CIV413 STRUCTURAL STEEL DESIGN

FLEXURAL
MEMBERS

Beams
1
Prof. Osama Ahmed Mohamed, Ph.D., P.E., M.ASCE
Goal

When the flexural member is 1) compact, 2)

continuously or appropriately laterally
supported:
Elastic vs. Plastic Moments

Notice Moment that will

how beam
is oriented cause initiation of
to bend yielding:
about
strong
axis
Yielding started

Elastic section
modulus:
Plastic Moment Plastic moment capacity can be achieved if the
beam doesn’t buckle due to compression

The plastic moment

capacity is:
A
M p = Fy Ac a = Fy At a = Fy a = Fy Z
2

Plastic section
modulus:

Plastic neutral axis (PNA)

divides a cross section into A
two equal areas Z = Ac a = At a = a
2
4
EXAMPLES

5
Example 5.1

Beam oriented to bend

about x axis (strong axis).
This produces higher plastic
moment capacity than
bending about the y-axis
(weak axis)

A
M p = Fy Ac a = Fy At a = Fy a = Fy Z 6
2
 Example 5.1: Calculate Moment of Inertia

Top
Bottom

6 0

Theorem of Parallel Axes to

find moment of inertia
Elastic neutral axis at 7
centroid of shape
Example 5.1: Yield Moment

8
Example 5.1: Locate centroid of upper half of the w-shape (T-shape)
A
M p = Fy a = Fy Z
2

=12 inch

9
 Example 5.1: calculate plastic moment capacity

Plastic moment
capacity is the
maximum moment the
beam can carry if it
doesn’t buckle locally
or globally

10
Example 5.2

Moment arm

A
M p = Fy a = Fy Z
2
11
Example 5.2: Final area and centroid of WT

A
M p = Fy a = Fy Z
2

Moment arm “a”

12
Example: Find “a”, plastic section modulus, and plastic moment

13
 Flexural Members in Steel Manual
• Beam Members:
• Chapter F: Flexural Strength
• Chapter G: Shear Strength
• Chapter I: Composite Member Strength Not covered in this first steel course

• Part 3: Design Charts and Tables

• Chapter B:Local Buckling Classification
Chapter F:
Flexural Strength

15
Flexural Strength

Φb = 0.90
 Flexural Strength

Specification assumes that the following

failure modes have minimal interaction and
can be checked independently from each
other:

Flexural capacity is the same as the

• Lateral Torsional Buckling(LTB) plastic capacity if LTB is not
• Flange Local Buckling (FLB) possible and shape is compact

• Shear
 Flexural Strength

• Local Buckling Criteria: Table B4.1

• Flexure Strength: Chapter F
• Shear Strength: Chapter G

When beam is compact for

flexure, local buckling is not a
problem, and plastic capacity
may be achieved
 Flexural Strength

Local Buckling
λ ≤ λp “compact”
Mp is reached and maintained before local buckling.
φMn = φMp

λp ≤ λ ≤ λr “non-compact”
Local buckling occurs in the inelastic range.
φ0.7My ≤ φMn < φMp

λ > λr “slender element”

Local buckling occurs in the elastic range.
φMn < φ0.7My 20
 Flexural Strength

Local Buckling Criteria

Slenderness of the flange and web, λ, are used as criteria to
determine whether buckling would control in the elastic or
inelastic range, otherwise the plastic moment can be obtained before
local buckling occurs.

Criteria λp and λr are based on plate buckling theory.

For W-Shapes
Determine whether
E E
shape is compact, non- FLB, λ = bf /2tf λpf = 0.38 , λrf = 1.0
compact, or slender for Fy Fy
flexure
E E
WLB, λ = h/tw λpw = 3.76 , λrw = 5.70
Fy Fy
 Local Buckling Criteria
Doubly Symmetric I-Shaped Members
Equation F3-1 for FLB:
  λ − λ pf 
M n =  M p − ( M p − 0.7 Fy S x )   

Mp = FyZx   λrf − λ pf  

Mr = 0.7FySx

Mn 0.9 Ekc S x
Equation F3-2 for FLB: M n =
λ2
λp λr λ
Note: WLB not shown. See Spec. sections F4 and F5.
22
 Local Buckling Criteria
Doubly Symmetric I-Shaped Members
Equation
Rolled W-shape F3-1sections
for FLB:are
  λ − λ pf  
dimensioned ( M p −that
M n =  M p −such S x ) webs are 
0.7 Fythe
 
Mp = FyZx compactand flanges are compact  λrf − λ pfin
 
most cases. Therefore, the full plastic
Mr = 0.7FySx moment usually can be obtained prior to
local buckling occurring.

Mn 0.9 Ekc S x
Equation F3-2 for FLB: M n =
λ2
λp λr λ
Note: WLB not shown. See Spec. sections F4 and F5.
23
Concrete floors with shear connectors for
composite beam actions
• https://youtu.be/q0HPQKtGcj8
• https://youtu.be/73Bwu5m_69A
 Example 5.3

Calculate flexural strength based on local

buckling criteria. Lateral Torsional Buckling will
not occur 25
 Example 5.3 solution: Local Buckling Criteria

Flange slenderness ratio

Web
slenderness
ratio

26
Example 5.3 solution continued

27
Example 5.3 solution continued

28
Example 5.3 solution continued

29
 Flexural Strength
When members are compact:
Only consider LTB as a potential failure mode
prior to reaching the plastic moment.

LTB depends on unbraced length, Lb, and can occur

in the elastic or inelastic range.

If the section is also fully braced against LTB,

Mn = Mp = FyZx Equation F2-1
 Lb Lateral Brace
Lateral Torsional Buckling
Strength for Compact
X X W-Shape Sections

M = Constant (Cb=1)

Mp Equation F2-2

Mr Equation F2-3 and F2-4

Mn Inelastic
Plastic LTB LTB Elastic LTB

Lp Lr Lb
If Lb ≤ Lp,
Mn = Mp

If Lp < Lb ≤ Lr,   Lb − L p 
M= n (
Cb  M p − M p − .7 Fy S x )  L − L   ≤ M p Equation F2-2
  r p  
Note that this is a straight line.

If Lb > Lr,
Mn = FcrSx ≤ Mp Equation F2-3
2
Cb π E
2
Jc  Lb 
Fcr = 1 + 0.078  

Where  Lb 
2
S x h0  rts  Equation F2-4
 r 
 ts 

Assume Cb=1 for now

 When LTB is a possible failure mode:
Mp = FyZx Equation F2-1
Mr = 0.7FySx
E
Lp = 1.76ry Equation F2-5
Fy
2
Lr = E Jc  .7 Fy S x ho  Equation F2-6
1.95rts 1 + 1 + 6.76  
0.7 Fy S x ho  E Jc 
rts2 = I y Cw Equation F2-7
Sx
ry =
Iy
A
For W shapes
c = 1 (Equation F2-8a)
ho = distance between flange centroids

Values of φMp, φMr, Lp and Lr are tabulated in Table 3-

2 (pages 3-11 to 3-19)
 Flexural Strength

Plots of φMn versus Lb for Cb = 1.0 are tabulated,

Table 3-10, pp. 3-96 to 3-131

Results are included only for:

• W sections typical for beams
• Fy = 50 ksi
• Cb = 1
Flexural Strength

To compute Mn for any moment diagram,

Mn = Cb(Mn(Cb1)) ≤ Mp
φMn = Cb(φMn(Cb1)) ≤ φMp
(Mn(Cb1)) = Mn, assuming Cb = 1

Cb, Equation F1-1

 Flexural Strength

X X Shown is the section of

MA Mmax MC
the moment diagram
MB between lateral braces.
Lb Lb Lb Lb
4 4 4 4

Mmax = absolute value of maximum moment in unbraced section

MA = absolute value of moment at quarter point of unbraced section
MB = absolute value of moment at centerline of unbraced section
MC = absolute value of moment at three-quarter point of unbraced
section
Rm = 1.0 for doubly symmetric members or single curvature
Example Flexural Strength
Consider a simple beam with differing lateral brace locations.

M
12.5M 12.5
= = = 1.31
( ) ( )
Cb
X X 2.5M + 3 M + 4 M + 3 M 9.5
2 2

M
12.5M 12.5
= = = 1.67
( ) ( ) ( )
X X X Cb
2.5M + 3 M + 4 M + 3 3M 7.5
4 2 4

X – lateral brace location

Note that the moment diagram is unchanged
by lateral brace locations.
 Flexural Strength
M
Cb=1.0 Mmax
M X X
M/2
Cb=1.25

M Mmax/Cb
Cb=1.67
X X
M
Cb=2.3

M Cb approximates an equivalent beam of

constant moment.
 Flexural Strength

Mp

Mr

Mn Cb=1

Lp Lr Lb
Lateral Torsional Buckling
Strength for Compact W-Shape Sections
Effect of Cb
 Flexural Strength

Mp

Mr
Cb>1
Mn Cb=1

Lp Lr Lb
Lateral Torsional Buckling
Strength for Compact W-Shape Sections
Effect of Cb
 Flexural Strength

Limited by Mp
Mp

Mr
Cb>1
Mn Cb=1

Lp Lr Lb

Lateral Torsional Buckling Strength for Compact W-Shape Sections

Effect of Cb
Example 5.4

47
Example 5.4 solution

Check the limit state of

local yielding

48
Flange and web need to be
classified for local buckling
limit state as:
Compact

Non-compact

Slender
• If both the flange and web are
compact, the entire section as
classified as compact.
• If the flange, web or both are slender,
the entire section is slender.
Example 5.4 solution continued

50
Example 5.4: With unbraced Length = 20ft
• The limiting length, LP :

• Laterally unbraced length Lb > Lp

Must now calculate Lr
Example 5.4 solution continued

Lb

52
Example 5.4 solution continued

53
Example 5.4 solution continued

54
Example 5.4 solution continued

55
Example 5.4 solution continued

56
Example 5.5

12.5M max
Cb = Rm ≤ 3.0
2.5M max + 3M A + 4 M B + 3M C

57
Example 5.5 solution

58
Example 5.5 solution continued

59
Example 5.6

What does this mean from

design point of view?
61
 Example 5.6 solution: Check Local Buckling

Check the limit state

(failure mode) of
local yielding:
Table B4.1

62
Example 5.6 solution continued

Accounts for residual stresses

63
Example 5.6 solution: Check LTB

64
Example 5.6 solution continued

66
Example 5.6 solution continued

67
Example 5.6 solution continued

68
3-10 is a useful table
for finding design
flexural strength for
many sections.