50m - pu = 211 ton - extop = 10.0 cm - exbottom = 5.0 cm

)Pu*e. ( Mx2 (top) = pu *extop = 211*0.10=21.10 t.m. Mx1 (bottom) = pu *exbottom = 211*0.05=10.55 t.m.


page No(- 18 -)

1- Check buckling:

About major: = . for un braced column (case(1)top ; case (3)bottom) get K from table (6-10) =1.60

= . .. =11.73 (10< b 23 un braced) long col. t = = . . =0.051m Mx add. = Pu* t = 211*0.051=10.88t.m

About minor: = . for un braced column (case(1)top ; case (3)bottom) get K from table (6-10) =1.60

= . .. =29.33 ( b > 23 un braced) un safe increase column dimension. take = 23 and get (b)

23.0= . . b= 0.38 m take b =0.40 m = . .. =22.0 (10< b 23 un braced) long col. b = = . . =0.097m My add. = Pu* b = 211*0.097=20.42t.m


page No(- 19 -)

UN braced column

1-Mx = Mx2 + Mxadd.= 21.10 + 10.88 = 31.98 t.m

2-My = My2 + Myadd.= 0.0 + 20.42 = 20.42 t.m

Convert bi axial moment to uni axial moment

t=t -2*cover = 75.0-5.0=65 cm

b=b -2*cover = 40.0-5.0=35 cm

= . =0. 49 & = . =0.58 >

From table (6-12-)using( . . = ^ =0.28)get =0.78 Myfinal=My+*Mx( )=20.42+0.78*31.98( )=33.85 t.m


page No(- 20 -)

2- Zone of design:

-get: =0.28> 0.04 (not zone D) -get: e (eccentricity)= = . = 0.16 m -get: = .. = 0.40

0.05 < < 0.50 (zone B design using i.D).

3- Design section of column:

/ :


get: = . ^ ^ =0.11 from( i.D page 428)using =0.11 & =0.28 get =7

get: = *fcu * 10^-5

= 7*250 *10^-5 = 0.0175

As= *b*t

= 0.0175*40*75 = 52. 50 cm2


page No(- 21 -)

check Asmin. ( long column):

Asmin=( . . .)*b*t= . . . )*40*75= 41.82 cm2 Asreq. >Asmin. use As = 52.50 cm2 =20 20

) 4( : .


page No(- 22 -)

Problem (8): Design an un braced column as shown for the following cases: 1- No buckling occurs. 2- Minimum dimensions. Given that: fcu=250 kg/cm2 & fy=3600 kg/cm2 - Case (2) & case (1) in plane - Case (1) & case (1) out plane - H0 =5.00m - pu = 200 ton

Solution Case (1) No buckling occurs: Means (short column) ( & ) 10.0 (un braced column). 1- get column dimensions:

About major: = . for un braced column (case(1)top ; case (2)bottom) get K from table (6-10) =1.30

10.0 = . . t = 0.65m =65cm About minor: = . for un braced column (case(1)top ; case (1)bottom) get K from table (6-10) =1.20


page No(- 23 -)

10.0 = . . b = 0.60m =60cm. )pu only.(

Design column for (Pu only) tied column:

Pu = 0.35 fcu . Ac +0.67 As. fy ............... eq. (1)

& Pu =200 t & Ac = (60 *65)cm2 get As

200 *10 ^3 = 0.35* 300 * (60 *65) + 0.67 *As * 3600

get As = -59.0 cm2 (-Ve)

take As = Asmin (short column)=0.006*b*t

= 0.006*65*60= 23.40 cm2 = 12 16


page No(- 24 -)

Case (2) minimum dimensions:

Means ( & ) =23.0 (un braced column). 1- get column dimensions:

= . 23.0 = . . t = 0.35m =35cm t = = . =0.093m Mx add. = Pu* t = 200*0.093=18.50 t.m

= . 23.0 = . . b = 0.30m =30cm. b = = . . =0.079m My add. = Pu* b = 200*0.079=15.87t.m

.

Convert bi axial moment to uni axial moment

t=t -2*cover = 35.0-5.0=30 cm

b=b -2*cover = 30.0-5.0=25 cm


page No(- 25 -)

= . =0. 62 & = . =0.64 >

from table(6-12-)using( . . = ^ =0.76)get = 0.60 Myfinal=My+*Mx( )=15.87+0.60*18.50( )=25.12 t.m 2- Zone of design:

-get: =0.76> 0.04 (not zone D) -get: e (eccentricity)= = . = 0.13 m -get: = .. = 0.42

0.05 < < 0.50 (zone B design using i.D).

3- Design section of column:

/ :


get: = . ^ ^ =0.32


page No(- 26 -)

from( i.D page 428)using =0.32 & =0.76 get =out of curve

.

PrGi-U-P-MfcuRe Dsec

fcol K:

3rd Civ

roblemiven:

Un bracPu=96 tMux=10u=300

RequiredDesign ection?

.

if sect

Mi(

if sect

ei(

for miolumn u

: UN br

ivil year

m (9)

ced colu t & H0.0t.m 0 kg/cmd: circula

bi axia(

tion su

( )

tion su

(

inimumuse raced c

Reinfo

umn Hc= 4.5

& -Mm2 & fy

ar colu

)i(

ial mom

ubject t

)=

ubject t

=(

m possi=18.0

column

orced Con

pa

50m (wMuy=6.0fy=400

umn fo

S

ment(

to (Mx+ to (ex + ible cro Fixed

n hing

ncrete - R

age No(-

with fix0t.m

00 kg/c

for its

Solutio

x & My

& ey):

oss -sec

d (case ged (cas

Revision

27 -)

xed-hin

/cm2

minim

on

My):

:

ction &

e 1) bott K fromase 3) to

n part (1)

nged) e

mum p

uni a (

& Un bra

ttom. m tabletop.

)-Notes N

end con

possible

axial m

raced ci

e (6-10)

No (9)

ndition

e cross

moment

ircular

0) =1.6

ns.

s -

)nt

60


page No(- 28 -)

= . =18.0 = . . get D =0.40m=40cm D = = . . = 0.064m Madd. = Pu* D = 96.0*0.064=6.22 t.m

for initial moment (Mx & My ) get Mi

Mi= + = 10.0 + 6.0 =11.66 t.m Mf = Mi +Madd. =11.66 +6.22 = 17.88 t.m Design circular column on (Pu=96 t & Mu=17.88 t.m)

/ : - Circular section -fy=4000 kg/cm2 (shaker page 432) - =0.90

get: = . ^ ^ =0.745 & = . ^ ^ =0.80 from( i.D page 433)using =0.745 & =0.80 get =9.0

get: = *fcu * 10^-5

= 9*300 *10^-5 = 0.027

As= * *R2

= 0.027*3.14*20^2 = 33. 93 cm2=8 25


page No(- 29 -)

Problem (10):

for the shown column given that:

fcu=250 kg/cm2 & fy=3600 kg/cm2

find its ultimate load resistance:

A-if it is applied at an eccentricity (ex=10.0 cm.)

B-if it is applied at an eccentricity (ey=6.0 cm.)

Solution

A- ex=10.0 cm

:

As= 12 16= 24.0 cm2

& As= *b*t 24.0= *40*60 get =0.01

= *fcu * 10^-5 0.01 = *250*10^-5 get =4.0

ex(about major)=10.0cm

= .. = 0.167


page No(- 30 -)

/ :



from chart using ( =4.0 & =0.167)

get . . =0.33 = ^ Pu =200 t.

B- ey=6.0cm

= .. = 0.15 from chart using ( =4.0 & =0.15)

get . . =0.37 = ^ Pu =220 t.
covere.pdf9-Mid term revision part _1_

9-mid term revision part _1

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