5.8 SHEAR STRENGTH
Where:
–
V
u
= maximum shear based on the controlling combination of
factored loads
–
Ø = resistance factor for shear = 0.90
–
V
n
= nominal shear strength
Consider
a
simple
beam
as
shown
in
Fig
.
a
.
At
a
distance
x
from
the
left
end
and
at
the
neutral
axis
of
the
cross
section,
the
state
of
stress
is
as
shown
in
Fig
.
d
.
Because
this
element
is
located
at
the
neutral
axis,
it
is
not
subjected
to
flexural
stress
.
From elementary mechanics of materials, the shearing stress is
This equation based on the assumption that the stress is constant
across the width b, and it is therefore accurate only for small
values of b, so the equation can not be applied to the flange of a
W shape in the same manner as for the web.
The average stress in the web w is V/ A
w
No big difference between the average stress and the maximum
stress
The web will completely yield long before the flanges begin to
yield.
The following Figure shows the shearing stress distribution for a W

shape.
We
can
write
the
equation
for
the
stress
in
the
web
at
failure
as
:
Where A
w
is the area of the web.
The nominal strength corresponding to this limit state is therefore
The relationship between shear strength and the width

thickness
ratio is analogous to that between flexural strength and the width
thickness ratio (for FLB or WLB) and between flexural strength
and unbraced length (for LTB)
w
A
y
0.60F
n
V
y
0.60F
w
A
n
V
v
f
This
relationship
is
illustrated
in
the
Figure
below
:
This
relationship
given
in
the
AISC
as
follows
:
w
y
n
y
w
A
F
V
and
y
instabilit
web
no
is
There
F
E
t
h
For
60
.
0
,
,
45
.
2
w
y
w
y
n
y
w
y
t
h
F
E
A
F
V
and
occur
can
buckling
web
Inelastic
F
E
t
h
F
E
For
45
.
2
60
.
0
,
07
.
3
,
45
.
2
2
52
.
4
260
,
07
.
3
w
w
n
w
y
t
h
E
A
V
t
h
F
E
For
buckling
web
elastic
is
state
limit
The
Where,
A
w
is
the
area
of
the
web
=
d*
t
w
And
d
is
the
overall
depth
of
the
web
.
If
h/
t
w
is
greater
than
260
,
web
stiffeners
are
required
.
Shear
is
rarely
a
problem
in
rolled
steel
beams
;
the
usual
practice
is
to
design
a
beam
for
flexure
and
then
to
check
it
for
shear
.
Example
5
.
7
Check
the
beam
in
Example
5
.
6
for
shear
.
Solution
:
From
Example
56
:
A
W
14
X
90
with
F
y
=
50
ksi
is
used
.
W
u
=
2
.
080
kips/ft
and
L=
40
ft
kips
41.60
2
40
*
2.080
2
L
u
w
n
V
From
the
Manual
0
.
59
50
29000
45
.
2
45
.
2
9
.
25
y
w
F
E
t
h
The
strength
is
governed
by
shear
yielding
of
the
web
y
w
F
E
2.45
t
h
since
(OK)
kips
41.6
kips
166
184.8
*
0.90
n
φV
kips
184.8
0.440
*
14.0
*
50
*
0.60
w
A
y
0.6F
n
V
The
shear
design
strength
is
greater
than
the
factored
load
shear,
so
the
beam
is
satisfactory
.
BLOCK SHEAR
To facilitate the connection of beams to other beams so that the
top flanges are at the same elevation, a short length of the top
flange of one of the beams may be cut away, or coped. If a coped
beam is connected with bolts as in Figure, segment ABC will tend
to tear out.
The applied load in this case will be
the vertical beam reaction.
Shear will occur along line AB and
there will be tension along BC. Thus
the block shear strength will be a
limiting value of the reaction.
We covered the computation of block shear strength in Chapter3
Example 5

8
5.9 Deflection
Steel beams are designed for the factored design loads. The moment
capacity, i.e., the factored moment strength (
φ
b
M
n
) should be greater
than the moment (M
u
) 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 beam, being serviceable usually means that the deformations,
primarily the vertical sag, 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.
Appropriate limits for deflection can be found from the governing
building code.
For the common case of a simply supported, uniformly loaded
beam such as that in the following Figure, the maximum vertical
deflection is:
Deflection limit:
Example 5

9 :
solution
Ponding is one deflection problem that does affect the safety of a
structure.
The AISC specification requires that the roof system have
sufficient stiffness to prevent ponding.
5
.
10
DESIGN
:
Beam design entails the selection of a cross

sectional shape that will have
enough strength and that will meet serviceability requirements.
The design process can be outlined as follows:
1.
Compute the factored load moment.
2.
Select a shape that satisfies this strength requirement. This can be
done in one of two ways:
Assume a shape, compute the design strength, and compare it with
the factored load moment.
Use beam design chart in part 5 at Manual (preferred).
3.
Check the shear strength.
4.
Check the defection.
Beam Design Charts
:
–
Many graphs, charts, and tables are available for the
practicing engineer, and these aids can greatly simplify the
design process.
–
To determine the efficiency, they are used in design offices.
–
You should approach their use with caution and not allow
basic principles to become obscured.
–
The curves of design moment versus un braced length given
in
Part 5
of the Manual.
–
All curves were generated with
F
y
=50
ksi
and C
b
= 1.0.
–
For other values of C
b
simply multiply the design moment
from the chart by C
b.
The following curve described the design moment
Ф
b
M
n
as a
function of unbraced length
L
b
for a particular compact shape.
Remember
that
design
strength
can
never
exceed
plastic
moment
Noncompact
shapes
may
fail
due
to
local
buckling
Two sets of curves are available, one for W

shapes and one for C

shapes and MC

shapes.
Example 5.12
Use A992 steel and select a rolled shape for the beam shown
below. The concentrated load is a service live load, and the
uniform load is 30% dead load and 70% live load. Lateral bracing
is provided at the ends and at mid span. There is no restriction on
deflection.
5.11
FLOOR AND ROOF FRAMING SYSTEMS
5.12
HOLES IN BEAMS
If beam connections are made with bolts, holes will be punched or
drilled in the beam web or flange.
Sometimes electrical conduits and ventilation ducts need large
holes.
Ideally, holes should be placed in the web only at section of low
shear, and holes should be made in the flanges at points of low
bending moment.
This is not always be possible, so the effect of the holes must be
accounted for.
The effect of small holes will be small, particularly for flexure.
AISC B
10
permits bolt holes in flanges to be ignored when:
0.75
F
u
A
fn
≥
0.90
F
y
A
fg
where,
A
fg
is the gross tension flange area
and,
A
fn
is the net tension flange area.
If this condition is not met, the previous equation is solved for an
effective tension flange area that satisfies that criterion.
fn
A
y
F
u
F
6
5
fe
A
or
y
0.9F
fn
A
u
0.75F
fe
A
fg
A
See examples
5.14
and
5.15
5.13
OPEN

WEB STEEL JOISTS
Open

web steel joists are prefabricated trusses of the type shown
in the Figure.
They are used in floor and roof systems.
For a given span, it is lighter in weight and its more easier for
electrical conduits and ventilation ducts than a rolled shape .
More economical than a rolled shape
5
.
14
Bearing
plates
and
column
base
plates
:
–
The function of the plate is to distribute a concentrated load to
the supporting material
–
Two types of beam bearing plates are considered
One that transmits the beam reaction to a support
One that transmits a load to the top flange of a beam.
–
The design of the bearing plate consists of three steps.
Determine dimension N so that web yielding and web
crippling are prevented
Determine Dimension B so that the area N
×
B is sufficient to
prevent the supporting material from being crushed in
bearing.
Determine the thickness t so that the plate has sufficient
bending strength.
Web Yielding
Web yielding is the compressive crushing of a beam web caused by a force
acting on the flange directly above or below the web
When the load is transmitted through a plate, web yielding is assumed to take
place on the nearest section of width
t
w
.
In rolled shape, this section will be at the toe of the fillet, a distance k from the
outside face of the flange
If the load is assumed to distribute itself at a slop of
1
:
2.5
as shown
The area at the support subject to yielding is (
2.5
k + N)
t
w
The nominal strength for web yielding at the support is:
The bearing length N at the support should not be less than k.
At the interior load, the length of the section subject to yielding is
The design strength is
ØRw
where
Ø =
1.0
Concrete
Bearing
Strength
Usually, concrete used as the material beam support.
This material must resist the bearing load applied by steel plate
If the plate covers the full area of the support,
The nominal strength is:
If the plate does not cover the full area of the support
Where
f
c
is the compressive strength of concrete after
28
days
A
1
is the bearing area
A
2
is the full area of the support
1
'
c
p
A
0.85f
P
1
2
1
'
c
p
A
A
A
0.85f
P
If area A
2
is not concentric with A
1
, then A
2
should be taken as the
largest concentric area that is geometrically similar to A
1
, as
illustrated in the following Figure
AISC also requires
The design bearing strength is
Ф
c
P
p
where
Ф
c
=
0.60
2
1
2
A
A
Plate
Thickness
Where
R
u
is
the
support
reaction
B
is
width
of
the
bearing
plate
N
is
length
of
the
bearing
plate
y
2
u
BNF
n
2.222R
t
Example
5
.
17
Design
a
bearing
plate
to
distribute
the
reaction
of
a
W
21
x
68
with
a
span
length
of
15
ft
10
inches
center
to
center
of
supports
.
The
total
service
load,
including
the
beam
weight
,
is
19
kips
with
equal
parte
dead
and
live
load
.
The
beam
is
to
be
supported
on
reinforced
concrete
wall
with
f
c
=
3500
psi
.
the
beam
made
of
A
992
steel
and
the
plate
is
A
36
.
Solution
The
factored
load
is
=
1
.
2
*
4
.
5
+
1
.
6
*
4
.
5
=
12
.
600
kips/ft
The
reaction
is
=
12
.
6
*
15
.
83
/
2
=
99
.
3
kips
Determine
the
length
of
bearing
N
required
to
prevent
web
yielding
.
R
u
=
2
.
5
k
+
N)
F
y
t
w
=
(
2
.
5
(
1
.
438
)
+
N)
50
*
0
.
430
≥
99
.
73
N
≥
1
.
044
Determine
the
length
of
bearing
N
required
to
prevent
web
crippling
.
Assume
N/d
>
0
.
2
N
≥
3
.
0
in
Check
the
assumption
N/d
=
3
.
0
/
21
.
1
=
0
.
14
<
0
.
2
(N
.
G
.
)
so
for
N/d
<
0
.
2
99.73
0.43
0.685
*
50
*
29000
1.5
0.685
0.43
0.2
21.1
4N
1
2
0.43
*
0.40
*
075
u
R
w
t
f
t
y
EF
1.5
f
t
w
t
0.2
d
4N
1
2
w
t
0.40
φ
99.73
0.43
0.685
*
50
*
29000
1.5
0.685
0.43
21.1
N
1
2
0.43
*
0.40
*
075
u
R
w
t
f
t
y
EF
1.5
f
t
w
t
d
N
1
2
w
t
0.40
φ
3
3
N
≥
2
.
59
in,
and
N/d
=
2
.
59
/
21
.
1
=
0
.
12
<
0
.
2
(OK)
Try
N
=
6
.
0
Assume
full
area
of
support
is
used,
the
required
plate
area
is
A
=
Ф
*
0
.
85
*
f
c
*
A
≥
R
u
A
=
0
.
60
*
0
.
85
*
3
.
5
*
A
≥
99
.
73
A
≥
55
.
87
in
2
B
=
55
.
87
/
6
=
9
.
31
in
The
flange
width
of
a
W
21
x
68
is
8
.
27
<
9
.
31
,
use
B
as
10
in
Compute
the
required
plate
thickness,
n
=
(B
–
2
k)/
2
=
(
10
–
2
*
1
.
19
)/
2
=
3
.
81
in
Use
a
PL
1
.
25
*
6
*
10
.
in
1.22
36
*
6
*
10
3.81
*
99.73
*
2.222
BNF
n
2.222R
t
2
y
2
u
Column
Base
Plates
Major differences between bearing and base plates are:
Bending in bearing plates in one direction, whereas column base
plates are subjected to two

way bending.
Web crippling and web yielding are not factors in column base
plate design.
Design steps:
Determine the allowable strength of the foundation
Determine the required column base plate area
Select column base plate dimension (not less than column
dimension)
Determine the thickness
Where,
y
u
0.9BNF
2P
L
t
p
c
u
f
f
f
P
P
b
d
db
X
X
X
b
B
n
2
4
1
1
1
2
2
8
.
0
2
0.95d

N
m
)
n
n
(m,
max
L
60
.
0
4
1
c
f
db
n
Pp = nominal bearing strength from
AISC equation
1
'
c
p
A
0.85f
P
1
2
1
'
c
p
A
A
A
0.85f
P
Example
5
.
18
A
W
10
x
49
is
used
as
a
column
and
is
supported
by
a
concrete
pier
as
shown
in
the
Figure
.
The
top
surface
of
pier
is
18
in
by
18
in
.
Design
an
A
36
base
plate
for
a
column
dead
load
of
98
kips
and
a
live
load
of
145
kips
the
concrete
strength
is
f
c
=
3000
psi
.
The factored load =
1.2
*
98
+
1.6
*
145
=
349.6
Compute the required bearing area (assume that plat area < pier
area)
(OK)
2
1.41
161.1
18
*
18
A
A
check
in
161.1
A
349.6
A
18
*
18
A
*
3
*
085
*
0.6
P
A
A
A
(0.85)f
φ
1
2
2
1
1
1
u
1
2
1
'
c
c
Also
the
plate
must
be
at
least
as
large
as
the
column,
so
B
f
d
=
10
*
9
.
98
=
99
.
8
<
161
.
1
in
2
(OK)
For
B
=
N
=
13
in,
A
1
provided
=
3
*
13
=
169
>
161
.
1
in
2
Check
the
assumption
Plate
area
=
169
<
pier
area
=
18
*
18
(OK)
Determine
the
required
thickness
in
2.497
10
*
9.98
4
1
db
4
1
n
in
2.5
2
8
13
2
0.8b
B
n
in
1.76
2
9.48
13
2
0.95d

N
m
f
f
in
0.893
6
*
13
*
13
*
0.9
349.6
*
2
2.5
0.9BNF
2P
L
t
y
u
As
a
conservative
simplification,
let
λ
=
1
,
giving
L
=
max(m,
n,
λ
n’)
=
max
(
1
.
76
,
2
.
5
,
2
.
497
)
=
2
.
5
in
The
required
plate
thickness
is
Use
a
PL
1
x
13
x
13
5
.
15
BIAXIAL
BENDING
Please
try
to
understand
it
alone
May
be
necessary
in
some
of
your
projects
.
If
there
is
a
time,
we
will
discuss
it
later
Any
problem,
you
can
see
me
in
my
office
within
the
office
hours
.
The
last
section
5
.
16
canceled
.
“
بر لله يتاممو يايحمو يكسن و يتلاص نا لق
نيملاعلا
*
نيملسملا لوأ انأو ترمأ كلذبو هل كيرش لا
"
ميظعلا الله قدص
Σχόλια 0
Συνδεθείτε για να κοινοποιήσετε σχόλιο