Ten Most Common Terms in Mechanical Calculation

Hey guys! Welcome to this ASP video over mechanical calculations. There are several different equations that you want to be familiar with when heading in to take the ASP.

In this video we will review a few different terms within mechanical calculations, and the equation each term is referencing.

The ten terms we will be looking at are:

Equilibrium Equations
Distributed Loads
Free Body Diagrams
Stress and Strain

Newton’s second law says that force is equal to mass times acceleration (F=m*a). When dealing with statics, the subset of mechanics that deals with rigid bodies in static equilibrium, the acceleration in Newton’s equation is zero. This means that the sum of all forces applied to non-moving objects is also equal to zero.

Vectors are quantities with both direction and magnitude. In two-dimensional space, vectors can have horizontal (x-direction) and vertical (y-direction) components. Vector x- and y-components are calculated by multiplying the magnitude of the factor by its angle from horizontal:

Ax= Axcos(theta)
Ay= Axsin(theta)
Ax is the horizontal component
Ay is the vertical component
θ is the angle between the positive direction on the horizontal
axis and the vector, measured counter clockwise.

To add vectors together, first sum the x-components and y-components.

Find the magnitude of the resultant vector by taking the square root of the sum of the components squared:

     R=Squrt(Rx^2 + Ry ^2)

Finally, determine the angle of the resultant vector by taking the inverse tangent of Ry divided by Rx:

     theta= tan^-1 * (Rsuby/Rsubx)

Although a moment, or torque, is the product of a particular force and distance, a moment can be either thought of as a rotational force or, more accurately, as a measure of how much a force will contribute to potential rotation. Moments may sound unnatural when described this way, but they are quite intuitive as a twisting force. The moment about a point is defined as the magnitude of the acting force multiplied by the distance between it and the line of action, given by the equation:

     M= F x d

Where M is the moment, F is the force, and d is the distance perpendicular to the force to the point. In the diagram below, a wrench is used to tighten a lug nut.

Equilibrium equations
If an object is not moving, all the forces on it must be balanced. They are not necessarily equal, but the sum of all the forces acting on the object is zero. This knowledge can be used to calculate missing forces in static situations. For example, if a block rests on the floor without motion, it is at equilibrium. Because the object has some force due to gravity pulling it down, the floor must be opposing the block with an equal, but opposite, force. Two-dimensional force equilibrium says that the sum of all forces in the x- and y-direction is zero, and the sum of all moments is zero:

So, Summation Fsubx=0, Summation Fsuby+0, and Summation M=0

An ideal cable has no mass, and tension is constant throughout the length. To determine the tension in cables, apply the equilibrium equations. For example, in the diagram below, two cables support a weight of 456 pounds.

To calculate the tension in each cable, write equilibrium equations for the system. The weight is not accelerating in the x direction, so the sum of the x-components of the cable tensions must be zero.

SummationFsubx = 0 = Tsub1x + Tsub2x

Once again, there is zero acceleration, so the sum of the weight and the y-components of the cable tensions must be zero.

SummationFsuby = 0 = Tsub1y + Tsub2y-W

The x- and y-components of the tension in both cables can be solved with trigonometry:

   Tsub1y = Tsub1 X sin(30)
Tsub1x= Tsub1 x cos (30)
Tsub2y= Tsub2 x sin (45)
Tsub2x= Tsub2 x cos (45)

Substituting these equations in for the original equilibrium equations results in:

Tsub1cos(30) + Tsub2 x cos(45) = 0
-456 + Tsub1 x sin(30) + Tsub2 x sin(45) = 0

Now T1 and T2 can be calculated by solving for T1 in the first equation and substituting it in the second equation:

          Tsub1= (Tsub2 x cos(45) / cos(30) )

A pulley is a wheel with a grooved rim around which a rope passes. It acts to change the direction of a force applied to the rope, and is chiefly used for lifting heavy weights. An ideal pulley has no mass or friction. In a simple configuration with two equal masses suspended from a single pulley, the force on both rope segments is equal to F, therefore the force on the cable holding the pulley up must be 2×F. More complex pulley systems can also be analyzed by applying the equations of static equilibrium.

Distributed loads
Unlike a point load, a distributed load acts upon a length or area of an object. To calculate the effect of a distributed load, it must first be reduced to a resultant load that acts on one point of the length or surface. The magnitude of the resultant force is equal to the area under the curve of the distributed load. The location of the force is at the centroid, or weighted center, of the distributed load.

For example, the beam in the diagram below has a rectangle-shaped distributed load on it.

The magnitude of the resultant force can be calculated by solving for the the area of the rectangle. We know that the area of a rectangle is A=b
l*w, if you didn’t already know that then there you go!

So, when we plug in our numbers and cancel our units, we get:

    Area subrectangle = l*w= 3ft * 2kip/ft = 6kip

The center of mass of this rectangle can be found by plugging the equation of the load into the following equation on the screen:

x=0Lx*w(x)dx0Lw(x)dx = 03x*34 xdx0334 xdx

Free-Body Diagrams
Free-body diagrams are essential to understanding the statics of rigid bodies. A correctly drawn diagram allows one to identify and calculate all of the unknown forces and moments acting on such a body. To begin drawing a free-body diagram, separate the member of interest from the rest of the system at the joints or points of contact with other members. Draw a sketch of the member, and include all applied forces and moments, as well as support reactions or joint forces. In the figure below, a block rests on a ramp. Here, a free-body diagram is drawn on the block, showing the three forces acting on it.

Stress and strain
In mechanics, stress is a quantity of force per unit area in or on an object. Tensile or compressive stress acts normal to the area of stress and is calculated with the following equation:

     Sigma= (Fsubn/ A)
σ is the compressive or tensile stress
Fn is the normal component force
A is the area

For example, if a reinforcement rod is pounded into the ground, it experiences compressive stress. If the rod is half an inch in diameter and is hit by a sledgehammer bearing 100 pounds of force, the stress can be calculated as:

    Sigmasubrod= (100lb/ pi * (0.25in)^2)

Strain is the deformation of an object due to stress and is expressed by the following equation:

Epsolon= delt L/ Lsub0
Ε is the unitless measure of strain
Δl is the change in length
l0 is the initial length

Hooke’s law states that stress and strain are directly proportional, and equal to the modulus of elasticity:

E=sigma/ epsilon
E is the modulus of elasticity (psi)
σ is the stress
ε is the strain

The most common types of slings used in rigging are wire rope, chain, synthetic web, round, and metal mesh slings. The most common types of hitches are vertical, choker, basket, and 3- or 4-legged hitches. All slings come from the manufacturer stamped or labeled with a working load limit (WLL). This is not the breaking threshold of the equipment, but the maximum load the manufacturer has designed it to carry on a regular basis. When slings or sling legs of any kind are used at an angle other than vertical, their WLL decreases accordingly. This reduction is called the sling angle factor. For basket and multi-legged slings, the sling angle factor is calculated by taking the sine of the angle between the load and the sling:

Sling angle factor = sin(angle between load and sling leg)

To calculate the total WLL for sling configurations, multiply the sling angle factor by the number of legs in the sling and by the vertical WLL of one leg:
Adjusted WLL = Vertical WLL x number of legs x sling angle factor

For example, consider a 4-legged wire rope sling configured with 45-degree angles between the load and each leg.

It is important to note that for chain slings, a 4-legged sling’s adjusted WLL uses 3 legs in the equation. This is because of the difficulty in rigging a chain sling to distribute the load equally among four legs. Riggers often apply the same reduction to 4-legged slings of any material for increased safety factor.
If each leg of the sling has a vertical WLL of 5,000 lb, the reduced WLL of this sling would be calculated as:
Adjusted WLL = 5000lb x 3legs x sin(45degrees) = 10, 606lb

It may be helpful to write down the equations we went over in this video, and familiarize yourself with them.

I hope this review was helpful. For further ASP study help, be sure to check out more of our video’s here.

See ya next time!



by Mometrix Test Preparation | Last Updated: June 15, 2020