Steel Design Notes CSA
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Transcript of Steel Design Notes CSA
Beam-Column Design (Non-Plate Girder Beams)
CHECK BEAM CLASS.
Class 1&2 Section follow below, Class 3 see next section.
Cy = AFy
Class 1&2 Sections:
Three failure modes must be checked:
1) Cross Section Strength
BI-AXIAL CHECK:
2) Overall In-Plane Member Strength
kL/r is always for strong axis for this check.
n= 1.34 normally, (W sections or Class C HSS)
n = 2.24 for WWF-shapes with flame cut flange edges or class H HSS sections.
NOTE: for this check U1x can be less than 1 for pin ends. For moment frames U1x = 1.0
BI-AXIAL CHECK:
3) Later-Tosional Buckling Stability
The check is exactly the same as Check 2 (In-Plane) except now:
kL/r is for the highest slenderness (x or y).
NOTE: U1x โฅ 1.0
Class 3 Sections:
Three failure modes must be checked:
1) Cross Section Strength
BI-AXIAL CHECK:
2) Overall In-Plane Member Strength
NOTE: for this check Uix can be less than 1 for pin ends.
kL/r is always for strong axis for this check.
BI-AXIAL CHECK:
3) Later-Tosional Buckling Stability
The check is exactly the same as Check 2 (In-Plane) except now:
kL/r is for the highest slenderness (x or y).
NOTE: U1x โฅ 1.0
Bearing and Base Plates: Bearing Plates:
1) Calculate acceptable Bearing stress on Concrete Column: ๐๐๐ฅ ๐๐ก๐๐๐ ๐ = 0.51๐๐
โฒ given in the following table: f'c 10 15 20 25 30 35 40 45 Bearing Stress 5.10 7.65 10.20 12.75 15.30 17.85 20.40 22.95
2) Find Useable Wall Width Wnet = Wall Width - 25mm
3) Find Required Bearing Plate Required Area AB = ๐ ๐ฅ๐๐ต๐๐๐๐๐๐ ๐๐ก๐๐๐ ๐ ๏ฟฝ
Where Rxn is the reaction force for the system.
4) Try a plate length of Bplate = Beam width + 25mm Ensure Plate is still shorter than the allowable wall width:
๐ด๐ต๐ต๐๐๐๐ก๐
๏ฟฝ โค ๐๐๐๐ก
If it is not, then take ๐ต๐๐๐๐ก๐ = ๐ด๐ต๐๐๐๐ก
๏ฟฝ
5) Increase all dimensions to easy numbers (multiples of 5 or 10 mm) 6) Compute Bending Length ๐ = ๐ต๐๐๐๐ก๐
2โ ๐, where k is a (vertical) property of the I beam
and comes from the property tables. It can be seen in the following diagram:
7) Computer ๐๐ = ๐ต๐๐๐๐๐๐ ๐๐ก๐๐๐ ๐ โ ๐2
2
8) Compute ๐๐ = ๐๐ก2 where t is the plate thickness.
Note that Q = ๐๐น๐ฆ
4๏ฟฝ which can be taken from the following table:
Fy 300 350 400 Phi Fy/4 = Q 67.5 78.75 90
9) Set Mf = Mr and solve for t Check that deflection limits hold by checking that ๐ก โฅ ๐ต๐๐๐๐ก๐โ ๐ต๐๐๐ ๐๐๐๐กโ
10
Base Plate Design: Case 1) No Eccentricities: For I-beams: 1) Find maximum bearing stress:
๐๐๐ฅ ๐๐ก๐๐๐ ๐ = 0.55๐๐โฒ given in the following table:
f'c 10 15 20 25 30 35 40 45 Bearing Stress 5.10 7.65 10.20 12.75 15.30 17.85 20.40 22.95
2) Find Required Bearing Plate Required Area AB = ๐ ๐ฅ๐๐ต๐๐๐๐๐๐ ๐๐ก๐๐๐ ๐ ๏ฟฝ
Where Rxn is the reaction force for the system. 3) Guess plate size as:
Beam Width + 200mm by Beam Height + 200mm (round down to multiples of 10mm)
4) Check that Area is APL > AB
5) Find m and n as given in the following diagram:
m = ๐ถโ0.95๐2
, n = ๐ตโ0.8๐2
and use the larger of the two in order to calculate Mf
6) Calculate BSA = ๐ ๐ฅ๐๐ด๐๐ฟ
๏ฟฝ
7) Find ๐๐ = ๐ต๐๐ด โ (๐ ๐๐ ๐)2
2
8) Compute ๐๐ = ๐๐ก2 where t is the plate thickness.
Note that Q = ๐๐น๐ฆ
4๏ฟฝ which can be taken from the following table:
Fy 300 350 400 Phi Fy/4 = Q 67.5 78.75 90
9) Set Mf = Mr and solve for t Check that deflection limits hold by checking that
๐ก โฅ ๐5
, ๐5
take larger of m and n. Note, for I-beams, anchor bolts have no large consequence for pinned supports provided plates are slightly larger than required area. For HSS Columns:
1) Find maximum bearing stress: ๐๐๐ฅ ๐๐ก๐๐๐ ๐ = 0.51๐๐
โฒ given in the following table: f'c 10 15 20 25 30 35 40 45 Bearing Stress 5.10 7.65 10.20 12.75 15.30 17.85 20.40 22.95
2) Find Required Bearing Plate Required Area AB = ๐ ๐ฅ๐๐ต๐๐๐๐๐๐ ๐๐ก๐๐๐ ๐ ๏ฟฝ
Where Rxn is the reaction force for the system. 3) Guess the plate Area and dimensions:
Case 1) No anchor rods required: Take square plate of width = ๏ฟฝ๐ด๐ต and round up to the nearest 10 mm
Case 2) Anchor rods are required: Use size of anchor rod to select wrench size, and take D value from that as minimum spacing from edge of column. (round the value up to the nearest 5mm)
Take Long Dimension of plate Lplate = ๐ถ๐๐๐ข๐๐ ๐๐๐๐กโ + 2๏ฟฝ๐ท๐๐๐ข๐๐๐๐ ๐ข๐ + 1.5๐๐๐๐โ๐๐ ๐๐๐๏ฟฝ (2 anchor rods assumed), again round 1.5danchor rod to the nearest 5mm.
4) Compute Width of the plate Wplate = ๐ถ๐๐๐ข๐๐ ๐๐๐๐กโ + 25, round down to nearest 5mm.
5) Find m = ๐ฟ๐๐๐๐ก๐โ๐ถ๐๐๐ข๐๐ ๐๐๐๐กโ+๐ป๐๐ ๐กโ๐๐๐๐๐๐ ๐
2
6) Compute reduced Bearing Stress Br = ๐ ๐ฅ๐๐๐๐๐๐ก๐๐ฟ๐๐๐๐ก๐
7) ๐๐ = ๐ต๐ โ (๐)2
2 , ๐๐ = ๐๐ก2
Fy 300 350 400 Phi Fy/4 = Q 67.5 78.75 90
8) Set Mf = Mr and solve for t 9) Check that deflection limits hold by checking that
๐ก โฅ ๐5
Eccentric Design:
1) Determine which case you are using :
e = ๐๐ ๐๐ ๐๐๐๐ข๐๐
๐ถ๐ ๐๐ ๐๐๐๐ข๐๐
Case 1: e < C/6 Case 2: C/6<e<C/2 Case 3: e>C/2
Case 1: 1) Compute effective Bearing Area:
An = Bโข(C-2e) 2) Check An > AB if yes, done, if not continue: 3) Increase C and or B. Then check m and n deflection limits again. (see previous
section) Find m and n as given in the following diagram: m = ๐ถโ0.95๐
2 , n = ๐ตโ0.8๐
2
4) Calculate BSA = ๐ ๐ฅ๐๐ด๐๐ฟ
๏ฟฝ
5) Find ๐๐ = ๐ต๐๐ด โ (๐ ๐๐ ๐)2
2
6) Compute ๐๐ = ๐๐ก2 where t is the plate thickness.
Note that Q = ๐๐น๐ฆ
4๏ฟฝ which can be taken from the following table:
Fy 300 350 400 Phi Fy/4 = Q 67.5 78.75 90
7) Set Mf = Mr and solve for t Check that deflection limits hold by checking that
๐ก โฅ ๐5
, ๐5
take larger of m and n.
Case 2:
1) Start with non eccentric procedure and find your plate dimensions. 2) If dimensions not given, increase C so e < C/6 3) Otherwise find a = 3 ๏ฟฝ๐ถ
2โ ๐๏ฟฝ
4) Calculate m = ๐ถโ0.95๐2
5) Select f From table below: (0.85*0.6*fcโ)
f'c 10 15 20 25 30 35 40 45 f 5.10 7.65 10.20 12.75 15.30 17.85 20.40 22.95
6) Calculate f2 = (๐)(๐โ๐)
๐
7) ๐ก2= 1๐
๏ฟฝ๐2๐2
2+ (๐ โ ๐2) ๐2
3๏ฟฝ โ 10โ3
Note that Q = ๐๐น๐ฆ
4๏ฟฝ which can be taken from the following table:
Fy 300 350 400 Phi Fy/4 = Q 67.5 78.75 90
8) Check that ๐ก โฅ ๐ ๐๐ ๐
5 take larger of m and n.
Case 3:
1) Anchor Rods needed, take f From table below: (0.85*0.6*fcโ)
f'c 10 15 20 25 30 35 40 45 f 5.10 7.65 10.20 12.75 15.30 17.85 20.40 22.95
2) Select (if not given) td the distance from the anchor rod to the center of the column.
3) Set sum of moments at Anchor rod to zero to find a.
๏ฟฝ ๐๐๐๐ = ๐ถ๐(๐ + ๐ก๐) โ ๐๐ต๐๐๐๐ก๐๐
2000 ๏ฟฝ๐ถ2 + ๐ก๐ โ
๐3๏ฟฝ = 0
Where all lengths are in mm So compute terms a1 b1 and c1: ๐1 = ๐๐ต๐๐๐๐ก๐
6000 , ๐1 = โ ๐๐ต๐๐๐๐ก๐
2000๏ฟฝ๐ถ
2+ ๐ก๐๏ฟฝ , ๐1 = ๐ถ๐(๐ + ๐ก๐)
So ๐ = โ๐1ยฑ๏ฟฝ๐1
2โ 4๐1 ๐1
2๐1 , where all units are mm and MPa.
4) Take the a value which is less than C (plate length). 5) Select a bar size and calculate T (tension force in the bar):
๐ = 0.526(๐ท โ 0.938๐)2๐น๐ข , Fu is almost always 450 MPa. (300W bars)
6) Find Rod Distance: drod = td - half column width + half flange width of column. (note this is the distance from the center of the rod to the edge of the column)
7) Compute Mrod = T drod 8) Compute Vrod = T
9) Calculate Mr = 0.45๐๐๐๐๐ก2๐น๐ฆ = 135๐๐๐๐๐ก2 ๐๐๐ 300๐ ๐ ๐ก๐๐๐ 10) First Check for t: Mr = Mf and solve for tmin . If Mr and Mf are not requested
explicitly:
๐ก๐๐๐ = ๏ฟฝ๐
0.45๐น๐ฆ= ๏ฟฝ ๐
135 for Fy = 300W grade rod.
11) Second check for t:
๐0.45๐น๐ฆ๐ก2 + ๐
1.188๐น๐ฆ๐๐๐๐๐ก = 1.0 solve for t. (N3-12).
Bar Size (D) (mm)
Pitch (P) (mm)
M16 16 2 M20 20 2.5 M22 22 2.5 M24 24 3 M27 27 3 M30 30 3.5 M36 36 4
12) Third check for plate thickness: Calculate m : m = ๐ถโ0.95๐
2
13) Calculate Mf2
๐๐2 = 0.5๐๐ ๏ฟฝ๐ โ ๐
3๏ฟฝ ๐ต ร 10โ3 all values in mm and MPa, answer in kNm.
14) Use Mr to find t. so
๐ก2 = 2.222๐๐ 1
๐น๐ฆ๏ฟฝ๐ โ ๐
3๏ฟฝ ร 10โ3 = 0.0047 ๏ฟฝ๐ โ ๐
3๏ฟฝ ร 10โ3 for 300W grade bar.
15) Deflection Check: make sure ๐ก โฅ ๐
5
16) Set t greater than the largest of those to a common plate thickness from the following table:
Lightly Loaded Base Plates: 1) Find Amin =
๐ถ๐
๐ร 103 in mm2
where f comes from the table below: f'c 10 15 20 25 30 35 40 45 f 5.10 7.65 10.20 12.75 15.30 17.85 20.40 22.95
2) Set ๐ด๐๐๐ = 4๐(๐ + 2๐) + 2๐๏ฟฝ๐ โ ๐ก๐๐๐๐๐๐ โ 2๐๏ฟฝ or
๐ด๐๐๐ = 2๐๏ฟฝ2๐ + ๐ โ ๐ก๐๐๐๐๐๐ + 2๐๏ฟฝ
3) Solve for m
4) Find ๐ก = ๏ฟฝ2๐ถ๐๐2ร103
0.9๐ด๐๐๐๐น๐ฆ๐๐
5) Plate Dimensions: 2m+0.95d x 2m+0.8b, round up to nice numbers. 6) Check deflection:
๐ก โฅ ๐5
.
Bolted Connections
Factored Tensile Resistance
Separation Load
To calculate Ap: 1) Get bolt spacing in terms of bolt diameters: Spacing = gauge/db = Coeffโขdb
2) Ap -
3) To comes from Table 7:
Prying Forces
Angles Tf = Pf + Q = Pf (1 + b/a) where Q < 0.3Pf
T-Sections
t is flange thickness Advanced method:
m โ number of shear planes
Slip Critical
Bolted Connections
Factored Tensile Resistance
Separation Load
To calculate Ap: 1) Get bolt spacing in terms of bolt diameters: Spacing = gauge/db = Coeffโขdb
2) Ap = ๐๐๐๐๐๐๐๐2
4 - ๐๐๐
2
4
3) To comes from Table 7:
Prying Forces
Angles Tf = Pf + Q = Pf (1 + b/a) where Q < 0.3Pf
T-Sections
t is flange thickness Advanced method:
m โ number of shear planes
Slip Critical
Eccentric Loads: 1) Find Centroid: xbar=
โ ๐ด๐โ๐ฅ๐๐ข๐๐๐๐ ๐๐ ๐ต๐๐๐ก๐ โ๐ด๐
ybar=โ ๐ด๐โ๐ฆ
๐๐ข๐๐๐๐ ๐๐ ๐ต๐๐๐ก๐ โ๐ด๐,
Assuming Bolts are of the same size. 2) Compute ra for each bolt: ra
2= (x - xbar)2 + (y - ybar)2 and โ ๐๐2
3) and then the force on each bolt: Rax=
๐โ๐โ(๐ฅโ๐ฅ๐๐๐)โ ๐๐
2 and Ray=๐โ๐โ(๐ฆโ๐ฆ๐๐๐)
โ ๐๐2
4) Find the total resultant shear due to direct shear and moment induced shear
Total Shear = VT =๏ฟฝ(๐ ๐๐ฅ + ๐๐ฅ)2 + ๏ฟฝ๐ ๐๐ฆ + ๐๐ฆ๏ฟฝ2
5) Calculate Vr and Br and compare to VT. ICR Method: 1) Calculate Pf = Vertical Force/Bolt Groups 2) Calculate Vr 3) Calculate C = Pf/Vr 4) Look up Callowable , from tables HB 3-29. Linear interpolation can be used between guage and values in needed. 5) Calculate connection shear capacity per side = Vr โข Callowable > Pf CHECK!
Composite Beams Prelim Checks - Check capacity of steel beam section without hardened concrete (ie concrete offers no flex resistance, use steel capacity only), under only dead load (Steel, concrete, formwork) and construction live load. รจ ie find Mu then find Mrโ etc----- see beam section.
Shear
1) Find n. Concrete slab must be transformed into equivalent steel units. f'c 20 25 30 35 40 n 9.93808 8.888889 8.114408 7.512482 7.027284
n = Es/(4500โข(fcโ)0.5) 2) Find b, effective slab width.
ยท Slab on both sides of steel beam (b is less of): i) 0.25 โข beam span ii) Avg. C-C spacing of the steel beams
ยท Slab on only one side of steel beam (b is lesser of): i) steel beam width + 0.1 โข beam span ii) steel beam width + 0.5 โข C-C steel beam spacing
3) Locate NA
find a (mm) = 0.9๐ด๐ ๐น๐ฆ
0.51๐๐โฒ๐
(mm and MPa) ,
if a < tslab then NA in the concrete, use Case 1 for full connectivity. if a > tslab then NA in the steel, use Case 2 for full connectivity. 4) Get qr and Qr min and calculate % connectivity, if โฅ 100%, stay in full connectivity otherwise move to case 3.
There are 3 cases: 1 โ full shear connectivity with NA in concrete 2 โ full shear connectivity with NA in steel 3 โ partial shear connectivity Case 1 โ Full shear connectivity with NA in concrete
i) find a = 0.9๐ด๐ ๐น๐ฆ
0.51๐๐โฒ๐
ii) find connectivity Qr min that gives benchmark for full connectivity:
lesser of: 0.9 AsFy x 10-3 OR 0.51 ๐๐โฒ b a x 10-3 (mm and MPa)
iii) compute qrs
x 10-3 (mm and MPa) gives qrs (kN)
Ec = 4500๏ฟฝ๐๐
โฒ iv) find length for n studs: from zero-moment to maximum-moment. Spacing = Lzero to max moment/n v) Find Qr = n qrs AND CHECK > Qr min
or if not given, find number of studs required, n= Qr min / qrs vi) Find Mrc eโ = d/2 + tslab โ a d is steel beam height Mrc (kNm) = 0.9AsFyeโ x 10-6 (mm and Mpa) Case 2 โ Full shear connectivity with NA in steel a > tslab i) Find Crโ = 0.51 fcโ b t where t is the concrete thickness, b is the effective width (above).
ii) Find Cr = 0.9๐ด๐ ๐น๐ฆโ10โ3โ๐ถ๐
โฒ
2
iii) Find Steel Compression Area Asc = 1000๐ถ๐
0.9๐น๐ฆ with Cr in KN, Fy in MPa.
iv) Check if NA is in flange: ๐๐ = ๐ด๐ ๐
๐๐ โค ๐ก๐, where w is the flange width, tf is the flange thickness.
if not ok, NA is in the web, so ๐๐ค = ๐ด๐ ๐โ๐๐โ๐ก๐
๐ก๐ค where tw is the web width.
NA is at ๐๐ from the top of the steel beam. v) Find Centroids of 3 sections. Case a) NA in flange (from base).
๐ฆ๏ฟฝ๐๐๐๐ ๐๐๐ ๐๐ก๐๐๐ =
๐ด๐ ๐2 โ ๐๐๐๐ ๏ฟฝ๐ โ
๐๐2 ๏ฟฝ
๐ด๐ โ ๐๐๐๐
so ๐ = ๐ โ ๐๐
2โ ๐ฆ๏ฟฝ๐ก๐๐๐ ๐๐๐ ๐ ๐ก๐๐๐
and ๐โฒ = ๐ + ๐ก๐๐๐๐๐๐๐ก๐ ๐ ๐๐๐2
โ ๐ฆ๏ฟฝ๐ก๐๐๐ ๐๐๐ ๐ ๐ก๐๐๐ case b) NA in web
๐ฆ๏ฟฝ๐๐๐๐ ๐๐๐ ๐๐ก๐๐๐ =
๐ด๐ ๐2 โ ๐ก๐๐๐ ๏ฟฝ๐ โ ๐๐
2๏ฟฝ ๏ฟฝ โ ๐ก๐ค๐๐ค ๏ฟฝ๐ โ ๐ก๐ โ ๐๐ค2๏ฟฝ ๏ฟฝ
๐ด๐ โ ๐ก๐๐๐ โ ๐ก๐ค๐๐ค
๐ฆ๏ฟฝ๐ถ๐๐๐๐๐๐ ๐ ๐๐๐ ๐๐ก๐๐๐ = ๐ก๐๐๐ ๏ฟฝ๐ โ ๐๐
2๏ฟฝ ๏ฟฝ + ๐ก๐ค๐๐ค ๏ฟฝ๐ โ ๐ก๐ โ ๐๐ค2๏ฟฝ ๏ฟฝ
๐ก๐๐๐ + ๐ก๐ค๐๐ค
so ๐ = ๐ โ ๐ฆ๏ฟฝ๐ถ๐๐๐๐๐๐ ๐ ๐๐๐ ๐๐ก๐๐๐ โ ๐ฆ๏ฟฝ๐ก๐๐๐ ๐๐๐ ๐ ๐ก๐๐๐ and ๐โฒ = ๐ + ๐ก๐๐๐๐๐๐๐ก๐ ๐ ๐๐๐
2โ ๐ฆ๏ฟฝ๐ก๐๐๐ ๐๐๐ ๐ ๐ก๐๐๐
vi) Compute Mrcomp = Cre + Crโeโ vii) find connectivity Qr min that gives benchmark for full connectivity:
lesser of: 0.9 AsFy OR 0.51 ๐๐โฒ b tslab
viii) compute qrs
ix) Find Qr = n qrs, or number of studs required, n= Qr min / qrs x) find length for n studs: from zero-moment to maximum-moment. Spacing = Lzero to max moment/n Case 3 โ Partial connectivity (NA always in the steel)
i) Find qr:
Ec = 4500๏ฟฝ๐๐
โฒ ii) find Qrmin lesser of: 0.9 AsFy OR 0.51 ๐๐
โฒ b tslab
iii) Find Percent Connectivity = nโขqrs / Qr min > 40% if flex controls or > 25% if defl. controls
iv) Find Crโ = nโขqrs where n is the number of studs in effective length.
v) Find a = ๐ถ๐โฒ
0.51๐๐โฒ๐
x 103 (kN, MPa and mm) where b is the effective slab width.
vi) Find Cr = 0.9๐ด๐ ๐น๐ฆโ10โ3โ๐ถ๐
โฒ
2
vii) Find Steel Compression Area
Asc = ๐ถ๐
0.9๐น๐ฆโ 103 with Cr in kN, Fy in MPa.
viii) Check if NA is in flange: ๐๐ = ๐ด๐ ๐
๐๐ โค ๐ก๐, where bf is the flange width, tf is the flange thickness.
if not ok, NA is in the web, so depth in the web is: ๐๐ค = ๐ด๐ ๐โ๐๐โ๐ก๐
๐ก๐ค where tw is the web width.
NA is at ๐๐ from the top of the steel beam. ix) Find Centroids of 3 sections. Case a) NA in flange (from base of steel):
๐ฆ๏ฟฝ๐๐๐๐ ๐๐๐ ๐๐ก๐๐๐ =
๐ด๐ ๐2 โ ๐๐๐๐ ๏ฟฝ๐ โ
๐๐2 ๏ฟฝ
๐ด๐ โ ๐๐๐๐
so ๐ = ๐ โ ๐๐
2โ ๐ฆ๏ฟฝ๐ก๐๐๐ ๐๐๐ ๐ ๐ก๐๐๐
and ๐โฒ = ๐ + ๐ก๐๐๐๐๐๐๐ก๐ ๐ ๐๐๐2
โ ๐ฆ๏ฟฝ๐ก๐๐๐ ๐๐๐ ๐ ๐ก๐๐๐ case b) NA in web (from base of steel):
๐ฆ๏ฟฝ๐๐๐๐ ๐๐๐ ๐๐ก๐๐๐ =
๐ด๐ ๐2 โ ๐ก๐๐๐ ๏ฟฝ๐ โ ๐๐
2๏ฟฝ ๏ฟฝ โ ๐ก๐ค๐๐ค ๏ฟฝ๐ โ ๐ก๐ โ ๐๐ค2๏ฟฝ ๏ฟฝ
๐ด๐ โ ๐ก๐๐๐ โ ๐ก๐ค๐๐ค
๐ฆ๏ฟฝ๐ถ๐๐๐๐๐๐ ๐ ๐๐๐ ๐๐ก๐๐๐ = ๐ก๐๐๐ ๏ฟฝ๐ โ ๐๐
2๏ฟฝ ๏ฟฝ + ๐ก๐ค๐๐ค ๏ฟฝ๐ โ ๐ก๐ โ ๐๐ค2๏ฟฝ ๏ฟฝ
๐ก๐๐๐ + ๐ก๐ค๐๐ค
so ๐ = ๐ โ ๐ฆ๏ฟฝ๐ถ๐๐๐๐๐๐ ๐ ๐๐๐ ๐๐ก๐๐๐ โ ๐ฆ๏ฟฝ๐ก๐๐๐ ๐๐๐ ๐ ๐ก๐๐๐ and ๐โฒ = ๐ + ๐ก๐๐๐๐๐๐๐ก๐ ๐ ๐๐๐
2โ ๐ฆ๏ฟฝ๐ก๐๐๐ ๐๐๐ ๐ ๐ก๐๐๐
x) Compute Mrcomp = [Cre + Crโeโ] x 10-3 (kN and mm) gives Mr comp in kNm
Behaviour Under Specified Loads (Deflection):
Case 1) Assume NA is in the steel:
๐ฆ๏ฟฝ =
๐ด๐ ๐2 +๐ก๐ ๐๐๐๐
๐ ๏ฟฝ๐+๐ก๐ ๐๐๐2๏ฟฝ ๏ฟฝ
๐ด๐ +๐ก๐ ๐๐๐๐๐
โค ๐ Otherwise Use Case 2 below
Find Ix-composite :
๐ผ๐ฅ๐ = ๐ผ๐ ๐ก๐๐๐ + ๐ด๐ ๏ฟฝ๐ฆ๏ฟฝ โ๐2
๏ฟฝ2
+๐๐ก๐ ๐๐๐
3
12๐+
๐๐ก๐ ๐๐๐
๐๏ฟฝ
๐ก๐ ๐๐๐
2+ ๐ โ ๐ฆ๏ฟฝ๏ฟฝ
2
Look up Ss in the section property tables, and calculate St = ๐ผ๐ฅ๐๐ฆ๏ฟฝ
.
Calculate loads M1 and M2: M1 = Self Weight + Concrete Slab + Formwork M2 = Additional Dead load + live loads. now check that: During construction, to make sure the tension flange doesnโt yield:
Case 2) if NA is in the concrete
h = โ๐ด๐ ยฑ๏ฟฝ๐ด๐
2+4โ ๐2๐๐ด๐ ๏ฟฝ๐
2+๐ก๏ฟฝ๐๐
๐ฆ๏ฟฝ =d + t โ h from the base of the steel member. Where, b is effective slab width (above); t is the concrete slab thickness; d is the depth of the steel member (total height). Find Ix-composite :
๐ผ๐ฅ๐ = ๐ผ๐ ๐ก๐๐๐ + ๐ด๐ ๏ฟฝ๐ฆ๏ฟฝ โ๐2
๏ฟฝ2
+๐โ3
12๐+
๐โ๐
๏ฟฝโ2
๏ฟฝ2
Look up Ss in the section property tables, and calculate St = ๐ผ๐ฅ๐๐ฆ๏ฟฝ
.
Calculate loads M1 and M2: M1 = Self Weight + Concrete Slab + Formwork M2 = Additional Dead load + live loads. now check that: During construction, to make sure the tension flange doesnโt yield:
Deflection Checks:
โ2 uses per permanent live load.
iii) โ3 is the same as โ2 BUT uses short term live load (no dead loads). Unless specified, use live load = 0.5โขtotal live load iv) ๐ฆ๏ฟฝ2 = d + tslab/2 - ๐ฆ๏ฟฝ
โs = 2๐ด๐๐ฟ2๐ฆ๏ฟฝ2
๐ผ๐ฅ๐โข10-6
Total Deflection
If the steel beam is simply supported, a truer value of the deflection is:
Compression Members
Step 1 โ Check beam class
The member must be class 3 or better:
โค remember to bring root Fy to the other side and check < 200
โค
Step 2 โ Slenderness
i) Check slenderness ratio
โค 200 Check for x and y axes, largest governs.
ii) Calculate lambda using maximum slenderness
Use Lambda to calculate Cr
x 10-3 Cr (kN) = (mm and MPa)
n= 1.34 normally, (W sections or Class C HSS)
n = 2.24 for WWF-shapes with flame cut flange edges or class H HSS sections.
Beam Resistances Moment Resistance Step 1 - Determine beam class (web and flange):
If Lu is unknown, check both of the following 2 cases, take the lower. Step 2 - For unbraced length, L < Lu (or beams that have full lateral bracing or weak axis bending):
x 10-6 (mm and MPa) gives M in kNm
Step 3 - For unbraced length, L > Lu :
G = 77000 MPa J = Shape tables HB 6-40 w2 = 1.0 when a) max moment is between braces b) no lateral support at ends
Shear Resistance
Check Deflection (HB 5-146) Uniform load: Delta = 5wL4/384EI
Tension and Bending:
1) Check section class:
2) If beam properties are not found in a table compute moment of inertia and section modulus of the beam:
3) Compute Mr as explained in the previous section. If web is class 4 see section on beam
columns to calculate Mr(4).
4) Use the following to check if the beam is acceptable:
in above check, take Z and S (ie 90 x 103 mm3) without 103 (ie take 90 mm3) Also check lazy Antoineโs forgotten check (cross-sect strength check): ๐๐
๐๐+ ๐๐
๐๐ โค 1.0 where Tr = 0.9AgFy x 10-3
= 0.765AnFu x 10-3 = 0.765AneFu x 10-3 least of (see tension section)
Plate Girders: If given no sizes, start by using the following preliminary sizing values:
โ = 540 ๏ฟฝ๐๐
๐น๐ฆ๏ฟฝ
13๏ฟฝ, ๐ด๐น๐ฟ = ๐๐
๐น๐ฆโ, and ๐ด๐ค = ๐๐
0.594๐น๐ฆ โ ๐ค = ๐๐
0.594โ๐น๐ฆ
as can be seen in the following diagram:
Checking Plate Girders: Bending:
1) Check Flange is class 3 or better and if web is class 3 or class 4:
Web: โ๏ฟฝ๐น๐ฆ
๐คโค 1900, class 3 if yes, class 4 if no
(1700 for classes 1 and 2)
Flange: ๐
2๏ฟฝ ๏ฟฝ๐น๐ฆ
๐กโค 200, if yes class 3, else must increase thickness.
(170 for classes 1 and 2) 2) Compute moment of inertia:
๐ผ = 2 ๏ฟฝ๐๐ก3
12 + ๐๐ก ๏ฟฝโ2 +
๐ก2๏ฟฝ
2
๏ฟฝ +๐คโ3
12
Where terms are defined in previous figure, b is the flange width, and t is the flange thickness.
3) Find Section modulus: ๐ = ๐ผ
๐ฆ๏ฟฝ or ๐ = 1
4[๐๐2 โ (๐ โ ๐ค)(๐ โ 2๐ก)2] (class 1, 2).
4) Compute Mr Class 1 or 2: ๐๐ = 0.9 ๐ ๐น๐ฆ Class 3: ๐๐ = 0.9 ๐ ๐น๐ฆ Class 4 web:
๐๐4 = 0.9 ๐ ๐น๐ฆ ๏ฟฝ1 โ 0.005 ๐ด๐ค๐๐
๐ด๐๐๐๐๐๐๐ ๐ ๐๐๐ ๐๐๐๐๐๐๏ฟฝโ
๐คโ 1900
๏ฟฝ๐น๐ฆ๏ฟฝ๏ฟฝ
5) Done, move on!!!!
Shear Stiffeners: No Stiffeners Specified:
Step 1: check if stiffeners are necessary.
1) Take kv = 5.34 for no stiffeners. 2) Check web type:
Compute Q = โ๏ฟฝ๐น๐ฆ
๐ค๏ฟฝ๐๐ฃ
And consult the appropriate case: Case 1: Q>621
1) Calculate Vr = 162000๐ค3๐๐ฃโ
x 10-3 2) If Vr > Vf then no stiffeners needed.
Case 2: 439<Q<621
1) Calculate Vr = 261๐ค2๏ฟฝ๐๐ฃ๐น๐ฆ x 10-3 2) If Vr > Vf then no stiffeners needed.
Case 3: Q<439
1) Calculate Vr = 0.594โ๐ค๐น๐ฆ x 10-3 2) If Vr > Vf then no stiffeners needed.
With Stiffeners:
1) Choose stiffener spacing a (done by guessing, or given to you) Compute a/h. In order to select a use the following criteria:
2) Calculate kv :
3) Compute Q =
โ๏ฟฝ๐น๐ฆ
๐ค๏ฟฝ๐๐ฃ
And consult the appropriate case:
Case 1: Q>621 1) Calculate Fs note that for end stiffeners, Ft = 0. Do for both end and central
2) Calculate Vr = 0.9 h w Fs x 10-3 3) If Vr > Vf then OK.
Case 2: 502<Q<621
1) Calculate Fs note that for end stiffeners, Ft = 0. Do for both end and central.
2) Calculate Vr = 0.9 h w Fs x 10-3 3) If Vr > Vf then OK.
Case 3: 439<Q<502
1) Calculate Fs
2) Calculate Vr = 0.9 h w Fs x 10-3 3) If Vr > Vf then OK.
Case 4: Q<439
1) Fs = 0.66Fy same as before, so Calculate Vr = 0.594โ๐ค๐น๐ฆ x 10-3
2) If Vr > Vf then OK.
Stiffener Design:
1) Write down a, Fy , h , w. 2) Calculate C:
3) Calculate Y =
๐น๐ฆโ๐๐๐๐๐๐
๐น๐ฆโ๐ ๐ก๐๐๐๐๐๐๐ usually 1.
4) Select D value from the following criteria:
5) Compute As:
6) Choose a standard thickness which allows it to fit inside the girder.
7) Make sure stiffener is at least class 3:
8) Find actual area: A = 2 t b > As for 2 stiffeners (always two for ends). 9) Compute moment of inertia:
๐ผ = 2 ๏ฟฝ๐๐ก3
12+ ๐๐ก
4(๐ + ๐ค)2๏ฟฝ for two stiffeners
And check: ๐ผ โฅ ๏ฟฝ โ50
๏ฟฝ4
Welds of Shear Stiffeners:
1) Compute Vweld = h Fy1.5x 10-4
2) Choose weld size:
Available weld sizes in next table:
3) Choose electrode type:
Note standard for 300W-350W steel is E49XX => Xu = 490 and Fu = 450 MPa.
4) Calculate VRL, take lesser of: 1. ๐๐
๐ฟ= 0.449๐ท๐น๐ข
2. ๐๐๐ฟ
= 0.317๐ท๐๐ข
5) Sub in Vf to find required weld length. Stitch Welds: i) Find max clear spacing = 16w ii) Calculate factored shear transfer per weld, VFL = Vweld/2s (N/mm/weld)
where s = number of stiffeners at that point (ie s = 2 for bear. stiff.) iii) Guess weld length L based on Lmin
iv) Guess clear spacing <300 mm or < 330๐ก๐ ๐ก๐๐๐
๏ฟฝ๐น๐ฆ for non-staggered welds
v) Check shear transfer resistance over weld height, VRH:
= ๐๐ ๐ฟโ๐ฟ๐/๐ถ
(N/mm) > VFL
where O/C = clear spacing + L
vi) Check clear spacing requirements: Non-Staggered Welds Staggered Welds
< 300 mm < 450
< 330๐ก๐ ๐ก๐๐๐
๏ฟฝ๐น๐ฆ <
525๐ก๐ ๐ก๐๐๐
๏ฟฝ๐น๐ฆ
< 16w < 16w < 4L < 4L
vii) Check O/C spacing <300 mm for non-staggered welds
< 450 mm for staggered welds viii) Specify spacing under stiffener >4w and <6w (mm) ix) Specify 25mm coping at top. M-V interaction: Check the following locations: Vf at x = 0.6Vr Maximum Mf Locations where flange reinforcements are placed. At each location do the following: If:
1) Vf > 0.6 Vr AND Mf > 0.75 Mr AND correct class (h/w > 502...) Then check the interaction equation:
at these locations. Bearing Stiffeners: Case 1: End of beam:
1) Required no matter what if โ๐ค
โฅ 1100
๏ฟฝ๐น๐ฆ , or Br < Rxnf
2) Check Crippling capacity:
i) ๐ต๐ = 0.75๐ค๐น๐ฆ(๐ + 4๐ก) ii) ๐ต๐ = 0.45๐ค2๏ฟฝ๐ธ๐น๐ฆ Where t is the flange thickness, w is the web thickness.
3) Select plates: use 2 plates both with ๐
๐กโค 200
๏ฟฝ๐น๐ฆ
4) Check Bearing Rest Capacity: ๐ด๐ต = ๐๐ข๐๐๐๐ ๐๐ ๐๐๐๐ก๐๐ ร ๐ก๐๐๐๐ก๐ ร ๏ฟฝ๐๐๐๐๐ก๐ โ 25๐๐ ๐๐๐๐๏ฟฝ
And ๐ต๐ = 1.35 ๐น๐ฆ๐ด๐ต ร 10โ3 ๐๐ ๐๐
5) Check compression resistance Calculate A = 2(tplatebplate) + 12w2
๐ผ๐ฅ = ๐ก๐๐๏ฟฝ2๐๐๐ + ๐ค๏ฟฝ
3
12 +๏ฟฝ12๐ค โ ๐ก๐๐๏ฟฝ๐ค3
12
r = ๏ฟฝ๐ผ๐ฅ๐ด
and finally
Where k = 0.75, L = h (height of the web) and the rest is as before.
6) Check that Cr > Rxnf
Case 2: Interior Bearing Stiffener 1) Required no matter what if โ
๐คโฅ 1100
๏ฟฝ๐น๐ฆ , or Br < Rxnf
2) Check Crippling capacity:
iii) ๐ต๐ = 0.80๐ค๐น๐ฆ(๐ + 10๐ก) iv) ๐ต๐ = 1.16๐ค2๏ฟฝ๐ธ๐น๐ฆ Where t is the flange thickness, w is the web thickness.
3) Select plates: use 2 plates both with ๐
๐กโค 200
๏ฟฝ๐น๐ฆ
4) Check Bearing Rest Capacity: ๐ด๐ต = ๐๐ข๐๐๐๐ ๐๐ ๐๐๐๐ก๐๐ ร ๐ก๐๐๐๐ก๐ ร ๏ฟฝ๐๐๐๐๐ก๐ โ 25๐๐ ๐๐๐๐๏ฟฝ
And ๐ต๐ = 1.35 ๐น๐ฆ๐ด๐ต ร 10โ3 ๐๐ ๐๐
5) Check compression resistance Calculate A = 2(tplatebplate) + 25w2
๐ผ๐ฅ = ๐ก๐๐๏ฟฝ2๐๐๐ + ๐ค๏ฟฝ
3
12 +๏ฟฝ25๐ค โ ๐ก๐๐๏ฟฝ๐ค3
12
r = ๏ฟฝ๐ผ๐ฅ๐ด
and finally
Where k = 0.75, L = h (height of the web) and the rest is as before.
6) Check that Cr > Rxnf
Weld of reinforcing plate to Top and Bottom Flanges: Transfers shear forces that flow along the beam length from the web to the flange.
1) qf = ๐๐โ(๐ด๐ด)(๐ด๐ด ๐ก๐ ๐๐ด)
๐ผ๐๐๐๐ (N/mm)
where AA is the area above or below the weld (ie. Area of the flange) where AA to NA is distance from centroid of AA to the the NA of the beam where Ibeam is the moment of inertia of the beam at that point NOTE: if the beam has several cross sections along its length, calculate q for each cross section and take the max q and design a weld with constant characteristics.
2) Do base metal and weld metal checks, take lesser of:
Calculate VRL, take lesser of:
๐๐๐ฟ
= 0.449๐ท๐น๐ข
๐๐๐ฟ
= 0.317๐ท๐๐ข
3) Guess weld length L based on Lmin 4) Guess clear spacing <300 mm for non-staggered welds
< 450 mm for staggered welds 5) Get O/C = clear spacing + L
6) Calculate and check: qr = 2๐๐ ๐ฟโ๐ฟ
๐/๐ถ (N/mm) > qf
Tension and Bending:
1) See Beam notes.
Single Storey Building Design
Load Calculations
For info on importance category (ie description) see HB 1-124
Snow:
S = Is[Ss(Cs Cw Cb Ca) + Sr] Sr and Ss in climactic data table (1/50 yrs) Find Cs = 1.0 for roofs < 30o
Find Cw = 1.0 Find Ca = 1.0 Find Cb =
Wind:
p = IwqCeCpCg (assume no internal pressure)
1) Find Ce: Open Terrain ร = (h/10)0.2 > 0.9 or take 0.9 Rough Terrain ร = 0.7(h/12)0.3 > 0.7 or take 0.7 2) Find z, lesser of: z = 0.1โขleast horizontal dimension = 0.4โขheight but not less than: 0.04โขleast horizontal dimension, or 1 m 3) Then use z to find y (length of edge section), greater of: y = 6 m = 2z 4) CpCg for , 1E, 4, 4E in Figure 1-7 Load Case 1, flat roof: 1 = 0.75 1E = 1.15 4 = -0.55 4E = -0.8 Now, calculate p for the four zones of interest: p = IwqCeCpCg 5) Distribute p-values onto the end windward and leeward wall columns (careful! break down edge effects) For edge effects: look at 4 cases: (each must be done on the leeward and windward side)
Windward Side: Case 1) Force = ๐ ๐๐๐
2๐1๐ธ
โ2
Case 2) Force =๏ฟฝ๐ฆ โ ๐ ๐๐๐2
๏ฟฝ ๐1๐ธโ2
+ ๏ฟฝ3 ๐ ๐๐๐2
โ ๐ฆ๏ฟฝ ๐1โ2
Case 3) Force =๐๐๐๐ โ ๐1โ2
Case 4) Force = ๐ ๐๐๐2
๐1โ2
For Leeward side simply replace 1 and 1E with 4 and 4E.
Design of Lateral Braces:
1) Using Wind loads found above, take sum of moments around lateral brace point equal to zero.
2) Use sum of forces equals zero to find reaction on other side of building.
3) Use larger reaction as controlling value, and divide by number of brace bays (Wmax).
4) Use tributary area of 1 brace bay (1/4 building for 4 bays) in order to find allocated gravity load.
5) Calculate FACTORED loads (kN) acting on a single brace bay (n bays):
Dead load per 1/n building: 1.25Dโข widthโขlength of total building/n Snow factor 1.5: 1.5Sโขwidthโขlength/n Snow factor 0.5: 0.5Sโข widthโขlength/n Live factor 1.5: 1.5Lโข widthโขlength/n Live factor 0.5: 0.5Lโข widthโขlength/n Wind factor 1.4: 1.4โขWmax Wind factor 0.4: 0.4โขWmax 6) Now we look at each load combination
Begin with: 1.25D + 1.4W + 0.5S, and calculate: i) โ ๐ถ๐ = ๐ ๐ข๐ ๐๐ ๐๐๐๐ก๐๐๐๐ ๐๐๐๐ฃ๐๐ก๐ฆ ๐๐๐๐๐ ๐๐ ๐๐๐๐ ๐๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐ ๐๐๐๐ ๐๐๐๐๐๐ ii) Notional Load = 0.005 โ โ ๐ถ๐ iii) Lateral Load โ ๐๐ = Notional Load + Factored Wind load (take correct load from above list, may be zero for some cases) 7) Select Brace Member (if given a member skip right to vi)) i) Find brace length = ๏ฟฝ๐ต๐๐ฆ ๐ค๐๐๐กโ2 + ๐ต๐๐ฆ ๐ป๐๐๐โ๐ก2 ii) Tft = โ ๐๐โข Full Brace Length / Full Bay width iii) Set Tr = 0.459 Ag Fu x 10-3 = Tft and solve for Ag. (for 300W material) iv) Find rmax = Effective Brace length / Fy , divide length by 2 if braces connected at midspan. v) Select Member based on rmax and Ag. vi) Check true Tr = 0.9AgFy > Tft = โ ๐๐โข Full Brace Length / Full Bay width 8) Calculate P-delta amplification factors:
i) โ๐น (๐๐) = ๐๐๐ก๐ฟ2
๐ธ๐ดโ๐น๐ข๐๐ ๐ต๐๐ฆ ๐๐๐๐กโ ร 103 (kN, MPa, mm)
ii) ๐2 = 1
1 โ โ ๐ถ๐โ๐น
โ ๐๐โ๐น๐ข๐๐ ๐ต๐๐ฆ ๐ป๐๐๐โ๐ก
< 1.4
9) Check amplified brace force (ABF) and amplified deflection (AD): i) ABF = ๐2๐๐๐ก < Tr ii) AD = U2โ๐น < h/500 (or use the deflection limit he gives)
10) Repeat steps 6i through 9, for each of the following load combinations, if he asks: 1.25D + 1.4W + 0.5L 1.25D + 1.5S + 0.4W 1.25D + 1.5L + 0.4W 1.25D + 1.5S + 0.5L
Seismic Fuse Member
Check fuse capacity and take HIGHEST capacity, as this will control. If a lower capacity is taken, the fuse will assume the behaviour of the overlooked higher capacity, and the fuse system will not bow out.
In the case where a brace member is used as the fuse element:
Fuse capacity is, higher of:
= 1.1 AgFy x 10-3 and
= 385 Ag x 10-3
Gerber Beam Design:
1) Find Full and Partial factored loading scenario distributed loads.
Full Load = 1.25D + 1.5S +0.5L Partial Load = 1.25D + 0.75S
Also put full load everywhere and calculate beam loads (V and M) if L>S Full Load = 1.25D+1.5L+0.5S Partial Load = 1.25D + 0.75S 2) Find purlin/owsj point loads on Gerber Beam and link spans.
4) Apply half link span load to each end of Gerber Beam
5) Analyze Gerber beam in normal manner (V and M diagrams... take max values) with this loading case; Design as a normal beam using both maximum positive and negative moments.
6) Repeat for multiple loading scenarios
+Tension Members (Notes 1 โ 3)
Shear Lag
1) For I beams with b > 2/3 d, where d is section depth.
Ane = 0.9 An
2) Angles with 4 or more bolts in one transverse line.
Ane = 0.8 An
3) Angles with 3 or less bolts in one transverse line.
Ane = 0.6 An
4) Other shapes with 3 or more bolts in a line.
Ane = 0.85 An
5) Other shapes with 1 or 2 bolts in a line.
Ane = 0.75 An
Welded Connections
Eccentric Loading: 1) Define Origin (x,y); Usually at bottom corner. 2) Calculate the centroid of the weld group assuming unit width, wrt the defined origin:
๏ฟฝฬ ๏ฟฝ =โ ๐๐๐๐ ๐ฟ๐๐๐๐กโ โ ๐ฅโ ๐๐๐๐ ๐ฟ๐๐๐๐กโ๐
๐ฆ๏ฟฝ =โ ๐๐๐๐ ๐ฟ๐๐๐๐กโ โ ๐ฆโ ๐๐๐๐ ๐ฟ๐๐๐๐กโ๐
3) Find Perpendicular distance, e, from line of force to centroid of weld group. 4) Compute Horizontal, Vertical and moment components of the Force: eg. Px = Pโขsinฮธ, Py = Pโขcosฮธ..., M=Pโขe 5) Compute Ixโ and Iyโ ; and Jโ= Ixโ+ Iyโ
Iโx-single weld = ๐โโ3
12+ ๐ โ โ โ (๐ฅ โ ๏ฟฝฬ ๏ฟฝ)2, where b is the width wrt the axis of interest, h is the
height wrt the axis of interest; the thickness of the weld is considered to be 1.
6) a) Look at edge points, determine which will have maximal shear due to moment and direct shear combination (ie if q and external shear in the same direction, and which one is farther from the centroid). b) Compute Shear at each edge point: ๐๐ฅ( ๐๐
๐๐) = ๐โ๐โ(๐ฆโ๐ฆ๏ฟฝ)
๐ฝ ๐๐ฆ( ๐๐
๐๐) = ๐โ๐โ(๐ฅโ๏ฟฝฬ ๏ฟฝ)
๐ฝ
7) Finally take qx+Px and qy+Py and compute the resultant. The edge point with the highest V governs:
Vmax =โ ๐ฟ๐ก๐๐ก๐๐ ๐ค๐๐๐๐ ๏ฟฝ๏ฟฝ๐๐ฅ + ๐๐ฅโ ๐ฟ๐ก๐๐ก๐๐ ๐ค๐๐๐๐
๏ฟฝ2
+ ๏ฟฝ๐๐ฆ + ๐๐ฆ
โ ๐ฟ๐ก๐๐ก๐๐ ๐ค๐๐๐๐ ๏ฟฝ
2
8) Calculate Vr and Br for the weld whose edge has the highest shear.
ICR Method: Look up the weld group in the table given in HB3-44. Pallowable = CDL Check that Pf < Pallowable
Note: D > 5 mm ALWAYS Complete Joint Penetration Groove Welds (CJPG) Shear Failure: i) Base Metal: Vr = 0.672 Am Fu
where Am is for the vertical face of the base metal. Aw = Am (most cases) ii) Weld Metal: Vr = 0.672 Aw Xu where Aw is vertical face of the base metal. Tension Failure: Match electrodes properly, then full tension capacity can be reached (only consider Xu). Tr = 0.67 Aw Xu
Partial Joint Penetration Groove Welds (PJPG) Shear Failure: i) Base Metal: Vr = 0.672 Am Fu where Am is for the net vertical face of the base metal (vertical fusion surface) ii) Weld Metal: Vr = 0.672 Aw Xu Tension Failure: less of Tr = 0.67 An Fu An is net section (ie fusion face). = 0.9 Ag Fy Ag is the gross sectional area of the plate. Weird Shape: