Portal-GS

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6. Design of Portal Frames(For class held on 20th, 26th and 27th March 07)

By Dr. G.S.Suresh, Professor, Civil Engineering Department, NIE, Mysore

(Ph:9342188467, email: gss_nie@ yahoo.com)

3.1 Introduction:A portal frame consists of vertical member called Columns and top member which

may be horizontal, curved or pitched. The vertical and top members built

monolithically are considered as rigidly connected. They are used in the constructionof large sheds, bridges and viaducts.

The base of portal frame may be hinged or fixed. The portal frames are spaced at

suitable distance and it supports the slab above the top members. Various forms ofRCC portal frames used in practice is shown in Fig.6.1

d) A Two Storeyed Portal

c) A Mill Bendb) For a Viaducta) For shed

Fig. 6.1

The portal frames have high stability against lateral forces such as wind and

earthquake and the moments in the top beam are also reduced. But at the same time,

large moments are induced in the columns which become more costly. A portal frameis a statically indeterminate structure.

In the case of buildings, the portal frames are generally spaced at intervals of 3 to 4m

with a reinforced concrete slab cast monolithically between the frames. Frames usedfor ware house sheds and workshop structures are provided with sloping of purlins

and asbestos sheet roofing between the portal frames. The base of the columns of the

portal frames are either fixed or hinged. Generally the columns having raft or pilesare considered as fixed for analysis purpose.

Analysis of frames can be done by any standard methods like i) Slope deflection

method, ii) Moment distribution method, iii) Strain energy method, iv) Kanis

method. Columns are designed for axial force and bending moment, whereas beam is

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designed for bending moment and shear force. These forces are obtained from the

analysis carried out on the frame. Limit state method of design is used for design ofmembers. Tables given in SP16 may be used for design.

3.2 Procedure for Analysis and design of Portal frames:

Step1:Design of slabs

Slabs are supported on beams and are designed as continuous. Generally these slabsare designed as one way slabs. Maximum bending moments and shear forces are

computed using the coefficients given in tables 12 and 13 respectively of IS456-2000.

For the assumed depth the required steel is computed from table 1 to4 or 5 to 44 ofSP16. Area of distribution steel are computed based on the minimum steel

requirement ie., 0.12% of gross area.

Step2:Preliminary design of beams and columns

Depth of the beam is generally decided on the basis of span to depth ratio. For lightlyloaded beams it is taken as 20 and 12 to 15 for heavily loaded beams. The width of

the beam depends on the architectural requirements. Generally the width of the beamkept equal to the width of the wall or column. The size of the column is decided basedon axial load calculated as reaction of beam or by experience.

Step3:Analysis

The forces on beams and at joints if any are first calculated and then forces incolumns and beams are calculated using any standard methods of analysis like slope

deflection method, moment distribution method etc., or tables given in SP43 can also

be used for finding the shear force and bending moment.

Step4:Design of beamsUsing the end moments and superposing simple support bending moment diagram,

the design moments at mid span and at ends are computed. The mid span section of

intermediate frame is designed as T-beam using the tables 57 to 59 of SP16. Thesections at ends of the beam are designed as rectangular beams. For the depth of the

beam used at mid span, the steel required is computed from finding steel percentage

using the tables 1 to 4 of SP16. These sections are also designed for shear using tables61 to 63 of SP16. The beam is checked for deflection using span to effective depth

ratio.

Step5:Design of Columns

The columns are designed for uniaxial moment using the charts 24 to 85 of SP 16.The tie reinforcement of the column is designed on the basis of recommendations

given in clause 26.5.3.2 of IS456-2000.

Step6:Design of footingsThe footings are designed for flexure, single shear and punching shear. The

reinforcement is generally provided on the basis of flexural requirement. If the base

of the columns is analised as hinged base, then the hinge is also designed consideringthe triaxial stresses.

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PROBLEMS:1. The roof of a 8m wide hall is supported on a portal frame spaced at 4m intervals.

The height of the portal frame is 4m. The continuous slab is 120 mm thick. Live

load on roof = 1.5 kN/m2, SBC of soil = 150 kN/m

2. The columns are connected

with a plinth beam and the base of the column may be assumed as fixed. Design

the slab, column, beam members and suitable footing for the columns of theportal frame. Adopt M20 grade concrete and Fe 415 steel. Also prepare the

detailed structural drawing.Solution:

Data given:

Spacing of frames = 4mSpan of portal frame = 8m

Height of columns = 4m

Live load on roof = 1.5 kN/m2

Thickness of slab = 120mmConcrete: M20 grade

Steel: Fe 415Three dimensional view of the frame with and without the slab is shown in Fig 6.2

8.00m

4.00m

4.00m

4.00m

4.00m

X

Y

Z

Fig. 6.2

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Step1:Design of slabSelf weight of slab = 0.12 x 24 = 2.88 kN/m2

Weight of roof finish = 0.50 kN/m2

(assumed)

Ceiling finish = 0.25 kN/m2

(assumed)

Total dead load wd = 3.63 kN/m

2

Live load wL = 1.50 kN/m2

(Given in the data)

Maximum service load moment at interior support =9

Lw

10

Lw 2L2

d = 8.5 kN-m

Mu=1.5 x 8.5 = 12.75 kN-m/m

Mulim=Qlimbd2= 2.76 x 1000 x 100

2/ 1 x 10

6= 27.6 kN-m > 12.75 kN-m (Qlim=2.76)

275.1100x1000

10x75.12

bd

M2

6

2

u

From table 2 of SP16 pt=0.384; Ast=(0.384 x 1000 x 100)/100= 384 mm2

Spacing of 10 mm dia bars = (78.54 x 1000)/384= 204.5 mm c/c

Provide #10 @ 200 c/cArea of distribution steel Adist=0.12 x 1000 x 120 / 100 = 144 mm

2

Spacing of 8 mm dia bars = (50.26 x 1000)/144= 349 mm c/c

Provide #8 @ 340 c/c. Main and dist. reinforcement in the slab is shown in Fig.6.3

Step2: Preliminary design of beams and columnsBeam:

Effective span = 8mEffective depth based on deflection criteria = 8000/12 = 666.67mmAssume over all depth as 700 mm with effective depth = 650mm, breadth b = 400mm

and column section equal to 400 mm x 600 mm.

Step3: AnalysisLoad on frame

i) Load from slab = (3.63+1.5) x 4 =20.52 kN/m

ii) Self weight of rib of beam = 0.4x0.58x24 = 5.56 kN/m

Total 27.00 kN/m

The portal frame subjected to the udl considered for analysis is shown in Fig. 6.4

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Fig. 6.4

The moments in the portal frame fixed at the base and loaded as shown in Fig. 6.4 are

analysed by moment distribution

IAB = 400 x 6003/12 = 72 x 10

8mm

4, IBC= 400 x 700

3/12 = 114.33 x 10

8mm

4

Stiffness Factor:

KBA= IAB/ LAB = 18 x 105 KBC= IBC/ LBC = 14.3 x 10

5

Distribution Factor:

55.0103.141018

1018

K

KD

55

5

BA

BABA

45.0103.141018

103.14

K

KD

55

5

BC

BCBC

Fixed End Moments:

MF

AB= MF

BA= MF

CD= MF

DC 0

MF

BC= -12

8x27

12

wL 22 =-144 kN-m and MFCB=

12

8x27

12

wL 22 =144 kN-m

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Moment Distribution Table

Joint A B C D

Members AB BA BC CB CD DC

DF - 0.55 0.45 0.45 0.55 -

FEM 0 0 -144 144 0 0Balance - 79.2 64.8 -64.8 -79.2 -

Carry

over

39.6 - -32.4 32.4 - -39.6

Balance - 17.82 14.58 -14.58 -17.82 -

Carry

over

8.91 - -7.29 7.29 - -8.91

Balance - 4 3.28 -3.28 -4 -

Carry

over

2 - -1.64 1.64 - -2

Balance - 0.90 0.74 -0.74 -0.9 -

Carryover

0.45 - -0.37 0.37 - -0.45

Balance - 0.20 0.17 -0.17 -0.2 -

Total 50.9651 102.11102 -102.11-102

102.11102 -102.11-102 -50.96-51

Bending Moment diagram

Fig. 6.5

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Design moments:

Service load end moments: MB=102 kN-m, MA=51 kN-mDesign end moments MuB=1.5 x 102 = 153 kN-m, MuA=1.5 x 51=76.5 kN-m

Service load mid span moment in beam= 27x82/8102 =114 kN-m

Design mid span moment Mu+=1.5 x 114 = 171 kN-m

Maximum Working shear force (at B or C) in beam = 0.5 x 27 x 8 = 108kNDesign shear force Vu = 1.5 x 108 = 162 kN

Step4:Design of beams:The beam of an intermediate portal frame is designed. The mid span section of this beam

is designed as a T-beam and the beam section at the ends are designed as rectangularsection.

Design of T-section for Mid Span :

Design moment Mu=171 kN-m

Flange width bf= fwo D6b

6

L , Here Lo=0.7 x L = 0.7 x 8 =5.6m

bf= 5.6/6+0.4+6x0.12=2mbf/bw=5 and Df/d =0.2 Referring to table 58 of SP16, the moment resistance factor isgiven by KT=0.459,

Mulim=KT bwd2

fck = 0.459 x 400 x 6002

x 20/1x106

= 1321.92 kN-m > Mu Safe

The reinforcement is computed using table 2 of SP16

Mu/bd2 = 171 x 106/(400x6002)1.2 for this pt=0.359

Ast=0.359 x 400x600/100 = 861.6 mm2

No of 20 mm dia bar = 861.6/(x202/4) =2.74Hence 3 Nos. of #20 at bottom in the mid span

Design of Rectangular-section for End Span :

Design moment MuB=153 kN-m

MuB/bd2= 153x106/400x6002 1.1 From table 2 of SP16 pt=0.327Ast=0.327 x 400 x 600 / 100 = 784.8

No of 20 mm dia bar = 784.8/(x202/4) =2.5Hence 3 Nos. of #20 at the top near the ends for a distance of o.25 L = 2m from face

of the column as shown in Fig 6.6

Check for Shear

Nominal shear stress = 675.0600400

10x162

bd

V 3uv

pt=100x 942/(400x600)=0.390.4Permissible stress for pt=0.4 from table 19 c=0.432 < v Hence shear reinforcementis required to be designed

Strength of concrete Vuc=0.432 x 400 x 600/1000 = 103 kNShear to be carried by steel Vus=162-103 = 59 kN

Spacing 2 legged 8 mm dia stirrup sv= 3671059

60050241587.0

V

dAf87.03

us

svy

Two legged #8 stirrups are provided at 300 mm c/c (equal to maximum spacing)

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Step5:Design of columns:Cross-section of column = 400 mm x 600 mm

Ultimate axial load Pu=1.5 x 108 = 162 kN (Axial load = shear force in beam)

Ultimate moment Mu= 1.5 x 102 = 153 kN-m ( Maximum)

Assuming effective cover d = 50 mm; d/D 0.1

053.060040020

10153

bDf

M2

6

2

ck

u

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033.060040020

10162

bDf

P 3

ck

u

Referring to chart 32 of SP16, p/fck=0.03; p=20 x 0.03 = 0.6

Minimum steel in column should be 0.8 %, Hence min steel percentage shall be adopted

Ast=0.8x400x600/100 = 1920 mm2

No. of bars required = 1920/314 = 6.1

Provide 8 bars of #208mm diameter tie shall have pitch least of the following

i) Least lateral dimension = 400 mm

ii) 16 times diameter of main bar = 320 mmiii) 48 times diameter of tie bar = 384

iv) 300mm

Provide 8 mm tie @ 300 mm c/c

Step6:Design of Footing:

Load:Axial Working load on column = 108 kN

Self weight of footing @10% = 11 kN

Total load= 119120 kNWorking load moment at base = 51 kN-m

Approximate area footing required = Load on column/SBC

= 108/150 =0.72 m2

However the area provided shall be more than required to take care of effect ofmoment. The footing size shall be assumed to be 2mx3m (Area=6 m2)

2m

1.2m

3m

0.6m

0.4m

X

X

600

Tie #8 @300 c/c8-#20

400

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Maximum pressure qmax=P/A+M/Z = 108/6+6x51/2x32

= 35 kN/m2

Minimum pressure qmin=P/A-M/Z = 108/6-6x51/2x32

= 1 kN/m2

Average pressure q = (35+1) = 18 kN/m2

Bending moment at X-X = 18 x 2 x 1.22/2 = 25.92 kN-m

Factored moment Mu39 kN-m

Over all depth shall be assumed as 300 mm and effective depth as 250 mm,312.0

2502000

1039

bd

M2

6

2

u

Corresponding percentage of steel from Table 2 of

SP16 is pt= 0.1%, Minimum pt=0.12%

Area of steel per meter width of footing is Ast=0.12x1000x250/100=300 mm2

Spacing of 12 mm diameter bar = 113x1000/300 = 376 mm c/c

Provide #12 @ 300 c/c both waysCheck for Punching Shear

Length of punching influence plane = ao= 600+250 = 850 mm

Width of punching influence plane = bo= 400+250 = 650 mmPunching shear Force = Vpunch=108-18x(0.85x0.65)=98 kN

Punching shear stress punch= Vpunch/ (2x(ao+bo)d =98x103/(2x(850+650)250)= 0.13 MPa

Permissible shear stress = 0.25fck=1.18 MPa > punch SafeCheck for One Way Shear

Shear force at a distance d from face of column

V= 18x2x0.95 = 34.2 kN

Shear stress v=34.2x103/(2000x250)=0.064 MPa

Referring to table 19 of IS456 this stress is very small and hence safe

Details of reinforcement provided in footing is shown in Fig.6.7

Fig.6.7

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LONGITUDINAL ELEVATION

Cross-Sections of Beam

Cross-Section of Column

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2. A portal frame hinged at base has following data:

Spacing of portal frames = 4mHeight of columns = 4m

Distance between column centers = 10m


RCC slab continuous over portal frames. Safe bearing capacity of soil=200 kN/m

2

Adopt M-20 grade concrete and Fe-415 steel. Design the slab, portal frame and

foundations and sketch the details of reinforcements.Solution:

Data given:

Spacing of frames = 4mSpan of portal frame = 10m

Height of columns = 4m


Concrete: M20 gradeSteel: Fe 415

Three dimensional view of the frame with and without the slab is shown in Fig 6.8

Fig. 6.8

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Step1:Design of slabAssume over all depth of slab as 120mm and effective depth as 100mm

Self weight of slab = 0.12 x 24 = 2.88 kN/m2

Weight of roof finish = 0.50 kN/m2

(assumed)

Ceiling finish = 0.25 kN/m

2

(assumed)Total dead load wd = 3.63 kN/m2

Live load wL = 1.50 kN/m2

(Given in the data)

Maximum service load moment at interior support =9

Lw

10

Lw 2L2

d = 8.5 kN-m

Mu=1.5 x 8.5 = 12.75 kN-m/mMulim=Qlimbd

2= 2.76 x 1000 x 1002/ 1 x 106 = 27.6 kN-m > 12.75 kN-m (Qlim=2.76)

275.1100x1000

10x75.12

bd

M2

6

2

u

From table 2 of SP16 pt=0.384; Ast=(0.384 x 1000 x 100)/100= 384 mm2

Spacing of 10 mm dia bars = (78.54 x 1000)/384= 204.5 mm c/c

Provide #10 @ 200 c/cArea of distribution steel Adist=0.12 x 1000 x 120 / 100 = 144 mm

2

Spacing of 8 mm dia bars = (50.26 x 1000)/144= 349 mm c/c

Provide #8 @ 340 c/c. Main and dist. reinforcement in the slab is shown in Fig.6.9

Step2: Preliminary design of beams and columnsBeam:Effective span = 10m

Effective depth based on deflection criteria = 10000/13 = 769.23mm

Assume over all depth as 750 mm with effective depth = 700mm, breadth b = 450mm

and column section equal to 450 mm x 600 mm.

Step3: AnalysisLoad on frame

i) Load from slab = (3.63+1.5) x 4 =20.52 kN/mii) Self weight of rib of beam = 0.45x0.63x24 = 6.80 kN/m

Total 28.00 kN/mHeight of beam above hinge = 4+0.1-075/2 =3.72 m

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The portal frame subjected to the udl considered for analysis is shown in Fig. 6.10

Fig. 6.10

The moments in the portal frame hinged at the base and loaded as shown in Fig. 6.10

are analysed by moment distributionIAB = 450 x 600

3/12 = 81 x 108 mm4, IBC= 450 x 7503/12 = 158.2 x 108 mm4

Stiffness Factor:

KBA= IAB/ LAB = 21.77 x 105

KBC= IBC/ LBC = 15.8 x 105

Distribution Factor:

5.0108.151077.21

1077.21

K

KDD

55

5

BA

BABCBA

Fixed End Moments:

MF

AB= MF

BA= MF

CD= MF

DC 0

MF

BC= -12

10x28

12

wL 22 =-233 kN-m and MFCB=

12

8x27

12

wL 22 =233 kN-m

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Moment Distribution Table

Joint A B C D

Members AB BA BC CB CD DC

DF - 0.5 0.5 0.5 0.5 -

FEM 0 0 -233 233 0 0Balance - 116.5 116.5 -116.5 -116.5 -

Carry

over

- - -58.25 58.25 - -

Balance - 29.13 29.13 -29.13 -29.13 -

Carry

over

- - -14.57 14.57 - -

Balance - 7.29 7.29 -7.29 -7.29 -

Carry

over

- -3.65 3.65 - -

Balance - 1.83 1.83 -1.83 -1.83 -

Carryover

- - -0.92 0.92 - -

Balance - 0.46 0.46 -0.46 -0.46 -

Total - 155.21156 -155.21-156

155.21156 -155.21-156 -

Bending Moment diagram

Fig. 6.11

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Design moments:

Service load end moments: MB=156 kN-m,Design end moments MuB=1.5 x 156 = 234 kN-m,

Service load mid span moment in beam= 28x102/8102 =194 kN-m

Design mid span moment Mu+=1.5 x 194 = 291 kN-m

Maximum Working shear force (at B or C) in beam = 0.5 x 28 x 10 = 140kNDesign shear force Vu = 1.5 x 140 = 210 kN

Step4:Design of beams:The beam of an intermediate portal frame is designed. The mid span section of this beam

is designed as a T-beam and the beam section at the ends are designed as rectangularsection.

Design of T-section for Mid Span :

Design moment Mu=291 kN-m

Flange width bf= fwo D6b

6

L , Here Lo=0.7 x L = 0.7 x 10 =7m

bf= 7/6+0.45+6x0.12=2.33mbf/bw=5.2 and Df/d =0.17 Referring to table 58 of SP16, the moment resistance factoris given by KT=0.43,

Mulim=KT bwd2

fck = 0.43 x 450 x 7002

x 20/1x106

= 1896.3 kN-m > Mu Safe

The reinforcement is computed using table 2 of SP16

Mu/bd2 = 291 x 106/(450x7002)1.3 for this pt=0.392

Ast=0.392 x 450x700/100 = 1234.8 mm2

No of 20 mm dia bar = 1234.8/(x202/4) =3.93Hence 4 Nos. of #20 at bottom in the mid span

Design of Rectangular-section for End Span :

Design moment MuB=234 kN-m

MuB/bd2= 234x106/450x7002 1.1 From table 2 of SP16 p t=0.327Ast=0.327 x 450 x 700 / 100 = 1030

No of 20 mm dia bar = 1030/(x202/4) =3.2Hence 4 Nos. of #20 at the top near the ends for a distance of o.25 L = 2.5m from

face of the column as shown in Fig 6.12

Check for Shear

Nominal shear stress = 67.0700450

10x210

bd

V 3uv

pt=100x 1256/(450x700)=0.390.4Permissible stress for pt=0.4 from table 19 c=0.432 < v Hence shear reinforcementis required to be designed

Strength of concrete Vuc=0.432 x 450 x 700/1000 = 136 kNShear to be carried by steel Vus=210-136 = 74 kN

Spacing 2 legged 8 mm dia stirrup

sv= 53.3411074

70050241587.0

V

dAf87.03

us

svy

Two legged #8 stirrups are provided at 300 mm c/c (equal to maximum spacing)

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Step5:Design of columns:Cross-section of column = 450 mm x 600 mmUltimate axial load Pu=1.5 x 140 = 210 kN (Axial load = shear force in beam)

Ultimate moment Mu= 1.5 x 156 = 234 kN-m ( Maximum)

Assuming effective cover d = 50 mm; d/D 0.1

07.060045020

10234bDf

M2

6

2

ck

u

04.060045020

10210

bDf

P 3

ck

u

Referring to chart 32 of SP16, p/fck=0.04; p=20 x 0.04 = 0.8 %Equal to Minimum percentage stipulated by IS456-2000 (0.8 % )

Ast=0.8x450x600/100 = 2160 mm2

No. of bars required = 2160/314 = 6.8

Provide 8 bars of #20

8mm diameter tie shall have pitch least of the followingv) Least lateral dimension = 450 mm

vi) 16 times diameter of main bar = 320 mmvii) 48 times diameter of tie bar = 384

viii) 300mm

Provide 8 mm tie @ 300 mm c/c

Step6:Design of Hinge:At the hinge portion, concrete is under triaxial stress and can withstand higher

permissible stress.Permissible compressive stress in concrete at hinge= 2x0.4fck=16 MPa

Factored thrust =Pu=210kN

Cross sectional area of hinge required = 210x103/16=13125 mm

2

Provide concrete area of 200 x100 (Area =20000mm

2

) for the hingeShear force at hinge = Total moment in column/height = 156/3.72=42

Ultimate shear force = 1.5x42=63 kN

Inclination of bar with vertical = = tan-1(30/50) =31o

Ultimate shear force = 0.87 fy Ast sin

2

o

3

st mm33931sin41587.0

1063A

Provide 4-#16 (Area=804 mm2)

600

Tie #8 @300 c/c8-#20

450

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Step7:Design of Foundations:Load:Axial Working load on column = 140 kN

Self weight of column=0.45 x 0.6 x3.72x 24 = 24

Self weight of footing @10% = 16 kN

Total load= 180 kNWorking moment at base = 42 x 1 =42 kN-m

Approximate area footing required = Load on column/SBC= 180/200 =0.9 m2

However the area provided shall be more than required to take care of effect of

moment. The footing size shall be assumed to be 1mx2m (Area=2 m2)

Maximum pressure qmax=P/A+M/Z = 180/2+6x42/1x22 = 153 kN/m2

Minimum pressure qmin=P/A-M/Z = 180/2-6x42/1x22

= 27 kN/m2

Average pressure q = (153+27)/2 = 90 kN/m2

Bending moment at X-X = 90 x 1 x 0.72/2 = 22 kN-m

Factored moment Mu33 kN-m

Over all depth shall be assumed as 300 mm and effective depth as 250 mm,528.0

2501000

1033

bd

M2

6

2

u

Corresponding percentage of steel from Table 2 of

SP16 is pt= 0.15% > Minimum pt=0.12%Area of steel per meter width of footing is Ast=0.15x1000x250/100=301 mm

2

Spacing of 12 mm diameter bar = 113x1000/375 = 376 mm c/c

Provide #12 @ 300 c/c both waysCheck for Punching Shear

Length of punching influence plane = ao= 600+250 = 850 mm

Width of punching influence plane = bo= 450+250 = 700 mm

Punching shear Force = Vpunch=180-90x(0.85x0.7)=126.5 kN

Punching shear stress punch= Vpunch/ (2x(ao+bo)d =126.5x103

/(2x(850+700)250)= 0.16 MPa

Permissible shear stress = 0.25fck=1.18 MPa > punch SafeCheck for One Way Shear

Shear force at a distance d from face of columnV= 90x1x0.45 = 40.5 kN

Shear stress v=40.5x103/(1000x250)=0.162 MPa

For pt=0.15 , the permissible stress c = 0.28 (From table 19 of IS456-2000)

2mX

1m

0.7m

0.6m

X

0.45m

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Details of reinforcement provided in footing is shown in Fig.6.13

Fig.6.13

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Cross-Sections of Beam

Cross-Section of Column

LONGITUDINAL ELEVATION

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Reference Books

N.Krishna Raju Advanced Reinforced concrete Design

Jaikrishna and O.P.Jain Plain and reinforced concrete Vol2

B.C.Punmia Reinforced Concrete Structures Vol2

Problems for Practice

1. A portal frame ABCD has fixed supports at A and D. The columns AB andCD are 5m in height while the beam BC is 10 m in length. The frames are

spaced at 3.5m intervals. The live load on the roof slab which is 100 mm thick

may be taken as 1.5 kN/m2. Design the beam, column and footing and sketch

the details of reinforcements. Adopt M-20 concrete, Fe-415 steel and

SBC=200 kN/m2

2. The roof of an assembly hall 30m long and 12 m wide between centres ofcolumns, consists of a continuous reinforced concrete slab over rectangular

portal frames spaced 3m apart. The columns are provided with independent

footings and hinged at the bottom. The ceiling height is 3.5m above the hingelevel. Adopting M-20 concrete and Fe-415 for steel, design the continuous

roof slab and the portal frame and foundation footing for the columns assume

safe bearing capacity of the soil as 150 kN/m2. Sketch the details of

reinforcements in the portal frame.

**************

Documents

Portal-GS