9-Mid Term Revision Part _1
-
Upload
alaa-altaie -
Category
Documents
-
view
235 -
download
0
description
Transcript of 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 :
-
3rd Civil year Reinforced Concrete - Revision part (1)-Notes No (9)
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.
-
3rd Civil year Reinforced Concrete - Revision part (1)-Notes No (9)
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
-
3rd Civil year Reinforced Concrete - Revision part (1)-Notes No (9)
page No(- 7 -)
/ :
-fy=3600 kg/cm2 - =0.90 (shaker page 428)
-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
-
3rd Civil year Reinforced Concrete - Revision part (1)-Notes No (9)
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
/ :
-fy=3600 kg/cm2 - =0.90 (shaker page 428)
-uniform steel distribution
From chart using =6.0 = 0.15 get
-
3rd Civil year Reinforced Concrete - Revision part (1)-Notes No (9)
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
-
3rd Civil year Reinforced Concrete - Revision part (1)-Notes No (9)
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
.
-
3rd Civil year Reinforced Concrete - Revision part (1)-Notes No (9)
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
-
3rd Civil year Reinforced Concrete - Revision part (1)-Notes No (9)
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.
-
3rd Civil year Reinforced Concrete - Revision part (1)-Notes No (9)
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
-
3rd Civil year Reinforced Concrete - Revision part (1)-Notes No (9)
page No(- 16 -)
/ :
-fy=3600 kg/cm2 - =0.90 (shaker page 428)
-uniform steel distribution
from chart using ( =6.30 & =0.106)
get . . =0.45 = ^ Pu =202.50 t.
-
3rd Civil year Reinforced Concrete - Revision part (1)-Notes No (9)
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.
-
3rd Civil year Reinforced Concrete - Revision part (1)-Notes No (9)
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
-
3rd Civil year Reinforced Concrete - Revision part (1)-Notes No (9)
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
-
3rd Civil year Reinforced Concrete - Revision part (1)-Notes No (9)
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:
/ :
-fy=3600 kg/cm2 - =0.90 (shaker page 428) -uniform steel distribution
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
-
3rd Civil year Reinforced Concrete - Revision part (1)-Notes No (9)
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( : .
-
3rd Civil year Reinforced Concrete - Revision part (1)-Notes No (9)
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
-
3rd Civil year Reinforced Concrete - Revision part (1)-Notes No (9)
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
-
3rd Civil year Reinforced Concrete - Revision part (1)-Notes No (9)
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
-
3rd Civil year Reinforced Concrete - Revision part (1)-Notes No (9)
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:
/ :
-fy=3600 kg/cm2 - =0.90 (shaker page 428) -uniform steel distribution
get: = . ^ ^ =0.32
-
3rd Civil year Reinforced Concrete - Revision part (1)-Notes No (9)
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
-
3rd Civil year Reinforced Concrete - Revision part (1)-Notes No (9)
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
-
3rd Civil year Reinforced Concrete - Revision part (1)-Notes No (9)
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
-
3rd Civil year Reinforced Concrete - Revision part (1)-Notes No (9)
page No(- 30 -)
/ :
-fy=3600 kg/cm2 - =0.90 (shaker page 428)
-uniform steel distribution
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_