Design of Rc Corbel

8 No. of 20 DIA BARS

35.00 tonne

500 mm

Input Data 5No. of 16 DIA BARS 346.50 mm 300 mm 478.46 mm 600.00 mm FAILURE PLANE(CL 40.5.1 IS 456)

30° d'=

450.46 mm

6No. of 16 DIA BARS

MT Vertical un factored load

Vertical Load =

Vu

516000 N

Horizontal Load =

Hu

113520 N

OK

500 mm mm

OK

Shear Span of Bracket= Width of Bracket =

400 mm

Overall Depth of Bracket (D) =

600 mm

Depth of Straight Portion(hv) =

300 mm

Grade of Concrete (f ck ) = M25 Grade of Steel (f y ) = Coefficient of Friction (=

25 Fe 415 1

Clear Cover (Cc ) =

20 mm

Primary Tensile Reinforcement Dia () =

20 mm

Dia of Horizontal Stirrup () =

16 mm

Dia of Vertical Stirrup () =

16 mm

Maximum Allowable Spacing =

300 mm

Minimum Allowable Spacing =

30 mm

Effective Depth (d) = D - Cc - PT/2 =

572 mm

Effective Depth (d) = D1 - Cc - PT/2 =

272 mm

28 days cylinder strength of concrete (fc') = 0.80 x fck = Vertical Shear Stress developed = Vu/ (bx0.8d) =

20 M Pa 2.82 M Pa

OK

Value of 0.15 x fc' =

3.00 M Pa

Shear friction reinforcement

(Avf)= Vu/(0.85 x fy x ) =

1462.792 mm

2

Direct tension reinforcement

(At)= Hu/(0.85 x fy ) =

321.8143 mm

2

1610.394 mm

2

Flexural tension reinforcement (Af)= [Vu x a + Hu (D – d)] / (0.85 x fy x d)= Percentage of steel(pt1)=

0.70 %

OK

OK

Total primary tensile reinforcement (As)= Maximum of ( (Af + At), (2/3Avf + At), (0.04f’c/fy)bd’) (Af + At)

2 1932.208 mm

(2/3Avf + At),

2 1297.009 mm

(0.04f’c/fy)bd’)

2 441.0602 mm

Max=

2 1932.208 mm

Percentage of steel(pt2)=

0.84 %

OK

Total Horizontal stirrup area (Ah)= maximum(0.5 x Af, 0.333 x Avf) =

805.1968 mm

2

C/S area of primary tensile reinforcement (APT)=x PT2 /4 =

314.1593 mm

2

C/S area of Horizontal Stirrup (Ah)=x h2 /4 =

201.0619 mm

2

C/S area of vertical Stirrup (Av)=x v2 /4 =

201.0619 mm

2

No. of bars required for primary tensile reinf (NPT)= As/APT =

8

Number of horizontal stirrup required (Nh)= Ah/Afh =

6

Number of Vertical stirrup

5

Minimum no of reinforcement bars considered = 2.0

Spacing of primary tensile reinforcement required (SPT)= (b - 2 x Cc )/(N PT- 1) =

51.43

Spacing of Horizontal stirrup required (Sh)= 2/3 x d' / Nh =

50.05

Spacing of Vertical stirrup required (Sv)= 2 x Av x fy x d/(Vu - 10 x b x d) =

145.68

Spacing of primary tensile reinforcement provided (SPTprov) =

50

Spacing of Horizontal stirrup provided (Shprov) =

50

Spacing of Vertical stirrup provided (Svprov) =

145