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ramu_se ...
Joined: 30 Oct 2017 Posts: 110
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Posted: Tue May 03, 2022 10:09 am Post subject: Percentage of reinforcement as per Concrete grade and steel grade(SP-16) |
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Dear All,
Reinforcement % for singly reinforced sections and doubly reinforced sections are given in SP-16 for the grade of concrete up to M30(fck) and grade of steel up to 500(fy).
How do we get the reinforcement % for M40 and above-grade concrete and 550-grade steel??
Thanks & Regards,
Ramu. |
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vikram.jeet General Sponsor
Joined: 26 Jan 2003 Posts: 3839
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Posted: Thu May 05, 2022 8:10 am Post subject: |
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Dear Er Ramu ji,
Going by limit stage design , I think following basic relations can be used :
Depth of NA Xu limiting value for any grade of steel :
As per IS 456
Strain in conc at extreme fiber Sc= 0.0035
Strain in steel at its centroid Ss = {Fy/(1.15 *Es)} + 0.002
Considering Strain diagram ( linear variation of Strain in section)
Xu = [Sc / (Sc +Ss) ] d
For Fe500 the value work to Xu = o.456 and rounded by IS 456 to Xu =0.46
Accordingly for Fe 550 Xu = 0.4435
You can also verify for Fe 250 as well as Fe 415
VALUE OF Es taken = 2 x 100000 mpa
To be cont'd _ _ _ |
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vikram.jeet General Sponsor
Joined: 26 Jan 2003 Posts: 3839
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Posted: Thu May 05, 2022 8:36 am Post subject: |
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Max % of reinf in singly reifd sections
This can be calculated by equating compressive force to tensile force at a balanced section
Compressive force in conc = 0.36Fck Xu B
Tensile force in steel = 0.87 Fy Ast
Xu = k*d and it's limiting value for Fe 550 as calculated in prev post = 0.4435 *d
Doing calcs for Fe 500 to verify k= 0.456 d
Hence using Fe 500 % reinf max for singh reifd section can be calculated for any conc grade be it M40
Equating
0.36Fck * k*d*B = 0.87Fy*Ast
Ast/Bd = 0.36 Fck*k / (0.87Fy)
For Fe 500 and M20 , the value works to 0.7548 and Sp16 charts
rounds to 0.755
Similarly u can check for others |
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ramu_se ...
Joined: 30 Oct 2017 Posts: 110
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Posted: Sun May 15, 2022 4:40 pm Post subject: |
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Thank you sir. |
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