Lecture 5  Page 1 of 9
Lecture 5 – Beams – Design for Shear & Deflection
Steel beams are usually designed solely on the basis of moment. This means
that bending stresses are the critical design factor. However, under certain
circumstances, shear and deflection must also be checked.
1. Design for Shear
Shear in steel beams generally does not control the design EXCEPT in
the following two situations:
• Reduced beam crosssectional area, as with “coped” beams
• Very heavy loads on shortspan beam
Short span
Reduced Shear plane
Normal beam
Shear plane
Coped beam
Very heavy loads
Lecture 5  Page 2 of 9
Shear in steel beams is assumed to be carried entirely by the area of the
web, A
w
:
Design for shear is dictated in AISC Spec. G p. 16.164 as follows:
LRFD Factored Design shear strength = φ
v
V
n
ASD Service Allowable shear strength =
v
n
V
Ω
where: φ
v
= 1.00 (LRFD)
Ω
v
= 1.50 (ASD)
V
n
= nominal shear strength
= 0.6F
y
A
w
C
v
A
w
= area of web (see sketch above)
= t
w
d
C
v
= Web shear coefficient
=
1.0 for webs of rolled “I” – shaped shapes (Conservative)
= see AISC Eq. G23, G24 and G25 p. 16.165
for other conditions
d
t
w
Beam XSect
A
w
= Shear area
in normal beam
(shaded)
A
w
= Shear area
in coped beam
(shaded)
Lecture 5  Page 3 of 9
Example 1 (LRFD)
GIVEN
: A W14x26 A992 steel beam.
REQUIRED
: Determine the FACTORED design shear strength, φ
v
V
n
for
the beam.
Step 1 – Determine area of web, A
w
:
A
w
= t
w
d
= 0.255 in(13.9 in)
= 3.54 in
2
Step 2 – Determine FACTORED design shear strength
:
φ
v
V
n
= 1.00(0.6F
y
A
w
)C
v
= 1.00(0.6(50 KSI)(3.54 in
2
)1.0)
φ
v
V
n
= 106.2 KIPS
(NOTE
: The values of φ
v
V
n
can be found directly in the “Maximum
Total Factored Uniform Load” Table 36 AISC p. 367)
d
t
w
Lecture 5  Page 4 of 9
Example 2 (ASD)
GIVEN
: The W12x30 A992 steel beam has a 4” cope and has the
SERVICE load as shown below. Disregard beam weight.
REQUIRED
:
1) Determine if the beam is acceptable on the basis of shear at the coped
end.
2) Determine if the beam is acceptable on the basis of shear at the
location of the point load.
W12x30
R1
R2
Lecture 5  Page 5 of 9
Step 1 – Determine the reaction at the left support R1
:
R1 =
'10
)'7(120KIPS
= 84 KIPS
Step 2 – Determine ALLOWABLE shear strength in coped web at R1
:
v
n
V
Ω
=
50.1
6.0
vwy
CAF
A
w
= Area of web
= t
w
h
From properties, t
w
= web thickness
= 0.260 in.
d = 12.3 in.
h = d – cope
= 12.3” – 4”
= 8.3”
A
w
= t
w
h
= 0.260”(8.3”)
= 2.16 in
2
v
n
V
Ω
=
50.1
)0.1)(16.2(50(6.0
2
inKSI
v
n
V
Ω
= 43.2 KIPS < 84 KIPS → UNACCEPTABLE
h = d  cope
d
t
w
Lecture 5  Page 6 of 9
Step 3 – Determine ALLOWABLE shear strength at point of load
:
Since the beam is not coped at this location, the design shear
strength
v
n
V
Ω
can be found from AISC Table 36 p. 371.
v
n
V
Ω
= 64.2 KIPS < 120 KIPS → UNACCEPTABLE
(Note
: This example looks ONLY at shear
. Very high loads also require a
detailed look at connections, which will be investigated later)
Possible Fixes for High Shear
:
By modifying any of the variables in the design shear equation, the
capacity may be increased. These include:
• Using a higher grade of steel (increase F
y
)
• Use a bigger beam (increase A
w
)
• Weld additional plates to the web (increase A
w
)
New plate welded
to web of beam
Lecture 5  Page 7 of 9
2. Design for Deflection
Building codes (such as the IBC) require that deflections in beams be held
to a minimum for occupancy comfort as well as to reduce likelihood of
cracking ceiling finishes such as plaster. It is considered to be a
“serviceability” check. Allowable deflection limits are dictated by the
codes, such as L/360 of the span.
Actual deflection is calculated using SERVICE LOADS in the formulas
given in the AISC p. 3208 thru 3226. These actual deflections are then
compared against the allowable deflection.
The following allowable deflection limits for steel construction are used by
the IBC:
Construction
Live Load
Snow or Wind
Roof member supporting plaster ceiling L/360 L/360
Roof member supporting nonplaster ceiling L/240 L/240
Roof member supporting no ceiling L/180 L/180
Floor members L/360 
It should be noted that the abovenoted allowable deflections are
minimums. Architects and engineers often reduce the deflection limit to
L/480 or even L/540 to ensure that floors are not “bouncy.”
Lecture 5  Page 8 of 9
Example 3
GIVEN
: The floor framing plan below. The total superimposed service
dead load = 86 PSF (not including beam weight) and the service live load
= 125 PSF.
REQUIRED
: Determine the maximum actual midspan deflection and
compare with a Live load limit = L/480 and a Dead load + Live load limit =
L/240 on the W18x35 steel beam.
Step 1 – Determine the actual max. deflection considering Live load only
:
From AISC p. 3211, the maximum deflection formula is:
∆
max
=
EI
wL
384
5
4
w = uniform live load on beam
= 8’(125 PSF)
= 1000 PLF
∆
max
=
)510)(29000000(384
)/"12'25(
12
1000
5
4
4
inPSI
ftx
PLF
⎟
⎠
⎞
⎜
⎝
⎛
= 0.59 in.
do
do
W18x35 beam
W24x62 girder
25’0”
4@8’0” = 32’0”
Lecture 5  Page 9 of 9
Step 2 – Determine allowable deflection of L/480 for live load only
:
∆
allow
=
480
L
=
480
)/"12("0'25 ft
−
= 0.625 in.
Since
∆
allow
= 0.625” > 0.59”
→
beam is acceptable
Step 3 – Determine max. defl. considering Deal load + Live load
:
∆
max
=
EI
wL
384
5
4
w = uniform dead load + live load on beam
= 8’(125 PSF + 86 PSF) + 35 PLF
= 1723 PLF
∆
max
=
)510)(29000000(384
)/"12'25(
12
1723
5
4
4
inPSI
ftx
PLF
⎟
⎠
⎞
⎜
⎝
⎛
= 1.02 in.
Step 4 – Determine allowable deflection of L/240 for LL + DL
:
∆
allow
=
240
L
=
240
)/"12("0'25 ft
−
= 1.25 in.
Since
∆
allow
= 1.25” > 1.02”
→
beam is acceptable
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