Beams
BEAMS
A structural member loaded in the
transverse direction to the longitudinal
axis.
Internal Forces:
Bending Moments and Shear
Beam Shapes
Section Properties AISC Section I
Beams
h
970

tw
fy
Plate Girders
h 970

tw
fy
• W- (eg W44x335)
Most commonly Used/Wide Flange I, pp 1-10:1-27
• S- (eg S24x121)
pp. 1-30:1-31
• M- (eg M12x11.8)
pp. 1-28:1-29
• Channels (eg C15x50)
pp. 1-34:1-39
Structural Steel - Sections
Built Up members
I Shapes
H Shapes
Box Shapes
Fillet Welding
Welding
Fillet
of Web
Web to
to
of
Flange Plates
Plates
Flange
Beams
E 29,0 0
 0.38 9.15  6.2
p  0.38
0
5
F
Depending on the use they are referred to as:
Joists
y
Support Floor Deck
Floor Beams
Beams that support joists
Girders:
Support most load in a floor system
Lintels
Over Windows and Door Openings
Support portion of wall above opening
Beams
Depending on the use they are referred to as:
Purlins
Support Roof Surface
Roof Beams
Support Purlins
Spandrel Beams
Support outside edges of
a floor deck and outside
walls of buildings
Structural Steel - Characteristics
Buckling: Instability due to slenderness
Elastic Buckling
Limit States
Load Deflection
500
450
Load (kips)
400
350
300
250
200
150
100
50
0
0
4
8
12
Deflection (in)
16
20
FEM
24
Test
Limit States
Limit States
Limit States
Limit States
• Flexure
•Elastic
•Plastic
•Stability (buckling)
•
•
•
•
Shear
Deflection
Fatigue
Supports
Flexure
LRFD
bM n  M u
b  0.90
ASD
Mn
 Ma
b
b  1.67
Elastic
Plastic
Stability (buckling)
Flexure - Elastic
My
f 
I
f max  M max
M max
c

I
S
S=I/c : Section Modulus (Tabulated Value)
Example
Compute My
Flexure - Plastic
Flexure - Plastic
C=T
Acfy=Atfy
Ac=At
Mp = Acfy = Atfy = fy (0.5A) a = Mp=Zfy
Mp/ My =Z/S
For shapes that are symmetrical about the axis of bending the plastic and elastic
neutral axes are the same
Z=(0.5A)a : Plastic Section Modulus (Tabulated Value)
Example
Compute Mp
Example
Flexure - Stability
A beam has failed when:
Mp is reached and section becomes fully plastic
Or
Flange Local Buckling (FLB) Elastically or Inelastically
Web Local Buckling (WLB) Elastically or Inelastically
Lateral Torsional Buckling (LTB) Elastically or Inelastically
Flexure - Stability
FLB
Slenderness Parameter
WLB
LTB
=bf/2tf
=h/tw
= Lb /ry
bf
tf
h
tw
Lb
Flexure - Stability
FLB and WLB (Section B5 Table B4.1)
Evaluate Moment Capacity for Different 
Mp
Mr
Non
Compact Compact
Slender
p
r
FLB
WLB
=bf/2tf
=h/tw
Slenderness Parameter - Limiting Values
AISC B5 Table B4.1
pp 16.1-16
Slenderness Parameter - Limiting Values
AISC B5 Table B4.1
pp 16.1-17
Slenderness Parameter - Limiting Values
AISC B5 Table B4.1
pp 16.1-18
Flexure - Stability
FLB and WLB (Section B5 Table B4.1)
Mp
Mr
Non
Compact Compact
Slender
p
r
FLB
WLB
=bf/2tf
=h/tw
Example
The beam shown is a W16X31 of A992 steel. It supports a reinforced
concrete slab that provides continuous lateral support of the compression
flange. Service dead load is 450 lb/ft (does not include weight of beam).
Service live load is 550 lb/ft. Does the beam have adequate moment
strength?
Example
Determine Nominal Flexural Strength
Flange Compactness
bf
 6.28
2t f
 p  0.38
E
29,000
 0.38
 9.15  6.28
Fy
50
Flange Compact
Web Compactness
h 15.7

 62.8
tw 0.25
 p  3.76
29,000
 90.55  62.8
50
Web Compact
Example
Shape is compact and continuously supported => Plastic Hinge Forms
Nominal Flexural Strength
Mn  M p  Fy Z x  5054.0  2,700in - kips 225ft.kips
Moment Demand
wD  450 31  481lb/ft
0.48130
MD 
 54.11 ft - kips
8
2
wL2
M max 
@ mid - span
8
wL  550lb/ft
0.5530
ML 
 61.88 ft - kips
8
2
Example
LRFD
M u  1.2M D  1.6M L  164 ft - kips
b M n  0.9225.0  203ft - kips  M u  164
OK
ASD
M a  M D  M L  54  61.88  116.0 ft - kips
M n 225

 135ft - kips  M a  116
b 1.67
OK