Lecture 4  Page 1 of 12
Lecture 4 – Beam Design for Moment
Beams, or sometimes referred to as “flexure” members, are designed on the
basis of moment.
In ASD and LRFD, the design strength of a beam in flexure is
called the “nominal flexural moment”, M
n
.
This
M
n
must be greater than the
maximum applied factored (LRFD
) or service (ASD)
moment.
Beams are designed on the basis of the following LRFD references:
•
AISC Part 3
•
AISC Spec Chapter F p. 16.144
1.
Beam Design Considering Yielding
:
Assuming a beam is
adequately laterally braced
, it will fail by yielding on
the compression flange.
Most beams are laterally braced by the metal
decking that is attached to the compression flange as shown below:
1) LRFD Beam Design
:
Design Flexural Strength =
φ
b
M
n
Where:
φ
b
= 0.90
M
n
= nominal flexural moment
= M
p
= Plastic moment
= F
y
Z
x
Z
x
= plastic section modulus
= from properties
Concrete slab
over metal
decking
Metal decking
puddlewelded to top
flange of beam
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2) ASD Beam Design
:
Allowable Flexural Strength =
b
n
M
Ω
Where:
Ω
b
= 1.67
M
n
= nominal flexural moment
= M
p
= Plastic moment
= F
y
Z
x
Z
x
= plastic section modulus
= from properties
GIVEN
:
A W16x26 steel beam using A992 steel is continuously laterally
braced, and experiences a
FACTORED
moment = 104 KIPFT.
REQUIRED
:
1) Determine the design flexural moment,
φ
b
M
n
for the beam.
2) Determine if the beam is adequate.
Step 1 – Determine
φ
b
M
n
for the beam
:
φ
b
M
n
= 0.90(F
y
Z
x
) since it is continuously laterally braced
= 0.90(50 KSI)(44.2 in
3
)
= 1989 KIPIN
φ
b
M
n
= 165.8 KIPFT
Step 2 – Determine if the beam is adequate
:
Since
φ
b
M
n
= 165.8 KIPFT > 104 KIPFT
→
beam is adequate
GIVEN
:
The W16x26 beam from Example 1.
REQUIRED
:
Determine the design flexural moment,
φ
b
M
n
for the beam
using the LRFD “Z
x
” Table 32 (see AISC p. 311 thru 319)
Step 1 – Refer to AISC p. 318 for W16x26
:
Look in the LRFD column
φ
b
M
px
= 166 KIPFT
From properties,
AISC p. 121
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 Fall '08
 HULTENIUS
 Column, Truss, Girder, AISC, AISC p.

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