Struc lec. no. 1
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Transcript of Struc lec. no. 1
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Theory of Structures(1)
Lecture No. 1
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Loads
Reactions
Supports
Conditional Equations
Stability and Determinancy
Chapter 1 : Loads and Reactions
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Chapter 2 : Statically Determinate Beams Types of Beams
Normal Force, Shearing Force, and Bending Moment
Relationship between Loads, Shearing Force
and Bending Moment
Standard Cases of S.F. and B.M.D(s) for Beams
Principal Of Superposition
Inclined Beams
A Beam: is a structural member subjected
to some external forces.
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Chapter 3 : Statically Determinate Rigid FramesInternal Stability and Determinacy
Method of Frames Analysis
A Frame: is a structure composed of a number of members
connected together by joints all or some are rigid.
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Chapter 4 : Statically Determinate Arches
Reactions , Thrust, Shearing Force and Bending Moment using Analytical Method Reactions Thrust, Shearing Force, and Bending Moment at any Point in the Arch using Graphical Method
An Arch :is a curved beam or two curved beams connected together by intermediate hinge.
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Chapter 5 : Statically Determinate Trusses
Stability and DeterminacyAnalysis for Trusses: Methods of: joints, Sections, and Stress Diagram
A truss: Consists of a number of straight members pin -
connected together and subjected to concentrated
loads at hinges.
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Chapter 1
Loads and Reactions
Loads classified according to:
Cause of Loading : dead ,live, and lateral Shape of Load : concen. and distrib. Rate of application : static and dynamic
Reactions: resistance by supports to counteract the action of loadsSupports : Roller, Hinge, Fixed, and Link
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Roller Hinged
Fixed Link
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Determination of the Reactions:
”External loads acting on the structure together
with the reactions at the supports must constitute a
system of non concurrent forces in equilibrium”
∑x = 0 (sum of horizontal forces) = 0
∑y = 0 (sum of vertical forces) = 0
∑M = 0 (sum of moments ) = 0
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YB
XA
YA
Solution of Example 1
1- ∑ X = 0 XA – 5 = 0 XA = 5t
2- ∑MB = 0 YA × 6 – 2 × 6 × 3 + (4 × 3)/2 × 1 = 0 YA= 5t
3- ∑ Y = 0 YA + YB = 2 × 6+ (4 × 3)/2=18 YB = 13t
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Solution of Example 2
1- ∑ X = 0 XA = 3t
2- ∑ MB = 0 YA× 4 + 8 + 4 × 1 – 3 × 1= 0 YA= -2.25t ↑
YA=2.25t ↓
3- ∑ y = 0 YA + 2× 2 = YB
2.25+ 4 = YB YB = 6.25 t ↑
YA
XA
YB
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Solution of Example 3
1- ∑ X = 0 XB – 4 = 0 XB = 4t
2 -∑MA = 0
10YB – 4 × 12 – 2 × 9 – 2 × 5× 2.5 – 4 × 3= 0
YB = 10.3 t
3- ∑ Y = 0 YA + YB = 4 + 2 + 10 = 16 YA = 5.7 t
YBYA
XB
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Solution of Example 4
1- ∑ X = 0 XA – 4 = 0 XA = 4t
2- ∑ Y = 0 YA – 1 × 6 – 5 – 5 – 2.5 = 0 YA = 18.5 t
3- ∑ M/A= 0 MA – 5 × 2 – 5 × 4 – 2.5 ×6 - 1×6×3 = 0
MA = 63 t.m
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Solution of Example 5
R1 = 1/2 × 4 × 8 = 16 t
R2 = 0.5 × 4 = 2 t
1- ∑ X = 0 XA – R2 = 0 XA = 2t
2- ∑ MA= 0 11 YB + 2 × 13 – R2 × 2 – 2 × 1 – R1 × 7= 0
YB = 8.36 t
3- ∑ Y = 0 YA+ YB+ 2 – 2 – 16 = 0 YA= 7.64t
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Solution of Example 6
1- ∑ X = 0 XB – 20 = 0 XB = 20t
2- ∑ MA = 0
6 YB + 20 × 3 – 10 × 6 – 20 × 6 = 0 YB = 20t
3- ∑Y = 0
YA + YB = 20 + 30 + 10 = 60 YA = 40t
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Solution of Example 7
∑MO = 0
3 × 0 + 8 × 1 + 1 × 3 – Rd × 5 = 0
Rd = 2.2 t ↑
∑Mp = 0
1 × 2 + 8 × 4 + 3 × 5 – Rb × 5 = 0
Rb = 6.93t
∑Mq = 0
1 × 2 + 8 × 4 + 3 × 5 – Rc × 5 = 0
Rc = 6.93t
2
2
8t
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Conditional Equations Some structures are
divided to several parts connected together by hinges, links or rollers.
Forces are transmitted through these connections
from one part to the others .
00 McC
BrightMcandMc
C
AleftCM
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Solution of Example 8
Method I1- Part FD
1) ∑ MD = 0 8 × 2 = 4 YF YF = 4t
2) ∑Y = 0 YD = 8 – 4 YD = 4t2- Part ECF 1) ∑ ME = 0 10×2.5+4×5= 4YC YC= 11.25t
2) ∑Y = 0 YE=10+4-11.25 YE=2.75t3- Part ABE 1) ∑MA=0 12×3+2.75×6=5YB YB=10.5t
2) ∑Y = 0 YA= 2.75 + 12-10.5 YA= 4.25t
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Solution of Example 8 “ Continued ”
Method II
1)∑MF(right)= 0 8 × 2 – 4 YD = 0 YD = 4t 2) ∑ME(right)= 0 18×4.5 – 4×9 – 4YC= 0 YC=11.25t
3) ∑MA = 0 30×7.5 – 4×15 – 11.25× 10 - 5YB = 0 YB=10.5t
4) ∑Y = 0 30 – 4 – 11.25 – 10.5 – YA=0 YA=4.25t
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c
c
d
M = 0 = Xd × 8 – 8 × 2
Xd = 2 t
b
dbM = 8 ×12 + 2.5 × 12 × 6
Xd
- 2 × 8 - 10 ×Yd=0
Yd = 26 t
∑Y = 0 = Ya + 26 – 2.5 × 12 – 8
Ya = 30 + 8 – 26 = 12 t.
∑X = 0 = Xa – 0.5 × 8 – 2
Xa = 4 + 2 = 6 t. ←
Solution of Example 9
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= 0 = Ma + 0.5 × 8 × 4 – 6 × 8 Ma = 48 – 16 = 32 m.t (anticlockwise)
Check : ∑Mc for part ac Xa Ya
32 + 0.5 × 8 × 4 + 2.5 × 12 × 4 – 6 × 8 – 12 × 10= 32 + 16 + 120 – 48 – 120 = 0
b
abM
c
aCM
Solution of Example 9 “Continued”
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1) ∑MF(right) = 0 RC cosα × 4.5 – 4 × 4.5 × 2.25 – 10 × 10.5 = 0 RC = 40.4 t
2) ∑MA = 0 YB×9 + XB × 2 + RC cosα × 18 + RC sinα × 8 –10×24
- 20×4.5 – 5 × 5 – 4 × 9 × 13.5 + 5 × 2 = 0
4.5 YB = 27.57 – XB (1)
3) ∑ME (right) = 0 4.5 YB - 6XB + RC cosα × 13.5 – 4 × 9 × 9 - 10 × 19.5 = 0
4.5 YB = 6XB + 82. 545 (2)
Solution of Example 10
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By solving equations (1) & (2) we get
YB = 7.87 t. XB= - 7.85t
4) ∑Y = 0 YA = 36 + 10 + 20 + 5 – 32.33 – 7.87
YA = 30.8 t.
5) ∑ME(left) = 0 4.5 YA – 8 XA – 5 × 3 – 5 × 6.5 = 0
XA = 11.4 t.
Solution of Example 10 “Continued”
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Stability and Determinancy
A. External concerned with the reactions.
B. Internal concerned with the internal forces and moments.
A Stable Structure is one that support: a system of loads. These loads together with the support reactions have to be in equilibrium and satisfy the three equilibrium equations in addition to conditional equations (if any).
Stability of Structures
Statically Unstable Structure is one in which: the number of the unknown reactions is less than the sum of the available equilibrium and conditional equations (if any).
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Externally Statically Unstable Structures
Geometrical Unstable Structures
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Statically Determinate Structure is one in which the reactions can be determined by application of the equations of
equilibrium in addition to conditional equations(if any).
Determinancy of Structures
Statically Indeterminate Structure is one in whichthe number of unknown reactions is more than the number of availableequations
Stable-Once Statically Indeterminate
Stable-Three Times Statically Indeterminate
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Check the stability and determinacy for the following structures. Then, show how to modify it for stability and determinancy
(a) (b)
(c)
(d)
Solution of Example 11
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Structure (a) :Is unstable because there are only two
unknown reactions.
The modification : Change either of the two roller supports to
a hinge.
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Structure (b) : Is unstable because the beam is rested on
four link members connected at a point.
The modification : Separate the link members
Modification for structure (b)
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Structure (C) : Is stable and three times statically indeterminate because there are six unknown support reactions.The modification : Add three intermediate hinges
Modification for structure (C)
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Structure (D) : Is stable and five times statically indeterminate because there are
8 unknown support reactions
The modification : One of fixed supports is changed to a roller In addition to three intermediate hinges
arranged as given in Struc. (c).