# Civil Engineering - Theory of Structures - Discussion

### Discussion :: Theory of Structures - Section 3 (Q.No.1)

1.

If a concrete column 200 x 200 mm in cross-section is reinforced with four steel bars of 1200 mm2 total cross-sectional area. Calculate the safe load for the column if permissible stress in concrete is 5 N/mm2 and Es is 15 Ec

 [A]. 264 MN [B]. 274 MN [C]. 284 MN [D]. 294 MN [E]. None of these.

Explanation:

No answer description available for this question.

 Nazeer said: (Mar 3, 2016) How is it done?

 Bapan Halder said: (Jul 25, 2016) P = 5 * {(40000 - 1200) + (1200 * 15)}.

 Mrinal Kanti Biswas said: (Aug 12, 2016) P = 5*(200*200-1200) + (1200*15) = 212000 N So the answer is E.

 Abhishek.Kumar said: (Sep 14, 2016) Sir. Plese tell me the correct answer with explanation.

 Nishant Jain said: (Oct 15, 2016) Ans is C. Find permissible stress[\$] in steel first as: Permissible stress in steel/permissible stress in concrete = Es/Ec. Permissible stress in steel = 5 * 15Ec/Ec = 75. Now solve: P = 5 * (200 * 200 - 1200) + (1200 * 75) = 284000N = 284.

 Shiva said: (Oct 1, 2017) Explain this with formula.

 Chulbul said: (Jan 31, 2018) As the elongations are equal, use the formula. Strain= stress/Young's modulus.

 Maniesh said: (Apr 7, 2018) Stress= load/ area, Stress in steel = load / area of steel and likewise same for concrete, Es= permissible stress in steel/ strain in steel and likewise same for concrete, strain in concrete and steel are same = permissible stress in steel/ Es= permissible stress in concrete/ Ec. Es= 15Ec. =permissible stress in steel= permissible stress in concrete/ Ec xEs =( 5 /Ec)x 15Ec= 75Mpa for steel. Load = permissible Stress in steel x area of steel +permissible stress in concrete x area of concrete. = 75 x 1200+ 5 x (200X200 -1200)= 284kN.

 Kisan Prasad Gaire said: (Mar 28, 2019) Excellent explanation, Thanks @Maniesh.

 Abhishek said: (Apr 10, 2019) Thank you for explaining.

 Raju said: (Jun 13, 2019) Thank you @Maniesh.

 Roy said: (Sep 15, 2019) So,modular ratio is m=Es/Ec=15. Stress in steel, t=mc=15*5 = 75mpa.

 Govind Singh said: (Dec 8, 2019) Thank you, Nishant Jain, for giving an answer with placid explanation.

 Akshat Kumar said: (Feb 27, 2020) As=1200. Ac=38800. 6c=5. Es=Ec. We know that, dL=PL/AE=6L/E. Here, dLs=dLc. 6sLs/Es=6cLc/Ec. Ls=Lc. 6s/Es=6c/Ec. 6s/15Ec=5/Ec. 6s=5/Ec*15Ec. 6s=75. Now, P=6sAs+6cAc. P=75*1200+5*38800. P=284000N.

 Nuuui said: (Mar 10, 2021) As = 1200. Ac = 38800. 6c=5. Es = Ec. We know that, dL = PL/AE = 6L/E. Here, dLs = dLc. 6sLs/Es = 6cLc/Ec. Ls = Lc. 6s/Es = 6c/Ec. 6s/15Ec = 5/Ec. 6s = 5/Ec * 15Ec. 6s = 75. Now, P = 6sAs + 6cAc. P = 75 * 1200 + 5 * 38800. P = 284000N.

 Avinash said: (May 16, 2021) It's 284KN not 284MN.