University of Sheffield Advanced Concrete Design Excel Sheets
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Transcript of University of Sheffield Advanced Concrete Design Excel Sheets
1)a Calculate the estimated elastic modulus at the age of 20 years (using Eurocode 2).For a concrete element (cement class N) at the age of 28 days,
(fck) 30 N/mm2t 40 years => 14600S 0.25 (Class N)e 2.71828βcc (t) 1.270
fcm 38 Mpafcm(t) = βcc(t) fcm 48.262
Ecm = 22 [(fcm)/10]^0.3 32.84 kN/mm^2
Ecm(t) = (fcm(t) / fcm)^0.3* Ecm 35.28 Gpa
2)a Determine the expected deformation if the concrete was at the age of 10 yearslateral elastic deflection 2.5 cm
(fck) N/mm2t 20 years => 7300S 0.2 (Class N) page 7 handout 2e 2.71828βcc (t) 1.206
fcm 8 Mpafcm(t) = βcc(t) fcm 9.651
Ecm = 22 [(fcm)/10]^0.3 20.58 kN/mm^2
Ecm(t) = (fcm(t) / fcm)^0.3* Ecm 21.77 Gpa
Ecm / Ecm(t) 0.945
deformation 2.36 Ecm / Ecm(t)*deflectionanswer:
another way 2.36 deflection/(βcc( t))^0.3
3) a Calculate deflection if M0 = 2Mcr and k unracked / k cracked= 0.4Mcr 1If M0 /Mcr <1 0.3 3.333 deformationwhen M0 /Mcr >1 1.5 so
k unracked / k cracked 0.5
Hence 10uncracked 33cracked 50.0 100.0
coefficient β 0.5 page 8 haout 2ς 0.778
answer: δ 88.9 mm
4)a Calculate the maximum value of point P that will lead to cracking (according toEurocode 2). (Material partial safety factors, γm = 1.0)
L 2 mCharacteristic cylinder strength fck 70 Mpa
γmb 300 mmwidth 300 mm
σcrack = fctmif fctm =0,30 fck^(2/3) ≤ C50/60 (in N/mm2)
σ crack 5.10if Not then fctm=2.12 ln(1+(fcm/10)) > C50/60 (in N/mm2)fctm 4.61moment I 675000000 mm^4
y 150Mcr 22929739.9053 20747131.423F 11464870 N 6915710.47432
F 11.46 KN 6.92
5) aCalculate the elastic (uncracked) defoδP 6000 N
Mean cylinder strength fcm 48 Mpafck 40 MpaEcm 35.220b 300 mmwidth 300 mmL 4 m
moment I 675000000 mm^4
K 0.3333 from tabledeformation δ 5.38353714E-06
δ 5.4 mm
6) a Calculate the final value of total free shrinkage strai(t=∞). sides 3f ck 60 Mpa
b 150 mmwidth 150 mmhumidity 50 % use table page 11
ε ca (un limit ) 0.000125Ac 22500u 450 make sure use perimeter of the sidesh0 100 from table Kh =
page 12 handout 2
ε cd0 0.42 Interpolationε cd1 0.00042 from page 11 tableε cd (un limit ) 0.000420
Hence,ε cs 0.000545
7) a calculate the compressive stress in the steel for the compositesection shown in the figure:
shrinkage strain εcs 0.0004Elastic modulus of concrete 40 KN/mm^2Elastic modulus of reinforcement 200 KN/mm^2
b 250 mmwidth 250 mm
4 steel bar diameter 20 mmAc 62500As 1256Fcs 100.48 KNεct 3.652224484E-05
εsc = εcs - εct εsc 0.000363477755Fs 91.31 KN
72.70 N/mm
8)a calculate short term deflection at mid span due to a load P.‐0.6 *fyk
L 2400 mmNeutral axis depth 70 mm
d 250 mmfyk 500 Mpa
Elastic modulus of reinforcement 200 KN/mmXs 180
300 MPa0.0015 N/mm^2
1/r K 8.333333333E-06 mmfrom table K (1/12) 0.083
δ 4.0 mm
εs
S
S
9)a For the following concrete section, calculate the drying shrinkage strain (εcd ) one year after the end of curing.one year 365 days
Characteristic concrete cylinder strength (fck) 80
Relative humidity (RH) 70table page 11 0.20
εcd,0 0.000195
εcd (t) =β (t, t ) kh* εcd,0 ?number of sides 4
b 300 mmwidth 300 mmAc 90000 mm^2
u 1200 mmh0 150 go to table page 12kh 0.93 go to table page 12β (t, t ) 0.83
εcd (t) 0.0001510
10)a For the following concrete section, calculate long term total shrinkage strain (ε cs ).‐Characteristic concrete cylinder strength (fck) 80
Relative humidity (RH) 70(t=∞). sides 4f ck 80 Mpa
b 300 mmwidth 300 mm
humidity 70 % use table page 110.000175
Ac 90000
u 1200 make sure use perimeter of the sides
h0 150 from table Kh = page 12 handout 2
ε cd0 0.2
ε cd1 0.0002 from page 11 table0.000186
Hence,ε cs 0.000361
11)a for following RC column, calculate long term deflection due to creep under compressive load P=500 KN applied at concrete age t0= 10 days.‐number of sides 4
P 500t0 10
Characteristic concrete cylinder strength (fck) 40Elastic modulus of concrete (Ec) 40
)(Ca
)(Cd
Cement is Class N s 0.25Relative humidity (RH) 50
b 200width 200
L 5Ac 40000u 800ho 100
from table page 15 2.50.0125
ε, cc 0.00078125
ε= x/L x 3.91
12)a For the following RC column, calculate long term deflection due to creep under compressive load P=1200 kN applied at concrete age t0= 10 days‐number of sides
Pt0
Characteristic concrete cylinder strength (fck)Mean concrete cylinder strength (fcm) at the time of loadingElastic modulus of concrete (Ec)
Cement is Class N sRelative humidity (RH)
bwidthLAcuh0
from table page 15
ε, cc
ε= x/L x
13)a For the following concrete cracked section, calculate long term total shrinkage curvature (1/r cs ).‐Characteristic concrete cylinder strength (f ck)Elastic modulus of concrete (Ecm)Creep coefficientDrying shrinkage strain (ε,cd)Moment of inertia for the cracked section (Icr)Elastic modulus of reinforcement, EsFirst moment of area of the reinforcement for the cracked section (S)
),( 0t
),( 0t
),( 0tK
K
long term total shrinkage curvature (1/r cs ).‐
14)a For the following RC beam, calculate total long term deflection at mid span. All loads are permanent‐L 3000
Characteristic steel yield strength (f yk ) 500Elastic modulus of reinforcement, Es 200Shrinkage curvature at mid span (1/rcs)= 6.00E-06Long term tension stress in reinforcement at mid 0.5span due to permanent loads (top Figure) 250effective depth d 500Neutral axis depth x 250
(d-x) 250K 0.104
strain= stress/Es => hence => ε 0.00125K 0.000005
Total K 1.10E-05
total long term deflection at mid span‐ 10.3
17)a Based on Eurocode 2 (2004), calculate the design flexural capacity of the section shown below (using material partial safety factors).NEW Euro Code 1
Characteristic steel yield strength (fyk) 500Characteristic concrete cylinder strength (fck) 33
steel bar diameter 20number of bars 2
b 150d 210h 250
1.151.5
434.7822
As 628Ac 37500
from Fs=FcFs 273043.48Fc 3300
Fs 273043.48 (=)Hence: X= 82.74
lever Arm Z= 168.63
M 46
totalK
SC
ydfCdf
18)a Based on Eurocode 2 (2004), calculate the predicted (i.e. actual) flexural capacity of the section shown below (No material partial safety factors).Characteristic steel yield strength (fyk) 500Standard deviation for steel yield strength (fyk) 30Characteristic concrete cylinder strength (fck) 50Mean concrete cylinder strength (fcm) 58
b 200d 250h 250
diameter of bar 25number of bars 2
As 981.25X 46.46
lever Arm Z 226.77
M 122
19) a Based on Eurocode 2 (2004), calculate the design flexural capacity of the section shown below (using material partial safety factors).NEW Euro Code 1
Characteristic steel yield strength (fyk) 500Characteristic concrete cylinder strength (fck) 33
steel bar diameter 22number of bars 3
b 400width of top rectangular w 150
cover+half dia 30d 570h 600
1.151.5
434.7822
As 1139.82Ac 240000
from Fs=FcFs 495573.91Fc 8800
Fs 495573.91 (=)Hence: X= 56.32
lever Arm Z= 541.84
M 269
21)a Based on Eurocode 2 (2004), calculate the design flexural capacity of the section shown below (using material partial safety factors).NEW Euro Code 1
SC
ydfCdf
Characteristic steel yield strength (fyk) 500Characteristic concrete cylinder strength (fck) 38
steel bar diameter 32number of bars 2
b 200width of top rectangular w 150
cover+half dia 30d 570h 600
1.151.5
fsd = 434.7825.3333333333
As 1607.68Ac 120000
from Fs=FcFs 698991.30Fc 5066.66666667
Fs 698991.30 (=)Hence: X= 137.96
lever Arm Z= 501.02M 350
22)a Based on Eurocode 2 (2004), the flexural capacity of the section shown below is 60 kN.m for the concrete cylinder strength (fcm) of 40 MpaEstimate the flexural capacity of the section if the concrete cylinder strength (fcm) becomes 30 MPa.
M 60 KNmLa Z 180 mm
Fs 333.333333333 Nfcm second (lower value) 30
stress block 74.07half of stress block 37.04
new lever arm 172.96b 150d 210h 300
ANSWER 57.65
27)a Based on Eurocode 2 2004), calculate the design shear capacity of the section shown below using material
SC
ydf
Cdf
Characteristic steel yield strength H20 (fyk)Characteristic concrete cylinder strength (fck)According to Eurocode, Min shear resistance (VRd,cmin)
buse slide 30 page 9 (exact formula) dNo link h
number of barsdiameter
As,l
Eq, 6.2 Vrd,c
28)a Based on Eurocode 2 (2004), calculate the design shear capacity of the section shown below (using material partial safety factors).Characteristic steel yield strength R8 (fyk) = 250Assume internal lever arm for shear reinforcement (z) 210
cot θ 2.5Characteristic concrete cylinder strength (fck) 50Design value of concrete compressive strength (fcd) 33
s 200use slide 34 page 11 d 8there is link cot 0
Sin 1
A,sw 50.24
Eq, 6.8 VRd,s 57
29)a Based on Eurocode 2 (2004), calculate the predicted shear capacity of the section shown below (No material partial safety factors).Mean steel yield strength H20 (fym)Characteristic concrete cylinder strength (fck)Mean concrete cylendrical strength (fcm)According to Eurocode, Min shear resistance (VRd,cmin)
buse slide 30 formula dinstead of fck use fcm hand instead of 0.12 use 0.18 diameter
number of barAs,l
VRd,s
30)a Based on euro code 2 2004 calculate the predicted shear capacity of the section shown below (No material partial safety factors).Mean steel yield strength R8 (fm) 280Internal lever arm for shear reinforcement 210
Characteristic concrete cylinder strength (fck) 50Mean concrete cylendrical strength (fcm) 58
cot θ 2.5b 200
use fcm instead of fywd d 250page 11 h 150
s 200diameter of link 8
cot 0Sin 1A,sw 50.24
VRd,s 74
31)a Based on Eurocode 2 (2004), calculate the spacing of the stirrups (s) to achieve the maximum shear capacity of the section shown below.Characteristic steel yield strength R8 (fyk) 500θ= 21.8 Assume cot θ => 2.5Characteristic concrete cylinder strength (fck) 50
b 200d 250
page 11, slide 36 hdiameter of link R8 8
diameter of Barnumber of bar
Asw 50.24cot 0Sin 1
VRd, s = VRd, min 248Z 225
fywd 435
S 99.1
32)a Based on Eurocode 2 (2004), calculate the minimum mean concrete cylinder strength (fcm) so that the section shown below does not require shearreinforcement to carry an unfactored shear load of 50 kN (no material partial safety factors).
Mean steel yield strength H20 (fym) 500VRd,c = P 90
b 200slide 30, d 250use fcm instead of fck h 250
diameter of the bar 25number of the bar 2
Asl 981.25
mean concrete cylinder strength fcm 75
36)a When the following RC section is subjected to the design ultimate moment, the neutral axis depthCalculate the strain in the longitudinal reinforcement at this stage:
Ultimate compressive strain in the concrete (ε cu ) 0.0035neutral axis depth (x) 300
hbottom flang 30
d -30
εs -0.00385
37)a Calculate the design bending moment for the following RC section. Neutral axis depth (x) is equal to 300mm when section is subjected to the designbending moment. 0.85
f yk 500 Mpaf ck 33 Mpa
Neutral axis depth (x) x 200 mm0.8 X = y 160 mm
top area b 300 mmh 400 mm
bottom flange 30 mmtop flange 100 mm
Web 100 mmd 370 mm
for calculating area 2 y2, 60 mmy1 110 mm
area 1 A1 30000 mm2y2 30 mm
area 2 A2 6000 mm2centroyed y 96.67 mm
lever arm Z 306.67 mmFc 673200
M 206 KNm
38) a For the following RC section, calculate the moment of inertia of the transformed cracked section (I cr ).Elastic modulus of concrete Ec 40 KN/mm2Elastic modulus of reinforcement Es 200 KN/mm2
b 200d 250h 300dc 1 assume
number of bars 2bars diameter 32
Ec/Es 5As 1608
100quadratic equ. A
2d
b d 8038.400c 2017638.4
quadratic equ. d1= 107.43 use this oned2= -188
moment of inertia Icr 24604.82 cm
39)a Calculate the cracking moment of the following section (Mcr).Elastic modulus of concrete 40 KN/mm2Elastic modulus of reinforcement 200 KN/mm2Characteristic concrete cylinder strength (fck) 40 Mpa
b 300 mmd 550 mmh 600 mm
bar diameter 32No of bars 4
area of 1 bar As 803.84d` 50 mm
depth Nutural Axis 300 mmx 250 mm
Es/Ec 5Ig 6404800000
0.3fck^(2/3) => 3.51
Mcr 75 KNm
40)a For the following RC section, at the service load of Mserv=35 kNm, the tension stress in the reinforcement is equal to 0.6f yk . Calculate the maximumcompressive stress in the concrete (sc) if an elastic concrete stress distribution (linear) is assumed.
Mser 61 KNm0.6f yk 300 Mpa
Characteristic steel yield strength (fyk) 500 Mpab 300 mmd 350 mmh 400 mm
bar diameter 20 mmNo, bar 2
As 628 mm2Fs 188400
lever arm Z 324X 26.22
Ac 11799.36
Fc=Ac* => 15.97 Mpa
2d
S
Cr
41) a For the following RC section, for serviceability limit states, calculate the amount of longitual reinforcement (As) for which the maximum compression stressin concrete and tension stress in reinforcement reach the Eurocode2 (2004) stress limits simultaneously (Assume triangular distribution for concretestress).
Characteristic steel yield strength (fyk) 500 MpaElastic modulus of reinforcemen Es 200 KN/mm2Characteristic concrete cylinder strength (fck) 33 MpaElastic modulus of concrete Ec 40 KN/mm2
b 300 mmd 350 mmh 0 mm
19.8300
0.0004950.002
X 69Fc 206.23 KN
As 516 mm2
42)a For the following RC section, at the service load, the tension stress in the reinforcement is equal to 0.6f yk . Calculate the curvature of the sectionCharacteristic steel yield strength (fyk) 500 MpaElastic modulus of reinforcement= 200 KN/mm2Neutral axis depth 65 mm
b 200 mmd 250 mmh 300 mm
0.6f yk 300 Mpa0.0015
X , tension 185
Curvature 8.108108108E-06
43)a For the following RC section, at the service load, the tension stress in the reinforcement is equal to 0.6f yk . Calculate the amount of longitual reinforcement(As) if an elastic concrete stress distribution is assumed
Neutral axis depth 65 mmCharacteristic steel yield strength (fyk) 500 MpaElastic Modulus of Reinforcement, Es 200 KN/mm2Elastic modulus of concrete, Ec 40 KN/mm2
b 200 mmd 250 mmh mm
S
CS
S
XrK C
1
C
S
300 Mpa0.0015
Fc=Fs => 0.00052702702721.08 Mpa
As 457 mm
44)a For the following RC section subjected to the service load, the tension stress in the reinforcement and compression stress in the concrete are 0.6f yk and
Characteristic steel yield strength (fyk) 500 MpaCharacteristic concrete cylinder strength (fck) 50 MpaElastic modulus of reinforcement 200 KN/mm2
b 200d 250h
bar diameter 18No, of Bar 2
3000.5fyk 25
As 508.68 mm2Fs 152604Xc 61.04 mm
0.0015Xs 188.96 mm
Curvature 7.938255193E-06
46)a For a confined column, the effective confinement ratio ( ) is equal to 0.15. CalcuMean concrete cylinder strength (f cm ) 35Strain at peak stress c,o)= 0.002Ultimate strain ( c,85) 0.0035
? 0.0539 Mpa
47)a For a confined column, the effective confinement ratio ( ) is equal to 0.15. Calculate the ultimate concrete strain for the confined concrete ( c.85 * ).Mean concrete cylinder strength (f cm ) 50 MpaUltimate strain ( c,85 0.0035
0.050.0085
48)a For a confined column, the effective confinement ratio ( ) is equal to 0.3. Calculate the strain at peak stress for the confined concrete ( c.0 * ).
0.5f ck respectively. Calculate the curvature of the section if triangular concrete stress distribution (elastic) is assumed.
S
CS
S
S
S
s
XrK
1
W*fcc*fcc
85,*CW
Mean concrete cylinder strength (f cm )= 50 MpaStrain at peak stress ( c,o 0.002Ultimate strain ( c,85) 0.0035
0.0556
0.00253125
49)a For the confined column shown below, calculate the volumetric ratio of confinement ( ). The whole column section is confined with an effectiveconfinement ratio 0.2The shear link spacing S 200 mm
b 350 mmwidth 350 mm
0.5100.6670.340
= 0.20.2 0.340 w
w 0.59
50) a For the confined column shown below, calculate the required cross section area for steel stirrups. The whole column section is confined.Volumetric confinement ratio ( )= 0.5The shear link spacing (S)= 200 mmYield stress of the stirrups (f sy)= 600 MpaThe concrete cylinder strength (f cc ) 35 Mpa
b 350 mmwidth 350 mm
Vc 24500000Vs 714583.3333333
one side 247.49 mmAs1 989.95As 299 mm
55) The curvature ductility of an RC column was found to be 5 and 15, for an unconfined section and a section with an effective confinement ratio ( ) of 0.1,respectively. Estimate the curvature ductility of the same column for the effective confinement ratio ( ) of 0.2.
/ 5 = 5
0.10.2
0.020.01
0*C
*fccW
W
Sn
W
W
85 0C85 0C
0C
W
0C
W
)01.05( 0c
1510
= 0.001 = 0.005
Curvature Ductility => 25
56)a From section analysis, the following information is provided. Calculate the curvature ductility for confined concrete.Unconfined sectionYield curvature (kyu) 3.50E-05Ultimate curvature (kuu) 4.50E-05Confined sectionUltimate curvature (kuc) 2.00E-04
5.714285714286
57) a From section analysis, the following information is provided for the section shown below. Calculate the curvature ductility for this section.Yield pointMaximum compressive strain in concrete (ε c ) 0.001Tension strain in reinforcement (ε s ) 0.0025Ultimate pointMaximum compressive strain in concrete (ε c )= 0.0035Tension strain in reinforcement (ε s )= 0.015
b 200d 400h 450
Yield point (x/y)=> 0.4y 286X 114
0.00000875
Ultimate point(x/y)=> 0.23
y 324.32X 75.68
0.00004625
Final K 5.3
58 - 59mu k 25 ax2+bx+c=0mu displacement 8a 12b -37c 7delta 32.1403173599764
Kyeild
Kultimate
0C
0C0C
85
K
)01.05( 0c
landa 1 2.88084655666568landa 2 0.20248677666765
455651
My Kn.m
Mult for unconfinedMult for confined
1)a Calculate the estimated elastic modulus at the age of 20 years (using Eurocode 2).
2)a Determine the expected deformation if the concrete was at the age of 10 years
page 7 handout 2
3) a Calculate deflection if M0 = 2Mcr and k unracked / k cracked= 0.4
deformation10 33
at this stage is = to Mcrk unracked < k cracked
4)a Calculate the maximum value of point P that will lead to cracking (according tofcm
78
N
KN
6) a Calculate the final value of total free shrinkage straiεInterpolation
RH1 RHcd RH240 50 60
use table page 11 ε 1 ε cd0 ε 2
0.46 0.42 0.38
make sure use perimeter of the sides1
h01 h0 h02100 100 200
k1 kh k2
1 1.00 0.85
7) a calculate the compressive stress in the steel for the composite
8)a calculate short term deflection at mid span due to a load P.‐
9)a For the following concrete section, calculate the drying shrinkage strain (εcd ) one year after the end of curing.
Mpa Interpolation
% RH1 RHcd RH260 70 80
ε 1 ε cd0 ε 2
0.24 0.20 0.15
Interpolationh01 h0 h02100 150 200
k1 kh k2
1 0.93 0.85go to table page 12go to table page 12
10)a For the following concrete section, calculate long term total shrinkage strain (ε cs ).‐Mpa Interpolation
% RH1 RHcd RH260 70 80
ε 1 ε cd0 ε 2
0.24 0.20 0.15Interpolation
use table page 11 h01 h0 h02100 150 200
k1 kh k2
make sure use perimeter of the sides 1 0.93 0.85
0.93
11)a for following RC column, calculate long term deflection due to creep under compressive load P=500 KN applied at concrete age t0= 10 days.‐KNdaysMpaKN/mm
page 7%
m
Look table
mm
12)a For the following RC column, calculate long term deflection due to creep under compressive load P=1200 kN applied at concrete age t0= 10 days‐4
200010 days50 Mpa58 Mpa40 KN/mm
0.25 page 750 %
200200
5 m40000
800100
20.86206897
503.71
4.63851457
23.19 mm
13)a For the following concrete cracked section, calculate long term total shrinkage curvature (1/r cs ).‐60 Mpa40 KN/mm^2
33.50E-044.00E+09 mm4
200 kN/mm25.00E+05 mm3
),( 0t
0.0001254.75E-04
10.0020.00
1.19E-06
14)a For the following RC beam, calculate total long term deflection at mid span. All loads are permanent‐mmMpakN/mm2
fyk
mmmmfrom table figur 6 , page 6
mm
17)a Based on Eurocode 2 (2004), calculate the design flexural capacity of the section shown below (using material partial safety factors).BASED ON EURO 2004
Mpa Characteristic steel yield strength (fyk)Mpa Characteristic concrete cylinder strength (fck)mm steel bar diameter
number of barsbdh
mm2 Asmm2 Ac
from Fs=FcFs
*X Fc3300 *X Fc Fs 426630.43
mm Hence: X=lever Arm Z=
KNm M
)(CaCSeffC .e
SC
ydfCdf
18)a Based on Eurocode 2 (2004), calculate the predicted (i.e. actual) flexural capacity of the section shown below (No material partial safety factors).MpaMpaMpaMpa
mm
mm2mm
KN
19) a Based on Eurocode 2 (2004), calculate the design flexural capacity of the section shown below (using material partial safety factors).BASED ON EURO 2004
Mpa Characteristic steel yield strength (fyk)Mpa Characteristic concrete cylinder strength (fck)mm steel bar diameter
number of barsbtf
mm coverdh
mm2 Asmm2 Ac
from Fs=FcFs
*X Fc8800 *X Fc Fs 330382.61
mm Hence: X=lever Arm Z=
KNm M
21)a Based on Eurocode 2 (2004), calculate the design flexural capacity of the section shown below (using material partial safety factors).BASED ON EURO 2004
SC
ydfCdf
Mpa Characteristic steel yield strength (fyk)Mpa Characteristic concrete cylinder strength (fck)mm steel bar diameter
number of barsbw
mm cover+half diadh
mm2 Asmm2 Ac
from Fs=FcFs
*X Fc5066.66667 *X Fc Fs 426630.43mm Hence: X=
lever Arm Z=KNm M
22)a Based on Eurocode 2 (2004), the flexural capacity of the section shown below is 60 kN.m for the concrete cylinder strength (fcm) of 40 Mpaflexural capacity of the section if the concrete cylinder strength (fcm) becomes 30 MPa.
27)a Based on Eurocode 2 2004), calculate the design shear capacity of the section shown below using material
SCydfCdf
500 Mpa70 Mpa
3.8 KN200 mm240 mm250 mm
220
628 mm2
49.7 KN
28)a Based on Eurocode 2 (2004), calculate the design shear capacity of the section shown below (using material partial safety factors).Mpamm Z=0.9D
MpaMpammmm
KN
29)a Based on Eurocode 2 (2004), calculate the predicted shear capacity of the section shown below (No material partial safety factors).500 Mpa
70 Mpa78 Mpa
4 KN200 mm240 mm250 mm
202
628 mm2
77 KN
30)a Based on euro code 2 2004 calculate the predicted shear capacity of the section shown below (No material partial safety factors).Mpamm
MpaMpa
mmmmmm
KN
31)a Based on Eurocode 2 (2004), calculate the spacing of the stirrups (s) to achieve the maximum shear capacity of the section shown below.Mpa
Mpammmmmmmmmm
KNmm
mm
32)a Based on Eurocode 2 (2004), calculate the minimum mean concrete cylinder strength (fcm) so that the section shown below does not require shear
MpaKNmmmmmm
mm2
Mpa
36)a When the following RC section is subjected to the design ultimate moment, the neutral axis depth
mmmmmmmm
37)a Calculate the design bending moment for the following RC section. Neutral axis depth (x) is equal to 300mm when section is subjected to the design
38) a For the following RC section, calculate the moment of inertia of the transformed cracked section (I cr ).
39)a Calculate the cracking moment of the following section (Mcr).
40)a For the following RC section, at the service load of Mserv=35 kNm, the tension stress in the reinforcement is equal to 0.6f yk . Calculate the maximumcompressive stress in the concrete (sc) if an elastic concrete stress distribution (linear) is assumed.
Linear
41) a For the following RC section, for serviceability limit states, calculate the amount of longitual reinforcement (As) for which the maximum compression stressin concrete and tension stress in reinforcement reach the Eurocode2 (2004) stress limits simultaneously (Assume triangular distribution for concrete
42)a For the following RC section, at the service load, the tension stress in the reinforcement is equal to 0.6f yk . Calculate the curvature of the section
43)a For the following RC section, at the service load, the tension stress in the reinforcement is equal to 0.6f yk . Calculate the amount of longitual reinforcement
44)a For the following RC section subjected to the service load, the tension stress in the reinforcement and compression stress in the concrete are 0.6f yk and
46)a For a confined column, the effective confinement ratio ( ) is equal to 0.15. Calcu Confined Concrete
47)a For a confined column, the effective confinement ratio ( ) is equal to 0.15. Calculate the ultimate concrete strain for the confined concrete ( c.85 * ).
48)a For a confined column, the effective confinement ratio ( ) is equal to 0.3. Calculate the strain at peak stress for the confined concrete ( c.0 * ).
respectively. Calculate the curvature of the section if triangular concrete stress distribution (elastic) is assumed.
W
W
W
49)a For the confined column shown below, calculate the volumetric ratio of confinement ( ). The whole column section is confined with an effective
50) a For the confined column shown below, calculate the required cross section area for steel stirrups. The whole column section is confined.
55) The curvature ductility of an RC column was found to be 5 and 15, for an unconfined section and a section with an effective confinement ratio ( ) of 0.1,respectively. Estimate the curvature ductility of the same column for the effective confinement ratio ( ) of 0.2.
W
W
56)a From section analysis, the following information is provided. Calculate the curvature ductility for confined concrete.
57) a From section analysis, the following information is provided for the section shown below. Calculate the curvature ductility for this section.
mmmmmm
mmmm
mmmm
9)a For the following concrete section, calculate the drying shrinkage strain (εcd ) one year after the end of curing.
11)a for following RC column, calculate long term deflection due to creep under compressive load P=500 KN applied at concrete age t0= 10 days.‐
look at the right side
12)a For the following RC column, calculate long term deflection due to creep under compressive load P=1200 kN applied at concrete age t0= 10 days‐
13)a For the following concrete cracked section, calculate long term total shrinkage curvature (1/r cs ).‐
14)a For the following RC beam, calculate total long term deflection at mid span. All loads are permanent‐
17)a Based on Eurocode 2 (2004), calculate the design flexural capacity of the section shown below (using material partial safety factors).0.85
500 Mpa50 Mpa25 mm
2200250300
1.151.5
434.7833.3333333
981.25 mm260000 mm2
426630.435666.66667 *X
(=) 5666.667 *X Fc75.29 mm
212.36
91 KNm
18)a Based on Eurocode 2 (2004), calculate the predicted (i.e. actual) flexural capacity of the section shown below (No material partial safety factors).
19) a Based on Eurocode 2 (2004), calculate the design flexural capacity of the section shown below (using material partial safety factors).0.85 0.85
500 Mpa 50050 Mpa 5022 mm 22
2 2300 b 300100 100
30 30370 d 370400 h 400
1.15 1.151.5 1.5
434.78 434.7828.3333333 22
759.88 mm2 As 759.88120000 mm2 Ac 120000
from Fs=Fc330382.61 Fs ###
8500 *X Fc 6600(=) 8500 *X Fc Fs 330382.61 (=)
38.87 mm Hence: X= 50.06350.57 lever Arm Z= 344.97
116 KNm M 114
21)a Based on Eurocode 2 (2004), calculate the design flexural capacity of the section shown below (using material partial safety factors).0.85
SC
ydfCdf
500 Mpa33 Mpa25 mm
2150100
30520550
1.151.5
434.7822
981.25 mm282500 mm2
426630.433300 *X
(=) 3300 *X Fc129.28 mm455.36
194 KNm this is right answers
22)a Based on Eurocode 2 (2004), the flexural capacity of the section shown below is 60 kN.m for the concrete cylinder strength (fcm) of 40 Mpa
27)a Based on Eurocode 2 2004), calculate the design shear capacity of the section shown below using material Shear Design
28)a Based on Eurocode 2 (2004), calculate the design shear capacity of the section shown below (using material partial safety factors).
29)a Based on Eurocode 2 (2004), calculate the predicted shear capacity of the section shown below (No material partial safety factors).
30)a Based on euro code 2 2004 calculate the predicted shear capacity of the section shown below (No material partial safety factors).
31)a Based on Eurocode 2 (2004), calculate the spacing of the stirrups (s) to achieve the maximum shear capacity of the section shown below.
32)a Based on Eurocode 2 (2004), calculate the minimum mean concrete cylinder strength (fcm) so that the section shown below does not require shear
section analysis
37)a Calculate the design bending moment for the following RC section. Neutral axis depth (x) is equal to 300mm when section is subjected to the design
38) a For the following RC section, calculate the moment of inertia of the transformed cracked section (I cr ).
40)a For the following RC section, at the service load of Mserv=35 kNm, the tension stress in the reinforcement is equal to 0.6f yk . Calculate the maximum
41) a For the following RC section, for serviceability limit states, calculate the amount of longitual reinforcement (As) for which the maximum compression stress
42)a For the following RC section, at the service load, the tension stress in the reinforcement is equal to 0.6f yk . Calculate the curvature of the section
43)a For the following RC section, at the service load, the tension stress in the reinforcement is equal to 0.6f yk . Calculate the amount of longitual reinforcement
44)a For the following RC section subjected to the service load, the tension stress in the reinforcement and compression stress in the concrete are 0.6f yk and
Confined Concrete
47)a For a confined column, the effective confinement ratio ( ) is equal to 0.15. Calculate the ultimate concrete strain for the confined concrete ( c.85 * ).
48)a For a confined column, the effective confinement ratio ( ) is equal to 0.3. Calculate the strain at peak stress for the confined concrete ( c.0 * ).
49)a For the confined column shown below, calculate the volumetric ratio of confinement ( ). The whole column section is confined with an effective
50) a For the confined column shown below, calculate the required cross section area for steel stirrups. The whole column section is confined.
55) The curvature ductility of an RC column was found to be 5 and 15, for an unconfined section and a section with an effective confinement ratio ( ) of 0.1,respectively. Estimate the curvature ductility of the same column for the effective confinement ratio ( ) of 0.2. W
W
56)a From section analysis, the following information is provided. Calculate the curvature ductility for confined concrete.Ductility
57) a From section analysis, the following information is provided for the section shown below. Calculate the curvature ductility for this section.
MpaMpamm
mm2mm2
*X6600 *X Fc
mm
KNm
40)a For the following RC section, at the service load of Mserv=35 kNm, the tension stress in the reinforcement is equal to 0.6f yk . Calculate the maximum
41) a For the following RC section, for serviceability limit states, calculate the amount of longitual reinforcement (As) for which the maximum compression stress
43)a For the following RC section, at the service load, the tension stress in the reinforcement is equal to 0.6f yk . Calculate the amount of longitual reinforcement
44)a For the following RC section subjected to the service load, the tension stress in the reinforcement and compression stress in the concrete are 0.6f yk and
47)a For a confined column, the effective confinement ratio ( ) is equal to 0.15. Calculate the ultimate concrete strain for the confined concrete ( c.85 * ).
48)a For a confined column, the effective confinement ratio ( ) is equal to 0.3. Calculate the strain at peak stress for the confined concrete ( c.0 * ).
55) The curvature ductility of an RC column was found to be 5 and 15, for an unconfined section and a section with an effective confinement ratio ( ) of 0.1,