Challenge

Week 3 Challenge

ANS 1. THE PRIMARY LOAD CASES TO BE CONSIDERED ARE LIVE LOAD, DEAD LOAD, SEISMIC LOADING (HORIZONTAL AND VERTICAL), ROOF LOAD, WIND LOAD, SNOW LOAD, SOIL, RAINWATER, TRAFFIC, TEMPERATURE ETC.

ANS 2. ONE WAY SLAB IS THAT SLAB WHERE THE RATIO OF LENGTH TO BREADTH IS GREATER THAN TWO. THESE SLABS HAVE ONE DIMENSION SLIGHTLY GREATER THAN THE OTHER DIMENSION WHICH MAKES THEM ONE WAY SLAB. WHILE THESE ARE SUPPORTED BY BEAMS AS THEIR DIMENSIONS NEEDS EXTRA SUPPORT.

ANS 3. THE UNIT WEIGHT OF STRUCTURAL STEEL IS 78.5 KN/mm2.

THE UNIT WEIGHT OF SOIL VARIES BETWEEN 14KN/mm2 to 18KN/mm2.

ANS 4. FEW SECTIONS THAT CAN BE DEFINED BY USING THE PROPERTY ICON ARE RECTANGULAR, CIRCULAR, TRAPEZOIDAL, TAPERED I, TAPERED TUBE, GENERAL ETC.

ANS 5. PRIMARY BEAMS ARE THOSE BEAMS WHICH ARE PRIMARY LOAD TAKERS WHEN IT COMES TO TAKING LOAD AND THEY ARE SUPPORTED ON THE COLUMNS AND ARE END TO END, WHICH MAKES THEM DIFFERENT FROM SECONDARY BEAMS WHICH ARE NOT END TO END.

ANS 2 1. THE FACTORS WHICH AFFECT THE PRESSURE INTENSITY WHICH WILL DETERMINE THE DESIGN OF THE BUILDING ARE SEISMIC LOCATION OF THE BUILDING, WIND SPEED OF THE REGION AND ALSO THE SURROUNDINGS OF THE BUILDING WHICH WILL TELL US WHETHER THE SURROUNDING BUILIDINGS ACT AS OBSTACLES TO THE FLOW OF WIND OR NOT. OTHER FACTORS SUCH AS PRECIPITATION NATURE AND ITS FORM ALSO PLAYS AN IMPORTANT ROLE IN THE DESIGN OF BUILDING.

ANS 2. IN RCC BUILDING, THE LOAD IS TRANSFERRED FROM THE TOP OF THE BUILDING TO THE PRIMARY BEAMS THEN ONTO THE SECONDARY BEAMS AND THEN TO THE COLUMNS AND FROM THERE IT GOES TO THE NEXT FLOOR. IN THIS WAY THE LOAD GETS TRANSFERRED FROM ONE FLOOR TO ANOTHER FLOOR AND FINALLY DOWN TO THE SOIL BENEATH. IN THIS WAY LOAD GETS TRANSFERRED FROM TOP TO THE SOIL STRUCTURE.

ANS 3. 1. THE MAJOR DIFFERENCE BETWEEN TWO WAY SLAB AND ONE WAY SLAB IS THE FACT THAT IN ONE WAY SLAB THE LENGTH TO BREADTH RATIO IS GREATER THAN 2 WHILE IN TWO WAY SLAB THE RATIO IS LESSER THAN OR EQUAL TO 2. THE ONE WAY SLABS ARE GENERALLY SUPPORTED BY THE BEAM STRUCTURE ON TWO SIDES. IN TWO WAY SLABS THE LOAD DISTRIBUTION IS ALONG THE FOUR DIRECTIONS. HENCE THE LOAD TRANSFER IS ALONG THE FOUR DIRECTIONS. THE MAIN REINFORCEMENT IS IN BOTH DIRECTIONS IN TWO WAY SLAB WHILE IN ONE WAY SLAB THE MAIN REINFORCEMENT IS IN ONE DIRECTION.

ANS 2. THE SECONDARY BEAMS ACT AS SECOND FIDDLE TO THE LOAD CARRYING MAIN BEAMS WHICH IS KNOWN AS PRIMARY BEAMS. IN THIS WAY THEY ACT AS EXTRA LOAD CARRYING AND LOAD DISTRIBUTION MACHENISM FOR THE STRUCTURE.

IN THE ABOVE DIAGRAM BE IS THE SECONDARY BEAM WHILE ALL OTHERS ARE PRIMARY BEAMS.

ANS 3. THE PRIMARY LOAD CASE INCLUDES THE FOLLOWING. THEY ARE-

1. DEAD LOAD- IS 875 PART 1 - THIS IS THE CODE FOR DEAD LOAD. IT IS DESIGNED IN CONSONANCE WITH THIS CODE THAT THE DEAD LOAD ON A STRUCTURE IS DETERMINED.

2. LIVE LOAD -IS 875 PART 2- THIS IS THE LIVE LOAD OR SUPERIMPOSED LOAD ON A STRUCTURE. LIVE LOAD ON A BUILDING IS DETERMINED ACCORDING TO THIS CODE.

3. WIND LOAD IS 875 PART 3- THIS IS USED TO DETERMINE THE WIND LOAD ON A STRUCTURE. THIS IS SPECIFICALLY USED FOR HIGH RISE BUILDINGS WHERE THE EFFECT OF WIND IS TOO MUCH TO IGNORE.

4. SEISMIC LOAD IS 875 PART 4- THIS IS USED FOR DESIGNING STRUCTURES BUILT IN EARTHQUAKE PRONE STRUCTURES.

5. SNOW LOAD IS 875 PART 4- THIS IS USED TO DETERMINE SNOW LOAD ON THE TOP OF ROOFS IN A BUILDING.

6. TEMPERATURE LOAD, SOIL LOAD, FLUID LOAD, ACCIDENTAL LOAD, FLOOD LOAD- IS 875 PART 5- FOR ALL THE LOADS OTHER THAN THOSE MENTIONED IN PART 1 TO 4 ARE DETERMINED BY PART 5 OF IS 875.

ANS 4. THE DETAILED STEPS INVOLVED IN ASSIGNING PROPERTY AND SUPPORT TO THE STRUCTURE ARE-

1. WHEN WE HAVE CREATED BEAM AND NODES WE WILL GO TO THE PROPERTY OPTION AND THEN CHOOSE THE PROPERTY. IT CAN BE STEEL , CONCRETE, ALUMINIUM ETC.

2. IN THIS OPTION WE CAN GIVE THE VALUE OF DEPTH AND BREADTH OF BEAM WHICH WILL HELP IN ASSIGNING PROPERTY TO THE BEAM.

3. WE CAN ADD SUPPORT BY CLICKING ON THE SUPPORT. OUR SUPPORT CAN VARY FROM FIXED TO HINGED TO ENFORCED TO ANY OTHER WE CAN SPECIFY ACCORDING TO OUR REQUIREMENT.

ANS 5. STEPS TO CREATE SECONDARY BEAM IN STAAD PRO.

1. CREATE TWO NODES AND CONNECT THEM WITH BEAM.

2. WE CAN MIRROR THIS BEAM AND IN THIS WAY WE HAVE GOT FOUR BEAMS ON ALL FOUR SIDES.

3. THEN WE CAN CONNECT THE TWO BEAMS FROM THE MIDDLE AND THESE BEAMS ARE KNOWN AS SECONDARY BEAMS.

4. ALWAYS REMEMBER SECONDARY BEAMS NEVER WILL WEIGH UPON THE COLUMNS DIRECTLY.

ANS PRACTICAL-

NOW WE HAVE TO DETERMINE THE WIND PRESSURE AT THE BUILDING IN DARJEELING.

Vz= Vb*k1*k2*k3.

FOR DARJEELING VALUE OF Vb is 47m/sec ACCORDING TO MAP 1 IN IS 875 PART 3.

K1= 1.07 (BASED ON THE FACT THAT LIFE OF HOSPITAL IS 100 YEARS)

K2= 0.76 (IT IS CATEGORY 4 TERRAIN AND CATEGORY B BUILDING)

K3= 1 (TOPOGRAPHY FACTOR).

Vz= 38.22m/sec.

DESIGN WIND PRESSURE Pz = 0.6 *Vz*Vz

Pz = 22.932 m2/s2.

ANS 2. AIM- TO CALCULATE THE DEAD LOAD AND LIVE LOAD IN THE BUILDING.

PROCEDURE- 1. OUR BUILDING HAS LENGTH OF 30M AND 20M IS ITS WIDTH, WHILE THE HEIGHT IS 20M.

2. NOW WE HAVE GOT 6 FLOORS IN THIS BUILDING WHICH ARE OF HEIGHT 3M EACH.

3. NOW WE HAVE A FRAMES STRUCTURE WHICH IS SUPPORTED UPON COLUMNS AND

WITH SLABS UPON IT.

4. HEIGHT OF COLUMN IS 2.5M, DIAMETER OF CIRCULAR COLUMN IS 0.5M.

5. LENGTH OF BEAM IS 100M ON EACH FLOOR, DEPTH IS 0.3M, WIDTH IS .25M.

6. THERE IS A SLAB ON TOP OF EACH OF THIS STRUCTURE.

7. WE WILL TAKE THE DENSITY OF CONCRETE TO BE 1380KG/M3.

NOW WE WILL CALCULATE THE DEAD LOAD OF THE ABOVE STRUCTURE.

VOLUME OF COLUMN= 22/7*0.5*0.5*2.5

= 1.96 m3

MASS OF THE COLUMN= 1.96*1380

= 2708.25 KG.

MASS OF 14 SUCH COLUMNS ON GROUND FLOOR= 37915.5KG.

TOTAL MASS OF BEAM= 0.3*0.25**100*1380.

= 10350 KG.

TOTAL MASS OF SLAB = 0.2*30*20*1380

= 165600 KG.

TOTAL MASS OF A SINGLE FLOOR= 213865.5 KG= 2138.655 KN.

TOTAL LOAD OF THE BUILDING= 2138.8*6

= 12834 KN.

NOW FOR CALCULATION OF LIVE LOAD,

WE HAVE 2 OFFICE ROOMS AT EACH FLOOR, ONE OPD ROOM, AND SEVERAL OTHER ROOMS.

TO MAKE OUR CALCULATION EASIER WE WILL TAKE UDL OF 3 KN/M2 FOR EVERY FLOOR.

OUR AREA OF FIRST FLOOR IS 600M2.

TOTAL LIVE LOAD ON FIRST FLOOR IS 600*3 = 1800KN.

WE HAVE 6 SUCH FLOORS = 1800*6= 10800 KN.

THIS IS THE TOTAL LIVE LOAD OF THE STRUCTURE.

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