Steel Design

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26 Νοε 2013 (πριν από 3 χρόνια και 10 μήνες)

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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
.




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