QUIZ : CIVIL ENGINEERING (12 MAY 2021)

Quiz: Civil Engineering
Exam: DFCCIL EXECUTIVE
Topic: REINFORCED CEMENT CONCRETE

Each question carries 1 mark
Negative marking: 1/4 mark
Time: 10 Minutes

Q1. The reduction coefficient of a reinforced concrete column with an effective length of 4.5 m and size 250×300 mm² us _:
(a) 0.8
(b) 0.85
(c) 0.9
(d) 0.95

Q2. In the case of columns of minimum dimension of 200 mm or under, whose reinforcing bars do not exceed 12 mm, a nominal cover may be provided –
(a) 40 mm
(b) 25 mm
(c) 20 mm
(d) 50 mm

Q3. A square column section of 350 mm × 350 mm is reinforced with four bars of 25 mm diameter and four bars of 16 mm diameter. The transverse reinforcement will be –
(a) 5 mm dia @ 240mm c/c
(b) 8 mm dia @ 300 mm c/c
(c) 6mm dia @ 350 mm c/c
(d) 8 mm dia @ 500 mm c/c

Q4. The load carrying capacity of a helically reinforced as compared to that of a tied column is about –
(a) 5% more
(b) 5% less
(c) 10% more
(d) 10% less

Q5. Minimum reinforcement in circular concrete column as per IS – 456 is –
(a) 4 bars of 12 mm
(b) Greater 0.80% of cross – sectional area 4 bars of 12 mm
(c) 6 bars of 12 mm
(d) Greater of 0.80% of cross – sectional area and 6 bars of 12 mm

Q6. The slenderness ratio of a vertical column of a square cross section of 2.5 cm sides and 300 cm length is:
(a) 200
(b) 240
(c) 360
(d) 416

Q7. A short column 300 mm × 300 mm is reinforced with 4 bars of 20 mm dia. (Fe-415 grade). If concrete is M-20 grade, the max. axial load P_u (kN) allowed on it is:
(a) 1059.0
(b) 1159.0
(c) 1173.0
(d) None of these

Q8. According to IS : 456 – 2000, the maximum reinforcement in a column is:
(a) 4%
(b) 2%
(c) 6%
(d) 8%

Q9. A column is a compression member, the effective length of which exceeds three times of its least lateral dimension. This is applicable to:
(a) Rectangular and circular sections.
(b) I section and circular sections.
(c) Rectangular, circular and I section sections.
(d) All the shapes of sections.

Q10. Column may be made of plain concrete if their least lateral dimension by
(a) Two times
(b) Three times
(c) Four times
(d) Five times

SOLUTION
S1. Ans.(c)
Sol. reduction coefficient (Cr) = 1.25 – l_eff/(48.b)
Cr=1.25-4500/(48×250)
▭(Cr=0.875≈0.9)

S2. Ans.(b)
Sol. → minimum, cover in reinforcing column is 40 mm (or) diameter of main bar.
→ In case of minimum dimension column of 200 mm or below 200 mm whoes reinforcing bar not exceed 12 mm then nominal cover is 25 mm.
→ column is contact with fire & water minimum cover is 50 mm.

S3. Ans.(b)
Sol.

Transverse reinforcement
→diameter ≮[(ϕmax\/4 )¦6mm or]
=[(25\/4=6.25)¦6mm or]

▭(deiameter=8mm)

→ spacing≯

→ spacing ≯

▭(spacing=300)

S4. Ans.(a)
Sol. in helical reinforced column load carrying capacity increased by 5% as compared to tied column.

S5. Ans.(d)
Sol. → Minimum % reinforcement in column = 0.8% of gross area
→ max^m % reinforcement = 4% of gross area (if bars are lapped)
= 6% of gross area (it bars are not lapped)
→ No. of bar’s in circular column equal to ‘6’
→ No. of bar’s in rectangular column equal to ‘4’

S6. Ans.(d)
Sol. slenderness ratio (λ) = (Effective length)/(least radious of gyration)
Moment’ of inertia (I) = Ar²
〖2.5〗^4/12=〖2.5〗^2×r^2
r= √(〖2.5〗^2/12)
▭(r= 0.7216 cm)
λ=300/0.7216
▭(λ=415.69)

S7. Ans.(d)
Sol. size of column = 300 mm ×300 mm
Reinforced with 4-20 mmϕ
Grade of steel = 415
M-20
Axial load (Pu)= ?
σ_cc = 5 Mpa (for M-20 grade concrete)
Asc = 4×π/4×20^2
=1256 mm^2
σsc=190 Mpa (for Fe 415 grade steel)
Ac=[(300×300)-1256]
Ac=88744 mm^2
Pu = σccAc+σscAsc
=(5×88744)+(190×1256)
Pu=682360 N
▭(Pu=682.360 KN)

S8. Ans.(c)
Sol. Max^m reinforcement in column = 4% of gross area
(if bars are lapped)
& 6% of gross area (if bars are not lapped)

S9. Ans.(a)
Sol. if effective length is greater than three times of least lateral dimension is called column. And this is applicable only for square, rectangular & circular cross – section column.

S10. Ans.(b)
Sol. if ▭(leff/(least lateral dimension)≤3 )
Is called pedestal and it’s made by using plain cement concrete.

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