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Transcript of 1-Introduction Model (1)
2014
Structural Analysis (I)
INTRODUCTION
PART ( 1 )
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PART (1)
2014
2014
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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.
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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
# END PART (1)