Subject
Year
: S0695 / STATICS
: 2008
Analysis of Statically Determinate Structures
Session 03 - 06
Analysis of Statically Determinate Structures
Structures must
be stable under all conditions
to support loads : their own weight & external loads
( wind, earthquake, peoples, etc )…………….
without
changing shape , large displacement or
collapsed
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Analysis of Statically Determinate Structures
To analyze structure we must consider :
1. Forces
2. Moment
3. Resultant of Planar Force system
4. Transmissibility
5. Loads Resultant
6. Support
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Analysis of Statically Determinate Structures
1. Forces :
Y
F
Fy = F.sin α
b
Fx = F.cos α
α
0
Z
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X
a
a
=
sin α
c
β
γ
b = c
sin β
sin γ
Analysis of Statically Determinate Structures
2. Moment :
Y
F
d
F
0
Z
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X
M = F.d
Analysis of Statically Determinate Structures
3. Transmissibility
a force
may be moved
along
its action line without changing the
external effect of it procedures in body
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Analysis of Statically Determinate Structures
3. Transmissibility
H
A
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Action line
B
3. Transmissibility
H
C
Action
line
Action
line
Analysis of Statically Determinate Structures
D
A
B
Action line
VA
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VB
F
90o
o
d Æ perpendicular distance
F
d
d
o
d
90
90
3. Transmissibility
So… for
F
Analysis of Statically Determinate Structures
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Analysis of Statically Determinate Structures
4. Resultant of Planar Force system
Resultant conditions :
a. Rx = Σ Fx
b. Ry = Σ Fy
c. Mo = R.d = Σ Fi.di + Σ Mi
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Analysis of Statically Determinate Structures
4. Resultant of Planar Force system
Resultant conditions :
Mo = R.d = Σ Fi.di + Σ Mi
R = Rx2 + Rx2
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d = perpendicular distance
Analysis of Statically Determinate Structures
5. Loads Resultant ( for distributed loads )
a. Simple shapes
q
q
L
1/3L
2/3L
w = q.L
w = ½ q.L
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Analysis of Statically Determinate Structures
5. Loads Resultant ( for distributed loads )
b. Parabolic, compute w by integrating the
area under parabolic
f(x)
q
w
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Analysis of Statically Determinate Structures
6. Support
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Sources :
Leet,Kenneth M, Chia Ming Huang. (2005). Fundamental of Structural Analysis. 2nd Ed. Mc Graw Hill. Singapore.
Analysis of Statically Determinate Structures
6. Support
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Sources :
Leet,Kenneth M, Chia Ming Huang. (2005). Fundamental of Structural Analysis. 2nd Ed. Mc Graw Hill. Singapore.
Analysis of Statically Determinate Structures
6. Support
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Sources :
Leet,Kenneth M, Chia Ming Huang. (2005). Fundamental of Structural Analysis. 2nd Ed. Mc Graw Hill. Singapore.
Analysis of Statically Determinate Structures
6. Support
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Sources :
Leet,Kenneth M, Chia Ming Huang. (2005). Fundamental of Structural Analysis. 2nd Ed. Mc Graw Hill. Singapore.
Analysis of Statically Determinate Structures
6. Support
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Sources :
Leet,Kenneth M, Chia Ming Huang. (2005). Fundamental of Structural Analysis. 2nd Ed. Mc Graw Hill. Singapore.
Analysis of Statically Determinate Structures
6. Support
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Sources :
Leet,Kenneth M, Chia Ming Huang. (2005). Fundamental of Structural Analysis. 2nd Ed. Mc Graw Hill. Singapore.
2.1. Idealized Structure
For analyze structure, engineer develop a
simple physical model of the
supports as well as
applied loads.
it represent by simple line drawing
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2.1. Idealized Structure
HA
A
B
VA
VB
q
HA
VA
VB
w
HA
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VA
VB
2.2. Determinacy & Stability
Determinacy Æ
Wheater or not the reactions and forces in a
structure can be analysed on static
equilibrium
It deals with wheater the forces on structure can
be determinated knowing only geometry
of the structure, or wheater the stiffness
component must be known.
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2.2. Determinacy & Stability
Determinacy consist of internal and external issues…
• externally determinate
If the reaction of structure can be solved based
on static equilibrium
• internally determinate
If the force of individual component can
be solved based on static equilibrium
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2.2. Determinacy & Stability
Assesing determinacy of single rigid body Æ
3 equations of static equilibrium….
If R = 3 Æ statically determinate
q
HA
VA
VB
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2.2. Determinacy & Stability
Assesing determinacy of single rigid body Æ
3 equations of static equilibrium….
If R > 3 Æ statically indeterminate
degree of indeterminacy = R-3
q
HA
VA
VC
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VB
2.2. Determinacy & Stability
Stability Æ
the geometric arrangement of members and
supports required for a stable structure, or the
structure ca resist load , without change the
structure shape or significant value for lateral
displacement, and the members must properly
held or constrained by their support.
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2.2. Determinacy & Stability
Stability Æ
If the structure is
matter if
unstable, it doesn’t
it is statically determinate or
indeterminate.
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2.2. Determinacy & Stability
Stability condition Æ
a. If R < 3n Æ unstable ; n = member
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Sources : RC Hibbeler. (2006). Structural Analysis. 6th ed. Pearson Prentice Hall. Singapore.
2.2. Determinacy & Stability
Stability condition Æ
b. If R > 3n Æ unstable if member reactions are
concurrent or parallel or some of components
form a collasible mechanism
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Sources : RC Hibbeler. (2006). Structural Analysis. 6th ed. Pearson Prentice Hall. Singapore.
2.2. Determinacy & Stability
Stability condition Æ
Sources : RC Hibbeler. (2006). Structural Analysis. 6th ed.
Pearson Prentice Hall. Singapore.
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2.3. Equation Equilibrium
Analysis Procedure :
1. Count the total number of unkowns Æ 3 equilbrium
equations
2. Moment equation Æ ΣMi = 0
3. Aplly the force equation Æ Σ Fx = 0
& Σ Fy = 0
4. If the solution Æ negative , it indicates its arrowhead
sense direction is opposite
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2.4. Reaction
Reaction Forces Æ
+
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-
-
Reaction Moment Æ
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2.4. Reaction
+
2.4. Reaction
Simple Beam :
P
q
A
B
L
Problem :
½L
P=1t
q = 1 t/m
L = 12 m
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1/4 L
1/4 L
2.4. Reaction
Simple Beam :
Step 1 :
Count the reaction forces / unknown forces
P=1t
q =1 t/m
HA
3 reactions
Determinate
structure
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A
B
VA
6
3
3
VB
2.4. Reaction
Simple Beam :
P=1t
Step 2 :
q =1 t/m
Eq. 1 Æ Moment
HA
Σ Mi = 0
ΣMA=0
w.3m + P.9m - VB.12m = 0
6t.3m + 1t.9m - VB.12m = 0
27 tm – 12 VB = 0
VB = 2 ¼ t
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A
B
w = q.6m = 6 t
VA
6
3
ΣMB=0
- P.3m- w.9m + VA.12m = 0
- 1t.3m - 6t.9m + VA.12m = 0
- 57 tm – 12 VA = 0
VA = 4 ¾ t
3
VB
2.4. Reaction
Simple Beam :
P=1t
Step 3 :
Eq. 2 & 3Æ
Forces
q =1 t/m
HA
Σ Fx = 0
B
w = q.6m = 6 t
VA
ΣFx=0
HA = 0
Σ Fy = 0
OK !!!
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A
6
3
3
ΣFy=0
VA + VB = P + w
4¾t+2¼t=1t+6t
7t=7t
VB