68402: Structural Design of Buildings II
61420: Design of Steel Structures
62323: Architectural Structures II
Design of Beams for Flexure
Monther Dwaikat
Assistant Professor
Department of Building Engineering
An-Najah National University
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Slide # 1
Design of Beams for Flexure

Introduction

Moment Curvature Response

Sectional Properties

Serviceability Requirements (Deflections)

Compact, Non-compact and Slender Sections

Lateral Torsional Buckling

Design of Beams
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Slide # 2
Beams under Flexure

Members subjected principally
to transverse gravity loading
•
•
•
•
•
Girders (important floor beams,
wide spacing)
Joists (less important beams,
closely spaced)
Purlins (roof beams, spanning
between trusses)
Stringers (longitudinal bridge
beams)
Lintels (short beams above
window/door openings)
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Slide # 3
Design for Flexure

Limit states considered
• Yielding
• Lateral-Torsional Buckling
• Local Buckling
• Compact
• Non-compact
• Slender
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Slide # 4
Design for Flexure – LRFD Spec.

Commonly Used Sections:
• I – shaped members (singly- and doubly-symmetric)
• Square and Rectangular or round HSS
• Tees and Double Angles
• Rounds and Rectangular Bars
• Single Angles
Will not be covered
in this course
• Unsymmetrical Shapes
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Slide # 5
Section Force-Deformation
Response & Plastic Moment (MP)


A beam is a
structural member
that is subjected
primarily to
transverse loads and
negligible axial
loads.
The transverse loads
cause internal SF
and BM in the beams
as shown in Fig. 1
w
P
x
V(x)
M(x)
Fig. 1- SF & BM in a SS Beam
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Slide # 6
Section Force-Deformation
Response & Plastic Moment (MP)

These internal SF & BM cause longitudinal axial stresses
and shear stresses in the cross-section as shown in the Fig.
2
dF = s b dy

s
y
d
M(x)
V(x)
b

s
d / 2

F   s b dy
Curvature =  = 2/d
d / 2
(Planes remain plane)
d / 2
M   s b dy y
d / 2
Fig. 2 - Longitudinal axial stresses caused by internal BM
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Slide # 7
Section Force-Deformation
Response & Plastic Moment (MP)

Steel material follows a typical stress-strain behavior as
shown in Fig 3 below. E = 200 GPa
su
sy
s
y

u
Fig 3 - Typical steel stress-strain behavior.
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Slide # 8
Section Force-Deformation
Response & Plastic Moment (MP)

If the steel stress-strain curve is approximated as a bilinear
elasto-plastic curve with yield stress equal to sy, then the
section Moment - Curvature (M-) response for
monotonically increasing moment is given by Fig. 4.

In Fig. 4, My is the moment corresponding to first yield and
Mp is the plastic moment capacity of the cross-section.
•
•
The ratio of Mp to My - the shape factor f for the section.
For a rectangular section, f = 1.5. For a wide-flange section, f ≈ 1.1.
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Slide # 9
Moment-Curvature (NEW)
•
Beam curvature  is related to its strain and thus to the
applied moment

y

(1)
(2)
(3)

M


y
EI
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Slide # 10
(4)
Moment-Curvature (NEW)
•
When the section is within elastic range
M y M
F x

Ix
Sx
M  Sx Fx
Where S is the elastic section modulus
•
•
When the moment exceeds the yield moment My
M y  Sx Fy
a
M p  Z x Fy
A
Z x  ( )a
2
Then
Where Z is the plastic section modulus
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 1.1 S
Slide # 11
Section Force-Deformation
Response & Plastic Moment (MP)
Mp
C
B
A
sy
Section Moment, M
My
sy
sy
A
sy
sy
y
y
E
D
sy
2y
2y
B
sy
sy
sy
10y
5y
5y
sy
10y
C
D
E
Curvature, 
A: Extreme fiber reaches y B: Extreme fiber reaches 2y C: Extreme fiber reaches 5y
D: Extreme fiber reaches 10y E: Extreme fiber reaches infinite strain
Fig. 4 - M- response of a beam section
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Slide # 12
Section Force-Deformation
Response & Plastic Moment (MP)

Calculation of MP: Cross-section subjected to either +sy or - sy at
the plastic limit. See Figure 5 below.
sy
Plastic centroid.
A1
sy A 1
y1
A2
y2
sy A 2
tic centroid.
A1
(b) Stress
distribution
y1
A
\ A1  A 2 
A
2
Where , y1 
\ M  sy
sy
(a) General crosssy
section
F  s y A1  s
(c) Force distribution
sy A 1
y2
2
Figure
5. Plastic
centroid and MP for
sy A 2
general cross-section.
sy
F  s y A1  s y A 2  0
\ A1  A 2  A / 2
A
´ ( y1  y 2 )
2
Where , y1  centroid of A1
\ M  sy
y 2  centroid of A 2
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Slide # 13
y2 
Moment-Curvature
•
When the whole section is yielding a plastic hinge
will be formed
plastic hinge
•
•
Structural analysis by assuming collapsing mechanisms of
a structure is known as “Plastic analysis”
The plastic moment Mp is therefore the moment needed at
the section to form a plastic hinge
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Slide # 14
Section Force-Deformation
Response & Plastic Moment (MP)

The plastic centroid for a general cross-section
corresponds to the axis about which the total area is
equally divided, i.e., A1 = A2 = A/2
•
•
The plastic centroid is not the same as the elastic centroid or
center of gravity (c.g.) of the cross-section.
As shown below, the c.g. is defined as the axis about which A1y1 =
A2y2.
y2
A1, y1
y1
c.g. = elastic N.A. A , y
2
2
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About the c.g. A 1 y 1 = A 2 y2
Slide # 15
Section Force-Deformation
Response & Plastic Moment (MP)
•



For a cross-section with at-least one axis of symmetry, the
neutral axis corresponds to the centroidal axis in the elastic
range. However, at Mp, the neutral axis will correspond to the
plastic centroidal axis.
For a doubly symmetric cross-section, the elastic and
the plastic centroid lie at the same point.
Mp = sy x A/2 x (y1+y2)
As shown in Figure 5, y1 and y2 are the distance from
the plastic centroid to the centroid of area A1 and A2,
respectively.
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Slide # 16
Section Force-Deformation
Response & Plastic Moment (MP)

A/2 x (y1+y2) is called Z, the plastic section
modulus of the cross-section. Values for Z are
tabulated for various cross-sections in the
properties section of the LRFD manual.
• bMp = 0.90 Z Fy
b - strength reduction factor
Mp - plastic moment, which must be  1.5 My for homogenous
cross-sections
My - moment corresponding to onset of yielding at the extreme
fiber from an elastic stress distribution = Fy S for homogenous
cross-sections and = Fyf S for hybrid sections.
Z - plastic section modulus from the Properties section of the
AISC manual.
S - elastic section modulus, also from the Properties section of
the AISC manual.
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Slide # 17
Ex. 4.1 – Sectional Properties

Determine the elastic section modulus, S, plastic section
modulus, Z, yield moment, My, and the plastic moment MP,
of the cross-section shown below. What is the design
moment for the beam cross-section. Assume A992 steel.
300
12 mm
in.
F1
15 mm
0.75
in.
W t = 0.5
10 mm
in.
w
F2
400in.
mm
16
25
mm
1.0
in.
400
15 mm
in.
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Slide # 18
Ex. 4.1 – Sectional Properties


Ag = 300 x 15 + (400 - 15 - 25) x 10 + 400 x 25 = 18100 mm2
Af1 = 300 x 15 = 4500 mm2
Af2 = 400 x 25 = 10000 mm2
Aw = 10 x (400 - 15 - 25) = 3600 mm2
distance of elastic centroid from bottom = y
y
4500(400  15 / 2)  3600 ´ 205  10000 ´12.5
 145.3 mm
18100
Ix = 400x253/12 +10000(12.5-145.3)2 + 10x3603/12 +3600(205145.3)2 + 300x153/12 +4500(392.5-145.3)2 = 503.7x106 mm4
Sx = 503.7x106 / (400-145.3) = 1977.5x103 mm3
My-x = Fy Sx = 680.2 kN-m.
Sx - elastic section modulus
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Slide # 19
Ex. 4.1 – Sectional Properties

distance of plastic centroid from bottom = y p
18100
\ 400 ´ y p 
 9050
2
\ y p  22.6 mm

y1
= centroid of top half-area about plastic centroid
18100 ´145.3  9050 ´ 22.6 / 2
=
 22.6  256.7 mm
9050

y2
= centroid of bottom half-area about plas. cent.
yp
= 22.6 / 2  11.3 mm

Zx = A/2 x (y1 + y2) = 9050 x (256.7 + 11.3) = 2425400 mm3

Zx - plastic section modulus
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Slide # 20
Ex. 4.1 – Sectional Properties


Mp-x = Zx Fy = 2425400 x 344/106 = 834.3 kN.m
Design strength according to AISC Spec. F1.1= bMp= 0.9 x 834.3
= 750.9 kN.m

Check = Mp  1.5 My

Therefore, 834.3 kN.m < 1.5 x 680.2 = 1020.3 kN.m
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- OK!
Slide # 21
Flexural Deflection of Beams Serviceability

Steel beams are designed for the factored design loads.
The moment capacity, i.e., the factored moment strength
(bMn) should be greater than the moment (Mu) caused
by the factored loads.

A serviceable structure is one that performs
satisfactorily, not causing discomfort or perceptions of
unsafety for the occupants or users of the structure.
•
•
For a steel beam, being serviceable usually means that the
deformations, primarily the vertical slag, or deflection, must be
limited.
The maximum deflection of the designed beam is checked at the
service-level loads. The deflection due to service-level loads
must be less than the specified values.
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Slide # 22
Serviceability Requirements
•
•
•
•
•
•
Steel beams need to satisfy SLS in addition to ULS
Serviceability limit states are usually checked using non-factored loads.
Deflection under live loads shall be limited to L/360
Dead load deflections can be compensated by cambering beams.
SLS might also include limiting stresses in bottom or top flanges if
fatigue is a concern in design (Will be further discussed with plate
girders).
Standard equations to calculate deflection for different load cases:
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Slide # 23
Flexural Deflection of Beams Serviceability

The AISC Specification gives little guidance other than a
statement, “Serviceability Design Considerations,” that
deflections should be checked. Appropriate limits for
deflection can be found from the governing building code
for the region.

The following values of deflection are typical max.
allowable deflections.
LL
DL+LL
•
•
•
•
Plastered floor construction
Unplastered floor construction
Unplastered roof construction
L/360
L/240
L/180
L/240
L/180
L/120
DL deflection – normally not considered for steel beams
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Slide # 24
Flexural Deflection of Beams Serviceability

In the following examples, we will assume that
local buckling and lateral-torsional buckling are
not controlling limit states, i.e, the beam section is
compact and laterally supported along the length.
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Slide # 25
Ex. 4.2 - Deflections
Design a 9 m long simply supported beam subjected to
UDL of 6 kN/m dead load and a UDL of 8 kN/m live load.
The dead load does not include the self-weight of the
beam.

•
•
Step I. Calculate the factored design loads (without self-weight).
wu = 1.2 wD + 1.6 wL = 20 kN/m
Mu = wu L2 / 8 = 20 x 92 / 8 = 202.5 kN.m (SS beam)
Step II. Select the lightest section from the AISC Manual design
tables.
Zx = Mu/(bFy) = 202.5x106/(0.9x344) = 654x103
select W16 x 26 made from A992 steel with bMp = 224 kN.m
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Slide # 26
Ex. 4.2 - Deflections
•
•
Step III. Add self-weight of designed section and check design
wsw = 0.38 kN/m
Therefore, wD = 6.38 kN/m
wu = 1.2 x 6.38 + 1.6 x 6 = 20.46 kN/m
Therefore, Mu = 20.46 x 92 / 8 = 207.2 kN.m < bMp of
W16 x 26.
OK!
Step IV. Check deflection at service loads.
w = 8 kN/m
D = 5 w L4 / (384 E Ix) = 5 x (8) x103 x (9)4 / (384 x 200x125)
D = 27.3 mm > L/360
•
- for plastered floor construction
Step V. Redesign with service-load deflection as design criteria
L /360 = 25 mm > 5 w L4/(384 E Ix)
Therefore, Ix > 136.7x106 mm4
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Slide # 27
Ex. 4.2 - Deflections
Select the section from the moment of inertia selection from
Section Property Tables – select W16 x 31 with Ix = 156x106 mm4

Note that the serviceability design criteria controlled the design
and the section
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Slide # 28
Ex. 4.3 – Beam Design

Design the beam shown below. The unfactored dead and live loads
are shown in Fig. 6 below.
40
kN (live load)
10 kips
10
kN/m
0.67
k/ft. (dead load)
11
kN/m
0.75
k/ft. (live load)
15
ft.m
4.5
930mft.

Step I. Calculate the factored design loads (without self-weight).
wu = 1.2 wD + 1.6 wL = 1.2 x 10 + 1.6 x 11 = 29.6 kN/m
Pu = 1.2 PD + 1.6 PL = 1.2 x 0 + 1.6 x 40 = 64 kN
Mu = wU L2 / 8 + PU L / 4 = 299.7 + 144 = 443.7 kN.m
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Slide # 29
Ex. 4.3 – Beam Design

Step II. Select W21 x 44
Zx = 1563x103 mm3
bMp = 0.9x1563x103x344/1000000 = 483.9 kN.m
Self-weight = wsw = 0.64 kN/m.

Step III. Add self-weight of designed section and check design
wD = 10 + 0.64 = 10.64 kN/m
wu = 1.2 x 10.64 + 1.6 x 11 = 30.4 kN/m
Therefore, Mu = 30.4 x 92/8 + 144 = 451.8 < bMp of
W21 x 44.
OK!
Step IV. Check deflection at service live loads.
Service loads
• Distributed load = w = 11 kN/m

•
Concentrated load = P = L = 40 kN
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Slide # 30
Ex. 4.3 – Beam Design
Deflection due to uniform distributed load = d = 5 w L4 / (384 EI)
Deflection due to concentrated load = c = P L3 / (48 EI)
Therefore, service-load deflection =  = d + c
 = 5x11x94x109/(384x351x106x200) +40x93x109/(48x351x106x200)
 = 13.4 + 8.7 = 22.1 mm
L = 9 m.
Assuming unplastered floor construction, max = L/240 = 9000/240 =
37.5 mm
Therefore,  < max
- OK!
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Slide # 31
Local Buckling of Beam Section –
Compact and Non-compact

Mp, the plastic moment capacity for the steel shape, is
calculated by assuming a plastic stress distribution (+ or - sy)
over the cross-section.

The development of a plastic stress distribution over the
cross-section can be hindered by two different length effects:
•
•

Local buckling of the individual plates (flanges and webs) of the
cross-section before they develop the compressive yield stress sy.
Lateral-torsional buckling of the unsupported length of the beam /
member before the cross-section develops the plastic moment Mp.
The analytical equations for local buckling of steel plates with
various edge conditions and the results from experimental
investigations have been used to develop limiting slenderness
ratios for the individual plate elements of the cross-sections.
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Slide # 32
Local Buckling of Beam Section –
Compact and Non-compact
M
M
Figure 7. Local buckling of flange due to compressive stress (s)
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Slide # 33
Local Buckling of Beam Section –
Compact and Non-compact

Steel sections are classified as compact, non-compact, or
slender depending upon the slenderness (l) ratio of the
individual plates of the cross-section.
•
•
•

Compact section if all elements of cross-section have   p
Non-compact sections if any one element of the cross-section has p  
 r
Slender section if any element of the cross-section has r  
It is important to note that:
•
•
•
If   p, then the individual plate element can develop and sustain sy for
large values of e before local buckling occurs.
If p    r, then the individual plate element can develop sy at some
locations but not in the entire cross section before local buckling occurs.
If r  , then elastic local buckling of the individual plate element occurs.
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Slide # 34
Classification of Sections
•
•
•
Classifications of bending elements are based on limits of local buckling
bf
The dimensional ratio l represents
Two limits exist p and r
h

tw

bf
2t f

p represents the upper limit for compact sections

r represents the upper limit for non-compact sections
tf
h
tw
  P compact
P    r non compact
  r slender
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Slide # 35

Thus, slender
sections cannot
develop Mp due to
elastic local
buckling. Noncompact sections
can develop My but
not Mp before local
buckling occurs.
Only compact
sections can
develop the plastic
moment Mp.
Co mpressive axial stress, s
Local Buckling of Beam Section –
Compact and Non-compact
Compact
sy
Non-Compact
Slender
Effective axial strain, 
Figure 8. Stress-strain response of plates
subjected to axial compression and local
buckling.
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Slide # 36
Local Buckling of Beam Section –
Compact and Non-compact

All rolled wide-flange shapes are compact with the
following exceptions, which are non-compact.
•
W21x48, W40x174, W14x99, W14x90, W12x65, W10x12, W8x10,
W6x15 (made from A992)
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Slide # 37
Classification of Sections
•
The limits are
bf
Flange


bf
2t f
 p  0.38
r  0.83
Web
E
Fy
h
tw
 p  3.76
E
Fy  68 .9
 r  5.70
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tf
h
tw
E
Fy
E
Fy
Slide # 38
Lateral-Torsional Buckling (LTB)
(a)
 The laterally unsupported
length of a beam-member
can undergo LTB due to
the applied flexural
loading (BM).
M
(b)
M
M
Figure 9. Lateral-torsional
buckling of a wide-flange beam
subjected to constant moment.
M
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Slide # 39
Lateral-Torsional Buckling (LTB)

LTB is fundamentally similar to the flexural buckling or
flexural-torsional buckling of a column subjected to axial
loading.
•
•

The similarity is that it is also a bifurcation-buckling type
phenomenon.
The differences are that lateral-torsional buckling is caused by
flexural loading (M), and the buckling deformations are coupled in
the lateral and torsional directions.
There is one very important difference. For a column, the
axial load causing buckling remains constant along the length.
But, for a beam, usually the LTB causing bending moment
M(x) varies along the unbraced length.
•
The worst situation is for beams subjected to uniform BM along the
unbraced length. Why?
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Slide # 40
Lateral-Torsional Buckling (LTB) –
Uniform BM

Consider a beam that is simply-supported at the ends and
subjected to four-point loading as shown below. The beam
center-span is subjected to uniform BM (M). Assume that
lateral supports are provided at the load points.
P
P
Lb


Laterally unsupported length = Lb.
If the laterally unbraced length Lb is less than or equal to a
plastic length LP then lateral torsional buckling is not a
problem and the beam will develop its plastic strength MP.
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Slide # 41
Lateral-Torsional Buckling (LTB) –
Uniform BM

Lp = 1.76 ry x
E / Fy - for I members & channels

If Lb is greater than LP then lateral torsional buckling will
occur and the moment capacity of the beam will be
reduced below the plastic strength MP as shown in Fig. 10.

As shown in Fig. 10, the lateral-torsional buckling moment
(Mn = Mcr) is a function of the laterally unbraced length Lb
and can be calculated using the equation:
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Slide # 42
Lateral-Torsional Buckling (LTB) –
Uniform BM

Mn = Mcr =
 2 EI y 
2
b
L
 GJ 

 2 ECw 
L2b


Mn - moment capacity
Lb - laterally unsupported length.
Mcr - critical lateral-torsional buckling moment.
E – 200 GPa;
G – 77 GPa
Iy - moment of inertia about minor or y-axis (mm4)
J - torsional constant (mm4) from the Section Property Tables.
Cw - warping constant (mm6) from the Section Property Tables.

This Eq. is valid for ELASTIC LTB only (like the Euler equation). This
means it will work only as long as the cross-section is elastic and no
portion of the cross-section has yielded.
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Slide # 43
Lateral-Torsional Buckling (LTB) –
Uniform BM
Mn = Mp
Zx Fy = Mp

 Lb  L p 

M n   M p  ( M p  M r )



 Lr  L p 
Moment Capacity, M n
Sx (F
= Mr
(0.7F
y – 10)
y)
Mn =
Plastic
No
Instability
Inelastic
LTB
Lp
 2 EI y 
L2b
GJ 

 2 ECw 
L2b
Elastic
LTB
Lr
Unbraced length, Lb
Fig. 10 Lateral Torsional Buckling (Uniform Bending)
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Slide # 44


Lateral-Torsional Buckling (LTB) –
Uniform BM

As soon as any portion of the cross-section reaches the
yield stress Fy, the elastic LTB equation cannot be used.
•
Lr is the unbraced length that corresponds to a LTB moment
Mr = Sx (0.7Fy).
•
Mr will cause yielding of the cross-section due to residual stresses.

When the unbraced length is less than Lr, then the elastic
LTB Eq. cannot be used.

When the unbraced length (Lb) is less than Lr but more
than the plastic length Lp, then the LTB Mn is given by the
Eq. below:
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Slide # 45
Lateral-Torsional Buckling –
Uniform BM
•
•
•
If Lp  Lb  Lr,
 Lb  L p 


 Lr  L p 

This is linear interpolation between (Lp, Mp) and (Lr, Mr)
See Fig. 10 again.
•
E
Lr  1.95rts
0.7 Fy
•
r 
•
•

then M n  M p  ( M p  M r )

2
ts
 0.7 Fy S x h0 
Jc

1  1  6.76
S x h0
Jc 
 E
2
I yCw
Sx
For a doubly symmetric I-shape: c = 1
h0 = distance between the flange centroids (mm)
68402
Slide # 46
Moment Capacity of Beams
Subjected to Non-uniform BM

As mentioned previously, the case with uniform bending
moment is worst for lateral torsional buckling.

For cases with non-uniform bending moment, the LTB
moment is greater than that for the case with uniform
moment.

The AISC specification says that:
•
•
The LTB moment for non-uniform bending moment case
Cb x lateral torsional buckling moment for uniform moment case.
68402
Slide # 47
Moment Capacity of Beams
Subjected to Non-uniform BM

Cb is always greater than 1.0 for non-uniform bending
moment.
•
•
Cb is equal to 1.0 for uniform bending moment.
Sometimes, if you cannot calculate or figure out Cb, then it can be
conservatively assumed as 1.0. for doubly and singly symmetric
sections
Cb 
•
12.5 M max
 3.0
2.5 M max  3 M A  4 M B  3 M c
Mmax - magnitude of maximum bending moment in Lb
MA - magnitude of bending moment at quarter point of Lb
MB - magnitude of bending moment at half point of Lb
MC - magnitude of bending moment at three-quarter point of Lb
Use absolute values of M
68402
Slide # 48
Flexural Strength of Compact
Sections
12.5M max
Cb 
 3.0
2.5M max  3M A  4M B  3M C
MC
@ three-quarter
MA
@ quarter
Mmax
MB
@ mid
Moments determined between bracing points
68402
Slide # 49
Values of Cb
3-1
68402
Slide # 50
Moment Capacity of Beams Subjected
to Non-uniform Bending Moments

The moment capacity Mn for the case of non-uniform
bending moment
•
•
•
Mn = Cb x {Mn for the case of uniform bending moment}  Mp
Important to note that the increased moment capacity for the nonuniform moment case cannot possibly be more than Mp.
Therefore, if the calculated values is greater than Mp, then you have
to reduce it to Mp
68402
Slide # 51
Moment Capacity of Beams
Subjected to Non-uniform BM
Mp
Moment Capacity, Mn
Non-uniform BM
Mr
Cb = 1.5
Cb = 1.2
Cb = 1.0
Uniform BM
Lp
Lr
Unbraced length, Lb
Figure 11. Moment capacity versus Lb for non-uniform moment case
Cb = 1.0 means uniform BM
68402
Slide # 52
Structural Design of Beams
•
Steps for adequate design of beams:
1) Compute the factored loads, factored moment and shear
2) Determine unsupported length Lb and Cb
3) Select a WF shape and choose Zx assuming it is a compact section
with full lateral support
Zx 
M n  M p  ZFy
Mu
b Fy
Mu  b Mn  0.9ZFy
4) Check the section dimension for compactness and determine bMn
M u  b M n
5) Use service loads to check deflection requirements
68402
Slide # 53
Ex. 4.4 – Beam Design

Use Grade 50 steel to design the beam shown below. The
unfactored uniformly distributed live load is equal to 40
kN/m. There is no dead load. Lateral support is provided at
the end reactions. Select W16 section.
wwLL== 40
kN/m
3 kips/ft.
24 ft.m
7.5
Lateral support / bracing
68402
Slide # 54
Ex. 4.4 – Beam Design
•
•
Step I. Calculate the factored loads assuming a reasonable selfweight.
Assume self-weight = wsw = 1.46 kN/m.
Dead load = wD = 0 + 1.46 = 1.46 kN/m.
Live load = wL = 40 kN/m.
Ultimate load = wu = 1.2 wD + 1.6 wL = 65.8 kN/m.
Factored ultimate moment = Mu = wu L2/8 = 462.3 kN-m.
Is BM uniform??
Yes
Cb =1.0
No
Go to Step II
Step II. Determine unsupported length Lb and Cb
There is only one unsupported span with Lb = 7.5 m
Cb = 1.14 for the parabolic bending moment diagram, See values of
Cb shown in Table 3-1.
68402
Slide # 55
Ex. 4.4 – Beam Design
•
Step III. Select a wide-flange shape
•
•
•
•
Compute Zx = 462.3*106/(0.9*344) = 1493x106 mm3.
Select W16 x 50 steel section
Zx = 1508x103 mm3
Sx = 1327x103 mm3
ry = 40.4 mm
Cw = 610x109 mm6
Iy = 15.5x106 mm4
J = 0.63x106 mm4
•
Lp  1.76ry E / Fy 
•
E
Lr  1.95rts
0.7 Fy
•
rts 
I y Cw
Sx

1.76´ 40.4 200000344
 1.71 m
1000
 0.7 Fy S x h0 
Jc

1  1  6.76
S x h0
E
Jc


2
15.5 ´106 ´ 610´109
 48.1 m m
3
1327´10
68402
Slide # 56
Ex. 4.4 – Beam Design
•
h0 = D - TF = 414 – 16 = 398 mm
2
2
•
 0.7 ´ 344 1327´103 ´ 398
 0.7 Fy S x h0 
  2.81
  1  6.76
1  6.76
6
Jc 
 E
 200000 0.63´10 ´1 
•
48.1 200000
0.63´106 ´1
Lr  1.95´
´
1  2.81  5.26 m
3
1000 0.7 ´ 344 1327´10 ´ 398
•
Lb > Lr
M n  Cb
 2 EI y 
 2 ECw 
2
b
2
b
 1.14
 GJ 

L
L


 2 ´ 200´15.5 ´106 
75002
 77 ´ 0.63´10 

6
 2 ´ 200´ 610´109 
75002


1508´103 ´ 344
 222´10 kN.m m  222 kN.m  M p 
 518.8 kN.m
6
10
3
68402
Slide # 57
Ex. 4.4 – Beam Design
•
•
•
Step IV. Check if section is adequate
•
Mu > Mn
Not OK
Step V. Try a larger section.
•
After few trials select W16 x 67
Mn = 497.7 > Mu OK
Step VI. Check for local buckling.
 = Bf / 2Tf = 7.7; Corresponding p = 0.38 (E/Fy)0.5 = 9.19
Therefore,  <  p - compact flange
 = H/Tw = 35.9;
Corresponding p = 3.76 (E/Fy)0.5 = 90.5
Therefore,  <  p
- compact web
Compact section.
- OK!


This example demonstrates the method for designing beams and
accounting for Cb > 1.0)
Values for Lr and Lp can be obtained from Tables too
68402
Slide # 58
Ex. 4.5 – Beam Design

Design the beam shown below. The concentrated live
loads acting on the beam are shown in the Figure. The
beam is laterally supported at the load and reaction points.
Use Grade 50 steel.
135
KN
30 kips
30 kips
135
KN
wsw = 0.1
1.5kips/ft.
KN/m
12
3.6ft.m
8 ft.m
2.4
10
3.0ft.m
309.0
ft. m
Lateral support / bracing
68402
Slide # 59
Ex. 4.5 – Beam Design
•
Step I. Assume a self-weight and determine the factored design
loads
Let, wsw = 1.5 kN/m
PL = 135 kN
Pu = 1.6 PL = 216 kN
wu = 1.2 x wsw = 1.8 kN/m
The reactions and bending moment diagram for the beam are
shown below.
68402
Slide # 60
Ex. 4.5 – Beam Design
216kips
KN
48
216
KN
48
kips
wsw = 1.8
0.12KN/m
kips/ft.
A
D
B
12m
ft.
3.6
C
8 ft.
2.4
m
10
3.0ft.m
46.6
kips
209.7
kN
A
53 kips
238.5
kN
30 kips
30 kips
wsw = 0.1 kips/ft.
B
12 ft.
C
8 ft.
550.6 kip-ft.
754.9
kN.m
524 kip-ft.
715.5
kN.m
D
10 ft.
30 ft.
Lateral support / bracing
68402
Slide # 61
Ex. 4.5 – Beam Design
•
Step II. Determine Lb, Cb, Mu, and Mu/Cb for all spans.
Span
Lb
(m)
Cb
Mu
(kN-m)
Mu/Cb
(kN-m)
AB
3.6
1.67
754.9
452.8
BC
2.4
1.0
(assume)
754.9
754.9
CD
3.0
1.67
715.5
429.2
It is important to note that it is possible to have different Lb and
Cb values for different laterally unsupported spans of the same
beam.
Cb – Table 3-1
68402
Slide # 62
Ex. 4.5 – Beam Design
•
Step III. Design the beam and check all laterally unsupported spans
Assume that span BC is the controlling span because it has the
largest Mu/Cb although the corresponding Lb is the smallest.
Required Zx = 754.9*106/(0.9*344) = 2438x103 mm3
After few trials select W21 x 68 from section property Table.
Lp = 1.94 m
Lr = 5.73 m (From Tables)

 Lb  L p 

For all members Lp < Lb < Lr M n  Cb M p  ( M p  M r )


 Lr  L p 
Check the selected section for spans AB, BC, and CD
Span
Lb (m)
Cb
bMn for Cb value
bMp limit
AB
3.6
1.67
1125.5 kN.m
811.8 kN.m
BC
2.4
1.0
773.6
CD
3
1.67
1208.7 kN.m
68402
811.8 kN.m
Slide # 63
Ex. 4.5 – Beam Design
Thus, for span AB, bMn = 811.8 kN.m > Mu
For span BC, bMn = 773.6 kN.m > Mu
For span CD, bMn = 811.8 kN.m > Mu
•
•
- OK!
- OK!
- OK!
Step IV. Check for local buckling
Bf / 2Tf = 6.04;
Corresponding p = 0.38 (E/Fy)0.5 = 9.19
Therefore compact flange
H/Tw = 43.6; Corresponding p = 3.76 (E/Fy)0.5 = 90.55
Therefore compact web
Compact section.
- OK!
This example demonstrates the method for designing beams with
several laterally unsupported spans with different Lb and Cb values.
68402
Slide # 64
Ex. 4.6 – Beam Design

Design the simply-supported beam shown below. The
uniformly distributed dead load is equal to 15 kN/m and
the uniformly distributed live load is equal to 30 kN/m. A
concentrated live load equal to 40 kN acts at the midspan. Lateral supports are provided at the end reactions
and at the mid-span. Use Grade 50 steel.
wD = 15
1.0 kips/ft.
wsw = 0.1kN/m
kips/ft.
wL =30
2.0kN/m
kips/ft.
40kips
kN
10
30 kips
30 kips
A
C
B
12 ft.
8 ft.
3.6
m
12 ft.
30 ft.
10 ft.
12 ft.
3.6
m
Lateral support / bracing
68402
Slide # 65
Ex. 4.6 – Beam Design
•
Step I. Assume the self-weight and calculate the factored design loads.
Let, wsw = 1.5 kN/m
wD = 15 + 1.5 = 16.5 kN/m
wL = 30 kN/m
wu = 1.2 wD + 1.6 wL = 67.8 kN/m
Pu = 1.6 x 40 = 64 kN
The reactions and the bending moment diagram for the factored loads
are shown below.
64 kips
kN
16
wu = 67.8
4.52 kips/ft.
kN/m
B
3.612mft.
12 ft.
3.6
m
62.24
276.1kips
kN
x
M(x)
x - 4.52
M(x)==62.24
276.1(x)
+ 67.8(x)2/2
x2 /2
68402
62.24 kN
kips
276.1
Slide # 66
Ex. 4.6 – Beam Design
•
Step II. Calculate Lb and Cb for the laterally unsupported spans.
Since this is a symmetric problem, need to consider only span AB
Lb =3.6 m, M(x) = 276.1 x – 67.8 x2/2
Therefore,
MA = M(x = 0.9 m) = 221 kN.m
- quarter-point along Lb = 3.6 m
MB = M(x = 1.8 m) = 387 kN.m
- half-point along Lb = 3.6 m
MC = M(x = 2.7 m) = 498 kN.m
- three-quarter point along Lb= 3.6 m
Mmax = M(x = 3.6 m) = 554.6 kN.m - maximum moment along Lb =3.6 m
Therefore, Cb = 1.36
68402
Slide # 67
Ex. 4.6 – Beam Design
• Step III. Design the beam section
Mu = Mmax = 554.6 kN.m
Lb = 3.6 m, Cb = 1.36
Required Zx = 554.6*106/(0.9*344) = 1791x103 mm3
After few trials, select W21 x 57 steel section
Lp = 1.46 m
Lr = 4.37 m
Lp < Lb < Lr
bMn = 699 kN.m > bMp = 639.3 kN.m
bMn= 639.3 >Mu
OK
68402
Slide # 68
Ex. 4.6 – Beam Design
•
Step V. Check for local buckling.
Bf / 2Tf = 7.87; Corresponding p = 0.38 (E/Fy)0.5 = 9.192
Therefore, compact flange
l = h/tw = 50.0; Corresponding p = 3.76 (E/Fy)0.5 = 90.55
Therefore, compact web
- OK!
Compact section.
• This example demonstrates the calculation of Cb and
the iterative design method.
68402
Slide # 69
Shear Capacity

The shear capacity of the beam is
Vu  vVn

Vn  0.6Fy AwCv
v  0.9
For I-shaped sections, the factors Cv and v are functions of the shear
buckling of the web and thus the ration h/tw
v  1.0
E
if h / t w  2.24
Fy
Cv  1.0
representing the case of no web instability.
Aw = dtw
68402
Slide # 70
Shear Capacity
For all other doubly and singly sym. sections and channels except
round HSS
 The first case represents the case of no web instability.
v  0.9
h / t w  1.10

v  0.9
Cv  1.0
The second case represents inelastic web buckling
1.10

kv E
Fy
kv E
k E
 h / t w  1.37 v
Fy
Fy
1.10 k v E F y
Cv 
h tw
v  0.9
The last case represents elastic web buckling
h / t w  1.37
kv E
Fy
Cv 
68402
1.51Ekv
h / tw 
2
Fy
v  0.9
Slide # 71
Shear Capacity
•
•
Aw = dtw
The web plate buckling coefficient, kv, is given
•
•
For unstiffened webs with h/tw <260, kv = 5 except for the stem of tee
shapes, kv =1.2
For stiffened webs
5


5



2


a
h


kv  
2
 260  

5
,
a
h

3
.
0
or
a
h

h t  

 w 

•
•
a = clear distance between transverse stiffeners
h = for rolled shapes, the clear distance between flanges less the fillet
of corner radii.
68402
Slide # 72
Beam Bearing Plates

Design of a beam bearing plate would require checking:
t
N
B

1- Web Yielding and Web crippling to determine N

2- Bearing capacity to determine B

3- Plate moment capacity to determine t
AISC Specifications: Chapter K
68402
Slide # 73
Beam Bearing Plates

When a bearing plate is used at beam end, two limit states shall be
considered

1. Web Yielding

  1.0
R  Rn
This represents yield of the web at the vicinity of the flange due
to excessive loading
R
CASE 1: At Support
N k
Rn  (2.5k  N ) Fytw
CASE 2: Interior Load
R
Rn  (5k  N ) Fytw
68402
Slide # 74
Beam Bearing Plates

2. Web Crippling

Web crippling represent the possible buckling of the web
CASE 1: At Support
R  Rn
1.5



N  t w  

2

Rn  0.4tw 1  3



 d  t f  


  0.75
E Fy t f
N
 0 .2
d
tw

 4N
 t
2
Rn  0.4t w 1  
 0.2  w

 d

tf

1.5








E Fy t f
tw
N
 0 .2
d
CASE 2: Interior Load
1.5



N  t w  

2

Rn  0.8tw 1  3



 d  t f  


E Fy t f
68402
tw
Slide # 75
Ex. 4.7 – Beam Design

Check the beam shown in the figure below for:
•
•
•
•
Shear capacity.
Web yielding.
Web crippling.
Assume the width of the bearing plate is 100 mm. Use Grade 50
steel.
40kips
kN
10
wD = 10
1.0kN/m
kips/ft.
wL =25
2.0kN/m
kips/ft.
A
C
B
W16x26
2m
12
ft.
12
ft.
2m
68402
Slide # 76
Ex. 4.7 – Beam Design
•
Step I. The section used from Example 4.6 is W21x57.
The self-weight wsw = 0.83 kN/m
wD = 10 + 0.38 = 10.38 kN/m
wL = 25 kN/m
wu = 1.2 wD + 1.6 wL = 52.5 kN/m
Pu = 1.6 x 40 = 64 kN
The reactions and the bending moment diagram for the factored loads
are shown below.
64 kips
kN
16
wu = 52.5
4.52 kips/ft.
kN/m
B
2 12
m ft.
12
ft.
2m
62.24
137 kips
kN
x
x2 /2
M(x) = 62.24
- 4.52kN
Vu =x 137
68402
62.24kN
kips
137
Slide # 77
Ex. 4.7 – Beam Design
•
Step II. h/tw = 56.8
2.24
E
200000
h
 2.24
 54   56.8
Fy
344
tw
Assume unstiffened web, h/tw <260, kv = 5
1.10
kv E
5 ´ 200000
h
 1.10
 59.3   56.8
Fy
344
tw
Cv  1   0.9
Assume unstiffened web, h/tw <260, kv = 5
Vn = 0.9*(0.6Fy)*d*tw*Cv
Vn = 0.9*(0.6x344)*399*6.4x10-3 = 474.4 kN> Vu
68402
Slide # 78
Ex. 4.7 – Beam Design
•
•
Step III. Web yielding
critical is support
k = 19 mm
R = (2.5k + N)*Fy*tw
R = 1x(2.5x19 + 100)x344x6.4/1000 = 324.7 kN
R > reaction = 137 kN
OK
Step IV. Web crippling
d = 399 mm
tw = 6.4 mm
N/d = 100/399 = 0.25 > 0.2
critical is support
tf = 8.8 mm
1.5

 EF t


t
4
N


y f
Rn  0.4t w2 1  
 0.2  w  
tw
  d
 t f  


68402
Slide # 79
Ex. 4.7 – Beam Design
1.5

 200000´ 344´ 8.8
4
´
100
6
.
4




2
Rn  0.4 ´ 6.4 1  
 0.2 
´103  238.7 kN
 
6.4
 8.8  
  399
R = 179 kN > reaction = 137 kN
68402
OK
Slide # 80