### What is a Force?

"May the

*Force*be with you"

"The Force is what gives a Jedi his power. It's an energy field created by all living things. It surrounds us and penetrates us. It binds the galaxy together."

Ok, well that may not be exactly what we're talking about when we're trying to describe a force... Let's retry this, in physics F = ma (thanks Sir Isaac Newton!), or force is equal to mass times acceleration. For Metric Units? this stands true where the mass of an object (kg) will be multiplied by it's acceleration (m/s^{2}) to get a force (N). For English Units (which is most often taught on this site), force is usually simplified into pounds (lbs). In almost all aspects of design, it would make sense not having to deal with English Units, which are often arbitrary at best. But it is much easier to deal with Pounds vs. Newtons in my opinion.

**To properly define a force:** is a push or pull that one body exerts onto another.

An example of a force is: if you have a large man standing on a W8x10 then his weight (lets say 220 lbs) acts as a force on the beam. Since gravity is pulling his mass towards the center of the earth (which we'll arbitrarily call down), and F=ma then his mass and the acceleration due to gravity, create a force on the beam.

**Note:** All members will have some force applied to them due to their own selfweight.

**A fun SideNote:** The weight of the atmosphere applies a constant pressure (that can be converted to a force) on all of us.

The examples below will help clarify what a force is and how it is an essential building block for all of structural engineering.

Examples of Force Calculations
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### Example #1: A Simple Force Calculation

You have a bridge with a group of W8x10's at 5'-0" o.c. that span 10'-0" (assume the car can only drive perpendicular to the beams). You are going to drive a car over the W members and plan on having 2 wheels on a set of W8x10's at a time. The wheel base of the car is 6'-0" (distance from the front axle to the back axle), and the track (distance CL left to CL right) is 4'-0". You also know that there is a 60%/40% distribution of weight between the front of the car and the back of the car respectively (due to the engine). And lastly the car weighs 3,200 lbs. What are the forces acting on one set of beams?

**First we have a few checks to make.**

Since the Vehicles wheel base is 6'-0" and the beams are only 5'-0" o.c. then only one axle can act on a beam at a time. Also since the wheels track is only 4'-0" and the length of the beam is 10'-0" then both wheels will be able to act on the beam at once.

**Now to find the Force acting on the Beam:**

**F =** 3,200 lbs * 60% / 2 wheels = **960 lbs**

**And your loads look like this:**

**Force Calculation Ex in Isometric View**

**Extra Credit:** Why did we center the wheel loads?

Centering the wheel loads like we did produced the maximum moment on the beam (which is most likely to govern).

### Example#2: A Medium Difficulty Force Calculation

You have W10x22's spaced at 12'-0" o.c. (length = 15'-0"). On top of that you have a 5" lightweight concrete slab. You plan on walking a giraffe (weight = 2,500 lbs) over the slab. What do the forces acting on the W10 look like?

This force has been broken up into two separate problems, A uniform load (caused by the concrete slab):**w =** Tributary Width * Load (psf) = 12'-0" * (5"/12") * 100 pcf **= 500 plf**

Once again we will assume the point load acts in the middle of the beam to find a worst case scenario for bending.**P** = weight of the giraffe **= 2,500 lbs**

**Force Calculation Ex in Isometric View**