Week 1 Challenge
1)AIM & INTRODUCTION: to find most effective truss system to carry gravity load
PROCEDURE: First one is Pratt truss which is most effective truss for vertical load and the second one is Howe truss which is opposite of pratt truss
RESULT:
TRUSS A IS MORE EFFECTIVE WHICH IS THE Pratt truss system
REPORT FORMAT:
Pratt trusses are most closely related to the howe truss ,where the diagonal members are in opposite direction causing an inverse compresssion/tension behaviour in members
2)AIM & INTRODUCTION: BENDING MOMENT FOR THE TWO DIFFERENT LOADINGS
REPORT FORMAT:
3) AIM & INTRODUCTION: TO FIND MOI AND WEIGHT AND AREA FOR GIVEN CTOSS SECTIONS:
PROCEDURE:
- DIVIDE SECTION INTO
SECTION 1 plate girder
b=160mm
h=10mm
y/=100mm
y1=10+160+10/2=175mm
I1=1/12(160*(10)3)=13.33*10^3 mm4
A1=160*10=1600mm2
d1 = 195-100 =95mm
section 2:
b=8mm
h=180mm
d2=10+180/2=100mm
I2=1/12(8*(180)3)=38.8*10^5mm4
A2=180*8 =1440mm4
d2=100-100 =0
section 3:
b=160mm
h=10mm
y/=100mm
y3=10/2=5mm
I3=1/12(160^3*(10))=3.41*10^6 mm4
A3=160*10=1600mm2
d3= 195-100 =95mm
total moment of inertia=1.33x10^4+1600*95^2 + 3.88x10^6 +1440x0 +3.41x10^6+1600x95^2
=2.88 x 10^11 mm4
total area =1600+1600+1440=4640 mm2
total weight = area x density
=4640*7850
=11.30x10^3 kg/m
b
c) from IS 808
ISBM 500
MOI=45200 X10^4 mm4
CSA =111 x 10^2 mm2
weight =86.9 kg/m
ISBM 300
MOI=313 X10^4 mm4
CSA =46.3 x 10^2 mm2
weight =36.3.9 kg/m
Total area = 157.3*10^3 mm2
Total weight = 123.2 kg/m
Centroid of i section y1=250mm
centroid of channel section y2 = (500 + 7.8) - 2.35
Y2=505.45 mm
centroid of built -up section(y[) =y1A1+y2A2/A1+A2
Y[ = (250*111*10^2)+(505.45 * 46.3*10^2)/(111*10^2*+46.3*10^2)
y[ =352.19 mm
Moment of inertia = (Moii + A xi^2 + Moic + A xC^2)
XC = (500 +7.8 - 2.35) - 325.19
XC =180.26mm
MOI = (45200 X 10^4 mm4 + 111 x 10^2 mm2 x 75.19^2 mm2 + (313 x 10^4 mm4+46.3 x10^2mm2 x 180.26 ^2 mm2)
MOI = 668.33 X10^6 mm4
b) AS PER IS 808 ISBM 300 - 300 mm X 140 mm X 7.7 mm
MOI = 8990 X 10^4 mm4
weight = 46 kg/m
for built up section:
moment of inertia:
A=12 X 160 =1920 mm2
x = 150 + (12/2) = 156 mm
MOI =8990 X 10^4 + 2(A X2)
=8890 X 10^4 + 2(1920 X 156^2)
MOI =183.35 X 10^4 mm4
CROSS SECTION AREA:
CSA = (58.6 X 10^2) + (12 X 150 X 2)
CSA= 5860+3840
CSA = 9700 mm2
weight :
weight = 46kg/m + (7050kg/m^3 x 3010 mm2)
weight = 76 kg/m
4) AIM & INTRODUCTION: CALCULATE THE LOAD
Load on Interior Beam :
Interior beam = 10 kn/m2 x 2 =20kn/m
Load on exterior Beam :
exterior beam = 10 kn/m2 x 1 =10kn/m
Load on interior Grider
interior grider = 20 kn/m2 x 4 =80kn/m
from two sides = 80 kn x 2
Interior grider = 160 kn spaced @ 2m
Load on exterior Grider :
exterior grider = 20 kn/m2 x 4 =80kn spaced @ 2m
5)
Given data
L=2000M
T = 12 X 10 ^-6/oC
change in lenth = 2000 x 12 x 10 ^-6 x (40 - (-20))
=1.44m
REPORT FORMAT:
