22
PROJECT : PAGE : CLIENT : DESIGN BY : JOB NO. : DATE : REVIEW BY : WF Simply Supported Beam Design with Torsional Loading Based on AISC 360-10/16 INPUT DATA & DESIGN SUMMARY BEAM SECTION = > W10X54 = > A d GRAVITY DISTRIBUTED LOAD w = 1.15 klf, (17 kN / m) 15.8 10.1 4.38 2.55 LATERAL POINT LOAD AT MID F = 5 kips, (22 kN) l TORSION AT MID SPAN T = 5.1 ft-kips, (7 kN-m) 103 20.6 0.01744 0.37 AXIAL LOAD P = 96 kips, (427 kN) BEAM LENGTH 15 ft, (4.57 m) BEAM YIELD STRESS 50 ksi, (345 MPa) VERTICAL BENDING UNBRACED LENGTH 15 ft, (4.57 m) AXIAL VERTICAL UNBRACED LENGTH 15 ft, (4.57 m) AXIAL HORIZONTAL UNBRACED LENGTH 7.5 ft, (2.29 m) ANALYSIS DETERMINE GOVERNING MOMENTS AT MIDDLE OF SPAN 32.3 ft-kips 18.8 ft-kips 22.7 ft-kips THE BEAM DESIGN IS ADEQUATE. 13.3 ft-kips 0.584 ,(Philip page 101) CHECK TORSIONAL CAPACITY (AISC 360 H3.3 & Philip page 100) 0.73 < 1.00 [Satisfactory] Where 21.93 ksi 29.94 ksi CHECK COMBINED COMPRESSION AND BENDING CAPACITY (AISC 360 H1) 1.08 < 4/3 [Satisfactory] Where 96 kips 109.7 ft-kips, (Sim. from Philip page 100) 18.8 ft-kips 721 / 1.67 = 431.971 kips, (AISC 360 Chapter E) > [Satisfactory] 252.623 / 1.67 = 151.271 ft-kips, (AISC 360 Chapter F) > [Satisfactory] 130.417 / 1.67 = 78.0938 ft-kips, (AISC 360 Chapter F) > [Satisfactory] DETERMINE DEFLECTIONS 0.22127 Where 11200 ksi 29000 ksi rx ry Iy Sy tw L = Fy = Lb = Lx = Ly = Mx = w L 2 / 8 = My = F L / 4 = M0 = T L / (4d) = MT = bM0 = fbx / Fnx = fbx = Mx / Sx + 2MT / Sy = Fnx = Fy / WT = Fy / 1.67 = Pr = Mrx = (Mx / Sx + 2MT / Sy) Sx = Mry = Pc = Pn / Wc = 3/4 Pr Mcx = Mn / Wb = Mrx Mcy = Mn / Wb = 3/4 Mry o , max twist angle at middle (Philip page 1 G = Es = 2 4 sinh 2 sinh L L L l b l l 2sinh 2 sinh 2 2 2 sinh L T L L GJ L l l l l l 8 , 0.2 9 , 0.2 2 ry r rx r c cx cy c ry r rx r c cx cy c M P M P for P M M P M P M P for P M M P

WF Beam with Torsion - Structural Design Softwareengineering-international.com/BeamWithTorsion.xls · 2018-04-06WF Beam with Torsion - Structural Design Software

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PROJECT : PAGE :

CLIENT : DESIGN BY : JOB NO. : DATE : REVIEW BY :

WF Simply Supported Beam Design with Torsional Loading Based on AISC 360-10/16

INPUT DATA & DESIGN SUMMARYBEAM SECTION = > W10X54 = > A dGRAVITY DISTRIBUTED LOAD w = 1.15 klf, (17 kN / m) 15.8 10.1 4.38 2.55 303 60

LATERAL POINT LOAD AT MID F = 5 kips, (22 kN) lTORSION AT MID SPAN T = 5.1 ft-kips, (7 kN-m) 103 20.6 0.01744 0.37 10.00 0.62AXIAL LOAD P = 96 kips, (427 kN)BEAM LENGTH 15 ft, (4.57 m)

BEAM YIELD STRESS 50 ksi, (345 MPa)

VERTICAL BENDING UNBRACED LENGTH 15 ft, (4.57 m)

AXIAL VERTICAL UNBRACED LENGTH 15 ft, (4.57 m)

AXIAL HORIZONTAL UNBRACED LENGTH 7.5 ft, (2.29 m)

ANALYSISDETERMINE GOVERNING MOMENTS AT MIDDLE OF SPAN

32.3 ft-kips

18.8 ft-kips

22.7 ft-kips THE BEAM DESIGN IS ADEQUATE.13.3 ft-kips

0.584 ,(Philip page 101)

CHECK TORSIONAL CAPACITY (AISC 360 H3.3 & Philip page 100)

0.73 < 1.00 [Satisfactory]Where 21.93 ksi

29.94 ksi

CHECK COMBINED COMPRESSION AND BENDING CAPACITY (AISC 360 H1)

1.08 < 4/3 [Satisfactory]

Where 96 kips

109.7 ft-kips, (Sim. from Philip page 100)

18.8 ft-kips

721 / 1.67 = 431.971 kips, (AISC 360 Chapter E)

> [Satisfactory]252.623 / 1.67 = 151.271 ft-kips, (AISC 360 Chapter F)

> [Satisfactory]130.417 / 1.67 = 78.0938 ft-kips, (AISC 360 Chapter F)

> [Satisfactory]

DETERMINE DEFLECTIONS

0.22127

Where 11200 ksi 29000 ksi

rx ry Ix Sx

Iy Sy tw bf tf

L =

Fy =

Lb =

Lx =

Ly =

Mx = w L2/ 8 =

My = F L / 4 =

M0 = T L / (4d) =

MT = bM0 =

fbx / Fnx =

fbx = Mx / Sx + 2MT / Sy =

Fnx = Fy / WT = Fy / 1.67 =

Pr =

Mrx = (Mx / Sx + 2MT / Sy) Sx =

Mry =

Pc = Pn / Wc =

3/4 Pr

Mcx = Mn / Wb =

Mrx

Mcy = Mn / Wb =

3/4 Mry

o , max twist angle at middle (Philip page 100)

G = Es =

24 sinh

2sinh

L

L L

l

bl l

2sinh

2 sinh2 22 sinh

LT LL

GJ L

lll

l l

8 , 0.29

, 0.22

ryr rx r

c cx cy c

ryr rx r

c cx cy c

MP M PforP M M P

MP M PforP M M P

J = 1.82 in4

(cont'd)

0.15 in = L / 1207 , vertical deflection at middle

Where 303

0.20 in = L / 885 , horizontal deflection at middle

Where 103

Technical References: 1. AISC: "Steel Construction Manual 14th Edition", American Institute of Steel Construction, 2010. 2. Philip H. Lin: "Simplified Design for Torsional Loading of Rolled Steel Members", Engineering Journal, AISC, 1977.

I3 = Ix sin2(90-) + Iy cos2(90-) = in4 , (AISC Manual Page 17-42)

I4 = Ix cos2(90-) + Iy sin2(90-) = in4 , (AISC Manual Page 17-42)

4

3

5384

wLvert E I

3

448F L

horiz E I

ROJECT : PAGE :

CLIENT : DESIGN BY : JOB NO. : DATE : REVIEW BY :

Plate Girder Design Based on AISC Manual 13th Edition (AISC 360-05)

INPUT DATA & DESIGN SUMMARYSTEEL YIELD STRESS 50 ksiSIMPLY SUPPORTED SPAN 15 ftSUPERIMPOSED UNIFORM DEAD LOAD kips / ftUNIFORM LIVE LOAD kips / ft

POINT DEAD LOAD kips

POINT LIVE LOAD kipsDISTANCE POINT LOAD TO END ft

TOP FLANGE WIDTH 10.00 in

TOP FLANGE THICKNESS 0.62 in

BOTTOM FLANGE WIDTH 10 in

BOTTOM FLANGE THICKNESS 0.615 in

WEB THICKNESS 0.37 inBEAM DEPTH 10.1 in Err:502UNBRACED LENGTH 15 ft FLANGE TO WEB WELDING USE 1/4 in - 24 in @ 2293 in o.c.

THE GIRDER DESIGN IS ADEQUATE.

ANALYSISCHECK LIMITING WIDTH-THICKNESS RATIOS FOR WEB (AISC 365-05 Table B4.1)

23.97 < 137.27

< 90.55Compact Web

where E = 29000 ksi

137.27

87.03

90.55

8.87 in 9.46 in

277.5 ft-kips 250.0 ft-kips

CHECK LIMITING WIDTH-THICKNESS RATIOS FOR FLANGES (AISC 365-05 Table B4.1)

8.13 < 25.09

< 9.15

Compact Flangeswhere 25.09

9.15

0.76

60 60

35 ksi, (AISC 360-05 Table note B4.1 & Eq F4-6)

DETERMINE CRITERIA FOR ALLOWABLE FLEXURAL STRENGTH (AISC 365-05 Table F1.1)

Required ConditionsChapter F Sections

F2 F3 F4 F5Double Symmetric x x

Compact Web x xx

Noncompact Web

Slender Web 151.3 ft-kipsCompact Flanges x ( from following analysis)

Noncompact FlangesSlender Flanges

Applicable Section ok

9.02 ft

33.66 ft

where 2.55 in 60

Fy = S =

DL = LL =

PDL =

PLL = c =

bf,top =

tf,top =

bf,bot =

tf,bot =

tw = d =

Lb =

hc / tw = lr =

lp =

lr = 5.7 (E / Fy)0.5 =

lp = (hc / hp) (E / Fy)0.5 / (0.54 Mp / My -0.09)2 = ,for Af,top ≠ Af,bot

lp = 3.76 (E / Fy)0.5 = ,for Af,top = Af,bot

hc = hp =

Mp = My =

0.5 bf,top / tf,top = lr =

lp =

lr = 1.0 (kc E / FL)0.5 =

lp = 0.38 (E / Fy)0.5 =

kc = Min [0.76 , Max (0.35 , 4 / (h / tw)0.5 )] =

Sxt = in3 Sxc = in3

FL =

Mallowable = Mn / Wb =

DETERMINE ALLOWABLE FLEXURAL STRENGTH , Mn / Wb , BASED ON AISC 365-05 Chapter F2

ry = Sx = in3

1.76p yy

EL r

F

2

0

0.7 01.95 1 1 6.760.7 0.7

r ts tsy x y

FE Jc Ey S hxL r rE JcS h FF

9.49 in 103h0 = Iy = in4

(cont'd)

2316.6 , (AISC 365-05 F2.2)

31.8102(Use J = 1.82

2.85 in

1.0 1.0 , (AISC Manual 13th Table 3-1, page 3-10)

101.524 ksi

252.6 ft-kips

151.3 ft-kips where 1.67 , (AISC 365-05 F1)

<== Not Applicable.

284.1 ft-kips

where 8.13

9.15 25.09

151.3 ft-kips

<== Not Applicable.

6.33 ft

33.85 ft

where 0.53

2.87 in

277.5 ft-kips

250 ft-kips 250 ft-kips

23.97

87.03 137.27

1.11

102.385 0.50 > 0.23, AISC 360-05 F4-5 )

245.2238 ft-kips

Cw = Iy h02 / 4 =

J = [tw d (tw2 + d2)] / 12 = in4, (not applicable if taken web only, EIT Manual page 26)

in4 )rts =[( Iy Cw)0.5/ Sx)]0.5 =

c = Cb =

Mallowable, F2 =Mn / Wb = Wb =

DETERMINE ALLOWABLE FLEXURAL STRENGTH , Mn / Wb , BASED ON AISC 365-05 Chapter F3

l = bf / (2 tf) =

lpf = lp = lrf = lr =

Mallowable, F3 =Min(Mn,F2 , Mn,F3) / Wb =

DETERMINE ALLOWABLE FLEXURAL STRENGTH , Mn / Wb , BASED ON AISC 365-05 Chapter F4

aw =hc tw/ (bfc tfc) =

Mp = Min [Zx Fy , 1.6Sxc Fy ] =

Myc = Sxc Fy = Myt = Sxt Fy =

l = hc / tw =

lpw = lp = lrw = lr =

ksi, (for Iyc / Iy =

22

20

1 0.078b bcr

tsxb

ts

E JcC LFrS hL

r

, 2

,

0.7 , ,

, ,

p b p

b pp p y p p b rb xn F

r p

cr p r bx

forM L L

L LMin forC SM M F M L L LM L L

Min forSF M L L

, 3

2

0.7 ,

0.9 ,

pfp p y x

rf pfn F

c x

for Noncompact FlangesSM M FM

Ek S for Slender Flanges

l ll l

l

1.1p ty

EL r

F

2

0

01.95 1 1 6.76 Lr t

L xc

E J S hF xcL r E JS hF

, 4.2

,

, ,

, ,

pc yc b p

b ppc yc pc yc L pc yc p b rb xcn F

r p

cr pc yc r bxc

forR M L L

L LMin forC SR M R M F R M L L LM L L

Min forSF R M L L

20

0

1126

fct

w

br

h had dh

, /

1 , , /

pc w pw

ycpc

p p pw pc w pw

yc yc ycrw pw

M for h tM

RM M MMin for h tM M M

l

l ll

l l

22

20

1 0.078b bcr

txcb

t

E JC LFrS hL

r

(cont'd)

277.5 ft-kips

1.11

146.8 ft-kips

<== Not Applicable.

21.62 ft

41.4981 ksi

50 ksi

1

124.2 ft-kips

9 24 , 3.7414.7 ft

5.00 1.000

112.11 kips

67.1317 kips 1.67 , (AISC 365-05 G1)

TOTAL SUPERIMPOSED GRAVITY LOAD

w = DL + LL = 0.000 kips / ft , 0.00 kips

CHECK EACH SECTION CAPACITIES

Section Left 0.00 S 0.00 S 0.00 S 0.00 S 0.00 S Point 0.17 S 0.33 S 0.50 S 0.67 S 0.83 S RightDistance 0 0.00 0.00 0.00 0.00 0.00 0.00 2.50 5.00 7.50 10.00 12.50 15.00

10 10 10 10 10 10 10 10 10 10 10 10 105 5 5 5 5 5 5 5 5 5 5 5 5

299 299 299 299 299 299 299 299 299 299 299 299 29953.0 53.0 53.0 53.0 53.0 53.0 53.0 53.0 53.0 53.0 53.0 53.0 53.00.40 0.40 0.40 0.40 0.40 0.40 0.00 -0.27 -0.13 0.00 0.13 0.27 0.40

0 0 0 0 0 0 0 1 1 1 1 1 0

Mallowable, F4 = Min( Mn,F4.2 , Mn,F4.3, RptMyt) / Wb =

DETERMINE ALLOWABLE FLEXURAL STRENGTH , Mn / Wb , BASED ON AISC 365-05 Chapter F5

Mallowable, F5 = Min( RpgFySxc , RpgFcr,F5.2Sxc , RpgFcr,F5.3Sxc , FySxt) / Wb =

DETERMINE ALLOWABLE SHEAR STRENGTH , Vn / Wv , BASED ON AISC 365-05 Chapter G2

h = d - tf,top - tf,bot = in , h / tw = Aw = in2 ,a =

Vn = 0.6 FyAwCv =

Vallowable = Vn / Wv = Wv =

P = PDL + PLL =

d (in)y (in)I (in4)Wt (plf)V (kips)M (ft-k)

255 , / 3/

5 , / 3v

for a ha hfor a h

k

2

1.0 , / 1.10

1.10 , 1.10 / 1.37/

1.51 , 1.37 //

vw

y

v v vwv

y y yw

v vw

y yw

Ekfor h tF

E E Ek k kfor h th t F F F

E Ek kfor h tF Fh t

C

0.7r t

y

EL r

F

, 5.2

2

2

,

0.3 , ,

, ,

y b p

b py y y p b rb

r p

cr F

by r b

forF L L

L LMin forC F F F L L LL L

F

ECMin forF L LLbrt

, 5.3

2

,

0.3 ,

0.9 ,

2

y

pfy y

rf pfcr F

c

f

f

for Compact FlangesF

for Noncompact FlangesF F

FEk for Slender Flanges

bt

l ll l

, 10

1 5.7 , 1.01200 300 , 10

w cpg

yw w

Min Ea hMinRMin a t F

, 5.3

2

,

,

0.9 ,

pc yc

pfpc yc pc yc L xc

n F rf pf

c xc

for Compact FlangesR M

for Noncompact FlangesSR M R M FM

Ek S for Slender Flanges

l ll l

l

, /

1 , , /

pc w pw

ytpt

p p pw pc w pw

yt yt ytrw pw

M for h tM

RM M MMin for h tM M M

l

l ll

l l

(cont'd)

1.491237 ft-kips @ 7.50 ft, from heel.

< 151.271 ft-kips[Satisfactory]

0.40 kips @ 15.00 ft, from heel.

< 67.1317 kips [Satisfactory]

DETERMINE DEFLECTION AT MID SPAN

0.01 in ( L / 25804 ) (for camber, self Wt included.)

where E = 29000 ksi w = 0.053 kips / ftI = 299 P = 0 kipsb = 0.3 ft L = 15.0 ft

0.00 #DIV/0!

where P = 0 kips w = 0.000 kips / ft

DETERMINE FLANGE TO WEB WELDING (AISC 360-05 J2.4 )

1/4 in

3/16

4/162.0

0.40 kips

29

0.04 kips / in

A = 24 in 2293 in. o.c.

USE 1/4 in - 24 in @ 2293 in o.c.

DESIGN STIFFENERS

2. CHECK LOCAL WEB YIELDING FOR THE CONCENTRATED LOAD. (AISC 360-05, J10.2)R = P = 0.00 kipsN = 0 in, bearing length, point.

0.87 in1.5

0.00 < [Satisfactory]

Mmax =

Mallowable =

Vmax =

Vallowable =

in4

w =wmin = in, < w

wmax = in, > wW =

Vmax =

Q = Af(d - y - 0.5 tf,top) = in3

vmax = Vmax Q / I =

1. BEARING STIFFENERS ARE REQUIRED AT EACH END SUPPORT. (AISC 360-05, J10.8)

k = tf,top + w =W =

Fy / W

-160-140-120-100-80-60-40-20

0

BENDING LOADS & CAPACITY

Length

Mom

ents

-68.00-67.50-67.00-66.50-66.00-65.50

SHEAR LOADS & CAPACITY

Length

Shea

r For

ces

4 3/ 2225 0.06415

384DLw PbL bL

EI EIL

4 3/ 2225 0.06415

384LLw PbL bL

EI EIL

0.6 0.707

max

EXX w AFBv

W

,5

,2.5

R for c dN kt w

R for c dN kt w

(cont'd)3. CHECK WEB CRIPPLING FOR THE CONCENTRATED LOAD. (AISC 360-05, J10.3)

2.0

85.01 > P [Satisfactory]

(Note : If item 2, local web yielding is Satisfactory, this item does not need to be checked.)

4. CHECK SIDESWAY WEB BUCKING FOR THE CONCENTRATED LOAD. (AISC 360-05, J10.4)

8.37 in

960000 ksi

1.261.76

171.48 > P [Satisfactory]

(Note : If item 2, local web yielding is Satisfactory, this item does not need to be checked.)

5. DETERMINE STIFFENER SIZE.

1/3 4 in

12.80[Satisfactory]

4.14 15

R = 0.4 kips

1.67 , (AISC 365-05 E1)

3.5

113

23867.77 ksi

123.9 kips, (AISC 360-05 E2)

> R [Satisfactory]

Techincal Reference: 1. AISC: "Steel Construction Manual 13th Edition", American Institute of Steel Construction, 2005.

W =

dc = d - 2k =

Cr =

(dc / tw ) / (l / bf) =W =

tw = in , bst =

bst / tw = < 0.56 (E / Fy)0.5 , (AISC 360-05 Table B4.1)

Ag = in2 , I = in4

Wc =

K l / r = 0.75 h / ( I / Aeff)0.5 =

Cc = 4.71 (E/ Fy)0.5 =

1.52

1.52

0.80 1 3 , 0.5

/ 1 /

0.40 1 3 , 0.5

yw fww

f w

yw fww

f w

tEFN t for c dtd t t

RntEFN t for c dt

d t t

W W

33

2

33

2

/ /0.4 , 1.7/ /

/ // 1 / 1 0.4 , 1.7 2.3/ /

/, 2.3/

r w f c w c w

f f

r w f c w c w

f f

c w

f

C t t d t d tforl lb bh

C t t d t d tforRn l lb bh

d tP forl b

W W W

/ /

,0.658

0.877 ,

F yy cFe

R An c g c

e c

klfor CFr

klfor CFr

W W

2

2/Fe

E

kl r

PROJECT : PAGE :

CLIENT : DESIGN BY : JOB NO. : DATE : REVIEW BY :

WF Simply Supported Beam Design with Torsional Loading Based on AISC Manual 9th

INPUT DATA & DESIGN SUMMARYBEAM SECTION = > W10X54 = > A dGRAVITY DISTRIBUTED LOAD w = 1.15 kips / ft 15.8 10.1 4.38 2.55 303 60

LATERAL POINT LOAD AT MID F = 5 kips lTORSION AT MID SPAN T = 5.1 ft-kips 103 20.6 0.01744 0.37 10.00 0.62AXIAL LOAD P = 96 kipsBEAM LENGTH 15 ft

BEAM YIELD STRESS 50 ksi

VERTICAL BENDING UNBRACED LENGTH 15 ft

AXIAL VERTICAL UNBRACED LENGTH 15 ft

AXIAL HORIZONTAL UNBRACED LENGTH 7.5 ft

ANALYSISCHECK LOCAL BUCKLING (AISC-ASD Tab. B5.1)

8.13 < 9.19[Satisfactory]

27.30 < 90.51 THE BEAM DESIGN IS ADEQUATE.[Satisfactory]

DETERMINE GOVERNING MOMENTS AT MIDDLE OF SPAN

32.3 ft-kips

18.8 ft-kips

22.7 ft-kips 0.584 ,(Philip page 101)

13.3 ft-kips

DETERMINE GOVERNING UNBALANCED SEGMENT LENGTH (AISC-ASD F1)

8.96 ft

20.30 ft

22.39 ft

Where 1.64

2.66

1.00

DETERMINE ALLOWABLE BENDING STRESSES (AISC-ASD F1)

= = N/A

= = 30.00

= = N/A

= = N/A

Where 25.85 ksi

16.67 ksi

30.00 ksi

CHECK VERTICAL FLEXURAL CAPACITY (AISC-ASD F & Philip page 100)

0.73 < 1.00 [Satisfactory]Where 21.93 ksi

rx ry Ix Sx

Iy Sy tw bf tf

L =

Fy =

Lb =

Lx =

Ly =

bf / (2tf ) = 65 / (Fy)0.5 =

d / tw = 640 / (Fy)0.5 =

Mx = w L2/ 8 =

My = F L / 4 =

M0 = T L / (4d) =

MT = bM0 =

Lc = MIN[76bf/(Fy)0.5 , 20000/(d/Af)Fy] =

Lu = MAX[rT(102000Cb/Fy)0.5 , 12000Cb/(d/Af)0.6Fy] =

L3 = rT(510000Cb/Fy)0.5 =

(d/Af) = in-1

rT =

Cb =

Fbx = {0.66Fy ksi, for Lb @ [0, Lc]

0.60Fy ksi, for Lb @ (Lc, Lu]

MAX(Fb1, Fb3) ksi, for Lb @ (Lu, L3]

MAX(Fb2, Fb3) ksi, for Lb @ (L3, Larger)

Fb1 = MIN{[2/3 - Fy(L/rT)2/(1530000Cb)]Fy , 0.6Fy} =

Fb2 = MIN[170000Cb/(L/rT)2, Fy/3] =

Fb3 = MIN[12000Cb/(Ld/Af), 0.6Fy] =

fbx / Fbx =

fbx = Mx / Sx + 2MT / Sy =

24 sinh

2sinh

L

L L

l

bl l

(cont'd)CHECK COMPRESSION CAPACITY (AISC-ASD E2)

0.24 < 1.33 [Satisfactory]Where 6.08 ksi

K = 1.0

29000 ksi 25.68

107 N/A

41.10 < 200 [Satisfactory]0.38

CHECK COMBINED STRESS (AISC-ASD H1)

0.24 > 0.15

1.33 < 1.33

Where 1.00

10.92 ksi

37.50 ksi

88.39 ksi 120.18 ksi

1.22 < 1.33

1.26 < 1.33 <== Not applicable.

[Satisfactory]

DETERMINE DEFLECTIONS

0.22127

Where 11200 ksiJ = 1.82

0.15 in = L / 1207 , vertical deflection at middle

Where 303

0.20 in = L / 885 , horizontal deflection at middle

Where 103

Technical References: 1. AISC: "Manual of Steel construction 9th", American Institute of Steel Construction, 1990. 2. Philip H. Lin: "Simplified Design for Torsional Loading of Rolled Steel Members", Engineering Journal, AISC, 1977.

fa / Fa =

fa = P / A =

Es =Fa = {

(1-F2/2)Fy / (5/3+3F/8-F3/8) = ksi, for Cc > (Kl/r)Cc = (22Es/Fy)0.5 = 122Es/[23(KL/r)2] = ksi, for Cc < (Kl/r)KL/r = MAX(KLx/rx, KLy/ry) =

F = (KL/ r) / Cc =

fa / Fa =

Cm =

fby = My / Sy =

Fby = 0.75 Fy =

o , max twist angle at middle (Philip page 100)

G =in4

I3 = Ix sin2(90-) + Iy cos2(90-) = in4 , (AISC-ASD Page 6-23)

I4 = Ix cos2(90-) + Iy sin2(90-) = in4 , (AISC-ASD Page 6-23)

1 1' '

fCf f myC bymxa bxfF a fa aFbx FbyFex Fey

212'2

23

EFex

K l xr x

212'2

23

EFey

Kl yr y

0.6

ff f bya bxF F Fbx byy

ff f bya bxF F Fa bx by

2sinh

2 sinh2 22 sinh

LT LL

GJ L

lll

l l

4

3

5384

wLvert E I

3

448F L

horiz E I