CRIOCM2012--Plastic

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26 Νοε 2013 (πριν από 3 χρόνια και 9 μήνες)

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PLASTIC SECTIONAL MODULUS



3
x
in
(z ),


Shape Property, Independent of material



Higher indicates more bending strength

i i
x
AY
z


1 1 2 2
x
A Y A Y
z
 
12x0.5 12x0.5
 
x
Z 12

3
in
12x1 12x1
 
x
Z 24

3
in
x
Z
1 1 2 2
x
A Y A Y
z
 
6 6
 
1
y 0.5"

2
y 0.5"

1
A 12

2
in
2
A 12

2
in
1
A 12

2
in
2
A 12

2
in
1
y 1"

2
y 1"

PNA

x
12x1.5 12x1.5
z
 
x
Z 36

3
in
x
12x3 12x3
z
 
x
Z 72

3
in
1
y 3"

18 18
 
36 36
 
2
y 3"

1
A 12

2
in
2
A 12

2
in
1
A 12

2
in
2
A 12

2
in
1
y 1.5"

2
y 1.5"

PNA

i i
x
AY
z


4 4
1 1 2 2 3 3
x
A Y A Y A Y A Y
z
   
8x4.5 4x2 4x2 8x4.5
   
36 8 8 36
   
x
Z 88

3
in
PNA

4
A 8

2
in
1
A 8

2
in
2
A 4

2
in
3
A 4

2
in
1
y

4
y

2
y

3
y

4 4
1 1 2 2 3 3
x
Z A Y A Y A Y A Y
   
PNA

(Plastic Neutral Axis)

4
1 2 3
A A A A A
   
TOTAL

10 2 2 10
   
24

10x4.5 2x2 2x2 10x4.5
   
45 4 4 45
   
98

3
in
2
in
1
y

4
y

2
y

3
y

1
A 10

2
in
2
A 2

2
in
3
A 2

2
in
4
A 10

2
in
4
1 2 3
A A A A A
   
TOTAL

10 2 2 10
   
24

2
in
4 4
1 1 2 2 3 3
x
Z A Y A Y A Y A Y
   
10x8.5 2x4 2x4 10x8.5
   
85 8 8 85
   
186

3
in
PNA

(Plastic Neutral Axis)

1
y

4
y

2
y

3
y

1
A 10

2
in
2
A 2

2
in
3
A 2

2
in
4
A 10

2
in
x
Z 88

3
in
x
Z 98

3
in
x
Z 186

3
in
x
Z 12

3
in
x
Z 24

3
in
x
Z 36

3
in
x
Z 72

3
in
FULLY LOADED

y
f
2
 
 
 


bd
2
y
f
2
 
 
 


bd
2


1 d
3 2


2 d
3 2
y
f
y
f
(INT)
M

FORCE X LEVERARM





2
y
y
f
bd 2d bd
x f
2 2 3 6
 
 
 
 
 
 
 
(INT)
M

y x
f.S
x
S
For rectangular sections=

2
bd
6


2 d
3 2


1 d
3 2


y
f


bd
2
(INT)
M

FORCE X LEVERARM







2
y
y
bd d bd
f x f
2 2 4
 
 
 
 
(INT)
M

y x
f.Z
i i
x
AY
z




y
f


bd
2


y
f


bd
2


y
f


bd
2
y
f
y
f
i i
x
AY
z


1 1 2 2
A Y A Y
 
bd d bd d
..
2 4 2 4
       
 
       
       




2 2 2
bd bd bd
8 8 4
  
(INT)
M

y y
8Fx9 4Fx4

M

y y
72F 16F

M

y
88F
M

x y
Z F
x
Z 88

3
in

SECTION ELEVATION

y
8F
y
4F
y
8F
y
4F
y
4F
y
4F
y
8F
y
8F
LATERAL
SUPPORT OF
STEEL BEAMS

47 kip
-
ft

28.5 kip
-
ft

CASE II

W10
X
12

CASE III

CASE I

CASE II

CASE III

CASE I

CASE I if lateral brace is spaced 0
-
2.75’

CASE II if lateral brace is spaced 2.75’
-
8’

CASE III if lateral brace is spaced more than 8’

W10X12

CASE I

Design Moment Strength = = 47 kip
-
ft


N
M
BRACING DISTANCE

BRACES

W10X12

b
0 L 2.75'
 
CASE II

Design Moment Strength reduces as increases

b
L
AT


b
2.75'
L
AT


b
'
L 8


N
M 47
kip
-
ft



N
M 28.5
kip
-
ft

Linear variation in &

b
L

N
M
W10X12

b
2.75'L 8'
 
LATERAL BRACES

b
L
b
L
AT


b
'
L 8
AT


b
'
L 18


N
M 28.5


N
M 8.25
Design Moment Strength reduces as increases

Should be avoided for load bearing floor beams.

kip
-
ft

kip
-
ft

CASE III

b
L
b
L 8'

CASE
I CONSTRUCTION DETAILS

b
L 0

x

x

X
-
X

CONCRETE

W
-
SHAPE

STEEL STUDS

Most Common Current Practice Using Metal Deck
and Shear Studs. Steel and Concrete Deck can
be designed as Composite or Non
-
Composite

Use

50

ksi

and

select

a

shape

for

a

typical

floor

beam

AB
.

Assume

that

the

floor

slab

provides

continuous

lateral

support
.

The

maximum

permissible

live

load

deflection

is

L
/
180
.

The

service

dead

loads

consist

of

a

5
-
inch
-
thick

reinforced
-
concrete

floor

slab

(normal

weight

concrete),

a

partition

load

of

20

psf
,

and

10

psf

to

account

for

a

suspended

ceiling

and

mechanical

equipment
.

The

service

live

load

is

60

psf
.



y
F


F
y

= 50ksi



Case 1






LIVE LOAD = 60psf

LL


DEAD LOADS


1) 5” Slab


2) Partition = 20psf


3) Ceiling, HVAC = 10psf

GIRDER

GIRDER

(PRIMARY BEAMS)

not to exceed

180
L
Figure 1

DESIGN LOAD (kips/ft) on AB =
w
u

x

TRIBUTORY AREA


LENGTH OF BEAM

  

1.2x(62.5 20 10) 1.6(60) x[6x30]
30
[ ]
x

1
1000
= 1.242 kips/ft

12” of Slab = 150 psf

6” of Slab = 75 psf

1” of Slab = 12.5 psf

5” of Slab = 62.5 psf

(12.5 psf for every inch of concrete
thickness)

*5
x

12.5 = 62.5


w
u

= 1.242 kips/ft

M
u

2
8
u
w L
2
1.242x30
8
=

=

= 139.7 ft
-
kips CASE 1

139.7

STRENGTH OF W14
X

26


= 150.7 ft
-
kips

STRENGTH OF W16
X

26


= 165.7 ft
-
kips

Page # 3
-
127

Page # 1
-
22

Page # 1
-
23

Page # 1
-
20

SELECT W16
X
26


SHAPE

AREA

W 14
x
26

7.69

245

W16
x
26

7.68

301

X
I
Page # 1
-
21

EXTRA SELF WEIGHT MOMENT
=

2
1.2x0.026x30
8
= 3.3 ft
-
kips

MOMENT STRENGTH APPLIED MOMENT





165.7 139.7 + 3.3


W16
X
26 is OK


CODE
30x12
180
360
180
ACTUAL
4
5
384
wl
EI
=

=

=
2”

=

Page # 3
-
211

w

=

(60)x(6x30)
30
= 360 lb/ft

0.360
12

4
5x0.03x(30x12)
384x29,000x301

= 0.360 kips/ft

= 0.03 kips/in

=

= 0.75” 2” OK

=

4
5
384
wl
EI
;

.
kips
in
 
 
 
4
( )
in
2
.
kips
in
2
in
in
=
in


4
( )
in
=


kips/in

Typical

Copes

for

a

shear

connection

of

a

large

girder

to

column

web
.

Note

that

duct

holes

have

to

be

strengthened

by

plates
.

Also,

holes

are

at

third

point

where

shear

&

moment

are

not

maximum
.

Cantilever construction
for projected balcony.

If

shear

studs

are

noticed

on

beams

and

column

then

those

members

have

to

be

encased

in

concrete

for

increasing

fire

resistance

of

steel
.

Details of web opening in steel girders for HVAC ducts.