flexural member demo xlsx
DESCRIPTION
Cold Form SteelTRANSCRIPT
-
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 = (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
=
= > 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 + ) =
For ho/bo 4b2 = be/2 = when > 0,236
b2 = be - b1 = when 0,236
)()(
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.3. Third iteration
Distance from top fiber Element lengthy = 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
=
= > 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 + ) =
For ho/bo 4b2 = be/2 = when > 0,236
b2 = be - b1 = when 0,236
)()(
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.4. Fourth iteration
Distance from top fiber Element lengthy = 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
=
= > 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 + ) =
For ho/bo 4b2 = be/2 = when > 0,236
)()(
1 FyytRy
fcg
cg
)()(
)(12 ftRy
tRyhf
cg
cgo
1
2
ff
Ef
th
k1052,1
-
in.
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.5. Fifth iteration
Distance from top fiber Element lengthy = 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
=
= > 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.
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.6. Sixth iteration
Distance from top fiber Element lengthy = 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
=
= > 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.
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.7. Seventh iteration
Distance from top fiber Element lengthy = 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
=
= > 0,673
Ycg - (R + t) = in.
in.
in.
For ho/bo 4b2 = be/2 = when > 0,236
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.
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)
e.8. Eighth iteration Leave this page, the web is fully effective
Distance from top fiber Element lengthy = 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
=
= > 0,673
Ycg - (R + t) = in.
in.
For ho/bo 4b2 = be/2 = when > 0,236
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) =
NOTE : Distance from top fiber for element y bottom = {ho - Ycg - (R + t) + b2}/2 + ineffective + b1 + (R + t)
e.9. Ninth iteration Leave this page, the web is fully effective
Distance from top fiber Element lengthy = 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
=
= > 0,673
Ycg - (R + t) = in.
in.
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 =
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) =
NOTE : Distance from top fiber for element y bottom = {ho - Ycg - (R + t) + b2}/2 + ineffective + b1 + (R + t)
e.10. Tenth iteration Leave this page, the web is fully effective
Distance from top fiber Element lengthy = 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
=
= > 0,673
Ycg - (R + t) = in.
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 =
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.
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)
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 + ) =
For ho/bo 4b2 = be/2 = when > 0,236
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,
-
COLD-FORMED STEELFLEXURAL MEMBER
Steel Structural Member, 2001 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
-
Page 2.
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
-
Page 2.
Figure 3.
(AISI, B2.3)
(AISI, B2.3)
(AISI, B2.3)
(AISI, B2.3)
f1
b2
b1
-
Page 3.
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
-
Page 4.
in.in.in.in.in.in.in.
in.
From the top of fiber, see Fig.2.
-
in.
Page 5.
in.in.in.in.in.in.in.
in.
From the top of fiber, see Fig.2.
-
in.
Page 6.
in.in.in.in.in.in.in.
in.
From the top of fiber, see Fig.2.
-
in.
Page 7.
in.in.in.in.in.in.in.
in.
From the top of fiber, see Fig.2.
-
in.
Page 8.
in.in.in.in.in.in.in.
in.
From the top of fiber, see Fig.2.
-
in.
Page 9.
in.in.in.in.in.in.in.
in.
From the top of fiber, see Fig.2.
-
in.
Page 10.
in.in.in.in.in.in.in.
in.
From the top of fiber, see Fig.2.
-
in.
Page 11.
in.in.in.in.in.in.in.
in.
From the top of fiber, see Fig.2.
-
in.
Page 12.
in.in.in.in.in.in.in.
in.
From the top of fiber, see Fig.2.
-
in.
Page 13.
in.in.in.in.in.in.in.
in.
From the top of fiber, see Fig.2.
-
in.
Page 14.
, Copyright 2001 by Marcel Dekker, Inc. All Rights Reserved.
-
Page 15.
-
Sumber : Cold-Formed Steel Design, 4 th, Wei-Wen Yu, Roger A. LaBoube,, John Wiley & Sons, Inc., 2010.
-
Sumber : Cold-Formed Steel Design, 4 th, Wei-Wen Yu, Roger A. LaBoube,, John Wiley & Sons, Inc., 2010.
-
COLD-FORMED STEELFLEXURAL MEMBER
Steel Structural Member, 2001 Edition.
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
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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
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,0.2357 in^20.1034 in^2 4.5599 in.
94.1500 in^27.4326 in^2 From the top of fiber, see Fig.2.
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/2 = L = 2{ho - 2(R + t)} =y = ho - (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.
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,0.2556 in^20.1034 in^2 4.4975 in.
94.1500 in^27.4326 in^2 From the top of fiber, see Fig.2.
20.8400 in^2 Ycg = 4.4975 in. < ho/2 =122.7816 in^2 The neutral axis is closer to the compression flange,
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/2 = L = 2{ho - 2(R + t)} =y = ho - (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
Where, k = 4.00 (Stiffened flange)
2.6877 > 0,673 0.34161
4.9242 in.
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,0.2585 in^20.1034 in^2 4.4883 in.
94.1500 in^27.4326 in^2 From the top of fiber, see Fig.2.
20.8400 in^2 Ycg = 4.4883 in. < ho/2 =122.7845 in^2 The neutral axis is closer to the compression flange,
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/2 = L = 2{ho - 2(R + t)} =y = ho - (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
Where, k = 4.00 (Stiffened flange)
2.6828 > 0,673 0.34218
4.93253 in.
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,0.2590 in^20.1034 in^2 4.4870 in.
94.1500 in^27.4326 in^2 From the top of fiber, see Fig.2.
20.8400 in^2 Ycg = 4.4870 in. < ho/2 =122.7850 in^2 The neutral axis is closer to the compression flange,
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/2 = L = 2{ho - 2(R + t)} =y = ho - (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
Where, k = 4.00 (Stiffened flange)
2.6821 > 0,673 0.34227
4.93375 in.
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,0.2590 in^20.1034 in^2 4.4868 in.
94.1500 in^27.4326 in^2 From the top of fiber, see Fig.2.
20.8400 in^2 Ycg = 4.4868 in. < ho/2 =122.7850 in^2 The neutral axis is closer to the compression flange,
the maximum stress occurs in the tension flange.
For Ycg = 4.4868 in.
40.6911 ksi
Where, k = 4.00 (Stiffened flange)
2.6819 > 0,673 0.34228
(Compression)
=
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 = 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 = (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 = 2b1 - 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
=
=
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.
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.3. Third iteration
Distance from top fiber Element lengthy = 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
=
=
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 =
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.
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.4. Fourth iteration
Distance from top fiber Element lengthy = 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
k = 4 + 2(1 + )3 + 2(1 + ) =
=
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 =
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.
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.5. Fifth iteration
Distance from top fiber Element lengthy = 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
k = 4 + 2(1 + )3 + 2(1 + ) =
=
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 =
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.
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.6. Sixth iteration
Distance from top fiber Element lengthy = 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
k = 4 + 2(1 + )3 + 2(1 + ) =
=
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 =
)()(
)(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.
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.7. Seventh iteration
Distance from top fiber Element lengthy = 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
k = 4 + 2(1 + )3 + 2(1 + ) =
=
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 =
)()(
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.
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)
e.8. Eighth iteration Leave this page, the web is fully effective
Distance from top fiber Element lengthy = 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
k = 4 + 2(1 + )3 + 2(1 + ) =
=
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 =
)()(
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.
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)
e.9. Ninth iteration Leave this page, the web is fully effective
Distance from top fiber Element lengthy = 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^2
k = 4 + 2(1 + )3 + 2(1 + ) =
=
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 =
)()(
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.
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)
e.10. Tenth iteration Leave this page, the web is fully effective
Distance from top fiber Element lengthy = 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
k = 4 + 2(1 + )3 + 2(1 + ) =
=
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 =
)()(
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 =
= 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.
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)
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 + ) =
For ho/bo 4b2 = be/2 = when > 0,236
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
-
COLD-FORMED STEELFLEXURAL MEMBER
Steel Structural Member, 2001 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
h
-
Page 1.
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
-
Page 2.
(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
-
Page 4.
4.8680 in.0.7536 in.
18.8300 in.0.7536 in.2.0950 in.
27.3002 in.
5.000 in.
-
Page 5.
4.9242 in.0.7536 in.
18.8300 in.0.7536 in.2.0950 in.
27.3564 in.
5.000 in.
-
Page 6.
4.9325 in.0.7536 in.
18.8300 in.0.7536 in.2.0950 in.
27.3647 in.
5.000 in.
-
Page 5.
4.9338 in.0.7536 in.
18.8300 in.0.7536 in.2.0950 in.
27.3660 in.
5.000 in.
-
Page 5.
-
in.in.in.in.in.in.in.
in.
From the top of fiber, see Fig.2.
-
in.
Page 5.
in.in.in.in.in.in.in.
in.
From the top of fiber, see Fig.2.
-
in.
Page 6.
in.in.in.in.in.in.in.
in.
From the top of fiber, see Fig.2.
-
in.
Page 7.
in.in.in.in.in.in.in.
in.
From the top of fiber, see Fig.2.
-
in.
Page 8.
in.in.in.in.in.in.in.
in.
From the top of fiber, see Fig.2.
-
in.
Page 9.
in.in.in.in.in.in.in.
in.
From the top of fiber, see Fig.2.
-
in.
Page 10.
in.in.in.in.in.in.in.
in.
From the top of fiber, see Fig.2.
-
in.
Page 11.
in.in.in.in.in.in.in.
in.
-
From the top of fiber, see Fig.2.
in.
Page 12.
in.in.in.in.in.in.in.
-
in.
From the top of fiber, see Fig.2.
in.
Page 13.
in.in.in.in.in.in.in.
-
in.
From the top of fiber, see Fig.2.
in.
Page 14.
, Copyright 2001 by Marcel Dekker, Inc. All Rights Reserved.
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Page 15.
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Sumber : Cold-Formed Steel Design, 4 th, Wei-Wen Yu, Roger A. LaBoube,, John Wiley & Sons, Inc., 2010.
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Sumber : Cold-Formed Steel Design, 4 th, Wei-Wen Yu, Roger A. LaBoube,, John Wiley & Sons, Inc., 2010.
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