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1. 1. State the primary load cases to be considered for design. DEAD LOAD [IS 875 -PART 1] LIVE LOAD [IS 875 -PART 2] WIND LOAD [IS 875 -PART 3] SNOW LOAD [IS 875 -PART 4] SEISMIC LOAD [IS 1893 -2016] 2. What is One – Way slab? One way slab is a slab which …
Bunti Parashar
updated on 21 May 2023
1. 1. State the primary load cases to be considered for design.
DEAD LOAD [IS 875 -PART 1]
LIVE LOAD [IS 875 -PART 2]
WIND LOAD [IS 875 -PART 3]
SNOW LOAD [IS 875 -PART 4]
SEISMIC LOAD [IS 1893 -2016]
2. What is One – Way slab?
One way slab is a slab which
longer span ration to shorter
span ration is grater than 2 .
-One way slab is a plane surface
of uniformly loaded which
deforms into into a cylindrical
shape ,
hense bending moments devlops in
one direction .
3. What is the value of unit weight of structural steel and soil?
the value of unit weight of structural steel = 7850 kg/m^3
and soil = 20
4. Name few sections that can be defined
using Properties tab in STAAD Pro.
rectangular section
circular , I -section , T -section
tapered -I , TRAPEZOIDAL ,
AND custom geometric section .
5. Define Primary Beams.
primary beams are beams directly
connected to column on both ends .
-fuction on this beam is directly transfer
load to column to foundation .
2. 1. State the factors and parameters
that influence the design pressure intensity
of a high-rise building.
ANS-
2. Mention the load transfer process of a Reinforced Concrete Building.
3. 1. Differentiate One-Way and
Two way slabs. Elaborate each slab type.
ONE WAY SLAB - TWO WAY SLAB
-THIS beam supported -supported by beam
all 4 sides
by beam on 2 opposite
sides .
-load is carried 1 - load is carried
direction perpendicular both direction .
to support beam .
-LENGTH /WIDTH - length to width
greater than equl to less than 2
2
-deflected shape is -deflected shape is
is cylindrical. saucer like shape .
2. Explain the concept of secondary
beams with proper sketch.
ans-
secondary beam primary connected to
primary beams .
this is a simply supported beams .
-this beams trasfer load to
primary beams than primary beam
trasfer load to column to foundation .
-primary beams are rests on column
but secondary beams rest or between
in primary beams .
-secondary beam provided at the
ends in pinned connection .
3. Mention the Primary load cases.
a. State the Indian Standards for each load case
ans-
DEAD LOAD [IS 875 -PART 1]
LIVE LOAD [IS 875 -PART 2]
WIND LOAD [IS 875 -PART 3]
SNOW LOAD [IS 875 -PART 4]
SEISMIC/earthquake LOAD [IS 1893 -2016]
b. Detail the parameters of each load case
some permament parameters of load cases are given below
-self weight -specify load case where self weight of structure can be defined.
-Standard -Specifies the load case where any load can be defined.
-Creep-Defines a special load case for specific calculation.
Prestress-This type is used for calculations of prestressed structures.
c. Brief each load case
1.dead load -
it is a self weight of
whole structure .
almost dead load cover 50 to 60%
of total load .
Live load is the second major load on building
From its name we can easily understand that
the live load is a load of live things,
such as humans loads, animals load and etc.
Live load should be taken from IS : 875 part 2
in which live load is changed as per
type of building and its uses.
Tall buildings is mostly affected by wind load.
Wind load is a type of horizontal load that is
acting on a various faces of building.
Wind load is depends on a wind
intensity which is keep changing as per location.
IS:875 part 3 is provide the various codal provision of wind load.
Snow load is also become a more
important in some cool places such as
jammu & kashmir, himalay, uttrakhand and other locations.
Is 875 part 4 includes the all calculation
provision of snow load on building.
A horizontal load is acting on a building during time of earthquake.
Earthquake load is like a sudden Impact load
on building due to movement or vibration of surrounding earth.
Earthquakes load become most dangerous
for building due to its sudden Impact with
high Intensity at leas time. But nowadays,
building is designed to resist earthquake load.
Building is designed accordingly IS 1893-2016
which includes the all details of seismic
calculation and consideration on building.
4. Detail the steps involved in assigning Properties
and Support to a model using STAAD Pro sequentially.
ans-steps involved in assigning Properties
-open staad pro
-goto properties
-open a dialouge box
click define option .
-next dialouge box will be
open ,
-there diffrent types section
are avilable like circle , reactangle , trapezoidal and t section .
-chosse any one section give dimesion and add on it
and assign to view than okay .
ASSIGNING Support -
-Go to support
-open a dialouge box > go to creat > diffrent
types of support open like fixed , fixed but ,pinned etc.
-click on desired support > click on add
-than go to assign on view . ok
5. Brief the steps to be followed to create
and assign secondary beams in STAAD Pro
ans-
-creat new analytical model on metric unit
-geometry tab creat diffrent nodes in
diffrent x , y , z direction .
-add the beams .
-next goto properties provide reactangle section
-next creat material
-next support and assign in view
-next creat secondary beam in here
we select a beam than go to right click on
mouse and goto insert node in beam than next click on
add mid point > ok
-add beam > perpendicular intersection
in this way we able to creat a secondary beam
1. Calculate the wind pressure for the given building
a. Building Dimension : 20m x 30m x 20m height RCC
b. Building usage : Hospital block in city centre
c. Location : Darjeeling
steps -
1.calculate design wind speed -
Vz =Vb . K1 .K2 .K3 .K4
K1= RISK COFFICIENT TAKE values is 875 part 3
2015 -table 1
k2 =terrain roughness and height factor
k3= topography factor is code 875 : part 3 2015 clause 6.3.3
k4 =importance factor .
acc. to IS CODE -
Vb =47m/s
k1= 1.07
k2=.8
k3=1
k4=1
calculation put values in above formula
Vz =Vb .K 1.K2 K3. K4
Vz=47 x1.07 x0.8 x1x1
=40.23 m/s
wind pressure -
Pz = 0.6 Vz ^2
=0.6 x [40.23]^2
=971.07 N
=.97107 KN
SIMILARLY calculate diffrent height
2. Calculate the Dead load and Live load for
the above building with reference to IS
standards ( assume suitable sections )
ans-
D.L . DUE TO SLAB -
assuming slab thickness = 150 mm
dead load due to slab = L X B XH X SLAB THICKNESS
=20 X30 X25 0.15
=225 KN .
IN This project total 5 floor slab
so total dead load due to slab
= 5 x225 = 1125 KN .
DEAD LOAD DUE TO BEAMS -
Assume beam cross section
350 x450 mm
-dead L. due to beams = size of beam x height
=350 x450 mm x0.25
=3.94 KN /M Take it 4 kn /m
super imposed load due to partition wall -
assume 150 mm thickness brickwork
height of brick work = floor ht.- depth of beam
3.75 -.45 =3.3 m
total dead load =3.3 x.15 x .18 = 8.91 KN/M .
--LIVE LOAD
USE CODE IS 875 -PART 2
GIVE LIVE LOADS VALUES FOR
DIFFRENT ROOMS
in here hospital building so
acc . to code
OCCUPANCY LIVE LOAD [KN/M^2]
-For ward room
dressing rooms 2
dormitries and lounges .
-kitchen and lab 3
-dining rooms ,cafeteria 3
-toilets and bathrooms 2
-x -ray and operating
rooms 3
-office and opd rooms 2.5
-corridoor and passage 4
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