Flexural Member Demo Xlsx

COLD-FORMED STEELFLEXURAL MEMBER

Steel Structural Member, 2001 Edition.

COMPANY :PROJECT :

1). INPUT DATAUnits : US unit

Fy E ho bo t Rksi ksi in. in. in. in.340 203402.5 10.000 5.000 0.105 3/16

2). SECTION PROPERTIES a. Parameter of section

R + t = 0.2925 in.R' = R + 1/2t = 0.2400 in.

w = bo - 2(R + t) = 4.4150 in.h = ho - 2(R + t) = 9.4150 in.

0.3768 in.c = 0.1553 in. corner, see Fig.2.

0.0396 in^2

b. Limitw/t = 42.0476 < 500 (Stiffened compression flange)h/t = 89.6667 < 200 (Unreinforced web)

c. Effective width of compression flange Where, k = 4.00 (Stiffened flange/section)

0.90425 > 0,673 0.83683

Nominal and allowable moment strength, based on initiation of yieldingStandard : AISI Standard, North Americans Specification for the Design of Cold-Formed

Lc = 1/2 R' =

Ac = 1/2 R t + 1/4 t2 =

=

Civil Engineering Tools For Learning And Working

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ho

h

R

wb/2 b/2

bo

xx

t

Efy

tw

k052,1

22

33

Rt)(R

Rt)(R4c

3

3.69461 in.

d. The neutral axis

Distance from top fiber Element length0.0525 in.0.1372 in.5.0000 in.9.8628 in.9.9475 in.

L n. Yn The neutral axis,0.1940 in^20.1034 in^2

94.1500 in^27.4326 in^2 From the top of fiber, see Fig.2.

43.9182 in^2145.7982 in^2

e. Effective width of web By using Fig.3, effectiveness of web calculate as follows,

= 320.596224 ksi

= 303.973173 ksi

= 0.9481

22.6839

= 0.78630173 > 0,673 0.91595

b = w =

y = 1/2 t = L = b =y = (R + t) - c = L = 2 Lc =

y = ho/2 = L = 2{ho - 2(R + t)} =y = ho - (R + t) + c = L = 2 Lc =

y = ho - 1/2 t = L = bo - 2(R + t) = Ln =

= = = = =

Ln . Yn =

(Compression)

(Tensile)

k = 4 + 2(1 + )3 + 2(1 + ) =

=

R1

2

Lc

3

2

R = R + 1/2t

t

R

bo

ho

Keterangan :

Lc = 1/2 R

w

h

b/2b/2

Ac = R t + t2

Ycg

X

Ln

Yn. LnYcg

)()(

1 FyytRy

fcg

cg

)()(

)(12 ftRy

tRyhf

cg

cgo

1

2

ff

Ef

th

k1052,1

Ycg - (R + t) = 4.8328 in.

8.6236 in.ho/bo = 2.0000 Ycg - (R + t) = 4,8328 in.The web is fully effective

If the web is fully effective, go to the next page.

be = h =

b1 = be/(3 + ) =

For ho/bo 4b2 = be/2 = when > 0,236

b2 = be - b1 = when 0,236

For ho/bo 4b2 = be/(1 + ) - b1 =

b1 + b2 =

ho

h

R

wb/2 b/2

bo

xx

t

Fy

f2

top web

bottom web

Ycg

ineffective web

Ef

th

k1052,1

f. Moment of Inertia and Section Modulus. The web is fully effective. Neutral axis from top fiber, Ycg = 5.1253 in. Distance from element c.g. to neutral axis, Element length (or area of corner),

y = Ycg - 1/2 t = 5.0728 in. L = b =y = Ycg - (R + t) + c = 4.9881 in. = 2 Ac =

L = 2 h =

y = ho - Ycg - (R + t) + c = 4.7375 in. = 2 Ac =y = ho - Ycg - 1/2 t = 4.8222 in. L = bo - 2(R + t) =

Moment of InertiaElement Ln Ln.t or Ac yn Ixo

in. in^2 in. in^2 in^4 in^4 3.6946 0.3879 5.0728 25.7332 9.98278633 0.0791 4.9881 24.8811 1.9688 18.8300 1.9772 58.4197

0.0791 4.7375 22.4440 1.7760 4.4150 0.4636 4.8222 23.2537 10.7798

Modulus of section,

16.180 in^3

g. Nominal Moment Strength.

5501.19 kip.in

h. Allowable Moment Strength.

3294.12 kip.in

1.67

0.95

5226.128 kip.in

yn2 Ln.t.yn2

Ix =Note : Ixo = 1/12 t . Ln3

Mn = Se . Fy =

h.1. ASD.

where, b =

h.1. LRFD.Stiffened element, b =

Mu = b . Mn =

cge y

IxS

b

na

MM

Leave this page, the web is fully effective e.1. First iteration

Distance from top fiber, Element length,y = 1/2 t = in. L = b =

y = (R + t) - c = in. L = 2 Lc =y top = b1/2 + (R + t) = in. L top = 2 b1 =

y bottom = in. L bot =2{ho-Ycg-(R+t)+b2} =y = ho - (R + t) + c = in. L = 2 Lc =

y = ho - 1/2 t = in. L = bo - 2(R + t) = Ln =

L n. Yn = in^2 = in^2 The neutral axis,

top = in^2bot = in^2 Ycg = Ln.Yn / Ln =

= in^2 = in^2 From the top of fiber, see Fig.2.

Ln . Yn = in^2

= ksi

= ksi

=

=

Ycg - (R + t) = in.

in.

in.

ho/bo = 2.0000 0,236

b2 = be - b1 = when 0,236

For ho/bo 4

)()(

1 FyytRy

fcg

cg

)()(

)(12 ftRy

tRyhf

cg

cgo

1

2

ff

Ef

th

k1052,1

in.

in.

The ineffective portion of the web, Ycg - (R + t) - (b1 + b2) =

NOTE : Distance from top fiber for element y bottom = {ho - Ycg - (R + t) + b2}/2 + ineffective + b1 + (R + t)

Leave this page, the web is fully effective e.2. Second iteration

Distance from top fiber Element lengthy = 1/2 t = in. L = b =



y = ho - 1/2 t = in. L = bo - 2(R + t) = Ln =




Ln . Yn = in^2

= ksi

= ksi

=

= > 0,673

Ycg - (R + t) = in.

in.

in.

ho/bo =

in.in.

b2 = be/(1 + ) - b1 =

b1 + b2 =

k = 4 + 2(1 + )3 + 2(1 + ) =

=

be = h =

b1 = be/(3 + ) =


b2 = be - b1 = when 0,236

)()(

1 FyytRy

fcg

cg

)()(

)(12 ftRy

tRyhf

cg

cgo

1

2

ff

Ef

th

k1052,1

in.

in.



Leave this page, the web is fully effective e.3. Third iteration




y = ho - 1/2 t = in. L = bo - 2(R + t) = Ln =




Ln . Yn = in^2

= ksi

= ksi

=

= > 0,673

Ycg - (R + t) = in.

in.

in.

ho/bo =

in.in.

For ho/bo 4b2 = be/(1 + ) - b1 =

b1 + b2 =

k = 4 + 2(1 + )3 + 2(1 + ) =

=

be = h =

b1 = be/(3 + ) =


b2 = be - b1 = when 0,236

)()(

1 FyytRy

fcg

cg

)()(

)(12 ftRy

tRyhf

cg

cgo

1

2

ff

Ef

th

k1052,1

in.

in.



Leave this page, the web is fully effective e.4. Fourth iteration




y = ho - 1/2 t = in. L = bo - 2(R + t) = Ln =




Ln . Yn = in^2

= ksi

= ksi

=

= > 0,673

Ycg - (R + t) = in.

in.

in.

ho/bo =

in.

For ho/bo 4b2 = be/(1 + ) - b1 =

b1 + b2 =

k = 4 + 2(1 + )3 + 2(1 + ) =

=

be = h =

b1 = be/(3 + ) =


)()(

1 FyytRy

fcg

cg

)()(

)(12 ftRy

tRyhf

cg

cgo

1

2

ff

Ef

th

k1052,1

in.

in.

in.



Leave this page, the web is fully effective e.5. Fifth iteration




y = ho - 1/2 t = in. L = bo - 2(R + t) = Ln =




Ln . Yn = in^2

= ksi

= ksi

=

= > 0,673

Ycg - (R + t) = in.

in.

in.

ho/bo =

b2 = be - b1 = when 0,236

For ho/bo 4b2 = be/(1 + ) - b1 =

b1 + b2 =

k = 4 + 2(1 + )3 + 2(1 + ) =

=

be = h =

b1 = be/(3 + ) =

For ho/bo 4

)()(

1 FyytRy

fcg

cg

)()(

)(12 ftRy

tRyhf

cg

cgo

1

2

ff

Ef

th

k1052,1

in.in.

in.

in.



Leave this page, the web is fully effective e.6. Sixth iteration




y = ho - 1/2 t = in. L = bo - 2(R + t) = Ln =




Ln . Yn = in^2

= ksi

= ksi

=

= > 0,673

Ycg - (R + t) = in.

in.

in.

ho/bo =

b2 = be/2 = when > 0,236b2 = be - b1 = when 0,236

For ho/bo 4b2 = be/(1 + ) - b1 =

b1 + b2 =

k = 4 + 2(1 + )3 + 2(1 + ) =

=

be = h =

b1 = be/(3 + ) =

)()(

1 FyytRy

fcg

cg

)()(

)(12 ftRy

tRyhf

cg

cgo

1

2

ff

Ef

th

k1052,1

in.in.

in.

in.



Leave this page, the web is fully effective e.7. Seventh iteration




y = ho - 1/2 t = in. L = bo - 2(R + t) = Ln =




Ln . Yn = in^2

= ksi

= ksi

=

= > 0,673

Ycg - (R + t) = in.

in.

in.


b2 = be - b1 = when 0,236

For ho/bo 4b2 = be/(1 + ) - b1 =

b1 + b2 =

k = 4 + 2(1 + )3 + 2(1 + ) =

=

be = h =

b1 = be/(3 + ) =

)()(

1 FyytRy

fcg

cg

)()(

)(12 ftRy

tRyhf

cg

cgo

1

2

ff

Ef

th

k1052,1

ho/bo =

in.in.

in.

in.



e.8. Eighth iteration Leave this page, the web is fully effective




y = ho - 1/2 t = in. L = bo - 2(R + t) = Ln =




Ln . Yn = in^2

= ksi

= ksi

=

= > 0,673

Ycg - (R + t) = in.

in.


b2 = be - b1 = when 0,236

For ho/bo 4b2 = be/(1 + ) - b1 =

b1 + b2 =

k = 4 + 2(1 + )3 + 2(1 + ) =

=

be = h =

)()(

1 FyytRy

fcg

cg

)()(

)(12 ftRy

tRyhf

cg

cgo

1

2

ff

Ef

th

k1052,1

in.

ho/bo =

in.in.

in.

in. The ineffective portion of the web,

Ycg - (R + t) - (b1 + b2) =


e.9. Ninth iteration Leave this page, the web is fully effective




y = ho - 1/2 t = in. L = bo - 2(R + t) = Ln =




Ln . Yn = in^2

= ksi

= ksi

=

= > 0,673

Ycg - (R + t) = in.

in.

b1 = be/(3 + ) =


b2 = be - b1 = when 0,236

For ho/bo 4b2 = be/(1 + ) - b1 =

b1 + b2 =

k = 4 + 2(1 + )3 + 2(1 + ) =

=

be = h =

)()(

1 FyytRy

fcg

cg

)()(

)(12 ftRy

tRyhf

cg

cgo

1

2

ff

Ef

th

k1052,1

in.

ho/bo =

in.in.

in.


Ycg - (R + t) - (b1 + b2) =


e.10. Tenth iteration Leave this page, the web is fully effective




y = ho - 1/2 t = in. L = bo - 2(R + t) = Ln =




Ln . Yn = in^2

= ksi

= ksi

=

= > 0,673

Ycg - (R + t) = in.

b1 = be/(3 + ) =


b2 = be - b1 = when 0,236

For ho/bo 4b2 = be/(1 + ) - b1 =

b1 + b2 =

k = 4 + 2(1 + )3 + 2(1 + ) =

=

)()(

1 FyytRy

fcg

cg

)()(

)(12 ftRy

tRyhf

cg

cgo

1

2

ff

Ef

th

k1052,1

in.

in.

ho/bo =

in.in.

in.


Ycg - (R + t) - (b1 + b2) =


REFERENCES1). AISI Standard, North Americans Specification for the Design of Cold-Formed Steel

Structural Member, 2001 Edition.

2).

3).Copyright 2010 John Wiley & Sons, Inc.

be = h =

b1 = be/(3 + ) =


b2 = be - b1 = when 0,236

For ho/bo 4b2 = be/(1 + ) - b1 =

b1 + b2 =

Gregory J. Hancock, Thomas M. Murray and Duane S. Ellifritt, Cold-Formed Steel Structuresto the AISI Specification, Copyright 2001 by Marcel Dekker, Inc. All Rights Reserved.

Wei-Wen Yu and Roger A. LaBoube, Cold-Formed Steel Design, Fourth Edition,



Figure 1.

corner, see Fig.2.

(AISI, B1.1)

(AISI, B1.2)

(AISI, B2.1)

(AISI, B2.1)

initiation of yielding : AISI Standard, North Americans Specification for the Design of Cold-Formed

Figure 2.

3.6946 in.0.7536 in.

18.8300 in.0.7536 in.4.4150 in.

28.4468 in.

5.1253 in.

From the top of fiber, see Fig.2.

(AISI, B2.3)

(AISI, B2.3)

R

1

3

4

4

t

c

Ln

Yn. LnYcg

Figure 3.

(AISI, B2.3)

(AISI, B2.3)

(AISI, B2.3)

(AISI, B2.3)

f1

b2

b1

Element length (or area of corner),3.6946 in.0.0791 in.

18.8300 in.

0.0791 in.4.4150 in.

Moment of InertiaIx

in^49.98281.9688

58.4197

1.776010.779882.9270

in.in.in.in.in.in.in.

in.


in.

Page 5.


in.


in.

Page 6.


in.


in.

Page 7.


in.


in.

Page 8.


in.


in.

Page 9.


in.


in.

Page 10.


in.


in.

Page 11.


in.


in.

Page 12.


in.


in.

Page 13.


in.


in.

Page 14.

, Copyright 2001 by Marcel Dekker, Inc. All Rights Reserved.

Sumber : Cold-Formed Steel Design, 4 th, Wei-Wen Yu, Roger A. LaBoube,, John Wiley & Sons, Inc., 2010.



COMPANY :PROJECT :

1). INPUT DATAUnits : US unit

Fy E ho bo b1 t Rksi ksi in. in. in. in. in.50 29500 10.000 15.000 1.340 0.105 3/16

2). SECTION PROPERTIES a. Parameter of section

R + t = 0.2925 in.R' = R + 1/2t = 0.2400 in.

w = bo - 2(R + t) = 14.4150 in.h = ho - 2(R + t) = 9.4150 in.

0.3768 in.c = 0.1553 in. corner, see Fig.2.

0.0396 in^2

b. Limitw/t = 137.2857 < 500 (Stiffened compression flange)h/t = 89.6667 < 200 (Unreinforced web)

c. Effective width of compression flange Where, k = 4.00 (Stiffened flange)

2.9729 > 0,673 0.31148

Nominal and allowable moment strength, based on initiation of yieldingStandard : AISI Standard, North Americans Specification for the Design of Cold-Formed

Lc = 1/2 R' =

Ac = 1/2 R t + 1/4 t2 =

=

Civil Engineering Tools For Learning And Working

www.thamrinnst.wordpress.com2013

ho

h

R

wb/2 b/2

b1

xx

t

bo

b1

Efy

tw

k052,1

22

33

Rt)(R

Rt)(R4c

3

4.4899 in.

d. Location of neutral axis d.1. First Approximation


L n. Yn The neutral axis,0.2357 in^20.1034 in^2 4.5599 in.


20.8400 in^2 Ycg = 4.5599 in. < ho/2 =122.7617 in^2 The neutral axis is closer to the compression flange,

the maximum stress occurs in the tension flange.Another trial is required, go to page 5.

d.5. Fifth ApproximationYcg = 4.4868 in.

f = 40.6911 ksi

2.6819

b = w =

y = 1/2 t = L = b =y = (R + t) - c = L = 2 Lc =


y = ho - 1/2 t = L = 2b1 - 2(R + t) = Ln =

= = = = =

Ln . Yn =

(Compression), top flange

=

R1

2

Lc

3

2

R = R + 1/2t

t

R

bo

ho

Keterangan :

Lc = 1/2 R

w

h

b/2b/2

Ac = R t + t2

Ycg

X

b1b1

Ln

Yn. LnYcg

b = 4.9339 in. (effective width of top flange)

e. Effective width of web By using Fig.3, effectiveness of web calculate as follows,

= 38.0384 ksi

= 47.3473 ksi

= 1.2447

31.1108

= 0.60728329 < 0,673 1.00000

Ycg - (R + t) = 4.1943 in.

9.4150 in.ho/bo = 0.6667 0,236

b2 = be - b1 = when 0,236

ho

h

R

wb/2 b/2

bo

xx

t

f

f2

top web

bottom web

Ycg

ineffective web

)()(

)(12 ftRy

tRyhf

cg

cgo

1

2

ff

Ef

th

k1052,1

)()(

1 fytRy

fcg

cg

in.

6.9255 in. > Ycg - (R + t) = 4,1943 in.The web is fully effective

If the web is fully effective, go to the next page.

f. Moment of Inertia and Section Modulus. The web is fully effective. Neutral axis from top fiber, Ycg = 4.4868 in. Distance from element c.g. to neutral axis, Element length (or area of corner),

y = Ycg - 1/2 t = 4.4343 in. L = b =y = Ycg - (R + t) + c = 4.3496 in. = 2 Ac =

L = h =

y = ho - Ycg - (R + t) + c = 5.3760 in. = 2 Ac =y = ho - Ycg - 1/2 t = 5.4607 in. L = 2b1 - 2(R + t) =

Moment of InertiaElement Ln Ln.t or Ac yn Ixo

in. in^2 in. in^2 in^4 in^4 4.9339 0.5181 4.4343 19.6628 10.1866 0.0791 4.3496 18.9189 1.4970 9.4150 0.9886 14.6049

0.0791 5.3760 28.9016 2.2869 2.0950 0.2200 5.4607 29.8195 6.5595

Modulus of section,

7.8308 in^3

g. Nominal Moment Strength.

391.5389 kip.in

h. Allowable Moment Strength.

234.4544 kip.in

1.67

0.95

371.9619 kip.in

For ho/bo 4b2 = be/(1 + ) - b1 =

b1 + b2 =

yn2 Ln.t.yn2

Ix =Note : Element , Ixo = 1/12 . (2t) . Ln 3

Mn = Se . Fy =

h.1. ASD.

where, b =

h.1. LRFD.Stiffened element, b =

Mu = b . Mn =

cge y

IxS

b

na

MM

d.2. Second Approximation

41.9098 ksi

Where, k = 4.00 (Stiffened flange)

2.7218 > 0,673 0.33771

4.86803 in.





the maximum stress occurs in the tension flange.

(Compression)

=

b = w =

y = 1/2 t = L = b =y = (R + t) - c = L = 2 Lc =


y = ho - 1/2 t = L = 2b1 - 2(R + t) = Ln =

= = = = =

Ln . Yn =

)(Fyyho

yf

cg

cg

Ef

tw

k052,1

Ln

Yn. LnYcg

d.3. Third Approximation

40.8671 ksi


2.6877 > 0,673 0.34161

4.9242 in.






(Compression)

=

b = w =

y = 1/2 t = L = b =y = (R + t) - c = L = 2 Lc =


y = ho - 1/2 t = L = 2b1 - 2(R + t) = Ln =

= = = = =

Ln . Yn =

)(Fyyho

yf

cg

cg

Ef

tw

k052,1

Ln

Yn. LnYcg

d.4. Fourth Approximation

40.7165 ksi


2.6828 > 0,673 0.34218

4.93253 in.






(Compression)

=

b = w =

y = 1/2 t = L = b =y = (R + t) - c = L = 2 Lc =


y = ho - 1/2 t = L = 2b1 - 2(R + t) = Ln =

= = = = =

Ln . Yn =

)(Fyyho

yf

cg

cg

Ef

tw

k052,1

Ln

Yn. LnYcg

d.5. Fifth Approximation

40.6944 ksi


2.6821 > 0,673 0.34227

4.93375 in.






For Ycg = 4.4868 in.

40.6911 ksi


2.6819 > 0,673 0.34228

(Compression)

=

b = w =

y = 1/2 t = L = b =y = (R + t) - c = L = 2 Lc =


y = ho - 1/2 t = L = 2b1 - 2(R + t) = Ln =

= = = = =

Ln . Yn =

(Compression)

=

)(Fyyho

yf

cg

cg

Ef

tw

k052,1

Ln

Yn. LnYcg

)(Fyyho

yf

cg

cg

Ef

tw

k052,1

4.93393 in.b = w =

Leave this page, the web is fully effective e.1. First iteration

Distance from top fiber, Element length,y = 1/2 t = in. L = b =



y = ho - 1/2 t = in. L = 2b1 - 2(R + t) = Ln =




Ln . Yn = in^2

= ksi

= ksi

=

=

k = 4 + 2(1 + )3 + 2(1 + ) =

=

)()(

1 FyytRy

fcg

cg

)()(

)(12 ftRy

tRyhf

cg

cgo

1

2

ff

Ef

th

k1052,1

Ycg - (R + t) = in.

in.

in.

ho/bo = 0.6667 0,236

b2 = be - b1 = when 0,236

For ho/bo 4b2 = be/(1 + ) - b1 =

b1 + b2 =

k = 4 + 2(1 + )3 + 2(1 + ) =

)()(

1 FyytRy

fcg

cg

)()(

)(12 ftRy

tRyhf

cg

cgo

1

2

ff

Ef

th

k1052,1

= > 0,673

Ycg - (R + t) = in.

in.

in.

ho/bo =

in.in.

in.

in.



Leave this page, the web is fully effective e.3. Third iteration




y = ho - 1/2 t = in. L = bo - 2(R + t) = Ln =




Ln . Yn = in^2

= ksi

= ksi

=

=

be = h =

b1 = be/(3 + ) =


b2 = be - b1 = when 0,236

For ho/bo 4b2 = be/(1 + ) - b1 =

b1 + b2 =

Ef

th

k1052,1

)()(

1 FyytRy

fcg

cg

)()(

)(12 ftRy

tRyhf

cg

cgo

1

2

ff

= > 0,673

Ycg - (R + t) = in.

in.

in.

ho/bo =

in.in.

in.

in.



Leave this page, the web is fully effective e.4. Fourth iteration




y = ho - 1/2 t = in. L = bo - 2(R + t) = Ln =




Ln . Yn = in^2

= ksi

= ksi

k = 4 + 2(1 + )3 + 2(1 + ) =

=

be = h =

b1 = be/(3 + ) =


b2 = be - b1 = when 0,236

For ho/bo 4b2 = be/(1 + ) - b1 =

b1 + b2 =

Ef

th

k1052,1

)()(

1 FyytRy

fcg

cg

)()(

)(12 ftRy

tRyhf

cg

cgo

1

2

ff

== > 0,673

Ycg - (R + t) = in.

in.

in.

ho/bo =

in.in.

in.

in.



Leave this page, the web is fully effective e.5. Fifth iteration




y = ho - 1/2 t = in. L = bo - 2(R + t) = Ln =




Ln . Yn = in^2

= ksi

k = 4 + 2(1 + )3 + 2(1 + ) =

=

be = h =

b1 = be/(3 + ) =


b2 = be - b1 = when 0,236

For ho/bo 4b2 = be/(1 + ) - b1 =

b1 + b2 =

1

2

ff

Ef

th

k1052,1

)()(

1 FyytRy

fcg

cg

)()(

)(12 ftRy

tRyhf

cg

cgo

= ksi

=

= > 0,673

Ycg - (R + t) = in.

in.

in.

ho/bo =

in.in.

in.

in.



Leave this page, the web is fully effective e.6. Sixth iteration




y = ho - 1/2 t = in. L = bo - 2(R + t) = Ln =




Ln . Yn = in^2

= ksi

k = 4 + 2(1 + )3 + 2(1 + ) =

=

be = h =

b1 = be/(3 + ) =


b2 = be - b1 = when 0,236

For ho/bo 4b2 = be/(1 + ) - b1 =

b1 + b2 =

)()(

)(12 ftRy

tRyhf

cg

cgo

1

2

ff

Ef

th

k1052,1

)()(

1 FyytRy

fcg

cg

= ksi

=

= > 0,673

Ycg - (R + t) = in.

in.

in.

ho/bo =

in.in.

in.

in.



Leave this page, the web is fully effective e.7. Seventh iteration




y = ho - 1/2 t = in. L = bo - 2(R + t) = Ln =




Ln . Yn = in^2

k = 4 + 2(1 + )3 + 2(1 + ) =

=

be = h =

b1 = be/(3 + ) =


b2 = be - b1 = when 0,236

For ho/bo 4b2 = be/(1 + ) - b1 =

b1 + b2 =

)()(

1 FyytRy

fcg

cg

)()(

)(12 ftRy

tRyhf

cg

cgo

1

2

ff

Ef

th

k1052,1

= ksi

= ksi

=

= > 0,673

Ycg - (R + t) = in.

in.

in.

ho/bo =

in.in.

in.

in.



e.8. Eighth iteration Leave this page, the web is fully effective




y = ho - 1/2 t = in. L = bo - 2(R + t) = Ln =



= in^2

k = 4 + 2(1 + )3 + 2(1 + ) =

=

be = h =

b1 = be/(3 + ) =


b2 = be - b1 = when 0,236

For ho/bo 4b2 = be/(1 + ) - b1 =

b1 + b2 =

)()(

1 FyytRy

fcg

cg

)()(

)(12 ftRy

tRyhf

cg

cgo

1

2

ff

Ef

th

k1052,1

= in^2 From the top of fiber, see Fig.2. Ln . Yn = in^2

= ksi

= ksi

=

= > 0,673

Ycg - (R + t) = in.

in.

in.

ho/bo =

in.in.

in.


Ycg - (R + t) - (b1 + b2) =


e.9. Ninth iteration Leave this page, the web is fully effective




y = ho - 1/2 t = in. L = bo - 2(R + t) = Ln =


top = in^2

k = 4 + 2(1 + )3 + 2(1 + ) =

=

be = h =

b1 = be/(3 + ) =


b2 = be - b1 = when 0,236

For ho/bo 4b2 = be/(1 + ) - b1 =

b1 + b2 =

)()(

1 FyytRy

fcg

cg

)()(

)(12 ftRy

tRyhf

cg

cgo

1

2

ff

Ef

th

k1052,1

bot = in^2 Ycg = Ln.Yn / Ln = = in^2 = in^2 From the top of fiber, see Fig.2.

Ln . Yn = in^2

= ksi

= ksi

=

= > 0,673

Ycg - (R + t) = in.

in.

in.

ho/bo =

in.in.

in.


Ycg - (R + t) - (b1 + b2) =


e.10. Tenth iteration Leave this page, the web is fully effective




y = ho - 1/2 t = in. L = bo - 2(R + t) = Ln =

L n. Yn = in^2

k = 4 + 2(1 + )3 + 2(1 + ) =

=

be = h =

b1 = be/(3 + ) =


b2 = be - b1 = when 0,236

For ho/bo 4b2 = be/(1 + ) - b1 =

b1 + b2 =

)()(

1 FyytRy

fcg

cg

)()(

)(12 ftRy

tRyhf

cg

cgo

1

2

ff

Ef

th

k1052,1

= in^2 The neutral axis,top = in^2bot = in^2 Ycg = Ln.Yn / Ln =


Ln . Yn = in^2

= ksi

= ksi

=

= > 0,673

Ycg - (R + t) = in.

in.

in.

ho/bo =

in.in.

in.


Ycg - (R + t) - (b1 + b2) =


REFERENCES1). AISI Standard, North Americans Specification for the Design of Cold-Formed Steel

Structural Member, 2001 Edition.

2).

3).Copyright 2010 John Wiley & Sons, Inc.

k = 4 + 2(1 + )3 + 2(1 + ) =

=

be = h =

b1 = be/(3 + ) =


b2 = be - b1 = when 0,236

For ho/bo 4b2 = be/(1 + ) - b1 =

b1 + b2 =

Gregory J. Hancock, Thomas M. Murray and Duane S. Ellifritt, Cold-Formed Steel Structuresto the AISI Specification, Copyright 2001 by Marcel Dekker, Inc. All Rights Reserved.

Wei-Wen Yu and Roger A. LaBoube, Cold-Formed Steel Design, Fourth Edition,

)()(

1 FyytRy

fcg

cg

)()(

)(12 ftRy

tRyhf

cg

cgo

1

2

ff

Ef

th

k1052,1



Figure 1.

corner, see Fig.2.

(AISI, B1.1)

(AISI, B1.2)

(AISI, B2.1)

(AISI, B2.1)

initiation of yielding : AISI Standard, North Americans Specification for the Design of Cold-Formed

h

Figure 2.

4.4899 in.0.7536 in.

18.8300 in.0.7536 in.2.0950 in.

26.9221 in.

5.000 in.

R

1

3

4

4

t

c

(AISI, B2.3)

(AISI, B2.3)

Figure 3.

(AISI, B2.3)

(AISI, B2.3)

(AISI, B2.3)

f1

b2

b1

(AISI, B2.3)

Page 3.

Element length (or area of corner),4.9339 in.0.0791 in^29.4150 in.

0.0791 in^22.0950 in.

Moment of InertiaIx

in^410.1866

1.497014.6049

2.28696.5595

35.1350

4.8680 in.0.7536 in.

18.8300 in.0.7536 in.2.0950 in.

27.3002 in.

5.000 in.

4.9242 in.0.7536 in.

18.8300 in.0.7536 in.2.0950 in.

27.3564 in.

5.000 in.

4.9325 in.0.7536 in.

18.8300 in.0.7536 in.2.0950 in.

27.3647 in.

5.000 in.

4.9338 in.0.7536 in.

18.8300 in.0.7536 in.2.0950 in.

27.3660 in.

5.000 in.


in.


in.

Page 5.


in.


in.

Page 6.


in.


in.

Page 7.


in.


in.

Page 8.


in.


in.

Page 9.


in.


in.

Page 10.


in.


in.

Page 11.


in.


in.

Page 12.


in.


in.

Page 13.


Sumber : Cold-Formed Steel Design, 4 th, Wei-Wen Yu, Roger A. LaBoube,, John Wiley & Sons, Inc., 2010.

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