Portal Frame Design with Example

Design Steps:

  1. Design of slabs
  2. Preliminary design of beams and columns
  3. Analysis
  4. Design of beams
  5. Design of Columns
  6. Design of footings

Problem:

A portal frame hinged at base has following data:

Spacing of portal frames = 4m

Height of columns = 4m

Distance between column centers = 10m

Live load on roof = 1.5 kN/m2

RCC slab continuous over portal frames. Safe bearing capacity of soil=200 kN/m2

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 = 4m

•Span of portal frame = 10m

•Height of columns = 4m

•Live load on roof = 1.5 kN/m2

•Concrete: M20 grade

•Steel: Fe 415

Step1: Design of slab

•Assume 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/m2 (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 = 8.5 kN-m

•Mulim=Qlimbd2 , where (Qlim=2.76)

= 2.76 x 1000 x 1002 / 1 x 106 = 27.6 kN-m > 12.75 kN-m

•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 Ф10 @ 200 c/c

•Area of distribution steel Adist=0.12 x 1000 x 120 / 100 = 144 mm2

•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

Step2: Preliminary design of beams and columns

•Beam:

•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: Analysis

Load on frame

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

•ii) Self weight of rib of beam = 0.45x0.63x24= 6.80 kN/m

•Total» 28.00 kN/m

•Height of beam above hinge = 4+0.1-(075/2 )=3.72 m

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

•The moments in the portal frame hinged at the base and loaded as shown in Fig. is analised by moment distribution

•IAB = 450 x 6003/12 = 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 Factors:

Fixed End Moments:

•MFAB= MFBA= MFCD= MFDC 0

•MFBC= -=-233 kN-m and MFCB= =233 kN-m

Fig: Moment Distribution Table

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= 28x102/8 – 102 =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 = 140kN

•Design 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 rectangular section.

•Design of T-section for Mid Span :

•Design moment Mu=291 kN-m

•Flange width bf=

•Here Lo=0.7 x L = 0.7 x 10 =7m

•bf= 7/6+0.45+6x0.12=2.33m

Step4: Design of T-beam:

•bf/bw=5.2 and Df/d =0.17 Referring to table 58 of SP16, the moment resistance factor is given by KT=0.43,

•Mulim=KT bwd2fck
= 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/(px202/4) =3.93

•Hence 4 Nos. of #20 at bottom in the mid span

Design of Rectangular beam:

•Design moment MuB=234 kN-m

•MuB/bd2= 234x106/450x7002 »1.1 From table 2 of SP16 pt=0.327

•Ast=0.327 x 450 x 700 / 100 = 1030

•No of 20 mm dia bar = 1030/(px202/4) =3.2

•Hence 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

Fig: Long Section of Beam

Fig: Cross Section

Check for Shear:

•Nominal shear stress =

•pt=100x 1256/(450x700)=0.39»0.4

•Permissible stress for pt=0.4 from table 19 tc=0.432 < tv Hence shear reinforcement is required to be designed

•Strength of concrete Vuc=0.432 x 450 x 700/1000 = 136 kN

•Shear to be carried by steel Vus=210-136 = 74 kN

•Nominal shear stress =

pt=100x 942/(400x600)=0.39»0.4

•Permissible stress for pt=0.4 from table 19 tc=0.432 < tv Hence shear reinforcement is required to be designed

•Strength of concrete Vuc=0.432 x 400 x 600/1000 = 103 kN

•Shear to be carried by steel Vus=162-103 = 59 kN

Spacing 2 legged 8 mm dia stirrup

•sv=

•Two legged Ф8 stirrups are provided at 300 mm c/c (equal to maximum spacing)

Step5: Design of Columns:

•Cross-section of column = 450 mm x 600 mm

•Ultimate 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

•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 dia20mm.

•8mm diameter tie shall have pitch least of the following

•Least lateral dimension = 450 mm

•16 times diameter of main bar = 320 mm

•48 times diameter of tie bar = 384

•300mm

•Provide 8 mm tie @ 300 mm c/c

Step6: Design of Hinges:

•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 mm2

•Provide concrete area of 200 x100
(Area =20000mm2) for the hinge

•Shear force at hinge = Total moment in column/height = 156/3.72=42

•Ultimate shear force = 1.5x42=63 kN

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

•Ultimate shear force = 0.87 fyAstsinq

•Provide 4-16dia (Area=804 mm2)

Step7: Design of Footings:

•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 kN

•Working 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 Mu»33 kN-m

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

•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=0.12x1000x250/100=300 mm2

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

•Provide #12 @ 300 c/c both ways

•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 tpunch
=Vpunch/(2x(ao+bo)d)
=126.5x103/(2x(850+700)250) = 0.16 MPa

•Permissible shear stress = 0.25Öfck=1.18 MPatpunch Safe

Check for One Way Shear

•Shear force at a distance ‘d’ from face of column

•V= 90x1x0.45 = 40.5 kN

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

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

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