Week 3 Challenge
SOLUTION 1
A ; PROBLEM
Z DIRECTION K FACTOR =0.8 L
Y DIRECTION K FACTOR =2.0 L
B ; PROBLEM
PICTURE 1
FRAME =BRACED, ASSUMING BASE TO BE PINNED SINCE NOT MENTIONED
K=1.0.
PICTURE 2
K=2.0
PICTURE 3
K= 1.2.
SOLUTION 2;
GIVEN DATA ;
HOT ROLLED SECTION=ISMB 300
FLANGE COVER PLATE =200*12 MM
LENGTH OF BEAM =3 M=3000 MM
FY = 250 N/MM^2
THE RELEVANT PROPERTIES OF ISMB 300
WEIGHT =46 KG/M= 451.10 N/M
AREA=58.6 CM^2= 5860 MM^2
IXX=8990 CM^4= 8990*10^4 MM^4
IYY= 486 CM^4= 486 *10^4 MM^4
RXX=12.4 CM= 124 MM
RYY=2.86 CM= 28.6 MM
AREA= 5860+(2*200*12)=10660 MM^2
Y=150+12/2=156 MM
IZZ FOR PLATES =2(I+AY^2)=2(200*12^3/12+(200*12)*156))=116870*10^3 MM^4
TOTAL IZZ=8990*10^4 MM^4+116870*10^3 MM^4=206770.4*10^3 MM^4
RZZ=SQUAREROOT(206770.4*10^3 /10660)=139.27 MM
AREA= 5860+(2*200*12)=10660 MM^2
IYY FOR PLATES =2*12*200^3/12=16000*10^3 MM^4
TOTAL IYY=486*10^4 MM^4+16000*10^3 MM^4=20860*10^3 MM^4
RYY=SQUAREROOT(20860*10^3 /10660)=44.2362 MM
The bottom of the column is Fixed and Top as Roller (Rotation Fixed, translation free).=
KL=1.20 L
KL=1.2*3000
KL= 3600 MM
LAMBDA=KL/R=3600/44.2362=81.3812
SINCE RY IS SMALL THE BUCKLING WILL BE ABOUT Y-Y AXIS
BUCKLING CURVE CLASSIFICATION
FROM Using Buckling curve classification & slenderness ratio using table 9c in code,
USING LAMBDA =KL/R=81.3812 AND FY =250 N/MM^2
FCD=133.9282 N/MM^2
DESIGN STRENGTH
P@D=AE*FCD=10660*133.9282=14276774.612 N OR 1427.67 KN
SOLUTION 3
CIRCULAR TUBE
ASSUMING DESIGN COMPRESSIVE STRESS FCD AS 140 MPA
L= 2M =2000 MM
FACTORED AXIAL LOAD =400 KN =400 *10^3 N
PD=AE*FCD
400*10^3 =AE*140
AE=2857.14 MM^2=28.57 CM^2
SELECTING HEAVY 2 CLASS STEEL TUBE OF NOMINAL BORE 150 MM
A=32 CM^2= 3200 MM^2
DIAMETER= 168.3 MM
THICKNESS= 6.3 MM
WEIGHT/M =25.2 KG/M=247.12 N/M
AREA=32 CM^2=3200 MM^2
I=1053 CM^4=1053*10^4 MM^4
R=5.73 CM=57.3 MM
CLASSIFICATION OF CROSS SECTION TABLE 2 OF IS800-2007
| epsilon |
ε">ε=SQUARE ROOT(250/FY)=SQUAREROOT(250/210)=1.091 |
D/T=150/6.3=23.80
CHECK
88 EPSILON^2=104.761
23.80<104.761
HENCE THE SECTION IS NON SLENDER.
RESISTANCE TO FLEXURAL BUCKLING=KL/R=1*2000/57.3=34.904
FROM TABLE 10 OF CODE WE HAVE TO USE CURVE B(ASSUMIMG THE SECTION IS COLD FORMED)
FROM TABLE 9B FOR KL/R=34.904 AND FY=210 MPA
AFTER INTERPOLATION WE GET
FCD=179.56 N/MM^2
PD=AE*FCD
PD=3200*179.56
PD=574615.04 N
PD=574.61 KN
PD=574.61> 400 KN
HENCE SAFE
SQUARE HOLLOW SECTION
ASSUMING DESIGN COMPRESSIVE STRESS FCD AS 140 MPA
L= 2M =2000 MM
FACTORED AXIAL LOAD =400 KN =400 *10^3 N
PD=AE*FCD
400*10^3 =AE*140
AE=2857.14 MM^2=28.57 CM^2
SELECTING SQUARE HOLLOW SECTION 150*150*6
A=33.63 CM^2= 3363 MM^2
DEPTH= 150 MM
THICKNESS= 6 MM
WEIGHT/M =26.40 KG/M=258.89 N/M
I=1145.91 CM^4=1145.91*10^4 MM^4
R=5.84 CM=58.4 MM
CLASSIFICATION OF CROSS SECTION TABLE 2 OF IS800-2007
| epsilon |
ε">ε=SQUARE ROOT(250/FY)=SQUAREROOT(250/210)=1.091 |
B=150-3*6=132 MM
b/t=132/6=22
CHECK
42 EPSILON^2=50
22<50
HENCE THE SECTION IS NON SLENDER
RESISTANCE TO FLEXURAL BUCKLING=KL/R=1*2000/58.4=34.24
FROM TABLE 10 OF CODE WE HAVE TO USE CURVE A(ASSUMIMG THE SECTION IS HOT ROLLED TYPE)
FROM TABLE 9B FOR KL/R=34.24 AND FY=210 MPA
AFTER INTERPOLATION WE GET
FCD=183.87 N/MM^2
PD=AE*FCD
PD=3363*183.87
PD=618377.38 N
PD=618.37 KN
PD=618.37> 400 KN
HENCE SAFE
SOLUTION 4;
DESIG OF LACED COLUMN
SINGLE LACING SYSTEM WITH INCLINATIONOF LACING BAR =45 DEGREE
GAUGE LENGTH FOR A 90 MM FLANGE =50 MM
FY = 250 N/MM^2
FU=410 MPA
ASSUMING DESIGN COMPRESSIVE STRESS, FCD AS 150 MPA
L= 10M =10,000 MM
FACTORED AXIAL LOAD =1000 KN =1000 *10^3 N
PD=AE*FCD
1000*10^3 =AE*150
AE=6666.67 MM^2=66.667 CM^2
AREA OF ONE SECTION 66.667/2=33.334 CM^2
SELECT TWO ISMC 300 AT 363 N/M
THE RELEVANT PROPERTIES OF ISMC 300 IS 808-1989 ARE
WEIGHT =36.3 KG/M=363 N/M
AREA=46.3 CM^2= 4630 MM^2
IZZ=6420 CM^4= 6420*10^4 MM^4
IYY= 313 CM^4= 313*10^4 MM^4
RZZ=11.8 CM=118 MM
RYY=2.6 CM=26 MM
T=13.6 MM
CYY=23.5 MM
AREA AVAILABLE FOR TWO SECTION=2*4630=9260 MM^2
ASSUMING THE COLUMN TO BE HINGED BOTH ENDS
K=1
KL=1*10000=10000
LAMBDA=KL/R=10,000/118=84.74
THE KL/R OF A BUILT OF COLUMN COLUMN SHOULD BE TAKEN AS 1.05*KL/R
1.05*10000/118=88.98
FROM Using Buckling curve classification & slenderness ratio using table 9c in code,
USING KL/R=88.98 AND FY=250 MPA
DOING INTERPOLATION
FCD=122.53 N/MM^2
PD=AE*FCD
PD=9260*122.53
PD=1134627.8 N
PD=1134.62 KN
PD=1134.62> 1000 KN
HENCE COLUMN IS SAFE
DESIGN OF LACING BAR
BACK TO BACK CONNECTED
CLEAR SPACING=S=D+2G
USING PARALLEL AXIS THEOREM
2*6420*10^4=2(313*10^4+4630(D/2+23.5)^2)
D=183.10 MM
S=D+2G
S=183.10+2*50
S=283.10 MM
ASSUME THETA =45 DEGREE
VERTICAL SPACING=C=2S TAN@THETA=2*283.1*TAN@45=566.2 MM
LENGTH OF LACING BAR =L=S/COS@THETA=283.1/COS@45=400.36 MM
THICKNESS OF LACING BAR=T=L/40=400.36/40=10.009 MM
CONSIDER THICKNESS AS 12 MM
WIDTH OF LACING BAR=B=3*D
ASSUME DIAMETER OF BOLT =16 MM
B=3*16=48 MM
ASSUME B=50 MM
PROVIDE 50ISF12
AREA OF LACING BAR A=B*T=50*12=600 MM^2
RADIUS OF GYRATION OF LACING BAR R=T/SQUAREROOT(12)=12/SQUAREROOT(12)-3.464 MM
LAMBDA=KL/R=400.36/3.464=115.57<145
HENCE SAFE
FROM Using Buckling curve classification & slenderness ratio using table 9c in code,
FOR KL/R=115.57 AND FY=250 N/MM^2
DOING INTERPOLATION WE GET
FCD=88.5287 N/MM^2
LOAD CARRYING CAPACITY
PD=AE*FCD
PD=600*88.5287
PD=53117.22 N
PD=53.1172 KN
V=VT/2
VT=2.5 PERCENT OF TOTAL LOAD
TRANSVERSE SHEAR =V=2.5/100*1134*10^3=28350 N OR 28.35 KN
COMPRESSIVE FORCE IN LACING BAR =(V/2 SIN45)=20046.47 N OR 20.04 KN
PL=53117.22 N> 20046.47 N
PL =53.1172 KN> 20.04 KN
HENCE LACING BAR IS SAFE
TENSILE STRENGTH OF LACING FLAT=0.9(B-d)T*FU/GAMMA@M1=0.9(50-18)*12*410/1.25=113356 N
F@Y*A@G/GAMMA@M0=250*50*12/1.1=136363 N
THUS THE TENSILE STRENGTH OF LACING OF LACING FLAT=113356 N>20045.47 N
HENCE THE LACING FLAT IS SAFE
CHECK SLENDERNESS RATIO
R MIN OF THE BUILT UP COLUMN=118 MM
L/R=10000/118=84.74
RMIN OF THE INDIVIDUAL CHORD=26.0 MM
L/R=566/26=21.77
LAMBDA OF BUILT OF COLUMN
LAMBDA E=SQUARE ROOT(84.74^2+3.14^2(9260/600)*400.36^3/(566*230)^2))
86.64<88.98
HENCE COLUMN IS SAFE
SOLUTION 5;
BATTEN SYSTEM
FY = 250 N/MM^2
FU=410 MPA
ASSUMING DESIGN COMPRESSIVE STRESS FCD AS 170 MPA
L= 5M =5000 MM
FACTORED AXIAL LOAD =2000 KN =2000 *10^3 N
PD=AE*FCD
2000*10^3 =AE*170
AE=11764.70 MM^2=117.647 CM^2
AREA OF ONE SECTION 117.647/2=58.8235 CM^2
SELECT TWO ISMC 400 AT 501 N/M
THE RELEVANT PROPERTIES OF ISMC 400 IS 808-1989 ARE
WEIGHT =50.1 KG/M=501 N/M
AREA=63.8 CM^2= 6380 MM^2
IZZ=15200 CM^4= 15200*10^4 MM^4
IYY= 508 CM^4= 508*10^4 MM^4
RZZ=15.4 CM=154 MM
RYY=2.82 CM=28.2 MM
T=15.3 MM
B@F=100 MM
CYY=24.2 MM
AREA AVAILABLE FOR TWO SECTION=2*6380=12760 MM^2
ASSUMING THE COLUMN TO BE HINGED BOTH ENDS
K=1
KL=1*5000=5000
LAMBDA=KL/R=5000/154= 32.46
THE KL/R OF A BUILT OF COLUMN COLUMN (BATTEN)SHOULD BE TAKEN AS 1.1*KL/R
1.1*5000/154=35.71
FROM Using Buckling curve classification & slenderness ratio using table 9c in code,
USING KL/R=35.71 AND FY=250 MPA FROM TABLE C
DOING INTERPOLATION
FCD=203.571 N/MM^2
PD=AE*FCD
PD=12760*203.571
PD=2597571.492 N
PD=2597.57 KN
PD=2597.57 KN> 2000 KN
HENCE COLUMN IS SAFE
DESIGN OF BATTEN
BACK TO BACK CONNECTED
USING PARALLEL AXIS THEOREM
SPACING BETWEEN CHANNEL
2*15200*10^4=2(508*10^4+6380(S/2+24.2)^2)
S=255.1 MM= 250 MM
THERFORE PROVIDE TWO ISMC 400 AT SPACING OF 250 MM BACK TO BACK
SPACING BETWEEN BATTENS;
L/RYY SHOULD BE LESS THAN 0.7*SLENDERNESS RATIO OF BUILT UP COLUMN
HENCE
L/RYY<0.7(L/R)
L<0.7(L/R)RYY
L<0.7*35.71*28.2
L<704.94 MM
ALSO
L/RYY SHOULD BE LESS THAN 50.THEREFORE
L/RYY<50
L<50*28.2
L<1210 MM
PROVIDE THE BATTEN AT A SPACING OF 700 MM
SIZE OF END BATTEN DE
DE=S+2*CYY> 2*B@F
DE=250+2*24.2=298.4= 298.4 MM=APPROX 300 MM
B@F= 100 MM FROM IS 808 1989
2*B@F=2*100=200 MM
DE=300 MM>2*B@F=200 MM
PROVIDE A 62 MM OVERLAP OF BATTEN ON CHANNEL FLANGE FOR WELDING
LENGTH OF BATTEN =250+2*62= 374 MM= APPROX 380 MM
THICKNESS OF BATTEN =1/50*380=7.60 MM=APPROX 8 MM
PROVIDE 380*300*8 MM END BATTEN
SIZE OF INTERMEDIATE BATTEN
OVERALL DEPTH =3/4*DE=3/4*300 = 225 MM
PROVIDE 380*225*8 MM INTERMEDIATE BATTEN
DESIGN FORCES
TRANSVERSE SHEAR=2.5 PERCENT OF AXIAL LOAD
V@T=2.5*2597.57*10^3/100=64939.25 N OR 64.9392 KN
LONGITUDINAL SHEAR=V@B=V@T*L/NS=64939.25*700/(2(250+2*62/2))=72848.51 N OR 72.8485 KN
MOMENT M=V@T*L/2N=64939.25*700/(2*2)=11.3643*10^6 N-MM OR 11.36 KN-M
CHECK
END BATTEN
SHEAR STRESS=72848.51/(300*8)=30.3535 N/MM^2<250/(SQUAREROOT(3)*1.1)=131.21 N/MM^2
HENCE SAFE
BENDING STRESS =11.3643*10^6*6/(8*300^2)=94.7025 N/MM^2<250/1.1=227 N/MM^2
HENCE THE BATTEN ARE SAFE
FOR INTERMEDIATE BATTEN
SHEAR STRESS=72848.51/(225*8)=40.47 N/MM^2<250/(SQUAREROOT(3)*1.1)=131.21 N/MM^2
BENDING STRESS =11.3643*10^6*6/(8*225^2)=168.36 N/MM^2<250/1.1=227 N/MM^2
HENCE THE BATTEN ARE SAFE
