Structural Analysis I Structural Analysis Trigonometry Concepts Vectors Equilibrium Reactions Static Determinancy and Stability Free Body Diagrams Calculating.

Structural Analysis I

• 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:

a2 + b2 = c2

a

b

c

Sine (sin) of an Angle

• In a right triangle, the angles are related to the lengths of the sides by the equations:

sinθ1 = =Opposite a

Hypotenuse c

sinθ2 = =Opposite b

Hypotenuse c

a

b

c

θ1

θ2

Cosine (cos) of an Angle

• In a right triangle, the angles are related to the lengths of the sides by the equations:

cosθ1 = =Adjacent b

Hypotenuse c

cosθ2 = =Adjacent a

Hypotenuse c

a

b

c

θ1

θ2

This Truss Bridge is Built from Right Triangles

a

b

c

θ1

θ2

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

• Suppose you hit a billiard ball with a force of 5 newtons at a 40o angle

• This is represented by a force vector

Θ = 40o

F = 5N

y

x

Vector Components

• Every vector can be broken into two parts, one vector with magnitude in the x-direction and one with magnitude in the y-direction.

• Determine these two components for structural analysis.

Vector Component Example

• The billiard ball hit of 5N/40o can be represented by two vector components, Fx and Fy F = 5N

Fx

Fy

θx

y

F = 5N

x

y

Fy Component Example

To calculate Fy, sinθ =

sin40o =

5N * 0.64 = Fy

3.20N = Fy

F = 5N

Fx

Fy

Θ=40o

Opposite Hypotenuse

Fy

5N

Fx Component Example

To calculate Fx, cosθ =

cos40o =

5N * 0.77 = Fx

3.85N = Fx

F = 5N

Fx

Fy

Θ=40o

Adjacent Hypotenuse

Fx

5N

What does this Mean?

Your 5N/40o hit is represented by this vector

F = 5N

Θ=40o

x

yFx = 3.85N

Fy=3.20N

x

y

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

F = 5N

Θ=40o

x

y

Force Name 5N at 40°

Free Body Diagram

x-component 5N * cos 40°

y-component 5N * sin 40°

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

• Represent the boat as a point at the (0,0) location

• Represent the pulling forces with vectors

Θm = 50oΘf = 70o

Fm = 150 lb

Ff = 100 lb

x

y

Boat Pull Solution (cont)

First analyse the force Ff

• x-component = -100 lb * cos70°• x-component = -34.2 lb

• y-component = 100 lb * sin70°• y-component = 93.9 lb

Separate force Ff into x and y components

Θf = 70o

Ff = 100 lb

-x x

y

Boat Pull Solution (cont)

Next analyse the force Fm

• x-component = 150 lb * cos50°• x-component = 96.4 lb

• y-component = 150 lb * sin50°• y-component = 114.9 lb

Separate force Fm into x and y components

Θm = 50o

Fm = 150 lb

x

y

Boat Pull Solution (cont)

Force Name Ff FmResultant

(Sum)

Vector Diagram

(See next slide)

x- component-100lb*cos70

= -34.2 lb150lb*cos50

= 96.4 lb62.2 lb

y-component100lb*sin70

= 93.9 lb150lb*sin50= 114.9 lb

208.8 lb

70o

100 lb

x

y

50o

150 lb

x

y

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

Fm

Ff

-x

y

xFTotalX

FTotalY

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