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Abt Wall

This document provides design calculations for the interior web of a cantilever beam for the Lyari Expressway project. The beam has a 1.35 m^2 cross-sectional area and is designed to resist an ultimate bending moment of 2965 kNm. Calculations are shown to check the nominal moment capacity, crack control, shear capacity, torsional resistance, and longitudinal reinforcement requirements. The provided reinforcement of 743 mm^2 is determined to satisfy the longitudinal reinforcement requirement.

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146 views9 pages

Abt Wall

This document provides design calculations for the interior web of a cantilever beam for the Lyari Expressway project. The beam has a 1.35 m^2 cross-sectional area and is designed to resist an ultimate bending moment of 2965 kNm. Calculations are shown to check the nominal moment capacity, crack control, shear capacity, torsional resistance, and longitudinal reinforcement requirements. The provided reinforcement of 743 mm^2 is determined to satisfy the longitudinal reinforcement requirement.

Uploaded by

edc1312
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as XLS, PDF, TXT or read online on Scribd
You are on page 1/ 9

LYARI EXPRESSWAY DESIGN OF CANTILEVER 1 x 42m SPAN

DESIGN OF INTERIOR WEB FOR LOCAL MOMENTS (OBTAINED FROM FRAME ACTION)
M V
KNm KN
Forces on the section SW 1.2 4.62
SDL 1.54 5.95
Ultimate Moment Mu = 2965 KNm LL 4 53
Service Moment Ms = 1977 KNm
Ultimate Shear Vu = 3391 KN Ult. 12 119
Ultimate Torsion Tu = 0 KNm Ser. 7 64

Section Definition Control Perimeters

Ax = 1.35 m2
F'c = 28.0 Mpa
Iz = 0.06328 m4 Fy = 414 Mpa
Yt = 0.375 m F = 0.9
Yb = 0.375 m Z = 23000 N / mm
h = 750 mm cover = 100 mm
bf = 1800 mm Es = 200000 Mpa
hf = 750 mm Fr = 3.3 Mpa
bw = 1800 mm Mcr = 563 KNm
1.2*Mcr = 675 KNm
1.33*Mu = 3943 KNm
Design Mu = 3270 KNm
b1 = 0.85
Ec = 25399 Mpa
h = 1.103

Check Nominal Moment Capacity Check Crack Control

Try Ms = 1977 KNm


As = 14730 mm2 Z = 23000 N / mm
A's = 0 mm2 As = 14730 mm2
F's = 0 Mpa r = 0.0126
d = 651 mm n = 7.9
d's = 100 mm j = 0.881
ecu = 0.003 dc = 56 mm
c = 167.5 mm A = 16800 mm2
a = 142.3 mm Fs = 234 Mpa OK
e's = 0.000 Fsa = 235 Mpa
F's = 0.0 Mpa
FMn = 3180 KNm N.G
Design Mu = 3270 KNm
c/d 0.257 (Under Reinforced)

Check for Shear Nominal Shear Capacity

Vu = 3740.273 KN Vc = 1102 KN
Mu = 3270.395 KNm Av = 904 mm2 OK
Tu = 0 KNm s = 100 mm OK
bv = 1800 mm Av/(s*bv) = 0.0050 mm2/ mm2
dv = 585 mm Vs = 3285 KN
Pu = 0 mm FVn = 3948 KN OK
Ao = 1 mm2 Vu = 3740 KN
Aoh = 1 mm2
Vu = 3740273 N Check for Torsion Resistance
v = 3.94 Mpa
v/ F'c = 0.141 Tcr = 620 KNm
q = 33.7 deree Acp = 1350000 mm2
b = 2.38 Pc = 5100 mm

ex =
1.42 x 10-3 At = 0 mm2
0.1*f'c*bv*dv = 2951 KN FTn = 0 KNm OK
ag 20 mm Tu = 0 KNm
sxe 570 mm
Check for Longitudinal Reinforcement

As*Fy = 6098 KN N.G


T = 9975 KN

Provide Vertical Flexural Reinforcement :


As = (1000/150)*(113) = 743 mm2

17:30:30
722656364.xls INT-WEB (2)
01/26/2024
LYARI EXPRESSWAY DESIGN OF CANTILEVER 1 x 42m SPAN

Formulas:

Determine unit shear stress:


For sections subjected to Torsion
v =(Vu-fVp)/(f*bv*dv) (5.8.2.9-1)
For sections not subjected to Torsion
v =(((Vu-fVp)/(f*bv*dv))^2+(Tu*Ph/(fAoh^2))^2)^0.5 (5.8.3.6.2-4)

Determine strain in reinforcement on tension side of the member:


ex = (Mu/dv + 0.5Nu +0.5(Vu-Vp)*cotq - Aps*fpo)/(2*(EsAs+EpAps)) <= 0.002 (5.8.3.4.2-1)
If ex < 0 (compression) then (Mu/dv + 0.5Nu +0.5(Vu-Vp)*cotq - Aps*fpo)/(2*(EcAc+EpAps+EsAs)) (5.8.3.4.2-3)
Where Vu = (Vu^2+(0.9*Ph*Tu/2/Ao)^2)^0.5 (For sections subjected to Torsion) (5.8.3.6.2-2)

From Table 5.8.3.4.2-1 find values of q and b as a function of concrete shear stress v and strain ex
Calculate shear capacity of concrete Vc
Vc = 0.083 b*sqrt(f'c)bv*dv (5.8.3.3-3)

Calculate stirrups shear capacity Vs


Vs = Av*fy*dv*cotq/S (5.8.3.3-4)

Calculate total shear capacity Vn, as


Lesser of:
Vn = Vc+Vs+Vp (5.8.3.3-1)
Vn = 0.25f'c*bv*dv+Vp (5.8.3.3-2)
fVn =min(Vn1, Vn2)

Maximum Spacing of Stirrups, S (5.8.2.7)

if Vu < 0.1 f'c*bv*dv then S<=0.8dv<=600 mm


if Vu >= 0.1 f'c*bv*dv then S<=0.4dv<=300 mm

Minimum shear reinforcing

Av/S >=0.083*Sqrt(f'c)*bv/fy (5.8.2.5-1)


Tn = 2*Ao*At*Fy*CotQ/s (5.8.3.6.2-1)
Longitudinal reinforcement requirements
As*fy+Aps*fps >= T
For sections not subjected to Torsion
As*fy+Aps*fps >= Mu/(dvfM)+0.5Nu/fN+(Vu/fV-0.5Vs-Vp)cotq (5.8.2.5-1)
For sections subjected to Torsion
As*fy+Aps*fps >= Mu/(dvfM)+0.5Nu/fN+ cotq*((Vu/fV-0.5Vs-Vp)^2+(0.45*Ph*Tu/2/Aof)^2))^0.5 (5.8.3.6.3-1)

17:30:30
722656364.xls INT-WEB (2)
01/26/2024
LYARI EXPRESSWAY DESIGN OF CANTILEVER 1 x 42m SPAN

Notations:

As Area of tensile steel


A's Area of compressive steel
h overall depth of member
d effective depth from extreme compression fiber to the centroid of tensile force
d's effective depth from extreme compression fiber to the centroid of compressive force
F's Stress in compressive steel at ultimate
Fs Stress in tensile steel at service
Fsa Allowable stress in tensile steel at service
dc Depth of the concrete meared from the extreme tension fiber to the centre of bar located closest thereto,
for the calculation purposees, the thickness of clear cover used to compute dc shall not be taken to be greater than 50 mm
A Area of the concrete having the same centroid as the principal tensile reinforcement and bounded by
the sufaces of the cross section and a straight lineparallel to the N.A., divided by the number of bars,
for the calculation purposees, the thickness of clear cover used to compute dc shall not be taken to be greater than 50 mm
r Reinforcement ratio
n Modular ratio
j Moment arm factor, for stress check at service
Av Area of shear reinforcing steel within S
bv Effective web width (5.8.2.7)
dv Effective shear depth taken as a distance between resultants of the tensile and compressive
forces due to flexure, but need not be taken less than greater of 0.9de or 0.72h
q angle of inclination of diagonal compressive stresses.
b concrete shear capacity factor
Ac area of concrete on the flexural tension side of the member (5.8.3.4.2 Fig. 3)
FTn Nominal torsion resistance
At Area of one leg of closed trans. torsion r/f
Vc Shear Capacity of concrete
Pu Perimeter of the CL of the closed trans reinf
Ao Area enclosed by shear flow path
Aoh Area enclosed by CL of ext closed trans torsion reinf
Ax Cross section Area
Iz Moment of inertia
Yt N.A. from top
Yb N.A. from bottom
bf Top flange width
hf Top flange thickness
bw Width of the web

17:30:30
722656364.xls INT-WEB (2)
01/26/2024
No. Rev.

By Date:
Project: Fly Over At Interchange on N-5 waqar
Chkd By Date
Subject: Abutment Wall waqar

DESIGN OF ABUTMENT WALL


Vertical Reinforcement

Forces on the section

Ultimate Moment Mu = 804 KNm


Service Moment Ms = 493 KNm
Ultimate Shear Vu = 317 KN
Ultimate Torsion Tu = 0 KNm

Section Definition Control Perimeters

Ax = 1 m2 F'c = 28.0 Mpa


Iz = 0.08333 m4 Fy = 414 Mpa
Yt = 0.5 m F = 0.9
Yb = 0.5 m Z = 23000 N / mm
h = 1000 mm cover = 75 mm
bf = 1000 mm Es = 200000 Mpa
hf = 1000 mm Fr = 3.3 Mpa
bw = 1000 mm Mcr = 556 KNm
Design Mu = 804 KNm
b1 = 0.85
Ec = 25399 Mpa

Check Nominal Moment Capacity Check Crack Control

Try Ms = 493 KNm


As = 2512 mm2 Z = 23000 N / mm
A's = 0 mm2 As = 2512 mm2
F's = 0 Mpa r = 0.0027
d = 925 mm n = 7.9
d's = 75 mm j = 0.938
ecu = 0.003 dc = 87.5 mm
c = 51.4 mm A = 3281 mm2
a = 43.7 mm Fs = 226 Mpa OK
e's = 0.000 Fsa = 248 Mpa
F's = 0.0 Mpa
FMn = 845 KNm OK
Design Mu = 804 KNm
c/d 0.056 (Under Reinforced)

Check for Shear Nominal Shear Capacity

Vu = 317 KN Vc = 793 KN
Pu = 0 mm FVn = 714 KN OK
Ao = 1 mm2 Vu = 317 KN
Aoh = 1 mm2
Vu = 317000 N Check for Torsion Resistance

v/ F'c = 0.014 Tcr = 434 KNm


q = 45 deree Acp = 1000000 mm2
b = 2 Pc = 4000 mm
ex = 0.00004 (LRFD_5.8.3.4.2-3)

ex = 1.04 x 10-3 At = 0 mm2


sxe 879 mm
No. Rev.

By Date:
Project: Fly Over At Interchange on N-5 waqar
Chkd By Date
Subject: Curtain Wall waqar

Forces on the section

Ultimate Moment Mu = 18 KNm


Service Moment Ms = 11 KNm
Ultimate Shear Vu = 21 KN
Ultimate Torsion Tu = 0 KNm

Section Definition Control Perimeters

Ax = 0.4 m2 F'c = 28.0 Mpa


Iz = 0.00533 m4 Fy = 414 Mpa
Yt = 0.2 m F = 0.9
Yb = 0.2 m Z = 23000 N / mm
h = 400 mm cover = 75 mm
bf = 1000 mm Es = 200000 Mpa
hf = 400 mm Fr = 3.3 Mpa
bw = 1000 mm Mcr = 89 KNm
Design Mu = 24 KNm
b1 = 0.85
Ec = 25399 Mpa

Check Nominal Moment Capacity Check Crack Control

Try Ms = 11 KNm
As = 791 mm2 Z = 23000 N / mm
A's = 0 mm2 As = 791 mm2
F's = 0 Mpa r = 0.0024
d = 325 mm n = 7.9
d's = 75 mm j = 0.941
ecu = 0.003 dc = 87.5 mm
c = 16.2 mm A = 3750 mm2
a = 13.8 mm Fs = 44 Mpa OK
e's = 0.000 Fsa = 248 Mpa
F's = 0.0 Mpa
FMn = 94 KNm OK
Design Mu = 24 KNm
c/d 0.050 (Under Reinforced)

Check for Shear Nominal Shear Capacity

Vu = 21.16 KN Vc = 279 KN
bv = 1000 mm Av/(s*bv) = 0.0000 mm2/ mm2
dv = 318 mm Vs = 0 KN
Pu = 0 mm FVn = 251 KN OK
Ao = 1 mm2 Vu = 21 KN
Aoh = 1 mm2
Vu = 21160 N Check for Torsion Resistance

v/ F'c = 0.003 Tcr = 99 KNm


q = 45 deree Acp = 400000 mm2
b = 2 Pc = 2800 mm
ex = 0.00001 (LRFD_5.8.3.4.2-3)

ex = 0.43 x 10-3 At = 0 mm2


sxe 310 mm
No. Rev.

By Date:
Project: Fly Over At Interchange on N-5 waqar
Chkd By Date
Subject: Wing Wall waqar

DESIGN OF ABUTMENT WALL


Horizontal Reinforcement

Forces on the section

Ultimate Moment Mu = 165 KNm


Service Moment Ms = 96 KNm
Ultimate Shear Vu = 104 KN
Ultimate Torsion Tu = 0 KNm

Section Definition Control Perimeters

Ax = 1 m2 F'c = 28.0 Mpa


Iz = 0.08333 m4 Fy = 414 Mpa
Yt = 0.5 m F = 0.9
Yb = 0.5 m Z = 23000 N / mm
h = 1000 mm cover = 75 mm
bf = 1000 mm Es = 200000 Mpa
hf = 1000 mm Fr = 3.3 Mpa
bw = 1000 mm Mcr = 556 KNm
Design Mu = 219 KNm
b1 = 0.85
Ec = 25399 Mpa

Check Nominal Moment Capacity Check Crack Control

Try 7 16 Dia Ms = 96 KNm


As = 1407 mm2 Z = 23000 N / mm
A's = 0 mm2 As = 1407 mm2
F's = 0 Mpa r = 0.0015
d = 925 mm n = 7.9
d's = 75 mm j = 0.952
ecu = 0.003 dc = 87.5 mm
c = 28.8 mm A = 3750 mm2
a = 24.5 mm Fs = 77 Mpa OK
e's = 0.000 Fsa = 248 Mpa
F's = 0.0 Mpa
FMn = 479 KNm OK
Design Mu = 219 KNm
c/d 0.031 (Under Reinforced)

Check for Shear Nominal Shear Capacity

Vu = 103.68 KN Vc = 802 KN
Pu = 0 mm FVn = 722 KN OK
Ao = 1 mm2 Vu = 104 KN
Aoh = 1 mm2
Vu = 103680 N Check for Torsion Resistance

v/ F'c = 0.005 Tcr = 434 KNm


q = 45 deree Acp = 1000000 mm2
b = 2 Pc = 4000 mm
ex = 0.00001 (LRFD_5.8.3.4.2-3)

ex = 0.41 x 10-3 At = 0 mm2


sxe 889 mm
No. Rev.

By Date:
Project: Fly Over At Interchange on N-5 waqar
Chkd By Date
Subject: Wing Wall waqar

DESIGN OF ABUTMENT WALL


Vertical Reinforcement

Forces on the section

Ultimate Moment Mu = 60 KNm


Service Moment Ms = 36 KNm
Ultimate Shear Vu = 317 KN
Ultimate Torsion Tu = 0 KNm

Section Definition Control Perimeters

Ax = 1 m2 F'c = 28.0 Mpa


Iz = 0.08333 m4 Fy = 414 Mpa
Yt = 0.5 m F = 0.9
Yb = 0.5 m Z = 23000 N / mm
h = 1000 mm cover = 75 mm
bf = 1000 mm Es = 200000 Mpa
hf = 1000 mm Fr = 3.3 Mpa
bw = 1000 mm Mcr = 556 KNm
Design Mu = 80 KNm
b1 = 0.85
Ec = 25399 Mpa

Check Nominal Moment Capacity Check Crack Control

Try 8 20Dia Ms = 36 KNm


As = 2512 mm2 Z = 23000 N / mm
A's = 0 mm2 As = 2512 mm2
F's = 0 Mpa r = 0.0027
d = 925 mm n = 7.9
d's = 75 mm j = 0.938
ecu = 0.003 dc = 87.5 mm
c = 51.4 mm A = 3281 mm2
a = 43.7 mm Fs = 16 Mpa OK
e's = 0.000 Fsa = 248 Mpa
F's = 0.0 Mpa
FMn = 845 KNm OK
Design Mu = 80 KNm
c/d 0.056 (Under Reinforced)

Check for Shear Nominal Shear Capacity

Vu = 317 KN Vc = 793 KN
Pu = 0 mm FVn = 714 KN OK
Ao = 1 mm2 Vu = 317 KN
Aoh = 1 mm2
Vu = 317000 N Check for Torsion Resistance

v/ F'c = 0.014 Tcr = 434 KNm


q = 45 deree Acp = 1000000 mm2
b = 2 Pc = 4000 mm
ex = 0.00002 (LRFD_5.8.3.4.2-3)

ex = 0.47 x 10-3 At = 0 mm2


sxe 879 mm
No. Rev.

By Date:
Project: Fly Over At Interchange on N-5 waqar
Chkd By Date
Subject: Wing Wall waqar

DESIGN OF ABUTMENT WALL


Horizontal Reinforcement

Forces on the section

Ultimate Moment Mu = 245 KNm


Service Moment Ms = 160 KNm
Ultimate Shear Vu = 200 KN
Ultimate Torsion Tu = 0 KNm

Section Definition Control Perimeters

Ax = 0.4 m2 F'c = 28.0 Mpa


Iz = 0.00533 m4 Fy = 414 Mpa
Yt = 0.2 m F = 0.9
Yb = 0.2 m Z = 23000 N / mm
h = 400 mm cover = 75 mm
bf = 1000 mm Es = 200000 Mpa
hf = 400 mm Fr = 3.3 Mpa
bw = 1000 mm Mcr = 89 KNm
Design Mu = 245 KNm
b1 = 0.85
Ec = 25399 Mpa

Check Nominal Moment Capacity Check Crack Control

Try 8 20 Dia Ms = 160 KNm


As = 2512 mm2 Z = 23000 N / mm
A's = 0 mm2 As = 2512 mm2
F's = 0 Mpa r = 0.0077
d = 325 mm n = 7.9
d's = 75 mm j = 0.902
ecu = 0.003 dc = 87.5 mm
c = 51.4 mm A = 3281 mm2
a = 43.7 mm Fs = 217 Mpa OK
e's = 0.000 Fsa = 248 Mpa
F's = 0.0 Mpa
FMn = 284 KNm OK
Design Mu = 245 KNm
c/d 0.158 (Under Reinforced)

Check for Shear Nominal Shear Capacity

Vu = 200 KN Vc = 266 KN
Pu = 0 mm FVn = 240 KN OK
Ao = 1 mm2 Vu = 200 KN
Aoh = 1 mm2
Vu = 200000 N Check for Torsion Resistance

v/ F'c = 0.026 Tcr = 99 KNm


q = 45 deree Acp = 400000 mm2
b = 2 Pc = 2800 mm
ex = 0.00008 (LRFD_5.8.3.4.2-3)

ex = 0.90 x 10-3 At = 0 mm2


sxe 295 mm
No. Rev.

By Date:
Project: Fly Over At Interchange on N-5 waqar
Chkd By Date
Subject: Wing Wall waqar

DESIGN OF ABUTMENT WALL


Vertical Reinforcement

Forces on the section

Ultimate Moment Mu = 125 KNm


Service Moment Ms = 81 KNm
Ultimate Shear Vu = 77 KN
Ultimate Torsion Tu = 0 KNm

Section Definition Control Perimeters

Ax = 0.4 m2 F'c = 28.0 Mpa


Iz = 0.00533 m4 Fy = 414 Mpa
Yt = 0.2 m F = 0.9
Yb = 0.2 m Z = 23000 N / mm
h = 400 mm cover = 75 mm
bf = 1000 mm Es = 200000 Mpa
hf = 400 mm Fr = 3.3 Mpa
bw = 1000 mm Mcr = 89 KNm
Design Mu = 125 KNm
b1 = 0.85
Ec = 25399 Mpa

Check Nominal Moment Capacity Check Crack Control

Try 8 12Dia Ms = 81 KNm


As = 1608 mm2 Z = 23000 N / mm
A's = 0 mm2 As = 1608 mm2
F's = 0 Mpa r = 0.0049
d = 325 mm n = 7.9
d's = 75 mm j = 0.919
ecu = 0.003 dc = 87.5 mm
c = 32.9 mm A = 3281 mm2
a = 28.0 mm Fs = 169 Mpa OK
e's = 0.000 Fsa = 248 Mpa
F's = 0.0 Mpa
FMn = 186 KNm OK
Design Mu = 125 KNm
c/d 0.101 (Under Reinforced)

Check for Shear Nominal Shear Capacity

Vu = 77 KN Vc = 273 KN
Pu = 0 mm FVn = 246 KN OK
Ao = 1 mm2 Vu = 77 KN
Aoh = 1 mm2
Vu = 77000 N Check for Torsion Resistance

v/ F'c = 0.010 Tcr = 99 KNm


q = 45 deree Acp = 400000 mm2
b = 2 Pc = 2800 mm
ex = 0.00004 (LRFD_5.8.3.4.2-3)

ex = 0.68 x 10-3 At = 0 mm2


sxe 303 mm

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