close

Вход

Log in using OpenID

embedDownload
Tutorial 02- Statics
(Chapter 3.0 – 5.3)
MECH 101 Spring 2009
F b 16th
Feb-16
Tutor: LEO
Contents
„
„
„
„
Reviews
Free Body Diagram (FBD)
Truss structure
Practices
Review
-Force vector, Moment, Equilibrium of Forces,
„
„
„
„
„
„
„
„
2D/3D Vectors (position, force)
Adding/Subtracting vectors (Parallelogram
law/trigonal)
Resolution of a vector using Cartesian vector
notation (CVN)
Additi using
Addition
i CVN
Multiplication of vectors (dot product)
Moment / torque (2D/3D)
MO = r × F
Cross product
Typical supports
Review
-Force vector, Moment, Equilibrium of Forces
„
„
„
The equivalent system
Moving a force from point A to O requires creating an additional
couple moment. Since this new couple moment is a “free” vector,
it can be applied at any point P on the body.
When several forces and couple
p moments act on a body,
y, you
y
can move each force and its associated couple moment to a
common point O. Afterwards, you can add all the forces and
couple moments together and find one resultant force-couple
force couple
moment pair.
Review -- GROUP PROBLEM SOLVING
Given: Handle forces F1 and F2 are
applied
li d to
t the
th electric
l t i drill.
d ill
Find: An equivalent
q
resultant
force and couple moment at
point O.
Plan:
a) Find FRO = Σ Fi
b) Find MRO = Σ MC + Σ ( ri × Fi )
Where
Where,
Fi are the individual forces in Cartesian vector notation (CVN).
MC are any free
f couple
l moments
t in
i CVN (none
(
in
i this
thi example).
l )
Ri are the position vectors from the point O to any point on the line
of action of Fi .
SOLUTION
F1 = {6 i – 3 j – 10 k} N
F2 = {{0 i + 2 j – 4 k}} N
FRO = {6 i – 1 j – 14 k} N
r1 = {0.15
{0 15 i + 0.3
0 3 k} m
r2 = {-0.25 j + 0.3 k} m
MRO = r1 × F1 + r2 × F2
MRO
i
j
= { 0.15 0
6
-33
k
0.3
-10
10
+
i
j
k
0 - 0.25 0.3
0
2
-44
} N·m
= {0.9
{0 9 i + 3.3
3 3 j – 0.45
0 45 k + 0.4
0 4 i + 0 j + 0 k} N·m
= {1.3 i + 3.3 j – 0.45 k} N·m
Rigid Body Equilibrium
„
THE PROCESS OF SOLVING RIGID BODY EQUILIBRIUM PROBLEMS
For analyzing an actual physical system, first we need to create an
idealized model.
Then we need to draw a free-body diagram showing all the external
(active and reactive) forces.
Finally, we need to apply the equations of equilibrium to solve for
any unknowns.
Rigid body:
y
The distance between any two points within the body is constant.
∑ Fx = 0
∑ Fy = 0
∑ MO = 0
Where point O is any arbitrary point.
Note: Newton’s 3rd LAW
F3
F4
F1
x
O
F2
Rigid Body Equilibrium
IMPORTANT NOTES
1 If we hhave more unknowns
1.
k
than
h the
h number
b off iindependent
d
d
equations, then we have a statically indeterminate situation.
W cannott solve
We
l these
th
problems
bl
using
i just
j t statics.
t ti
2. The order in which we apply equations may affect the
simplicity of the solution.
solution For example,
example if we have two
unknown vertical forces and one unknown horizontal force,
then solving ∑ FX = O first allows us to find the horizontal
unknown quickly.
g
number,,
3. If the answer for an unknown comes out as negative
then the sense (direction) of the unknown force is opposite to
that assumed when startingg the pproblem.
4. Two force members; Three force members.
Rigid Body Equilibrium
„
Free Body Diagram – FBD
PROCEDURE FOR DRAWING A FREE BODY G
DIAGRAM 1. Draw an outlined shape. Imagine the body to be isolated
or cut “free” from its constraints and draw its outlined shape.
2. Show all the external forces and couple moments. These
typically include: a) applied loads, b) support reactions, and,
c) the weight of the body (if applicable) .
3. Label loads and dimensions: All known forces and couple
moments should be labeled with their magnitudes and directions. For the
unknown forces and couple moments, use letters like Ax, Ay, MA, etc..
Indicate any necessary dimensions.
dimensions
Force direction in free body diagram
Whether a force positive or negative depends on your own d fi iti
definition.
y
x
=0
⎧ FAx
⎪ A
⇒⎨
1
⎪⎩ FAy = 2 P
=0
⎧ FBx
⎪ B
⎨F = 1 P
⎪⎩ By 2
⎧ F Ax = 0
⎪
⇒⎨
1
F
=
P
A
Ay
⎪⎩
2
⎧ FBx = 0
⎪
⎨F = 1 P
B
⎪⎩ By
2
⎧ FAx = 0
⎪
⇒⎨
1
−
F
=
P
Ay
⎪⎩
2
⎧ FBx = 0
⎪
⎨F = − 1 P
⎪⎩ By
2
Moment direction in free body
y diagram
g
∑ M = 0 ⇒ −M
O
+ M1 − M 2 + M 3 = 0
∑M = 0⇒ M
O
− M1 + M 2 − M 3 = 0
∑M
= 0 ⇒ M 0 + M1 − M 2 + M 3 = 0
∑ M = 0 ⇒ −M
0
− M1 + M 2 − M 3 = 0
A General Case
Aim: Find the reaction force on point A and B.
FAx
FAy
FBy
Equations of Equilibrium
∑F
=0
⇒ FAx = 0
∑F
Y
=0
⇒ FAy − qa + FBy − F = 0
B
(F ) = 0
5
⇒ qa a − 2aFAy − M − Fa = 0
2
X
∑M
1
M 5
FAy = − ( F +
− qa )
2
a 2
1
M 1
FAx = 0, FBy = (3F +
− qa)
2
a 2
Force Analysis in a Truss Structure
Zero‐force Members:
In Engineering mechanics,
mechanics a zero force member refers to a
member in a truss which, given a specific load, is at rest ‐‐ neither in
tension nor in compression.
p
In a truss a zero force member is often
found at pins (connections within the truss) where no external load
is applied and three or fewer truss members meet.
(Wikipedia: zero-force member)
How to identify a zero‐force member?
¾Two principles:
1) The geometry of a truss structure is fixed.
) Th f i fi d
a) Rigid body b) Triangles
2) Inspect those potential two‐force members first, then the others.
Force Analysis
y in a Truss Structure
Zero-force Members in a Truss Structure
¾ Two principles:
1) The geometry of a truss structure is fixed.
a) Rigid body b) Triangles
2) Inspect those potential two‐force members first, then the others.
Example: Indicate all zero-force
zero force members.
members
Answer: AB,BC,CD,DE,HI, and GI
JJoint Method
+
G
⊗
E
Dy
D
Ay
A
B
Ax
1m
C
1m
0
+
1m
100 kN
¾ Simplified
Si lifi d the
h structure to a frame
f
to solve
l out the
h support reactions.
i
Force balance in X-direction:
Force balance in Y-direction:
Moment balance at C:
Ax=0
Ay + Dy -100=0
Ay= 33.33 kN, Dy=66.67 kN
2Ay -1Dy=0
G
E
Ay=33.33
Dy=66.67
B
A Ax=0
C
1m
1m
Joint A:
0
D
1m
100 kN
Definition: The forces are applied
pp
to the jjoints by
y the beams.
∑F
∑F
x
= 0 ⇒ FAB − FAG cos 45° = 0
y
= 0 ⇒ 33.33 − FAG sin 45° = 0
FAB = 33.3kN
Joint B:
,
FAG = 47.1kN
(C)
(T)
FBG = 0 , FBC = 33.3 kN
(T)
G
E
Ay=33.33
Dy=66.67
B
A Ax=0
1m
1m
Joint D:
C
0
∑F =0
∑F =0
1m
100 kN
x
⇒ − FDC + FDE cos 45° = 0
y
⇒ 66.67 − FDE sin 45° = 0
FDC = 66.67 kN=66.7 kN
FDE = 94.29 kN=94.3 kN
Joint E:
D
∑F
∑F
(T)
(C)
⇒ FEG − 94.29sin 45° = 0
94 29 cos 45° = 0
= 0 ⇒ − FEC + 94.29
=0
x
y
FEC = 66.67 kN=66.7 kN
FEG = 66.67 kN=66.7 kN
G
((T))
(C)
E
Ay=33.33
Dy=66.67
B
A Ax=0
1m
1m
0
Joint C:
∑F
y
C
D
1m
100 kN
66 67 − 100 = 0
= 0 ⇒ FCG cos 45° + 66.67
FCG = 47.1
47 1 kN
((T))
Joint Method
1.
Joint Method
∑M
A
+
= 0 ⇒ 20 × 1 + 30 × 2 − RD cos 30°× 3 = 0
RD = 30.79kN
∑F
y
=0
⇒ Ay − 10 − 20 − 30 + 30.79 cos 30° = 0
Ay = 33.34kN
∑F
x
=0
⇒ Ax − 30.79sin
30 79sin 30° = 0
Ax = 15.4kN
⊗
+
Joint Method
+
⊗
+
15.4 kN
30.79 kN
33.34 kN
Joint A:
∑F
y
= 0 ⇒ 33.34 − 10 −
FAB = 33kN (C)
∑ Fx = 0 ⇒ 15.4 + FAF −
FAF = 7.93kN (T)
1
FAB = 0
2
1
× 33 = 0
2
Joint Method
+
⊗
+
15.4 kN
33.34 kN
Joint B:
1
× 33 − FBF = 0
y
2
FBF = 23.33kN=23.3kN (T)
∑F
=0 ⇒
1
× 33 − FBC = 0
2
= 23.33kN=23.3kN (C)
∑ Fx = 0
FBC
⇒
30.79 kN
+
Joint F:
⊗
+
⊗
+
1
FFC − 20 + 23.33 = 0
2
FFC = 4.714kN=4.71kN (C)
1
∑ Fx = 0 ⇒ FFE − 7.93 − 2 × 4.714 = 0
FFE = 11.26kN
11 26kN = 11.3kN
11 3kN (T)
∑ Fy = 0 ⇒ −
Joint E:
∑F
= 0 ⇒ FEC = 30kN (T)
∑F
= 0 ⇒ FED = 11.26kN=11.3kN (T)
y
x
+
Joint C:
∑ Fx = 0 ⇒
1
1
× 4.714 + 23.33 −
FCD = 0
2
2
⇒ FCD = 37.71kN = 37.7kN (C)
∑ Fy = 0 ⇒
1
1
× 4.714 − 30 +
37.71 = 0 (check)
2
2
JJoint Method
The Approach of Joint Method:
1) FBD ‐‐ Free body diagram. Remove the boundaries; add back support reactions.
back support reactions
2) Use EoE equilibrium equations (forces & moments) to solve out the support reactions.
solve out the support reactions
3) Inspect the force condition in each joints. Figure out the forces.
forces
4) Indicate the members are in tension or compression.
Section Method
1.
Support Reactions:
∑M
+
+
= 0 ⇒ Ey × 9 − 8 × 6 − 6 × 3 = 0
E y = 7.333kN
7 333kN
Moment balance at C
∑ MC = 0
⇒ 7.33 × 4.5 − 8 × 1.5 − FFG × 3 × sin 60° = 0
FFG = 8.08kN (T)
Moment balance at F
∑ M F = 0 ⇒ 7.33 × 3 − FCD × 3 × sin 60° = 0
Y
A
FCD = 8.47kN
X
(C)
Force balance along Y direction
∑ Fy = 0 ⇒ FCF sin 60° + 7.333 − 8 = 0
FCF = 0.770kN (T)
Stress condition
2.
+
Y
+
X
Force balance along
g Y direction
3
4
⇒
F
−
× 1500 = 0
F
=
0
DF
∑ y
5
5
FDF = 2000kN (C)
Moment balance at D
∑M
D
=0
4
3
⇒ × 1500 × 12 + × 1500 × 3 − FGF × 3 = 0
5
5
FGF = 5700kN (C)
Moment balance at F
4
⇒
× 1500 × 16 − FDE × 3 = 0
M
=
0
∑ F
5
FDE = 6400kN (T)
Section Method
Notes when using section method:
1. Imaginary cutting shall through the cross‐section of the member you want to analyze.
y
y
2. The cutoff rule is not necessarily straight.
3. The section must be continuous to split the structure into Th ti t b ti
t lit th t t i t TWO parts.
1/--pages
Пожаловаться на содержимое документа