steel project
DESCRIPTION
Steel analysis and designTRANSCRIPT
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ASIAN INSTITUTE OF TECHNOLOGY
SCHOOL OF ENGINEERING TECHNOLOGY
DESIGN OF STEEL STRUCTURES
TERM PROJECT
Sai Than Lwin (20000251)
Ni Ni Aung (20000238)
Nway Nandar Win (20000241)
May Hnin Ou (20000231)
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Introduction
This project is about the design and analyze of a steel building which consists of four stores with
one story below grade.
i. LOCATION
Williamsport, Pennsylvania
41 N latitude and 77 W longitude.
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ii. ARRANGEMENT OF THE STRUCTURE
Floor Area for each floor = 25 500 ft2
Floor total area =102 000 ft2
Bays
East west direction 30ft bays throughout
North south direction 45ft, 30ft and 45ft
14 6
6 13
13 6
15 6
4 th story
3 rd story
2 nd story
1 st story - Below grade
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Figure Schematic plan
Figure Complete 3 dimensional computer model
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iii. LATERAL LOAD RESISTING SYSTEM
East west direction Moment frames
North south direction Braced frames
iv. OTHER FEATURES
Faade is a lightweight metal curtain wall that extends above the roof surface.
6 ft screen wall around the middle bay of the roof to conceal mechanical equipment and roof access.
A two story atrium in the first floor
An opening in the second floor (An area bounded by A C 4 5)
Figure Lateral load resisting system
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1) LOADS ON COLUMNS BASED ON TRIBUTARY AREAS
Figure - Arrangement of columns in the 1 st , 3 rd and 4 th floors
Figure - Arrangement of columns in
the 2nd floor
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RC
Column Tributary
area
(ft2)
Dead load
(D)
(psf)
Live load
(Live
roof)(psf)
Snow load
(psf)
1.2D+0.5S
(ksf)
Load
(Kips)
Total
load
(kips)
RC1 168.75 20 20 102 75 12.65 12.65
RC2 675 20 20 102 75 50.62 50.62
RC3 562.5 20 20 102 75 42.18 42.18
RC4 337.5 20 20 102 75 25.3 25.3
RC5 1380 20 20 102 75 103.5 103.5
TC
Column Tributary
area
(ft2)
Dead load
(D)
(psf)
Live load
(Live
roof)(psf)
Reduced
live load
(Lred)(psf)
1.2D+1.6Lred
(ksf)
Load
(Kips)
Total
load
(kips)
TC1 168.75 80 80 66.46 202.39 34.15 46.80
TC2 675 80 80 43.1 164.96 111.35 161.97
TC3 562.5 80 80 45.3 168.48 94.77 136.95
TC4 337.5 80 80 52.66 180.256 60.86 86.16
TC5 1380 80 80 36.15 153.84 212.3 315.8
RC
TC
SC
BC
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SC
Column Tributary
area
(ft2)
Dead load
(D)
(psf)
Live load
(Live
roof)(psf)
Reduced
live load
(Lred)(psf)
1.2D+1.6Lred
(ksf)
Load
(Kips)
Total
load
(kips)
SC1 168.75 80 80 66.46 202.39 34.15 80.95
SC2 675 80 80 43.1 164.96 111.35 273.32
SC3 562.5 80 80 45.3 168.48 94.77 231.72
SC4 337.5 80 80 52.66 180.256 60.86 147.02
SC5 1380 80 80 36.15 153.84 212.30 528.10
FC
Column Tributary
area
(ft2)
Dead load
(D)
(psf)
Live load
(Live
roof)(psf)
Reduced
live load
(Lred)(psf)
1.2D+1.6Lred
(ksf)
Load
(Kips)
Total
load
(kips)
FC1 168.75 80 80 66.46 202.39 34.15 115.10
FC2 675 80 80 43.1 164.96 111.35 384.67
FC3 562.5 80 80 45.3 168.48 94.77 326.49
FC4 337.5 80 80 52.66 180.256 60.86 207.88
FC5 1380 80 80 36.15 153.84 212.30 740.40
FC6 287.5 80 80 55.39 184.624 53.08 200.1
FC7 168.75 80 80 66.19 201.904 34.07 562.17
** Reduced live load Lred = Live load * (0.25 + 15/(A KLL))
K depends on the arrangement of the columns
Values for k can be obtained by tables in ASCE 7-10
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2) Load on beams based on tributary areas
B1 B1 B1 B1 B1 B1 B1
B2 B2 B2 B2 B2 B2 B2
B3 B3 B3 B3 B3 B3 B3
B4 B4 B4 B4 B4 B4 B4
B6
B5
B7
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ROOF
Joist/Column Tributary
area
(ft2)
Dead
load (D) (psf)
Live load (Live roof)(psf)
Snow load
(psf)
1.2D+0.5S
(psf)
Load
(Kips)
UDL
(kips/ft)
J1 270 20 20 102 75 20.25 0.45
J2 180 20 20 102 75 13.5
0.45
B5 135 20 20 102 75 10.125
0.225
B6 90 20 20 102 75 6.75
0.225
For beams B1, B2, B3 & B4
B4 10.125
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For Floors 3,2,1
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.
Joist Tributary
area (ft-sq)
Dead load
(D) (psf)
Live load
(psf)
Reduced
Live load
(psf)
1.2D+1.6L
(ksf)
Load
(kips)
UDL
(kips/ft)
J1 270 80 80 71.64 210.624 56.87 1.26
J2 180 80 80 83.25 229.2 41.26 1.38
B5 135 80 80 80 224 30.24 0.672
B6 90 80 80 80 224 20.16 0.672
Beam R
B1 28.35
B2 49.05
B3 49.05
B4 28.35
2. Design lateral bracing
L = 258 in Fy = 36 ksi Fu = 58 ksi A36 Steel
Ag (req) = PU / FY = 242.3/0.9(36) = 7.47 in2
Ae (req) = PU / FY = 242.3/0.75(58) = 5.57 in2
L/300 = 258/300 = 0.86 in
P U = 242.3 kips ( From Etabs )
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2nd Floor
F = 88 kips h = 13 6
From similar triangles
PU = 1 x 88 x (21.57/15) = 126.54 kips
3rd Floor
F = 93 kips h = 14 6
From similar triangles
PU = 1 x 93 x (20.15/15) = 124.93 kips
4th Floor
F = 102 kips h = 15 6
From similar triangles
PU = 1 x 102 x (20.15/5) = 137.02 kips
Roof
F = 54 kips h = 14 6
From similar triangles
PU = 1 x 102 x (20.86/15) = 75.1 kips
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Design brace frame for maximum PU
Ag (min) = PU / FY = 137.02/0.9(36) = 4.23 in2
Ae (min) = PU / FY = 137.02/0.75(58) = 3.15 in2
Select L-section: L5 x 5 x (Ag = 4.75 in2 , rx =1.53 )
Select C-section: C9 x 15 (Ag = 4.41 in2 , rx =3.40 )
Select I-section: W6 x 16 (Ag = 4.74 in2 , rx = 2.60 )
Ae / Ag = 0.745
By using maximum L = 21.57 ft.
L-section: L / rx = 169.18 300
C-section: L / rx = 76.13 300
W-section: L / rx = 99.55 300
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4. Design of columns
As single columns
In order to carry out a conservative design all end conditions were considered as pin supports.
Hence the k value is taken as 1.
RC
Column Pu (Kips) KL(ft) Shape
RC1 12.65 14.5 W8x31 (230 kips)- least weight section
W10x33(233 kips)
W12x40(280 kips)
W14x48(332 kips)
RC2 50.02 14.5 W8x31 (230 kips)
RC3 42.18 14.5 W8x31 (230 kips)
RC4 25.3 14.5 W8x31 (230 kips)
RC5 103.5 14.5 W8x31 (230 kips)
RC
TC
SC
BC
6 14
6 13
13 6
15 6
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TC
Column Pu (Kips) KL(ft) Shape
TC1 46.8 13.5 W8x31 (248 kips)- least weight section
W10x33(253 kips)
W12x40(304kips)
W14x48(361 kips)
TC2 161.97 13.5 W8x31 (248 kips
TC3 136.95 13.5 W8x31 (248 kips
TC4 86.16 13.5 W8x31 (248 kips
TC5 315.8 13.5 W8x40 (322 kips)- least weight
section w10x45(358 kips) w12x45
(343 kips)
W14x48(361 kips)
SC
Column Pu (Kips) KL(ft) Shape
SC1 80.95 13.5 W8x31 (248 kips)- least weight section
W10x33(253 kips)
W12x40(304kips)
W14x48(361 kips)
SC2 273.32 13.5 W8x35 (280 kips) least weight section
W10x39 (305 kips)
W12x40 (304 kips)
W14x 48 (361 kips)
SC3 231.72 13.5 W8x31 (248 kips)
SC4 147.02 13.5 W8x31 (248 kips)
SC5 528.10 13.5 W8x67 (560 kips)- least weight
section w10x60(581 kips) w12x58
(553 kips) W14x61(572 kips)
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FC
Column Pu (Kips) KL(ft) Shape
FC1 115.10 15.5 W8x31 (212 kips)- least weight section
W10x33(213 kips)
W12x40(257 kips)
W14x48(304 kips)
FC2 384.67 15.5 W8x58 (417 kips)
W10x49(428 kips) - least weight section
W12x53(452 kips)
W14x61(515 kips)
FC3 326.49 15.5 W8x48 (340 kips)
W10x49(428 kips) - least weight section
W12x53(452 kips)
W14x53(338 kips)
FC4 207.88 15.5 W8x31 (212 kips) - least weight section
W10x33(213 kips)
W12x40(257 kips)
W14x48(304 kips)
FC5 740.4 15.5 W8 none
W10x88(789 kips)
W12x79(781 kips) - least weight section
W14x90(978 kips)
FC6 200.1 15.5 W8x31 (212 kips)
FC7 562.17 15.5 W8 none
W10x68(602 kips)
W12x65(639 kips) - least weight section
W14x68(576 kips)
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Design of two story columns from 2nd floor to the 4th floor
Column Pu
(Kips)
KLx(ft) KLy(ft) Shape Checks
C1 (Braced
in both x
and y
direction)
80.95 13.5 13.5 W8x31 (248 kips)-
least weight section
W10x33(253 kips)
W12x40(304 kips)
W14x48(361 kips)
C2 (Braced
in both x
and y
direction)
273.32 13.5 13.5 W8x35 (280 kips)
W10x39(305 kips) -
least weight section
W12x40(304 kips)
W14x48(361 kips)
C3 (Braced
in both x
and y
direction)
231.72 13.5 13.5 W8x31 (248 kips)-
least weight section
W10x33(253 kips)
W12x40(304 kips)
W14x48(361 kips)
C4 (Braced
only in y
direction)
147.02 27 13.5 W8x31 (248 kips)-
least weight section
W10x33(253 kips)
W12x40(304 kips)
W14x48(361 kips)
KLx/[(rx/ry)]=27/1.72=15.69>kLy
Therefore use KL=16.69 and
W8x31to obtain Pn
Pn = 212 > Pu (147.02)
Therefore W8x31 is ok
C5 (Braced
in both x
and y
direction)
528.10 13.5 13.5 W8 x67 (560kips)
W10x60(581 kips)
W12x58(553 kips) -
least weight section
W14x61(572 kips)
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5. Design of compression bracing members
Compression(max) force is equal to Tension(max) force which are indcued in section (3). It is
assumed, therefore, that the sections used to design the tension bracing will be same as
the section used for compression.
6. Design of Beams
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If Wfloor is the floor load,
W = 6(Wfloor) / 1000 = 0.006 kips/ft.
R1 = R2 = 0.003 x 45W = 0.135 kips
For joist type 2,
If Wfloor is the floor load,
W = 6(Wfloor) / 1000 = 0.006 kips/ft.
R1 = R2 = 0.003 x 30W = 0.09 kips
For beams B1, B2, B3 and B4
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Bending Moment Diagram,
Mmax = 18R
Bending Moment Diagram
Shear Force Diagram
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Check for sections
W 16 x 31 :
Zx = 54.0 Lp = 4.13 Lr = 11.9 Lb = 6 ; Lp < Lb < Lr
Mp = 0.9 x 50 x 54/12
= 202.5 kips-ft
Mu =MpBF(Lb-Lp)
= 202.5 10.2 ( 6 4.13 )
= 183.43 kips-ft ; OK
Story Load (Psf) UDL (kips/ft) WL/2 (kips)
Roof 75 0.45 10.125 J1
75 0.45 6.75 J2
Beam R (kips) Mmax (kips-ft) Vmax (kips) Sections
RB1 10.125 182.25 20.25 W 16 x 31
RB2 16.875 303.75 33.75 W 21 x 44
RB3 16.875 303.75 33.75 W 21 x 44
RB4 10.125 182.25 20.25 W 16 x 31
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Shear check
Vu = (1.0) (0.6Fy) (d) (tw) = 1.0 x 0.6 x 50 x 15.9 x 0.275 = 131.75 kips ; OK
W 21x 44 :
Zx = 95.4 Lp = 4.45 Lr = 13.0 Lb = 6 ; Lp < Lb < Lr
Mp = 0.9 x 50 x 95.4/12 = 357.75 kips.ft
Mu =MpBF(Lb-Lp) = 357.75 16.8 ( 6 4.45 ) = 331.71 kips.ft ; OK
Shear check
Vu = (1.0) (0.6Fy) (d) (tw) = 1.0 x 0.6 x 50 x 20.7 x 13 = 672.75 kips ; OK
For Floors 3,2,1
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Bending Moment Diagram
10R 10R
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Story Load (Psf) UDL (kips/ft)
WL/2 (kips)
Joist
3 to 1
210.624 1.26 28.35 J1
224 1.34 20.1 J2
Beam V(x) (kips)
M(x) (kips.ft)
Sections
B1 47.475 475.75 W 21 x 68
B2 33.6 336 W 21 x 48
B3 33.6 336
W 21 x 48
B4 47.475 475.75 W 21 x 68
Check for sections W 21 x 68 :
Zx = 160 Lp = 6.36
Mp = 0.9 x 160 x 50 /12= 600 kips.ft
Mu =MpBF(Lb-Lp)
= 600 18.8 ( 10 6.36 )
= 531.56 kips.ft ; OK
Lr = 18.7 Lb = 10 ; Lp < Lb < Lr
Shear check
Vu =273 kips ; OK
W 21x 48 :
Zx = 107, Lp = 6.09, Lr = 16.6, Lb =10 ; Lp < Lb < Lr
Mp = 0.9 x 50 x 107/12= 401.25 kips.ft
Mu =MpBF(Lb-Lp) = 401.25 14.7 (10 6.09) = 343.773 kips.ft ; OK
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Story Beam Floor Load (Psf) UDL (kips/ft) Vx (kips)
Mx (kips.ft)
Sections
Roof
B5 75 0.225 2.53 14.24 W 8 x 28
B6 75 0.225
3.38
25.31
W 10 x 33
3 to 1
B5 210.624 1.05 11.81 66.44 W 8 x 31
B6 224 1.12
16.8
126
W 10 x 49
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7. Redesign section with beams spanning in opposite direction in AC67 area
Dead load (D)
(psf)
Live load (L)
(psf)
Tributary area
(At) (ft2)
Lred (psf) 1.2D+1.6Lre
d
Wu (UDL)
(Kips/ft)
80 80 30 x 15 = 450 60 192 2.8
Dead load (D)
(psf)
Live load (L)
(psf)
Tributary area
(At) (ft2)
Lred (psf) 1.2D+1.6Lre
d
Wu (UDL)
(Kips/ft)
80 80 45 x 10 = 450 60 192 1.92
AC-6
Maximum bending moment Mu= moment due to the two point loads + moment due to
the uniform load = 43.2 x 15 + 0.96 x452/8
= 243 + 648 = 891 kip-ft
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Unbraced length =15ft Shape = W 30 x 99 (Mn=918 kip-ft > Mu)
From the previous arrangement of joists the A67 =J1.
It has a UDL of 1.26 kips/ft. Hence the maximum moment is 319 kip-ft.
Therefore a lesser moment occurs when the joist are arranged in the previous way.
A67
Dead load (D)
(psf)
Live load (L) (psf) Tributary area
(At) (ft2)
Lred (psf) 1.2D+1.6Lred Wu (UDL)
(Kips/ft)
80 80 30 x (45/6) = 225 76.56 218.5 1.64
Maximum Bending moment Mu =1.64 x 302/8 = 184.5 kips-ft
Unbraced length = 30ft Shape=12x58(Mn >Mu)
From the previous arrangement of beams the A67 beam =B2 W 21 x 48.
It requires a more light section.
C67
Dead load (D)
(psf)
Live load (L) (psf) Tributary area
(At) (ft2)
Lred (psf) 1.2D+1.6Lred Wu (UDL)
(Kips/ft)
80 80 30 x (45/6) = 225 76.56 218.5 1.64
Mu = moment due to the two point loads + moment due to the uniform load
= 28.8 x 10 + 1.64 x302/8
= 288 + 184.5 = 472.5 kip-ft
Unbraced length = 10 ft Shape=W18x97 (Mn =782kipft>Mu)
From the previous arrangement of beams the C67 beam =B1 W 21 x 68.
Hence, more heavy section is required.
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8. JOIST SECTION (KCS)
Required moment for L = 45 ft. span 1366 kips.in 20kcs4 1371 kips.in
Bridging section no.10 Required moment for L = 30 ft. span 607.5 kips.in 14kcs3 642 kips.in
Bridging section no.6
Both joists for spacing 30
5 = 6 ft.
9. SHEAR CONNECTIONS FOR BEAMS
pply 34 449()
= v= 68 0.441 = 30 ips
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Yield Failure
= 0.9 36 14.3 = 463.325 ips
Rupture Failure
= 0.75 58 14.03 = 610.3 ips
For Single Angle Bearing and Tearout For the edge
For the middle
lc = 2 (0.8125) = 1.19 n
Tearout governs
= 1.2 lctFu = 0.75 1.2 [1.2 + 1.19] 5/46 58
= 48.78 ips >
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Block Shear
= 0.6 58 0.53 + 58 0.8 = 69.84 ips
= 0.6 36 0.53 + 58 0.8 = 66.70 ips
= 0.75 64.84 = 48.63 ips > K
The block shear & tear out was done for single angle however since > for both cases the L section passes.
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Plate Design
We considered the plate to have similar bolt arrangement.
If yield governs
tw 6 n = 1.37 n2
tw = 0.23 n
If rupture governs
tw 6 n 0.875t = 1.02 n2
t= 0.20 n
From available thickness,
t=
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Check For Tearout and
Bearing
lc (from center) = 2 (0.8125) = 1.1875 in < 2
Tearout governs
Therefore use a greater
thickness, consider t = 3/8
Block Shear Check
v = 3 7/16 = 1.3125 n2
v = 1.3125 1.5(0.875) 7/16 = 0.738 n2
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t= 7/16 3 0.5(0.875)] 7/16 = 1.12 n2
= 0.6 58 0.738 + 58 1.12 = 90.64 ips
= 0.6 36 1.3125 + 58 1.12 = 93.31 ips = 0.75 90.64 = 67.98 ips > K
FINAL DESIGN
Double Angle: 2Lx6x6x5/16
Rectangular Plate: 4x6x7/8
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