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Unit 1: Staircases (Limit state method) Multiple Choice Questions 1. The effective span of the staircase supported at top and bottom risers by beams spanning in parallel with the riser is A. Clear distance between beams B. Distance between c/c of beams C. Clear distance between beams + d D. None of the above ANS: B 2. If R and T are rise and tread of a stair spanning horizontally, the steps are supported by a wall on one side and by a stringer beam on the other side, the steps are designed as beams of width A. R + T B. T – R C. √ ( R 2 +T 2 ) D. R - T ANS: C 3. If T and R are tread and rise respectively of a stair, then A. 2R + T = 60 B. R + 2T = 60 C. 2R + T = 30 D. R + 2T= 30 ANS: A 4. The minimum head room over a stair must be A. 200 cm B. 205 cm C. 210 cm D. 230 cm ANS: C 5. The number of treads in a flight is equ
DESIGN OF CONCRETE STRUCTURES - I MCQ Unit 1: Introduction 1) The maximum compressive strain in concrete in axial compression is taken as A. 0.001 B. 0.002 C. 0.003 D. 0.040 ANSWER: B 2) The flexural strength of concrete is given by A. 0.6√fck B. 0.7√fck C. 0.5√fck D. 0.8√fck ANSWER: B 3) Tolerance on placing the reinforcement for effective depths 200 mm or less is A. +/-5mm B. +/-6mm C. +/-8mm D. +/-10mm ANSWER: D 4) Tolerance on placing the reinforcement for effective depths more than 200 mm is A. +/-5mm B. +/-6mm C. +/-8mm D. +/-10mm ANSWER: A 5) The minimum frequency of sampling of concrete of each grade for 1-5m3 shall be A. 1 B. 2 C. 3 D. 4 ANSWER: A 6) The minimum frequency of sampling of concrete of each grade for 51m^3 and above shall be……samples. A. 4 + one additional for each 50m3 B. 5 + one additional for each 50m3 C. 6 + one additional for each 50m3 D. 4 - on
How to calculate load from two-way slab on beam? As per Clause No 24.5 of IS 456: 2000 , the loads on beams supporting solid slabs spanning in two directions at right angles and supporting uniformly distributed loads, may be assumed to be in accordance with following Fig. Let us consider the following example: ABCD is a two-way slab ( L/B = 1.5<2 ) Assume: Dead load of slab : 0.125 x 25 = 3.1255 kN/m^2 (100 mm thick slab) Live load on slab : 3 kN/m^2 Total load = 5.5 kN/m^2 Now for finding force on beam BC the load will come from the triangular hatched part of slab So we multiply the load by vertical length of triangle to get load on beam per unit length 2 ∗ 5.5 = 11 k n / m 2 ∗ 5.5 = 11 k n / m fig (b) shows the disribution of load on beam BC ( same will be for beam DC ) (If we convert this load on beam as uniform load we will get maximum bending moment of 11 kN/m but now as we have considered this load in form of triangle so we will get a max
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