Statics Using 2 index cards: Create a structure or system of structures that will elevate two textbooks at least 1.5cm off your desk

Statics

Using 2 index cards:Create a structure or system of structures that will elevate two textbooks at least 1.5cm off your desk

Statics

What is Statics?Branch of Mechanics that deals with objects/materials that are stationary or in uniform motion. Forces are balanced.

Examples:1. A book lying on a table (statics)2. Water being held behind a dam (hydrostatics)

Chicago

Kentucky & Indiana Bridge

Dynamics

Dynamics is the branch of Mechanics that deals with objects/materials that are accelerating due to an imbalance of forces.

Examples:1. A rollercoaster executing a loop (dynamics)2. Flow of water from a hose (hydrodynamics)

1. Total degrees in a triangle:2. Three angles of the triangle below:3. Three sides of the triangle below:4. Pythagorean Theorem:

x2 + y2 = r2

180

A

B

C

x, y, and r

y

x

r

HYPOTENUSE

A, B, and C

Trigonometric functions are ratios of the lengths of the segments that make up angles.

Q

y

x

r

sin Q = =opp. y hyp. r

cos Q = =adj. x hyp. r

tan Q = =opp. y adj. x

sin A = opposite

hypotenuse

cos A = adjacent

hypotenuse

tan A = opposite adjacentsin A = 1

2

cos A =

tan A =

√3 2

12

3A

B

C

1 √3

For <A below, calculate Sine, Cosine, and Tangent:

ac

A

B

Cb

Law of Cosines:c2 = a2 + b2 – 2ab cos C

Law of Sines:sin A sin B sin C a b c

= =

1. Scalar – a variable whose value is expressed only as a magnitude or quantityHeight, pressure, speed, density, etc.

2. Vector – a variable whose value is expressed both as a magnitude and directionDisplacement, force, velocity, momentum, etc.

3. Tensor – a variable whose values are collections of vectors, such as stress on a material, the curvature of space-time (General Theory of Relativity), gyroscopic motion, etc.

Properties of Vectors

1. MagnitudeLength implies magnitude of vector

2. DirectionArrow implies direction of vector

3. Act along the line of their direction4. No fixed origin

Can be located anywhere in space

Magnitude, Direction

Vectors - Description

45o40 lb

s

F = 40 lbs 45o

F = 40 lbs @ 45o

magnitude direction

Hat signifies vector quantity

Bold type and an underline F also identify vectors

1. We can multiply any vector by a whole number.2. Original direction is maintained, new magnitude.

Vectors – Scalar Multiplication

2

½

1. We can add two or more vectors together. 2. 2 methods:

1. Graphical Addition/subtraction – redraw vectors head-to-tail, then draw the resultant vector. (head-to-tail order does not matter)

Vectors – Addition

Vectors – Rectangular Components

y

x

F

Fx

Fy

1. It is often useful to break a vector into horizontal and vertical components (rectangular components).

2. Consider the Force vector below. 3. Plot this vector on x-y axis.4. Project the vector onto x and y axes.


y

x

F

Fx

Fy

This means:

vector F = vector Fx + vector Fy

Remember the addition of vectors:


y

x

F

Fx

Fy

Fx = Fx i

Vector Fx = Magnitude Fx times vector i

Vector Fy = Magnitude Fy times vector j

Fy = Fy j

F = Fx i + Fy j

i denotes vector in x direction

j denotes vector in y direction

Unit vector


y

x

F

Fx

Fy

Each grid space represents 1 lb force.

What is Fx?

Fx = (4 lbs)i

What is Fy?

Fy = (3 lbs)j

What is F?

F = (4 lbs)i + (3 lbs)j


If vector

V = a i + b j + c k

then the magnitude of vector V

|V| =


F

Fx

Fy

cos Q = Fx / F

Fx = F cos Qi

sin Q = Fy / F

Fy = F sin Qj

What is the relationship between Q, sin Q, and cos Q?

Q


y

x

F Fx +

Fy +

When are Fx and Fy Positive/Negative?

FFx -

Fy +

FFFx -Fy -

Fx +Fy -


Complete the following chart in your notebook:

III

III IV

1. Vectors can be completely represented in two ways:1. Graphically2. Sum of vectors in any three independent directions

2. Vectors can also be added/subtracted in either of those ways:1.

2. F1 = ai + bj + ck; F2 = si + tj + uk

F1 + F2 = (a + s)i + (b + t)j + (c + u)k

Vectors

A third way to add, subtract, and otherwise decompose vectors:

Use the law of sines or the law of cosines to find R.

Vectors

F1 F2

R45o

105o

30o

Brief note about subtraction1. If F = ai + bj + ck, then – F = – ai – bj – ck

2. Also, if

F =

Then,

– F =

Vectors

Resultant Forces

Resultant forces are the overall combination of all forces acting on a body.

1) find sum of forces in x-direction

2) find sum of forces in y-direction

3) find sum of forces in z-direction

3) Write as single vector in rectangular components

R = SFxi + SFyj + SFzk

Resultant Forces - ExampleA satellite flies without friction in space. Earth’s gravity pulls downward on the satellite with a force of 200 N. Stray space junk hits the satellite with a force of 1000 N at 60o to the horizontal. What is the resultant force acting on the satellite?

1. Sketch and label free-body diagram (all external and reactive forces acting on the body)

2. Decompose all vectors into rectangular components (x, y, z)

3. Add vectors

Statics Newton’s 3 Laws of Motion:

1. A body at rest will stay at rest, a body in motion will stay in motion, unless acted upon by an external force

This is the condition for static equilibrium

In other words…the net force acting upon a body is Zero

Now on to the point…

Newton’s 3 Laws of Motion:

2. Force is proportional to mass times acceleration:F = ma

If in static equilibrium, the net force acting upon a body is Zero

What does this tell us about the acceleration of the body?It is Zero

Newton’s 3 Laws of Motion:

3. Action/Reaction

Statics Two conditions for static equilibrium:

1.

Individually.

Since Force is a vector, this implies

Two conditions for static equilibrium:1.

Two conditions for static equilibrium:

Why isn’t sufficient?


2. About any point on an object,

Moment M (or torque t) is a scalar quantity that describes the amount of “twist” at a point.

M = (magnitude of force perpendicular to moment arm) * (length of moment arm) = (magnitude of force) * (perpendicular distance from point to force)


MP = F * x MP = Fy * x

M = (magnitude of force perpendicular to moment arm) * (length of moment arm) = (magnitude of force) * (perpendicular distance from point to force)

PF

x

P

F

x

Moment Examples:1. Tension test apparatus – unknown and reaction forces?2. If a beam supported at its endpoints is given a load F at its

midpoint, what are the supporting forces at the endpoints?

Find sum of moments about a or b.

Ra Rb

Watch your signs – identify positive

Moment Examples:3. An “L” lever is pinned at the center P and holds load F at the

end of its shorter leg. What force is required at Q to hold the load? What is the force on the pin at P holding the lever?

What is your method for solving this problem? Remember,

Trusses

Trusses: A practical and economic solution to many structural engineering challenges

Simple truss – consists of tension and compression members held together by hinge or pin joints

Rigid truss – will not collapse

Trusses Joints:Pin or Hinge (fixed)

Trusses Supports:Pin or Hinge (fixed) – 2 unknowns

Reaction in x-direction

Reaction in y-direction Rax

Ray

Trusses Supports:Roller - 1 unknown

Reaction in y-direction only

Ray

Assumptions to analyze simple truss:

1. Joints are assumed to be frictionless, so forces can only be transmitted in the direction of the members.

2. Members are assumed to be massless. 3. Loads can be applied only at joints (or nodes). 4. Members are assumed to be perfectly rigid.

2 conditions for static equilibrium:1. Sum of forces at each joint (or node) = 02. Moment about any joint (or node) = 0

Start with Entire Truss Equilibrium Equations

Truss Analysis Example Problems:

1. A force F is applied to the following equilateral truss. Determine the force in each member of the truss shown and state which members are in compression and which are in tension.

Truss Analysis Example Problems:

2. Using the method of joints, determine the force in each member of the truss shown. Assume equilateral triangles.

Static determinacy and stability:

Statically Determinant: All unknown reactions and forces in members can be determined by the methods of statics – all equilibrium equations can be satisfied.

m = 2j – r (Simple Truss)

Static Stability:The truss is rigid – it will not collapse.

Conditions of static determinacy and stability of trusses:

Materials Lab Connections:• Tensile Strength = Force / Area• Compression is Proportional to 1 / R4

Problem Sheet solutions due Monday

Documents

Statics Using 2 index cards: Create a structure or system of structures that will elevate two textbooks at least 1.5cm off your desk