Download Week 5

Structural Analysis I
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Structural Analysis
Trigonometry Concepts
Vectors
Equilibrium
Reactions
Static Determinancy and Stability
Free Body Diagrams
Calculating Bridge Member Forces
Learning Objectives
• Define structural analysis
• Calculate using the Pythagoreon Theorem,
sin, and cos
• Calculate the components of a force vector
• Add two force vectors together
• Understand the concept of equilibrium
• Calculate reactions
• Determine if a truss is stable
Structural Analysis
• Structural analysis is a mathematical
examination of a complex structure
• Analysis breaks a complex system down to
individual component parts
• Uses geometry, trigonometry, algebra, and
basic physics
How Much Weight Can This
Truss Bridge Support?
Pythagorean Theorem
• In a right triangle, the
length of the sides are
related by the
equation:
a 2 + b2 = c 2
c
a
b
Sine (sin) of an Angle
• In a right triangle, the
angles are related to the
lengths of the sides by
the equations:
c
Opposite
sinθ1 =
=
Hypotenuse
θ1
a
c
Opposite
b
sinθ2 =
=
Hypotenuse
c
θ2
a
b
Cosine (cos) of an Angle
• In a right triangle, the
angles are related to the
lengths of the sides by the
equations:
c
Adjacent
b
cosθ1 =
=
Hypotenuse
c
θ1
Adjacent
a
cosθ2 =
=
Hypotenuse
c
θ2
a
b
This Truss Bridge is
Built from Right Triangles
θ2
c
a
θ1
b
Trigonometry Tips for
Structural Analysis
• A truss bridge is constructed from members
arranged in right triangles
• Sin and cos relate both lengths AND magnitude of
internal forces
• Sin and cos are ratios
Vectors
• Mathematical quantity that has both
magnitude and direction
• Represented by an arrow at an angle θ
• Establish Cartesian Coordinate axis
system with horizontal x-axis and
vertical y-axis.
Vector Example
y
• Suppose you hit a
billiard ball with a force
of 5 newtons at a 40o
angle
F = 5N
Θ = 40o
x
• This is represented by a
force vector
Vector Components
• Every vector can be broken into two parts,
one vector with magnitude in the xdirection and one with magnitude in the ydirection.
• Determine these two components for
structural analysis.
Vector Component Example
y
F = 5N
• The billiard ball hit
of 5N/40o can be
represented by two
vector components,
Fx and Fy
x
y
F = 5N
Fy
θ
Fx
x
Fy Component Example
To calculate Fy,
sinθ =
sin40o
F = 5N
Fy
Θ=40o
Fx
Opposite
Hypotenuse
Fy
=
5N
5N * 0.64 = Fy
3.20N = Fy
Fx Component Example
To calculate Fx, cosθ =
Adjacent
Hypotenuse
cos40o =
F = 5N
Fy
Θ=40o
Fx
Fx
5N
5N * 0.77 = Fx
3.85N = Fx
What does this Mean?
y
y
Fx = 3.85N
F = 5N
Θ=40o
Your 5N/40o hit is
represented by this vector
x
Fy=3.20N
x
The exact same force and
direction could be achieved if
two simultaneous forces are
applied directly along the x
and y axis
Vector Component Summary
Force Name
5N at 40°
y
Free Body
Diagram
F = 5N
Θ=40o
x-component
5N * cos 40°
y-component
5N * sin 40°
x
How do I use these?
• Calculate net forces on
an object
• Example: Two people
each pull a rope
connected to a boat.
What is the net force
on the boat?
She pulls with
100 pound force
He pulls
with 150
pound
force
Boat Pull Solution
y
• Represent the boat
as a point at the
(0,0) location
• Represent the
pulling forces with
vectors
Fm = 150 lb
Ff = 100 lb
Θf = 70o
Θm = 50o
x
Boat Pull Solution (cont)
Separate force Ff into x and y components
y
First analyse the force Ff
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x-component = -100 lb * cos70°
x-component = -34.2 lb
y-component = 100 lb * sin70°
y-component = 93.9 lb
Ff = 100 lb
Θf = 70o
-x
x
Boat Pull Solution (cont)
Separate force Fm into x and y components
Next analyse the force Fm
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x-component = 150 lb * cos50°
x-component = 96.4 lb
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y-component = 150 lb * sin50°
y-component = 114.9 lb
y
Fm = 150 lb
Θm = 50o
x
Boat Pull Solution (cont)
Force Name
Ff
y
Vector
Diagram
y-component
y
x
-100lb*cos70
= -34.2 lb
100lb*sin70
= 93.9 lb
Resultant
(Sum)
150 lb
(See next slide)
100 lb
70o
x- component
Fm
50o
x
150lb*cos50
= 96.4 lb
150lb*sin50
= 114.9 lb
62.2 lb
208.8 lb
Boat Pull Solution (end)
• White represents
forces applied
directly to the boat
• Gray represents the
sum of the x and y
components of Ff
and Fm
• Yellow represents
the resultant vector
y
FTotalY
Fm
Ff
-x
x
FTotalX
Equilibrium
• Total forces acting on an object is ‘0’
• Important concept for bridges – they
shouldn’t move!
• Σ Fx = 0 means ‘The sum of the forces in
the x direction is 0’
• Σ Fy = 0 means ‘The sum of the forces in
the y direction is 0’ :
Reactions
• Forces developed at
structure supports to
maintain equilibrium.
• Ex: If a 3kg jug of water
rests on the ground, there is
a 3kg reaction (Ra) keeping
the bottle from going to the
center of the earth.
3kg
Ra = 3kg
Reactions
• A bridge across a river
has a 200 lb man in
the center. What are
the reactions at each
end, assuming the
bridge has no weight?
Determinancy and Stability
• Statically determinant trusses can be
analyzed by the Method of Joints
• Statically indeterminant bridges require
more complex analysis techniques
• Unstable truss does not have enough
members to form a rigid structure
Determinancy and Stability
• Statically determinate truss: 2j = m + 3
• Statically indeterminate truss: 2j < m + 3
• Unstable truss: 2j > m + 3
Acknowledgements
• This presentation is based on Learning
Activity #3, Analyze and Evaluate a Truss
from the book by Colonel Stephen J.
Ressler, P.E., Ph.D., Designing and Building
File-Folder Bridges