SOLUTION 1
P@U=1000 KN
F@Y =250 MPA
D=1200 MM
d=1150 MM
T@F=25 MM
T@W=10 MM
TAU@CR,E=K@V*PI^2*E/(12*(1-MU^2)*(d/T@W)^2)=5.35*PI^2*2*10^5/(12*(1-0.3^2)*(1150/10)^2
TAU@CR,E=73.12 N/MM^2
LAMBDA=SQRT(F@YW/(SQRT(3)*TAU@CR,E))=SQRT(250/(SQRT(3)*73.12))=1.404
TAU@B=F@YW/SQRT(3)*)(LAMBDA@W)^2=250/SQRT(3)*)(1.404)^2=73.12 N/MM^2
V@CR=A@V*TAU@B=D*T@W*TAU@B=1150*10*73.12= 840936.19 N OR 840.93 KN
V>V@CR
1000 KN> 840.96 KN
HENCE THE WEB IS INADEQUATE HENCE INCREASE THE WEB THICKNESS
SOLUTION 2
P@U=2000 KN
F@Y =250 MPA
D=1500 MM
d=1440 MM
T@F=30 MM
T@W=12 MM
TAU@CR,E=K@V*PI^2*E/(12*(1-MU^2)*(d/T@W)^2)=5.35*PI^2*2*10^5/(12*(1-0.3^2)*(1440/12)^2
TAU@CR,E=67.1580 N/MM^2
LAMBDA@W=SQRT(F@YW/(SQRT(3)*TAU@CR,E))=SQRT(250/(SQRT(3)*67.1580))=1.4660
TAU@B=F@YW/SQRT(3)*)(LAMBDA@W)^2=250/SQRT(3)*)(1.4660)^2=67.1580 N/MM^2
V@CR=A@V*TAU@B=D*T@W*TAU@B=1440*12*67.1580= 1160490 N OR 1160.49 KN
V>V@CR
2000 KN> 1160.49 KN
HENCE THE WEB IS IN -ADEQUATE
HENCE WE NEED TO PROVIDE THE STIFFNER
TAU@B=V@CR/A@V=2000/(1440*12)=115.74 N/MM^2
LAMBDA=SQRT(F@YW/(SQRT(3)*TAU@B))=SQRT(250/(SQRT(3)*115.74))=1.1167
TAU@CR,E=F@YW/SQRT(3)*LAMBDA=250/SQRT(3)*1.1167=129.25 N/MM^2
K=12*(1-MU^2)*(d/T@W)^2*TAU@CR,E/(PI^2*E)=12*(1-0.3^2)*(1440/12)^2*129.25/(PI^2*2*10^5)
K=10.29
K=4+5.35/(C/d)^2
10.29=4+5.35/(C/1440)^2
C=1328 MM
HENCE PROVIDE STIFFNER AT 1328 MM SPACING
SOLUTION 3.
P@U=2000 KN
F@Y =250 MPA
ASSUMING d/T@W=67*EPSILON
1000/67=14.92
ADOPT T@W= 16MM
ADOPT d=1000 MM
TAU@CR,E=K@V*PI^2*E/(12*(1-MU^2)*(d/T@W)^2)=5.35*PI^2*2*10^5/(12*(1-0.3^2)*(1000/16)^2
TAU@CR,E=247.571 N/MM^2
LAMBDA@W=SQRT(F@YW/(SQRT(3)*TAU@CR,E))=SQRT(250/(SQRT(3)*247.571))=0.7635
TAU@B=F@YW/SQRT(3)*)(LAMBDA@W)^2=250/SQRT(3)*)(0.7635)^2=247.571 N/MM^2
V@CR=A@V*TAU@B=D*T@W*TAU@B=1000*16*247.571= 3961136 N OR 3961.136 KN
V<V@CR
2000 KN< 3961.136 KN
HENCE THE WEB IS ADEQUATE
HENCE ASSUME T@W=16 MM AND d=1000 MM
SOLUTION 4.
SPAN = 15 M
FACTORED UDL=100*1.5 =150 KN/M
V@A=1125 KN
V@B=1125 KN
MAX BENDING MOMENT=W*L^2/8=150*15^2/8=4218.75 KNM
DEPTH OF GIRDER
d=(M*K/F@Y)^0.33=(4218.75*10^6*67/250)^0.33=971.84 MM
ADOPT d=1000 MM
T@W=(M/F@Y*K^2)^0.33=(4218.75*10^6/250*67^2)^0.33=15.12 MM
ADOPT T@W=16 MM
DESIGN REQUIREMENT REGARDING THE MINIMUM WEB THICKNESS FOR THE CONDITION OF NO INTERMEDIATE STIFFNERS
CLAUSE 8.6.1.1 AND 8.6.1.2
d/T@W=1000/16=62.5<200*EPSILON=200*1=200 (FOR SERVICEABILITY)
d/T@W=1000/16=62.5<345*EPSILON^2=345*1^2=345 (TO AVOID FLANGE BUCKLING)
MINIMUM WEB THICKNESS CHECK
CLAUSE 8.4.2.1
HENCE T@W=16 MM
TO SATISFY THE REQUIREMENT OF PLASTIC
THUS B/T@F SHOULD BE LESS THAN 8.4 EPSILON FOR PLASTIC SECTION
ASSUMING B@F=0.3*1000=300 MM
ADOPT B@F=550 MM
T@F>B@F/(2*8.4)=550/(2*8.4)=17.85 =32.73 MM
ADOPT T@F=40 MM
NOTE =B@F AND T@F VALUE IS INCREASED TO SATISFY MOMENT CAPACITY
b/T@F=275/40=6.875<8.4*EPSILON
SHEAR CAPACITY
V<V@N
V@N=A@V*F@YW/(SQRT(3)*GAMMA@M0)=d*T@W*F@YW/(SQRT(3)*GAMMA@M0)
V@N=1000*16*250/(SQRT(3)*1.1*10^3)=2099.45 KN
V@N=2099.45 KN>1125 KN=V@Z
HENCE SAFE .
ALSO
0.6*V@N=0.6*2099.45=1259.67 KN >1125 KN=V@Z
HENCE SAFE IN SHEAR
MOMENT CAPACITY
Z@P=2*B@F*T@F*(D-T@F)/2=2*550*40*(1080-40/2)=22.88*10^6
M@D=1*22.86*10^6*250/(1.1*10^6)=5200 KN-M
M@D=5200 KNM> 4218.75 KNM
HENCE SAFE TO CARRY THE APPLIED MOMENT
CHECK FOR SHEAR BUCKLING OF WEB
TAU@CR,E=K@V*PI^2*E/(12*(1-MU^2)*(d/T@W)^2)=5.35*PI^2*2*10^5/(12*(1-0.3^2)*(1000/16)^2
TAU@CR,E=247.57 N/MM^2
LAMBDA@W=SQRT(F@YW/(SQRT(3)*TAU@CR,E))=SQRT(250/(SQRT(3)*247.57))=0.7635
TAU@B=F@YW/SQRT(3)*)(LAMBDA@W)^2=250/SQRT(3)*)(0.7635)^2=247.57 N/MM^2
V@CR=A@V*TAU@B=D*T@W*TAU@B=1000*16*247.57= 3961145.843 N OR 3961.14 KN
V@CR >V@Z
3961.14 KN> 1125 KN
HENCE SAFE