# 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 mm

^{2}total cross-sectional area. Calculate the safe load for the column if permissible stress in concrete is 5 N/mm^{2}and*E*is 15_{s}*E*_{c}Discussion:

16 comments Page 1 of 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)

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.

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)

Akshat kumar said:
4 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)

Chulbul said:
6 years ago

As the elongations are equal, use the formula.

Strain= stress/Young's modulus.

Strain= stress/Young's modulus.

Govind Singh said:
5 years ago

Thank you, Nishant Jain, for giving an answer with placid explanation.

Roy said:
5 years ago

So,modular ratio is m=Es/Ec=15.

Stress in steel, t=mc=15*5 = 75mpa.

Stress in steel, t=mc=15*5 = 75mpa.

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.

Bapan Halder said:
8 years ago

P = 5 * {(40000 - 1200) + (1200 * 15)}.

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