all steel ppt

Timber and Steel DesignTimber and Steel Design

Welcome to

3 2549

Lecture 1 - Introduction

Mongkol JIRAVACHARADET

S U R A N A R E E INSTITUTE OF ENGINEERINGUNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

-

-

- ...

-

Conduct of CourseConduct of Course

Midterm ExamMidterm Exam 40 %40 %

Final ExamFinal Exam 40 %40 %

Projects/Assignment/QuizzesProjects/Assignment/Quizzes 20 %20 %

Grading Policy

100 - 90Final Score Grade

A89 - 85 B+84 - 80 B79 - 75 C+74 - 70 C69 - 65 D+64 - 60 D59 - 0 F

1) > 20% = 5

2)

3) Project :- 3 - rejectreject

What is Structural Engineering ?What is Structural Engineering ?

in such a way that nobody suspect the extent of our ignorance.

Structural Engineering is:

the art of molding materials we do not entirely understand

into shapes we cannot precisely analyse,

so as to withstand forces we cannot really assess,

STABILITYSTABILITY SAFETYSAFETY ECONOMYECONOMY

Frame structural systemFrame structural system

LinganLingan Island Island PowerplantPowerplant,, Skeleton FrameSkeleton Frame

Nashville Convention Nashville Convention CentreCentre,, tower overview tower overview

Pan Pacific Hotel and Convention Pan Pacific Hotel and Convention CentreCentre,, VancouverVancouver,, main tower main tower

Suspension structural systemSuspension structural system

RAMA IX Bridge, Bangkok ThailandRAMA IX Bridge, Bangkok Thailand

RAMA VIII Bridge, Bangkok ThailandRAMA VIII Bridge, Bangkok Thailand

Steel Applications

STEEL is one of most widely used materials for structures

What is STEEL ?What is STEEL ?

manganese (Mn), silicon (Si),

aluminum (Al), nickel (Ni),

chromium (Cr), molybdenum (Mo),

copper (Cu), vanadium (V),

niobium (Nb), titanium (Ti) and

boron (B).

Steel is an alloy of iron (Fe) and carbon (C). Depending on the desired steel properties, one or more of other alloying elements are also added to the steel. The important ones are:

Iron Ore

STEEL Manufacturing STEEL Manufacturing

Advantages of Steel

4 High Strength to Weight

4 Uniformity (properties not change with time)

4 Highly Ductile

4 Good Fracture Toughness

4 Easily Constructed and Modified Structures

4 Easily Recycled

Disadvantages of Steel

4 Maintenance Cost

4 Requires Fireproofing

4 Slender Members Susceptible to Buckling

4 Fatique

4 Brittle Fracture at very low temperature

Steel Casting Steel Casting && Hot RollingHot Rolling

Continuous Caster

Blooms

Billets

Slabs

Rolling Mill

W section S section Unequal-legangle

Equal-legangle

Channel Zee Tee

Flange

Web 0-5% slope 16.67% slope

16.67% slope

HotHot--Rolled Steel ShapeRolled Steel Shape

ColdCold--Formed SteelFormed SteelCold-formed steel shapes are formed at room temperature.

Angle Channel Stiffenedchannel

Zee StiffenedZee

Hat StiffenedHat

D

D

Stress Stress -- Strain RelationshipsStrain Relationships

su

strain hardening

ultimatestress

sf

necking

fracturestress

spl

elasticregion

elasticbehavior

elastic limit

proportional limit

sY

plastic behavior

yielding

yield stress

true fracture stresssf

e

s

Low-carbon steel

ASTM A36JIS G3101 SS400JIS G3106 SM400TIS SM400

4000 - 5200

DESIGNATION

YIELD STRESS(kg/cm2) TENSILE STRENGTH

(kg/cm2)Thickness (mm)16 or under over 16

2500 2400

JIS G3101 SS490 5000 - 62002900 2800

JIS G3106 SM490TIS SM490 5000 - 62003300 3200

ASTM A572JIS G3106 SM520TIS SM520

5300 - 65003700 3600

JIS G3106 SM570 5800 - 73004600 4500

In this course:

Yield stress, Fy = 2500 kg/cm2

Tensile Strength Fu = 4000 kg/cm2

Properties of Structural SteelsProperties of Structural SteelsASTM = American

JIS = Japanese

TIS = Thailand

Steel Sections : Steel Sections : Wide Flange : WFWide Flange : WF

Channel Sections :Channel Sections :

Angle Sections :Angle Sections :

C Sections :C Sections :

Round Tube :Round Tube :

Square Tube :Square Tube :

Rectangular Tube :Rectangular Tube :


Lecture 2 - Specification, Loads andDesign Methods

< Specifications and Building Codes

< Design Loads

< Wind Load

< Structural Members

< Load Transfer between Structural Members

< Design MethodsMongkol JIRAVACHARADET

S U R A N A R E E INSTITUTE OF ENGINEERINGUNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

Specifications

Developed by organizations such as AISC, ACIASCE, and EIT

Recommendations of good practice based onthe accepted body of knowledge

NOT legally enforceable

OrganizationsEIT = Engineering Institute of Thailand

AISC = American Institute of Steel Construction

ASCE = American Society of Civil Engineers

AASHTO = American Association of State Highwayand Transportation Officials

UBC = Uniform Building Code

BOCA = Building Officials & Code Administrators

ATTC = American Institute of Timber Construction

American Institute of American Institute of Steel Construction SpecificationSteel Construction Specification

Latest: Steel Construction Manual, Thirteenth Edition

The following specifications, codes, and standards are printed in this Manual:

2005 AISC Specification for Structural Steel Buildings

2004 RCSC Specification for Structural Joints Using ASTM A325 or A490 Bolts

2005 AISC Code of Standard Practice for Steel Buildings and Bridges

- The 3rd Edition (2001) Load and Resistance Factor Design (LRFD) Manual.

This Manual replaces both :This Manual replaces both :

- The 9th Edition (1989) Allowable Strength Design (ASD) Manual and

Visite AISC Website @

http://www.aisc.org

... : http://www.eit.or.th

...... 1015-40

1020-46

487 39 () . 10310 0-2319-2410~3 0-2319-2710~11 E-mail: [email protected]

((.....).)

Building Codes

-- ....

- -

-

ArchitecturalFunctional Plans

DesignDesign ProcessProcess

Select StructuralSystem

Trial Sections,Assume Selfweight

Analysis for internalforces in member

Member Design

Acceptable?Redesign

NG

Final Design& Detailing

OK

Design LoopDesign Loop

AllowableAllowable StreStrengthngth Design Design ((ASDASD))

Design MethodsDesign Methods

ASDASD methodmethod remainsremains practicalpractical andand isis stillstill widelywidely usedused

allowable strength of each structural component equals or exceeds the

required strength

Design shall be performed in accordance with :

W /na RR

whereRa = required strength (ASD)

Rn = nominal strength

W = safety factor

Rn / W = allowable strength

... 1015-40

Load and Resistance Factor Design Load and Resistance Factor Design ((LRFDLRFD))

LRFDLRFD willwill becomebecome thethe dominantdominant methodmethod inin thethe futurefuture

design strength of each structural component equals or exceeds the

required strength

Design shall be performed in accordance with :

nu RR f

where

Ru = required strength (LRFD)

Rn = nominal strength

f = resistance factor

f Rn = design strength

Factor of Safety (FS) =Strength of member

Strength due to load

1) Material strength: initial variation, creep, corrosion and fatigue

2) Method of analysis

3) Disaster (hurricane, earthquake)

4) Fabrication and erection stress

5) Technology change live load ; traffic load

6) Live load estimation

7) Other --> residual stress, stress concentration and variation

in dimension

Uncertainties affecting F.S.

Design Loads

< Dead Loads - self weight not known before design

< Live Loads - change in position and magnitude

< Wind Load

< Seismic Load

< Snow/Rain Load

< Earth Pressure

Dead Loads

Caused by the weight of structure

Include both the load bearing and non-load

bearing elements in a structure

Generally can be estimated with reasonable

certainty

kg/m3 2,400 2,320 500-1,200 7,850 kg/m2 14 50, 5 10-30 5 180-360 100-200

Live Loads

Floor Loads

Snow and Ice: 50 - 200 kg/sq.m.

Traffic Load & Pedestrian Load for Bridges

Impact Loads

Lateral Loads: Wind & Earthquake

6 (.. 2527) ... .. 2522

(kg/m2)

(1)

(2) (3)

(4)

(5)

(6) ()

()

30

100150

200

250

300

300

() 6 (.. 2527) ... .. 2522

(kg/m2)

(7) ()

()

(8) ()

()

(9)

400

500

500

600

(10)

500

800

Wind Loads

25.0 Vq r=

q = stagnation pressure (./.2)

V = basic wind speed 10 (./.)

ASCE 7-98 200483.0 VKq =

K = 10

10 5010 < h < 20 8020 < h < 40 120 40 160

() (./..)

WIND DIRECTION

Windwardside

Leew

ard

side

0 m

10 m

20 m

30 m

Step wind loading

... .. 2522

19

50(8) 40(7) 30(6) 20(5) 10(4) 0(3) 0(2) 0(1)

.. = 20 .. = 20

.. = 20 .. = 20

.. = 20 .. = 20

.. = 20 .. = 20

.. = 20 .. = 20

.. = 20 .. = 20

.. = 20 .. = 20

.. = 20 .. = 20

98.8%

97%

95%

92.9%

= 8(20)+3(20)+0.9(20)+0.8(20)+0.7(20)+0.6(20)+ 0.5(20) = 290 = 90.625%

100%

100%

= 120 = 100%

88.9%87.5%

Load Transfer between Structural Members

Roof + Dead load

Snow, Rain, Windand Construction load Floor loads

Slab + Dead load Wall load

Beam + Dead load

Foundation

Wind load Column + Dead load

Tributary area = 0.5SL sq.m

Load on beam = 0.5wSL kg/m

Floor load = w kg/sq.m

LoadLoad from from PrecastPrecast Concrete SlabConcrete Slab

S

L

Example: CPAC Hollow Core Slab HC100Example: CPAC Hollow Core Slab HC100

600 mm

100 mm

SLAB WEIGHT 296 KG/M2

PC WIRE 6 4 MM.

SPAN 4 M.

LIVE LOAD 300 KG/M2

L

L : Beam span

w = ? kg/m Superimposed DL = 80 KG/M2

Total Load = 296 + 300 + 80 = 676 KG/M2

Load on Beam = 676 4 / 2 = 1,352 KG/M4 m

2 m 2 m

1 m 1 m

Load Tributary Area on Beam

2 m DISTRIBUTIVEWIDTH TO INT.FLOOR BEAM

Example 2.1: Compute load on structural membersExample 2.1: Compute load on structural members

EXTE

RIO

R G

IRD

ER

EXTERIORCOLUMN

INTE

RIO

R G

IRD

ER

INTERIOR COLUMN

EXTERIOR FLOOR

BEAM

INTERIOR FLOOR

BEAM

A

B

8 m

2 m TYP.

1 2 34 m 4 m

INDICATESCONCENTRATEDLOAD POINT ATTRANSFER TOGIRDER

Live Load = 500 kg/m2

Dead Load = 600 kg/m2

2 m

2 m

4 m

INTERIOR FLOOR BEAM :INTERIOR FLOOR BEAM :

w = 2200 kg/m

4 mR R

w = 2 (500 + 600) (2) / 2 = 2,200 kg/m

R = (2,200) (4) / 2 = 4,400 kg/m

Reaction @ Beam End :Reaction @ Beam End :

INTERIOR GIRDER :INTERIOR GIRDER : 21 3

R R

R R

R R

P = 2 R = 2 4,400 = 8,800 kg/m

Point Load on Girder :Point Load on Girder :

P = 8,800 kg

8 mR R

P P

R = 3 P / 2 + P1/2 = 3 8,800 / 2 + 4,400 = 17,600 kg/m

Reaction transfer to Column :Reaction transfer to Column :

P1/2P1/2


< Allowable Tensile Strength< Net Areas< Staggered Holes< Effective Net Area< Block Shear

Rama 8 Bridge

Lecture Lecture 33 -- Analysis of Tension Members

Mongkol JIRAVACHARADETS U R A N A R E E INSTITUTE OF ENGINEERINGUNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

T

S W

Types of Tension Members

Tension Members subjected to axial tensile forcesTension Members subjected to axial tensile forces

- truss members

- bracing for buildings and bridges

- cables in suspended roofand bridges

P

Stress in an axially loaded tension member:

The stress P/A must be less than a limiting stress F or

f =PA

< FPA

P < FAThus the load P must be less than FA or

The stress in a tension member is uniform throughout the cross-section except:

- near the point of application of load, and

- at the cross-section with holes for bolts or other discontinuities, etc.

P

Gusset plate

22 mm hole

20 x 1 cm. bar

For example, consider an 20 x 1 cm. bar connected to a gusset plate and loaded in tension as shown below.

b b

a a

Section b-b

Section a-a

Area of bar = 20 x 1 = 20 cm2

Area of bar = (20 2 x 2.2) x 1

= 15.6 cm2

From f = P/A, the reduced area of Section b b will be subjected to higher stresses.

However, the reduced area and therefore the higher stresses will be localizedaround Section b b.

Section b-b

P P

TENSILE STRENGTHTENSILE STRENGTH

Use the minimum of

P = 0.60 P = 0.60 FFyy AAggYield Yield StrengthStrength ::

P = 0.50 FP = 0.50 Fu u AAeeUltimate Ultimate StrengthStrength ::

0.50 Fu Ae 0.50 Fu Ae0.60 Fy Ag

AAggGrossGrossAreaArea

AAnnNetNet

AreaArea

AAeeEffectiveEffective

AreaArea

The unreduced area of the member = cross-section total area

The reduced area of the member = Ag hole area

which may be equal An or smaller

Hole = bolt + punched(1/16 or 1.5 mm)

+ damaged metal (1/16 or 1.5 mm)

= bolt + 3 mm

T b

d

TNet area

t

(Net Area), (Net Area), AAnn

Ag = Gross Area =

An = Net Area =

An = Ag -

3-1 36 AISC A36 19 ..

20 cm

36 ton36 ton 1.5 cm

Ag = 20 .1.5 . = 30 .2

An = 30 .2 2(1.9 . + 0.3 .)(1.5 .)= 23.4 .2

f = 36(1,000) .. / 30 .2 = 1,200 ../.2

0.60Fy = 0.60(2,500) = 1,500 ../.2 < 1,200 ../.2 OK f = 36(1,000) .. / 23.4 .2 = 1,539 ../.2

0.50Fu = 0.50(4,000) = 2,000 ../.2 < 1,539 ../.2 OK

3-2 25 19 .. A36

15 cm

25 ton25 ton t

Ag = 25 /0.60Fy = 25/(0.62.5) = 16.67 .2

t = 16.67/15 = 1.11 . 12 ..

An = 25 /0.50Fu = 25/(0.54.0) = 12.5 .2

t

An = 15 t (1.9 + 0.3) t = 12.8 t

12.5 = 12.8 t

t = 0.977 . 10 ..

3-3 P A36 22 ..

Bar 15 cm1.2 cm

P

P = 0.60Fy Ag = 0.60(2.5)(15)(1.2) = 27

An = 1.2(152.20.3) = 15 .2 = Ae

P = 0.50Fu Ae = 0.50(4.0)(15) = 30

P 27 n

Example 3.4 A 15 x 1 cm bar of A572 Gr. 50 steel is used as a tension member. It is connected to a gusset plate with six 22 mm. diameter bolts as shown in below. Assume that the effective net area Ae equals the actual net area An and compute the tensile design strength of the member.

P

Gusset plate

22 mm bolt

15 x 1 cm. bar

Solution

Gross section area = Ag = 15 x 1 = 15 cm2

Net section area = An = (15 2 x (2.2+0.3)) x 1

= 10 cm2

Gross yielding design strength = 0.6 Fy Ag

= 0.6 x 3.5 x 15 = 31.5 ton

Fracture design strength = 0.5 Fu Ae

Fracture design strength = 0.5 x 4.5 x 10 = 22.5 ton

Assume Ae = An (only for this problem)

Design strength of the member in tension = smaller of 31.5 ton and 22.5 ton

Therefore, design strength = 22.5 ton (net section fracture controls) Ans

Staggered HolesStaggered Holes A

B

A

B

++-= g

sdnbtAn 4)3.0(

2

min

s

gs = pitch distance

g = gage distance

Possible failure paths: ABC or ABDE

A

C

B

E

D

Compute net area of each path:

Select the minimum net areaSelect the minimum net area

T T

10 cm

10 cm

10 cm

10 cm

A

B

D

C

E

F

10 cm

40 2(2.2) 35.6 ABCD cm= - =

An = 35.6(1.2) = 42.7 cm2

(control)

21040 3(2.2) 35.94(10)

ABCEF cm= - + =

21040 2(2.2) 36.854(20)

ABEF cm= - + =

Plate thickness = 12 mm

Bolt diameter = 19 mm

3-4

TT

5 cm

5 cm

5 cm

s s

AD

B

G C

E

F

0

1

2

3

4

5

0 5 10 15 20 25 30

, g, .

, s

^2/4

g,

.

s = 5 .

s = 10 . s = 15 .

s = 20 .

3-5 Minimum pitch (smin)

ABC = 15 (1)(2.2) = 12.8 .2 2

15 (2)(2.2) 10.6(4)(5) 20

s sDEFG = - + = +

s ABC = DEFG2

12.8 10.620s

= +

s = 6.63 . n

3-6 A36 19 ..

L100x75x10

60 mm100 mm

45 mm

Unfolding

g = 100 + 75 - 10 - 30 - 40 = 95 mm

5 @ 50 mm

A

B

C

DE 30 mm

40 mm

ABC = 100 + 75 10 22 = 143 ..

DEBC = 100 + 75 10 2(22) + 502/(495) = 127.6 ..

An = (12.76)(1.0) = 12.76 .2 n

tw

g

g1g + g1 tw

tw

g

g1g/2 + g1 tw/2

3.5

2 2

2

8 8 1.6 1.0569.39 2(2.2)(1.6 1.05) (1.05) (2)4(20) 4(13.68) 2

61.6 cm

+ = - + + +

=

ABCDEF

3-7 ABCDEF C38010054.5 (Ag = 69.39 .2, tw = 10.5 .., tf = 16 ..) 19 ..

g + g1 tw60+90-10.5 = 139.5 mm

40 mm

200 mm

80 mm

139.5 mm

40 mm

A

B

C

D

E

F

tw = 10.5 mm

g = 60 mm

g1 = 90 mm

100 mm

380 mm

90 mm

200 mm

tf = 16 mm

16 mm

10.5 mm

Effective Net Areas, Effective Net Areas, AAee

Bolted or Riveted Members: Ae = U An

P

P

Shaded areastressed very little

shear lag

U = 0.75

III

U = 0.85 I

II

U = 0.90 W, S

I

3.2 U

2d/3(min.)

dd

2d/3(min.)

3.7

Load

3 Bolts per lineU = 0.90 or 0.85

Load

2 Bolts per lineU = 0.75

Welded Members: Ae = U Ag

2. Longitudinal welded:

L > 2w U = 1.00

2w > L > 1.5w U = 0.87

U = 0.751.5w > L > w

- Flat plate and Bar:

Short connection fittings: splice plate, gusset plate, and beam-to-column fitting

1. Transverse welded: Ae = Connected area

L

w

Ae = An 0.85 Ag

TABLE D3.1: Shear Lag Factors for Connections to Tension Members

Case Description of Element Shear Lag Factor, U Example

1 All tension members where the tensionload is transmitted directly to each ofcross-sectional elements by fastenersor welds (except in Cases 3, 4, 5 and 6)

U = 1.0

2 All tension members, except plates andHSS, where the tension load is trans-mitted to some but not all of the cross-sectional elements by fasteners or lon-gitudinal welds (Alternatively, for W, M,S and HP, Case 7 may be used.)

LxU /1-= x

x

U = 1.0

and

An = area of the directlyconnected elements

3 All tension members where the tensionload is transmitted by transverse weldsto some but not all of the cross-sectio-nal elements.

= the distance from the centroid of the connected area to the plane of the connection.x

L = length of connection.

LxU -= 1

The definition of was formulated by Munse and Chesson (1963).x

If a member has two symmetrically located planes of connection,

Munse, W.H. and Chesson, E., Jr. 1963. Riveted and Bolted Joints: Net Section Design. Journal of the Structural Division, ASCE 89 (no. ST1): 107-126.

is measured from the centroid of the nearest one-half of the area.x

x x

x

x x

Case Description of Element Shear Lag Factor, U Example

4 Plates where the tension load is trans-mitted by longitudinal welds only.

L 2w U = 1.0

2w > L 1.5w U = 0.87

1.5w > L w U = 0.75 L

w

5 Round HSS with a single concentric gusset plate

L 1.3D U = 1.0

D > L 1.3D DLxU /1-=p/Dx =

6 Rectangular HSS L H

B

LxU /1-=

)(422

HBBHBx

++

=

with a single con-centric gusset plate

H

L H

B)(4

2

HBBx

+=

with two side gusset plates

HLxU /1-=

w = plate width

B = overall width of rectangular HSS measured 90o to the plane of the connection

H = overall height of rectangular HSS measured in the plane of the connection

U = 0.70

Case Description of Element Shear Lag Factor, U Example7 bf 2/3d U = 0.90W, M, S or HP

shapes or tees cut from these shape. (If U is calculated per Case 2, the larger value is per-mitted to be used)

with flange con-nected with 3 or more fasteners per line in direction of loading

bf < 2/3d U = 0.85

with web connected with 4 or more fasteners per line in the direction of loading

8 with 4 or more fas-teners per line in the direction of loading

Single angles (If U is calculated per Case 2, the larger value is permitted to be used)

with 2 or 3 fas-teners per line in the direction of loading

U = 0.80

U = 0.60

Eccentric Bracing ConnectionSydney Airport Domestic Terminal - Sydney, Australia

3-8 W250 66.5 19 .. A36

W 250 x 66.5 T

T/2

T/2

W25066.5 (Ag = 84.70.2, d = 248 .., bf = 249 ..)T = 0.60FyAg = 0.60(2.5)(84.70) = 127

An = 84.70 4(2.2)(1.3) = 73.26 .2

U = 0.90 bf > 2/3dAe = UAn = 0.90(73.26) = 65.93 .2

T = 0.50FuAe = 0.50(4.0)(65.93) = 132 T = 127 n

Example 3.2 A single-angle tension member, an L90 x 90 x 10 mm, is connected to gusset plate with 22-mm-diameter bolts. A36 steel is used. The service loads are 16 ton dead load and 8 ton live load. Investigate this member for compliance with the AISC Specification. Assume that the effective net area is 85% of the computed net area.

L90 x 90 x 10 mm

Section

Solution First compute tensile strengths.

Gross section: Ag = 17.1 cm2 ( .5)P = 0.6 Fy Ag = 0.6(2.5)(17.1) = 25.7 tonYielding design strength:

An = 17.1 1 x (2.2+0.3) = 14.6 cm2Net section:

Ae = 0.85An = 0.85(14.6) = 12.4 cm2Effective area:

The design strength is the smaller value: P = 24.8 ton (Fracture)

Dead Load + Live Load = 16 + 8 = 24 ton

Since the design strength 24.8 ton > 24 ton required by loads,the member is satisfactory. Ans.

Fracture design strength: P = 0.5 Fu Ae = 0.5(4.0)(12.4) = 24.8 ton

Single Angle ConnectionsSydney Convention &Exhibition Centre - DarlingHarbour, Australia

Example 3.4 Determine the effective net area for the tension member

L90 x 90 x 10 mm

Section

Gross section: Ag = 17.1 cm2 ( .5)An = 17.1 1 x (2.2+0.3) = 14.6 cm2Net section:

Only one leg of the cross section is connected, so the net area must be reduced.

.5, the distance from the centroid to the outside face of an L90x90x10 is

cm 58.2=x

cm 58.2=x

7.5 cm7.5 cm

The length of the connection is L = 7.5 + 7.5 = 15 cm

828.01558.211 =-=-=\

LxU

Ae = UAn = 0.828(14.6) = 12.09 cm2Effective area:

Alternative U : this angle has 3 bolts in the direction of load.

L90 x 90 x

10 mm

the reduction factor U can be taken as 0.60

Either U value is acceptable, and the code permits the larger one to be used.

Computed U is more accurate.

Alternative U can be useful during preliminary design.

Ae = UAn = 0.60(14.6) = 8.76 cm2Effective area:

Block Shear

Tbs = 0.30Fu Av + 0.50Fu AtBlock Shear

PP

Tension area

Shear area

P P

Shear areaP

3-11 A36 19 ..

T

T

L150x100x12 mm, A = 28.56 cm29 cm 6 cm

5 cm

10 cm

10 cm

:T = 0.30(4.0)(25-2.5(2.2))(1.2)+0.50(4.0)(6-0.5(2.2))(1.2) T = 39.8

:T = 0.60FyAg = 0.60(2.5)(28.56) = 42.8 An = 28.56 2.2(1.2) = 25.92 .2

U = 0.85 AISC T = 0.50FuAe = 0.50(4.0)(0.8525.92) = 44.1 T = 39.8 n

3-12

T T

10 cm

PL1.2 20 cm

:T = 0.30(4.0)(2)(101.2) + 0.50(4.0)(201.2)

= 76.8 :

T = 0.60(2.5)(201.2) = 36.0 T = 36.0 n


< Selection of Sections

< Rods and Cables

< Sag Rods

Lecture Lecture 44 -- Design of Tension Members


Selection of Sections

Slenderness ratio:(tension members) 300r

L

T = 0.60 FyAgy

g FTA

60.0min =

T = 0.50 FuAeu

e FTA

50.0min =

UFT

UAA

u

en 50.0

minmin ==

+=UF

TAu

g 50.0min

300min Lr =

4-1 W300 8 A36 100 22 ..()

Load

Ag :

U = 0.90 W300 tw = 14 ..

2100min 4(2.2 0.3)(1.4) 69.6 cm0.50(4.0)(0.90)g

A = + + =

2100min 66.7 cm0.60 0.60(2.5)g y

TAF

= = =

8(100)min 2.67 cm300 300Lr = = =

W30065.4 (Ag = 83.36 .2, d = 298 .., bf = 201 .,tf = 14 ., ry = 4.77 .)

: T = 0.60FyAg = 0.60(2.5)(83.36) = 125.0 > 100 OKT = 0.50FuUAn U = 0.90 bf /d > 2/3

An = 83.36 4(2.5)(1.4) = 69.36 .2

T = 0.50(4.0)(0.90)(69.36) = 124.9 > 100 OK8(100) 168 3004.77

Lr

= = < OK

Check Block Shear Strength: From block shear path, Four blocks will separate from the tension member (two

from each flange).

5 cm

5 cm

5cm 8cm 8cm

W300X65.4

At = 4 x [5 0.5(db+0.3)] x tf

At = 4 x (5 2.5/2)(1.4) = 21.0 cm2

Av = 4 x [5+8+8 2.5(db+0.3)] x tf

Av = 4 x (21 2.5(2.5))(1.4) = 82.6 cm2

Calculate block shear strength:

Tbs = 0.30FuAv + 0.50FuAt

Tbs = 0.30 x 4.0 x 82.6 + 0.50 x 4.0 x 21.0 = 141.1 tons

Summary of solution:

Member W300x65.4

Design load 100 ton

Yield strength 125.0 ton

Ultimate strength 124.9 ton

Block shear strength 141.1 ton

W300 x 65.4 is adequate for T = 100 tons and the given connection Ans.

Design strength = 124.9 tons (ultimate strength governs)

2cm33.21)5.2(60.0

3260.0

min ===y

g FTA

232min 18.82 cm0.50 0.50(4.0)(0.85)

3(100)min 1.0 cm300 300

nu

TAF U

Lr

= = =

= = =

3, U = 0.85

4-2 3 32 22 .. () A36 (Fy = 2,500 ../.2, Fu = 4,000 ../.2)

(..)

25.. (.2)

(.2)

9 2.25 21.07 150 x 100 x 9(A=21.84, r=2.15)

10 2.50 21.32 120 x 120 x 10(A=23.20, r=2.36)

12 3.00 21.82 100 x 100 x 12(A=22.70, r=1.94)

13 3.25 22.07 125 x 75 x 13(A=24.31, r=1.60)

L150 x 100 x 9 ..

4-3 BC 20 A36 2 19 ..

C

B 220min 13.3 cm

0.60 0.60(2.5)g y

TAF

= = =

3(200)min 2.0 cm

300 300Lr = = =

12 .. Ae U = 0.85220min 11.8 cm

0.50 0.50(4.0)(0.85)n u

TAF U

= = =

min Ag = 11.8 + 2(1.9 + 0.3)(1.2) = 17.1 cm2

2L 75 75 12 .. (Ag = 33.4 .2, ry = 2.22 .) n

Rods and CablesRods and Cables

4.1

CableOpen socket

Open socket connection

RodWeld

Weld

Sag Rod

Anchorsocket

Steel plateor washer

StructuralmemberCable

0.33D uTA

F=

Example 3.14 A threaded rod is to be used as a bracing member that must resist a service tensile load of 3.6 tons. What size rod is required if A36 steel is used?

Solution: Required area = 2cm 73.20.433.0

6.333.0

=

==u

D FTA

From AD = ,4

2dp

Required d = cm 86.1)73.2(4

=p

Use a 19-mm-diameter threaded rod (AD = 2.84 cm2) Ans.

To prevent damage during construction, use rod min. of 16 mm.

CableCable--Stayed BridgesStayed Bridges

CableCable--Stayed RoofsStayed Roofs

Manchester City Stadium

Olympic stadium roof - Munich

Lisbon Expo 1996 - and 1998

Strand & Wire ropeStrand & Wire rope

Wire Strand Wire rope

Low Relaxation Prestressed Concrete Steel Strand (PC Strand)

BracingSag rod

Tension Members Tension Members In Roof TrussesIn Roof Trusses

PinnedSupport

Span

RollerSupport

Purlin

Truss

Truss

Bay Truss Members :Truss Members : -- TrussTruss

-- PurlinPurlin

-- Sag rodSag rod

-- BracingBracing

y wwxwy

x

q

T

q

s/2

s/2

s s/2

Truss

Tributary area fordesign of rod ab

a b

Purlin

Sag rod

T1

T2

RT1

T2

R

Design of Sag RodDesign of Sag Rod

31

6 @ 4 m = 24 m

4 m

12.65 m

C 200 24.6 purlins3.16

6 mSag rods

Truss

Truss

Truss

6 m

4-5 6.0 A36 16 .. 80 ./.2 100 ./.2

: = 80.00 ../.2

= (7 24.6 ) / 12.65 = 13.61 ../.2

= 100.00 ../.2

= 80 + 13.61 + 100 = 193.61 ../.2

T1 = (1/3.16)(193.61) = 61.27 ../.2 1

33.16

W

T1W

T1

Details of sag rod connections

4 m

24 m

12.65 m

2 m

:ST1 = 12.65(2.0)(61.27) = 1,550 ..

ST1ST121.550 1.17 cm

0.33 0.33(4.0)D u

TAF

= = =

TTOP

:TTOP = (3.16/3)(1550) = 1633 ..

21.633 1.24 cm0.33(4.0)D

A = = 16 . ( AD = 2.01 .2 )

< Sections used for Columns

< Critical Load (Eulers Formula)

< Allowable Stress

< Effective Length


Lecture Lecture 55 -- Analysis of Compression Members


Tall BuildingsTall Buildings

World Trade Center, WTCNew York

WTC on September 11, 2001

Frame Buckling MechanismFrame Buckling Mechanism

How can column fail?How can column fail?

Buckling Buckling TypesTypes

Sections used for ColumnsSections used for Columns

Double angleWide flange Tube Pipe

Built-up sections

Critical LoadCritical Load

P

L

B

A

P

L

B

A

Buckling of column due toaxial compressive load P

Buckling = Type of failure which occurs when column is slender

ShortYielding

LongBuckling

IntermediateYielding & Buckling

Columns with Pinned EndsColumns with Pinned Ends

P

L

B

A

Ideal column: - perfectly straight- linearly elastic material- no imperfections

P

L/2

B

A

yx u

umax

P

P

M = -Pu

By

x

u P

2

2

dEIdx

u=

Bending-momentequation

2

2 0d P

EIdxu

u+ =Differential equationof deflected curve

Critical load:p

=2

2crEIP

LEulers formula

Buckling about the Weakest AxisBuckling about the Weakest Axis p=2

2From crEIP

L

I = least moment of inertia

Which axis give the least moment of inertia?a

a

b

b

Pcr

For symmetry section,

use minimum of Ix and Iyx

y

Angle sectionequal legs

45oIx

Imin ImaxIy

Angle sectionUnequal legs

q Ix

IminImax

Iy

Critical StressCritical StressDividing critical load by cross-sectional area

2

2

ALEI

APcr

crp

s ==

L/r

scr

Eulers curve

spl Proportional limit

0

AIr /=Radius of gyration

2

2

)/( rLE

crp

s =Critical stress

rL / Slenderness ratio

5-1 (a) W250x175 5.0 F.S. = 2 Proportional Limit = 2,500 ../.2 (b) (a) 2.5 (a) W250175 (A = 56.24 .2, rx = 10.4 ., ry = 4.18 .)

r ry = 4.18 .L/r = 5.0(100)/4.18 = 119.6

( )( )

2 6

2

2.1 101,448

119.6PA

p = = ../.2 < 2,500 ../.2 OK

(b) L/r = 2.5(100)/4.18 = 59.81

( )( )

2 6

2

2.1 105,791

59.81PA

p = = ../.2 < 2,500 ../.2 NG

Pipe or tubecolumn Base plate

welded to column

Welded or boltedconnection

Stiffening plateswelded to flange

(Inflection Point) KL

KL=L

K = 1.0(a)

K = 0.5(b)

KL = 0.5L LKL = 0.7L

K = 0.7(c)

L

K 0.5 0.7 1.0 1.0 2.0 2.0

K 0.65 0.8 1.2 1.0 2.1 2.0

Effective Length Factor Effective Length Factor ((K K ))

Buckling AxisBuckling Axis

P

yy

x

x

x x

y

y

Deflectedpositions

Buckling about y-y axisin the x-x plane

Buckling about x-x axis in the y-y plane

Columns normally buckle laterally (the cross-section moves sideways).

This is usually the y-y axis, but both must be considered.It occurs about the axis which has the highest value of L / r.

Usually, rx > ryx-x = strong axis

y-y = weak axis

But . . .But . . .

x - x = Strong axis ()

y - y = Weak axis ()

= Lateral support against

buckling about weak axis

x x

y

y

2 m

2 m

KL = 2 m

2 m

2 m

KL = 4 m

Column Buckling Strength from ExperimentColumn Buckling Strength from Experiment

Weight

a b

Eulers formula

Fy

L/r

Pu/A

Short Column fails by Yielding, Long Column fails by Buckling

So, What happen to Intermediate Column ? Fails by ???

Buckling Buckling strength strength vs. slendernessvs. slenderness

MAX 200MAX 200

A

A

KLr

Elasticbuckling

Inelasticbuckling

Failu

re s

tress Band of

test data

Eulersformulas =

crcr

PA Where is section AWhere is section A--AA

Cc

KL/r

P/A

p=

2

2( / )crP EA KL r

Eulers formula:

ElasticElastic vs. vs. InelasticInelastic BucklingBuckling

= -

2

2

( / )12

cry

c

P KL rFA C

Elastic Buckling = 0.50Fy ( )

2 2

2 2/ c

E ECL r

p p= =

Fy

0.50Fy

yc F

EC22p

=

( )

( ) ( )

-

=

+ -

2

2

3

3

/1

2

3 / /53 8 8

yc

a

c c

KL rF

CF

KL r KL rC C

< cKL Cr

( )p

=2

212

23 /a

EFKL r> c

KL Cr

Inelastic Buckling:

Elastic Buckling:

p=

22c

y

ECF

0.60Fy Fa

/ 223/12

yF

AISCAISC FormulasFormulasAllowable Compressive StressAllowable Compressive Stress

F.S. Variable

Cc

KL/r

P/A

p=

2

2( / )crP EA KL r

= -

2

2

( / )12

cry

c

P KL rFA CFy

F.S. = 23/12

5-2 P W300 94

P

P

5 mW300x106

W30094 ../.(A = 119.8 .2, ry = 7.51 .) - 5.1 K = 0.8

2 62 (2.1 10 ) 128.82,500c

C p = =

0.8(500) 537.51

KLr

= =

KL/r < Cc (5.8)3

35 3(53) (53)F.S. 1.813 8(128.8) 8(128.8)

= + - =

2

22

(53)1 2,5002(128.8)

1,264 kg/cm1.81a

F

-

= =

P = FaAg = (1.26)(119.8) = 151 n

30 cm

PL1.2 50 cmy

39.2 cmMC380 100

67.3 kg/m

5-3 .1 P KL = 6.0

= .12.1542.231

)2.20)(71.85(2)6.0)(50(2.1=

+y

MC380100 67.3 ./.

(A=85.71 .2, Ix=17,600 .4,

Iy = 671 .4, cy = 2.50 .)

A = 1.2(50) + 2(85.71) = 231.42 .2

Ix = 2(17,600) + 2(85.71)(20.2-15.12)2 + (1/12)(50)(1.2)3

+ (1.2)(50)(15.12-0.6)2

= 52,281 .4

Iy = 2(671) + 2(85.71)(15+2.5)2 + (1/12)(1.2)(50)3 = 66,339 .4

.03.1542.231

281,52==r

600 4015.03

KLr

= =

.1 Fa = 1,337 ../.2

P = Fa Ag = 1.337(231.42) = 309.4

5-4 W25066.5 7.2 x-x 3.6 y-y A36

W25066.5 (A = 84.7 .2, rx = 10.8 ., ry = 6.29 .)1(360) 576.29

y

y

KLr

= = ()

()0.8(720) 5310.8x

x

KLr

= =

.1 Fa = 1,263 ../.2

P = (1.263)(84.7) = 107 n

< Column Design Tables

< Local Buckling

< Alignment Chart

< Column Base Plate


Lecture Lecture 66 -- Design of Compression Members


Use of structural steel column-tree assemblies prefabricated in the shop to avoid welding at the critical regions of frame structures.

Overall view of the erection of a structural steel- framed structure usingprefabricated tree-column assemblies.

Column Design TablesColumn Design Tables

.1 2,500 ./.2

KL/r vary from 1 to 200(Max)

Fa vary from 1,497 ksc to 270.3 ksc

50

1,281 ksc

.2 W, Fy = 2,500 ./.2

.3 .4-5 .6 Strength Reduction Factor (SRF)

6-1 W350 A36

KL/r = 500/8.78 = 57Fa = 1,238 ./.2

P = 1,238(146)/1,000 = 181 > 170 OK

W350 x 115

5.0 m

170 ton

170 ton

KL/r = 50 Fa = 1,281 ./.2 .1Areqd = 170(1,000)/1,281 = 132.7 .2

W350 x 106 (A = 135.3 . 2, ry = 8.33 .)KL/r = 500/8.33 = 60

Fa = 1,219 ./.2

P = 1,219(135.3)/1,000 = 165 < 170 NG W350 x 115 (A = 146 . 2, ry = 8.78 .)

- Large part of load transferred through bearing between columns

Shop weld

Field weld

(a) Columns from same W series (dlower < 5 cm greater than dupper )

Erectionbolts

Spliceplate

d for uppercolumn

d for lowercolumn

Erectionclearance

Column SplicesColumn Splices

- 60-90 cm above floor levels to keep from interfering with beam

Clip angles

Erection bolts

Bearing or butt plate

Shop weld

(b) Columns from different W series

Local Buckling of the Beam Flange Local Buckling of the Beam Flange

Classification of Sections for Local BucklingClassification of Sections for Local Buckling

Sections are classified as :

Compact sections

Noncompact sections

Slender-element sections

lp = width-thickness ratio

h t w

Point of fixity

Point of fixity

t f

b f / 2

STIFFENED ELEMENT = WEB; WIDTH/THICKNESS RATIO = h / twUNSTIFFENED ELEMENT = FLANGE; WIDTH/THICKNESS RATIO = 1/2 bf / tf

bb bb

tt

t

welds

Unstiffened Elements

b

bt

welds

t bt

b

t

Stiffened Elements

Stiffened and Stiffened and UnstiffenedUnstiffened ElementsElements

6.1

N.A.h / tw

N.A.1/2 bf / tf

1/2 bf / tf

b / t

b / t

yF/544 yF/795

yF/795

yF/100,2

TABLE B4.1 Limiting Width-Thickness Ratios for Compression Elements

Case Description ofElement

WidthThick-nessRatio

Unstiffened Elements:Flexure in flange of rolled W-shaped sections andchannels

1

Limiting Width-Thickness Ratio

lpcompact

lrnoncompact

Example

b

tb / t yFE /38.0 yFE /0.1

Unstiffened Elements:Flexure in flange of doubly And singly symmetric W-shaped built-up sections

2

b

tb / t yFE /38.0

][],[/95.0

ba

Lc FEk

Unstiffened Elements:Uniform compression in flange of rolled W-shapedsections

3

b

tNAb / t yFE /56.0




Unstiffened Elements:Uniform compression in flange of built-up W-shapedsections and plates or angle legs projecting frombuilt-up W-shaped sections

4


lpcompact

lrnoncompact

Example

b

tNAb / t

][/64.0

a

yc FEkt

b

Unstiffened Elements:Uniform compression inlegs of single angles, legsof double angles with seperators, and all otherstiffened element

5

b

tNAb / t yFE /45.0

Unstiffened Elements:Flexure in legs of singleangles

6 b / t yFE /91.0

b

tyFE /54.0




Unstiffened Elements:Flexure in flanges of tees7


lpcompact

lrnoncompact

Example

b

tb / t yFE /0.1yFE /38.0

Unstiffened Elements:Uniform compression instems of tees

8 dt

NAd / t yFE /75.0

Stiffened Elements:Flexure in webs of doublysymmetric W-shapedsections and channels

9 h / tw yFE /70.5yFE /76.3 h tw

Stiffened Elements:Uniform compression in webs of doubly simmetricW-shaped sections

10 NAh / tw yFE /49.1 h tw




Stiffened Elements:Flexure in webs of singly-symmetric W-shapedsections

11


lpcompact

lrnoncompact

Example

hc / tw yFE /70.5r

y

p

yp

c

MM

FE

hh

l

-

2

09.054.0

b

tw

2ch

cg2

phpna

Stiffened Elements:Uniform compression inflanges of rectangular boxand hollow structuralsections of uniformthickness subject tobending or compression;flange cover plates and diaphragm plates betweenlines of fasteners or welds

12 b / t yFE /40.1yFE /12.1b

t

Stiffened Elements:Flexure in webs of rectangular HSS

13 h / t yFE /70.5 ht

yFE /42.2




Stiffened Elements:Uniform compression in allother stiffened elements

14


lpcompact

lrnoncompact

Example

NAb / t yFE /49.1

b

t

Stiffened Elements:Circular hollow sections

In uniform compression

In flexure

15

yFE /07.0

t

D / t

D / t

NA

yFE /31.0

yFE /11.0D

,/4][

wc

a

thk = but shall not be taken less than 0.35 nor greater than 0.76 for calculation purpose.

(See Cases 2 and 4)[b] FL = 0.7Fy for minor-axis bending, major axis bending of slender-web built-up W-shaped members, andmajor axis bending of compact and noncompact web built-up W-shaped members with Sxt / Sxc 0.7;FL = FySxt / Sxc 0.5Fy for major-axis bending of compact and noncompact web built-up W-shaped memberswith Sxt / Sxc < 0.7. (See Case 2)

(a) Diagonal bracing (b) Shear wall

AISC (conservative) K = 1.0

SideswaySidesway Inhibited FramesInhibited Frames

bb

c

c

LILI

LEI

LEI

GS

S=

S

S=

beamsfor 4

columnsfor 4

0

0.1

0.2

0.3

0.40.50.60.70.80.91.0

2.03.05.0

10.050.0

GA

0

0.1

0.2

0.3

0.40.50.60.70.80.91.0

2.03.05.010.050.0

GB

0.5

0.6

0.7

0.8

0.9

1.0

K

(a) Sidesway prevented (b) Sidesway uninhibited

GA

GBK

0

1.0

2.0

3.0

4.05.06.07.08.09.0

10.0

20.030.050.0

100.0

0

1.0

2.0

3.0

4.05.06.0

8.010.0

30.0

100.0

7.09.0

20.0

50.0

1.0

1.5

2.0

3.0

4.05.010.020.0

Alignment ChartsAlignment Charts

K :1.

3. GA GB K

2. GA GB

K = 1.6

Alignment Charts:1. , G G = 102. , G G = 1.03. , 6.2

6.2 Condition at far end Sidesway Prevented Sidesway UninhibitedPinned 1.5 0.5Fixed against rotation 2.0 0.67

6-2

6.0 m 9.0 m

3.0 m

3.5 m

A

HEB

GD

C IF

W20

0x30

.6W

200x

30.6

W20

0x30

.6W

200x

30.6

W450x76.0 W450x124

W400x56.6 W400x94.3

W20

0x56

.2W

200x

56.2

Ix L Ix/L

AB W200x30.6 2690 350 7.686BC W200x30.6 2690 300 8.967DE W200x56.2 4980 350 14.23EF W200x56.2 4980 300 16.60GH W200x30.6 2690 350 7.686HI W200x30.6 2690 300 8.967BE W450x76.0 33500 600 55.83CF W400x56.6 20000 600 33.33EH W450x124 56100 900 62.33FI W400x94.3 33700 900 37.44

Stiffness factors:

G factors for each joint:

JOINT (Ic/Lc)/ (Ib/Lb) G

A 10.0B (7.686+8.967)/55.83 0.298C 8.967/33.33 0.269D 10.0E (14.23+16.60)/(55.83+62.33) 0.261F 16.60/(33.33+37.44) 0.235G 10.0H (7.686+8.967)/62.33 0.267I 8.967/37.44 0.240

COLUMN GA GB K

AB 10.0 0.298 0.77BC 0.298 0.269 0.63DE 10.0 0.261 0.77EF 0.261 0.235 0.61GH 10.0 0.267 0.77HI 0.267 0.240 0.61

Column K factors from chart:

weld

Anchor bolts

(a) (b)

Concretefooting

Column Base PlatesColumn Base Plates

A1 = = B x N

A2 =

Transfer bearing to concrete footing

B

NA1

A2

A1 = A2,

A1 A2,

Allowable Bearing Pressure, Allowable Bearing Pressure, FFppA1 A2Fp =

fc =

0.35p cF f =

2

1

0.35 0.7p c cAF f fA

=

Fp

A2/A10 1 2 3 4

0.35fc

0.7fc2

12

10.35 c

PAA f

=

2

1 1

0.35 cAP f

A A=

Minimum Thickness of Column Base PlateMinimum Thickness of Column Base Plate

P

fp

t

B or N

1 cm

1 cm

bfB

d N

m

0.95d

m

n n0.80bf

fp = P / ( B N )

1 . 0.80bf 0.95d

:2

2 2p

p

f nnM f n= = 2

2 2p

p

f mmM f m= =

1 cm

t 1 . t

3 21/ (1)( ) /( / 2) / 612

S I c t t t= = =

M/S Fb

( )2 22 2

/ 2 3/ 6

p pb

f m f mMFS t t

= = =23 p

b

f mt

F=

,23 p

b

f nt

F=

AISC Fb = 0.75Fy:

2 py

ft m

F= 2 p

y

ft n

F=

Design of Column Base PlatesDesign of Column Base Plates

1 cm

1 cm

bfB

d N

m

0.95d

m

n n0.80bf N B m = n

1N A= + D

A1 =

D = 0.5(0.95d 0.80bf)

N B B = A1/N

6-4 A36 W30094.0 (d= 30 ., bf = 30 .) 160 2.5 2.5 = 210 ./.2

A1

( )

22

21 ' 2

2

1 1 160 75.8 cm0.35 250 0.35 0.210c

PAA f

= = =

21 '

160 1,088 cm0.7 0.7(0.210)c

PAf

= = =

:0.5(0.95 0.80 ) 0.5(0.95(30) 0.80(30)) 2.25 cmfd bD = - = - =

1 1088 2.25 35.2 cmN A= + D = + = ( 35 .)1 1088 31.1 cm

35ABN

= = = ( 32 .)

:2160(1000) 143 kg/cm

(32)(35)pPf

B N= = =

m n :0.95 35 0.95(30) 3.25 cm2 2

N dm - -= = =

0.80 32 0.80(30) 4.00 cm2 2

fB bn- -

= = =

2

' 22

1

2500.35 0.35(210) 549 kg/cm(32)(35)P c

AF fA

= = =

' 20.7 147 kg/cmP cF f= =

fp < Fp OK

:

1432 2(4.0) 1.9 cm2,500

p

y

ft n

F= = = ( 2 .)

PL 2 32 35 . n

!"#

$%"" %##$#"&"%'%""%'

$%"#(" %&&'( &&&"%'%""%'

!

) *

) +,

) -*

-..+

/

+0 /...+

&1234567

8

8

9

:

9

+

8

8

:

8 0

: 0 ;

: < ++

+0*

+0

0 *

8

9

:

"# "#

"# "#

$

! %&

'()*

%+ ,&-.

$%/"-.

%+&"+&0,&

()*

%+ ,&-.

$% /"%-.

,&

"

0

0

+

1

0

;

0;

1 %

% %

0

0 =

= 0;

23'444

!" #$%'$ ( &) $

*+,-./0.

>

>+/

!/..?

!/..?

1#' "

2-../.

%5.65&7%.#"

- %.#"688&6.5#%6955:0

34

567 89:4;

1" #--./. ! $*+,-./.

.@@++

.

.

A

0>>

A

>>

( A( O

;..+ A

6&1 &1&)(&) !"

( A P/ P/ '('"

#"" & :#&

#90## &

4 AA--D>&L =KQ RS 2*,--../0.2

1--D> -0. T -.. PU -0. U ,.. ) *+-0.A&)>

?AC

:#

#""

% /+

>

8.: 8.888"

1

1

;A

++/--

-)

.++...

-/..

D

8

+/

D!!

9

:

1

1

&8889"8896

505

"

"

/+

/+

55.8

JA!='A? &L# ! "#$ #% =K ! &! "#$ #%Q =K ! 4! "#$ #% =KQ =KJ ?

! "#$ #% =K !

&! "#$ #%Q =K !

8.: 8.888"

1

1

4! "#$ #% =KQ =KJ ?

J!(&) % %0V$A! &L '' ' ''

%"#$$ # Q ' 8.58 %"#$$A!"% "#$(J! !'

J% !%(&) % %"#$$$ & $

#$9)

$:

?

#$9)

#$9

% % Q %(

'W %X #QA%$(&) % % !=""#$V$$( % %

%*

%* "#$$A!(V"%% !%&J0 "960 68 09#7# 68

$

()**++,+

" 8.580 68958 68

$"09#7# 68

()**+,+

"

696:# 68 8.58

%*

%*

7"09588 8.58 1

$%%?& % AB(JO"#$ #J!( # %!.-YV2.=$!A$ "#$B$Q %6.8

>

:

C8C: 8

&

E

:F 8

&

'$

9

E

9

: 8C> 8

'

&

E

9

88G 8

&

'

>

'#$

8E $

&

FE:>

$

&

'#

%*

%*

%*

%*

'

0?+ .+/..

(>

)+

)+

.

.

'

'

>>

>

'

'

*

*

'

% "#$(0960 68

69"889888 &

1

55.8

58.8

58.8

>>

>

'

'

*

*

'

*!$

'

'

-

0 >

/+

/+

- A='A? %"#$$(%.--D>&L#$.,*+.- !D.-Q # %(RS 2*,--../0.2

+1/%.--D> 01 -0. 2T -.. PU -0. 2U ,.. ZS +*3,-0.A.%@088:" ! !

0 :F9

>>

0 FE

>

RP -.++RS -.++02*,--*+,-../0.2[[ 8445:8445: \[6 8447&:

RP -.+-RS -.+-02*,--*,--../0.2[6 8447&: \ [[ 8-4:8-4: \[] 874:

[] 874: \

[[ 84:84:

% !%

8/'W D--/@.2+-D

[] 874: \

[[ 84:84:

='A?=""#$%&%"# % % 6

+,

"

-"

""

=

4/(

1 &(1 4

1/? 6

+,

;LCD=MGNN?GABH

&'9( "&)4..3V3%2WX,-&$%&$ 2'Y&Z!$$T''$3[SR/\

;

= =

*+&)4..3V3 ], Q31RR^]- Q00RR^.Q0]- Q1R1SR^/Q_VR0SR

"$$!2&$`"&$

1&0+('..., *&0/(!%+ 1/4 , (/0+(4 1/4 0/0,"

= 5

./33- 7./33+1'4..,7('34.RR+SR0

*&0/(RR+SR0 a ('34.RR+SR0 @A

"$$!2&

(/4

11 1 ( & -" "" "

=

(/41 (4 (/0 1'4..+1/1,1 1+(/0, ( &

)*/1 1/1 (/0

700'3&3RR Q 00/3"& b &0"& @A

0+12&;cC>dcC>MGABe?>3CPH

.

"#6

=

./&4

1

(

./&4 ./)

$6

1

61 %

1 1

& 1

+ % 1,+ % 1, &( % (1

%

& 7 7./)4-1&

1./)4

- -

&'94"&)1[.3.-/Q1,R1SR 5- Q0VRVSR ], Q3.RR ]- Q3[RR2'.Q,RVSR fZ&$%#&"2W!2-g6 Q03.RR+SR0$$$#"2WX,-!h!2,."&Y#&0.SR !"$

*

&.+('..., ./33+1'4..,+ 1/4 , (/.+ 1/4 &/*,"

= 5 =

" =7 /0&SR a 1.SR @A

(/4

1 (/. 1'4..+(/4,&.+('..., 1 1+(/., ( &0&/0 (/4 (/.=

" =7(*/(SR a 1.SR @A

1

(&.+('..., 0.

./&4 ./&4+1(.,

7(= 70.%1.71./0SR

( 7+1.,+14,74..SR0>0.SR0 @A

714%1B &/*7/3SR

7&.+('...,%4..73.RR+SR0

1 1& &+3.,+/3, 1/33./)4+1'4..,

77;iAPJddC MG=>?cCAKGABH

/0

/01

:

1 4

1

1

/01

1

1

/04

1

4

/0

U!'""2&2'- 7 'C

%#"UC2'-#'%$#"U&

-C

C -

%&"&" j C C (/.CC

%&"&" $ - - (/.--

#2&"&/3R.^ (/.-CC -

(/../33 ./)4

-C

- -

f!%"&W&"

&'()*+,5L%&"2&''#f&Q [R."Y2W&

2'#f&

-.7'/7

=?

1

.>

?@;

23('.%&" $#fQ 03R110-.1Q41RR+R%&" $2 Q 03R1100[1Q,[RR+R

1 1

)/4&4 0 ((./)%)/4 &/3

C"

1 1

&/3 0 &3/&4%)/4 &/3

-"

.

>

.

>

.

>

:

=?

A

!"#

$ %&'

(

)

)(*+),)),

()%-*+

&)**,- &***+%),)),

! .!

/0

/0

1

2

3

4

!

!

!

5

#

5

6

6

77

8!39:3::34

" #$%&'($")#*+,-

&+$.&'$$$/ ,#(*(.&$$/.&$/&*&) &'&)&

&'#$ #$ &'($ 0

"&*&))

!")1))2,+*$3.)*#&/)4"&'&)

"#.))*#/"(*30%

"5)%(".(*3/.+*$/)%(",#(*(0

!"# $%&'( )*+ ,- ./-%0 )+ & 123+++4 '56-%0 7

$*&

6

6

5

#

5 $*&

!

!

6

6

6

&*$

6 6

6

6

5 %6 - /- , $*&,

77

8-,-7

956 ! & 95,'-:& &*$

& . % / 6

,-$ 6 $6, 76% - ; )8%&)

)

)&)

)8 %

96:

&*$& &

! !

!

!

6 6 66 6

%- < % -'! 75 %6 ; $*&,

&*$

& &

! !

!

!

6 6 66 6

/40# 25 %6 ; $*&,

/;0""5

&*$$*3$

!

!

6 6 6

/30!5 %6 $*&,

&*$ ! !

6 6 6

"

"

5

5

2>2

1:.=9

2< >5

/0

/0

/0

"

3

"

4

"

3

?"

4

/0

"

3

"

4

"

3

?"

4

/0

/0

&

)

$*3 $*#

&*$& %

6

$@@@

$@@@

6 =>"#?#"! @ A B 4+4 CD4E F$ B 4EG)+ CD%E H$ B 3*% CDE HI B J*& CDE '( B 1*1J DE 'K B 4*4) DL

!") %6 *%3+3& M.N60&-,-7 9 +,6, -%G ,%9/- - .N,M - O$ B *&4 & OI (9 &-56 , *+

3::

3::

4:6

4:6

9

"&$$.&'$$$/%)$)"#, *PM-*4

: ".&*$$/.,$$/%(*",3*&(

:!!! ".&*)/.,$$/%&,*8"3)*+, ./0

6 "&')$+*PM-*4

% 6 "#,%&')$+"$*#&;$*&,

N+%- @&*GL & @&*)L

")$.&'$$$/.&$$/%)'3+$"+#**PM-*4

))

)&) )'+#30%

)8 % ! ! !

96:

79 0- + "$*(,$*(, &*$# &*$$

"

>

>

!"6 % 6( 9 6 *4+++*G (9, Q,$

0 (. 0 R ( 7,6 G %- 'N696 $/- S M. 9 @TL %% (. R &N6 0 3 $8 %--0!6 - &.-

9

;4

;4

:

3

3.34

3.34

/0

>?7 "8).&'$$$/%8*$&"()$0

6 "4++#+*G @A B $*+ CD4E H$ B J*+ CDE HI B *G CDE F$ B 4)) CDE FI B G)*G CDE % B *$ DL

:

3

4

;

AB4.:134.:

;.:

/0

: "&*$

:!!! ".&*$/.))#/%(*8$")+

: ".&*$/.&&)/%8*3&"8&

* 6 "&'8(8 *PM-*4

6 "()$%&'8(8"$*,;$*&,

N&%- @$*GL & @$*)L

) 3

)&) .)*& &$ / '(8#0

)8 )+!6

'()* >+

% 6 7-

.N U, B +*J3

% 6 7- /- .N U, B *+

,V ".$*(,2&*$/%)"$*),

"#".,/.)*)#/%#")*($

",%8)",.,/.)*)#/%8)

"&*+,

;4

;4

9

!

:

3

;4

;4

9

!

:

3

;4

;4

9

!

:

3

;4

;4

9

!

:

3

&,V W B @4*J+ X *)3L P 4 B 4*4J -,-

$*&+*$

(8#'()$&

),*$

&

6

@

0.&- B )*)(.&'$$$/.&$$/%)++"&'$$#*PM-*4

,V.-,-7B .$28&3/%)"&*$,-,-

&-B &*$,.&'$$$/.&$$/%)++"8+*PM-*4

()$ .&*$/.&'$$#/ &*)3 &*$&'8(8 &',$$

&

6 6

6

ABCCC

/012134/50

666

///Y/Y

D !D ! CCCCCC

!"#$% ""&$&'

!"#$% ""&$&'

C /

0

/0DC

'

#.

")2

! "

- & "% 66 # Z/(95/ * ;8 - &-5N&& " (*, & " " 8

"&$$2.+/.,*+/2.#/.,*+/.)*3)/")$$

!"#$%&(*,%'"$8

"&$$2.+/.(*,/2.#/.(*,/.8/")3&*,

%()**+,-.&&/

0 1234 15

$# 6 7570 1234 122

"&$$2.+/.3*3/2.#/.3*3/.)*3&/")&,8%!&,)98:;

!"7 ,$6 " ,%0,'$ - [ , ++ &.-,-7 W$ B ) -,- & WI B 1 -,- %--0! O B 3 ,- U,$ B U,I B +*J3 &&,6 G

"&$$.&'$$$/%&+8*3",+30

:!!! ",$$%&,*)"8)*(

: ",$$%(*(#",3*,3""

03'(*8))8

/&$&*).&))

3)

!6

08($'8,3*,3)8

/&$&*).&))

3)

6

&*$$*&$)*$

3',+3&

(,*$

&

!

!

6

$*&$),*&

8($'8,+3&

(,*$

&

6

'(

%8 -/&.%'-/.

$*(*$(+,'&

/,&,/.$),*&.,$$'&

/8$#/.$*&.)#&'&,+3

'(

,+3 8$# ,&, $*(#8 &*$&',$$ &'3,$ &'(+,

'(

@D ) *

< Types of Roof

< Bracing of Truss

< Loading on Truss & Purlin

< Secondary Truss


Lecture Lecture 1010 Roof Design


Types of RoofTypes of Roof

(a) Flat roof

(d) Gable

(b) Lean to

(e) Hip

(c) Butterfly

(f) Curve

RoofRoof RafterRafter

(RUN)

(SPAN)

(RISE)

..

1.0 1.2 .15 ../..10, 15, 20 ()

0.32 0.34 .50 ../..30 - 35 ()

1.0 2.0 .5 ../..MIN. 1 - 2 (Metal sheet)

(RUN) (SPAN)

(RISE)

P/2P

PP

P/2 RR

RL

10-1

4 m0.8 m 0.8 m

1.2 m

= 30 ../. = 50 ../. = 80 ../.

0.8 m

0.8 m

2 m

2 m

1 .

1 .

1.0 1.0 = 801.0 = 80 ../. = 10 ../. = 90 ../.

2 . 0.8 .

0.8 m RL RR2 m

w = 90 kg/m

72 kg 75.6 kg

104.4 kg

0.84 m

31.8 kg-m

28.8 kg-m

RL = (1/2) (90) (2.8)2 / 2.0 = 176.4 ..RR = (90) (2.8) 176.4 = 75.6 ..

Mmax = 31.8 ..-.Fb = 0.60(2,500) = 1,500 ../.2

S = Mmax/Fb = 31.8(100)/1,500 = 2.12 .3

5025 .. 1.6 .. (Sx = 2.81 .3)

10-2 1

4 m

4 m

0.5 m

0.5 m

4 m 4 m0.5 m

0.5 m

90 ../..

L

L

45

2 24 4= + = 5.66

2 20.5 0.5= + = 0.707 L = 4.0/2 = 2.0

wmax = 2L() = 2(2.0)(90)= 360 ../.

= 3(0.5)2(90) = 67.5 ..= (0.5)2(90) = 22.5 ..

:

:

RRRL5.66 m0.707 m

wmax = 360 kg/m67.5 kg

RRRL5.66 m0.707 m

wmax = 360 kg/m22.5 kg

RoofRoof TrussTruss

Truss

Free Support

Purlins

Top chord

Bottom chord

Sag rod

Truss

Fixed Support

Types of TrussesTypes of Trusses

King Post Truss Pratt Truss

Howe Truss

Fink Truss

Fan Truss

Howe TrussPratt Truss

King Post This is the most basic truss type. Primarily used for simple structures with short spans.

Fink Capable of longer spans than the King Post, adding further design flexibility.

Flat - Using this truss can reduce the amount of wall area that needs to be sheathed.

Double Pitch or Dual Pitch An asymmetrical truss used where the designer wants a change in roof appearance.

Gambrel This truss is often used in agricultural buildings and barns.

Hip This is one section of a hip roof system.

T-2

Sag rodPurlins Bracing

T-1

Bracing of TrussesBracing of Trusses

Roof Plan

1 25.00

35.00

45.00

55.00

65.00

75.00

85.00

95.00

105.00

A

B

10.0

0M

C

10.0

0M

Loading on TrussesLoading on Trusses

Purlins

Prim

ary

Trus

sS

econ

dary

Tru

ss

Trib

utar

y A

rea

=5

x 20

=10

0S

q.M

.

Loading on Loading on PurlinPurlin

s s

Primary Truss

C Purlins

L

s

s

s

Tributary Area

+ = ./.

s L

= x s ./.

= 6 ./.

q

q

sin,8

,

cos,8

,

2

0

2

wwlwMSM

f

wwwlw

MSMf

xx

yy

yby

yy

xx

xbx

===

+===

0.175.066.0

+=+y

by

y

bx

by

by

bx

bx

Ff

Ff

Ff

Ff

q

wwy

wx

Design of Design of PurlinsPurlins

Secondary TrussSecondary Truss

Primary Truss under Loading

Lower chordin Tension

Upper chordin Compression Lateral Buckling of upper

chord in primary trussLa

tera

l win

d lo

ad

Secondary Truss helps Primary Truss for

- Lateral wind load

- Lateral buckling of upper chord

- StabilityL

L

L/r

: = 30 ./.

= 14 ./.

= 44 ./.

1.0 5.0

= 44 x 1.0 = 44 ./.

= 6 ./.

= 44 + 6 = 50 ./.

m-kg1538

549kg/m,4904.14cos50

m-kg388

512kg/m,1204.14sin50

20

20

=

===

=

===

xy

yx

Mw

Mw

3cm27.9)500,2(66.0

)100(153===

bx

xx F

MS

C125x50x20x2.3 . (Sx = 21.9.3, Sy = 6.22.3, Ix =137.4, Iy=20.6.4, 4.51./.)

x :

y :

2kg/cm 6.6989.21

)100(153===

x

xbx S

Mf

2kg/cm 9.61022.6

)100(38===

y

yby S

Mf

OK( ) ( ) 0.175.0500,275.0

9.610500,266.06.698

75.066.0=+=+

y

by

y

bxF

fF

f

:)137)(101.2(384

)500)(100/49(53845

6

44

max

==DEI

wl

OK

==

= 60 x 5 x 0.5 = 150 .

T-1:

T-1 = 5

= 30 ../.

= 14 ../.

(+) = 6 ../.

() = 10 ./.

= 60 ../.

= 60 60 1 = 300 .. 150 ..

300 ..

(.) (.)L1L2 = L12L13 0 1.00L2L3 = L11L12 2200(T) 1.00L3L4 = L10L11 3000(T) 1.00L4L5 = L9L10 3240(T) 1.00L5L6 = L8L9 3200(T) 1.00L6L7 = L7L8 3000(T) 1.00

U1U2 = U12U13 2267(C) 1.03U2U3 = U11U12 3092(C) 1.03U3U4 = U10U11 3339(C) 1.03U4U5 = U9U10 3298(C) 1.03U5U6 = U8U9 3092(C) 1.03U6U7 = U7U8 2783(C) 1.03

Lower Chord:

Upper Chord:

VerticalBracing:

DiagonalBracing:

(.) (.)

L1U1 = L13U13 1800(C) 0.50L2U2 = L12U12 1100(C) 0.75L3U3 = L11U11 600(C) 1.00L4U4 = L10U10 240(C) 1.25L5U5 = L9U9 50(T) 1.50L6U6 = L8U8 300(T) 1.75

L7U7 1050(T) 2.00

L2U1 = L12U13 2460(T) 1.12L3U2 = L11U12 1000(T) 1.25L4U3 = L10U11 339(T) 1.41L5U4 = L9U10 64(C) 1.60L6U5 = L8U9 360(C) 1.80L7U6 = L7U8 604(C) 2.02

L4L5 L9L10 = 3240 . (T) 1.0

, Ft = 0.60Fy = 0.60(2,500) = 1,500 ./.2

, Ag = 3240/1500 = 2.16 .2

L50 x 50 x 4 . (Ag = 3.89 .2, rmin = 0.98 .)

A307 12 .

, Ae = 0.85Ag = 0.85(3.89) = 3.31

= 0.5FuAe = 0.5(4,000)(3.31)

= 6,620 . > 3,240 . OK

, L/r = 100/0.98 = 102 < 300 OK

:

U3U4 U10U11 = 3,339 ..() 1.03

, Fa = 1,000 ../.2

, A = 3,339/1,000 = 3.34 .2

L50 x 50 x 4 . (A = 3.89 .2, rmin = 0.98 .)

, L/r = 103/0.98 = 105

, Fa = 876.2 ../.2

= (3.89)(876.2) = 3,408 .. > 3,339 .. OK

:

L2U1 L12U13 = 2,460 .. 1.12

, Ft = 0.60Fy = 0.60(2,500) = 1,500 ../.2

, Ag = 2460/1500 = 1.64 .2

L40 x 40 x 3 . (A = 2.35 .2, rmin = 0.78 .)

:

A307 12 ..

, Ae = 0.85Ag = 0.85(2.35) = 2.00

= 0.5(4,000)(2.00) = 4,000 .. > 2,460 .. OK

, L/r = 112/0.78 = 143.6 < 300 OK

1) :

L7U6 L7U8 = 604 . 2.02

L65 x 65 x 6 . (A = 7.53 .2, rmin = 1.27 .)

= (4.70)(387.7) = 1,822 . > 600 . OK

L/r = 202/0.78 = 259 > 200 NG

2 L40 x 40 x 3 . (A = 2(2.35) = 4.70 .2)y

y

x x

Ix = 2 ( 3.45 ) = 6.90 .4

Iy = 2 ( 3.45 ) + 4.70 ( 1.07 )2 = 12.28 .4Control

min 6.90 / 4.70 1.21 cmr = =

L/r = 202/1.21 = 167 < 200 OK

, Fa = 387.7 ./.2

2) :

150 kg300 kg

300 kg300 kg

300 kg300 kg

300 kg

150 kg300 kg

300 kg300 kg

300 kg300 kg

12 m

FIXFREE

Max. Deflection:

From analysis,

Dmax = 1.64 cm

< [1200/300 = 4.0 cm] OK

Free (Roller) Support Design:

slot

= 12

a = 1210-6 /oc DT = 40 oCaDTL = 1210-6401,200 = 0.576 cm

Slot length = 2 0.576 + 1.6 = 2.75 cm USE 5 cm

BOLT 16 MM

5 cm

2 cm

! " #$

% & $$ &%$'$&(&$$&(

%&!$)'$"&'*'() "''*'!*$&(&$$&(

+ ,-. /

0

+ 1

+11

,

+

1,

2 11

2

2 01

'3 ,

1

,14

2

2 1

,/

516

7

*

+ 8

,

,

799*7

8:

8:

,

+ 8:

8:

8:

!"#$""%

"'"() "

8:

8:

;

-. -,.

-.-.

-. !" -.

,2,

>

=

?

01,/,

=1

111

8:

8:

11

,

22,=11,

( 22,=11,

$ 1,

1!

+"#@2:1

#$%$ ,/,

1 +1

?AB

>

>BC

!

"

(# ")*+,-")*+.-/01"/2 /01"/2

34546789:;:?@A BC@@ A@@>?2D*EF- ?B@@ B2@@ BCG@>?2D*!"EF- ?B@@ B2@@ 2B@@>CH@*EF- ?G@@ BCG@ BHA@>CH@*!"EF- ?G@@ BCG@ 2G@@

#$

+:

2

2

:

:

D@

+ =

!"#$%&'%&%()&* +,-./,-01#%0022/ (#%3/('45/#'216/& 6

::11,

1

1

1

7&%8 !9% &'()*(+&*(:20% &*(, ")"'-+)'(+&"-:20 &%. &('/()"'-+)*('(+&!-. &('-()!'(+)"-'(+&-(

;!&45&&&'%9&459 . &('"- )"'"+")"'+)!+&*'0*&&'9 . &)'(+)"'"+)'"+)!'(+)!+&!"'"!

=>?@ A!" #B

$!%!,-01#%&22 (#%3:8

('45$%&'' &%%()C%! +,-.&$ !#'& )DE)#! %

+

+E

11,- +:E1

:

.

:1

1

+

1

7(;!&45&&%' (9'&45 8F 0(02)G*(02HH*F2&- @

%#

( %#

( 6K0226K22

6K022 6K0H226K022 0216

(3!*X( V

( YX1!*X( V

!

!

F:11

-.

FE11

-,.G

F11

-.

F:11

:@

-.

F:11

2%&3

=,11,/

1,1H,H ,/

,I,1/11,11

J:

$IEJ$()>?2D"BH"/"/EF"KEFLMJFB2"/"/

@=@= ==

:E

!

:

@=@=E

E

@

=

@

=

E

N O>?2DEF

P +. Q2RB@@/01"/2

P + QB/2*CR@@@-QCRG@@/01"/2

J

"#$"

: :

: : :

%

S

())@=@='EE>?2DBH"/"/ B

: :-. E E:

)Q 2@0BB/HBQB/TG

"#$"

U S ())==E

: :

: : :

%

E

: :-. E E:

)Q 2G0BB/HBQ2/D?J

"#$"

V S ())@=@=E

: :

: @@

%

E

E E

)Q G0D/HDQB/?D

"#$"

W S ())@=@=E

: :

: @@

%

E

E E

)Q A0D/HDQB/BG

J@

!

: 11

"$&& '"$(

2 11,-@

.

2 )11,

@=@=

==

@=@=E

@

=

@

=

E

:@ 11

#*#$+*#

JK

>

9(

"2'/

*-'9

"-(0("

?""'

!" O#"$%&%$'()!%

$ &%$*+#&,- &%$.!*+#&,-

*(2

*"-

!0(

!((9'9"(

"" *0'!*(9( "" -'"*(9(

"" 2-'**29( "" 9"'*29(

!"#

$% %$&'%(%%(

$% )&'!"%&&() !"&&&'! %(%%(

!

**

**

** **

"

#

$

#

$

##$$%%

"&

#

#

$

$

#

#$

$

'''

&

&

&

&

()*+

++**

, ,

, ,

-

, ,

. .

, ,

-

- -

.

,-.+

-

,

**

,

!"#$% &'()**

/ "0

1

1

1

1

"23

"23 "23

"23"23

"23

"23

"23

**

,/0

/0

!"#$%&' ! ()*$!*&*

. . .

-

. . . .

12/3 42/ ,/ 3 ,5.66

.

.

.462,/3

7*6

,.66

.462/3

,*6

,.66

81 ,. 81 ,*/

-

+"(*,)()*-)

. .

7*6 .*/ 7*9

:,/;/:.60

: :2,.32.63:.46

+!,( #$ !- *.$'!,( #$ /$(0123#$ 1245

66 '7

'*89"#$% &'()**

/=10

:460

,>0

: ;

all steel ppt

Documents