Week 10 Challenge
1)
2)
AIM & INTRODUCTION:
from the STAAD file attached and for given loading taper the beam for the BMD envelope and check utility ratio within limit .
Given data:
- Staad file of the model
- Loading
- Design parameters assigned to the members
Procedure in Details:
From the attached file of the model following utilisation factor of various members as indicated in the table given below found in the post processing command after analysing the structure . print screen of model indicating utilization factors is also attached
Section of various members have been optimized using Taper section as per Bending Moment diagram to such an extent that all the members are utilized more than 90% . Print screen after optimization of design is also attached below:
print screen of BMD and utility Ratio after optimization is given below-
3d rendering view of the model after optimization of sections is given below:
RESULT: As per challenge Beams have been tapered as pe BMD envelope and utiility ratio of various members have been checked and are with in limit .staad file after tapering the beam is attached
3)
AIM & INTRODUCTION:
Wind pressure on the wall of given industrial building is to be calculated:
Given data:
Dimension of building = 10 m x 5 m
Height = 6m
Bay spacing = 2.5 m
Wall opening = 15 %
Builsing is situated in chennai
Assuming -1) Terrain Category 3
2) Design life of the structure 50 years
3)Topography :plane with upwind slope 0<3degree
Roof angle = 5 degree
PROCEDURE IN DETAILS:
Model with given data has been prepared as given vbelow:
4)
AIM & INTRODUCTION:
Mono sloped structure (20 degree ) of width 10 m & length 30 m with bay spacing of 6 m having np wall sheeting is to moddeled in staad pro .dead load ,live load and wint load on the structure is to be applied . static check is also to be performed to see if the total load applied on the structure is captuared in support reaction for each load case .
given data :
self weight factor= 1.1
roof sheet loading = .5kn/m^2
purlin load = 6kn/m^2
live load on roof = 1.5 kn/m^2
wind load on rafter(pressure) = 2 kn/m^2
PROCEDURE IN DETAILS :
Model with given data has been prepared using staadpro and loading is also assigned as per given data.Here it is to mention that loading on thf Gable frame is half of main frame.Load have been multiplied with bay width in order to arrive loading in KN/m for assigning it on members . Staad file for reference is attached .
Print screen of model with self load-
PRINT screen of model with DEAD load:
PRINT screen of model with DEAD load:
PRINT SCREEN OF MODEL WITH DEAD LOAD:
PRINT SCREEN OF MODEL WITH WIND LOAD:
RUN ANALYSIS COMMAND HAS BEEN PERFORMED SUCCESFULLY WITH 0 ERROR 0 NWARNINGS & 0 NOTE AS SHOWN BELOW:
STATIC LOAD ANALYSIS:
LOAD CASE -1(DL)
SUMMATION FORCE-Y = -7493.65 KN
SUMMATION FORCE-Y = 7493.65 KN
LOAD CASE -2(LL)
SUMMATION FORCE-Y = -478.87 KN
SUMMATION FORCE-Y = 478.87 KN
LOAD CASE -3(WL)
SUMMATION FORCE-Y = -360.00 KN
SUMMATION FORCE-Y = 360.00 KN
LOAD CASE -4(EQUIVALENT LOAD AS PER IS 456)
SUMMATION FORCE-Y = -11958.79 KN
SUMMATION FORCE-Y = 11958.79 KN
LOAD CASE -5(EQUIVALENT LOAD AS PER IS 456)
SUMMATION FORCE-X = 432.00 KN
SUMMATION FORCE-Y = -9567.03 KN
SUMMATION FORCE-X = -432.00 KN
SUMMATION FORCE-Y = 9567.03 KN
LOAD CASE -6(EQUIVALENT LOAD AS PER IS 456)
SUMMATION FORCE-X = -432.00 KN
SUMMATION FORCE-Y = -9567.03 KN
SUMMATION FORCE-X = 432.00 KN
SUMMATION FORCE-Y = 9567.03 KN
LOAD CASE -7(EQUIVALENT LOAD AS PER IS 456)
SUMMATION FORCE-Y = -9567.03 KN
SUMMATION FORCE-Y = 9567.03 KN
LOAD CASE -8(EQUIVALENT LOAD AS PER IS 456)
SUMMATION FORCE-X = 540.00 KN
SUMMATION FORCE-Y = -11958.79 KN
SUMMATION FORCE-X = -540.00 KN
SUMMATION FORCE-Y = 11958.79 KN
LOAD CASE -9(EQUIVALENT LOAD AS PER IS 456)
SUMMATION FORCE-X = -540.00 KN
SUMMATION FORCE-Y = -11958.79 KN
LOAD CASE -10(EQUIVALENT LOAD AS PER IS 456)
SUMMATION FORCE-X = 540.00 KN
SUMMATION FORCE-Y = -7175.27 KN
SUMMATION FORCE-X = -540.00 KN
SUMMATION FORCE-Y = 7175.27 KN
LOAD CASE -11(EQUIVALENT LOAD AS PER IS 456)
SUMMATION FORCE-X = -540.00 KN
SUMMATION FORCE-Y = -7175.27 KN
SUMMATION FORCE-X = 540.00 KN
SUMMATION FORCE-Y = 7175.27 KN
LOAD CASE -12(EQUIVALENT LOAD AS PER IS 456)
SUMMATION FORCE-Y = -7175.27 KN
SUMMATION FORCE-Y = 7175.27 KN
RESULT
1) MODEL IS PREPARED
2) LOADS ARE APPLIED .THUS STATIC EQUILIBRIUM IS MAINTAINED