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 β₯ 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 β₯ 1.0
Bearing and Base Plates: Bearing Plates:
1) Calculate acceptable Bearing stress on Concrete Column: πππ₯ ππ‘πππ π = 0.51ππ
β² 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 = π π₯ππ΅ππππππ ππ‘πππ π οΏ½
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:
π΄π΅π΅ππππ‘π
οΏ½ β€ ππππ‘
If it is not, then take π΅ππππ‘π = π΄π΅ππππ‘
οΏ½
5) Increase all dimensions to easy numbers (multiples of 5 or 10 mm) 6) Compute Bending Length π = π΅ππππ‘π
2β π, 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 ππ = π΅ππππππ ππ‘πππ π β π2
2
8) Compute ππ = ππ‘2 where t is the plate thickness.
Note that Q = ππΉπ¦
4οΏ½ 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 π‘ β₯ π΅ππππ‘πβ π΅πππ ππππ‘β
10
Base Plate Design: Case 1) No Eccentricities: For I-beams: 1) Find maximum bearing stress:
πππ₯ ππ‘πππ π = 0.55ππβ² 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 = π π₯ππ΅ππππππ ππ‘πππ π οΏ½
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 = πΆβ0.95π2
, n = π΅β0.8π2
and use the larger of the two in order to calculate Mf
6) Calculate BSA = π π₯ππ΄ππΏ
οΏ½
7) Find ππ = π΅ππ΄ β (π ππ π)2
2
8) Compute ππ = ππ‘2 where t is the plate thickness.
Note that Q = ππΉπ¦
4οΏ½ 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
π‘ β₯ π5
, π5
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: πππ₯ ππ‘πππ π = 0.51ππ
β² 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 = π π₯ππ΅ππππππ ππ‘πππ π οΏ½
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 = οΏ½π΄π΅ 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 = πΆπππ’ππ ππππ‘β + 2οΏ½π·πππ’ππππ π’π + 1.5ππππβππ ππποΏ½ (2 anchor rods assumed), again round 1.5danchor rod to the nearest 5mm.
4) Compute Width of the plate Wplate = πΆπππ’ππ ππππ‘β + 25, round down to nearest 5mm.
5) Find m = πΏππππ‘πβπΆπππ’ππ ππππ‘β+π»ππ π‘βππππππ π
2
6) Compute reduced Bearing Stress Br = π π₯ππππππ‘ππΏππππ‘π
7) ππ = π΅π β (π)2
2 , ππ = ππ‘2
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
π‘ β₯ π5
Eccentric Design:
1) Determine which case you are using :
e = ππ ππ ππππ’ππ
πΆπ ππ ππππ’ππ
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 = Bβ’(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 = πΆβ0.95π
2 , n = π΅β0.8π
2
4) Calculate BSA = π π₯ππ΄ππΏ
οΏ½
5) Find ππ = π΅ππ΄ β (π ππ π)2
2
6) Compute ππ = ππ‘2 where t is the plate thickness.
Note that Q = ππΉπ¦
4οΏ½ 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
π‘ β₯ π5
, π5
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 οΏ½πΆ
2β ποΏ½
4) Calculate m = πΆβ0.95π2
5) Select f From table below: (0.85*0.6*fcβ)
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 = (π)(πβπ)
π
7) π‘2= 1π
οΏ½π2π2
2+ (π β π2) π2
3οΏ½ β 10β3
Note that Q = ππΉπ¦
4οΏ½ which can be taken from the following table:
Fy 300 350 400 Phi Fy/4 = Q 67.5 78.75 90
8) Check that π‘ β₯ π ππ π
5 take larger of m and n.
Case 3:
1) Anchor Rods needed, take f From table below: (0.85*0.6*fcβ)
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.
οΏ½ ππππ = πΆπ(π + π‘π) β ππ΅ππππ‘ππ
2000 οΏ½πΆ2 + π‘π β
π3οΏ½ = 0
Where all lengths are in mm So compute terms a1 b1 and c1: π1 = ππ΅ππππ‘π
6000 , π1 = β ππ΅ππππ‘π
2000οΏ½πΆ
2+ π‘ποΏ½ , π1 = πΆπ(π + π‘π)
So π = βπ1Β±οΏ½π1
2β 4π1 π1
2π1 , 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):
π = 0.526(π· β 0.938π)2πΉπ’ , 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 = 0.45πππππ‘2πΉπ¦ = 135πππππ‘2 πππ 300π π π‘πππ 10) First Check for t: Mr = Mf and solve for tmin . If Mr and Mf are not requested
explicitly:
π‘πππ = οΏ½π
0.45πΉπ¦= οΏ½ π
135 for Fy = 300W grade rod.
11) Second check for t:
π0.45πΉπ¦π‘2 + π
1.188πΉπ¦πππππ‘ = 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 = πΆβ0.95π
2
13) Calculate Mf2
ππ2 = 0.5ππ οΏ½π β π
3οΏ½ π΅ Γ 10β3 all values in mm and MPa, answer in kNm.
14) Use Mr to find t. so
π‘2 = 2.222ππ 1
πΉπ¦οΏ½π β π
3οΏ½ Γ 10β3 = 0.0047 οΏ½π β π
3οΏ½ Γ 10β3 for 300W grade bar.
15) Deflection Check: make sure π‘ β₯ π
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 =
πΆπ
πΓ 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 π΄πππ = 4π(π + 2π) + 2ποΏ½π β π‘ππππππ β 2ποΏ½ or
π΄πππ = 2ποΏ½2π + π β π‘ππππππ + 2ποΏ½
3) Solve for m
4) Find π‘ = οΏ½2πΆππ2Γ103
0.9π΄ππππΉπ¦ππ
5) Plate Dimensions: 2m+0.95d x 2m+0.8b, round up to nice numbers. 6) Check deflection:
π‘ β₯ π5
.
Bolted Connections
Factored Tensile Resistance
Separation Load
To calculate Ap: 1) Get bolt spacing in terms of bolt diameters: Spacing = gauge/db = Coeffβ’db
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 β 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 = Coeffβ’db
2) Ap = ππππππππ2
4 - πππ
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 β number of shear planes
Slip Critical
Eccentric Loads: 1) Find Centroid: xbar=
β π΄πβπ₯ππ’ππππ ππ π΅πππ‘π βπ΄π
ybar=β π΄πβπ¦
ππ’ππππ ππ π΅πππ‘π βπ΄π,
Assuming Bolts are of the same size. 2) Compute ra for each bolt: ra
2= (x - xbar)2 + (y - ybar)2 and β ππ2
3) and then the force on each bolt: Rax=
πβπβ(π₯βπ₯πππ)β ππ
2 and Ray=πβπβ(π¦βπ¦πππ)
β ππ2
4) Find the total resultant shear due to direct shear and moment induced shear
Total Shear = VT =οΏ½(π ππ₯ + ππ₯)2 + οΏ½π ππ¦ + ππ¦οΏ½2
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 β’ 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. Γ¨ ie find Mu then find Mrβ 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/(4500β’(fcβ)0.5) 2) Find b, effective slab width.
Β· Slab on both sides of steel beam (b is less of): i) 0.25 β’ beam span ii) Avg. C-C spacing of the steel beams
Β· Slab on only one side of steel beam (b is lesser of): i) steel beam width + 0.1 β’ beam span ii) steel beam width + 0.5 β’ C-C steel beam spacing
3) Locate NA
find a (mm) = 0.9π΄π πΉπ¦
0.51ππβ²π
(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 β₯ 100%, stay in full connectivity otherwise move to case 3.
There are 3 cases: 1 β full shear connectivity with NA in concrete 2 β full shear connectivity with NA in steel 3 β partial shear connectivity Case 1 β Full shear connectivity with NA in concrete
i) find a = 0.9π΄π πΉπ¦
0.51ππβ²π
ii) find connectivity Qr min that gives benchmark for full connectivity:
lesser of: 0.9 AsFy x 10-3 OR 0.51 ππβ² b a x 10-3 (mm and MPa)
iii) compute qrs
x 10-3 (mm and MPa) gives qrs (kN)
Ec = 4500οΏ½ππ
β² 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 eβ = d/2 + tslab β a d is steel beam height Mrc (kNm) = 0.9AsFyeβ x 10-6 (mm and Mpa) Case 2 β Full shear connectivity with NA in steel a > tslab i) Find Crβ = 0.51 fcβ b t where t is the concrete thickness, b is the effective width (above).
ii) Find Cr = 0.9π΄π πΉπ¦β10β3βπΆπ
β²
2
iii) Find Steel Compression Area Asc = 1000πΆπ
0.9πΉπ¦ with Cr in KN, Fy in MPa.
iv) Check if NA is in flange: ππ = π΄π π
ππ β€ π‘π, where w is the flange width, tf is the flange thickness.
if not ok, NA is in the web, so ππ€ = π΄π πβππβπ‘π
π‘π€ where tw is the web width.
NA is at ππ from the top of the steel beam. v) Find Centroids of 3 sections. Case a) NA in flange (from base).
π¦οΏ½ππππ πππ ππ‘πππ =
π΄π π2 β ππππ οΏ½π β
ππ2 οΏ½
π΄π β ππππ
so π = π β ππ
2β π¦οΏ½π‘πππ πππ π π‘πππ
and πβ² = π + π‘πππππππ‘π π πππ2
β π¦οΏ½π‘πππ πππ π π‘πππ case b) NA in web
π¦οΏ½ππππ πππ ππ‘πππ =
π΄π π2 β π‘πππ οΏ½π β ππ
2οΏ½ οΏ½ β π‘π€ππ€ οΏ½π β π‘π β ππ€2οΏ½ οΏ½
π΄π β π‘πππ β π‘π€ππ€
π¦οΏ½πΆππππππ π πππ ππ‘πππ = π‘πππ οΏ½π β ππ
2οΏ½ οΏ½ + π‘π€ππ€ οΏ½π β π‘π β ππ€2οΏ½ οΏ½
π‘πππ + π‘π€ππ€
so π = π β π¦οΏ½πΆππππππ π πππ ππ‘πππ β π¦οΏ½π‘πππ πππ π π‘πππ and πβ² = π + π‘πππππππ‘π π πππ
2β π¦οΏ½π‘πππ πππ π π‘πππ
vi) Compute Mrcomp = Cre + Crβeβ vii) find connectivity Qr min that gives benchmark for full connectivity:
lesser of: 0.9 AsFy OR 0.51 ππβ² 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 β Partial connectivity (NA always in the steel)
i) Find qr:
Ec = 4500οΏ½ππ
β² ii) find Qrmin lesser of: 0.9 AsFy OR 0.51 ππ
β² b tslab
iii) Find Percent Connectivity = nβ’qrs / Qr min > 40% if flex controls or > 25% if defl. controls
iv) Find Crβ = nβ’qrs where n is the number of studs in effective length.
v) Find a = πΆπβ²
0.51ππβ²π
x 103 (kN, MPa and mm) where b is the effective slab width.
vi) Find Cr = 0.9π΄π πΉπ¦β10β3βπΆπ
β²
2
vii) Find Steel Compression Area
Asc = πΆπ
0.9πΉπ¦β 103 with Cr in kN, Fy in MPa.
viii) Check if NA is in flange: ππ = π΄π π
ππ β€ π‘π, 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: ππ€ = π΄π πβππβπ‘π
π‘π€ where tw is the web width.
NA is at ππ from the top of the steel beam. ix) Find Centroids of 3 sections. Case a) NA in flange (from base of steel):
π¦οΏ½ππππ πππ ππ‘πππ =
π΄π π2 β ππππ οΏ½π β
ππ2 οΏ½
π΄π β ππππ
so π = π β ππ
2β π¦οΏ½π‘πππ πππ π π‘πππ
and πβ² = π + π‘πππππππ‘π π πππ2
β π¦οΏ½π‘πππ πππ π π‘πππ case b) NA in web (from base of steel):
π¦οΏ½ππππ πππ ππ‘πππ =
π΄π π2 β π‘πππ οΏ½π β ππ
2οΏ½ οΏ½ β π‘π€ππ€ οΏ½π β π‘π β ππ€2οΏ½ οΏ½
π΄π β π‘πππ β π‘π€ππ€
π¦οΏ½πΆππππππ π πππ ππ‘πππ = π‘πππ οΏ½π β ππ
2οΏ½ οΏ½ + π‘π€ππ€ οΏ½π β π‘π β ππ€2οΏ½ οΏ½
π‘πππ + π‘π€ππ€
so π = π β π¦οΏ½πΆππππππ π πππ ππ‘πππ β π¦οΏ½π‘πππ πππ π π‘πππ and πβ² = π + π‘πππππππ‘π π πππ
2β π¦οΏ½π‘πππ πππ π π‘πππ
x) Compute Mrcomp = [Cre + Crβeβ] x 10-3 (kN and mm) gives Mr comp in kNm
Behaviour Under Specified Loads (Deflection):
Case 1) Assume NA is in the steel:
π¦οΏ½ =
π΄π π2 +π‘π ππππ
π οΏ½π+π‘π πππ2οΏ½ οΏ½
π΄π +π‘π πππππ
β€ π Otherwise Use Case 2 below
Find Ix-composite :
πΌπ₯π = πΌπ π‘πππ + π΄π οΏ½π¦οΏ½ βπ2
οΏ½2
+ππ‘π πππ
3
12π+
ππ‘π πππ
ποΏ½
π‘π πππ
2+ π β π¦οΏ½οΏ½
2
Look up Ss in the section property tables, and calculate St = πΌπ₯ππ¦οΏ½
.
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 doesnβt yield:
Case 2) if NA is in the concrete
h = βπ΄π Β±οΏ½π΄π
2+4β π2ππ΄π οΏ½π
2+π‘οΏ½ππ
π¦οΏ½ =d + t β 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 :
πΌπ₯π = πΌπ π‘πππ + π΄π οΏ½π¦οΏ½ βπ2
οΏ½2
+πβ3
12π+
πβπ
οΏ½β2
οΏ½2
Look up Ss in the section property tables, and calculate St = πΌπ₯ππ¦οΏ½
.
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 doesnβt yield:
Deflection Checks:
β2 uses per permanent live load.
iii) β3 is the same as β2 BUT uses short term live load (no dead loads). Unless specified, use live load = 0.5β’total live load iv) π¦οΏ½2 = d + tslab/2 - π¦οΏ½
βs = 2π΄ππΏ2π¦οΏ½2
πΌπ₯πβ’10-6
Total Deflection
If the steel beam is simply supported, a truer value of the deflection is:
Compression Members
Step 1 β Check beam class
The member must be class 3 or better:
β€ remember to bring root Fy to the other side and check < 200
β€
Step 2 β Slenderness
i) Check slenderness ratio
β€ 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 Antoineβs forgotten check (cross-sect strength check): ππ
ππ+ ππ
ππ β€ 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:
β = 540 οΏ½ππ
πΉπ¦οΏ½
13οΏ½, π΄πΉπΏ = ππ
πΉπ¦β, and π΄π€ = ππ
0.594πΉπ¦ β π€ = ππ
0.594βπΉπ¦
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: βοΏ½πΉπ¦
π€β€ 1900, class 3 if yes, class 4 if no
(1700 for classes 1 and 2)
Flange: π
2οΏ½ οΏ½πΉπ¦
π‘β€ 200, if yes class 3, else must increase thickness.
(170 for classes 1 and 2) 2) Compute moment of inertia:
πΌ = 2 οΏ½ππ‘3
12 + ππ‘ οΏ½β2 +
π‘2οΏ½
2
οΏ½ +π€β3
12
Where terms are defined in previous figure, b is the flange width, and t is the flange thickness.
3) Find Section modulus: π = πΌ
π¦οΏ½ or π = 1
4[ππ2 β (π β π€)(π β 2π‘)2] (class 1, 2).
4) Compute Mr Class 1 or 2: ππ = 0.9 π πΉπ¦ Class 3: ππ = 0.9 π πΉπ¦ Class 4 web:
ππ4 = 0.9 π πΉπ¦ οΏ½1 β 0.005 π΄π€ππ
π΄πππππππ π πππ πππππποΏ½β
π€β 1900
οΏ½πΉπ¦οΏ½οΏ½
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 = βοΏ½πΉπ¦
π€οΏ½ππ£
And consult the appropriate case: Case 1: Q>621
1) Calculate Vr = 162000π€3ππ£β
x 10-3 2) If Vr > Vf then no stiffeners needed.
Case 2: 439<Q<621
1) Calculate Vr = 261π€2οΏ½ππ£πΉπ¦ x 10-3 2) If Vr > Vf then no stiffeners needed.
Case 3: Q<439
1) Calculate Vr = 0.594βπ€πΉπ¦ 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 =
βοΏ½πΉπ¦
π€οΏ½ππ£
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 = 0.594βπ€πΉπ¦ x 10-3
2) If Vr > Vf then OK.
Stiffener Design:
1) Write down a, Fy , h , w. 2) Calculate C:
3) Calculate Y =
πΉπ¦βππππππ
πΉπ¦βπ π‘πππππππ 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:
πΌ = 2 οΏ½ππ‘3
12+ ππ‘
4(π + π€)2οΏ½ for two stiffeners
And check: πΌ β₯ οΏ½ β50
οΏ½4
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. ππ
πΏ= 0.449π·πΉπ’
2. πππΏ
= 0.317π·ππ’
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 < 330π‘π π‘πππ
οΏ½πΉπ¦ for non-staggered welds
v) Check shear transfer resistance over weld height, VRH:
= ππ πΏβπΏπ/πΆ
(N/mm) > VFL
where O/C = clear spacing + L
vi) Check clear spacing requirements: Non-Staggered Welds Staggered Welds
< 300 mm < 450
< 330π‘π π‘πππ
οΏ½πΉπ¦ <
525π‘π π‘πππ
οΏ½πΉπ¦
< 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 βπ€
β₯ 1100
οΏ½πΉπ¦ , or Br < Rxnf
2) Check Crippling capacity:
i) π΅π = 0.75π€πΉπ¦(π + 4π‘) ii) π΅π = 0.45π€2οΏ½πΈπΉπ¦ Where t is the flange thickness, w is the web thickness.
3) Select plates: use 2 plates both with π
π‘β€ 200
οΏ½πΉπ¦
4) Check Bearing Rest Capacity: π΄π΅ = ππ’ππππ ππ ππππ‘ππ Γ π‘ππππ‘π Γ οΏ½πππππ‘π β 25ππ πππποΏ½
And π΅π = 1.35 πΉπ¦π΄π΅ Γ 10β3 ππ ππ
5) Check compression resistance Calculate A = 2(tplatebplate) + 12w2
πΌπ₯ = π‘πποΏ½2πππ + π€οΏ½
3
12 +οΏ½12π€ β π‘πποΏ½π€3
12
r = οΏ½πΌπ₯π΄
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 β
π€β₯ 1100
οΏ½πΉπ¦ , or Br < Rxnf
2) Check Crippling capacity:
iii) π΅π = 0.80π€πΉπ¦(π + 10π‘) iv) π΅π = 1.16π€2οΏ½πΈπΉπ¦ Where t is the flange thickness, w is the web thickness.
3) Select plates: use 2 plates both with π
π‘β€ 200
οΏ½πΉπ¦
4) Check Bearing Rest Capacity: π΄π΅ = ππ’ππππ ππ ππππ‘ππ Γ π‘ππππ‘π Γ οΏ½πππππ‘π β 25ππ πππποΏ½
And π΅π = 1.35 πΉπ¦π΄π΅ Γ 10β3 ππ ππ
5) Check compression resistance Calculate A = 2(tplatebplate) + 25w2
πΌπ₯ = π‘πποΏ½2πππ + π€οΏ½
3
12 +οΏ½25π€ β π‘πποΏ½π€3
12
r = οΏ½πΌπ₯π΄
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 = ππβ(π΄π΄)(π΄π΄ π‘π ππ΄)
πΌππππ (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:
πππΏ
= 0.449π·πΉπ’
πππΏ
= 0.317π·ππ’
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 = 2ππ πΏβπΏ
π/πΆ (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 Γ = (h/10)0.2 > 0.9 or take 0.9 Rough Terrain Γ = 0.7(h/12)0.3 > 0.7 or take 0.7 2) Find z, lesser of: z = 0.1β’least horizontal dimension = 0.4β’height but not less than: 0.04β’least 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 = π πππ
2π1πΈ
β2
Case 2) Force =οΏ½π¦ β π πππ2
οΏ½ π1πΈβ2
+ οΏ½3 π πππ2
β π¦οΏ½ π1β2
Case 3) Force =ππππ β π1β2
Case 4) Force = π πππ2
π1β2
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: 1.25Dβ’ widthβ’length of total building/n Snow factor 1.5: 1.5Sβ’widthβ’length/n Snow factor 0.5: 0.5Sβ’ widthβ’length/n Live factor 1.5: 1.5Lβ’ widthβ’length/n Live factor 0.5: 0.5Lβ’ widthβ’length/n Wind factor 1.4: 1.4β’Wmax Wind factor 0.4: 0.4β’Wmax 6) Now we look at each load combination
Begin with: 1.25D + 1.4W + 0.5S, and calculate: i) β πΆπ = π π’π ππ ππππ‘ππππ ππππ£ππ‘π¦ πππππ ππ ππππ ππππππππ‘πππ πππππ ππππ ππππππ ii) Notional Load = 0.005 β β πΆπ iii) Lateral Load β ππ = 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 = οΏ½π΅ππ¦ π€πππ‘β2 + π΅ππ¦ π»πππβπ‘2 ii) Tft = β ππβ’ 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 = β ππβ’ Full Brace Length / Full Bay width 8) Calculate P-delta amplification factors:
i) βπΉ (ππ) = πππ‘πΏ2
πΈπ΄βπΉπ’ππ π΅ππ¦ ππππ‘β Γ 103 (kN, MPa, mm)
ii) π2 = 1
1 β β πΆπβπΉ
β ππβπΉπ’ππ π΅ππ¦ π»πππβπ‘
< 1.4
9) Check amplified brace force (ABF) and amplified deflection (AD): i) ABF = π2πππ‘ < Tr ii) AD = U2βπΉ < 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 β 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:
οΏ½Μ οΏ½ =β ππππ πΏππππ‘β β π₯β ππππ πΏππππ‘βπ
π¦οΏ½ =β ππππ πΏππππ‘β β π¦β ππππ πΏππππ‘βπ
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 = Pβ’sinΞΈ, Py = Pβ’cosΞΈ..., M=Pβ’e 5) Compute Ixβ and Iyβ ; and Jβ= Ixβ+ Iyβ
Iβx-single weld = πββ3
12+ π β β β (π₯ β οΏ½Μ οΏ½)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: ππ₯( ππ
ππ) = πβπβ(π¦βπ¦οΏ½)
π½ ππ¦( ππ
ππ) = πβπβ(π₯βοΏ½Μ οΏ½)
π½
7) Finally take qx+Px and qy+Py and compute the resultant. The edge point with the highest V governs:
Vmax =β πΏπ‘ππ‘ππ π€ππππ οΏ½οΏ½ππ₯ + ππ₯β πΏπ‘ππ‘ππ π€ππππ
οΏ½2
+ οΏ½ππ¦ + ππ¦
β πΏπ‘ππ‘ππ π€ππππ οΏ½
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: