DESIGN OF PLATED COLUMNS SUBJECTED TO AXIAL COMPRESSION & BENDINGDesign input:Member No: 1Axial load P 25.00 t

Moment about major axis Mz-z 15.00 t-m y-

Moment about major axis My-y 2.00 t-m 150

Shear force , q = 8.00 tLength between the node or height from supports=L= 5.00 m 1800Effective length about major axis Lz-z Kz= 1.00 5.00 m4 1700Effective length about minor axis Ly-y Ky= 1.00 5.00 m z- Eff. length of compression flange L K = 1.00 5.00 m

250 M Pa 435 150

D = Depth of the section 1800 mmT = thickness of flange 50 mm 250t = thickness of web 50 mm 920B = Width of flange 920 mm y - distance of cg of each componentAllow. stress ratio 1.00 from axis 1-1 z - distance of cgTo find Moment of inertia of section: Unit-mmComponent. Comp. No Area y1-1 A . y A . y2 I self (z) z2-2 A .z A . z2 I self (y)

Abf x tf 1 920 50 46000 25 1.2E+06 2.9E+07 9.6E+06 460 2.1E+07 9.7E+09 3.24E+09h x tw 2 50 1700 85000 900 7.7E+07 6.9E+10 2.0E+10 460 3.9E+07 1.8E+10 1.77E+07bf x tf 3 920 50 46000 1775 8.2E+07 1.4E+11 9.6E+06 460 2.1E+07 9.7E+09 3.24E+09

plate top(R/S) 4 150 12 1800 150 2.7E+05 4.1E+07 2.2E+04 75 1.4E+05 1.0E+07 3.38E+06plate top(L/S) 5 150 12 1800 150 2.7E+05 4.1E+07 2.2E+04 560 1.0E+06 5.6E+08 3.38E+06Plate bottom(R/S) 6 150 12 1800 1650 3.0E+06 4.9E+09 2.2E+04 75 1.4E+05 1.0E+07 3.38E+06Plate bottom(L/S) 7 150 12 1800 1650 3.0E+06 4.9E+09 2.2E+04 560 1.0E+06 5.6E+08 3.38E+06

184200 1.66E+08 2.24E+11 2.05E+10 8.37E+07 3.86E+10 6.52E+09I1-1 = Iself-z + ( A.y2 ) = 2.44E+11 mm4

y- = A.y / A = 900.00 mm

Iz-z = I1-1 - A . (y-)2 = 9.50E+10 mm4

Zz-z = Iz-z / y- = 1.06E+08 mm3

rz = sqrt(Iz-z / A) = 718.07 mm2

I2-2 = Iself-y+ (A.z2 ) = 4.51E+10 mm4

z- = A.z / A = 454.43 mm

Iy-y = I2-2 - A . (z-)2 = 7.08E+09 mm4

Remarks

Yield stress of steel fy

Section

bf

tf

2

2

Z Z

Y

Y

1 1

tW

Zy-y = Iy-y / z- = 1.56E+07 mm3

ry = sqrt (Iy-y / A) = 196.11 mm2

Permissible direct stress in compression: ac (N/mm2)fy= 230

Axial load P in t = As per IS:800-1984 Cl.3.5.2.2,(a) 25.001440t/sqrt(fy)= 4747.54

90t = 4500.00Depth of web d1 in mm = 1700.00d1<Max. of (1440t/sqrt(fy) or 90t) Thickness of web is OK of each component800t/sqrt(fy)= 2637.52 mm from axis 2-250t = 2500.00 mm I1-1 = MI about 1-1

I2-2 =MI about 2-2Effective length about major axis Lz-z in m = 5.00 m

Effective length about minor axis Ly-y in m = 5.00 m

z-z = Lz-z / rz = 6.96 m

y-y = Ly-y / ry = 25.50 mMaximum slenderness ratio = 25.50 <180

SAFEIf d1<Max of (800t/sqrt(fy) or 50t), Anet = Aelse, Anet=A-{[d1-(Min.of (800t/sqrt(fy), 50t)]x t}Anet = #########ac cal = P/Anet (N/mm2) = 1.36 N/mm2

Modulus of elasticity of steel E = 2.00E+05 N/mm2

Elastic critical stress in compression = fcc(N/mm2) = 2E / 2 = 3036.61 N/mm2

Yield stress of steel in N/mm2 fy = 250 N/mm2

ac = ( 0.6 . fcc . fy) / [(fcc)n + (fy)

n]1/n = 146.83 N/mm2

( n = 1.4) Ref. IS:800-1984Permissible bending compressive stress :bcz (N/mm2) Cl.5.1.1

15.00bcz.cal = Mz-z / Zz-z 1.42 N/mm2

Effective length of compression flange L 5.00 mY = (26.5 x 10 5)/(L/ry)

2 = 4076.67 mm Refer IS:800-1984T = Mean thickness of compression flange = 50 mm Cl. 6.2.4D = Depth of the section = 1800 mmX = Y. Sqrt [1+1/20(L.T/ry.D)2 = 4127.47 mm

c1 = 900.00c2 = 900.00

Moment about major axis Mz-z in t-m =

k1 = 1k2 = 0

Elastic critical stress in bending in N/mm2 = fcb = k1(X+k2 .Y)c2/c1 = 4127.47 N/mm2

T = thickness of flange = 50t = thickness of web = 50T/t = 1d1=D-2T= 1700 Refer IS:800-1984 fcb = (fcb is increased by 20% if T/t<2.0 & d1/t <1344/SQRT(fy) 4952.96

d1 / t = 34 Cl. 6.2.4.1 Elastic critical stress in compression = fccz(N/mm2) = 2E / zz

2 = 40711.891344/SQRT(fy) = 85.00bcz = ( 0.66 . f cb . fy) / [(fcb)

n + (fy)n]1/n = 163.22 N/mm2 ( n = 1.4 ) Refer IS:800-1984

Permissible bending compressive stress :bcy (N/mm2) Cl. 6.2.3

Moment about minor axis My-y in t-m = 2.00bcy.cal = My-y / Zy-y (N/mm2) = 1.28 N/mm2

bcy = 0.66 fy = 165.00 N/mm2

Elastic critical stress in compression = fccy(N/mm2) = 2E /yy2 = 3036.61 N/mm2

Check for combined stresses:Refer IS:800-1984Cl. 7.1.1(a)

( Cm = 0.85 ) = 0.02 < SAFE 1.00 Cl. 7.1.3

Check for shear:Shear force (q) in t = 8 Refer IS:800-1984Depth of web d1 in mm = 1700 Cl. 6.4.2Thickness of web t in mm = 50Shear stress va.cal (N/mm2) = q / d1 . t = 0.941 < 0.4.fy SAFE

92 SAFEChecks for welded plate girders:The thickness of flange plates have been so choosen that the projection of the plate beyond its connection to the web (x)mm is kept within 256 tf /sqrt(fy) subject to a maximum of 16T.fy 230x =B/2 - t/2 = 435256 tf /sqrt(fy)= 844.01 Refer IS:800-1984

(ac.cal / ac) + (Cmx . bcz.cal ) / [{1 - (ac.cal / 0.60 . fccz)} bcz] + (Cmy . bcy.cal ) / [{1 - (ac.cal / 0.60 . fccy)} bcy]

16tf = 800 Cl.3.5.2x <= Max. of (256tf/sqrt(fy) and 16tf) = 844.01 OKFor unstiffened webs the thickness of web plate shall be maximum of the following: d1. sqrt(fy)/1344 = 19.18 Refer IS:800-1984d1. sqrt(va.cal)/816 = 2.02 Cl.6.7.3.1 (a)d1/85 = 20.00t > Max of the above values = OK