simply supported beam

Topics: Beam, Flange, I-beam Pages: 7 (674 words) Published: November 19, 2013
SIMPLY SUPPORTED FLANGED BEAM
DESIGN SIMPLY SUPPORTED FLANGED BEAM

bf

1) Load Analysis
- N= 1.35gk + 1.5qk
2) SFD and BMD
- consider type of load

hf

h

*min diameter bar provided is 12mm
*min diameter link provided is 8mm

d

d = h – Cnom – Ølink – Øbar/2

Neutral Axis Lies in Flange
Design as a rectangular section
Size of beam (bf X d)

Z = d (0.5+(0.25 – (k/1.134))1/2 0.95 d, use 0.95d as z value

Asreq = M/0.87fykZ
Provide main reinforcement
Asmin = 0.26fctmbd/fyk < 0.0013bd
Asmax = 0.04Ac

Neutral Axis Below Flange
Design as a flange section

Muf = 0.167fckbwd2+ 0.567 fck(bf–bw) (d-(hf/2))
If M < Muf – singly reinforced
If M > Muf – doubly reinforced

K = M/fckbd2
If k < 0.167 (singly reinforced)
If k > 0.167 (doubly reinforced)
SINGLY

d’ = Cnom + Ølink + Øbar/2

Mf = (0.567fckbfhf)(d-(hf/2))
M < Mf – neutral axis lies in flange
M > Mf – neutral axis below flange

bw

SINGLY

DOUBLY

Compression reinforcement:
As’ = (K-Kbal)fckbd2/0.87fyk(d-d’)
d’ = Cnom + Ølink – Øbar
Provide compression reinforcement

Tension reinforcement:
As = (Kbalfckbd2/0.87fykZ) + As’
Provide tension reinforcement
Asmin = 0.26fctmbd/fyk < 0.0013bd
Asmax = 0.04Ac

DOUBLY

Asreq= M + 0.1 fckbwd (0.36d –hf)/0.87fyk(d-(hf/2))
Provide main tension reinforcement
Asmin = 0.26fctmbd/fyk < 0.0013bd
Asmax = 0.04Ac

As’ = (M-Muf)/0.87 fyk(d-d’)
Provide main compression reinforcement
Asmin = 0.26fctmbd/fyk < 0.0013bd
Asmax = 0.04Ac

As = [0.2 fckbwd + 0.567 fckhf(bf-bw) ] + As’
0.87 fyk
Provide main tension reinforcement
Asmin = 0.26fctmbd/fyk < 0.0013bd
Asmax = 0.04Ac

*IF FORMULA IN APPENDIX IS DIFFERENT, YOU MAY FOLLOW APPENDIX.

MSAH/FKA/UiTMPP

Page 1

SIMPLY SUPPORTED FLANGED BEAM
SHEAR REINFORCEMENT DESIGN:
VEd = Vmax – from SFD
VRd,c = 0.12k(100ρ1fck)1/3bwd
k = 1 + (200/d)1/2 < 2.0mm if more than 2.0 used 2.0 as k
ρ1 = Asl/bwd Asl = Area of tension reinforcement provided at mid-span Vmin =(0.035k3/2fck1/2)bwd
Compare VRd,c and Vmin– use largest value to compare with VEd If VEd > VRd,c or Vmin – shear reinforcement is required
If VEd < VRd,c or Vmin – shear reinforcement is not required

VRd,max = (0.36bwd(1-(fck/250))fck)/(cotθ + tanθ) – 22ᵒ < θ < 45ᵒ i)

If VEd < VRd,max – (θ=22ᵒ) - design shear links

ii) If VEd > VRd,max – (θ=22ᵒ) recalculate VRd,max with θ=45ᵒ and compare with VEd
-

If VEd < VRd,max – recalculate an angle
θ = sin-1(VEd/(0.18bwdfck(1-(fck/250)))

-

If VEd > VRd,max – redesign the section

Provide shear reinforcement:

Use new angle to
calculate Asw/S

Asw/S = VEd/(0.87fykdcotθ)
Area of link:
Try H8(Ølink)= (π(Ølink)2/4)X 2
S = (Asw/S)/Area of link
-

Provide links

Minimum link:
Asw/S =(0.08bw(fck)1/2)/fyk

MSAH/FKA/UiTMPP

Page 2

SIMPLY SUPPORTED FLANGED BEAM
Deflection Check:
i)

Calculate (l/d)basic.

-

If ρ < ρ0 , used equation 7.16a in Appendix

-

If ρ > ρ0 , used equation 7.16b in Appendix
Where;
1/2
-3
ρ = As required tension at mid-span
ρ0 = (fck) x 10
K = from Table 7.4N in Appendix or Eurocode 2
ρ’= As required compression at mid-span

*if the beam is singly reinforced, ρ’ = 0

ρ= Asl/bwd
Asl = area of tension reinforcement

ρ’= Asl/bwd
ii)

Calculate (l/d) allowable = (l/d)basic x MFreinforcement x MFelement span -

MFreinforcement = (As provided)/(As required)
Where:
As provided = Area of main reinforcement provided
As required = Area of main reinforcement required

-

Asl = area of comp reinforcement

MFelement span ; if element span less than 7 meter (MF = 1.0) ; if element span more than 7 meter (MF = 7/length of span)

iii)

Calculate (l/d)actual = (l = length of span) (d = depth of beam)

iv)

Conclusion
if (l/d)actual < (l/d)allowable – deflection check pass
if (l/d)actual > (l/d)allowable – deflection check fail

CRACK CHECK:

S1 = (b – 2cover -2Ølink – 3Øbar)/2


S1

If the bar is arrange in one spacing...
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