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
Discussion:
16 comments Page 1 of 2.
Akshat kumar said:
5 years ago
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.
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.
(2)
Nuuui said:
3 years ago
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.
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.
(2)
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.
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.
(1)
Nazeer said:
9 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.
So the answer is E.
Abhishek.kumar said:
8 years ago
Sir. Plese tell me the correct answer with explanation.
Nishant Jain said:
8 years ago
Ans is C.
Find permissible stress[$] in steel first as:
Permissible stress in steel/permissible stress in concrete = Es/Ec.
Permissible stress in steel = 5 * 15Ec/Ec = 75.
Now solve:
P = 5 * (200 * 200 - 1200) + (1200 * 75) = 284000N = 284.
Find permissible stress[$] 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:
7 years ago
Explain this with formula.
Chulbul said:
7 years ago
As the elongations are equal, use the formula.
Strain= stress/Young's modulus.
Strain= stress/Young's modulus.
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