PROJECT : PAGE :

CLIENT : DESIGN BY :
JOB NO. : DATE : REVIEW BY :
Column Deformation Compatibility Design using Finite Element Method

INPUT DATA & DESIGN SUMMARY

COLUMN SECTION (WF, HSS tube & pipe) W10X60

STORY HEIGHTS Habove = 14 ft, (4.3 m)

Hbelow = 16 ft, (4.9 m)
H= 14 ft, continued top & bot
( 4.3 m)
COLUMN AXIAL LOAD, ASD PDL = 20 kips, (89.0 kN)   
PLL = 18 kips, (80.1 kN)

EXPECTED DEFORMATION D= 4.2 in, (107 mm)

(ASCE 7-16/10 12.12.4 )

STRONG AXIS BENDING ? (1=Yes, 0=No)

=> 1 yes, strong axis, x-x, bending.

THE DESIGN IS ADEQUATE.

ANALYSIS
DETERMINE JOINT DEFLECTIONS AND SECTION FORCES
Fy = 50 ksi E = 29000 ksi G= 11154 ksi
A= 17.6 in2 I= 341 in4

JOINT DEFLECTIONS UNDER D ONLY THE DESIGN COLUMN SECTION FORCES UNDER D ONLY
Note X (in) Y (in) q (deg) Note V (k) P (k) M (ft-k)
N1 0 0 0.496192 N2 -33.09 0.000 225.24
N2 0 0.0000 -1.007175 N3 33.09 0.000 238.00
N3 4.2 0.0000 -0.932621
N4 4.2 0.0000 0.457379

Pu = 1.2 PDL + 1.0 PLL = 42.00 kips

Mu = (PDL + PLL) D + M = 251.30 ft-kips
Vu = V = 33.09 kips

CHECK COMBINED COMPRESSION AND BENDING CAPACITY (AISC 360 H1)

 P r 8  M rx M ry  P r  0.2
     , for
 c
P 9 M M P 1.00 < 1.0 [Satisfactory]
 cx cy  c
 
 Pr  M ry 
  M rx  , for P r  0.2

2 P  M cx M cy  Pc
 c  
Where KLx = 14 ft, for x-x axial load. KLy = 14 ft, for y-y axial load.
(KL / r)max = 65 < 200 [Satisfactory]
Pr = 42.00 kips
Mrx = 251.30 ft-kips
Mry = 0.00
Pc = f P n = 0.9 x 643.43 = 579.09 kips, (AISC 360 Chapter E)
> Pr [Satisfactory]
Mcx = f Mn = 0.9 x 289.99 = 260.99 ft-kips, (AISC 360 Chapter F)
> Mrx [Satisfactory]
Mcy = f Mn = 0.9 x 145.833 72.92 ft-kips, (AISC 360 Chapter F)
> Mry [Satisfactory]

CHECK SHEAR CAPACITY (AISC 360 G)

f Vn = 115.67 kips > Vu [Satisfactory]

Note: The governing case is always that only the design story column has Story Drift and other stories no drift, since the other
story at same direction drifts, including PD effects, will reduce the section forces of design story column. The opposite story
drifts are not Expected Deformation, because Expected Deformation can not be developed from Linear Modes Analysis
after CQC or SRSS.
 PROJECT : PAGE :
CLIENT : DESIGN BY :
JOB NO. : DATE : REVIEW BY :
Tube / Pipe Column Design Based on AISC Manual 13th Edition (AISC 360-05)

INPUT DATA & DESIGN SUMMARY

COLUMN SECTION (Tube or Pipe) HSS12X12X1/2 Tube

COLUMN YIELD STRESS Fy = 46 ksi

DIMENSION H= 14 ft

AXIAL LOAD, ASD P= kips

STRONG AXIS BENDING ? (1=Yes, 0=No) => 1 yes, strong axis, x-x, bending.
UNIFORM LATERAL LOAD, ASD w= k / ft
CONCENTRATED LATERAL LOAD, ASD F= kips at ft from bottom

THE DESIGN IS ADEQUATE.

ANALYSIS
CHECK COMBINED COMPRESSION AND BENDING CAPACITY (AISC 360-05, H1)
 P r 8  M rx M ry  Pr
     , for  0.2
 P c 9  M cx M cy  Pc
  0.00 < 1.0 [Satisfactory]
 P r  M rx M ry  P r  0.2
    , for
2 P  M M Pc
 c  cx cy 

Where KLx = 14 ft, for x-x axial load. KLy = 14 ft, for y-y axial load.
(KL / r)max = 36 < 200 [Satisfactory]
Pr = 0 kips
Mrx = 0.00
Mry = 0.00
Pc = P n / W c = 881 / 1.67 = 527.812 kips, (AISC 360-05 Chapter E)
> Pr [Satisfactory]
Mcx = Mn / Wb = 343.467 / 1.67 = 205.669 ft-kips, (AISC 360-05 Chapter F)
> Mrx [Satisfactory]
Mcy = Mn / Wb = 343.467 / 1.67 = 205.669 ft-kips, (AISC 360-05 Chapter F)
> Mry [Satisfactory]

CHECK LATERAL DEFLECTION

D max = 0.00 in, at 14.00 ft from bottom

< L/ 240 = 0.70 in [Satisfactory]

Where Es = 29000 ksi
Ix = 457 in4 I= 457 in4
Iy = 457 in 4
 PROJECT : PAGE :
CLIENT : DESIGN BY :
JOB NO. : DATE : REVIEW BY :
WF Column Design Based on AISC Manual 13th Edition (AISC 360-05)

INPUT DATA & DESIGN SUMMARY

COLUMN SECTION W10X60

COLUMN YIELD STRESS Fy = 50 ksi

DIMENSIONS H= 14 ft

AXIAL LOAD, ASD P= kips

STRONG AXIS BENDING ? (1=Yes, 0=No) => 1 yes, strong axis, x-x, bending.
UNIFORM LATERAL LOAD, ASD w= k / ft
CONCENTRATED LATERAL LOAD, ASD F= kips at ft from bottom

THE DESIGN IS ADEQUATE.

ANALYSIS
CHECK COMBINED COMPRESSION AND BENDING CAPACITY (AISC 360-05, H1)
 P r 8  M rx M ry  P r  0.2
     , for
 c
P 9 M
 cx M cy  P c
  0.00 < 1.0 [Satisfactory]
 P r  M rx M ry 
  , for P r  0.2

2 P  M cx M cy  Pc
 c  

Where KLx = 14 ft, for x-x axial load. KLy = 14 ft, for y-y axial load.
(KL / r)max = 65 < 200 [Satisfactory]
Pr = 0 kips
Mrx = 0.00
Mry = 0.00
Pc = Pn / W c = 643 / 1.67 = 385.288 kips, (AISC 360-05 Chapter E)
> Pr [Satisfactory]
Mcx = Mn / Wb = 289.986 / 1.67 = 173.644 ft-kips, (AISC 360-05 Chapter F)
> Mrx [Satisfactory]
Mcy = Mn / Wb = 145.833 / 1.67 = 87.3253 ft-kips, (AISC 360-05 Chapter F)
> Mry [Satisfactory]

CHECK LATERAL DEFLECTION

D max = 0.00 in, at 14.00 ft from bottom

< L/ 240 = 0.70 in [Satisfactory]

Where Es = 29000 ksi
Ix = 341 in4 I= 341 in4
Iy = 116 in4 A= 17.6 in2
 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 SUMMARY

STEEL YIELD STRESS Fy = 50 ksi
SIMPLY SUPPORTED SPAN S= 14 ft
SUPERIMPOSED UNIFORM DEAD LOAD DL = kips / ft
UNIFORM LIVE LOAD LL = kips / ft
POINT DEAD LOAD PDL = kips
POINT LIVE LOAD PLL = kips
DISTANCE POINT LOAD TO END c= ft
TOP FLANGE WIDTH bf,top = 10.1 in
TOP FLANGE THICKNESS tf,top = 0.68 in
BOTTOM FLANGE WIDTH bf,bot = 10.1 in
BOTTOM FLANGE THICKNESS tf,bot = 0.68 in
WEB THICKNESS tw = 0.42 in
BEAM DEPTH d= 10.2 in 1/4 x 4 STIFFENER, E. SIDES, AT SUPPORTS ONLY.
UNBRACED LENGTH Lb = 14 ft FLANGE TO WEB WELDING USE 1/4 in - 24 in @ 2202 in o.c.

THE GIRDER DESIGN IS ADEQUATE.

ANALYSIS
CHECK LIMITING WIDTH-THICKNESS RATIOS FOR WEB (AISC 365-05 Table B4.1)
hc / tw = 21.05 < lr = 137.27
< lp = 90.55
Compact Web
where E= 29000 ksi
lr = 5.7 (E / Fy)0.5 = 137.27
lp = (hc / hp) (E / Fy)0.5 / (0.54 Mp / My -0.09)2 = 87.60 ,for Af,top ≠ Af,bot
lp = 3.76 (E / Fy)0.5 = 90.55 ,for Af,top = Af,bot
hc = 8.84 in hp = 9.20 in
Mp = 310.8 ft-kips My = 277.9 ft-kips

CHECK LIMITING WIDTH-THICKNESS RATIOS FOR FLANGES (AISC 365-05 Table B4.1)
0.5 bf,top / tf,top = 7.43 < lr = 25.09
< lp = 9.15
Compact Flanges
where lr = 1.0 (kc E / FL)0.5 = 25.09
lp = 0.38 (E / Fy)0.5 = 9.15
kc = Min [0.76 , Max (0.35 , 4 / (h / tw)0.5 )] = 0.76
Sxt = 67 in3 Sxc = 67 in3
FL = 35 ksi, (AISC 360-05 Table note B4.1 & Eq F4-6)

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

Chapter F Sections
Required Conditions F2 F3 F4 F5
Double Symmetric x x
Compact Web x x x
Noncompact Web
Slender Web Mallowable = Mn / W b = 173.6 ft-kips
Compact Flanges x ( from following analysis)
Noncompact Flanges
Slender Flanges
Applicable Section ok

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

E 9.07 ft
L p  1.76r y 
Fy

2
E Jc  0.7 F y S x h 0  E
L r  1.95r ts 1  1  6.76
 E    r ts  36.58 ft
0.7 F y S x h0  Jc  0.7 Fy
where ry = 2.57 in Sx = 67 in3
h0 = 9.52 in Iy = 116 in4
 (cont'd)
Cw = Iy h02 / 4 = 2628.28 , (AISC 365-05 F2.2)
J = [tw d (tw2 + d2)] / 12 = 37.2053 in4, (not applicable if taken web only, EIT Manual page 26)
(Use J= 2.48 in4 )
rts =[( Iy Cw)0.5/ Sx)]0.5 = 2.88 in
c= 1.0 Cb = 1.0 , (AISC Manual 13th Table 3-1, page 3-10)

2
C b 2E Jc  L b 
F cr  2
1  0.078    119.864 ksi
 Lb  S x h 0  r ts 
 
 r ts 

 M p , for L b  L p

    Lb  L p  
M n, F 2   Min  C b  M p   M p  0.7 F y S x  
   , M p  , for L p  L b  L r
  290.0 ft-kips
 L
 r  L p  
  


 Min  F S
cr x , M p  , for L r  L b

Mallowable, F2 =Mn / Wb = 173.6 ft-kips where Wb = 1.67 , (AISC 365-05 F1)

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

<== Not Applicable.

     pf  
  M p   M p  0.7 F y S x    , for Noncompact Flanges
    rf   pf  
M n, F 3      323.4 ft-kips
 0.9 E k c S x
 , for Slender Flanges
 2
where l = bf / (2 tf) = 7.43
lpf = lp = 9.15 lrf = lr = 25.09

Mallowable, F3 =Min(Mn,F2 , Mn,F3) / Wb = 173.6 ft-kips

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

<== Not Applicable.

E 6.42 ft
L p  1.1r t 
Fy

2
E J  S h 
L r  1.95r t 1  1  6.76 F L xc 0   36.96 ft
FL S xc h 0  E J 

where aw =hc tw/ (bfc tfc) = 0.54

b fc
rt   2.91 in
 1 2 
12  h 0  a w h 
 d 6 h 0d 

Mp = Min [Zx Fy , 1.6Sxc Fy ] = 310.833 ft-kips

Myc = Sxc Fy = 277.917 ft-kips Myt = Sxt Fy = 277.917 ft-kips
l = h c / tw = 21.05
lpw = lp = 87.60 lrw = lr = 137.27

Mp
 , for h c / t w   pw
 M yc
R pc   1.11844
 Min  M p   M p  1     pw  , M p  , for h c / t w   pw
      
  M yc  M yc    rw   pw  M yc 

2
C b 2E 1  0.078 J  L b  
F cr  2   121.731 ksi, (for Iyc / Iy = 0.50 > 0.23, AISC 360-05 F4-5 )
 Lb  S xc h 0  r t 
 
 rt 

 R pc M yc , for L b  L p

    Lb  L p  
M n, F 4.2   Min  C b  R pc M yc   R pc M yc  F L S xc      , R pc M yc  , for L p  L b  L r
  281.9633 ft-kips
    L r L p   

 Min  F S
cr xc , R pc M yc  , for L r  L b
 (cont'd)
 R pc M yc , for Compact Flanges

     pf 
 R pc M yc   R pc M yc  F L S xc     , for Noncompact Flanges

M n, F 5.3     rf   pf  310.833 ft-kips
  

 0.9 E k c S xc
 , for Slender Flanges
 2
M p
 , for h c / t w   pw
 M yt
R pt   1.11844
 Min  M p   M p  1     pw  , M 
p
for h c / t w   pw
       ,
  M yt  M yt    rw   pw  M yt 

Mallowable, F4 = Min( Mn,F4.2 , Mn,F4.3, RptMyt) / Wb = 168.8 ft-kips

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

<== Not Applicable.
E
Lr   rt  21.91 ft
0.7 F y

 F y , for L b  L p

    L b  L p  
 Min  C b  F y   0.3F y   
 , F y  ,

for L p  L b  L r
  
  L r L p  
 


F cr , F 5.2     42.6577 ksi
  
  C b 2E 
 Min  2
, F y  , for L r  L b
   Lb  
  r  
  t  

 F y , for Compact Flanges


     pf  
  F y   0.3F y     , for Noncompact Flanges
    rf   pf  
F cr , F 5.3    50 ksi
 0.9 E k c , for Slender Flanges
 2
  b f 
  2t f 
 
 Min  a w , 10   hc E  
R pg  Min 1    5.7  , 1.0   1
 1200  300 Min  a w , 10   t w 
Fy  
  
Mallowable, F5 = Min( RpgFySxc , RpgFcr,F5.2Sxc , RpgFcr,F5.3Sxc , FySxt) / Wb = 142.0 ft-kips

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

h = d - tf,top - tf,bot = 9 in , h / tw = 21 , Aw = 4.28 in2 ,

a= 13.7 ft 
1.0 , for h / t w1.10 k vE
 5  Fy
5  2
, for a / h 3 
k v    a / h  5.00 Cv
 1.10 k vE

E E
, for 1.10 k v  h / t w1.37 k v  1.000
 h /tw F y Fy Fy
5 , for a / h 3 
Vn = 0.6 FyAwCv = 128.52 kips  1.51E k v , for 1.37 k vE h /
 h/ tw
  t w F y
2
 Fy

Vallowable = Vn / Wv = 76.9581 kips Wv = 1.67 , (AISC 365-05 G1)

TOTAL SUPERIMPOSED GRAVITY LOAD

w = DL + LL = 0.000 kips / ft , P = PDL + PLL = 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 Right
Distance 0 0.00 0.00 0.00 0.00 0.00 0.00 2.33 4.67 7.00 9.33 11.67 14.00
d (in) 10 10 10 10 10 10 10 10 10 10 10 10 10
y (in) 5 5 5 5 5 5 5 5 5 5 5 5 5
I (in4) 336 336 336 336 336 336 336 336 336 336 336 336 336
Wt (plf) 59.4 59.4 59.4 59.4 59.4 59.4 59.4 59.4 59.4 59.4 59.4 59.4 59.4
V (kips) 0.42 0.42 0.42 0.42 0.42 0.42 0.00 -0.28 -0.14 0.00 0.14 0.28 0.42
M (ft-k) 0 0 0 0 0 0 0 1 1 1 1 1 0
 (cont'd)

BENDING LOADS & CAPACITY

Moments
0

-50

-100

-150

-200
Length

Mmax = 1.4546725 ft-kips @ 7.00 ft, from heel.

< Mallowable = 173.644 ft-kips
[Satisfactory]
S h ear F o rces

SHEAR LOADS & CAPACITY

-75.50
-76.00
-76.50
-77.00
-77.50
Length

Vmax = 0.42 kips @ 14.00 ft, from heel.

< Vallowable = 76.9581 kips [Satisfactory]

DETERMINE DEFLECTION AT MID SPAN

5w L 4 0.06415 Pb
 L 2 b 2  
3/ 2
 DL   0.01 in ( L / 31891 )
384 EI EIL (for camber, self Wt included.)

where E= 29000 ksi w= 0.059 kips / ft

I= 336 in4 P= 0 kips
b= 0.3 ft L= 14.0 ft

5w L 4 0.06415 Pb
 L 2 b 2  
3/ 2
 LL   0.00 #DIV/0!
384 EI EIL

where P= 0 kips w= 0.000 kips / ft

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

w= 1/4 in
wmin = 3/16 in, < w
wmax = 5/16 in, > w
W= 2.0
Vmax = 0.42 kips
Q = Af(d - y - 0.5 tf,top) = 33 in3
vmax = Vmax Q / I = 0.04 kips / in

A= 24 in B
 0.6 F EXX   0.707 w A  2202 in. o.c.
v max 
USE 1/4 in - 24 in @ 2202 in o.c.

DESIGN STIFFENERS

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

2. CHECK LOCAL WEB YIELDING FOR THE CONCENTRATED LOAD. (AISC 360-05, J10.2)
R=P= 0.00 kips
N= 0 in, bearing length, point.
k = tf,top + w = 0.93 in
W= 1.5

 R
 , for c  d
 t w  N  5k 
  0.00 < Fy / W [Satisfactory]
 R
, for c  d
 t w  N  2.5 k 

 (cont'd)
3. CHECK WEB CRIPPLING FOR THE CONCENTRATED LOAD. (AISC 360-05, J10.3)

W= 2.0
  1.5
 2 N  t w   EF ywt f , for c 0.5 d
0.80t w13 d  
 t f   tw
  
Rn /   1 /    108.11 > P [Satisfactory]
  1.5
0.40t 2w13 N  t w   EF ywt f , for c 0.5 d
   
 d  t f  tw

  

(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)
dc = d - 2k = 8.34 in
Cr = 960000 ksi
(dc / tw ) / (l / bf) = 1.19
W= 1.76
  3
 C r t wt f 0.4 d c / t w   , for d c / t w 1.7
3

 h 2   l / b f   l/b f
   

 C r t 3wt f   
3 
Rn /   1 /   10.4 d c / t w   , for 1.7  d c / t w 2.3  239.31 > P [Satisfactory]
 h
2   l /b f   l /b f
   

 dc /tw
P , for l /  2.3
 bf



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

5. DETERMINE STIFFENER SIZE.

tw = 1/3 in , bst = 4 in

bst / tw = 12.80 < 0.56 (E / Fy)0.5 , (AISC 360-05 Table B4.1)

[Satisfactory]
Ag = 4.62 in2 , I= 16 in4

R= 0.4 kips
Wc = 1.67 , (AISC 365-05 E1)
Kl/r= 0.75 h / ( I / Aeff)0.5 = 3.6
Cc = 4.71 (E/ Fy)0.5 = 113

 2E  21968.64 ksi
Fe 
 kl / r  2
 Fy kl
 0.658 F e  F y , for  Cc
 r
R n / c  A g / c     138.1 kips, (AISC 360-05 E2)
 kl
 0.877 F e , for r  Cc > R [Satisfactory]

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