Civil Engineering - Theory of Structures - Discussion

Discussion Forum : 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
264 MN
274 MN
284 MN
294 MN
None of these.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
16 comments Page 1 of 2.

Nazeer said:   8 years ago
How is it done?

Bapan Halder said:   8 years ago
P = 5 * {(40000 - 1200) + (1200 * 15)}.

MRINAL KANTI BISWAS said:   8 years ago
P = 5*(200*200-1200) + (1200*15) = 212000 N

So the answer is E.

Abhishek.kumar said:   7 years ago
Sir. Plese tell me the correct answer with explanation.

Nishant Jain said:   7 years ago
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:   6 years ago
Explain this with formula.

Chulbul said:   6 years ago
As the elongations are equal, use the formula.

Strain= stress/Young's modulus.

Maniesh said:   6 years ago
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:   5 years ago
Excellent explanation, Thanks @Maniesh.

Abhishek said:   5 years ago
Thank you for explaining.


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