plastic sectional modulus

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PLASTIC SECTIONAL MODULUS . Shape Property, Independent of material Higher indicates more bending strength. PNA. PNA. PNA. TOTAL. PNA (Plastic Neutral Axis). TOTAL. PNA (Plastic Neutral Axis). FULLY LOADED. FORCE X LEVERARM. For rectangular sections=. FORCE X LEVERARM. - PowerPoint PPT Presentation

TRANSCRIPT

PLASTIC SECTIONAL MODULUS

3x in(z ),

• Shape Property, Independent of material• Higher indicates more bending strength

i ix A Yz

1 1 2 2x A Y A Yz

12x0.5 12x0.5

xZ 12 3in

12x1 12x1

xZ 24 3in

xZ

1 1 2 2x A Y A Yz

6 6

1y 0.5"

2y 0.5"

1A 12 2in

2A 12 2in

1A 12 2in

2A 12 2in

1y 1"

2y 1"

PNA

x 12x1.5 12x1.5z

xZ 36 3in

x 12x3 12x3z

xZ 72 3in

1y 3"

18 18 36 36

2y 3"

1A 12 2in

2A 12 2in

1A 12 2in

2A 12 2in

1y 1.5"

2y 1.5"

PNA

i ix A Yz

4 41 1 2 2 3 3x A Y A Y A Y A Yz

8x4.5 4x2 4x2 8x4.5 36 8 8 36

xZ 88 3in

PNA

4A 8 2in

1A 8 2in2A 4 2in

3A 4 2in

1y

4y

2y

3y

4 41 1 2 2 3 3xZ A Y A Y A Y A Y

PNA(Plastic Neutral Axis)

41 2 3A A A A A TOTAL

10 2 2 10 24

10x4.5 2x2 2x2 10x4.5 45 4 4 45

98 3in

2in1

y

4y

2y

3y

1A 10 2in2A 2 2in

3A 2 2in

4A 10 2in

41 2 3A A A A A TOTAL

10 2 2 10 24 2in

4 41 1 2 2 3 3xZ A Y A Y A Y A Y

10x8.5 2x4 2x4 10x8.5 85 8 8 85

186 3in

PNA(Plastic Neutral Axis)

1y

4y

2y

3y

1A 10 2in

2A 2 2in

3A 2 2in

4A 10 2in

xZ 88 3in

xZ 98 3in

xZ 186 3in

xZ 12 3in

xZ 24 3in

xZ 36 3in

xZ 72 3in

FULLY LOADED

yf2

bd

2

yf2

bd

2

1 d3 2

2 d3 2

yf yf

(INT )M FORCE X LEVERARM

2y

y

f bd 2d bdx f2 2 3 6

(INT )M y xf .S

xS For rectangular sections=2bd

6

2 d3 2

1 d3 2

yf bd2

(INT )M FORCE X LEVERARM

2

y ybd d bdf x f2 2 4

(INT )M y xf .Z i ix A Yz

yf bd2 yf bd

2

yf bd2

yf yf

i ix A Yz

1 1 2 2A Y A Y

bd d bd d. .2 4 2 4

2 2 2bd bd bd8 8 4

(INT )M y y8F x9 4F x4

M y y72F 16F

M y88F

M x yZ F xZ 88 3in

SECTION ELEVATION

y8F

y4F

y8F

y4F

y4F y4F

y8F y8F

LATERAL SUPPORT OF STEEL BEAMS

47 kip-ft

28.5 kip-ft

CASE II

W10X12

CASE IIICASE I

CASE II CASE IIICASE 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 NM

BRACING DISTANCE

BRACES

W10X12

b0 L 2.75'

CASE II

Design Moment Strength reduces as increasesbL

AT b 2.75 'LAT b 'L 8

NM 47kip-ft

NM 28.5 kip-ft

Linear variation in & bL NM

W10X12b2.75' L 8'

LATERAL BRACES

bL

bL

AT b 'L 8AT b 'L 18

NM 28.5

NM 8.25

Design Moment Strength reduces as increases

Should be avoided for load bearing floor beams.

kip-ft

kip-ft

CASE III

bL

bL 8'

CASE I CONSTRUCTION DETAILS

bL 0

x

xX-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.

yF

• Fy = 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 = wu x TRIBUTORY AREA LENGTH OF BEAM

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

[ ] x1

1000= 1.242 kips/ft

12” of Slab = 150 psf6” of Slab = 75 psf1” of Slab = 12.5 psf5” of Slab = 62.5 psf(12.5 psf for every inch of concrete thickness)*5 x 12.5 = 62.5

wu = 1.242 kips/ft

Mu2

8uw L 21.242x30

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 W16X26

SHAPE AREA

W 14x26 7.69 245

W16x26 7.68 301

XI

Page # 1-21

EXTRA SELF WEIGHT MOMENT = 21.2x0.026x30

8= 3.3 ft-kips

MOMENT STRENGTH APPLIED MOMENT 165.7 139.7 + 3.3

W16X26 is OK

CODE 30x12

180

360180

ACTUAL45

384

wl

EI= = = 2” =

Page # 3-211

w =(60)x(6x30)

30= 360 lb/ft

0.36012

45x0.03x(30x12)384x29,000x301

= 0.360 kips/ft

= 0.03 kips/in

= = 0.75” 2” OK

= 45

384wl

EI;

.kipsin

4( )in

2 .kipsin

2inin

= 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.

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