Moment Capacity for Flanged beam with neutral axis in the web

When the neutral axis falls below the flange then we revert back to the Simplified rectangular stress block in BS8110.

Reference: Kong & Evans: Reinforced and Prestressed Concrete, 3rd Edition. Fig 4.4-5    

(where K1 = 0.45 and k2 = 0.9/2 = 0.405)

Example  Mapp = 19324.6 kNm

Basic Data

          Design to          BS 8110: 1997                    

          Fcu , Fy , Fyv , v crit          40, 460, 460, 5.0

          Main Top Outer Steel          12-T12          12-T12                    

          Main Bottom Inner Steel          14-T32          14-T32                    

          Main Bottom Outer Steel          14-T32          14-T32                    

          Shear Reinforcement (legs)          14 sets of 8-T16@150,   Nominal 8-T16@250,  17 sets of 8-T16@175          

In-Span Steel @ 5000 mm. Sagging

          Note: Flanged beam with N/A in web.          Using BS8110 Simplified Rectangular Stress Block                    

          X/d=Fn(As, Fy , Beff , Bweb , d, Fcu)          42726, 460, 3000, 2000, 1234.0, 40          0.33          

          Mu =Fn(Z, Beff , Bweb , X, Fcu)          1060.44, 3000, 2000, 401.27, 40          19800 kNm          

          Mapp/Mu          19324.6 / 19799.5          0.976          OK

          As Required (mm2)           (to simplified method Cl 3.4.4.4)          top 46633          

Hand Calculation for X = 401.27 mm as suggested by program.

X correct if Fc=Ft. Lets test for this value of X.

X>350 mm >>>> web in compression

As = 2 x 17 No 40 dia          = 2 * 17 * pi * 20 * 20                     = 42,726 mm2

d   = H – cvr – lnk – bars/2          = 1350 – 40 – 16 -40-20                    =   1,234 mm

Ft = 0.95 * Fy * As                     = 0.95 * 460 * 42726 * 1E-3             =   18,671 kN

Limit flange depth utilised to 0.9 to be slightly conservative.  If we don’t the 0.9 X limit will only reduce the area of the web and would be wrong.

Hf = 0.9 * FlgDepth            = 0.9 * 350                                         = 315.00 mm

Hw= 0.9 * X - Hf                      = 0.9 *401.27 – 315                     =   46.14 mm

Fcf = .45 * Fcu * Bf * Hf          = 0.45 * 40 * 3000 * 315                     = 17,010 kN

Fcw = .45 * Fcu * Bw * Hw          = 0.45 * 40 * 2000 * 46.14                    =   1,661 kN

Assumed X correct if Ft = Fcf + Fcw

Fc = Fcf + Fcw                     = 17,010 + 1,661                              = 18671 kN = Ft            .

% error= 100* (Fc-Ft)/Ft     = 100 *(18671-18,671)/18,671           = 0.00% error  < 1%  X is correct.

Moment Capacity

Mc = Mcf + Mcw

Mcf   = Fcf  * (d-hf/2)          = 17,010 * (1234 – 315/2)                    =  18,311 kNm

Mcw = Fcw * (d-hf-hw/2)          =   1,661 * (1234 – 315 – 46.1/2)          =    1,488 kNm

Mc = Mcf + Mcw                                                                      = 19,799 kNm  QED