A WORKSHOP ON ADVANCED CONCRETE MIX DESIGN

A WORKSHOP ON ADVANCED CONCRETE MIX DESIGN


Course Material



Comments

Post a Comment

Popular posts from this blog

Design of Concrete Structures - II: Multiple Choice Questions (MCQ) & Answers

Image
  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.   ...
Read more

Some Books for Engineering Competitive Exam Preparation

Image
 Some Books for Engineering Competitive Exam Preparation  Engineering Graphics (Includes Free Drawing Sketch Book) For SPPU Sem 2 First Year Savitribai Phule Pune University  Hardcover – 1 January 2019 by  Panchal A Rajkumar   (Author),  Kulkarni J.V   (Author) Building Planning and Design by Dr Amol d Pawar
Read more

What is the formula for taking load from two way slab for beam design?

Image
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...
Read more