1-Introduction Model (1)

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Structural Analysis (I)

INTRODUCTION

PART ( 1 )

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Course Contents:1st Term (50%):

Ch(1) Introduction + Reactions of Plane Structures.Ch(2) Internal Forces for Beam structures.Ch(3) Analysis of Rigid Frame structures.Ch(4) Analysis of Arches.

2nd Term (50%):

Ch(5) Analysis of Trusses.Ch(6) Influence Lines (Beams - Frames - Trusses).Ch(7) Moving Loads.Structural Analysis

Study of

Stresses and Strains. Reactions and Internal Forces.

External Forces (Loads).

Structural Analysis (I) Introduction

Connections

(FX, FY and M)

(FX and FY only)

1

2

1

2

2 2

11

Types

1- Rigid Connection

2- Pin Connection

Real Member

A BIn Analysis

CL

Idealized Structure

CL A B

Structural member

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Structural Analysis (I) Introduction

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Define1- Structural Members.2- Connections / Supports.3- Loads.4- Types of structures.

Types1- Column / Tie Element.2- Beam Element.3- Beam-Column Element.4- Slab Element.

Types

1- Roller Support (One Unknown)

2- Hinged Support (Two Unknowns)

3- Fixed Support (Three Unknowns)

4- Link Support (One Unknown)

RY

RX

RRY

RY

RX

RY

RX

RY

RX

RY

RX

In TensionIn ComprssionF FF F

MM

Loads

34

1- Concentrated Loads( )

Force

10t

5t6t

Moment

6t.m 10t.m

(Dead Loads - Live Loads)

* Magnitude* Position* Direction

Types

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Structural Analysis (I) Introduction

Supports (Unknown Reactions )

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ــل محم غير ــل محم غير

PIN PIN

ه . ب ــع القط عند إال تظهر وال ــاه إتج فــي ــون تكـ الداخليـــه Linkالقــوه

2- Distributed Loads ( )

Resultant Force

Magnitude = AreaPosition = C.G of AreaDirection (Given)

a- Uniform Dist. Loads

WLWt/m`

WL

Wt/m`

Wt/

m`

Wh

Wt/m`

WL`

Wt/m`(H.P)

WL

Wt/m`

WL`

b- Non-uniform Dist. Loads

WL2

Wt/m`

W1L

W1t/m`

W2t/m`(W2-W1)L2

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Structural Analysis (I) Introduction

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1- Point Stability2- Stability Equations - ∑ FX = 0.0

- ∑ FY = 0.0

Stability Equations

Structures

1- Plane Structures:

Studied in (X-Y) plane - Beams. - Rigid Frames. - Trusses. - Arches.

2- Space Structures:

Studied in (X-Y and Z) planes - Grid. - Space Frames. - Space Trusses. - Domes.

Types

3- Stability Equations

6- Stability Equations

6- Stability Equations

3- Stability Equations - ∑ FX = 0.0 - ∑ FY = 0.0 - ∑ M @ any point = 0.0

2- Body Stabilitya- Two - Dimensions (X and Y)

b- Three - Dimensions (X, Y and Z)

Moments - ∑ M X = 0.0

- ∑ MY = 0.0- ∑ MZ = 0.0

Forces - ∑ FX = 0.0 - ∑ FY = 0.0 - ∑ FZ =

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Structural Analysis (I) Introduction

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(b)

Simple Frame Cantilever Frame 3- Hinged Frame

Pin Conn.

Link memberh

3- Hinged Arch

Curved shape ReducesBending Moment ( M≈ 0)

Problems

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Plane Structures

1- Beams

Simple BeamCantilever Beam Overhanging Beam

Continuous Beam Compound Beam

2- Frames

Rigid Con.

Beam

Column

3- Trusses (2nd Term)

4- Arches

Simple Truss

Moment = 0.0

2nd Degree Curve

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Structural Analysis (I) Introduction

→ Given ( Structure + Loads)→ Require ( Reactions + Internal Forces )

Steps of Analysis1 - Unknown Reactions @ Supports

2 - Loads

3- Stability Equations.

e

Get 3- Unknons

Wt/m`* Magnitude* Position* Direction

PPcos Θ

Psin Θ

Θ

Condition Equtions

e

M= 0 M= 0 M= 0 M= 0

M= 0

M= 0

M= 0C= 1

C= 2 C= 1 C= 1

Internal Hinge

∑ Me left = 0.0 ∑ Me Right = 0.0Number of Condition Equations (C)

C = Number of Members - 1.0

M

M

+

ــة. المجهول الفعــل ردود إليجـــاد ــزان اإلت ــادالت مع #تســــتخدم

∑ FX = 0.0

∑ M b= 0.0

∑ FY = 0.0وال أ الواحد المجهــول ــاه اتج فــي التعـــويض يتــم

ــل مجاهي عدد ــبر اك تجمــع ــه نقط د عن العزم ــة معادل

ــوب المحسـ الفعــل رد إشارة ــت كان #إذا

ــروض← المف اإلتجـــاه ــس نف فــي الفعــل رد ــون يكـ +VE

ــروض← المف اإلتجـــاه ــس عك الفعــل رد ــون يكـ -VE

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Structural Analysis (I) Introduction

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Primary equations

∑ M@ I.H (Right or Left) = 0.0

Number of Unknowns (U) > 3.0

a

Ya

Xa 4t

5t

b

Yb5t

8t.m 5t3

4 3t

4t

3

45

cos Θ = 35

Θ

sin Θ = 45

4t/m`

16t

UNKNOWNSWe have two Supports → Hinge @ point a

→ Roller @ point b

Ya

Xa

Distributed Load

Structure C.L. Concentrated Moment Inclined Conc. Load

Dimensions

Yb

ــوي للق الموجــب اإلتجـــاه مع المجاهيــل ــرض تف #

LOADSDistributed loads → Resultant Force

4t/m`16t * Magnitude = Area

* Position = C.g* Direction = as Given

Lw

Concentrated load → in X-Y directions5t

34 5(35)t

5(45)t

STABILITY EQUATIONS

ResultantUnknowns

∑ FX → = 0.0 → Xa = 3t

∑ M b = 0.0 → Ya = 12 t

∑ M a =-8(6) + 4(10) - 8 + 16(2) = 0.0 ....... OK.CHECK

Ya(8) - 16(6) - 8 + 4(2) = 0.0

∑ FY ↑ = 0.0 → Yb = (16+4) - 12 = 8tLoads Ya

GIVEN → Structure( Beam - Frame - Truss) + loads + Supports.REQUIRE → REACTIONS @ SUPPORTS

# Structural problem(1)

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Structural Analysis (I) REACTIONS

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Force moment arm

BEAMS

BEAM STRUCTURES CHAPTER 2.

# Structural problem(2)

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PLANE FRAMES

RIGID FRAMES CHAPTER 3.

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3t/m`

18t

4t/m` 30t

18t

24t

7t

0.8t

5.2t

36.74t

12.26t

12t

24m

33

4.5

6m 6m2

Frame Type → THREE HINGED FRAME.

a

b

UNKNOWNSWe have two Supports → Hinge @ point a

Ya

Xa

→ Hinge @ point bYb

Xb

LOADSDistributed loads → Resultant Force

3t/m`

6m

18t

4t/m`

30t

5m4t/m

`

24t

18t

STABILITY EQUATIONS

∑ M b= 0.0Ya(12)-Xa(3) -12(1)-7(14)-18(9)-24(3)-18(5.25) = 0.0

∑ FY = 0.0

Get Ya, Xa

∑ M e= 0.0 Left Part (a-e)

Xa = 0.80 tGet Yb, Xb

Ya = 36.74 t

∑ M e= 12.26(6)+5.2(7.5)-30(3.75) = 0.0 ....... OK.

12Ya-3Xa = 438.5 .......(1)

Ya(6)-Xa(10.5) -12(8.5)-7(8)-18(3) = 0.06Ya- 10.5Xa = 212 .......(2)

Solve equations (1) and (2) get

Yb =(24+18+7) - 36.74 = 12.26 t ∑ FX = 0.0 Xb = 18 - (12+0.8) = 5.20 t

Ya

XaYb

Xb

3 3

2.25

د عن ــيزتين الركــ من أي د عن األفعـــال ردود إليجـــادــالي. كالتــ مجهــولين فــي معــــادلتين تكــــوين يتــم

X & Ya & b

Structural Analysis (I) REACTIONS

3t/m`

18t

4t/m` 30t

18t

24t

7t

12t

a

b

Ya

XaYb

Xb

CheckRight Part (b-e)

# Structural problem(3)

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ARCHES

ARCHES CHAPTER 4.

1/10

14

m

4 m 4 m 8 m

FLink

2t/m`10t

b

11.5t 14.5tYa Yb

Xb0.0t

16t

UNKNOWNSWe have two Supports → Hinge @ point b

→ Roller @ point aYb

Xb

Ya

LOADSDistributed loads → Resultant Force

2t/m`16t

Link member

يتــم ــيزتين الركــ د عن األفعـــال ردود حساب ــد بعــي ف القــوه حساب

a & bLink member F

STABILITY EQUATIONS

∑ FX → = 0.0 → Xb = 0.0 t

∑ M b = 0.0 → Ya = 11.5 t

∑ M a =-14.5(16) +16(12) +10(4) = 0.0 ....... OK.CHECK

Ya(16) - 10(12) - 16(4) = 0.0

∑ FY ↑ = 0.0 → Yb = (16+10) - 11.5 = 14.5t

Get Reactions @ supports a & b

Get Force in link member

4 m 4 m

FA

10t

11.5t

a

e

e

FLink

∑ M @ e= 0.0 Left Part (a-e)

FLink (3)-11.5(8) +10(4) = 0.0

3

→ FLink = 17.33t

Structural Analysis (I) REACTIONS

2t 4t 2t 3t 2t5t

∑ M b= 0.0

Ya(12) - 2 (12) - 4(9) - 2(6) - 3(3) - 5(3) = 0.0

-Yb(12) - 5(3) + 2(12) + 3(9) + 2(6) + 4(3) = 0.0

Check ∑ FY = (8 + 5) - (2+4+2+3+2) = 0.0 ..... OK.

∑ FX = 0.0

Ya = 8 t

Yb = 5 t

Xa = 5 t

∑ M a= 0.0

a

Ya

Xa

Yb

2t

4t3t 4t 5tEx(10) Truss

∑ M b= 0.0Ya(4.5) - 2 (10) - 4(6) - 3(4.5) - 4(3) - 5(1.5) = 0.0

∑ FX = 0.0

Ya = 17.11 t

Xb = 0 t

a

b

b

Yb

Xb

Ya

∑ FY = 0.0 Yb = (2+4+3+4+5) - 17.111 = 0.889 tCheck

∑ M a= -0.889(4.5) +5(3) + 4(1.5) - 4(1.5) - 2(5.5) = 0.0 .......OK.

5t8t

5t

0t

0.9t

17.1t

# Structural problem(4)

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PLANE TRUSSES

TRUSSES ( PIN-JOINTED FRAMES) CHAPTER 5.

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LOADSConcentrated loads on pin joints → as shown

Structural Analysis (I) REACTIONS

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