Statics

Let's say that you're using a cantilever beam to support a point load.

Cantilever beams are a great way to ensure static equilbrium: the fixed connection has the capacity to develop a moment reaction.

Unfortunately, cantilever beams are also prone to large amounts of deformation, a change in shape due to loading.

This beam is curving visibly under the load. Is that a problem? Well, it depends on the scenario. But for this example, let's say that we need to reduce the deformation.

Let's revise the design. We add another member (2) to help prop up the horizontal beam (1).

We have created a simple structural frame. It is comprised of two members: a horizontal beam element and an inclined brace or strut.

We no longer require a fixed connection at B for static equilibrium, so we simplify it to a common pinned connection.

Nodes A, B, and C each have a pin (the through-bolt, shown as a circle), but only B and C are pinned (prevented from translating).

Member 1 will still curve under the load, but much less than the cantilever beam we started with.

To solve multi-member structures, we often use exploded FBDs. 

Member (1) has had the bolts removed at A and B. The forces on that FBD are the actions of the missing components on the body.

Member (2) has had bolts removed at A and C, and replaced with their effects on the body.

Take care to apply Newton's Third Law (N3L) between adjacent FBDs as shown.

You may have noticed that the forces at B are expressed in terms of components while FAC  is expressed as a resultant force.

That was intentional; you'll learn why as this lesson progresses.

A two-force member (2FM) is defined as one that:

• has pins at both ends (can be internal pins, pinned connections, or rollers)

• does not have any internal or intermediate connections to other members

• does not have any applied forces or moments along the member's length.

You can also call a 2FM an axially-loaded member. 

They are loaded in either axial tension or in axial compression.

The longitudinal axis runs in the long direction of the member.

Let's determine whether or not each member is a 2FM in this unusual multi-member structure.

• Member 1 is not a 2FM due to the internal pin at B.

• Member 2 is not a 2FM due to the internal pin at D.

• Member 3 is a 2FM.

• Member 4 is not a 2FM due to load P1.

• Member 5 is not a 2FM due to load M1.

• Member 6 is not a 2FM due to pin K and load P2.

• Member 7 is a 2FM.

• Member 8 is not a 2FM, because of the fixed support at P.

Two-force members are easy to analyze, because the x-direction and y-direction components of the axial force are coupled. This occurs because the line of action of the resultant force is known to be the longitudinal axis of the member.

For notation, it's common to use F to represent force in a 2FM.

The x- and y-direction force components are similar to the x-and y-direction length components of the member's geometry. Use those ratios to solve for unknowns.

Let's say that you know force component Fx  and want to solve component Fy . All you have to do is divide Fx  by Lx , and then multiply that by Ly .

When you need to compute the axial force (the tension or compression along the longitudinal axis), you can solve it with ratios or with the Pythagorean Theorem.

You can illustrate the axial force by cutting through the member as shown. 

Before this lesson, we only knew how to cut through a cable and expose the internal force.

Now we know that we can cut through a 2FM and expose the internal tension or the internal compression.

WARNING! You're not ready to cut through anything else at this time! Once we reach Lesson 16 Internal Forces and Moments we'll take off the training wheels! For now, only cut through cables and 2FM.

Let's analyze a two-member structure that is comprised of a horizontal beam and an inclined cable.

Recall from Lesson 09 - Cables that it is possible to use several different FBDs on a given Statics problem:

• You can make a FBD of the entire structure, and use the equations of equilibrium to solve unknowns.

• You can also break the structure into pieces; each piece must also be in static equilibrium.

What does this mean? It means that the big learning outcome in this unit is to learn how to be strategic with the FBD you select for analysis.

We can create a FBD of the entire structure by disconnecting the pin at C (and replacing it with x- and y- component forces) and disconnecting the pin at B (and replacing it with x- and y- component forces).

Since we did not disconnect the pin at A to create this FBD, we do not include any forces at node A in this FBD.

This FBD has 4 unknowns (Cx, Cy, Bx, and By). With only 3 E.o.E. in effect (force summation x-direction, force summation y-direction, and moment summation about a z-axis), it's not solvable as shown...

... until you recognize that Cx and Cy are coupled because Member 1 is a two-force member.

Even so, the solution would be a little tedious if you chose this FBD.

We could also isolate a FBD of the cable by disconnecting the pins at A and C and replacing them with their effects on the body.

There are four unknowns, but a pattern we have seen before: equal and opposite vertical forces and equal and opposite horizontal forces. The couple moment correlated to Ay and Cy must be equal and opposite to the couple moment correlated by Ax and Cx.

Unfortunately, this FBD is not solveable, because it doesn't have any numeric values.

This leads us to a key takeaway: in order to solve numeric answers, we need a numeric load in the FBD we choose.

We could also isolate the horizontal member by cutting through Cable 1 and disconnecting the pin at B.

This FBD is the big winner. We have three unknowns (Bx, By, and T1), plus a numeric value (300#).

To solve reactions at B, simply:

1. Sum moments about A to solve By = 150# ↑

2. Sum forces in the y-direction to solve T1 sin 40° = 300-150 = 150#↑; so T1 = 233.4# of tension.

3. Solve T1 cos 40 = 178.8# which makes Bx = 178.8# ←

Note that instead of cutting through the cable to create the FBD, you could also take out the pin at A. This FBD is also acceptable. 

Note that if you construct it this way you'll have to remember that Ax and Ay are coupled (due to the 2FM). With that substitution in effect, this FBD is solvable.