RCC Design Test 3

Result

RCC Design Test 3

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

If the unsupported length of a column is 3 m and its sides are 240 mm × 360 mm the minimum eccentricity so that the column is designed as an axially loaded column is _______

A
e_x = 18 mm, e_y = 14 mm

B
e_x = 14 mm, e_y = 18 mm

C
e_x = 20 mm, e_y = 20 mm

D
None of the above

Solution

Concept:
Minimum eccentricities:
\({e_{\;x\;min}} = {\rm{max}}\left\{ {\begin{array}{*{20}{c}} {\frac{{{\ell _x}}}{{500}} + \frac{D}{{30}}}\\ {20\;mm} \end{array}} \right.\)
\({e_{y\;min}} = {\rm{max}}\left\{ {\begin{array}{*{20}{c}} {\frac{{{\ell _y}}}{{500}} + \frac{b}{{30}}}\\ {20\;mm} \end{array}} \right.\)
ℓ_x, ℓ_y = Unsupported length
Calculation:
ℓ_x = ℓ_y = 3m = 3000 mm
D = 360 mm, b = 240 mm
\({e_{x\;min}} = \max \left\{ {\begin{array}{*{20}{c}} {\frac{{3000}}{{500}} + \frac{{360}}{{30}}}\\ {20\;mm} \end{array}} \right. = 18\;mm\)
= 20 mm
\({e_{y\;min}} = \max \left\{ {\begin{array}{*{20}{c}} {\frac{{3000}}{{500}} + \frac{{240}}{{30}}}\\ {20\;mm} \end{array}} \right. = 14\;mm\)
= 20 mm
Question 2
1 / -0

The reduction coefficient of a long RC column with effective length of 6 m and size 250 × 400 mm is

A
0.25

B
0.50

C
0.75

D
1.00

Solution

Concept:
The strength reduction coefficient method is used for the design of a long column. It is valid for Working Stress Method only.
Strength Reduction coefficient is given as
\({C_r} = 1.25 - \frac{{{\ell _{eff}}}}{{48\;b}}\)
Where, b = least lateral dimension of core and for the circular column with helical reinforcement it is the diameter of the core.
Calculation:
ℓ_eff = 6 m = 6000 mm, b = 250 mm
\(\therefore {C_r} = 1.25 - \frac{{6000}}{{48 \times 250}} = 0.75\)
Question 3
1 / -0

A column is subjected to direct compression and bending compression both such that the stress developed in direct compression is 2.5 N/mm² and the stress developed in concrete in bending compression is 3.6 N/mm². If M - 20 concrete is used in the column. Then identify which of the following statement is correct

A
Design is done by neglecting the bending compression.

B
Design is done on the basis of cracked section.

C
Design is done on the basis of uncracked section.

D
design is done on the basis of either cracked or uncracked section.

Solution

Concept:
If \(\frac{{\sigma \begin{array}{*{20}{c}} '\\ {cc} \end{array}}}{{{\sigma_{cc}}}} + \frac{{\sigma\begin{array}{*{20}{c}} '\\ {cbc} \end{array}}}{{{\sigma_{cbc}}}} \le 1\) (Design is done on the basis of uncracked section)
If \(\frac{{\sigma\begin{array}{*{20}{c}} '\\ {cc} \end{array}}}{{{\sigma_{cc}}}} + \frac{{\sigma\begin{array}{*{20}{c}} '\\ {cbc} \end{array}}}{{{\sigma_{cbc}}}} > 1\) (Design is done on the basis of cracked section)
\(\sigma\begin{array}{*{20}{c}} '\\ {cc} \end{array}\)⇒ Developed (Calculated) stress in concrete in direct compression
\(\sigma\begin{array}{*{20}{c}} '\\ {cbc} \end{array}\)⇒ Developed (Calculated) stress in concrete in bending compression
σ_cc⇒ Permissible stress in concrete in direct compression
σ_cbc⇒ Permissible stress in concrete in bending compression
Calculations:
For M - 20, concrete
σ_cbc = 7 N/mm²
σ_cc = 5 N/mm²
Now, \(\sigma\begin{array}{*{20}{c}} '\\ {cc} \end{array} = 2.5\;N/m{m^2}\)
\(\sigma\begin{array}{*{20}{c}} '\\ {cbc} \end{array} = 3.6\;N/m{m^2}\)
\(\frac{{2.5}}{5} + \frac{{3.6}}{7} = 1.014 > 1\)
Hence design is done on the basis of cracked section
Question 4
1 / -0

An isolated footing of 2 m × 3 m along the plane of bending moment is provided under the column. If the net maximum and minimum pressure on the soil under the footing are 95 kN/m² and 55 kN/m² respectively and the axial load is 450 kN, then the bending moment at its base will be (The centre of gravity of column and footing coincides)

A
60 kNm

B
50 kNm

C
70 kNm

D
80 kNm

Solution

Area of footing = 2 × 3 = 6 m²
Section modulus,
\(Z = \frac{{b \times {d^2}}}{6} = \frac{{2 \times {3^2}}}{6} = 3 {m^3}\)
Stress, \(\sigma = \frac{P}{A} \pm \frac{M}{Z}\)
\(95 = \frac{P}{A} + \frac{M}{Z}\) -------------- (i)
\(55 = \frac{P}{A} - \frac{M}{Z}\) -------------- (ii)
From (i) and (ii), we get
\(\begin{array}{l} 40 = 2 \times \frac{M}{Z}\\ M = 2 \times \frac{M}{3} \end{array}\)
⇒ M = 60 kNm