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 ≥ 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 ≥ 1.0

Bearing and Base Plates: Bearing Plates:

1) Calculate acceptable Bearing stress on Concrete Column: 𝑀𝑎𝑥 𝑆𝑡𝑟𝑒𝑠𝑠 = 0.51𝑓𝑐

′ 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 = 𝑅𝑥𝑛𝐵𝑒𝑎𝑟𝑖𝑛𝑔 𝑆𝑡𝑟𝑒𝑠𝑠�

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:

𝐴𝐵𝐵𝑃𝑙𝑎𝑡𝑒

� ≤ 𝑊𝑛𝑒𝑡

If it is not, then take 𝐵𝑃𝑙𝑎𝑡𝑒 = 𝐴𝐵𝑊𝑛𝑒𝑡

�

5) Increase all dimensions to easy numbers (multiples of 5 or 10 mm) 6) Compute Bending Length 𝑛 = 𝐵𝑃𝑙𝑎𝑡𝑒

2− 𝑘, 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 𝑀𝑓 = 𝐵𝑒𝑎𝑟𝑖𝑛𝑔 𝑆𝑡𝑟𝑒𝑠𝑠 ∙ 𝑛2

2

8) Compute 𝑀𝑟 = 𝑄𝑡2 where t is the plate thickness.

Note that Q = 𝜑𝐹𝑦

4� 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 𝑡 ≥ 𝐵𝑃𝑙𝑎𝑡𝑒− 𝐵𝑒𝑎𝑚 𝑊𝑖𝑑𝑡ℎ

10

Base Plate Design: Case 1) No Eccentricities: For I-beams: 1) Find maximum bearing stress:

𝑀𝑎𝑥 𝑆𝑡𝑟𝑒𝑠𝑠 = 0.55𝑓𝑐′ 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


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 = 𝐶−0.95𝑑2

, n = 𝐵−0.8𝑏2

and use the larger of the two in order to calculate Mf

6) Calculate BSA = 𝑅𝑥𝑛𝐴𝑃𝐿

�

7) Find 𝑀𝑓 = 𝐵𝑆𝐴 ∙ (𝑛 𝑜𝑟 𝑚)2

2




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

𝑡 ≥ 𝑚5

, 𝑛5

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: 𝑀𝑎𝑥 𝑆𝑡𝑟𝑒𝑠𝑠 = 0.51𝑓𝑐

′ 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


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 = �𝐴𝐵 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 = 𝐶𝑜𝑙𝑢𝑚𝑛 𝑊𝑖𝑑𝑡ℎ + 2�𝐷𝑟𝑜𝑢𝑛𝑑𝑒𝑑 𝑢𝑝 + 1.5𝑑𝑎𝑛𝑐ℎ𝑜𝑟 𝑟𝑜𝑑� (2 anchor rods assumed), again round 1.5danchor rod to the nearest 5mm.

4) Compute Width of the plate Wplate = 𝐶𝑜𝑙𝑢𝑚𝑛 𝑊𝑖𝑑𝑡ℎ + 25, round down to nearest 5mm.

5) Find m = 𝐿𝑝𝑙𝑎𝑡𝑒−𝐶𝑜𝑙𝑢𝑚𝑛 𝑊𝑖𝑑𝑡ℎ+𝐻𝑆𝑆 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠

2

6) Compute reduced Bearing Stress Br = 𝑅𝑥𝑛𝑊𝑝𝑙𝑎𝑡𝑒𝐿𝑝𝑙𝑎𝑡𝑒

7) 𝑀𝑓 = 𝐵𝑟 ∙ (𝑚)2

2 , 𝑀𝑟 = 𝑄𝑡2

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

𝑡 ≥ 𝑚5

Eccentric Design:

1) Determine which case you are using :

e = 𝑀𝑓 𝑖𝑛 𝑐𝑜𝑙𝑢𝑚𝑛

𝐶𝑓 𝑖𝑛 𝑐𝑜𝑙𝑢𝑚𝑛

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 = B•(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 = 𝐶−0.95𝑑

2 , n = 𝐵−0.8𝑏

2

4) Calculate BSA = 𝑅𝑥𝑛𝐴𝑃𝐿

�

5) Find 𝑀𝑓 = 𝐵𝑆𝐴 ∙ (𝑛 𝑜𝑟 𝑚)2

2




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

𝑡 ≥ 𝑚5

, 𝑛5

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 �𝐶

2− 𝑒�

4) Calculate m = 𝐶−0.95𝑑2

5) Select f From table below: (0.85*0.6*fc’)

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 = (𝑓)(𝑎−𝑚)

𝑎

7) 𝑡2= 1𝑄

�𝑓2𝑚2

2+ (𝑓 − 𝑓2) 𝑚2

3� ∙ 10−3



Fy 300 350 400 Phi Fy/4 = Q 67.5 78.75 90

8) Check that 𝑡 ≥ 𝑚 𝑜𝑟 𝑛

5 take larger of m and n.

Case 3:

1) Anchor Rods needed, take f From table below: (0.85*0.6*fc’)

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.

� 𝑀𝑟𝑜𝑑 = 𝐶𝑓(𝑒 + 𝑡𝑑) − 𝑓𝐵𝑝𝑙𝑎𝑡𝑒𝑎

2000 �𝐶2 + 𝑡𝑑 −

𝑎3� = 0

Where all lengths are in mm So compute terms a1 b1 and c1: 𝑎1 = 𝑓𝐵𝑝𝑙𝑎𝑡𝑒

6000 , 𝑏1 = − 𝑓𝐵𝑝𝑙𝑎𝑡𝑒

2000�𝐶

2+ 𝑡𝑑� , 𝑐1 = 𝐶𝑓(𝑒 + 𝑡𝑑)

So 𝑎 = −𝑏1±�𝑏1

2− 4𝑎1 𝑐1

2𝑎1 , 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):

𝑇 = 0.526(𝐷 − 0.938𝑃)2𝐹𝑢 , 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 = 0.45𝑑𝑟𝑜𝑑𝑡2𝐹𝑦 = 135𝑑𝑟𝑜𝑑𝑡2 𝑓𝑜𝑟 300𝑊 𝑠𝑡𝑒𝑒𝑙 10) First Check for t: Mr = Mf and solve for tmin . If Mr and Mf are not requested

explicitly:

𝑡𝑚𝑖𝑛 = �𝑇

0.45𝐹𝑦= � 𝑇

135 for Fy = 300W grade rod.

11) Second check for t:

𝑇0.45𝐹𝑦𝑡2 + 𝑇

1.188𝐹𝑦𝑑𝑟𝑜𝑑𝑡 = 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 = 𝐶−0.95𝑑

2

13) Calculate Mf2

𝑀𝑓2 = 0.5𝑓𝑎 �𝑚 − 𝑎

3� 𝐵 × 10−3 all values in mm and MPa, answer in kNm.

14) Use Mr to find t. so

𝑡2 = 2.222𝑓𝑎 1

𝐹𝑦�𝑚 − 𝑎

3� × 10−3 = 0.0047 �𝑚 − 𝑎

3� × 10−3 for 300W grade bar.

15) Deflection Check: make sure 𝑡 ≥ 𝑚

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 =

𝐶𝑓

𝑓× 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 𝐴𝑚𝑖𝑛 = 4𝑚(𝑏 + 2𝑚) + 2𝑚�𝑑 − 𝑡𝑓𝑙𝑎𝑛𝑔𝑒 − 2𝑚� or

𝐴𝑚𝑖𝑛 = 2𝑚�2𝑏 + 𝑑 − 𝑡𝑓𝑙𝑎𝑛𝑔𝑒 + 2𝑚�

3) Solve for m

4) Find 𝑡 = �2𝐶𝑓𝑚2×103

0.9𝐴𝑚𝑖𝑛𝐹𝑦𝑚𝑚

5) Plate Dimensions: 2m+0.95d x 2m+0.8b, round up to nice numbers. 6) Check deflection:

𝑡 ≥ 𝑚5

.

Bolted Connections

Factored Tensile Resistance

Separation Load

To calculate Ap: 1) Get bolt spacing in terms of bolt diameters: Spacing = gauge/db = Coeff•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 – 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 = Coeff•db

2) Ap = 𝜋𝑆𝑝𝑎𝑐𝑖𝑛𝑔2

4 - 𝜋𝑑𝑏

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 – number of shear planes

Slip Critical

Eccentric Loads: 1) Find Centroid: xbar=

∑ 𝐴𝑏∙𝑥𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝐵𝑜𝑙𝑡𝑠∙𝐴𝑏

ybar=∑ 𝐴𝑏∙𝑦

𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝐵𝑜𝑙𝑡𝑠∙𝐴𝑏,

Assuming Bolts are of the same size. 2) Compute ra for each bolt: ra

2= (x - xbar)2 + (y - ybar)2 and ∑ 𝑟𝑎2

3) and then the force on each bolt: Rax=

𝑉∙𝑒∙(𝑥−𝑥𝑏𝑎𝑟)∑ 𝑟𝑎

2 and Ray=𝑉∙𝑒∙(𝑦−𝑦𝑏𝑎𝑟)

∑ 𝑟𝑎2

4) Find the total resultant shear due to direct shear and moment induced shear

Total Shear = VT =�(𝑅𝑎𝑥 + 𝑉𝑥)2 + �𝑅𝑎𝑦 + 𝑉𝑦�2

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 • 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. è ie find Mu then find Mr’ 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/(4500•(fc’)0.5) 2) Find b, effective slab width.

· Slab on both sides of steel beam (b is less of): i) 0.25 • beam span ii) Avg. C-C spacing of the steel beams

· Slab on only one side of steel beam (b is lesser of): i) steel beam width + 0.1 • beam span ii) steel beam width + 0.5 • C-C steel beam spacing

3) Locate NA

find a (mm) = 0.9𝐴𝑠𝐹𝑦

0.51𝑓𝑐′𝑏

(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 ≥ 100%, stay in full connectivity otherwise move to case 3.

There are 3 cases: 1 – full shear connectivity with NA in concrete 2 – full shear connectivity with NA in steel 3 – partial shear connectivity Case 1 – Full shear connectivity with NA in concrete

i) find a = 0.9𝐴𝑠𝐹𝑦

0.51𝑓𝑐′𝑏

ii) find connectivity Qr min that gives benchmark for full connectivity:

lesser of: 0.9 AsFy x 10-3 OR 0.51 𝑓𝑐′ b a x 10-3 (mm and MPa)

iii) compute qrs

x 10-3 (mm and MPa) gives qrs (kN)

Ec = 4500�𝑓𝑐

′ 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 e’ = d/2 + tslab – a d is steel beam height Mrc (kNm) = 0.9AsFye’ x 10-6 (mm and Mpa) Case 2 – Full shear connectivity with NA in steel a > tslab i) Find Cr’ = 0.51 fc’ b t where t is the concrete thickness, b is the effective width (above).

ii) Find Cr = 0.9𝐴𝑠𝐹𝑦∙10−3−𝐶𝑟

′

2

iii) Find Steel Compression Area Asc = 1000𝐶𝑟

0.9𝐹𝑦 with Cr in KN, Fy in MPa.

iv) Check if NA is in flange: 𝑑𝑓 = 𝐴𝑠𝑐

𝑏𝑓 ≤ 𝑡𝑓, where w is the flange width, tf is the flange thickness.

if not ok, NA is in the web, so 𝑑𝑤 = 𝐴𝑠𝑐−𝑏𝑓∙𝑡𝑓

𝑡𝑤 where tw is the web width.

NA is at 𝑑𝑓 from the top of the steel beam. v) Find Centroids of 3 sections. Case a) NA in flange (from base).

𝑦�𝑇𝑒𝑛𝑠𝑖𝑜𝑛 𝑆𝑡𝑒𝑒𝑙 =

𝐴𝑠𝑑2 − 𝑑𝑓𝑏𝑓 �𝑑 −

𝑑𝑓2 �

𝐴𝑠 − 𝑑𝑓𝑏𝑓

so 𝑒 = 𝑑 − 𝑑𝑓

2− 𝑦�𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑠𝑡𝑒𝑒𝑙

and 𝑒′ = 𝑑 + 𝑡𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 𝑠𝑙𝑎𝑏2

− 𝑦�𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑠𝑡𝑒𝑒𝑙 case b) NA in web


𝐴𝑠𝑑2 − 𝑡𝑓𝑏𝑓 �𝑑 − 𝑏𝑓

2� � − 𝑡𝑤𝑑𝑤 �𝑑 − 𝑡𝑓 − 𝑑𝑤2� �

𝐴𝑠 − 𝑡𝑓𝑏𝑓 − 𝑡𝑤𝑑𝑤

𝑦�𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑆𝑡𝑒𝑒𝑙 = 𝑡𝑓𝑏𝑓 �𝑑 − 𝑏𝑓

2� � + 𝑡𝑤𝑑𝑤 �𝑑 − 𝑡𝑓 − 𝑑𝑤2� �

𝑡𝑓𝑏𝑓 + 𝑡𝑤𝑑𝑤

so 𝑒 = 𝑑 − 𝑦�𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑆𝑡𝑒𝑒𝑙 − 𝑦�𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑠𝑡𝑒𝑒𝑙 and 𝑒′ = 𝑑 + 𝑡𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 𝑠𝑙𝑎𝑏


vi) Compute Mrcomp = Cre + Cr’e’ vii) find connectivity Qr min that gives benchmark for full connectivity:

lesser of: 0.9 AsFy OR 0.51 𝑓𝑐′ 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 – Partial connectivity (NA always in the steel)

i) Find qr:

Ec = 4500�𝑓𝑐

′ ii) find Qrmin lesser of: 0.9 AsFy OR 0.51 𝑓𝑐

′ b tslab

iii) Find Percent Connectivity = n•qrs / Qr min > 40% if flex controls or > 25% if defl. controls

iv) Find Cr’ = n•qrs where n is the number of studs in effective length.

v) Find a = 𝐶𝑟′

0.51𝑓𝑐′𝑏

x 103 (kN, MPa and mm) where b is the effective slab width.

vi) Find Cr = 0.9𝐴𝑠𝐹𝑦∙10−3−𝐶𝑟

′

2

vii) Find Steel Compression Area

Asc = 𝐶𝑟

0.9𝐹𝑦∙ 103 with Cr in kN, Fy in MPa.

viii) Check if NA is in flange: 𝑑𝑓 = 𝐴𝑠𝑐

𝑏𝑓 ≤ 𝑡𝑓, 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: 𝑑𝑤 = 𝐴𝑠𝑐−𝑏𝑓∙𝑡𝑓

𝑡𝑤 where tw is the web width.

NA is at 𝑑𝑓 from the top of the steel beam. ix) Find Centroids of 3 sections. Case a) NA in flange (from base of steel):


𝐴𝑠𝑑2 − 𝑑𝑓𝑏𝑓 �𝑑 −

𝑑𝑓2 �

𝐴𝑠 − 𝑑𝑓𝑏𝑓

so 𝑒 = 𝑑 − 𝑑𝑓


and 𝑒′ = 𝑑 + 𝑡𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 𝑠𝑙𝑎𝑏2

− 𝑦�𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑠𝑡𝑒𝑒𝑙 case b) NA in web (from base of steel):


𝐴𝑠𝑑2 − 𝑡𝑓𝑏𝑓 �𝑑 − 𝑏𝑓

2� � − 𝑡𝑤𝑑𝑤 �𝑑 − 𝑡𝑓 − 𝑑𝑤2� �

𝐴𝑠 − 𝑡𝑓𝑏𝑓 − 𝑡𝑤𝑑𝑤

𝑦�𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑆𝑡𝑒𝑒𝑙 = 𝑡𝑓𝑏𝑓 �𝑑 − 𝑏𝑓

2� � + 𝑡𝑤𝑑𝑤 �𝑑 − 𝑡𝑓 − 𝑑𝑤2� �

𝑡𝑓𝑏𝑓 + 𝑡𝑤𝑑𝑤

so 𝑒 = 𝑑 − 𝑦�𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑆𝑡𝑒𝑒𝑙 − 𝑦�𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑠𝑡𝑒𝑒𝑙 and 𝑒′ = 𝑑 + 𝑡𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 𝑠𝑙𝑎𝑏


x) Compute Mrcomp = [Cre + Cr’e’] x 10-3 (kN and mm) gives Mr comp in kNm

Behaviour Under Specified Loads (Deflection):

Case 1) Assume NA is in the steel:

𝑦� =

𝐴𝑠𝑑2 +𝑡𝑠𝑙𝑎𝑏𝑏

𝑛 �𝑑+𝑡𝑠𝑙𝑎𝑏2� �

𝐴𝑠+𝑡𝑠𝑙𝑎𝑏𝑏𝑛

≤ 𝑑 Otherwise Use Case 2 below

Find Ix-composite :

𝐼𝑥𝑐 = 𝐼𝑠𝑡𝑒𝑒𝑙 + 𝐴𝑠 �𝑦� −𝑑2

�2

+𝑏𝑡𝑠𝑙𝑎𝑏

3

12𝑛+

𝑏𝑡𝑠𝑙𝑎𝑏

𝑛�

𝑡𝑠𝑙𝑎𝑏

2+ 𝑑 − 𝑦��

2

Look up Ss in the section property tables, and calculate St = 𝐼𝑥𝑐𝑦�

.

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 doesn’t yield:

Case 2) if NA is in the concrete

h = −𝐴𝑠±�𝐴𝑠

2+4∙ 𝑏2𝑛𝐴𝑠�𝑑

2+𝑡�𝑏𝑛

𝑦� =d + t – 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 :

𝐼𝑥𝑐 = 𝐼𝑠𝑡𝑒𝑒𝑙 + 𝐴𝑠 �𝑦� −𝑑2

�2

+𝑏ℎ3

12𝑛+

𝑏ℎ𝑛

�ℎ2

�2

Look up Ss in the section property tables, and calculate St = 𝐼𝑥𝑐𝑦�

.

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 doesn’t yield:

Deflection Checks:

∆2 uses per permanent live load.

iii) ∆3 is the same as ∆2 BUT uses short term live load (no dead loads). Unless specified, use live load = 0.5•total live load iv) 𝑦�2 = d + tslab/2 - 𝑦�

∆s = 2𝐴𝑐𝐿2𝑦�2

𝐼𝑥𝑐•10-6

Total Deflection

If the steel beam is simply supported, a truer value of the deflection is:

Compression Members

Step 1 – Check beam class

The member must be class 3 or better:

≤ remember to bring root Fy to the other side and check < 200

≤

Step 2 – Slenderness

i) Check slenderness ratio

≤ 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 Antoine’s forgotten check (cross-sect strength check): 𝑇𝑓

𝑇𝑟+ 𝑀𝑓

𝑀𝑟 ≤ 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:

ℎ = 540 �𝑀𝑓

𝐹𝑦�

13�, 𝐴𝐹𝐿 = 𝑀𝑓

𝐹𝑦ℎ, and 𝐴𝑤 = 𝑉𝑓

0.594𝐹𝑦 → 𝑤 = 𝑉𝑓

0.594ℎ𝐹𝑦

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: ℎ�𝐹𝑦

𝑤≤ 1900, class 3 if yes, class 4 if no

(1700 for classes 1 and 2)

Flange: 𝑏

2� �𝐹𝑦

𝑡≤ 200, if yes class 3, else must increase thickness.

(170 for classes 1 and 2) 2) Compute moment of inertia:

𝐼 = 2 �𝑏𝑡3

12 + 𝑏𝑡 �ℎ2 +

𝑡2�

2

� +𝑤ℎ3

12

Where terms are defined in previous figure, b is the flange width, and t is the flange thickness.

3) Find Section modulus: 𝑆 = 𝐼

𝑦� or 𝑍 = 1

4[𝑏𝑑2 − (𝑏 − 𝑤)(𝑑 − 2𝑡)2] (class 1, 2).

4) Compute Mr Class 1 or 2: 𝑀𝑟 = 0.9 𝑆 𝐹𝑦 Class 3: 𝑀𝑟 = 0.9 𝑍 𝐹𝑦 Class 4 web:

𝑀𝑟4 = 0.9 𝑆 𝐹𝑦 �1 − 0.005 𝐴𝑤𝑒𝑏

𝐴𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑓𝑙𝑎𝑛𝑔𝑒�ℎ

𝑤− 1900

�𝐹𝑦��

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 = ℎ�𝐹𝑦

𝑤�𝑘𝑣

And consult the appropriate case: Case 1: Q>621

1) Calculate Vr = 162000𝑤3𝑘𝑣ℎ

x 10-3 2) If Vr > Vf then no stiffeners needed.

Case 2: 439<Q<621

1) Calculate Vr = 261𝑤2�𝑘𝑣𝐹𝑦 x 10-3 2) If Vr > Vf then no stiffeners needed.

Case 3: Q<439

1) Calculate Vr = 0.594ℎ𝑤𝐹𝑦 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 =

ℎ�𝐹𝑦

𝑤�𝑘𝑣

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.


Case 3: 439<Q<502

1) Calculate Fs


Case 4: Q<439

1) Fs = 0.66Fy same as before, so Calculate Vr = 0.594ℎ𝑤𝐹𝑦 x 10-3

2) If Vr > Vf then OK.

Stiffener Design:

1) Write down a, Fy , h , w. 2) Calculate C:

3) Calculate Y =

𝐹𝑦−𝑔𝑖𝑟𝑑𝑒𝑟

𝐹𝑦−𝑠𝑡𝑖𝑓𝑓𝑒𝑛𝑒𝑟 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:

𝐼 = 2 �𝑏𝑡3

12+ 𝑏𝑡

4(𝑏 + 𝑤)2� for two stiffeners

And check: 𝐼 ≥ � ℎ50

�4

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. 𝑉𝑟

𝐿= 0.449𝐷𝐹𝑢

2. 𝑉𝑟𝐿

= 0.317𝐷𝑋𝑢

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 < 330𝑡𝑠𝑡𝑖𝑓𝑓

�𝐹𝑦 for non-staggered welds

v) Check shear transfer resistance over weld height, VRH:

= 𝑉𝑅𝐿∙𝐿𝑂/𝐶

(N/mm) > VFL

where O/C = clear spacing + L

vi) Check clear spacing requirements: Non-Staggered Welds Staggered Welds

< 300 mm < 450

< 330𝑡𝑠𝑡𝑖𝑓𝑓

�𝐹𝑦 <

525𝑡𝑠𝑡𝑖𝑓𝑓

�𝐹𝑦

< 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 ℎ𝑤

≥ 1100

�𝐹𝑦 , or Br < Rxnf

2) Check Crippling capacity:

i) 𝐵𝑟 = 0.75𝑤𝐹𝑦(𝑁 + 4𝑡) ii) 𝐵𝑟 = 0.45𝑤2�𝐸𝐹𝑦 Where t is the flange thickness, w is the web thickness.

3) Select plates: use 2 plates both with 𝑏

𝑡≤ 200

�𝐹𝑦

4) Check Bearing Rest Capacity: 𝐴𝐵 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑙𝑎𝑡𝑒𝑠 × 𝑡𝑝𝑙𝑎𝑡𝑒 × �𝑏𝑝𝑙𝑎𝑡𝑒 − 25𝑚𝑚 𝑐𝑜𝑝𝑒�

And 𝐵𝑟 = 1.35 𝐹𝑦𝐴𝐵 × 10−3 𝑖𝑛 𝑘𝑁

5) Check compression resistance Calculate A = 2(tplatebplate) + 12w2

𝐼𝑥 = 𝑡𝑝𝑙�2𝑏𝑝𝑙 + 𝑤�

3

12 +�12𝑤 − 𝑡𝑝𝑙�𝑤3

12

r = �𝐼𝑥𝐴

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 ℎ

𝑤≥ 1100

�𝐹𝑦 , or Br < Rxnf

2) Check Crippling capacity:

iii) 𝐵𝑟 = 0.80𝑤𝐹𝑦(𝑁 + 10𝑡) iv) 𝐵𝑟 = 1.16𝑤2�𝐸𝐹𝑦 Where t is the flange thickness, w is the web thickness.

3) Select plates: use 2 plates both with 𝑏

𝑡≤ 200

�𝐹𝑦

4) Check Bearing Rest Capacity: 𝐴𝐵 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑙𝑎𝑡𝑒𝑠 × 𝑡𝑝𝑙𝑎𝑡𝑒 × �𝑏𝑝𝑙𝑎𝑡𝑒 − 25𝑚𝑚 𝑐𝑜𝑝𝑒�

And 𝐵𝑟 = 1.35 𝐹𝑦𝐴𝐵 × 10−3 𝑖𝑛 𝑘𝑁

5) Check compression resistance Calculate A = 2(tplatebplate) + 25w2

𝐼𝑥 = 𝑡𝑝𝑙�2𝑏𝑝𝑙 + 𝑤�

3

12 +�25𝑤 − 𝑡𝑝𝑙�𝑤3

12

r = �𝐼𝑥𝐴

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 = 𝑉𝑓∙(𝐴𝐴)(𝐴𝐴 𝑡𝑜 𝑁𝐴)

𝐼𝑏𝑒𝑎𝑚 (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:

𝑉𝑟𝐿

= 0.449𝐷𝐹𝑢

𝑉𝑟𝐿

= 0.317𝐷𝑋𝑢

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 = 2𝑉𝑅𝐿∙𝐿

𝑂/𝐶 (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 à = (h/10)0.2 > 0.9 or take 0.9 Rough Terrain à = 0.7(h/12)0.3 > 0.7 or take 0.7 2) Find z, lesser of: z = 0.1•least horizontal dimension = 0.4•height but not less than: 0.04•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 = 𝑠𝑝𝑎𝑛

2𝑃1𝐸

ℎ2

Case 2) Force =�𝑦 − 𝑠𝑝𝑎𝑛2

� 𝑃1𝐸ℎ2

+ �3 𝑠𝑝𝑎𝑛2

− 𝑦� 𝑃1ℎ2

Case 3) Force =𝑆𝑝𝑎𝑛 ∙ 𝑃1ℎ2

Case 4) Force = 𝑠𝑝𝑎𝑛2

𝑃1ℎ2

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: 1.25D• width•length of total building/n Snow factor 1.5: 1.5S•width•length/n Snow factor 0.5: 0.5S• width•length/n Live factor 1.5: 1.5L• width•length/n Live factor 0.5: 0.5L• width•length/n Wind factor 1.4: 1.4•Wmax Wind factor 0.4: 0.4•Wmax 6) Now we look at each load combination

Begin with: 1.25D + 1.4W + 0.5S, and calculate: i) ∑ 𝐶𝑓 = 𝑠𝑢𝑚 𝑜𝑓 𝑓𝑎𝑐𝑡𝑜𝑟𝑒𝑑 𝑔𝑟𝑎𝑣𝑖𝑡𝑦 𝑙𝑜𝑎𝑑𝑠 𝑖𝑛 𝑙𝑜𝑎𝑑 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑏𝑒𝑖𝑛𝑔 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑒𝑑 ii) Notional Load = 0.005 ∙ ∑ 𝐶𝑓 iii) Lateral Load ∑ 𝑉𝑓 = 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 = �𝐵𝑎𝑦 𝑤𝑖𝑑𝑡ℎ2 + 𝐵𝑎𝑦 𝐻𝑒𝑖𝑔ℎ𝑡2 ii) Tft = ∑ 𝑉𝑓• 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 = ∑ 𝑉𝑓• Full Brace Length / Full Bay width 8) Calculate P-delta amplification factors:

i) ∆𝐹 (𝑚𝑚) = 𝑇𝑓𝑡𝐿2

𝐸𝐴∙𝐹𝑢𝑙𝑙 𝐵𝑎𝑦 𝑊𝑖𝑑𝑡ℎ × 103 (kN, MPa, mm)

ii) 𝑈2 = 1

1 − ∑ 𝐶𝑓∆𝐹

∑ 𝑉𝑓∙𝐹𝑢𝑙𝑙 𝐵𝑎𝑦 𝐻𝑒𝑖𝑔ℎ𝑡

< 1.4

9) Check amplified brace force (ABF) and amplified deflection (AD): i) ABF = 𝑈2𝑇𝑓𝑡 < Tr ii) AD = U2∆𝐹 < 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 – 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:

�̅� =∑ 𝑊𝑒𝑙𝑑 𝐿𝑒𝑛𝑔𝑡ℎ ∙ 𝑥∑ 𝑊𝑒𝑙𝑑 𝐿𝑒𝑛𝑔𝑡ℎ𝑠

𝑦� =∑ 𝑊𝑒𝑙𝑑 𝐿𝑒𝑛𝑔𝑡ℎ ∙ 𝑦∑ 𝑊𝑒𝑙𝑑 𝐿𝑒𝑛𝑔𝑡ℎ𝑠

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 = P•sinθ, Py = P•cosθ..., M=P•e 5) Compute Ix’ and Iy’ ; and J’= Ix’+ Iy’

I’x-single weld = 𝑏∙ℎ3

12+ 𝑏 ∙ ℎ ∙ (𝑥 − �̅�)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: 𝑞𝑥( 𝑘𝑁

𝑚𝑚) = 𝑃∙𝑒∙(𝑦−𝑦�)

𝐽 𝑞𝑦( 𝑘𝑁

𝑚𝑚) = 𝑃∙𝑒∙(𝑥−�̅�)

𝐽

7) Finally take qx+Px and qy+Py and compute the resultant. The edge point with the highest V governs:

Vmax =∑ 𝐿𝑡𝑜𝑡𝑎𝑙 𝑤𝑒𝑙𝑑𝑠 ��𝑞𝑥 + 𝑃𝑥∑ 𝐿𝑡𝑜𝑡𝑎𝑙 𝑤𝑒𝑙𝑑𝑠

�2

+ �𝑞𝑦 + 𝑃𝑦

∑ 𝐿𝑡𝑜𝑡𝑎𝑙 𝑤𝑒𝑙𝑑𝑠�

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:

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Steel Design Notes CSA