rcd beam analysis and design
DESCRIPTION
Reinforced Concrete Beam Design and AnalysisTRANSCRIPT
IPA
WPHP* :a
%^ pprORCED CONCRETE DESIGN1e XJ
229
BEAM REINFORCEDFOR TENSION
ao ,V
U -F**
•O-O-O- -
0 85/’
C=0.85/tai
JW,
C = 0.85 fc' ab(compressive force of concrete)
T = As fy
(tensile force of steel reinforcem ;nt)
IFh = CT = CAs fy = 0.85 fc' ab
Asfy0.85 fc' b
Mu = 0 C (d - a/2)Mu = 0 0.85 fc' ab (d - a/2)
.. o 0.85 fc' As fy bMu " — SSfc-V-(d ' a /2)
As * p bd
(tension reinforcement index )
fc’ p bd fyMu = 0(0.85) 0 85 fc-
[d - AS fy ]0.85 fc' b(2)p bd fyr . P bd fy
Mu = 0 fc' co bd [d - 2 fc- (Q.85) br p b2d2 fy TMu = 0 fc' co[bd2 - 2 (0.85) b Jr co bd2 l
Mu = 0 fc' co[bd - 2 (0.85) JAAu = 0 fc' co bd2 ( l - YJ )Mu = o fc' © bd2 (1 - 0.59 co)
Rn = fc' © (1 - 0.59 co)Rn = coefficient of resistance
Mu = 0 bd2 RnRn ' 0 bd2
Hi[®5
(steel ratio)
Mu * o As fy (d - a /2)
(ultimate moment capacity)
Mn = Asfy (d - a /2)
(nominal moment capacity)
1
230 REINFOR^SIA. Design for dimension and steel
reinforcements.Givendata :
Ultimate moment M ,Concrete strength fe'Yield strength of steel . . . . f
Required:a) Width of beam bb) Effective depth of beam . . . dc) Steel reinforcement needed At
dcU*30.003 £, + 0.003
c6 0 003sin
e + 0.0030.003 CO'
"-L + 0.003E.
c„ 0,003(200,000)7 ” f + 0.003(200,000)
n,Sek
600 d A,c* =6 f + 600
i
C = T0.85 f ’ ab = f,0-85 f.’ R cb b = Af
N=Is
* 0003 o 85fc -y i l 1 Ste
c A ft.p X6 bd
0.85 f/ Bctb =p() bdf>.0.85 f ’act
b f, d_ 0.85r 8 (600) d
Pb "
*, <f, + 600)
ft./ P
i» •
»
8. h)vStepl. Compute balanced steelratio:In a balanced section, steel starts yieldingwhen concrete just reaches its ultimate straincapacity and commences to crush. At thestart of failure , the permissible extreme fibercompressive strain is 0.003, while the tensilestrain in the steel equals the yield strain
f
fcr(Ur_ 0.85 f/ R 600
- Pt (Pif (fy + 600)
(balanced steel ratio)
- 0 85 Vs 600Pt ~ f, (600 + fy )Assume p - 0.5 pb
pt(0 = — lr
Wir&
E.Cofie* OSSfc -
Step 2. Assume min. thickness of be*'1S«hply supported ~z
«
-BT,. A 16
LOne endcontinuous
So<hendcontinuous
Cantilever beams
fia’ ix - T -** /,21
* eiNWBCI rONCRETE DESIGN 231
U»o steel covering 62.5 mh *d + 62.5d *h - 62.5
Compute for "b"u8ing:
Mu * 0 f,' bdJ (o (1 - 0.59(u)
Step 3. Steelreinforcement neededA, =pbdSelect no. of bars:
A( = — d2 N
N =no. of barsd = diameter of bars
Step 4. Check for pmt andp^P = 075P*Pim > Pntn
1.4P > P =T
B. Investigation of Beams Reinforcedfor Tension if Steel will yield:(Under Reinforced)(pmln < p < pmax)
Givendata:
b 0.003 0.85fc’
Ac
(d-a/2 )
£t T=*Sf,a) TensionReinforcement index
p = -F bd
b) Depthof compressionblockCompare withp
^p < pmax steel bars will yieldp >psteelbars will not yield0 =1 *
0.85 fc' ab = A fyMa
0.85 fc' ab
c) Ultimatemomenta = lic
aC _
IJWidth of beam bEffective depth of beam . . dSteel reinforcement AtConcrete strength f 'Yield strength of steel . . . . fSpan of beam L
f^l Required:a) Tension reinforcement index , . w
^ Depth of compression block . ac) Ultimate moment Muf Coefficient of resistance R„®) Safe concentrated load P
e, _ 0.003d - c cWhen:£, > 0.005 XUse 0 = 0.90IWhen:£, > 0.002£, < 0.005
»*0.7 + <e,-0.002)25<V3
0 a 0.65 (£ -0.002)250/3Other
carolled coo®
Use 0 = 0.65 + (£, - 0.002)
When:£, < 0.002Use 0 = 0.65
REINFORCEDn232
t
n
Nominal Moment Capacity
Mn = A, fy (d - 3/2)Ultimate moment capacity:Ma = 0 At (d - a/2)
d) Coefficient of resistanceMu = 0r <u bd2 (1- 0.59<a)
MR
"~
0 bd2
e) Safe concentratedload
M8 4
C. Investigation of Beams Reinforcedfor Tension if Steel will not yield:(Over Reinforced)(pmln ^ Pmtx< p)
Givendata :Width of beam bEffective depth of beam . . . dSteel reinforcement AsConcrete strength f 'Yield strength of steel . . . . fSpan of beam L
Required:a) Balanced Steel ratio .b) Depth of compression block . . ac) Ultimate moment y
SHiZXf "1 |C=° 85/C'«4
*Dre* T=AsFy
s) Balanced SteelratioLdMIrCsai* m
005ft'- 30) TrJbut not less than 0.65 :
B = 0.85 -
Pb_ 0.85 f ' B 600 /lr
U600*',)
b) Depthof compressionblockA,
'Sp < pmt steel bars will not yield
Use: f_ 600 (d- c)
cC = T0.85 f ' ab = A f
cSolve fore: (quadratic equation)a = Bc
c) UltimatemomentMu = 0 A f (d - a/2)
D. Balanced Condition
Givendata :Width of beam bEffective depth of beam. . . dSteel reinforcement A,Concrete strength f 'Yield strength of steel . . . . fSpan of beam L
Required:a) Depth of compression block 3
b) Balanced steel areac) Ultimate moment
for balanced condition
t
I
a) Depth
a= Gc (d
b) BalantC=TO^Sf ' ab
0.8
ks t
^^'mai
E*P*®
C1^5
17,
V
m b 0.85 fc '
&30) r(d-al2 )5 r
cssionbi iOCk
will not yield
3 80Od
»)(d-c)cequation)
. . b. . d
• A,. . V. .
L
fr /E, 0.003
a) Depth of compressionblock :
f,+ 600
a = lie (depth of compression block)
b) Balanced steelareaC = T0.85T ab = As6 fyAs^ OSSVab
As# = balanced steel area
c) UltimateMoment for BalancedCondition:
Max. A -0.75 Asb*J!t»nate moment for balanced condition:M = 0 A fy (d- a/2)Check:n .0.85 f ' li (600)p> "T(MOH )
(baiancedsteelrali0)
1=^bd
233
Tension Steel Yields
3. Problem:
A rectangular concrete beam has a width of250 mm and a total depth of 450 mm. It isreinforced with a total steel area of 1875 mm2
placed at an effective depth of 375 mm.fc' = 27.6 MPa, fy = 414.7 MPa.
® Determine the depth of compressionblock.
® Determine the moment capacity reductionfactor.
(D Determine the safe live load that the beamcould carry in addition to a dead load of20 kN/m if it has a span of6 m.
Solution:© Depth of compression block:
>'< «•
A,=1873'
Assuming the steel yields:T = CA,fy = 0.85 fc’ab1875(414.7) = 0.85(27-6)(a)<250)a = 132.58 mm
@ Moment capacity reduction factor.a = UC132.58 = 0.85 CC = 155.98
M
X/r-
0.001
C =155.98!- -»5 •VI
:i9 o:
t / 75Av
M.
%V
ii«
234 REINFORCED CONCRETE
^E, 0.00391Q 09 IRSQft
E{ = 0.0042125 > 0.002 but < 0.005
=1e, E.C,= 414.7
200 000£y = 0.0020735 < 0.0042125
Steel yields:
0 = 0.7 + (£,-0.002)250/3
0.90
Spiralc 0.70
0.65Other
/0 = 0.65 + (£ -0.002)250/3
Compression ' Transition 1 Tensioncontrolled controlled
SI cV=0.002 e,== 0.60 c/d = 0.375
= 0.005
Since e, is between 0.002 and 0.005,tiiis value is within the transition rangebetween compression controlled sectionand tension controlled section.
Use 0 = 0.65 + (£, - 0.002)^0 = 0.65 + (0.0042125 - 0.002) 2503
0 = 0.834
© Live load it could support:
M u = 0M y (d")Mu = 0.834(1875)(414.7)(375 -
Mu = 200.2 x 106 N.mm
Mu = 200.2 kN.m
_ WUL2
8Wu (6)2
8W„= 44.48 kN/m
Mu
200.2
Wu =1.4 DL + 1.7 LL44.48 = 1.4(20) + 1.7 LLLL = 9.69 k N / m
Tension SteelDoes Not Yield
4. Problem:
A rectangular beam has a width of 280 mmand an effective depth of 500 mm. It has areinforcement area of 4072 mm2 at the bottom.Balanced steel ratio p t> = 0.026. fc' = 25 MPa,fy = 414 MPa.
© Determine the depth of compressionblock.
© Determine the tensile force carried by thesteel bars.
® Determine the resisting moment capacityof the beam, checked for moment capaotyreduction factor.
Solution:© Depth of compression block.
d=500
0.S5fc
ifS
I f,v.500*s
I TXT,
rrPmPmp -
st»
c
f :
f :
C =0.8!
0.8!
505c2
c =a =a =a s
i
r.
DESIGN 235
P'- hp =
bd1072 :0.029
L
n Steelot Yield
280(500)
Pimx = ^ '7$ PbPnB = 0.75(0.026) = 0.0195
P > P m x (over reinforced)
Steel does not yield:0.003. £,
c d- c0.003(d-c)
f, = £sE._ 0.003(d - c)(200,000)
c
as a width of 2S3) of 500 mm. IthL.1072 mm2 atthebcq|-
, = 0.026. fe' - Z
f
f,c_ 600 (d - c)
c
jepth of con?®fsile force carriedPisssd
C = T0.85 f ‘ab = A f
0.85(25)(0.85)c(280) = 4072(6-00ic
5057.50 c2 =122160000 - 2443200cc2 + 483.08c - 241542.3 = 0c = 306.08 mma =IJca = 0.85(306.08)a = 260.17 mm (depth of compression block)
ssion bloC^o#
® Tensile force carried by the steel bars.
0.003a:
f
-iM' ‘T
f
f
f
.600 (d - c)c
. 600(500 - 306.08)306 08
= 380.14 MPa < 414 MPa
Tensile force:T = As fsT = 4072(380.14)T = 1547930 NU 15*7.93 kN
® Resisting moment capacity of thebeam, checked for moment capacityreduction factor.Mu = 0 As fy (d - a/2)Mu = 0 T (d - a/2)
£, _ 0.003193.92 "
306.08£, = 0.0019 < 0.002Use 0 = 0.65
= 0 T (d - a/2)
Mu = 0.65 (1547930) (500 - 260 171
= 372,2 kN.m2
)
Balanced Condition ofBeams Reinforced for
Tension
5. Problem:
500
300
;
Aab
295.86
0.003 0.85fc‘
Ml?.v. 14
Et f^ab/y
M
A beam has a width of 300 mm and aneffective depth of 500 mm. fc' = 28 MPa.,fy = 414 MPa. Es = 200,000 MPa.
© Determine the depth of compression blockfor a balanced condition.
® Determine the balanced steel arearequired.
® Determine the moment capacity formaximum steel area requirements for abalanced condition.
Solution:© Depth of compression block for a
balanced condition.
Si* mr
236 REINFORCED CONCRETE DES,^c 600d
f 900600(600)
C ~ 414 + 600C » 295.86
aa (ic
a * 0.85(295.86)
a - 251.48 mm
U'I Balanced steel area required.C = T0.85 fc ab = Asb fy0.85(28)(300)(251.48) = Asb (414)Asi ~ 4337 mm2
(max. steel area for balanced condition)A, = 0.75(4337)
At = 3253 mm2
£ = £, = 0.002 < 0.005
Use 0 = 0.65
Mu = 0 A, f (d - a/2)
Mu = 0.65(3253)(414) (500 - — ~)Mu = 327.62 kN.m
(3) Moment capacity for maximum steelarea requirements for a balancedcondition.A = 0.75 Asti
5-A Problem:
A rectangular beam having a width of 300^and an effective depth of 450 mm.dliu 01» w.WM.w -ryy ^reinforced with 4-36 mm in diameterfc’ = 30 MPa fy = 270 MPa.Es = 200000 MPa.
© Compute the balanced steel ratio of febeam.
© Compute the nominal moment capacityithe beam.® If the value of fc' is reduced by 50X,
compute the percentage of the reducednominal moment capacity of the beam.
Solution:© Balanced steel ratio.
0.85 fc' R (600)pb " fy (600 + fy)
0.85(30)(0.85)(600)pb ~ 270(600 + 270)
Pb = 0.0544
0.850.853 si
Mn =
Mn =
Mn =
Mn =
p Percemorm
© Nominal moment capacity.
0-85/c’
Tp =5P = 0j
p = 0.
Pb* S;
!0,7
C = T
J1 <** fl«>lP 0.85(30)(a)(300) = J(36) 2(4)(270)
0.85 fc' ab = As fy
143.70a
v .
= Asfy (d -|)
-ORCED CONCRETE DESIGN 236-A
Anientis reducedJ Mn = f (36) 2(4)(270)(450 -
rrtage ofoadty of ihetsy j Mn = 415.7 x 106 N.mm
143.702
Mn = 415.7 kN.m.
m'0 )
@ Percentage of the reduced nominalmoment capacity of the beam.
AsP = Mbd
f(36)2(4)P = 300(450)
p = 0.03016 < 0.75(0.0554)
p = 0.04155 (under reinforced)
larity-
0.85/r '
#1
r
Pb0.85 fc' (J 600fy(600 + fy)
0.85(15)(0.85)(600)270(600 + 270)
Pb = 0.0277
Pmax " 0.75 Pb
pmax = 0.75(0.0277)
Pmax = 0.0208
AsP = bd
f(36)2(4)p = 300(450)
p = 0.03016 > 0.0208 steel does notyield
Since the steel will not yield, locate theneutral axis from the top of the beam. :i
M
0.003
c
Z4!
450 - c
£s=/y/Es= 270/200000= 0.00135
0.003 0.00135c = 450 - c
1.35 - 0.003c = 0.00135c
0.00435c = 1.35
c = 310.34
a = lie
pm
REINFORCED CONCRETE’236-B
Note: 6 = 0.85 for fc' 5 30 MPa3 = 0.85(310.34)a = 263.79 mm.
Mn = Asfy(d -|)Mn = J(36) 2(4)(270)(450 - |Zi)Mn = 349.7 x 106 N.mmMn = 349.7 kN.m
Percentage of the reduced nominalmoment capacity:
349 7Percentage = x 100
Percentage = 84.1%&
5-6 Problem:
ft*»m
t**ft
A reinforced concrete beam has a width of400 mm and an effective depth of 600 mm. Itis reinforced for tension with 4 - 28 mm 0bars. fc’= 20.7 MPa, fy = 414.6 MPa.
© Determine the percent increase in nominalmoment if the depth is increased to700 mm.
© Determine the percent increase in nominalmoment if fc’ is increased to 27.6 MPa.
© Determine the percent increase in nominalmoment if the steel is change to4-32 mmo.
Solution:© Percent increase in nominal moment if the
depth is increased to 700 mm.
600
- 4-28,mm*
0003 0.85fc’
C
Es
c
d-a!2
T=Asfy
k
C = T0.85 fc' ab = A, fy
K0.85(20.7)0(400) = - (28f (4,(4^Ia = 145.09 mma =Bc145.09 = 0.85 cc = 170.69 mm
0.003
170.69
£.s
0.003
429.31
429.31 170.69£,= 0.00754
e’ " i£, 414.6
200,000£y = 0.00207£s > £y (steel yields)
Nominal moment if d = 600mm
M. =« A, ft (d -|) JM„ = 0.90(7i:/4)(28)2 (4K41
M„ = 484755959 N.mmM„ =484756kN.m
I4.6|H
A
400
/ >%..
%
EggCONCRETE DESIGN 236-C
%
Honing moment if d = 700 mm
Mr * eA, fy (d - a/2)
M =0.90(7rM)(28)2 (4)(414.6) (700 -
Ma = 576660663 N.mm
= 576661 kN.m
Percent increase in nominal moment:
(576661 - 484756)% inCre3Se = 484756% increase = f8.96%
100
@ Percent inaease in nominal moment if fc’is increased to 27.6 MPa.
C=T0.85 fc* ab = As fy
0.85(27.6)(a)(400) = J (28)2 (4)(414.6)
a = 108.82 mm
M. = ® A, C (d f)= 0.90(I)(28)’(4)(414.6)(600- ®?)
Mn = 501423048.1 N.mmM„= 501423 kN.m
Percentage inaease in nominal moment:
W,cf - (501423 - 484756)li? -:% incr
484756ease= 3.44%
® Percent increase in nominal moment if thesteel is change to 4 - 32 mm 0.
C = T0.85 fc' ab = A, fy
0.85(20.7)(a)(400) = — (32)J(4X414.6)4
a = 189.51 mm
Mn = 0 AS fy (d -|)M„=0.90(^)(32)2(4X414.6) (600 -^M„= 606490 kN.m
% inaease in nominal moment:'
(606490 - 484756)% increase= — <00
% inaease = 25.f%
REINFORCED CONCRETE DES I^236-D
:5-C Problem® Amount of steel needed.
C = T0.85 fc’ ab = As fy
0.85{20.7)(183.28)(250) = As (414.6)
:A rectangular beam has a width of 250 mmand an effective depth of 575 mm. The beamis reinforced for tension to cause the strain inthe tension steel to be 0.005 just as themaximum strain in the concrete reaches0.003.
ic
As = 1945 mm2
«40fc = 20.7 MPaf,= 414.6 MPa
© Determine the depth of compressionblock.
® Determine the amount of steel needed.® Determine the ultimate strength of the
beam.
Ultimate strength of the beam -3
M u = 0 A s f y (d -|) ASvJ"
C = TMu = 0.90(1945)(414.6) (575 -Mu = 350.8 kN.m
0.85 f/Solution:
i , . © Depth of compression block B - z- A, = 141.85(27:a = 104
0.003 0.85fc’
r* -1J 4,.,
5-D Problemd -af2 HeA rectangular beam has a width of 250and a total depth of 640 mm. The beamreinforced with 3 - 25 mm 0 placed with *Bp122steel covering of 65 mm from the bottom of tl ^B^beam. fc’ = 27.6 MPa. Assume A 615 gra* E60 steel with fy = 414.7 MPa.
AsT=A fy
b=250 mm
By ratio and proportion0.003 _ 0.005
c~
575 - c© Determine the distance from
compression fiber to the neutral axis ofbeam.1.725 = 0.008 c
c = 215.63 mm © Determine the nominal moment of ^beam.ucam.
u® Determine the percentage increase jo jjl
nominal strength of the beam if|increase the steel reinforcementl°03a = 8 c
3 =0.85 (215.63)a = 183.28 mm
J
100%
{REINFORCED CONCRETE DESIGN 236-E
offr«beani.
Solution:
Q Distance from extreme compression fiberto the neutral axis of the beam.
0.003 0.85fc'
r
S 6»C
3-25 mm 9
Es
d -an
T=* sfy
f)250 mm
? Assume steel will yield:C = T
K414.6)(575-H 0.85 fc'ab = As f,71 ; A = j(25)J (3)
4
As =1472.6 mm2
0.85(27.6)(a)(250) = 1472.6(414.7)a = 104.12 mm
& a = fic1 ofSi 104.12=0.85c
c=12150 mm (distance from extremecompression fiber to neutral axis)
(tB-
Checkifsteel wiflyield: 0.003
dlSt3nC!e^ 122.50
/T.„,esn o f - £sl^so '^<^|c
* = 0-0111 > e = = 0.0021of «VHI- 290,000^ «S<*lyields
® Nominal moment of the beam.M.= A,l, (d - 5)M.= 1472.6(414.6) (575
M u = 319.4 kN.m
® Percentage increase in the nominalstrength of the beam if we increase thesteel reinforcement by 100%,
0.003 0 851V 1
575
C
E.S
C
d 41/2
250— Ar=As /y
C = T0.85 fc‘ab = A, fy0.85(27.6)(a)(250) = 2945.2(414.7)a = 208.25 mm
M u = A s f y (d -1)M„= 2945.2(414.6) (575
Mu = 575.11 kN.m
% increase = 80%
M
i
REINFORCED CONCRETE DESI^A rectangular reinforced concrete beam has awidth of 400 mm with an effective depth of600 mm. It is reinforced with 4 - 28 mm 0
bars at the bottom, fc’ = 20.7 MPa,fy = 414.6 MPa.
© Determine the percentage increase innominal moment if the steel reinforcementis changed to 4 - 32 mm a.
® Determine the percentage increase innominal moment if the effective depth isincreased to 700 mm.
® Determine the percentage increase innominal moment if fc is increased to27.6 MPa.
*
mm
0236-F
jit ''”iA, (d - f ) 5 # TM P
0.9Mn = 2463(414.6) (600 - 145.09)
M„= 538.62 kN.m C** ;
P-466Mu= 466
When A, = - (32)2 (4)
A, = 3217 mm2
Check form
Mu = —;| 40(!HI 8Mu = 405C = T
0.85 fc' ab = A, fy0.85(20.7) a (400) = 3217(414.6)
a = 189.51mm
Irate rrftebeai
Solution:© Percentage increase in nominal moment if
the steel reinforcement is changed to4-32 mm a.
9 Cut-ofsuppo
2». 0.003 0.85fc’ 9r i
Mn = 3217(414.6) (eoo -^)cw
*23 ,
* V ' M = 673.88 kN.m/ -a :
3-28mm. Percentage increase in M.T-Asfy- o/0 increase = < »°°53862
40
\ = ~ (28)2 (4)
A, = 2463 mm2&% inaease = 25.11%
2)a.38.!
C = T0.85 fc' ab = A, fy0.85(20.7){aK400) = 2463(414.6)a = 145.09 mm
\ 0;
*
iH8B1
PIETE DESIGN 237
a2
6) <600.'•.m
145.09)
(32)J (4)
2
(2) Ultimate moment capacity of the beam:
Mu = 0l(d -1)Mu = 0.90 Asfy (d - ~ )Mu = 0.90( j)(25)2 (4)(414.7)(675 - — H)Mu = 466.4 x 10® N.mmMu = 466.4 kN.mCheck for actual moment:
WuL2
Mu =
Mu:8
40(9)2
8Mu = 405 kN.m < 466.4 kN.m (safe)
fr lIHii
I00) = 3217(414.6) |Ultimate moment capacity of
the beam is Mu = 466.4 kN.m
Cut-off point of two of the bars from thesupport:
- )2'
.6) (eool.m
189^)"T
450
O; p0.85 fc' 0.003
ease inM»:
573538-62
,11%
2-25 mm 0
-c
dJl2/fT~
/ rf-c=655.72
T=Asfy Ej
c=38.57
. 53862) ,00 ^= 981.75 mm2
C = T0-85 fc’ ab = As fy085(27.6a = 38.57Mu
Remaining steel bars is 2 - 25 mm aAs= j(25)2 (2)
°-85(27.6)(a)(450) = 981.75(414.7)
:0 As fy (d - a/2)Mu = 0.90(981 75)(414 7) (675 -
= 240.3 x 106 N.mmMu = 240.3 kN.m
Wu»40 kN/m
Z*~
X ^180
Mu = 180(x) -240.3 = 180x - 20x2
x2 - 9x + 12.02 = 09 ± 5.74
n180
X =2
x = 1.63 m. (cut off point of 2 bars)
Alternate Solution:»'„=40kN/m
1 L |X
180
2-25 mm 0 j. J
9 m 180
iv„=40kN/m 2-25 mm 0
/L 2-25 mm a
x
9 m
</ a
164.7
240.3
4.5 ' 4.5Using squaredproperty of parabola:
405 _ 1647(4.5)2 "
a2
a = 2.87 m. from centerx = 4.5 - 2.87x = 1.63 m. (from support)
•i
.
'
238
S*»
*IfJkuf . .
REINFORCED CONCRETE DEs|g
Determination WhetherCompression Bars or
Tension Bars is needed
Q|Q3s SZZESE12u3lA 12 m.simply supported beam is provided byan additional support at mid-span. The beamhas a width of b = 300 mm and a total depthh= 450 mm. It is reinforced with 4 - 25 mm 0at the tension side and 2 - 25 mm 0 at thecompression side with 70 mm cover tocentroid of reinforcements. Fc’ = 30 MPa,fy = 415 MPa. Use 0.75 pt,= 0.023.
P -1963.5
300(380)p- 0.017pma - 0.75 pb = 0.023p < pmax
Only tension bars is needed.
© Determine the depth of the rectangularstress block.
© Determine the nominal bending moment,Mn.
® Determine the total factored uniform loadincluding the beam weight consideringmoment capacity reduction of 0.90.
Solution:© Depth of the rectangular stress block:
Therefore the beam needs only ^bars as specified in the problem.C = T0.85 fc' ab = As fy0.85 (30)(a)(300) = ( /4) (25)2 (4X415)a = 106.52 mm
(depth of rectangular stress block)
® Nominal bending moment:
Mn = Asfy(d-Mn = - (25)J (4)(415) (380 - ^y?Mn = 266.2 x 106 N.mm
266.2 kN.mMn :
</=380
0003 0.85fe'
50J e* T
unmniHiiuuH
Check if compression bars is needed.As
P a TTbd
c310 - A
S|sR
' kI.
*8=1^.384 El
8. "L48 El
A> = f (25)2 (4
A,= 1963.5 mm2 fv.
Po0ft|
® Total factored uniform load includinjbeam weight:
REINFORCED CONCRETE DESIGN 239
0 023
'S '5 '****bsamrf 'n the proi^lify0) = (R/4) (25)iI
angular stress
ing moment:
~ )2
(4)(415)(380
0* N mm.m
j uniform
5 wL4
P =
PLa3M ® " 48 El
SwL8
2R + P = wl, 5 wL
2R = wL • — g—OD 3 wL2Rl _r
3R = — wL
M, =R (|)- W (|) (t)3 wL2 wL2
16(2) 83 wL2 - 4 wL2
MB 32wL2
MB = -Mu = 0.90 MnMu = 0.90(266.2)Mu = 239.58 kN.m
‘ 239.58 = 32.w(12)2
239.5832
w = 53.24 kN1 m102-
CURVATURE ofBEAMS
ri
A rectangular reinforced concrete beam has awidth of 300 mm and an effective depth of 450mm It is reinforced for. tension at the bottom
with a total steel area of 962.5 mm'v fc = 24.2MPa fy = 345 6 MPa. Es = 200000 MPa.
® Evaluate the curvature vfy due,0moment My which produces initial yielding
°f the tension steel in radians per m.
© Evaluate the curvature % due to nominalflexural strength Mn of the cross section in
radians per m.® Determine the magnitude of the cross
sections curvature ductility ratio
450
300
-45=962.5.mm2
-o
£c
A.L315.10 w 450-x
er=? I
Solution:© Curvature y/y due to moment My which
produces initial yielding of tension steel:
Esn'"Ec
2000004700 >/fc'
2000004700 V242
n = 8.65 say 9
|M = n/1s (450 - x)
150x2 = 9(962.5)(450 - x)150x2 + 8662.5X - 3898125 = 0x2 - 57.75x - 25987.5 = 0x = 134.90 mm450 - x = 315.10 mm
345.6£y " 200000ey = 0.001728
mm
40
1*IR
REINFORCED
e.Curvature y/y = 45Q - x
0.001728Vy ~ 315.10y/y = 0.00000548 radians/mm
\ffy = 0.00548 rad/m.
© Curvature y/u due to nominal flexuralstrength Mn:
450
2 £*=0-003 0-8S/C-As=9«JSmrai
c
T=Asf,T = CAsfy = 0.85 fc'ab
962.5(345.6) = 0.85(24.2)(a)(300)a = 53.90a = Rc
53.90 = 0.85cc = 63.42 mm
0.003W/ = —Vu
0.00363.42
if /u = 0.0000473 rad/mm
y/u = 0.0473 rad/m
%3 Curvature ductility ratio = —
Wy
CO* -fMkCDR - 8.63
BEAM REINFORCEDFOR COMPRESSION
A. Design for dimension and steelreinforcements.
d'
Givendata :Design moment MuWidth of beam bEffective depth of beam . . . dConcrete strength f 'Yield strength of steel . . . . f
Required:a) Moment tobe carried by
compression bars M2b) Total steel area in tension . . Asc) Steel area in compression . . As'
0.85fc'Cj=4s7s'
&
C,
id«n\
Tr*sif, TrzAsi!>
a) Moment carried by the compressionkTo obtain a reduction factor of 0.90, themax. reinforcement ratio correspondingto a net tensile strain of 0.005 is equal to:
0.003P = 0.85 ft V / 0003 jf '0.003 + 0.005f '0.003 + 0.005'
From the strain diagram:
y0.003
C
: d
Et=0.005 0.003*1
c
o.oo
P' \
pbd
P - 7
Assun
A„= /C = T
°.85 f 1
(d-d )
I © a = —= 0
, = M
b) Totsa = GCa 'Frorrdied
"Sfon
BSpRCEDCONCRETE DESIGN 241
C d0.003 0.003 + 0.005£ = 0.003
beam .
teel .
HV
df;
. fi
riedbyrs IIIitension . . Aipression . . A
d 0.003 + 0.005As fy = 0.85 fc' a b
A# fy =0.85 fc'R cb
p =^H bdpbdfy = 0.85 f ’lic b
0.85 T lie
0.85 fc’B (0.003)
fy (0.003 + 0.005)
P
P
Assumed: p = 0.85liM— — — )fy '0.003 + 0.005'
0.85fc’Cf*'+Ci
(d -aP-\&
As1 =pbdC = T0.85 f ' ab = A , fy
a = - _L_0.85 f ' b
r,= i/A
&tgram
00540.^i
M, = 0 A , fy (d - a/2)
M2 = MU - M,
bj Tota/ sfee/ area in tension :a = lic
ac = -li
From the strain diagram,' check whether compression bars will yield,
g ' _ 0.003c - d' ce . _ 0.003 (c - d1)
* c
1) When e;>iE.
Compression bars will yield(V =f,)
2) When e;<1E.
Compression bars will not yield600 (c - d1)
cUse: r
M2 = 0 As2 fy (d - d1). M2A , = 2
82 0 fy (d - d1)A, = A„+ Aa
c) Steel area in compression :If compression bars will yieldAS2\ = As' VAS2 =a;If compression bars will not yieldAs. f^vr
600 (c - d')c
where: f ' =
A ' = i i’ V
B. Compression Bars Will YieldGivendata:
Total steel area in tension . . AtSteel area in compression . . At’Width of beam : bEffective depth of beam . . . dConcrete strength f 'Yield strength of steel . . . . fSpan of beam L
242
aM
REINFORCED CO
Required:a) Depth of compression blockb) Ultima!’moment capacityc) Safe uniform liveload
it axiki carry t-t
c) Safe uniformliveloadW, l }
M„
rf’Tt**
i
L- •V
0.85fr‘— r r, C£gJ>rrfT
wu = 1.4D + 1.7L
C. Compression Bars Will Not Yield
(rf^/2
©{d -d’ )
©T,=ASIf } Ti=Asif,
a) Depthof compressionblock :Check if compression bars is really needed.
K bdP , --0.75p,IfP >P (compression.bars are needed)
( repression bars will yield if theblowing condition occurs:
^0.85 fc'Rd' 600
P fy d (600 - fy )
F bdf,' = fy (steel in compression yields) -T = C, + C2
As fy = 0.85 fc’ab + As’ fySolving for a:
0.85 fc' b
b) Ultimate moment capacity :a =Bc
M„=C1 (d-|) +CJ (d-d'jC, = 0.85 fc' a bC2 = As’fyNominal moment capacity:
Mn = 0.85 fc'ab(d -|) +A,' fy (d- d')Ultimate moment capacity:Mu = oMn
'' A
Givendata:Total steel area in tension . . AfSteel area in compression . . At 'Width of beam b 'Effective depth of beam . . . dConcrete strength f 'Yield strength of steel . . . . fSpan of beam L
Required:a) Depth of compression block . . ab) Ultimate moment capacity . . . Muc) Safe uniform liveload
it could carry LLb ' 0.85fc'
%(i4\(<d-a/21© i 8
r.T=V>
a) Depthof compressionblock :Check if compression bars is really needed pOfep - h,
bdPnax = 0.75 pb,fP >P (compression bars are needed)Compression bars will not yield if thefollowing condition occurs:n.r> * .
M5 fc' Rd' 600fy d (6d0 -g
fs * fy (steel in compression does not
UseV =Hii I ®N
ft i
T =C, - C?
A, f, = 0.85 f{ ’ a b + A,' f,'
Solving faa:A, f - A;Vh
0.85 fc' bI
:-P
Vj
fy
. . . LL
fc
l -all )
*SI'>
NCRETE DESIGN 243
b) Ultimate moment capacity :
a=Sc(d-|) *C2 (d- (f)
C, = 0.85 fc’a bC2 = A,' VNominal moment capacity:
Mn = 0.85 fe' ab(d -|) + A,' f/ (d - <T)
Ultimate moment capacity:Mu =0 Mn
c) Safe uniformiiveload:W„L?
8Wu =1.40 + 1.7L
D. Balanced Condition for BeamsReinforced for Compression andTension:
Givendata:Total steel area in tension . . A,Steel area in compression . . A,'Width of beam bEffective depth of beam . . . dConaete strength f 'Yield strength of steel . . . • fySpan of bean L
Required:a) Depth of compression block . a
, b) Balanced Steel area
^v
•" ' c) Maximum Amount of steel
area permitted:
wms
*1d
b 0.85fc-
V
-4-i-a Cl
C2~AS fy
(d-a/2 )
&(d-d )
&TrASlfy TflAS2f,
a) Depth of compressionblock :Balanced Steel Ratio,(1) If compression bar: will yield:- 0.85 fc'R (600)Pb fy (600 + fy ) P
(2) If compression bars will not yield:— _ 0.85 fc' R (600)
t, VPb ~
^ (600 + fy )+ P fy
c = 600 d600 + fy
B = 0.85 - 0.05 (fc' - 30)7
(but not less than 0.65)8 = 0.85 for fc’ < 30 MPaa = IJc
b) BalancedSteel Area :C, + C2 = T0.85 fc ' ab + As ' fy = Asfy(A, - A,') fy = 0.85 fc' ab •
AS - A5' = AS1_ 0.85 fc' ab- r-
Atf = AS'
Ast = Asi + AS2
c) Maximum Amount of steel area permitted:Max. As = 0.75 ft bd + As'
_ 0.85 fc' R (600)fy (600 + fy )
TOP*
REINFORCED CONCRETE DESl243-A
A simply supported beam has a width of300 mm and an effective depth of 330 mm. Itis reinforced at the bottom with a steelreinforcement area of 2464 mm2. Assumingsteel covering to the centroid of thereinforcement is 70 mm. fc' = 34 MPa,fy = 415 MPa.
© Determine the ratio of the depth ofcompression block to the distance of theneutral axis from the top of the beam.
© Determine the balanced steel ratio.© Determine the depth of compression
block.© Determine the total compressive force
carried by concrete.® Determine the value of the moment
reduction factor 0.© Determine the design strength of the
beam.® Determine the distance of the neutral axis
from the top of the beam if it is reinforcedwith a steel area pf 1232 mm2, at the topin addition to the existing steel area of2464 mm2 at the bottom.
© Determine the stress of the compressionbar.
© Determine the design strength of thebeam using moment reduction factor 0 =0.90.
® Determine the concentrated live load itcould carry at the mid-span in addition to adead load'of 20 kN/m including its ownweight for a span of 6 m.
Solution:© Ratio of depth of compression block t,J
distance of the neutral axis from the tar 'the beam.
b 0.003T
d
\
At .
fN.A.
L7
0 s
}c1 'c11.
- = ftc
^0.85.0.05(34 - 30)8 = 0.85 - 7
II = 0.82
Pb
© Balanced steel ratio:_ 0.85 fe‘II600“ fy (600 + fy )_ 0.85(34)(0.82)(600)
415(600 + 415)Pb = 0.034
Pb
® Depth of compression block:
</=330
70'
300 0.85/c’
yt,=>246*ran?
/1
3= *B
0.003
143.83 /l*0\1 JJ ,
T* A, f, £s
C = T0.85 fc' a b = As fy0.85(34)(a)(300) = 2464(415)a = 117.94 (depth of compres ^i
i Value
a = B <117.94|c=14
186.1]
le, = o
B =0.(
0 = 0.1|8 = 0,
% 0esigr
k «,h
f:.
* 1NF0RCED CONCRETE DESIGN 243-B
’SIJ%3
|* compressive force carried by concrete:
C s 0.85 tc’ a b| C = 0.85(34X117.94)(300)
i C = 1022540 N
i C* 1022.54 kN
If Value of moment reduction factor 0:
0.003
C=143.83
330-C=186.17
a =Bc117.94 = 0.82 cc = 143.83
£, _ 0.003186.17 '143.83£, = 0.0039 <0.005
0 = O.65 + (£, - 0.002) (250)3
0 =0.65 + (0.0039 - 0.002)^0 = 0.81
tilMy
* Design strength of the beam:
M.= BA. f,(d-|)
,AM K =0.81(2464)(415) (330 -=224.5 x 10* N.mm
- 224.5 kN.m
® Distance of neutral axis from the top of thebeam if additional steel bars are placed atthe top:
330
70
300
LfTS-rJli-:
A,*2464 nun?
:ip0.85/f ’ 0.003
PLA +
C2 c
300-C
T, Tl
Check whether compression bars will yield:
Compression bars will not yield.
, 0.85 fc'B d' 600whenp -p <
f, d (600 - fy)
Compression bars will yield., 0.85 fc’Bd' 600
A.P=b d
P = 2464300(330)
p = 0.025
M
I*
i!
••
5 '
TLm.-3L
% m
REINFORCED CONCRETE DESIG^ I J243-Ctoo 0,85/r '
bd1232
P = 300(330)
p' = 0.0124
p - p* = 0.025 - 0.0124
p - p' = 0.0126
If , a c,_.
_• N A __
;rf =330 > d -an.)
A,»246* nun'70 r
0.003(c - 70)(200,000)V: C
600 (c - 70)V0.85 fc’ (3 d' 600 _ 0.85(34)(0.82)(70)(600)
f d (600 - f ) 415(330)(600 - 415)G
C, + C2 = T, + T2
C1 + C2 = T
0.85 fe' a b + A,' f,' = As fy0.85(34)(a)(300) + 1232 f,' = 2464(415)
8670a + 1232 f,' = 1022560
a = (3 c
; 0.85 f/ l3 d' 600 = 0.0393H fyd (6O0 -g0.0126 < 0.0393
:
*8’
Iif Steel in compression does not yield:
V * fy0.003
a = 0.82 c
8670(0.82) c + 1232(60°HC :3 = 10c
c2 - 39.86c - 7278.25 = 0
c = 107.54
rE, '70
cC -70I/H-
I*
® Stress of compression bar0.003 _ g;c c - 7 0
600 (c - 70)0.003 (c - 70)£‘ 3 V c
600 (107.54 - 70)5 107.54Vs209.45 MPa < 415 MPa
RCED CON 243-D
N Ac, '
® Design strength using reduction factor0 = 0.90
r
(200,000)
MlJ' 71
70U .—A,'-2464 ram’rf=330
70
300 0.85/c’“a
^=2464 mro? .(d-an)
T l
C2
(d-d' )
T2
‘a= Bc
3 =0.82 (107.54)
a = 88.18
:=\\+ 1232 f,' = 2«d= 1022560
C, = 0.85 fc' a b
C, = 0.85(34)(88.18)(300)
C, = 764521 N
232(60008.25' °
C2 =A;VC2 = 1232(209.45)
C2 = 258042
b3r:
Mn = C, (d - j) + C2 (d - d1)
M„= 764521 (330 -M)}SSi°n + 258042(330 - 70)
M„= 285.68 x 10s N.mm
M„= 285.68kN.m
MU = 0 M„M„= 0.90(285.68)
Mu =257.11kNjn
® Live load it could carry.
1wDL =20 kNVm
nmiinmii iHi
6 m
H
U'&w.m L v .4)
M_ P(6)(1.7) , 20(36)(1.4)
Mu =51.42 kN
i
wm
P
244
&
::l
Stir I Areas Are Required0, Problem:A rectangular beam that must carry a servicelive load ot 36 kN/m and a calculated deadload ol 16.4 kN/m on a 5,4 m, simple span islimited In cross section for architecturalreasons to 250 mm and 50 mm total depth.Assume that the tension steel centroid will be100 mm above the bottom face of the beamand that compression steel if required will beplaced 82,5 mm below the beams top surface,
f,/ * 27,6 MPa, f„« 414.7 MPa .
<D Determine the remaining moment to becarried by the compression steel.
® Determine the steel area In compression.(3) Determine the steel area in tension
Solution:CD Remaining moment to be carried by the
compression steel.
BWg ' 1.4 DL + 1.7 LiW„ " 30(1,7) + 1.4(15.4)W„ * 72.56 kN/m
400
250
a, •B C,d a n <&<5,
r/ "Vr
d d
WUL?
8M.M„*264.48 kN.m
To obtain a reduction factor of 0,90, the max.reinforcement ratio correspondingto a net tensile strain of 0.005 is equal to:
“ [ L J
REINFORCED C
Assume value of p :
^_ 0.85(0.85)(27.6)(0.003)
P ~414.7(0.008)
p - 0.018
A,, -pbdA„= 0.018(250)(400)A„= 1800 mm2
C = T0.85 fc' a b = A„ f y0.85(27.6)(a)(250) = ,1800(414.7)a = 127.27 mm
M, = 0 A„fy (d -1)M, = 0.90(1800)(414.7) (4OO -~M, = 226 kN.m
M? = MU - M,Mj = 264.48 - 226M2 = 38.48 kN.m
© Steel area in compression.
a = Rc127.27 = 0.85cc = 149.73
0.003
149.7:
C, _ 0,003250.27 ' 149.73C, = 0.005
87
250.27
e,e; _ 0.003
87.23 " 149.73e, ’ = 0.00175 < ey = ~
414.7£y 200 000et = 0.00207
' A
Co presswV = fs' E‘f - = 0.00175!•= 350 MP
Mj = 0 Atf f37.48 x 106 :
A* = 306 mi
m i - o2A, f, = As' f306(414.7) =\' = 363 mn
® Steel area irAs = Asi + AsAs = 1800 +:
I As = 2108 mi
tegular be
N?*nsio"at
Aiii>mPressioni®
I
tf
th,
Pan '
245! Value of p .
i84,47fo35ipS;
b d
°18<250)(400)300 mm*'
ab=M•6)(a)(250) = 1800(4147:.27 mm
BH&SRCEP"CONCRETE DESIGN
*-• *
400
250 0 S5fc
••
— # —
1c?
j-a n *<5.
S'.
\M, dd
Solution:Depth of compression block.
300
rW 5!]Tt -^Sjfy T2 = AS2fj
600.
90(1800)(414.7) (400 -26 kN.m
64.48 - 2268.48 kN.m
el
Compression steel does not yield
V = £s’E,v = 0.00175 (200 000)f,’= 350MPa
M,= oAs, f, (d - tf)37.48 x 10* = 0.90 A*, (414.7){400 - 62.5)A^ = 306 mm1
T;=C,M S V V306(414.7)= AS' (350)A,'= 363 mm:
@ Steel area in tensionAs = As1 + AS2As = 1800 + 306As = 2f06 mm2
Compression Bars will yield
l - 2t m
6 - fnrfi *I I I1A
e
0003
A, = 7 (32)2 (6)4
A, = 4825 mm2
P bd4825
rea in compress'0"'
a |A rectangular beam has a width of 300 mm„ and an effective depth of 600 mm. The steel,49* area of tension at the bottom is 4762 mm2 and
= 0.85 c1.73
0.003= 14073
005
_ 0.003= 149731.00175 < " £,
4147 _!00 000.00207
that of the compression side at the top is987.5 mm2. Steel covering is 62.50 mm fromthe compression face of the beam.A = 0.034, fc' = 34.6 MPa, ^ = 414.6 MPa
© Determine the depth of compressionblock.
® Determine the design strength using areduction factor of 0.90.
® Determine the concentrated live load atthe mid-span in addition to a dead load of20 kN/m including its own weight if it has aspan of 6 m.
P 300(600)p = 0.0268
P-„= 0.75p„0.85V 6600
Pt‘ '
6 = 0.85 -8 = 0.817
f, (600 + f,)0.05 (V - 30)
7
300 0.003 0.85fcV
62
600
V7•— hv Y /a
’- — f i x'
:-i-4±\ >r*(d-afl ) W’
e*
(d-u/2 )
, © tP>
0.85(34 6)(0.817)(600)414.7(600 + 414.7)
pt = 0.03426P = 0.75(0.03426)P = 0 025695
Compression bars is needed:
P > P0.0268 > 0.025695
7/ > REINFORCED CONCRETE D£s
Mu = oMuMu = 0.90(1031)
= 927.9 kN.m
(D Concentrated live load at the mid-spanin addition to a dead load of 20 kN/mincluding its own weight if it has a spanof 6 m.
® Pes'9n stre"9th using a reductionfactor of 0.90.0.003 no«-„-
d-d l i l tbo
s#
2472 ^ + c? (d ^BBJfCED CONCRETE DESIGN
/ ^Iab (d - 2. Ja b * '
W'6)(18064K3(yf87 N
5(414.7)I2
Solution:I ^ $ Depth of compression block
As = -• (28)2 (4)
As - 2463 mm2
As' = -- (28f (2)4
As' = 1231 5 mm*m (6oo - 1®^,+ 407132 (&302- 62i;x 106 N.mm
A..o - — -K fcd
P ~- 2463330(330;
p =0.0248I />«,-075p,
.p_ =0.75(0031).
pw = 0.023
its own weight if i la
3 L; 1.7 PL!
+ 44(20)(6f ,“~
8’ 4
rkN
1031)IkN.m
pmaxited live load attteil to a dead load ofII ‘ Therefore reinforcemerit for comoression
is needed.
A(p ='d
Compression bars wiinot / ed if -. 055Id8 <7600p - p <
f/ d (600 - fy)0.85 (30)(0.85)(70/yX3)
^ 415 (330) (600- 415)p -p’< 0.036
jorted beam ^ p -p> (0.036t the bottom #w 0.0248 - C.C124 <0.036 *
le beam. 0.0124 < 0,036nforcementf the beam .w>I. Therefore compression bars wfl not yield.OOmmaod^Vl * * fy
= 415 MP3. | 0003 =J1 Jk C
‘c - 70
£. W03(c- 70)
' - *0#^rthedes^ 5 ' / V=0OoM
rr-Sht-
c c TO
L .....A
o.m£*Z
. sv/ zq WJi . ^. __F'
r
T = Ci CaA* fr = 085If ab r As V2463(415) = 0>85(36)(a^300)
me - 70)12315c
1022145 =7650a + 738900(c - 70)c
1022145 = 7650(0.85c) 73890^c:^157.19c = cr^ + 113.63(c - 70)c* - 4356c - 7954.10 =0
43.56 ± 183.61
c
1
c = - 2 = 113.59
a =8 c3=085(11359)a = 96.55 (deptfi of compression block)
% Design strength using 0.90 as reductionfactor:
Tl"300
. «qr~ri_
„,i'
OtH-•rqC/h@ ^
7 /
© ay/'
7?
C. =055 f ' abC, =055(30X9655X300)C, = 738 60750 N
C2 = A,' Vf,* = £,'Ft
...600(113.59 - 76;’
113.59f '= 23C 25« 41-3
*
(
4
i
m'
248 REINFORCED CONCRETElia l ^|Np(
c,C, = 1231.5(230.25)C, = 283553
M, = C,(d -|)M, = 738607.5 (330 -^)M, = 208 x 10® N.mmM2 = C2 (d- d1)M2 = 283553 (330 - 70)Mj = 73.7 x 10' N.mmMU = 0 (M, + M2 )Mu = 0.90(208 + 73.7)
= 253.53 kN.m
® Concentrated live load at mid-span:PL WL2
Mu4
(1-7)8
(I'4) .253.53 =^(1.7) + 11 (1.4)
4 8P = 50.01kN
Balanced Condition ofBeams Reinforced for
Compression
12. Problem:A reinforced concrete beam has a width of250 mm and an effective depth of 625 mm. itis reinforced for compression having a steelarea As’ = 1250 mm2 with a steel covering of62.5 mm measured from the center of thesteel reinforcement to the top most fibers ofthe beam, fc = 20.68 MPa, fy = 400 MPaE.= 200 000 MPa.
Determine the depth of compression blockfor a balanced condition.
© Determine the area of balanced steel Asbfor the given cross section.
© Determine the maximum area of flexuralsteel in tension permitted in the givencross section as required by the NSCPSpecifications.
Solution:© Depth of compression block fw fbalanced condition.
250
62.5|
</=625 mm._ _
0.003
c
JE0.85fc*
625-ces
m’ E,
400£> =
200 000ey = 0.002
0.003 0.002c 625 - c
0.005 c = 0.003(625)c = 375 mm
Check:600 dc =
fy +600
c_ 600(625)
400 + 600c = 375 mma =Rca = 0.85 ca = 0.85(375)a = 318.75 mm
© Area of balanced steel Asb for thecross section.
62.5 l
<*=625 mm
250 0.003 0-8lfI'
312.5 375
u£,' _ 0.003
312.5 "375e;= 0.0025 > £y = 0.002
(steel in compression yields)
C.Ci +0.85 fc a0.85(20-
+47 :ASb =
MaximurrpermittedMax. As :
0.8£
*‘T= 0.0Z
Max. As =Max. As :
|A rectangularI fd has an eff|fc = 25 MPa,
P» 2***iE f°°0 mm2M factor.cssmine
ix "ne!SSi(
*
MffiPORCEP CONCRETE DESIGN 248-A
%' * fy = 400 MPaCi + Cz = T0.85 fe‘ ab + A,' fy = A8b fy1,85(20,68)(318.75)(250)
+ 1250(400) = Asb (400)ASb = 4752 mm2
Maximum amount of tensile steelpermitted.Max. A, = 0.75 (\ bd + A,'.0.85 fc' ft (600)
fy (600 + fy )_ 0.85(20.68)(0.85)(600)’
400(600 + 400)I ft = 0.0224
Max. As = 0.75(0.0224)(250)(625) + 1250Max. As = 3875 mm2
12-A Problem:A rectangular beam has a width of 375 mmand has an effective depth of 450 mm.
i V * 25 MPa, fy = 300 MPa.
if 0 Determine the moment capacity if it isf reinforced with steel area in tension As =
6000 mm2. Use 0.9 as moment reductionfactor.
I ® Determine the depth of stress block if steelf in' compression As' = 3000 mm2 is added
.. jiff, with steel covering of 60 mm below the topIf of the beam.i ® Determine the moment capacity of the'% beam reinforced for tension and1» compression,
Solution:' 1 Moment capacity if At * 6000 mm2.
T «CA,1
f = 0,5 (fab6000(300) *Q.85(25)(a)(375)a * 225,88 mma * IJc225,88 = 0,85 cc = 265.74 mm
e, _ 0,003184.26 265.74et = 0.00208
AV300
t y "
200,000Ey = 0.0015 < 0,00208 (steel yields)f.HMu = 0 A, fy (d - a/2)
Mu = 0,90(6000)(300) (450 -Mu = 546 kN.m
Q) Depth of compression block if As' = 3000mm2 is at the top of the beam.
4[
450
_2Z*_
A,*6aoo.
0.85/,'|B "<>•Iif" *
©
V1s'
(d -a /2 ) &Tl‘ASI /, Tl‘*st f,
(d-d’ )
6000y bd 375(450)
3000A 'd = — *- =P bd 375(450)
0.036
= 0.0178
0.85 f * a d* 600when p - p’ > r
1 H fy d (600 - fy )y
0.85(25)(0.85)(60)(600)0.036 - 0.0178 >
300(450)(600 - 300)
a
A0.003575 am
II 265.74
id am 0.0178 > 0,0161I»4 26A,+*600fi
r,
ipy§
248-B REINFORCED CONCRETE D
Compression bars will yield;f ' = f
a*K ', *V,Atf = 3000 mm2
*“ «000 = A„+ 3000Af1 = 3000 mm2
WVT,0.85 T ab + A,' fy = A fy0.85(25)a(375) + 3000(300) = 6000(300)a = 112.94 mm
® Moment capacity of the beam reinforcedfor tension and compression.Ml = Ast fy (d- a/2)Mi = 3000(300)(450- 112.94/2)Mi = 354.18 kN.mM2 = AS2 fy (d — d )M2 = 3000(300)(450- 60)M2 = 351 kN.mMu = 0 (Mi + M2)Mu = 0.90(354.18 + 351)Mu = 634.66 kN.m
12-B Problem:A rectangular beam has a width of 280 mmand an effective depth of 510 mm. Steel areain tension As = 3000 mm2 and steel incompression As’ = 1000 mm2. Steel coveringfor steel in compression is 50 mm. fc’ = 35MPa, fy = 400 MPa.(D Determine the depth of compression
block.® Determine the actual stress of
compression bar.@ Determine the design moment capacity of
the beam. Use 0.90 as moment reductionfactor.
510
»a— 1A,’*1000
I, Mg -
000.
0,85fc.'
cr so!0003
©p-41/2)10\d3n VA
Solution:Depth of compression block
A 3000^ M
= 280S=°®
,0.007pbd 280(510). 0,85 ft'S(T 600
' (steel in compression will not yield)0.05 (f/ - 30)
7B = 0.85 -6 = 0.85 -6 = 0.81
0.05 (35 - 30)7
0021- 0007 < 0 85<35)(0-8g400(510X600 - 400)
0.014 < 0.0177V * f,e; _ 0.003
C .^O.OOstc - SO)es =c
V = e.' E,,, 0.003(c -50X200 000)ft =,, 600 (c - 50)'• = .— :—cT = C, + C2A, fy = 0.85 fc' ab + Al' fs'3000(400) = 0.85(35)(0.81) c (280)
x 1000(600)(c - 50)c
1200000c = 6747.3c3 + 600000c -6747.3c2 - 600000c - 30,000,000= 0
c2 - 88.92c - 4446.22 = 088.92 ± 160.29c =
2c = 124.60 mma = 6 c3 = 0.81(124.60)a = 100.93 mm (depth of compresS*£in
/
-X -
glNFORCnD CONCRETE DESIGN 249
® Actual stress of compression bar.,, 600(c - 50)
m f*
=' c
V600(124.60 - 50)
!0124.60
fs' = 359.23 MPa
) 1@ Design moment capacity of the beam.
M2 = A;fs' (d - d')M2 = 1000(259.23)(510 - 50)M2 =165.25 kN.m
11X600 - 41]
M, = C, (d - a/2)M, = 0.85 fc' ab' (d - a/2)M. = 0.85(35)(100.93)(280)(510 - 100.93/2)M, = 386.35 kN.m
Mu = 0 (M, + M2)Mu = 0.90(165.25 + 386.35)Mu = 496.44 kN.m
ANALYSIS OFT-BEAMS
A. Steel Area required with NeutralAxis on the Flange:
C|2f»Givendata:
Design moment MuThickness of flange tWidth of web bwEffective depth of beam . . dConcrete strength f 'Yield strength of steel . . . \Span of beam L
0«»Xio f \ \1
Required:a) Effective width of flange bt) Depth of compression block . ac) Steel area required As
a) Effective Width of Flange:
A. FORT-SECTIONS:
Limits of the effective flange with"b" as controlled by NSCP:
Use the smallest value of "b"
Slab ' Flangel/\ Uv- b
u 5 MLWeb orstem —
b'Spacing o( Beams |
d
i
1© b = ^ of span
© b = 16t + b„© b = center to center spacing
of beams1'i
B. FOR L-SHAPED FLANGE:
d
b
-MM-Use smallest value of "b"
© b = J2 sPan + b
© b = b' + 6tU - u T ui
© b = y center to center spacing of
beams