RECTANGULAR CONCRETE BEAM/SECTION ANALYSIS BEAM Per IS-456:2000 & SP-16 Code CE 692 [ Detailing of RC and Steel Structures ] Design a RCC Beam Subject:
Job Name: Job No.:
1
Shyamal Ghosh
Originator:
Date:
23-02-2013
Checker: Arnab Deb
DESIGN A SIMPLY SUPPORTED BEAM WHERE, EFFECTIVE SPAN, SUPPORT CONDITIONS, fCK, fY, BREDTH, OVERALL DEPTH AND IMPOSED LOAD(s) ARE GIVEN.
DATA Input:Effective Span (L) = Breadth (b) = Overall Depth (D) = f ck ck = f y = Initial Assumption:Dia of Main Bar = Dia of Stirrups = Nominal cover (n c) =
8
m = 8000 mm
Imposed Load =
35.4 KN/m 3
300
mm
Unit Wt. Of Concrete =
25 KN/m
700 M 20
mm
Tensile Stress = Self wt. of beam =
230 N/mm 5.25 5. 25 KN KN/m /m
2
Fe 415 25 8 25
mm mm mm
Effective Depth (d) = 637. 63 7.5 5 mm Clear Cover (d') = 62..5 mm 62 d'/d = 0.098039 Let, d'/d = 0.1
Output :-
General Design:Imposed Load = Self wt. of beam = Total =
35.40 KN 5.25 KN 40.650 KN
Hence, Factored Load = 60.975 KN Factored Moment (M u) = 487.80 487.80 KN-m KN-m Factored Shear (V u) =
[ From, Table D of SP-16, for Fe 415 & M 20 ]
Hence,
2
Thence,
2.76 N/mm2 336.50 336.504 4 KN-m KN-m
Mu,lim/bd = Mu,lim =
Therefore, Mu,lim
243.90 KN
<
Mu
It should be designed as Doubly Reinforced Reinforced beam. 4.00 N/mm2
Mu/bd =
[ From, Table 50 of SP-16, for M 20 & Fe 415 ]
Reqd. Tension Reinforcement:Reinforcement:6 No N os. of 25TOR bar(s).
pt =
1.337 % Ast =
pc =
0.466 % Asc =
2557.013 mm2 891.225 mm2
Reqd. Compression Compression Reinforcement:Reinforcement:2 No N os. of 25TOR bar(s).
Shear Check:2 2 Tauv = 1.2752941 = 1. 1.28 Vu/bd = N/mm Tauc,max = 2.8 N/mm2 [ From, Table J of SP-16 for M20 ] Tau Tau Hence, < Therefore, Shear Design is reqd. v c,max
Shear Design:Tauc = 0.689044 0.6890443 3 = 0.689 0.689 Shear Capacity = tau_c * b * d = Vus = Vu - Tauc * b * d = Vus / d =
[ From, Table 61 of SP-16 ]
131.780 KN
300
112.12 KN
0.175874 0.1758745 5 = 0.18 KN/mm KN/mm = 1.8 KN/cm KN/cm
[ From Table 62 of SP-16, assuming 2 legged Vertical Stirrups of 8TOR dia ]
Spacing =
200 200 mm c/c c/c
2 # 25 TOR 2 Lgd8mm @ 200mm c/c 6 # 25 TOR
Deflection Check:f sc =
240.7
pt =
1.337 %
N/mm
2
MFt =
0.96
[Ref. IS 456:2000 Fig 4]
MFc =
1.1398
[Ref. IS 456:2000 Fig 5]
(L/d)basic =
20
(L/d)maxm =
21.88416
Now, Span / d =
12.54902
[From Cl. 23.2.1 of IS 456:2000, for Simply Supported Beam]
<
21.88416
Hence, Deflection Check is OK
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