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B F K K M D: 1. Design of Beam A. Reinforcement Calculation

The document summarizes the key steps in the limit state design of a reinforced concrete beam: 1. Calculate reinforcement based on bending moment capacity and required depth. 2. Check minimum reinforcement requirements. 3. Check the span to depth ratio against modification factors that account for tension steel, compression steel, and beam geometry. 4. Calculate allowable shear stress considering factors like concrete strength and steel ratio.

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100% found this document useful (1 vote)
139 views2 pages

B F K K M D: 1. Design of Beam A. Reinforcement Calculation

The document summarizes the key steps in the limit state design of a reinforced concrete beam: 1. Calculate reinforcement based on bending moment capacity and required depth. 2. Check minimum reinforcement requirements. 3. Check the span to depth ratio against modification factors that account for tension steel, compression steel, and beam geometry. 4. Calculate allowable shear stress considering factors like concrete strength and steel ratio.

Uploaded by

Prayas Subedi
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Limit state design

1. Design of beam
a. Reinforcement calculation
0.0035
k
fy
(0.0055  0.87 )
E
M
d lim 
k (1  0.42k ) f ck b

 f y Ast 
Msteel  0.87 f y Ast  d  
 f ck b 

Mconcrete  0.36 f ck bd d  0.42 x 

b. Check minimum reinforcement


0.85
A min  bd
fy

c. Check span depth ratio L/d=R*F1*F2*F3


F1,f2,f3 from graphs or following formula

Modification factor for


tension steel
F1 1/(0.225+0.00322fs+0.625logpt)<=2.0
fs=0.58*fy*As,req/As,provided*β
pt=100*As provided/b/D
redistribution factor, β
Modification factor for
compression steel
1+(pc/(3+pc))<
F2 =1.5
pc=100*As
provided/b/D

Correction of flanged
beams
0.8+2/7*(bw/bf-
F3 0.3)
d. Shear calculation

Maximum permissible shear stress for this section τmax=0.62*sqrt(fck) <= 4


Allowable design shear stress τc=0.85 sqrt(0.8fck)sqrt(1+5β-1)/6β
β=0.8fck/6.89pt
τc=0.79pt^(1/3)*(400/d)^(1/4)*(1/SF)*(fck/25)^(1/3)

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