Week 1 Challenge
SOLUTION 1;
Truss type reasoning
PRATT TRUSS
1. IT CONSIST OF VERTICAL AND DIAGONAL MEMBERS THAT FORM AN "N" SHAPE OF PATTERN.
2. THE DIAGONAL MEMBERS ARE ARRANGED TO DEVELOP TENSILE FORCES
3. ALSO THEY CAN BE DESIGNED TO RESIST AXIAL TENSION FORCES ONLY.
4. THE VERTICAL MEMBER DEVELOP COMPRESSION FORCES.HENCE THEY ARE MORE CAPABLE OF RESISTING BUCKLING.
HOWE TRUSS
1. The Howe truss includes vertical members and diagonals that slope up towards the center, the opposite of the Pratt truss
2. In contrast to the Pratt truss, the diagonal web members are in compression and the vertical web members are in tension.
CONCLUSION.
PRATT TRUSS THAT IS TRUSS A IS BETTER THAN TRUSS B HOWE TRUSS
The Pratt truss disspipated the load more efficiently than the Howe truss, although both truss bridges dissipated the force significatnly more effectively than the beam bridge. In addition, the Pratt truss deflected the least and held the most, on average, while the beam bridge deflected the most and held the least. This brings one to the conclusion that the more rigid the structure is, the stonger it is and the more load it can hold. Therefore, the Pratt truss dissipated the force of the load the most effectively, due to its rigid design.
The Pratt truss will have its diagonals in tension, and so there will be a little less steel in the diagonals.
The Howe truss will have diagonals in compression and so, due to slenderness ratio considerations, there might be a little extra steel in the diagonals.
The maximum top chord compressive force will be slightly more than the bottom chord tensile force in the Pratt truss and you may need more steel for the top chord. You may not need that in the Howe truss.But when trusses have more tension members they have a tendency to deflect more, so the Pratt truss might record a slightly higher mid span deflection.
Solution 2. Deflection shape & Bending Moment Drawing (Qualitative approach)
REFER SHEET
Solution 3; Determine MOI, Weight, C/S area of sections below
1.
X1=150
X2=150
X3=150
Y1=15
Y2=530
Y3=1045
A1=9000
A2=8000
A3=9000
X**=A1*X1+A2*X2+A3*X3/A1+A2+A3
X**=9000*150+8000*150+9000*150/9000+8000+9000
X**=150
Y**=A1*Y1+A2*Y2+A3*Y3/A1+A2+A3
Y**=9000*15+8000*530+9000*1045/9000+8000+9000
Y**=530
I-XX-1=IG1+A1*(Y1-Y**)^2 =300*30^3/12+9000*(15-530)^2= 2.38777*10^9 MM^4
I-XX-2=IG2+A2*(Y2-Y**)^2 =8*1000^3+8000*(530-530)^2=666.6667*10^6 MM^4
I-XX-3=IG3+A3*(Y3-Y**)^2 =300*30^3/12+9000*(1045-530)^2= 2.38777*10^9 MM^4
I-XXX= 2.38777*10^9 MM^4+666.6667*10^6 MM^4+ 2.38777*10^9 MM^4
I-XXX=5.4420*10^9 MM^4
I-YY-1=IG1+A1*(X1-X**)^2=30*300^3/12+9000*(150-150)^2=67.5*10^6 MM^4
I-YY-2=IG2+A2*(X2-X**)^2=1000*8^3/12+8000*(150-150)^2=42.667*10^3 MM^4
I-YY-3=IG3+A3*(X3-X**)^2=30*300^3/12+9000*(150-150)^2=67.5*10^6 MM^4
I-YY=67.5*10^6 MM^4+42.667*10^3 MM^4+67.5*10^6 MM^4
I-YY=135.0426*10^6 MM^4
SOLUTION 3
2.
X1=40
X2=4
X3=40
Y1=5
Y2=100
Y3=195
A1=800
A2=1440
A3=800
X**=A1*X1+A2*X2+A3*X3/A1+A2+A3
X**=800*40+1440*4+800*40/800+1440+800
X**=22.9473
Y**=A1*Y1+A2*Y2+A3*Y3/A1+A2+A3
Y**=800*5+1440*100+800*195/800+1440+800
Y**=100
I-XX-1=IG1+A1*(Y1-Y**)^2 =80*10^3/12+800*(5-100)^2= 7.2267*10^6 MM^4
I-XX-2=IG2+A2*(Y2-Y**)^2 =8*180^3+1440*(100-100)^2=3888000 MM^4
I-XX-3=IG3+A3*(Y3-Y**)^2 =80*10^3/12+800*(195-100)^2= 7.2267*10^6 MM^4
I-XXX= 7.2267*10^6 MM^4+3888000 MM^4+7.2267*10^6 MM^4
I-XXX=18.3413*10^6 MM^4
I-YY-1=IG1+A1*(X1-X**)^2=10*80^3/12+800*(40-22.9473)^2=659.3023*10^3 MM^4
I-YY-2=IG2+A2*(X2-X**)^2=180*8^3/12+1440*(4-22.9473)^2=524.640*10^3 MM^4
I-YY-3=IG3+A3*(X3-X**)^2=10*80^3/12+800*(40-22.9473)^2=659.3023*10^3MM^4
I-YY=659.3023*10^3 MM^4+524.640*10^3 MM^4+659.3023*10^3MM^4
I-YY=1.8432*10^6 MM^4
SOLUTION 3.
3.
X1=80
X2=80
X3=80
X4=80
X5=80
Y1=6
Y2=18.55
Y3=162
Y4=305.45
Y5=318
A1=1920
A2=1834
A3=2108.26
A4=1834
A5=1920
X**=A1*X1+A2*X2+A3*X3+A4*X4+A5*X5/A1+A2+A3+A4+A5
X**=1920*80+1834*80+2108.26*80+1834*80+1920*80/1920+1834+2108.26+1834+1920
X**=80
Y**=A1*Y1+A2*Y2+A3*Y3+A4*Y4+A5*Y5/A1+A2+A3+A4+A5
Y**=1920*6+1834*18.55+2108.26*162+1834*305.45+1920*318/1920+1834+2108.26+1834+1920
Y**=162
I-XX-1=IG1+A1*(Y1-Y**)^2 =160*12^3/12+1920*(6-162)^2= 46748160 MM^4
I-XX-2=IG2+A2*(Y2-Y**)^2 =140*13.1^3+1834*(18.55-162)^2=37766100.91 MM^4
I-XX-3=IG3+A3*(Y3-Y**)^2 =7.7*273.8^3/12+2108.26*(162-162)^2= 13170728.9 MM^4
I-XX-4=IG4+A4*(Y4-Y**)^2 =140*13.1^3/12+1834*(305.45-162)^2=37766100.91 MM^4
I-XX-5=IG5+A5*(Y5-Y**)^2 =160*12^3/12+1920*(318-162)^2= 46748160 MM^4
I-XXX= 46748160 MM^4+37766100.91 MM^4+13170728.9 MM^4+37766100.91 MM^4+46748160 MM^4
I-XXX=182.199*10^6 MM^4
I-YY-1=IG1+A1*(X1-X**)^2=12*160^3/12+1920*(80-80)^2=4096000 MM^4
I-YY-2=IG2+A2*(X2-X**)^2=13.1*140^3/12+1834*(80-80)^2=2995533.333 MM^4
I-YY-3=IG3+A3*(X3-X**)^2=273.8*7.7^3/12+2108.26(80-80)^2=10416.56128MM^4
I-YY-4=IG4+A4*(X4-X**)^2=13.1*140^3/12+1834*(80-80)^2=2995533.333 MM^4
I-YY-5=IG5+A5*(X5-X**)^2=12*160^3/12+1920*(80-80)^2=4096000 MM^4
I-YY=4096000 MM^4+2995533.333 MM^4+10416.56128MM^4+10416.56128MM^4+4096000 MM^4
I-YY=14.1934*10^6 MM^4
SOLUTION 3
4.
Y1=8.6
Y2=482.8
Y3=491.4
Y4=503.9
Y5=462.8
Y6=462.8
A1=3096
A2=4749.12
A3=3096
A4=2127.84
A5=1224
A6=1224
Y**=A1*Y1+A2*Y2+A3*Y3+A4*Y4+A5*Y5+A6*Y6/A1+A2+A3+A4+A5+A6
Y**=3096*8.6+4749.12*482.8+3096*491.4+2127.84*503.9+1224*462.8+1224*462.8/3096+4749.12+3096+2127.84+1224+1224
Y**=389.639
I-XX-1=IG1+A1*(Y1-Y**)^2 =180*17.2^3/12+3096*(8.6-389.639)^2= 449586794.4 MM^4
I-XX-2=IG2+A2*(Y2-Y**)^2 =10.2*465.6^3+4749.12*(482.8-389.639)^2=127011661.7MM^4
I-XX-3=IG3+A3*(Y3-Y**)^2 =180*17.2^3/12+3096*(491.4-389.639)^2= 32136338.99 MM^4
I-XX-4=IG4+A4*(Y4-Y**)^2 =272.8*7.8^3/12+2127.84*(503.9-389.639)^2=27790965.24MM^4
I-XX-5=1-XX-6=IG5+A5*(Y5-Y**)^2 =13.6*90^3/12+1224*(462.8-398.639)^2= 7377699.071MM^4
I-XXX= 449586794.4 MM^4+127011661.7MM^4+32136338.99 MM^4+27790965.24MM^4
+7377699.071MM^4+7377699.071MM^4
I-XXX=651.281*10^6 MM^4
I-YY-1=17.2*180^3/12=8359200 MM^4
I-YY-2=465.6*10.2^3/12=41174.8704 MM^4
I-YY-3=17.2*180^3/12=8359200MM^4
I-YY-4=7.8*272.8^3/12=13196126.03 MM^4
I-YY-5=I-YY6=90*13.6^3/12=18865.92 MM^4
I-YY=8359200 MM^4+41174.8704 MM^4+8359200MM^4+13196126.03 MM^4+18865.92 MM^4+18865.92 MM^4
I-YY=29.993*10^6 MM^4
WEIGHT OF I SECTION =204.1 KG
WEIGHT OF C SECTION =22.608 KG
WEIGHT OF I SECTION WITH PLATE=75.4876 KG
WEIGHT OF I SECTION AND C SECTION =121.8081 KG
CROSS SECTION OF I SECTION = 26000 MM^2
CROSS SECTION OF C SECTION= 3040 MM^2
CROSS SECTION OF I SECTION WITH PLATE= 9616.26 MM^2
CROSS SECTION OF I SECTION AND C SECTION= 15516.96 MM^2
SOLUTION 4. Load calculation problem
GIVEN DATA : TOTAL LOAD =10KN/M^2
SLAB =(8*2)=ASPECT RATIO=8/2=4
HENCE ONE WAY SLAB
LOAD INTENSITY ON BEAM=TOTAL LOAD ON SLAB*TRIBUTARY WIDTH OF BEAM
TRIBUTARY WIDTH OF B1=0.5*(2+0)=1 M
TRIBUTARY WIDTH OF B2=0.5*(2+2)=2 M
UNIFORM LOAD ON BEAM B1= LOAD ON FLOOR *TRIBUTARY WIDTH
=10 KN/M^2*1 M
= 10 KN/M
UNIFORM LOAD ON BEAM B2= LOAD ON FLOOR *TRIBUTARY WIDTH
=10 KN/M^2*2 M
= 20 KN/M
EXTERIOR GIRDER= 160 KN
INTERIOR GIRDER= 320 KN
SOLUTION 5. Expansion joint size determination for temperature stated problem
L*=2000 M
T*=-20 C
T=40 C
ALPHA= 12*10^-6/C
TEMPERATURE DIFFERENCE=DELTA @T=T-T*=40-(-20)=40+20=60 C
LENGTH DIFFERENCE=DELTA L=ALPHA*(L*)*DELTA @T
=12*10^-6*2000*60=1.44 M
FINAL LENGTH
L=L*+DELTA L=2000+1.44=2001.44 M
