Architecture and Planning Miscellaneous-topic
- Super plasticizer is added in a concrete mix to
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Water reducers, retarders, and super - plasticizers are admixtures for concrete, which are added in order to reduce the water content in a mixture or to slow the setting rate of the concrete while retaining the flowing properties of a concrete mixture.
Correct Option: A
Water reducers, retarders, and super - plasticizers are admixtures for concrete, which are added in order to reduce the water content in a mixture or to slow the setting rate of the concrete while retaining the flowing properties of a concrete mixture.
- The planning norms for provision of schools in a given town is shown in the table below
Schools Population norm Land requirement per school
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104 to 104
Correct Option: B
104 to 104
Direction: A semi-circular stonearch of thickness 30 cm is provided over an opening in a brick wall. The wall has length 3.0 m, width 30 cm and height 3.0 m. The opening has span 1.0 m and height 2.0 m.
- The quantity of brickwork in the wall (in cu.m) is
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Quantity of Brickwork = (Area of wall × depth of wall) – (Area of opening × depth of opening)cum
= [(3 × 3 × 0.3) – (Volume of rectangular opening + Quantity of stone work from above + Volume of arched opening)
= 2.7 – [(1.5 × 1 × 0.3) + 0.183 + (3.14 × 0.5 × 0.5/2)] cum
= 2.7 – [0.45 + 0.183 + 0.39] cum
= 2.7 – [0.63 + 0.39] cum = 2.7 – 1.02 cum
= 1.68 cum Nearly 1.7 cumCorrect Option: D
Quantity of Brickwork = (Area of wall × depth of wall) – (Area of opening × depth of opening)cum
= [(3 × 3 × 0.3) – (Volume of rectangular opening + Quantity of stone work from above + Volume of arched opening)
= 2.7 – [(1.5 × 1 × 0.3) + 0.183 + (3.14 × 0.5 × 0.5/2)] cum
= 2.7 – [0.45 + 0.183 + 0.39] cum
= 2.7 – [0.63 + 0.39] cum = 2.7 – 1.02 cum
= 1.68 cum Nearly 1.7 cum
- The quantity of stone work in the semi-circular arch (in cu.m) IS
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π 
r + 1 
× t × 2 2 
π 0.5 + 0.3 × 0.3 × 0.3 = 0.18m³ 2 Correct Option: B
π r + 1 × t × 2 2 π 0.5 + 0.3 × 0.3 × 0.3 = 0.18m³ 2
Direction: A cantilever beam XY of 2.5 m span is supported at P and is subjected to 40 kN point load at free end Y.
- A uniformly distributed load (in kN/m) that will result in the same value of bending moment at the fixed end is
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Mmax (at P) = WL² 2 100 = W × (2.5)² 2
200 = 6.25 W
⇒ W = 32 KN/mCorrect Option: C
Mmax (at P) = WL² 2 100 = W × (2.5)² 2
200 = 6.25 W
⇒ W = 32 KN/m
