Forces in the Structures: Mechanical Engineering

“The Force can have a strong influence on the weak-minded”

- Ben Obi-wan Kenobi, explaining to Luke Skywalker how he made the famous “these aren’t the Droids you’re looking for” Jedi Mind Trick work.

Forces

Forces acting on objects are vectors that are characterized by not only a magnitude (e.g. Newtons or pounds force) but also a direction. A force vector F (vectors are usually noted by a boldface letter) can be broken down into its components in the x, y and z directions in whatever coordinate system you’ve drawn:

F = F_xi + F_yj + F_zk                                                     Equation 11

Where Fx, Fy and Fz are the magnitudes of the forces (units of force, e.g. Newtons or pounds force) in the x, y and z directions and i, j and k are the unit vectors in the x, y and z directions (i.e. vectors whose directions are aligned with the x, y and z coordinates and whose magnitudes are exactly 1 (no units)).

Forces can also be expressed in terms of the magnitude = (Fₓ2+ F_y2+ F_z2)1/2 and direction relative to the positive x-axis (= tan-1(F_y/F_x) in a 2-dimensional system). Note that the tan-1(F_y/F_x) function gives you an angle between +90° and -90° whereas sometimes the resulting force is between +90° and +180° or between -90° and -180°; in these cases you’ll have to examine the resulting force and add or subtract 180° from the force to get the right direction.

Degrees of Freedom

Imagine a one-dimensional (1D) world, i.e. where objects can move (translate) back and forth along a single line but in no other way. For this 1D world there is only one direction (call it the xdirection) that the object can move linearly and no way in which it can rotate, hence only one force balance equation is required. For the field of dynamics this equation would be Newton’s Second Law, namely that the sum of the forces F_x,1 + F_x,2 + F_x,3 + … + F_x,n = max where m is the mass of the object and ax is the acceleration of the object in the x direction, but this chapter focuses exclusively on statics, i.e. objects that are not accelerating, hence the force balance becomes simply

So a 1D world is quite simple, but what about a 2D world? Do we just need a second force balance equation for translation in the y direction (that is, ΣF_y = 0) and we’re done? Well, no. Let’s look at a counter-example (Figure 3). The set of forces on the object in the left panel satisfies the requirements ΣF_x = 0 and ΣF_y = 0 and would appear to be in static equilibrium. In the right panel, it is also true that ΣF_x = 0 and ΣF_y = 0, but clearly this object would not be stationary; instead it would be rotating clockwise. Why is this? In two dimensions, in addition to the translational degrees of freedom in the x and y directions, there is also one rotational degree of freedom, that is, the object can rotate about an axis perpendicular to the x-y plane, i.e., an axis in the z-direction. How do we ensure that the object is not rotating? We need to account for the moments of force (M) (discussed in the next sub-section) in addition to the forces themselves, and just as the forces in the x and y directions must add up to zero, i.e. ΣF_x = 0 and ΣF_y = 0, we need to have the moments of force add up to zero, i.e. ΣM = 0. So, to summarize, in order to have static equilibrium of an object, the sum of all the forces AND the moments of force must be zero. In other words, there are two ways that a 2-dimensional object can translate (in the x and y directions) and one way that in can rotate (with the axis of rotation perpendicular to the x-y plane.) So, there are 3 equations that must be satisfied in order to have equilibrium,

where the number of forces in the x direction is n, the number of forces in the y direction is m and p = n + m is the number of moments of force calculated with respect to some point A in the (x,y) plane. (The choice of location of point A is discussed below, but the bottom line is that any point yields the same result.

Two sets of forces on an object, both satisfying ΣF_x = 0 and ΣF_y = 0, but one (left) in static equilibrium, the other (right) not in static equilibrium.

This is all fine and well for a 2D (planar) situation, what about 3D? For 3D, there are 3 directions an object can move linearly (translate) and 3 axes about which it can rotate, thus we need 3 force balance equations (in the x, y and z directions) and 3 moment of force balance equations (one each about the x, y and z axes.) Table 1 summarizes these situations.

# of spatial dimensions	Maximum # of force balances	Minimum # of moment of force balances	Total # of unknown forces & moments
1	1	0	1
2	2	1	3
3	3	3	6

Table: Number of force and moment of force balance equations required for static equilibrium as a function of the dimensionality of the system. (But note that, as just described, moment of force balance equations can be substituted for force balance equations.)

Moments of Forces

Some types of structures can only exert forces along the line connecting the two ends of the structure, but cannot exert any force perpendicular to that line. These types of structures include ropes, ends with pins, and bearings. Other structural elements can also exert a force perpendicular to the line (Figure 4). This is called the moment of force (often shortened to just “moment”, but to avoid confusion with “moment” meaning a short period of time, we will use the full term “moment of force”) which is the same thing as torque. Usually the term torque is reserved for the forces on rotating, not stationary, shafts, but there is no real difference between a moment of force and a torque.

The distinguishing feature of the moment of force is that it depends not only on the vector force itself (F_i) but also the distance (d_i) from that line of force to a reference point A. (I like to call this distance the moment arm) from the anchor point at which it acts. If you want to loosen a stuck bolt, you want to apply whatever force your arm is capable of providing over the longest possible d_i. The line through the force F_i is called the line of action. The moment arm is the distance (d_i again) between the line of action and a line parallel to the line of action that passes through the anchor point. Then the moment of force (M_i) is defined as 

Force, line of action and moment of force (= Fd) about a point A. The example shown is a counterclockwise moment of force, i.e. force F_i is trying to rotate the line segment d counterclockwise about point A.

M_i = F_id_i                                                                Equation 14

where F_i is the magnitude of the vector F. Note that the units of M_o is force x length, e.g. ft lbf or N m. This is the same as the unit of energy, but the two have nothing in common – it’s just coincidence. So one could report a moment of force in units of Joules, but this is unacceptable practice – use N m, not J.

Note that it is necessary to assign a sign to M_i depending on whether the moment of force is trying to rotate the free body clockwise or counter-clockwise. Typically we will define a clockwise moment of force as positive and counterclockwise as negative, but one is free to choose the opposite definition – as long as you’re consistent within an analysis.

Note that the moment of forces must be zero regardless of the choice of the origin (i.e. not just at the center of mass). So one can take the origin to be wherever it is convenient (e.g. make the moment of one of the forces = 0.) Consider the very simple set of forces below:

Force diagram showing different ways of computing moments of force

Because of the symmetry, it is easy to see that this set of forces constitutes an equilibrium condition. When taking moments of force about point ‘B’ we have:

ΣF_x = +141.4 cos(45°) lbf + 0 - 141.4 cos(45°) lbf = 0

ΣF_y = +141.4 sin(45°) lbf - 200 lbf + 141.4 sin(45°) = 0

ΣM_B = -141.4 lbf * 0.707 ft - 200 lbf * 0 ft +141.4 lbf * 0.707 ft = 0.

But how do we know to take the moments of force about point B? We don’t. But notice that if we take the moments of force about point ‘A’ then the force balances remain the same and

ΣM_A = -141.4 lbf * 0 ft - 200 lbf * 1 ft + 141.4 lbf * 1.414 ft = 0.

The same applies if we take moments of force about point ‘C’, or a point along the line ABC, or even a point NOT along the line ABC. For example, taking moments of force about point ‘D’,

ΣM_D = -141.4 lbf * (0.707 ft + 0.707 ft) + 200 lbf * 0.5 ft +141.4 lbf * 0.707 ft = 0

The location about which to take the moments of force can be chosen to make the problem as simple as possible, e.g. to make some of the moments of forces = 0.

Example of “why didn’t the book just say that…?” The state of equilibrium merely requires that 3 constraint equations are required. There is nothing in particular that requires there must be 2 force and 1 moment of force constraint equations. So one could have 1 force and 2 moment of force constraint equations:

where the coordinate direction x can be chosen to be in any direction, and moments of force are taken about 2 separate points A and B. Or one could even have 3 moment of force equations:

Also, there is no reason to restrict the x and y coordinates to the horizontal and vertical directions. They can be (for example) parallel and perpendicular to an inclined surface if that appears in the problem. In fact, the x and y axes don’t even have to be perpendicular to each other, as long as they are not parallel to each other, in which case ΣF_x = 0 and ΣF_y = 0 would not be independent equations.

Read Next Part Of This Article Here: Types of Forces and Moments of Force