Steel Design Sheets

RAFTER ID :- ( R-1 )

Case a/b Combination

1)- APPLIED FORCES :-

M+ive = 13.00 mt a

M-ive = 20.00 mt a

N = 0.00 t

Q = 0.00 t

2)- DIM. OF SECTION :-

The section is BUS

bFL.U = 170.0 mm

tFL.U = 12.7 mm

hWEB = 334.6 mm

tWEB = 8.0 mm

bFL.L = 170.0 mm

tFL.L = 12.7 mm

If part of the section is used as haunch:-

hWEB = 335.0 mm

tWEB = 8.0 mm

bFL.L = 170.0 mm

tFL.L = 12.7 mm

3)- RAFTER DATA :- Notes:-

Total length of rafter (Lg) = 24.00 m 1) +ive direction of all reactions are

Purlin spacing (S) = 2.00 m indicated in the above figure, the

Length of column (Lc) = 8.00 m reverse will be -ive

Uniform load on rafter (W) = 0.50 t/m 2) Values of W, H & V are those

Horizontal reaction of column (H) = 2.50 t accompaning the value of M-ive

Vertical reaction of column (V) = 6.00 t

Angle of inclination of rafter (a) = 5.71

In case of crane :-

Length of bracket (LB) = 0.00 m

Py reaction of crane (h) = 0.00 t

Px reaction of crane (u) = 0.00 t

Length of col. above bracket (L) = 0.00 m

4)- PROPERTIES OF SECTION :-

Y = 18.00 cm

A = 69.95 cm2

Ix = 15523.83 cm4

Iy = 1041.35 cm4

Sx = 862.44 cm3

rx = 14.90 cm

hWEB

Y Mx

bFL.U

bFL.L

tFL.U

tFL.L

X

Y

ry = 3.86 cm

5)- CHECK COMPACTNESS :-

dw/tw = 39.825

C/tf = 6.378

The sec is Compact

6)- CHECK STRESSES :-

A)- Section at M+ive or Mhunch:-

fbcx = 1.51 t/cm2

Cb =

fca = 0.00 t/cm2

( 20.bf ) / (√ fy ) = 2.195 m

( 1380.Af ) / ( fy.d ) x Cb = 4.82 m

(Lateral torsional buckling of comp.flange ) No Ltb

Fltb = 3.636 t/cm2

Fbcx = 1.536 t/cm2

Lbin = 15.60 m

Lbout = 2.00 m

lin = 104.72

lout = 51.83

lmax = 104.72 < 180 SAFE

Fc = 0.68 t/cm2

fca / Fc = 0.00

A1 = 1.00

Applying the interaction equation :

( fca / Fc ) + ( fbcx / Fbcx ) A1 = 0.981 <

B)- Section at M-ive :-

PROPERTIES OF SECTION :-

Y = 35.39 cm

A = 118.34 cm2

Ix = 73326.68 cm4

Iy = 1562.73 cm4

Sx = 2072.05 cm3

rx = 24.89 cm

ry = 3.63 cm

To get length of haunch we must satisfy that M-ive = M+ive

Length of haunch = 1.290 m Taken 1.50 m

To get length of compression flange we must satisfy that Mx = 0

Length of comp. Flange ( Luact ) = 4.272 m

Lu act.of comp. Flange from sap =4.27 m

Lumax =

Use Knee Bracing

fbcx = 0.97 t/cm2

fca = 0.00 t/cm2

( 20.bf ) / (√ fy ) = 2.195 m

( 1380.Af ) / ( fy.d ) x Cb = 4.82 m

(Lateral torsional buckling of comp.flange SAFE

Fltb = 1.536 t/cm2

fbcx = 1.536 t/cm2

Lbin = 15.60 m

Lbout = 2.00 m

lin = 62.67

lout = 55.04

lmax = 62.67 < 180 SAFE

Fc = 1.14 t/cm2

fca / Fc = 0.00

A1 = 1.00

Applying the interaction equation :

( fca / Fc ) + ( fbcx / Fbcx ) A1 = 0.628 <

Use Knee Bracing

Lumax =

X

1.300

no

1.000 SAFE

2.195 m

yes

1.000 SAFE

2.195 m

t-y = 18.00 cm

x1= 35.382

compact non com slend

81.978 122.644 122.644 1.000 0.000 0.000

9.876 13.555 13.555 1.000 0.000 0.000

2.000 0.000 0.000

fsec= 1.536

0.000

At = 26.051 cm2

F1 = 3.355

Iyt = 520.197 cm4

F2 = 1.400

Rt = 4.469 cm F2 = 0.000

Af = 21.590 cm F2 = 0.000

Lu/Rt = 44.757 F3 = 1.400

√(Cb/Fy) = 0.736 Fltb = 3.636

Length of haunch : Length of comp. flange :

a b c a b c

0.250 -5.750 7.000 0.250 -5.750 20.000

X1 = 21.710 X1 = 18.729

X2 = 1.290 X2 = 4.272

all +ive= 1.000

At = 30.52 cm2

F1 = 1.572

Iyt = 520.44 cm4

F2 = 1.400

Rt = 4.130 cm F2 = 0.000

Af = 21.590 cm F2 = 0.000

Lu/Rt = 44.757 F3 = 1.400

√(Cb/Fy) = 0.736 Fltb = 2.105

all -ive= 1.000

COLUMN ID :- ( C-2 )

DESIGN CASE :- (D.L + L.L+W ) a


Mx1 = 4.00 mt

N = 7.00 t

Mx2 = 0.00 mt

My = 0.00 mt

2)-DIM. OF SECTION :-

The section is Bus

bFL.U = 150.0 mm

tFL.U = 10.7 mm

hWEB = 280.0 mm

tWEB = 7.1 mm

bFL.L = 150.0 mm

tFL.L = 10.7 mm

3)-COLUMN DATA :-

Total length of column = 4.00 m

Lu act.of comp. Flange = 4.00 m

Length subject to buckling in plan = 4.00 m

Length subject to buckling out plan = 4.00 m

Length of girder = 7.00 m

Ix (rafter) = 8360.00 cm4

In case of crane :-




A = 51.98 cm2

Y = 15.07 cm

X = 7.50 cm

Ix = 8083.54 cm4

Iy = 602.71 cm4

Sx = 536.399 cm3

Sy = 80.361 cm3

rx = 12.47 cm

ry = 3.41 cm


dw/tw = 36.423

C/tf = 7.009

The sec is Compact

hWEB X

Y

M

b

tFL.U

tFL.L

X

bFL.U

bFL.L

Mx

Y

My

M

Lc

Lg/2

Lh

1

M2

M

Lc

Lg/2

L

B

1

2M

L


fbcx = 0.75 t/cm2

fbcy = 0.00 t/cm2

Fbcy = 1.73 t/cm2

fca = 0.13 t/cm2

a = 0.00

Cb = 1.75

( 20.bf ) / (√ fy ) = 1.936 m

( 1380.Af ) / ( fy.d ) x Cb = 5.77 m

(Lateral torsional buckling of comp.flange Ltb

Fltb = 1.536 t/cm2

yes

Fbcx = 1.536 t/cm2

GA = 10 (according to base type)

GB = 1.69

K = 2.10 E.C.O.P (P.100)

Lbin = 8.40 m

Lbout = 4.00 m

lin = 67.36

lout = 117.47

lmax = 117.47 < 180 SAFE

Fc = 0.54 t/cm2

fca / Fc = 0.25

Cmx = 0.85

A1 = 1.00

Cmy = 1.00

A2 = 1.33

APPLYING THE INTERACTION EQUATION :

( fca / Fc ) + ( fbcx / Fbcx ) A1 +( fbcy / Fbcy ) A2 = 0.733 < 1.000 SAFE

m

Use Knee Bracing

Lumax = 1.936

web

flange

M

Lc

Lg/2

Lh

1

M2

M

Lc

Lg/2

L

B

1

2M

L

t-y = 15.07 cm

xx= 0.00 cm

x1= 0.355 cm

x2= 7.500 cm

x3= 0.355 cm

x4= 7.500 cm

alpha= 0.573 f1= 0.611

f2= -0.880

epsi -0.694


69.914 93.865 93.865 1.000 1.000 1.000

10.910 14.800 14.800 1.000 1.000 1.000

2.000 2.000 2.000

fsec x = 1.536

M

Lc

Lg/2

L

B

1

2M

L

fsec y = 1.728

At = 19.363 cm2

F1 = 2.006

Iyt = 301.077 cm4

F2 = 0.000

Rt = 3.943 cm F2 = 0.000

Af = 16.050 cm F2 = 0.666

Lu/Rt = 177.521 F3 = 0.666

√(Cb/Fy) = 0.854 Fltb = 1.400

all= 1.000




Mx1 = 0.00 mt

N = 64.30 t

Mx2 = 14.65 mt

My = 21.50 mt

2)-DIM. OF SECTION :-

The section is bus

bFL.U = 220.0 mm

tFL.U = 19.0 mm

hWEB = 562.0 mm

tWEB = 12.0 mm

bFL.L = 220.0 mm

tFL.L = 19.0 mm

Add Part For My

hWEB1 = 257.8 mm

tWEB1 = 11.1 mm

hWEB2 = 257.8 mm

tWEB2 = 11.1 mm

bf1 = 210.0 mm

tf1 = 17.2 mm

bf2 = 210.0 mm

tf2 = 17.2 mm

3)-COLUMN DATA :-




Length subject to buckling out plan = 11.00 m

Length of girder = 6.00 m

Ix (rafter) = 44333.00 cm4

In case of crane :-




hWEB

Y Y

Mx

bFL.

tFL.U

tFL.L

X

bFL.U

bFL.L

Mx

Y

tweb1 tweb2

bf2

M

Lc

Lg/2

Lh

1

M2

M

Lc

Lg/2

L

B

1

2M

L

A = 280.51 cm2

Y = 30.00 cm

X = 11.00 cm

Ix = 90980.65 cm4

Iy = 68258.67 cm4

Sx = 3032.688 cm3

Sy = 2429.134 cm3

rx = 18.01 cm

ry = 15.60 cm


dw/tw = 21.833 dw/tw = 23.225

C/tf = 5.789 C/tf = 6.105

The main sec is Compact The add. sec is Compact


fbcx = 0.00 t/cm2

fbcy = 0.89 t/cm2

Fbcy = 1.536 t/cm2

fca = 0.23 t/cm2

a = 0.00

Cb = 1.75

( 20.bf ) / (√ fy ) = 2.840 m

( 1380.Af ) / ( fy.d ) x Cb = 7.48 m

(Lateral torsional buckling of comp.flange No Ltb

Fltb = 1.400 t/cm2

Fbcx = 1.536 t/cm2

GA = 1 (according to base type)

GB = 2.46

K = 2.10 E.C.O.P (P.100)

Lbin = 10.50 m

Lbout = 11.00 m

lin = 58.30

lout = 70.52

lmax = 70.52 < 180

Fc = 1.08 t/cm2

fca / Fc = 0.21

A1 = 1.00

A2 = 1.00

Lumax =


( fca / Fc ) + ( fbcx / Fbcx ) A1 +( fbcy / Fbcy ) A2 = 0.790 <

t-y =

xx=

x1=

x2=

x3=

x4=

X bf1

web1

My

alpha= 0.699 f1= -0.229

f2= -0.229

epsi 1.000

I1 compact non com slend

web 37.400 41.300 41.300 1.000 1.000

flange 10.910 14.800 14.800 1.000 1.000

2.000 2.000

fsec x = 1.536

I2 compact non com slend

web 37.400 41.300 41.300 1.000 1.000

flange 10.910 14.800 14.800 1.000 1.000

2.000 2.000

fsec y = 1.536

At = 53.040 cm2

F1

Iyt = 1687.282 cm4

F2

Rt = 5.640 cm F2

Af = 41.800 cm F2

Lu/Rt = 106.380 F3

√(Cb/Fy) = 0.854 Fltb

all= 1.000

SAFE

m2.840

1.000 SAFE

30.00 cm

0.00 cm

28.100 cm

11.000 cm

28.100 cm

28.100 cm

1.000

1.000

2.000

act

1.000 hw1/tw1 23.225 c/tf1 6.105

1.000 hw2/tw2 23.225 c/tf2 6.105

2.000

= 10.413

= 0.000

= 1.219

= 0.000

= 1.219

= 1.400

CONNECTION ID :- ( C-1 )

1)- APPLIED FORCES : case a/b

M = 20.00 mt a

Q = 6.00 t

Text = 0.00 t

2)- H.S.B USED ARE :

M 24 Q 10.9

AS = 2.45 cm2

PS = 7.11 t

T = 22.23 t

3)- DIMS. OF HEAD PLATE USED ARE :

H = 76 cm

B = 17 cm

tp = 20 mm

4)- NO. OF BOLTS REQUIRED :

n = 2.466 bolts

Take ntot = 6 bolts

5)- ARRANGEMENT OF BOLTS :

y1 = 69 cm y4 = 0 cm

y2 = 61 cm y5 = 0 cm

y3 = 0 cm y6 = 0 cm

6)- CHECK FORCES ON BOLTS :

Text,b1,m = 9.971 t < 0.8T SAFE

Text,b2,m = 9.766 t < 0.8T SAFE

Text,b3,m = 0.000 t < 0.8T SAFE

Text,b4,m = 0.000 t < 0.8T SAFE

Text,b5,m = 0.000 t < 0.8T SAFE

Text,b6,m = 0.000 t < 0.8T SAFE

Text,b = Text/ntot 0.000 t < 0.6T SAFE

(Text,b) / 0.6T + (Text,b,m) / 0.8T = 0.561 < 1.00 SAFE

Q / ntot = 1.000 < Ps ( 1 - (Text,b ) / T ) = 7.110 SAFE

No Prying effect

7)- CHECK HEAD PLATE THICKNESS :

(Ft + F1)x B x (X1)/4 =

(F1 + F2)x B x (X2)/4 =

(F2 + F3)x B x (X3)/4 =

(F3+ F4)x B x (X4)/4 =

=

(F4 + F5)x B x (X5)/4 =

(F5 + F6)x B x (X6)/4 =

H

B

Y2

e = 40 mm

Mplate = 39.884 cmt

tp = √(6Mplate/(0.864B)) = 4.036 cm 42.00 mm UNSAFE

Text,b1,m x e =

X 1

X 2

X 3

x1

x2

x3

ft

f1

f2

I=

all=

Y1 Y2

F1

Fc

Ft

F2

X1

X2

= 11.270 X 4 = 0.000

= 26.730 X 5 = 0.000

= 0.000 X 6 = 0.000

= 7.270 x4 = 0.000 x7

= 8.000 x5 = 0.000

= 60.730 x6 = 0.000

= 0.122 f3 = 0.000

= 0.086 f4 = 0.000

= 0.000 f5 = 0.000

f6 = 0.000

621882.667

1.000

= -38.000



M = 10.00 mt a

Q = 5.00 t

Text = 5.00 t


M 20 Q 10.9

AS = 2.45 cm2

PS = 4.93 t

T = 15.43 t


H = 54 cm

B = 20 cm

tp = 20 mm


n = 3.040 bolts

Take ntot = 6 bolts


y1 = 46 cm y4 = 0 cm

y2 = 38 cm y5 = 0 cm

y3 = 0 cm y6 = 0 cm


Text,b1,m = 9.371 t < 0.8T SAFE

Text,b2,m = 4.518 t < 0.8T SAFE

Text,b3,m = 0.000 t < 0.8T SAFE

Text,b4,m = 0.000 t < 0.8T SAFE

Text,b5,m = 0.000 t < 0.8T SAFE

Text,b6,m = 0.000 t < 0.8T SAFE




=

(Ft + F1)x B x (X1)/4 =

(F1 + F2)x B x (X2)/4 =

(F2 + F3)x B x (X3)/4 =

(F3+ F4)x B x (X4)/4 =

(F4 + F5)x B x (X5)/4 =

(F5 + F6)x B x (X6)/4 =

H

B

Y2

No Prying effect


e = 45 mm

Mplate = 42.167 cmt

tp = √(6Mplate/(0.864B)) = 3.826 cm 40.00 mm UNSAFE

Text,b1,m x e =

X 1

X 2

X 3

x1

x2

x3

ft

f1

f2

I=

all=

Y1 Y2

F1

Fc

Ft

F2

X1

X2

= 11.600 X 4 = 0.000

= 15.400 X 5 = 0.000

= 0.000 X 6 = 0.000

= 7.600 x4 = 0.000 x7

= 8.000 x5 = 0.000

= 38.400 x6 = 0.000

= 0.103 f3 = 0.000

= 0.059 f4 = 0.000

= 0.000 f5 = 0.000

f6 = 0.000

262440.000

1.000

= -27.000



M = 4.17 mt a

Q = 10.00 t

Text = 1.00 t


M 20 Q 8.8

AS = 2.45 cm2

PS = 3.45 t

T = 10.81 t


H = 40 cm

B = 15 cm

tp = 22 mm


n = 2.163 bolts

Take ntot = 6 bolts


y1 = 36 cm y4 = 0 cm

y2 = 27 cm y5 = 0 cm

y3 = 0 cm y6 = 0 cm


Text,b1,m = 5.128 t < 0.8T SAFE

Text,b2,m = 2.691 t < 0.8T SAFE

Text,b3,m = 0.000 t < 0.8T SAFE

Text,b4,m = 0.000 t < 0.8T SAFE

Text,b5,m = 0.000 t < 0.8T SAFE

Text,b6,m = 0.000 t < 0.8T SAFE




(Ft + F1)x B x (X1)/4 =

(F1 + F2)x B x (X2)/4 =

(F2 + F3)x B x (X3)/4 =

(F3+ F4)x B x (X4)/4 =

(F4 + F5)x B x (X5)/4 =

(F5 + F6)x B x (X6)/4 =

=

b

H

B

Y2

Prying effect

a = 4.00 cm Sw = 1.00 cm

b = 3.00 cm

w = 7.50 cm P = 1.640 t

P + (Text,b) + (Text,b,m) = 6.935 t < 0.8T SAFE


M1 = 6.561 cmt

M2 = -9.324 cmt

Mplate = max of M1 and M2 = 9.324

tp = √(6Mplate/(0.864B)) = 2.078 cm 22.00 mm SAFE

P x a - Text,b1,m x b =

P x a =

X 1

X 2

X 3

x1

x2

x3

ft

f1

f2

I=

all=

Y1 Y2

F1

Fc

Ft

F2

X1

X2

= 8.268 X 4 = 0.000

= 11.733 X 5 = 0.000

= 0.000 X 6 = 0.000

= 4.000 x4 = 0.000 x7

= 8.535 x5 = 0.000

= 27.465 x6 = 0.000

= 0.104 f3 = 0.000

= 0.061 f4 = 0.000

= 0.000 f5 = 0.000

f6 = 0.000

80000.000

1.000

= -20.000

CONNECTION ID:- ( C-3 )


M = 10.50 mt

Q = 5.70 t

T = 0.00 t

2)-DIM. OF STEEL SECTION :-

The section is : BUS

bFL.U = 375.0 mm

tFL.U = 20.0 mm

hWEB = 500.0 mm

tWEB = 8.0 mm

bFL.L = 375.0 mm

tFL.L = 20.0 mm

If part of the section is used as haunch:-

hWEB = 0.0 mm

tWEB = 0.0 mm

bFL.L = 0.0 mm

tFL.L = 0.0 mm

3)- DIM. OF WELD :-

L1 = 375.0 mm S1 = 6.0 mm

L2 = 150.0 mm S2 = 6.0 mm

L3 = 400.0 mm S3 = 6.0 mm

L4 = 0.0 mm S4 = 0.0 mm

L5 = 0.0 mm S5 = 0.0 mm

L6 = 90.0 mm S6 = 6.0 mm

L7 = 90.0 mm S7 = 6.0 mm

CU = 10.0 mm

CL = 10.0 mm

4)- PROPERTIES OF WELD :-

AVL = 69.60 cm2

AHZ = 81.00 cm2

ATOT = 150.60 cm2

y = 27.60 cm

IX = 85287.18 cm4


fhz = T / ATOT + M / IX * y1 = 0.340 t/cm2

< 0.72 t/cm2

SAFE

fvl = T / ATOT + M / IX * y2 = 0.463 t/cm2

qvl = Q / AVL = 0.082 t/cm2

feq(vl) = (fvl)2+3(qvl)

2= 0.484 t/cm

2< 0.792 t/cm

2SAFE

L1

L2

L3

L4

L5

L2

L1

Y

X X

L6

L7

Y

y

CU

CL

yhz = 27.60 cm

yvl1 = 37.60 cm

yvl2 = 20.00 cm

yvl = 37.60 cm



22.000

1)- APPLIED FORCES :- 9.500

Mx = 5.00 mt

N = 1.40 t

2)-DIM. OF SECTION :- 12.000

The section is bus 500.000

bFL.U = 6.0 mm

tFL.U = 9.8 mm

hWEB = 12.0 mm

tWEB = 6.2 mm

bFL.L = 120.0 mm

tFL.L = 9.8 mm

3)-COLUMN DATA :-




Length subject to buckling out plan = 2.50 m 10.000

10.000

6.000


Y = 90.00 cm 10.000

A = 100.00 cm2

10.000

Ix = 10.00 cm4

Iy = 10.00 cm4

Sx = 0.11 cm3

rx = 0.32 cm

ry = 0.32 cm


dw/tw = -1.226

c/tf = 0.306

The sec is Compact


hWEB

Y

M

b

tFL.U

tFL.L

X

bFL.U

bFL.L

Mx

Y

fbcx = 4500.00 t/cm2

fca = 0.01 t/cm2

Cb = 1.13

( 20.bf ) / (√ fy ) = 0.077 m

( 1380.Af ) / ( fy.d ) x Cb = 3.18 m


Fltb = 1.400 t/cm2

no

Fbcx = 1.400 t/cm2

K = 1.00 E.C.O.P (P.100)

Lbin = 9.20 m

Lbout = 2.50 m

lin = 2909.30

lout = 790.57

lmax = 2909.30 < 180 Unsafe

fc = 0.00 t/cm2

fca / fc = 15.80

A1 = 1.00


( fca / fc ) + ( fbcx / Fbcx ) A1 = 3230.085 < 1.000

0.077

Use Knee Bracing

Lumax =

t-y = -86.84

alpha= 0.892 f1= 4499.986

f2= -4500.014

epsi -1.000


42.581 122.580 122.580 1.000 1.000 1.000

10.910 14.800 14.800 1.000 1.000 1.000

2.000 2.000 2.000

fsec= 1.536

X

At = 0.712 cm2

F1 =

Iyt = 0.022 cm4

F2 =

Rt = 0.174 cm F2 =

Af = 0.588 cm F2 =

Lu/Rt = 1434.932 F3 =

√(Cb/Fy) = 0.686 Fltb =

Unsafe

m

cm

1.772

0.000

0.000

0.007

0.007

1.400

Crane Girder ID :- (C.G.-1 )

DESIGN CASE :- (D.L + L.L )

22.000

1)- Crane Properties :- 9.500

5.000

Max Wheel Load = 9.20 t

Crane Girder Span = 10.00 m

Wheels Spacing = 2.50 m

Impact Factor = 25.00 %

Steel used Fy = 12.00 t/cm2

500.000

6.00

2)- Loads On Crane Girder :-

12.00

a)-Dead Load :-

Wd (.15-.2) = 0.20 t/m'

Md = 2.50 mt

Qd = 1.00 t

b)-Live Load :-

Mll (with impact) = 44.02 mt

Qll (with impact) = 20.13 t

c)-Design Values :-

Mmax = = 46.52

My = 100.000 = 3.52

Qmax = = 21.13

10.000

3)- Proposed Section :-

Symetric Sec. Steel37

Sx = 4446.34 cm3

4)- Dim. Of Section :-

The section is BUS

bFL.U = 450.0 mm

tFL.U = 16.0 mm

hWEB = 750.0 mm

tWEB = 8.0 mm

90.000

10.000

Wheel spacing

P P

bFL.L = 300.0 mm

tFL.L = 16.0 mm


Y = 44.21 cm

A = 180.00 cm2

Ix = 197918.669 cm4

Iy = 15753.20 cm4

Sxu = 5822.28 cm3

Sxl = 4477.12 cm3

Syu = 540.00 cm3


dw/tw = 89.750

C/tf = 14.063

The sec is Non Compact


Cb = 1.35

Luact = 10.00 m

( 20.bf ) / (√ fy ) = 5.809

( 1380.Af ) / ( fy.d ) x Cb = 7.15


Fltb = 1.630 t/cm2

Fbcx = 1.400 t/cm2

Fbcy = 1.400 t/cm2

Fbtx = 1.400 t/cm2

Point 1 : subjected to Mx only, Compresion flange

f1 = 0.799 t/cm2

< 1.400

Point 2 : subjected to Mx only, Tension flange

f2 = 1.039 t/cm2

< 1.400

Point 3 : subjected to Mx and My, Compression flange

= 1.037 < 1.200

Check Shear :

Lumax =

qact = 0.35 t/cm2

< 0.35xFy t/cm2

Check Deflection :

dl.l = 0.84 cm < Span/800

10.000

10.000

6.000

10.000

10.000

t

Unsymetric Sec

hw = 750.00 mm

tw = 8.00 mm

tflanges = 18.85 mm

bu = 376.98 mm

bl = 188.49 mm

Y Mx

3

Wheel spacing


82.000 122.650 122.650

10.910 14.800 14.800

0.000

m

m At = 82.000

Iyt = 12150.533

Rt = 12.173

Af = 72.000

Lu/Rt = 82.150

√(Cb/Fy) = 0.750

SAFE

SAFE

SAFE

5.809 m

hWEB X

Y

bFL.

bFL.L

tFL.U

tFL.L

X

Y

2

1

3

SAFE

SAFE

t1= 0.800 f1 43.125

t2= 0.067 f2 64.688

t3= 0.612 f3 107.813

m 350.391

t= 63.950

a= 53.291

au= 71.055

al= 35.528

tu= 1.885

bu= 376.976

tl= 1.885

bl= 188.488

t-y = 33.99 cm

0.000 1.000 1.000

0.000 1.000 1.000

0.000 2.000 2.000

fsecx= 1.400

fsecy= 1.400

cm2

F1 = 0.994

cm4

F2 = 0.000

cm F2 = 1.291

cm F2 = 0.000

F3 = 1.291

Fltb = 1.630

Documents

Steel Design Sheets