محاضرات انشاءات معدنية د.سعد الدين مصطفي

١٩١٩ PagesPages ١١ســـعد الدين مصـطفى ســـعد الدين مصـطفى . . دد..أأ

رئيس قســم االنشـــــاءات رئيس قســم االنشـــــاءات

بسم اهللا الرحمن الرحيمبسم اهللا الرحمن الرحيم))وأنزلنا الحديَد فيه بأٌس شديٌد ومنافُع للناس وأنزلنا الحديَد فيه بأٌس شديٌد ومنافُع للناس ((

))٢٥٢٥--الحديدالحديد((سورة سورة صدق اهللا العظيم صدق اهللا العظيم

STEEL (III)STEEL (III)سعد الدين مصطفى سعد الدين مصطفى / / د د ..أأ

أستاذ ورئيس قسم الهندسة اإلنشائية أستاذ ورئيس قسم الهندسة اإلنشائية ))إبراهيم أسامةإبراهيم أسامة((شكر خاص للمهندس شكر خاص للمهندس

لمساهمته في إعداد وإخراج المحاضرات لمساهمته في إعداد وإخراج المحاضرات

١٩١٩ PagesPagesســـعد الدين مصـطفى ســـعد الدين مصـطفى . . دد..أأ

رئيس قســم االنشـــــاءات رئيس قســم االنشـــــاءات ٢٢

DESIGN OF STEEL DESIGN OF STEEL STRUCTURESSTRUCTURES

CHAPTER(1)INTRODUCTION


رئيس قســم االنشـــــاءات رئيس قســم االنشـــــاءات ٣٣

Types Of Steel Structures:Types Of Steel Structures:--

Trussed -Frames

Beams

Trusses

Frames


رئيس قســم االنشـــــاءات رئيس قســم االنشـــــاءات ٤٤

Types Of Steel Structures:Types Of Steel Structures:--

Bridges

Tall Buildings Towers


رئيس قســم االنشـــــاءات رئيس قســم االنشـــــاءات ٥٥

Advantages And Disadvantages Of Advantages And Disadvantages Of Steel Structures:Steel Structures:--

Advantages:-1-High carrying capacity.2-Lighter In Weight.3-Small Sections.4-High Ductile Material.5-No need Formwork.6-Faster Construction.7-Ability Of changing And Adding To Sections.8-Weldability.


رئيس قســم االنشـــــاءات رئيس قســم االنشـــــاءات ٦٦

Disadvantages : -1-Cost Of Maintenance.2-Tendency To Buckling.3-Less Resistance To Fires.4-Problems Of Instability.


رئيس قســم االنشـــــاءات رئيس قســم االنشـــــاءات ٧٧

StressStress--Strain Curve For Steel:Strain Curve For Steel:--

STRAIN STRAIN

STRESS

Elastic Plastic Strain Hardening failure

Fy

FuST 37

ST 44

ST 52


رئيس قســم االنشـــــاءات رئيس قســم االنشـــــاءات ٨٨

Mechanical Properties Of Steel:Mechanical Properties Of Steel:--

1-Mass density γ

= 7.85 t/m3.2-Modulus of elasticity E = 2100 t/cm2.3-Shear Modulus G = 810 t/cm2.4-Poisson's ratio μ

= 0.3

5-Coefficient of thermal expansion α

= 1.2x10-5 co.


رئيس قســم االنشـــــاءات رئيس قســم االنشـــــاءات ٩٩

Grades Of Steel:-1-

Steel 37.

2-

Steel 44.3-

Steel 52.

t > 40mmT < 40mmGrade of steel FUFYFUFY

3.42.153.62.4St 37

4.12.554.42.8St 44

4.93.355.23.6St 52


رئيس قســم االنشـــــاءات رئيس قســم االنشـــــاءات ١٠١٠

--Design Procedure:Design Procedure:

1-2-

Determination of service Load.

3-

Determination of internal forces in theStructural components.

4-

Design of section.5-

Drawing and Detailing


رئيس قســم االنشـــــاءات رئيس قســم االنشـــــاءات ١١١١

Codes and specification:-

1-

BSS

436,449 (British Standard Specifications).2-

Euro-Code 1, 2, 3 & 4.

3-

AISC

(American Institute Of Steel Construction).4-

AISI (American Iron Of Steel Institute).

5-

AASHJ6-

ECP. (Egyptian Cod Of Practice for Steel Construction and Bridges)


رئيس قســم االنشـــــاءات رئيس قســم االنشـــــاءات ١٢١٢

The Aim Of Structural Designer Is The Aim Of Structural Designer Is To Produce:To Produce:--1-Asafe And Economical Structural To Achieve A certain Function.2-A Structure That Is Aesthetically Pleasing.3-A Structure That is Easy To Build And Maintain.


رئيس قســم االنشـــــاءات رئيس قســم االنشـــــاءات ١٣١٣

The Structural Systems:The Structural Systems:--

TrussesOR

Frames


رئيس قســم االنشـــــاءات رئيس قســم االنشـــــاءات ١٤١٤

General Layout of project:General Layout of project:--

Mansoura Uni.

Faculty of Eng

Steel Project

Student’s NameData

Elevation (Transverse W.Br.)

Scale(1:100)

Cross Section of Structura

System (Truss).

Scale(1:50)

Plane (Upper&Lower W.Br.)

Scale(1:100)

Isometric

Scale(1:500)

General Layout of projectGeneral Layout of project


رئيس قســم االنشـــــاءات رئيس قســم االنشـــــاءات ١٥١٥

The Given Area:The Given Area:--


رئيس قســم االنشـــــاءات رئيس قســم االنشـــــاءات ١٦١٦

General Layout.General Layout.Elements Of Structure (Isometric):-


رئيس قســم االنشـــــاءات رئيس قســم االنشـــــاءات ١٧١٧

Main System Arrangement:Main System Arrangement:--


رئيس قســم االنشـــــاءات رئيس قســم االنشـــــاءات ١٨١٨

Main System (Elevation):Main System (Elevation):--


رئيس قســم االنشـــــاءات رئيس قســم االنشـــــاءات ١٩١٩

Plane & (S.V):Plane & (S.V):--

١٩ Pagesســـعد الدين مصـطفى . د.أ

١ رئيس قســم االنشـــــاءات

Cont. CH(1)

Loads on Trusses


٢ رئيس قســم االنشـــــاءات

*Types of loads:-

: Primary Loads-1(Dead Load - Live Load - superimposed Load –

Dynamic effects –

centrifugal forces).

Secondary Load:-2(Wind Loads –

Earth quake Loads –

Braking

forces –

lateral chock effect –

changing of temperature –

frictional Resistance of bearing

settlement of support) .


٣ رئيس قســم االنشـــــاءات

-Primary Loads:-1) Dead Loads:-a

).WstWeight of steel Structure (-[Truss members –

Purlins –

Connections –

Wind bracing

members …]Assumed Wst [20 - 40] kg/m2.

).WcWeight of covering material (-: Flexible Roof*

Corrugated steel sheets. Wc= [5 - 8] kg/m2.Sandwich panel = [10 - 15] kg/m2. Asbestos Wc = [15 - 25] kg/m2.

Tiels Wc = [30 - 50] kg/m2.Glass Wc = [20 - 40] kg/m2.

:*Rigid RoofConcrete Wc = γ. ts γ

= 2.5 t/m2.


٤ رئيس قســم االنشـــــاءات

-Calculating of Dead loads:

.. CstLD PPP +=

ststW PS x a⎯⎯ →⎯

cCW PS x a. ⎯⎯ →⎯ ′

S x ax sts WP =

S x ax .. ′= CC WP

S x ax S x ax .. ′+= CstLD WWP


٥ رئيس قســم االنشـــــاءات

-Dead loads distribution:


٦ رئيس قســم االنشـــــاءات

b-) Live Loads:

KgSaWP LLLL .....**... ==

6.040)tan6.0(20. α−+=LLW

)Flexible–**(Inaccessible Roof

0.1 0.2 0.3 0.4 0.50.60.7 0.8 0.90.0

10

2030

40

50

6070

80

90

100110

120

130

140

150

160

170

180

190

200

tan

WL.L

Accessible Roof

InAccessible Roof (Rigid Roof)

Live Load on Inclined Roof

Kg/m 2

InAccessible Roof (Flexible Roof)

50 kg/m2

20 kg/m2

From Curve get WL.L

.accordingTo Roof type

)Rigid–Inaccessible Roof **(WLL

= 50 + (0.6 –

tanα) (50/0.6)

)Rigid–Accessible Roof **(WL.L

. =50+ (0.6-tan α)(150/0.6)

Then,


٧ رئيس قســم االنشـــــاءات

-Live loads distribution:


٨ رئيس قســم االنشـــــاءات

-Secondary Loads:-1Wind Loads:The external pressure or suction of wind

forces is calculated using the following equation: qk x x ew cW =


٩ رئيس قســم االنشـــــاءات

-s parameters:’Equation

C معامل توزيع ضغط أو سحب الرياح الخارجي علي أسطح المبني ويعتمد علي الشكل الهندسي

K اإلرتفاع معامل التعرض يتغير معq ضغط الرياح األساسي ويعتمد علي الموقع

٢م / الجغرافي للمبني وتؤخذ قيمته من الكود آجم


١٠ رئيس قســم االنشـــــاءات

-(C):Value Of Coefficient


١١ رئيس قســم االنشـــــاءات

-s curve:’Wind Load


١٢ رئيس قســم االنشـــــاءات

-Some special cases:


١٣ رئيس قســم االنشـــــاءات

-Some special cases:


١٤ رئيس قســم االنشـــــاءات

-(K):Value Of Coefficient

Coefficient "K"Height "m"

1.00 to 10

1.110 to 20

1.320 to 30

1.530 to 50

1.750 to 80

1.980 to 120

2.1120 to 160

2.3More than 160


١٥ رئيس قســم االنشـــــاءات

Basic Wind Pressure at Different Location in Egypt:

q"Kg/m2"Location

90Marsa Matrouh,El-Dabaa,Ras- Sedr

80Alexandria,El-saloum,Abu Sower, Hurghda and all

costal Area

70Cairo , Asyot , Belbis

60Fayoum , Menya , Aswan ,

Modereyat El-Tahrir

50Tanta , Mansoura , Damanhour


١٦ رئيس قســم االنشـــــاءات

-Wind loads distribution:


١٧ رئيس قســم االنشـــــاءات

-Wind loads distribution:


١٨ رئيس قســم االنشـــــاءات

-Wind loads calculation:

)/(w..)/( S 2 mtqkcmtW ww ⎯ →⎯=

mtxSxxW w /.......05.00.18.01 =+=

mtxSxxW w /.......05.01.18.02 =+=

mtxSxxW w /.......05.00.15.03 =−=

mtxSxxW w /.......05.01.15.04 =−=

txSaxxxPw .......05.01.18.01 =′−=

txSaxxxPw .......05.01.15.02 =′−=

)(P..)/( S x a2 tonqkcmtW ww ⎯⎯ →⎯= ′


١٩ رئيس قســم االنشـــــاءات

-Structural analysis:1-Method of sections.2-Method of joints.3-Stress diagram Method.ORUsing Computer Programs (Sap2000, Broeken, Ansys, Abakus, Staad III Provion).

٢٣ Pagesســـعد الدين مصـطفى . د.أ

رئيس قســم االنشـــــاءات ١

Ch(2) Design of Tension “

and Compression"


رئيس قســم االنشـــــاءات ٢

The Type of loading which acts on structural member classified the steel members into four categories:

1-Tension Members:

2-Compression Members:

3-Flexural Members:

4-Combined Members:

allTfANf −≤

+=

allcfANf −≤

−=

by

y

x

x fZM

ZMf ≤+=

ally

y

x

x fZM

ZM

ANf ≤++=


رئيس قســم االنشـــــاءات ٣

Sections Used in truss members:

a

a

v

v u

u

t iv =0.2xa1 one equal angle

a

b

t

iv =0.1xb

1 one unequal angle

b:a=2:1iv =0.14xbb:a=1.5:1


رئيس قســم االنشـــــاءات ٤

Cont. Sections Used in truss members:

t

Two angle back to back

a

t

ix =0.3xbb:a=1:1ix =0.28xbb:a=1.5:1ix =0.26xbb:a=2:1

u

u

v

v

t iv =0.385xaStar shape

t

a


رئيس قســم االنشـــــاءات ٥

Cont. Sections Used in truss members:

XX

Y

Yix =

2<ix 1<

Iy2< = 2 [I y 1< + A

2]

1<.X

iy2< = Iy 2<

A2<

X=e + t G.Pl

2

Xe

XX

Y

Y

X

ex

ix =2<

iy1<

Iy2< = 2 [I X1< + A

2]

1<.X

iy2< = Iy 2<

A2<

X=e + t G.Pl

2x

(From table)

(From table)

(From table)

(From table)


رئيس قســم االنشـــــاءات ٦

Allowable stresses in tension:-

2.6.2(p.13)

} yy

yy FF

FmmtmmFmmt

F get , 1.4 clause rom 10040

4058.0Ft →

⎭⎬⎫

→≤<→≤

=

Ft (t/cm2)Grade of Steel

100mm>t > 40mmT<40mm

1.31.4St 37

1.51.6St 44

2.02.1St 52


رئيس قســم االنشـــــاءات ٧

Before designing of any member you have to ask about:

1-Type of force:

2-Position of Member:

3-Type of connection:

⎪⎩

⎪⎨⎧

=→

=→2

2

/2.14.1)(

/4.1)(

cmtxFBCase

cmtFACase

t

t

⎩⎨⎧

<<→<→

shapestar or 1or b.b 2)&(.2)&(

VDInnerbbLUOuter

⎩⎨⎧

member) in tension Area gross (UseConnection Weldedmember) in tension AreaNet (UseConnection Bolted


رئيس قســم االنشـــــاءات ٨

Design of Tension Members

. .

A-SymmetricalSections

B-UnsymmetricalSections


رئيس قســم االنشـــــاءات ٩

Design Procedure :

A-Symmetrical Sections

Step (1): Choose of Cross Section:-

a) Stiffness condition:

300≤xi

L

1:2:26.01:5.1:28.0

1:1:3.0

=→==→==→=

abbiabbi

abai

x

x

x


رئيس قســم االنشـــــاءات ١٠

Cont. A-Symmetrical Sections

b) Construction condition:(In Case of bolted

connection only)

ta +Φ≥ 3

cmMcmM5.5

4

16

12

→→

c) Stress condition

)85.02.1(58.0 xFFA

y

=

21AA =<

For Bolted case (B)

yFFA

58.0= For Welded case (A)

Choose The Section witch satisfy The above Three conditions.


رئيس قســم االنشـــــاءات ١١

Cont. A-Symmetrical Sections

Step (2): Check of the Chosen Member:-

a)Check of Actual Stress:

⎩⎨⎧

→→

≤=)(2.158.0

)(58.0AreaNet BxF

AFForcefY

Yact

WeldedAA →= < ][2 1

BoltedxtdAA hnet →−= < ][2 1

clearancedh +Φ=

:Clearance1412 &1 MMmm →24162 MMmm −→

erLMmm arg&3 27→

p95


رئيس قســم االنشـــــاءات ١٢

b) Check of Stiffness:

60≤dL


رئيس قســم االنشـــــاءات ١٣

B-Unsymmetrical Sections (one angle)

a) Stiffness condition:

300≤vi

L

1:2:1.01:5.1:14.0

1:1:2.0

=→==→=

=→=

abbiabbi

abai

v

v

v

Step (1): Choose of Cross Section:-


رئيس قســم االنشـــــاءات ١٤

Cont. B-Unsymmetrical Sections (one angle)

b) Construction condition:(In Case of bolted

connection only)

ta +Φ≥ 3

cmMcmM5.5

4

16

12

→→

c) Stress condition

)6.085.02.1(58.0 xxFFA

y

= For Bolted case (B)

)6.0(58.0 yFFA = For Welded case (A)


رئيس قســم االنشـــــاءات ١٥

Cont. B-Unsymmetrical Sections (one angle)

Step (2): Check of the Chosen Member:-

a)Check of Actual Stress:

⎩⎨⎧

→→

≤=)(2.158.0

)(58.0Area Reduced BxF

AFForcefY

Yact

141.3

3

).().(

).(

21

121

2

1

1

pAA

AAAA

ttaAWeldedtaA

BoltedtdaA

red

h

→⎟⎟⎠

⎞⎜⎜⎝

⎛+

+=

−=→=

→−=

A 1

A 2


رئيس قســم االنشـــــاءات ١٦

b) Check of Stiffness:

60≤dL


رئيس قســم االنشـــــاءات ١٧

Example 1:

Design the lower chord tension member “A”

shown in the Figure. The force in the member is 30t (Case II) and the bolts used in the connections are 16 non- pretensioned bolts.


رئيس قســم االنشـــــاءات ١٨

Solution:

Data: Force = 30 t (Case II ) Length = 3.00 m Lx= 3.0 m and Ly = 9.0 m

Step (1): Estimation of the cross section:Stiffness condition:

Stress condition:

A req. =A req. = 30 / ( 0.85 x 1.4 x 1.20 ) = 21.0 cm2.

Choose 2 angles back-to-back for the lower chord. A of 1L = 21.0/2= 10.50 cm2

300≤vi

L

Force

0.58fy


رئيس قســم االنشـــــاءات ١٩

Cont. Solution:

Knowing that for 2 angles back-to-back with equal legs:

rx=0.3 a and ry = 0.45 aλx= Lx / (rx =0.3a)=300/0.3 a= 300 a req.=3.33 cmλy= Ly / (ry=0.45 a)= 900/(0.45x300) = 6.67 cm.

Construction Condition :a-t > 3d = 4.8 cm.From the tables choose 2Ls 80x80x8


رئيس قســم االنشـــــاءات ٢٠

Cont. Solution:

Step (2): Check:1. Strength:A net. = A gross – A Holes

A net. = 2 ( 12.3 - (1.6+0.2) * 0.8 ) = 21.72 cm2.

F act. = 30 / 21.72 = 1.38 t/cm2 < 1.4x1.2( case II ) =1.68 t/cm2 Safe

2. Stiffness:l/d = 300 / 8 = 37.5 < 60 OK


رئيس قســم االنشـــــاءات ٢١

Example 2:

Design the same member of Example 1 with welded connections.

Solution:Step (1): Estimation of the cross section:Stiffness Condition:

Stress condition:A req. = 30 / (1.4 x 1.2) = 17.86 cm2Choose 2 angles back-to-back for the lower chord.

A of 1L = 17.86/2 = 8.92 cm2

300≤vi

L


رئيس قســم االنشـــــاءات ٢٢

Cont. Solution:

For 2 angles back-to-back with equal legs:rx=0.3 a and ry=0.45 arx = Lx / (rx=0.3 a)=300/(0.3x300) = 3.33 cmry = Ly / (ry=0.45 a) = 900/(0.45x300) = 6.67 cm

From the tables choose 2Ls 70x70x7


رئيس قســم االنشـــــاءات ٢٣

Cont. Solution:

Step (2): Check:1. Strength:A net = Agross = 2 x 9.4= 18.8 cm2.f ca = 30 / 18.8 = 1.595 t/cm2< 1.4x1.2( case II ) =1.68 t/cm2. Safe

2. Stiffness:l/d = 300/7 = 42.85 < 60 O.K. Safe

٢٠ Pagesســـعد الدين مصـطفى . د.أ


Cont. Ch.(2) Design of Compression Member“



of Compression LenghtsBuckling Members

Lbx = Buckling Length in-plane of TrussLby = Buckling Length in-out-of-plane of TrussBuckling Lengths depend on selendernessratio()

See Code table(4-4) p.56&57

LL L



Euler Column Formula

•Maximum Srtrenght of Axially Loaded MemberEuler Column Formula = Fall.c *Area Strength is determined by Euler Formula (Pin-ended Column) -Using Euler Formula = Pe = (TEI/L2) FCritical = Pe/Area = (TEI/L2*A)= (TE/L2)(I/A)= TE/L2*(r2)

rLwhereEF

rL

rLratiosslendernes

bcr =⇒=

=⇒==

λλ

π

λλ

2

2

.

2

*

)(2



Relation between and Ultimate Stress Fu

.

λ

λ

λ>100<100 λ

Fu

Euler Curve



:CompresionAllowable stresses in -

100 <==⇔r

Klr

lorlFor byxbλ

100 ≥==⇔rkl

rorrorll

Foryx

bybxλ2/7500 λ=→ cF

AIorI

rorr yxyx =⇒

2.6.4(p15) Axial compression

Fc

t/cm2Grade of Steel

40 mm < t ≤

100 mmt ≤

40 mm

1.3-0.000055λ21.4-0.000065λ2St 371.5-0.000075λ21.6-0.000085λ2St 442.0-0.000125λ22.1-0.000135λ2St 52



Design of Compression

Members

. .

A-SymmetricalSections

B-UnsymmetricalSections



Design Procedure : A-Symmetrical Sections (Two angle)Data Given : Lbx , Lby , Fdesign

Step (1) : Choose of Cross Section:-

a) Stiffness Condition:

b) Stress Condition

c) Construction Design

(In Case of bolted connection only)From three Conditions choose the greatest two angles

angleonearanglestwoar

rl

rl

xx

y

byy

x

bxx

⇒=⇒=

≤==≤==

2.0,3.0

180.....180... λλ ο

( )2

.

2

3.17.0

......

cmtFAssume

cmA

Call

FF

req Call

Design

−=

==

ta +≥ φ3



Step (2): Checks the Satisfy of theChosen Section:

1. Check of Stresses:

2. Check The class of section (compact ,non-compact &Slender)

callDesign

act

call

call

y

byy

x

bxx

fA

Ff

fif

fif

takerl

rl

.2

max2.max

max2

.max

max

......

7500100

000065.04.1100

≤=⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎠⎞

⎜⎝⎛=⇒≥

−=⇒⟨

====

⟨

λλ

λλ

λλλ

anglesUnequalforNoncompactft

hb

anglesequalforNoncompactft

b

y

y

⇒≤=⎟⎠⎞

⎜⎝⎛ +

⇒≤=⎟⎠⎞

⎜⎝⎛

17.....2

23......



Step 3 :Design of Tie Plates:

Tie plate is used to prevent the occurrence of local buckling and to prevent any of the two angles forming the cross section to buckle individually.

mL

angleoneofaxisorrrL

rL

vv

5.1......

min

*

*

2

1max

≤=

⇒⇒=

=

<

<λλ



Design Procedure : B-Unsymmetrical Sections (one angle)

Data Given : Lbx , Lby , Fdesign

Step (1) : Choose of Cross Section:-

a) Stiffness Condition:

b) Stress Condition

c) Construction Design

(In Case of bolted connection only)From three Conditions choose the greatest two angles

greatery

byy

x

bxx take

rl

rl λλλ ==== ο

( )2

.

2

9.04.0

......

cmtFAssume

cmA

Call

FF

req Call

Design

−=

==

ta +≥ φ3



Step (2): Checks the Satisfy of theChosen Section:

1. Check of Stresses:

2. Check The class of section (compact ,non-compact &Slender)

callDesign

act

call

call

y

byy

x

bxx

fA

Ff

fif

fif

takerl

rl

.1

max2.max

max2

.max

max

*6.0......

7500100

000065.04.1100

≤=⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎠⎞

⎜⎝⎛=⇒≥

−=⇒⟨

====

⟨

λλ

λλ

λλλ

anglesUnequalforNoncompactft

hb

anglesequalforNoncompactft

b

y

y

⇒≤=⎟⎠⎞

⎜⎝⎛ +

⇒≤=⎟⎠⎞

⎜⎝⎛

17.....2

23......



Example 1:

Design Force = -18 t (A), length of member = 160 cm (Upper member), the steel used (37), use bolts M16 in the connections.

Solution:Step (1): Choose of Cross Section:-

1- Stress Condition:-Assume Fc=1.0t/cm2.

From table choose two equal angle (70*70*7)

221

2

.

0.92

0.180.10.18 cmAAcm

FF

Acall

Design ==→=== <<



Cont. Solution:

2- Stiffness condition:-

3- construction condition:-

From the above conditions choose the greatest two equal angle (70*70*7)

89.0180160

180180 =≥→≥→≤ x

bxx

x

bx iLiiL

89.0180160

180180 =≥→≥→≤ y

byy

y

by iL

iiL

163 Mta →+≥ φ

cma 5.57.06.1*3 ≥+≥



Cont. Solution:

Step (2): Check the satisfy of the chosensection:-

1- Check the class of section(compact, non-compact, slender):-

2- Check of Stress:-

compact.-non issection

angle e8.14231010

100 qualFt

b

y

→=≤==

( ) okiLcmtableii

x

bxxyx →≤===→==⇒ << 1805.75

12.216012.2)(12 λ



Cont. Solution:

( ) oki

cmI

iL

AI

iteAtableII

yy

y

y

byy

yy

GPyyy

→≤==→==→

=++=

≤=→=→++=⇒

<

<

<

<<<<<

1800.4926.3

16026.38.185.199

5.199])20.197.1(*4.94.42[2

180])2

(*)([2

2

422

2

22

2112

λ

λ

22

2

/03.1))5.75(000065.04.1(

)000065.04.1(

1005.75

cmt

Fc

cr

=−

=−=→

<=⇒

λ

λ



Cont. Solution:

OK. Safe.3- Design of Tie Plate:-

Use one tie plate in the middle.

222 /03.1))5.75(000065.04.1()000065.04.1(

1005.75

cmtFc

cr

=−=−=→

<=⇒

λ

λ

2/03.1/96.04.9*2

0.18A

2

2

cmtcmtF

f Designact π===

∠

;2/4.1035.75*37.1

/L maxgreater-b

Ll

rv

>==

≤ λ



Example 2:

Design Force = -3 t (A), length of member = 200 cm (Vertical member), the steel used (37), use bolts M16 in the connections.(one angle)

Solution:Step (1): Choose of Cross Section:-

1- Stress requirement:-Assume Fc=0.5t/cm2.

From table choose two equal angle 60*60*6

2

..

0.65.00.3 cm

FF

Acall

Design ===



Cont. Solution:

2- Stiffness Condition:-

3- construction requirement:-

From the above conditions choose the greatest angles (60*60*6)

11.1180200

180180 =≥→≥→≤ x

bxx

x

bx iLiiL

11.1180200

180180 =≥→≥→≤ y

byy

y

by iL

iiL

163 Mta →+≥ φ

cma 5.57.06.1*3 ≥+≥


رئيس قســم االنشـــــاءات ١٩

Cont. Solution:

Step (2): Check the satisfy of the chosensection:-

1- Check the class of section(compact, non-compact, slender):-

2- Check of Stresses:-

compact.-non issection

8.1423106

60=≤==

yFtb

( ) okiLcmLLv

xbybx →≤===→==⇒ 1809.17017.1

200200 λ


رئيس قســم االنشـــــاءات ٢٠

Cont. Solution:

The section is unsafe try another angle (80*80*8)

222 /154.0))9.170/(7500(6.0)/7500(*6.0

1009.170

cmtFc

cr

===→

>=⇒

λ

λ

( )

222 /27.0))129/(7500(6.0)/7500(*6.0

100129

18012955.1

200200

cmtF

okiLcmLL

c

cr

vxbybx

===→

>=⇒

→≤===→==⇒

λ

λ

λ

2/154.0/43.091.60.3

A2

1

cmtcmtF

f Designact >===

∠

2/27.0/244.03.12

0.3A

2 cmtcmtForcef act <=== OK Safe

١٨ pagesســـعد الدين مصـطفى . د.أ


Ch(3) "Design of Steel

”Connections



Types of steel Connection:

(I) Riveted Connections

(II) - Bolted connections.a) Ordinary Bolted Connectionsb) High Strength Bolted Connections

Two Classes of Bolts used in Structures 1 - Pretensioned Bolts, 2 - Non-pretensioned Bolts.

(III) - Welded Connections.



-Ordinary Bolted Connections:

Diameter of Bolts:-M12, M16, M20, M22, M24, M27, M30, M36.

Grade of Bolts:-

10.98.86.85.85.64.84.6Bolt grade

9.06.44.84.03.04*0.8=3.2

4*0.6=2.4

Fybt/cm2

10865544Fubt/cm2

Ordinary BoltsMild Steel

High Strength BoltsHeat-treated Steel



Modes of Failure in Bolted Connections Subjected -to Shear :

Either in Bolts or in Connected PlatedOccuresFailure

(1) – Single Shear Failure (2) -Double Shear ailureResistance of Bolt (Rs.sh.) Resistance of Bolt (Rd.sh.)

... sbshs AqR =

( ) ( ) ( )8.8,6.5,6.4gradesbolt 25.0 forFq ubb →=( ) ( ) ( ) ( )9.10,8.6,8.5,8.4gradesbolt 2.0 forFq ubb →=

P.106 6.3) (Tablebolt of area stressA s →

... *2 shsshd RR =



, Resistance of PlateBearing Failure-(3) Connected, (Rb .):

Where:

∑= tFR bb min..φ

⎪⎪⎩

⎪⎪⎨

⎧

→

→

→

=→==2

2

2

U

/2.552

/4.444

/6.337

F*stress bearing Allowable

cmtsteel

cmtsteel

cmtsteel

FF Ub α



= factor depends on end distance α (e)(P102).

bolt ofdiameter =φ

∑ =tmin

End distance

1.21.00.80.6α

φ5.1≥e φ0.3≥eφ0.2≥e φ5.2≥e



-Required Number of Bolts:

FailureShearDoubleRorRofSmallerR

FailureShearSingleRorRofSmallerR

bshdLeast

bshsLeast

⇒=

⇒=

.

.

)(2.1*.

BcaseRFN

least

D

→=φ



(II) Resistance of Bolts in Bolted Connection Subjected to Tension:

•Tensile Resistance of a Single Bolt is given by:

)33.0(4

*7.0

**7.0

.

2..

boltUTension

bolttallboltTension

FR

FAR

πφ=

=



(III) Resistance of Bolts in Bolted Connection Subjected to Tension and Shear:

•Bolts subjected to Combined action of Shear and Tension should Satisfy the interaction Equation :

oneBoltTensionActualnT

boltoneShearActualnQ

Where

Rn

T

Rn

Q

tensionLeast

=

=

≤=⎟⎟

⎠

⎞

⎜⎜

⎝

⎛+

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

)(

)(

:

0.1.....



Positioning Holes for Bolts:

Spacing in Tension andCompression Member

Min.Edge distance = e1 & e2 > 1.5 diameterMax.Edge

distance = e1 & e2 < 12t

Staggered Spacing in Compression Member

Spacing Bet. Bolts = Pitch =SSmin = gmin = 3 diameterSma x= gmax = 14t or 20cm (Least)

Staggered Spacingin Tension Member

mmandts 20014 ≤≤

e1

e2g

s

s

s

2sOuter row

Inner row

T or C

C C

TT



Design Procedure of Bolted Connections:.Data given-1)

Diameter of Bolts, Thickness of Guesst

plate,Sections of members & Design forces

Resistance of Bolts.:-2

Required Number of Bolts.-3Drawing.-4

FailureShearDoubleRorRofSmallerR

FailureShearSingleRorRofSmallerR

bshdLeast

bshsLeast

⇒=

⇒=

.

.

MembereachfortFR bb ⇒= ∑min..φ

... sbshs AqR =

... *2 shsshd RR =

)(2.1*.

BcaseRFN

least

D

→=φ



Solved Example.:

3



-Solved Example:Design the connection shown in figure using non-pretensionbolts grade (4.6) M16, Gusset plate thickness is 10 mm.

N3N2

N1

N5

N4F= -12t(B) & L=1.60m & 2<60x6

F= +12t(B) & L=1.60m & 2<55x5F= -2.0t(B) & L=2.5m & 1<80x8

F= -2

.5t(B

) & L

=2.5m

& 1<

80x8 F= +4.5t(B) &

L=2.0m &

1<55x5

-2.0 t

+4.5t

+12.0 t

-12.0 t

-2.5 t



Solution:-1) Data given.Diameter of Bolts (M16),Grade of Bolts (4.6),Thickness of Gusset plate (10mm),Sections of members & Design forces (as in fig.)

2) Resistance for bolts:(Rleast): Rleast*1.2 Case (B)

tRtR

nnnAqR

SD

SS

sbsh

14.357.1

57.1*57.1*4*25.0..

.

.

=→=→

===

∑ ∑∑ === minminmin 6.4*6.1*6.3*8.0... tttFR ub φα



Cont. Solution:-3- Required Number of Bolts.

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( ) 1642.32.1*14.3

0.120.12

1642.32.1*14.3

0.120.12

1621.12.1*57.1

0.20.2

1634.22.1*57.1

5.45.4

16233.12.1*57.1

5.25.2

0.1&.

55

0.1&.

44

8.0&.

33

5.0&.

22

8.0&.

11

MRorRR

FN

MRorRR

FN

MRorRR

FN

MRorRR

FN

MRorRR

FN

tbSDboltleast

tbSDboltleast

tbSSboltleast

tbSSboltleast

tbSSboltleast

≈====

≈====

≈====

≈====

≈====

=−

=−

=−

=−

=−

- 2 .0 t

+ 4 .5 t

+ 1 2 .0 t

- 1 2 .0 t

- 2 .5 t



خطوات الرسم **لكل الوصلة) CL(S)( رسم ال -١سواء علوية أو سفلية ثم رسم األعضاء ) Cords( رسم ال -٢

بقدر اإلمكان وعدم ) Cords(الداخلية بحيث تكون قريبه من ال .ترك مسافات بينهما بقدر اإلمكان

أوال على ال Cord)(األآبر عددا في ال ) Bolts( رص ال -٣Smin ثم المائل الذي يحتوى على أآبر عدد على الSmin ثم بعد

بشكل مناسب ) G.P(ذلك نشكل ال الداخلية (في األعضاء اآلخرى ) Bolts( فرض بقية ال -٤

بقياس المسافة وقسمتها على عدد المسامير ونوجد ) والخارجية تقع بين القيم العليا والدنيا بحيت pitch ال

( )mmtS 200&143 min≤≤φ



-Drawing (Connection):

2 < 6 0x6

2 < 5 5x5

1<80x8

1<80

x8

1<55

x5

1 0m mG .P l

S c a le 1 :1 0



-Drawing (Gusset Plate):

46,77

5,5

22,16

5,516,08

8

11,6

8

6

2 ,79 5,57 5,57 5,577,58

5,57 5,57 5,57 2,97

2,75

32,

585,

155,

152 ,75

3,77

7,52

2,45

4,9

7,26

١٢ Pagesسعد الدين مصطفى ./ د.أ

١ رئيس قسم اإلنشاءات

Cont. Ch.3 "Design of steel

“connections

III- Welded Connections


٢ رئيس قسم اإلنشاءات

Types of Weld:Types of Weld:(1) - Butt Weld (Groove weld):

Allowable Stress:-

Square Butt Weld

Double "V" Butt weld

Single "V" Butt weld

Single "U" Butt Weld

Excellent WeldGood WeldKind of StressType

1.1 Fc1.0 FcCompression

Butt Weld 1.0 Ft0.7 FtTension

1.1 qall1.0 qallShear


٣ رئيس قسم اإلنشاءات

(2) - Fillet Weld:

SS

S

SL

S S

L

e

SS

L

L

F

F1

F2

a


٤ رئيس قسم اإلنشاءات

Notes on Fillet Weld:Notes on Fillet Weld:Min. and Max. Thickness and Lengths of Welds-(1)

S Size of weld (S) Smin= 4mm & Smax ≤ tminL Length of weld (L) Lmin.=4S or 5 cm & Lmax ≤ 70S

Area of Weld = L * S(2) - Allowable stress:-

All Kind of stresses Fw ≤

0.2 Fu Fu =3.6 t/cm2 (St.37)

L1

L2

X>2X >2XX


٥ رئيس قسم اإلنشاءات

(3) (3) -- Welding Symbols :Welding Symbols :-

S = 4 mm

S = 4 mm

Purlin

S = 4 mmS = 4 mm

S = 4 mm


٦ رئيس قسم اإلنشاءات

-Distribution of Forces in Fillet Weld-(4)

m = 1.0 (one angle) &m = 2.0 (two angle)

e SS

L1

L2

F

F1

F2

a

( ) ( )aeFF

aeaFF =

−= 21 &

( ) ( )U

weldallweld

FLSmFFAForce

2.0***

11

.

==

( )

( )70S L cm 5or 4S22.0**

70S L cm 5or 4S22.0**

22

11

≤≤→+=

≤≤→+=

SFSm

FL

SFSm

FL

u

u


٧ رئيس قسم اإلنشاءات

**Steps of Design:**Steps of Design:1- Data Given:Design force. , Section of members. , Allowable Stress

& Assume Size of Weld (S=4mm).

2- Design Procedure:a)- Determine F1 &F2 In fillet weld.

b)- Determine The lengths of weld L1 &L2 .

c)- Draw the connection.

e SS

L1

L2

F

F1

F2

a

( ) ( )aeFF

aeaFF =

−= 21 &

( )

( )70S L cm 5or 4S.....22.0**

2

70S L cm 5or 4S....22.0**

1

2Pr

1Pr

≤≤→=+=

≤≤→=+=

cmSFSm

FL

cmSFSm

FL

uactice

uactice


٨ رئيس قسم اإلنشاءات

Solved Example:Design the connection shown in figure as Welded Connection,

Gusset plate thickness is 10 mm.

F= -12t(B) & L=1.60m & 2<60x6

F= +12t(B) & L=1.60m & 2<55x5

F= -2.0t(B) & L=2.5m & 1<80x8

F= -2

.5t(B

) & L

=2.5m

& 1<

80x8

F= +4.5t(B) & L=2.0m

& 1<55x5

-2.0 t

+4.5t

+12.0 t

-12.0 t

-2.5 t D1D2

V1

U2

U1


٩ رئيس قسم اإلنشاءات

Cont. Solution:-1- Data Given:Design forces, Section of members, Allowable Stress (0.2Fu)& Assume Size of Weld (S=5mm).

2- Design Procedure:a)- Determine F1&F2 In fillet weld.b)- Determine The lengths of weld L1&L2.For Diagonal and Vertical: m=1.0 one angle 1< For Diagonal and Vertical: m=1.0 one angle 1<

Case(BCase(B) ) FweldFweld=0.2Fu*1.2=0.2Fu*1.2

LL 22 (cm)(cm)LL 11 (cm)(cm)Size of Size of weld (S) weld (S) FF 22 =F=Fmm *e/a*e/aFF 11 =F=Fmm *(a*(a--e)/ae)/aaa--eee (table)e (table)FF mmSectionSectionmembermember

2.645.140.50.711.795.742.26-2.5(B)1<80*8D1

2.294.330.50.561.445.742.26-2.0(B)1<80*8D2

3.858.550.51.233.263.981.52+4.50(B)1<55*5V1


١٠ رئيس قسم اإلنشاءات

Cont. Solution:-For Upper Members: m=2.0 one angle 2< For Upper Members: m=2.0 one angle 2<

Case(BCase(B) ) FweldFweld=0.2Fu*1.2=0.2Fu*1.2

Hint:Hint:Any Length ≤

5cm Take it = 5cm

LL 22 (cm)(cm)LL 11 (cm)(cm)Size of Size of weld (S)weld (S)FF 22FF 11aa--eee (table)e (table)FF mmSectionSectionmembermember

4.9111.00.53.388.624.311.69-12.0(B)2<60*6U1

4.8411.050.53.328.683.981.52-12.0(B)2<55*5U2


١١ رئيس قسم اإلنشاءات

Cont. Solution:-c)- Draw the connection.

240x5

240x5

240x5

167x5

167x5

240x5

S cale 1 :10

68x5


١٢ رئيس قسم اإلنشاءات

Cont. Solution:-

c)- Draw the connection.

Scale 1:10

50,21

22,96

50,21

22,96

90°

90°

90°

90°

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