Week 4 Challenge

SOLUTION 1

A.

EPISILON VALUE FOR ALL SECTIONS=SQUAREROOT(250/F@Y)=SQUAREROOT(250/250)=1

ISMB 500

D=500

T@F==17.2

B@F=180

T@W=10.2

R=17

d=D-2(t@f+R)=500-2(17.2+17)=431.6

B@F/T@F=180/17.2=10.46<10.5*EPISILON=10.5*1=10.5

d/T@W=431.6/10.2=42.31<84*EPISILON=84*1=84

HENCE THE SECTION IS COMPACT

B.

ISMC-200

D=200

B@F=75

T@W=6.1

T@F=11.4

R=11.0

d=D-2(T@F+R)=200-2(11.4+11.0)=155.2

B@F/T@F=75/11.4=6.578<9.4*EPISILON=9.4*1=9.4

d/T@W=155.2/6.4=24.25<42*EPISILON=42*1=42

HENCE THE SECTION IS PLASTIC

C.

PLATE GIRDER

1300*450*20*12

D=1300

B@F=450

T@W=12

T@F=20

d=1260

B@F/T@F=450/20=22.5>13.6*EPISILON=13.6*1=13.6

d/T@W=1260/12=105<126*EPISILON=125*1=126

HENCE THE SECTION IS SLENDER

D.

DOUBLE TO DOUBLE ANGLE

2*ISA 75*75*6

D/T=75/6=12.5<15.7 EPISILON

HENCE THE SECTION IS SEMI COMPACT

E.

RHS

• RHS - 150 x 100 x10
• B=100-3*T=100-3*10=70
• B/T=70/10=7<29.3*EPISILON=29.3*1=29.3
• HENCE THE SECTION IS PLASTIC

G.

BUILT UP

1000*450*20*12

B=450-3T=450-3*12=414

D=1000-3*T=1000-3*20=940

B/T@F=414/20=20.7<29.3*EPISLON=29.3=29.3

D/T@W=940/12=78.33<84*EPISILON=84*1=84

HENCE THE SECTION IS PLASTIC

SOLUTION 2

KL =0.85*40=34 MM ALONG X-X AXIS

KL=1*10.36 =10.36 ALONG Y-Y AXIS

SOLUTION 3.

A.

SIGMA V=0

V@A+V@B=100+100+42

V@A+V@B=242 N

SIGMA M@B=0

V@A*7-100*5-100*2-6*7*7/2=0

V@A*7=500+200+147

V@A=121 N

V@B=121 N

CONSIDER A SECTION X-X AT A DISTANCE FROM LEFT HAND SUPPORT

TAKING MOMENT OF ALL FORCES AT SECTION X-X

M@XX=EID^2Y/DX^2=V@A*X-100*(X-2)-100*(X-5)-6*X*X/2

INTEGRATING ONCE WE GET

EI*DY/DX=121*X^2/2-100/2*(X-2)^2-100/2*(X-5)^2-6/2*X^3/3+C1

EI*DY/DX=121*X^2/2-100/2*(X-2)^2-100/2*(X-5)^2-6/6*X^3+C1

INTEGRATING AGAIN WE GET

EIY=121/6*X^3-100/6*(X-2)^3-100/6*(X-5)^3-6*X^4/24+C1X+C2

THEREFORE TO FIND THE CONSTANT USING BOUNDARY CONDITION

AT X=0,Y=0

X=7,Y=0

PUTTING 1 BOUNDARY CONDITION IN EQUATION 2

0=0+0+0+C2

C2=0

PUTTING 2 BOUNDARY CONDITION IN EQUATION 2

0=121*7^3/6-100/6*(7-2)^3-100/6*(7-5)^3-6*7^4/24+C1*7+C2

C1=-585.75

PUTTING C1 AND C2 IN EQUATION 2 WE GET

EIY=121/6*X^3-100/6*(X-2)^3-100/6*(X-5)^3-6*X^4/24-585.75 X+0

PUT X=3.5 IN EQUATION 4

EIY=121*3.5^3/6-100/6*(3.5-2)^3-100/6*(3.5-5)^3-6*3.5^4/24-585.75*3.5

EI*Y=-1222.99

Y=-1222.99/EI=-1222.99/20*10^6

Y=-0.00006 M

B.

CONVERTING ALL LOAD OF KN IN N

SIGMA V=0

V@A+V@B=2000+5000+7500

V@A+V@B=14500 N

SIGMA M@B=0

V@A*5-2000*3.5-5000*2.5-1500*5*5/2=0

V@A*5=7000+12500+18750

V@A=7650 N

V@B=6850 N

CONSIDER A SECTION X-X AT A DISTANCE FROM LEFT HAND SUPPORT

TAKING MOMENT OF ALL FORCES AT SECTION X-X

M@XX=EID^2Y/DX^2=V@A*X-2000*(X-1.5)-5000*(X-2.5)-1500*X*X/2

INTEGRATING ONCE WE GET

EI*DY/DX=7650*X^2/2-2000/2*(X-1.5)^2-5000/2*(X-2.5)^2-1500/6*X^3+C1

INTEGRATING AGAIN WE GET

EIY=7650/6*X^3-2000/6*(X-1.5)^3-5000/6*(X-2.5)^3-1500*X^4/24+C1X+C2

THEREFORE TO FIND THE CONSTANT USING BOUNDARY CONDITION

AT X=0,Y=0

X=5,Y=0

PUTTING 1 BOUNDARY CONDITION IN EQUATION 2

0=0+0+0+C2

C2=0

PUTTING 2 BOUNDARY CONDITION IN EQUATION 2

0=7650*5^3/6-2000/6*(5-1.5)^3-5000/6*(5-2.5)^3-1500*5^4/24+C1*5+C2

C1=-18599.99

PUTTING C1 AND C2 IN EQUATION 2 WE GET

EIY=7650*X^3/6-2000/6*(X-1.5)^3-5000/6*(X-2.5)^3-1500*X^4/24-18599.99X+0

PUT X=2.5 IN EQUATION 4

EIY=7650*2.5^3/6-2000/6*(2.5-1.5)^3-5000/6*(2.5-2.5)^3-1500*2.5^4/24-18599.99*2.5

EI*Y=-29352.834

Y=-29352.834/EI=-1222.99/25*10^6

Y=-0.001174 M

C.

SIGMA V=0

V@A+V@B=10*10

V@A+V@B=100 KN

SIGMA M@B=0

V@10-10*10^2/2=0

V@A=50 KN

V@B=50 KN

CONSIDER A SECTION X-X AT A DISTANCE FROM LEFT HAND SUPPORT

TAKING MOMENT OF ALL FORCES AT SECTION X-X

M-XX=EID^2Y/DX^2=V@A*X-100*X*X/2

INTEGRATING ONCE WE GET

EI*DY/DX=50*X^2/2-100/2*X^3/3+C1

EI*DY/DX=50*X^2/2-100/6*X^3+C1

INTEGRATING AGAIN WE GET

EIY=50/6*X^3-100/24*X^4+C1*X+C2

THEREFORE TO FIND THE CONSTANT USING BOUNDARY CONDITION

AT X=0,Y=0

X=10,Y=0

PUTTING 1 BOUNDARY CONDITION IN EQUATION 2

0=0+0+0+C2

C2=0

PUTTING 2 BOUNDARY CONDITION IN EQUATION 2

0=50/6*10^3-100/24*10^4+C1*10+0

C1=3333.34

PUTTING C1 AND C2 IN EQUATION 2 WE GET

EI*Y=50/6*X^3-100/24*X^4+3333.34*X+0

PUT X=5 IN EQUATION 4

EI*Y=50/6*5^3-100/24*5^4+3333.34*5

EI*Y=15104.207

Y=15104.207/5000

EI=3.02 KN M^2=3020 N M^2

SOLUTION 4

DEAD LOAD=5 KN/M^2*3=15 KN/M

FACTORED DEAD LOAD =15*1.5 =22.5 KNM

LIVE LOAD=6 KN/M^2*3=18 KN/M

FACTORED LIVE LOAD =18*1.5=27.0 KNM

TOTAL DEAD AND LIVE LOAD=22.5+27=49.5 KNM

CONCENTRATED LOAD=100*1.5=150 KN

BENDING MOMENT =WL^2/8+WL/4=49.5*6^2/8+150*6/4

B.M=222.75+225=447.75 KNM

CALCULATION OF REQUIRED PLASTIC SECTION MODULUS

Z@P=M*GAMMA*M0/F@Y=447.75*10^6*1.1/250=1970.1*10^3 MM^3

SELECTION OF SUITABLE SECTION

CHOOSE TRIAL SECTION OF [email protected] KN/M

THE PROPERTIES OF THE SECTION ARE AS FOLLOWS

OVERALL DEPTH(H)=550 MM

WIDTH OF FLANGE(B)=190 MM

THICKNESS OF FLANGE(T@F)=19.3 MM

DEPTH OF WEB(D)=H-2(T@F+R)=550-2(19.3+18)=475.4 MM

THICKNESS OF WEB(T@W)=11.2 MM

MOMENT OF INERTIA ABOUT MAJOR AXIS I@ZZ=64900*10^4 MM^4

ELASTIC SECTION MODULUS (Z@E)=2359.8*10^3 MM^3

PLASTIC SECTION MODULUS (Z@P)=2711.98*10^3 MM^3

SECTION CLASSIFICATION

EPISILON=SQUAREROOT(250/F@Y)=SQUARERROOT(250/250)=1

B/T@F=95/19.3=4.922<9.4*EPISILON=9.4

D/T@W=475.4/11.2=42.45<84*EPISILON=84

HENCE THE SECTION IS CLASSIFIED AS A PLASTIC SECTION

SELF WEIGHT OF THE SECTION (FACTORED)=1.5*1.037==1.555 KN/M

MAXIMUM BENDING MOMENT=447.75+1.5*1.037*6^2/8=454.75 KNM

REQUIRED PLASTIC SECTION MODULUS=454.75*10^6*1.1/250=2000.8989*10^3 MM^3

HENCE THE CHOSEN IS ADEQUATE

CALCULATION OF DESIGN SHEAR FORCE

V=1.555*6/2+49.5*6/2+150/2=228.165 KN

DESIGN SHEAR STRENGTH OF THE SECTION

DESIGN SHEAR STRENGTH

V@D=F@Y*H*T@W/GAMMA*M0*SQUAREROOT(3)=250*550*11.2/1.1*SQUAREROOT(3)=808.29 KN

ALSO 0.6*V@D=485 KN

V<0.6*V@D

CHECK FOR DESIGN CAPACITY OF THE SECTION

D/T@W=42.45 <67EPISILON=67

M=BETA@B*Z@P*F@Y/GAMMA*M0<1.2*Z@E*F@Y/1.10

B@B=1.0 SINCE THE SECTION IS PLATIC

M@D=1.0*2711.98*10^3*250/1.1=616.36 KNM

<1.2*2359.8*10^3*250/1.1=643.58 KNM

HENCE THE DESIGN CAPACITY OF MEMBER IS MORE THAN THE BENDING MOMENT

M@D (616.38 KNM>454.75 KNM)

CHECK FOR DEFLECTION

ACTUAL DEFLECTION

DELTA=5WL^3/384*EI+W*L^3/48EI

DELTA=(5*49.5*10^3*6000^3/384*2*10^5*64900*10^4)+(150*10^3*6000^3/48*2*10^5*64900*10^4)=6.28 MM

ALLOWABLE DEFLECTION DELTA MAX=L/300=6000/300=20 MM

CHECK FOR WEB BUCKLING AT SUPPORT

B1=50 MM

DEPTH OF WEB D=475.4 MM

I@EFF WEB=B*T@W^3/12=50*11.2^3/12=5853.86 MM^4

A@EFF WEB=B1*T@W=50*11.2=560 MM^2

R=SQUAREROOT(I@EFFWEB/A@EFFWEB)=SQUAREROOT(5853.86/560)=3.23 MM

EFFECTIVE LENGTH OF WEB D@EFF=0.7 TIMES THE DEPTH OF WEB(D)=

LAMBDA=D@EFF/R=0.7*475.4/3.233=102.9

THERE FORE

FROM TABLE 9C OF CODE

FCD=103.4 N/MM^2

N1=550/2=275 MM

B1+N1=50+275=325 MM

A@B=325*11.2=3640 MM^2

BUCKLING RESISTANCE=FCD*A@B=103.44*3640/1000=376.52 KN>228.165 KN

HENCE THE WEB IS SAFE

CHECK FOR WEB BEARING

F@W=(B1+N2)*T@W*F@Y/GAMMA*M0

B1=50 MM

N2=2.5*(ROOT RADIUS+FLANGE THICKNESS)=2.5*(18+19.3)=93.25 MM

SINCE ANGLE OF DISPERTION IS 1:2:5

T@W=11.2 MM

F@Y=250 N/MM^2

F@W=(50+93.25)*11.2*250/1.10=428.27 KN>268 KN

HENCE THE WEB IS SAFE

SOLUTION 5

DEAD LOAD =1.4 KN/M^2*2=2.8 KN/M

LIVE LOAD = 0.75 KN/M^2*2=1.5 KN/M

WIND LOAD =1 KN/M^2*2 =2 KN/M SUCTION

W@DZ=2.8*COS20 DEGREE=2.6311 KN/M

W@IZ=1.5*COS20 DEGREE=1.4095 KN/M

W@WZ=-2 KN/M

W@DY=2.8*SIN20 DEGREE=-0.9576 KN/M

W@IY=1.5*SIN20 DEGREE=0.5130 KN/M

FACTORED LOAD COMBINATION

Z DIRECTION

WL+DL+LL=(1.2*2)+(1.2*2.6311)+(1.2*1.4095)=7.248 KN/M

D.L+L.L=(1.5*2.6311)+(1.5*1.4095)=6.0609 KN/M

Y DIRECTION

D.L+L.L=(1.5*0.9576)+(1.5*0.5130)=2.2059 KN/M

BENDING MOMENT AND SHEAR FORCE

SINCE CHANNEL SECTION PURLIN IS PROVIDED WITH A SAG ROD AT MIDSPAN THE BENDING MOMENT IN Y DIRECTION IS REDUCED CONSIDERABLY

M@Z=7.248*4^2/8=14.496 KN-M

M@Y=2.2059*4^2/32=1.10295 KN-M

V@Z=7.248*4/2=14.496 KN

V@Y=2.2059*4/2=4.4118 KN

REQUIRED SECTION MODULUS =

(M@Z*GAMMA*M0/F@Y)+2.5*(D/B)*(M@Y*GAMMA*M0/F@Y)

REQUIRED

(14.496*10^6*1.1/250)+2.5*(100/50)*(1.10295*10^6*1.1/250)=88.04*10^3 MM^3

PROVIDE ISMC 150 WITH THE FOLLOWING SECTION PROPERTIES

DEPTH OF SECTION H=150 MM

R@Y=22 MM

WIDTH OF FLANGE B=75 MM

THICKNESS OF FLANGE T@F=9 MM

THICKNESS OF WEB T@W=5.7 MM

DEPTH OF WEB D=150-2(9+10)=112

ELASTIC SECTION MODULUS Z@EZ=105*10^3 MM^3

ELASTIC SECTION MODULUS Z@EY=19.5*10^3 MM^3

PLASTIC SECTION MODULUS Z@PZ=119.82*10^3 MM^3 >88.04*10^3 MM^3

MOMENT OF INERTIA I@Z=788*10^4 MM^3

SECTION CLASSIFICTION

B/T@F=75/9=8.33<9.4

D/T@W=112/5.7=19.65<42

HENCE THE SECTION IS PLASTIC

CALCULATION OF SHEAR CAPACITY OF THE SECTION

Z DIRECTION

V@D=F@Y*H*T@W/GAMMA@M0*SQUAREROOT(3)=250*150*5.7/1.1*SQUAREROOT(3)=112.18*10^3

0.6*V@D=0.6*112.18*10^3=67.31 KN> 14.196 KN

Y DIRECTION

250*2*75*9/1.1*SQUAREROOT(3)=177.14*10^3 KN>4.4118 KN

DESIGN CAPACITY OF THE SECTION

M@DZ=(Z@PZ*F@Y/GAMMA*M0)=(119.82*10^3*250/1.1*10^6)=27.23 KNM

<(1.2*Z@EZ*F@Y/GAMMA@M0)=(1.2*105*10^3*250)/(1.1*10^6)=28.63

Z@PY=2T@F*B@F^2/4+(H-2*T@F)*T@W^2/4=2*9*75^2/4+(150-2*9)*5.7^2/4=26384.6 MM^3

M@DY=Z@PY*F@Y/GAMMA*M0=26384.6*250/(1.1*10^6)=6 KNM

<1.5*Z@EY*F@Y/GAMMA*M0=1.5*19.5*10^3*250/(1.1*10^6)=6.6 KNM

HENCE THE SECTION IS SAFE

OVERALL MEMBER STRENGTH DUE TO LIVE LOAD

M@Z/M@DZ+M@Y/M@DY<1.0

14.496/27.23+1.10295/6=0.716<1.0

HENCE THE MEMBER IS SATISFACTORY

CHECK FOR DEFLECTION DUE TO LIVE LOAD

DELTA=5*W*L^4/384*E*I=5*1.4095*4000^4/(384*2*10^5*788*10^4)=2.98 MM<22.22 MM

CHECK FOR WIND SUCTION

FACTORED WIND LOAD=W@Z=0.9*2.6311-1.5*2=-0.63201 KN/M

W@Y=0.9*0.9576=0.86184 KN/M

M@Z=-0.63201*4^2/8=-1.26402 KNM

M@Y=0.86184*4^2/8=1.72368 KNM

KL/R=4000/22=181.8

H/T@F=150/9=16.67

F@CR=(1473.5/181.8)^2*(1+(1/20)(181.8/16.67)^2)^0.5=173.1 N/MM^2

FROM TABLE 13 A CODE

F@BD=120 N/MM^2

M@DZ=Z@PZ*F@BD=119.82*10^3*120/10^6=14.38 KNM

OVERALL BUCKLING STRENGTH

M@Z/M@DZ+M@Y/M@DY<1.0

1.26402/14.38+1.7236/6=0.375<1.0

HENCE SATISFACTORY