EN3: Introduction to Engineering and Statics

 

 

                      

 

   Division of Engineering

    Brown University

 

 

9. Internal forces and moments in slender members

 

There are two reasons why engineers need to calculate forces. 

1.      To design a machine that will apply a controlled force to something, e.g. a motor, or a machining process;

2.      To design components that can support forces without failure.

 

Failure occurs when internal forces in part of a structure get too large. To design against failure, we need not only to calculate external and reaction forces in a system, but we also need a way to calculate internal forces in parts of a structure.

 

A full study of internal forces is way beyond the scope of this course – it will be covered in EN31 and EN175.  But many design calculations can be made with only a few simple ideas.

 

We will focus here on what is known in the trade as slender members (no, not members of weight watchers…).  These are components that are long and thin, and have a more or less uniform cross section.  Examples include

1.      Ropes and cables

2.      Most structural members, e.g. I beams, girders, etc

3.      Pipes

4.      Screws, bolts

5.      Shafts, eg the camshaft or driveshaft in your car.

 

There are two general classes of slender members: (a) solid shafts, which are stiff, and resist bending, twisting, shearing, stretching and compression, and (b) ropes and cables, which are flexible, and resist only stretching. (There are a few weird cases like bicycle chains that can be bent in some directions but not others). 

 

We can deduce the general nature of internal forces in these solids by following the same reasoning we used to deduce the nature of reaction forces.

 

 

 

 

9.1 Internal forces in ropes and cables

 

Let’s think first about internal forces in a rope, chain or cable, which are infinitely simpler to deal with than general solid shafts.

 

Of course, the only thing you can do to a rope is to pull on it.  So visualize a rope, pulled at the two ends. Then, create an imaginary cut through some representative cross-section of the rope, and ask what kind of forces and moments are required to resist relative motion of the two adjacent pieces of the cable

 

 

We know that

(i)                  The adjacent parts of the cable can’t separate.  This requires a tension force T that acts to hold the adjacent faces of the cut together.

(ii)                The cable can bend, shear and twist freely.  Therefore, there can’t be any transverse forces, or any moments acting on the adjacent faces of the cut.

 

Because you can only stretch a cable (you can’t compress it), the tension force T must be positive T>0. (This is actually because the cable has no bending stiffness – it buckles under compression.  We’ll discuss buckling later.)

 

Thus, The internal force in a cable, rope or chain consists of a positive tension force T>0 that acts perpendicular to the surface of any imaginary cut perpendicular to the cable.

 

 

Notice that the tension force is a bit strange, because it appears to act in two directions at the same time… On section AB, it acts in the negative i direction; while on section BC it acts in the positive i direction.  This obeys the rules governing tension forces – they must always act perpendicular to the cut surface.  Internal forces don’t really behave like vectors, they are actually tensors.  Their direction depends on the orientation of the cut surface.

 

 

 

 

9.2 Example problems involving tension forces in ropes and cables

 

Solving problems involving cables is a breeze.  The objective is generally to calculate the tension in the cable, so you know how strong it needs to be.  Two tricks are helpful (i) when isolating part of the system to draw a FBD, you can always create a cut through any cable, and draw the tension force acting on the cut surface in the FBD, and (ii) a cable can be treated just like a 2 force member – you know the direction of the reaction force at the joints on each end of the cable.

 

Example 1:

 

Here’s a Mickey Mouse problem to illustrate the first point.  Mickey’s weight is W.  Find the tension in the cable

 

 

We cut Mickey loose as shown (hooray!)  and draw a free body diagram for the isolated system.  There’s no need to do a force-balance table – it’s obvious that T=W for forces to balance in the vertical direction. Force balance in the horizontal direction and moment balance are satisfied automatically.

 

 

 

 

 

 

 

 

 

Example 2

 

Here’s a more complex 3D example.  A weight W is suspended from three cables attached to three vertical poles with length L as shown.  Calculate the tension in each cable.

 

 

 

We can cut through each cable to isolate part of the system as shown. We draw a FBD showing tension forces acting on each cut surface. Note the convention used to denote tension forces – this notation is standard.

 

We now do a force balance for the isolated part of the system (there’s no need to do a moment balance – remember that if all the forces acting on a system pass through the same point they automatically exert no resultant moment.  Here, the forces all act through D).  Each tension force needs to be expressed as a vector.  To do this, we need to find unit vectors , , , pointing in the direction of each force.  The forces are then

 

You can find the unit vectors using elementary geometry, or if you prefer you can use the position vectors of the end points

 

 

 

We can do the force balance in a table

 

 

This give three equations for the three unknown tensions

These are readily solved (we set  in the Maple worksheet below)

so that

 

 

Notice that we computed the cable tensions in the preceding problem by cutting out and balancing forces on the junction between the three cables.  To find the tension in each of a complex network of cables, you will need to cut out each junction, and perform a force balance on each.

 

 

 

 

9.3 Cable and Pulley problems

 

Systems of cables and pulleys can be used to construct force amplifiers.  The goal is to devise a system where you can apply a huge force to something by applying only a modest force to a rope.  They are often used in cranes and ships.  They are also popular Fundamentals of Engineering examination questions…

 

Cable and pulley systems can be analyzed using the method described in the preceding sections. 

Two additional results help to simplify the calculations

 

1.      The tension force in a light rope or cable that passes over a frictionless pulley is constant

 

`Frictionless’ can either mean that the pulley has a frictionless bearing; or alternatively the contact between rope and pulley could be frictionless (unlikely in practice!).

`Light’ means that the weight of the cable is small compared with the tension force.

 

To see why this is the case we can do a force and moment balance for a simple rope/pulley combination

 

 

 

We cut the cables where they just come into contact with the pulley, isolate the pulley from its bearing and draw a free body diagram for the pulley and the segment of cable that contacts the pulley together.  Since the bearing at C is frictionless, it prevents motion but allows free rotation.  The reaction forces consistent with this constraint are shown.

 

 

 

 

 

 

 

A force balance table for the system is shown below

 

 

 

Moment balance shows that , as promised.

 

 

 

 

2. The tension force in a light cable that contacts a frictionless surface is constant

 

This statement can be proved by doing a force balance for an infinitesimal segment of the cable at some point inside the contact region.  We won’t work through the details here.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Example 1

 

We’re now in a position to solve some cable-pulley problems.  Here’s a simple Bugs Bunny problem.  Calculate the tension in the cable and hence deduce the force Bugs must apply to raise the carrot of weight W.  Friction, and the weight of all other components, may be neglected.

 

 

Cable-pulley problems are always solved by cutting through some of the cables, drawing free body diagrams showing forces acting on the cut cables, and then doing a force balance. We also make use of the fact that the tension in the cable is constant.  In this case it’s best to cut out the carrot…

 

 

We don’t need to do any fancy calculations to see that  …  Since the tension is constant in the cable, Bugs must pull on the rope with a force T=W/2.  The pulley system gives a mechanical advantage of 2: you (or Bugs) can lift the weight W with a force equal to only half the weight.

 

 

 

 

Example 2

 

More complex pulley systems can give a larger mechanical advantage

 

 

 

 

This time there are two separate cables, so we need to isolate two parts of the system to get enough equations to solve for the unknown cable tensions.  There are various choices.  One pair is shown below. 

 

When we draw the free body diagram, we assume that the tension is constant in each cable.  The forces in the two separate cables have been denoted by  and .  Balancing forces in the vertical direction for the two isolated systems gives two equations for the unknown tension forces

which are easily solved to see that .  This pulley design gives a mechanical advantage of 4.

 

 

9.4 Internal forces in straight 2-force members

 

Internal forces in stiff solid shafts are ugly.  We’ll put off discussing them as long as possible… Fortunately straight two-force members, which are a common structural component, are not much more difficult to deal with than a rope.

 

Consider a straight, solid bar; pinned at the top end, and subjected to a force W at the other end.  We’ve already seen that the reaction force at both ends of the member must act parallel to the bar, and must be equal and opposite.

Now, create an imaginary cut through any cross section of the bar as shown.  The bar is stiff, and so constrains both relative rotation and relative displacement of the material surfaces created by the cut.  This generally requires 3 moment components, and 3 force components, as discussed below.  However, it’s easy to see that for a two force member, only one nonzero force component is required to hold each part of the bar in place.  Thus

 

The internal force in a solid, straight two-force member consists of a tension force T acting parallel to the bar.

 

Unlike a cable, the tension force in a straight two-force member can be positive or negative.  By convention, a positive force is tensile, and acts perpendicular to any surface created by an imaginary cut perpendicular to the bar’s axis.

 

Once again, the tension force appears to act in two directions at the same time – in the picture, it acts downwards on the top part of the bar, and upwards on the bottom.

 

 

 

9.5 Calculating internal forces in structures made of two-force members

 

It’s possible to build very complicated 3D frames in which every member is a two-force member.  To do so, the structure must be designed so that that the members only have joints at their ends.  In addition, each joint must be designed to impose only forces (and no moments) on the bars.  Therefore, the joints need to all be pin joints in a 2D frame, or ball-and-socket type joints in a 3D frame.  Such structures are known as pin jointed frames, or pin jointed structures.

 

Not many real structures are made with ball-and-socket joints.  But it’s so easy to analyze forces in a pin jointed frame that we often assume that the joints allow relative rotation, just to be able to do design calculations quickly.

 

One can calculate tension or compression forces in the members of a pin jointed structure using exactly the same procedure we followed when analyzing cables. 

 

Example 1

 

It’s best to illustrate this using a simple example.  Calculate the internal forces in each member of the frame shown in the picture.

 

There’s a simple recipe to follow here

1.      Cut out each joint, and draw positive tension forces acting on the cut ends of the members, as well as any reaction or external forces;

2.      Balance forces for each joint (there’s no need to balance moments);

3.      Solve the equations resulting from step (2) for unknown member forces.

 

Let’s apply this to the example

 

Force balance for joint A
Reaction force
Member AD
Member AB
Sum (=0)

 

 

 

 

 

 

Force balance for joint B
External force
Member BC
Member BD
Member BA
Sum (=0)

 

Force balance for joint C
Reaction force
Member BC
Member CD
Sum (=0)

 

 

Force balance for joint D
Member BD
Member CD
Member AD
Sum (=0)

 

 

Collecting together the force balance equations

We have eight equations, and eight total unknowns (3 reaction components and 5 member tension).  The equations are readily solved

 

 

 

so the reactions and member forces are

 

A few short remarks on what’s been done in this example are in order

1.      Note that reaction forces at external joints have been treated exactly as in all our earlier joint force calculations.  The reaction forces act to prevent motion along constrained directions at the joint.

2.      We’ve followed the proper sign convention when drawing tension forces – tension forces always act away from a cut end of a bar.  Some members turned out to have negative tension forces in them – this tells us that they are in compression.  But we don’t try to guess which way the forces act when we draw the free body diagram; by convention, internal forces are always drawn in the positive sense.

3.      This procedure is straightforward, but a big pain.  But if you’re a structural engineer, it will pay the bills….

 

 

 

Example 2

 

Let’s try a 3D problem.  Conceptually, nothing changes, but of course there’s a lot more work involved to get a solution.

 

In the structure shown, joint A is pinned, joint B is on a roller that allows motion in the i direction only, and joint C is on a roller that allows motion in both i and j directions.  Calculate the forces in all the members, in terms of W and a.

 

 

 

We will need to work through joint-by-joint, drawing free body diagrams and writing down force balance equations in the i, j and k directions for each isolated joint.  The tension forces will need to be expressed as vectors.  For this purpose, it is helpful to compile a table of unit vectors pointing parallel to each member. In the table below, unit vectors are specified pointing from the first index to the second, e.g. in the first row the unit vector points from A to B.  When we need the unit vector from B to A, we can just change the sign of the one shown.

 

 

 

 

We can now set up the necessary force balance tables

Force balance for joint A
Reaction
Member AB
Member AC
Member AD
Sum (=0)

 

 

 

 

 

 

Force balance for joint B
Reaction
Member AB
Member BC
Member BD
Member BE
Sum (=0)

 

 

 

 

 

 

Force balance for joint C
Reaction
Member AC
Member BC
Member CD
Member CE
Sum (=0)

 

 

 

Force balance for joint D
Member AD
Member BD
Member CD
Member DE
Sum (=0)

 

 

 

 

 

 

 

Force balance for joint E
External load
Member CE
Member BE
Member DE
Sum (=0)

 

 

 

We now have all the governing equations.  They are listed below

We have 15 equations (3 for each joint), and 15 unknowns (6 reactions and 9 member tensions), so we can proceed to solve them.  The solution is shown below (but I am way too lazy to type up all the member forces…)

 

 

A few remarks on general procedure

1.      A table of unit vectors parallel to each member is very helpful for 3D problems

2.      Reaction forces are treated using the usual rules.  Forces (and no moments) act at any joints that are fixed.  The force components act so as to enforce the constraint associated with the joint.

3.      When we draw free body diagrams for the joints, we always follow the correct sign convention for tension forces: a tension force must always act perpendicular to a cut surface.

4.      Equilibrium equations in 3D problems will have i,j,k components

5.      We get huge numbers of equations to solve in 3D problems.  A computer is almost essential to solve them.  We will show you later how to program a magic EXCEL spreadsheet that analyzes 3D structures automatically.

6.      This procedure is still a pain.

 

 

9.6 Necessary conditions for being able to calculate forces in pin-jointed frames using force-balance:  Classifying structures as determinate, indeterminate or mechanisms.

 

In both the examples we just solved, we found exactly the right number of equation for the unknown member forces and unknown reaction forces.  Does this always happen?

 

Unfortunately, no.  In fact, force balance will only provide the correct number of equations in very special structures.  Here, we specify the conditions necessary for this to happen.  We also discuss what it means when there aren’t the same number of unknowns as equations.

 

 

2D structure	3D structure
M = 5 members	M = 9 members
R = 3 reaction force components	R = 6 reaction force components
J = 4 joints	J = 5 joints
No. unknowns = M+R = 8	No. unknowns = M + R = 15
No. equations = 2J=8	No. equations = 3J = 15

 

 

Number of unknowns and equations

 

Let’s first think about the number of unknowns in a generic pin jointed frame problem.

1.      There will be an unknown tension force for each member in the structure.

2.      There will be a whole bunch of unknown reaction force components (but no moment components).  The precise number of unknown reaction components will vary from one problem to another. –(Often 3 unknown reactions for a 2D structure, and 6 for a 3D structure, but this is not always the case).

 

As for equations, we will always find two equilibrium equations for each joint in a 2D structure, or 3 equilibrium equations at each joint for a 3D structure.

 

Maxwell’ s conditions.

 

We can therefore devise a simple technique to determine whether there will be the same number of equations as unknowns.

Let M denote the number of members in the structure

Let R denote the number of unknown reaction force components

Let J denote the number of joints in the structure.

 

In a 2D structure, we will have the same number of equations as unknowns if M+R=2J

 

In a 3D structure, we will have the same number of equations as unknowns if M+R=3J.

 

These are known as Maxwell’s conditions.

 

Classifying structures as statically determinate, statically indeterminate, or mechanisms.

 

 

A structure is said to be Statically determinate if all member forces and reaction forces can be calculated using equilibrium equations for each joint.  Two conditions must be satisfied for this to be the case:

1.      Maxwell’s condition must be satisfied, i.e. M+R=2J for 2D structures or M+R=3J for 3D structures.

2.      The equations must have a unique solution.  There’s no simple way to check for this – you either have to actually solve the equations, or use some advanced linear algebra techniques.  But in most sensible structures, the equilibrium equations can be solved if Maxwell’s condition is satisfied.

 

 

A structure is said to be statically indeterminate if the equilibrium equations can be satisfied, but the member forces or reaction forces cannot be determined uniquely.  Two conditions must be satisfied for this to be the case

1.      There must be more member forces and reaction components than equations: M+R>2J for 2D structures or M+R>3J for 3D structures.

2.      The equations must have more than one solution.  This is almost always the case if (1) is satisfied.

 

 

A structure is said to be a mechanism if it is impossible to satisfy the equilibrium equations at every joint.  Two conditions must be satisfied for this to be the case:

1.      There must be fewer member forces and reaction components than equilibrium equations: M+R<2J for 2D structures; or M+R<3J for 3D structures

2.      The external loading must be such that equilibrium is not satisfied at one or more joints.  This is almost always the case if (1) is satisfied.

 

 

Some 2D examples where Maxwell’s equation is sufficient to classify the structures are shown below

 

 

An example where Maxwell’s equation doesn’t work is shown below

Maxwell’s condition suggests that since M+R=2J, the structure is statically determinate.  But this structure satisfies Maxwell’s rule, but violates the second condition necessary to guarantee that the structure is statically determinate. It’s worth listing the governing equations to see what goes wrong.  The equation numbers below specify which joint and which equilibrium condition was used to derive it: for example (Ax) is the equilibrium equation in the x direction at joint A.  You might like to check the equations as a simple practice problem.

 

It’s clearly impossible to find a value for  that will satisfy both (Cx) and (Dx), so the structure must in part be a mechanism (equilibrium can’t be satisfied).    At the same time, the remaining equations can be satisfied by many different choices of the remaining variables (there are 7 remaining variables, and only 6 equations).  So another part of the structure must be statically indeterminate.  That’s why it fools poor professor Maxwell.

 

There’s no easy way to check for situations like this even in simple 2D structures, and it’s even more difficult in general 3D structures.  You can sometimes spot them using your physical intuition.  But the only method that is guaranteed to work is to go through the tedious process of assembling all the equilibrium equations, and then using matrix methods to examine their character.

 

 

 

Physical significance of indeterminate, or determinate, structures and mechanisms.

 

Why do we care whether a structure is determinate, indeterminate or a mechanism?  It turns out that the three cases behave in rather different ways.

 

The characteristic feature of a mechanism is that it will move when subject to forces (not just small deflections, but huge shape changes).  This, of course, is because if  then the structure must accelerate.

 

For example, the mechanism shown below can collapse as indicated, since the joints allow relative rotation of its members.

 

Some engineering systems are intentionally designed to be mechanisms, of course.  But for structural applications, it is essential to ensure that such motion cannot occur, otherwise the structure will be highly unsafe.

 

 

Statically indeterminate structures are not unsafe, but they do have some drawbacks.  One characteristic feature of an indeterminate structure is that it is hard to assemble.  For example, if the last member in the structure shown below were slightly too long, it wouldn’t fit in the space available for it.  Even without this member, the structure is determinate, and therefore doesn’t move easily.  So the only way you can fit the last member in is to apply big forces either to shorten the member itself, or to cause the structure to deform.

 

Other disadvantages of indeterminate structures are

1.      They can contain large, (and unknown) forces in the members even if the structure is not subject to external loads;

2.      They are hard to analyze.  It is possible to calculate forces in an indeterminate structure (we’ll develop a computer method for this purpose shortly), but the calculations are more difficult;

3.      Small motions at the constrained joints will cause large internal forces.  This would be undesirable in a building whose foundation would be expected to settle somewhat over time, for example.

4.      If part of the structure is heated, e.g. by sunlight, some members may expand slightly.  This will induce large forces in an indeterminate structure.

A statically determinate structure avoids all these drawbacks.  It has just enough members to hold it in place: no more, and no less.  It’s easy to analyze, assemble, and behaves in a predictable manner.  It’s not affected by small motions at constrained joints, and it’s insensitive to thermal expansion.

 

So how can you be sure that you have a determinate structure?  Maxwell’s rule is not fool-proof but it’s a good indication.  To recapitulate: for a structure with M members, R unknown reaction components and J joints:

1.      If M+R>2J for a 2D structure, or  M+R>3J for a 3D structure, it’s almost certainly statically indeterminate

2.      If M+R<2J for a 2D structure, or M+R<3J for a 3D structure, it’s almost certainly a mechanism

3.      If M+R=2J for a 2D structure, or M+R=3J for a 3D structure, there’s a good chance that it will be statically determinate, but you can’t be certain until you analyze it in detail.

 

 

 

9.7 Short-cuts for analyzing pin jointed frames

 

Zero Force Members

 

Sometimes, you can see by inspection that forces in some members of a structure must be zero. These members can then be removed from the calculation.

 

Here’s how to spot zero force members:

 

1. If only two members meet at a joint, and no external force acts on the joint, the forces in both members must be zero.  For example, this rule would tell us that the forces in members AD and CD of the structure must be zero.  We could remove these members before starting calculations.

 

 

2.  If (i) only three members meet at a joint, and (ii) two of the members are parallel, and (iii) no external force acts on the joint, the force in the non parallel member must be zero.  This rule would tell us that BD in the structure below is a zero force member.

So why the heck would anyone design a structure with zero force members in it?  To torture engineering students, mostly.  But there are other reasons too.

 

One reason is that most structures are designed to withstand many different kinds of force (gravity, wind loading, etc).  You might be analyzing only the effect of one force (weight, perhaps).  The weight may not induce forces in some members, but these members could be crucial in providing strength against wind loading.

 

A more subtle reason is that zero force members are often added to prevent the structure from collapsing by buckling.  We will discuss failure mechanisms in structural members in the next section. We will see that if the compression force in a member exceeds a critical value, the member will buckle.  The critical force depends on the length of the member – a short member is much stronger than a long one.  In fact, halving the length of a member will increase its buckling resistance by a factor of four.

 

For example, the structure shown below will almost certainly fail by buckling in member BC.

You could increase the strength of the structure by a factor of four by adding the zero-force member AD shown below

 

 

 

The method of sections

 

The method of sections is a fantastic short cut for calculating forces in a few members of a structure.  It is valuable for checking computer calculations, for example.

 

The idea is very simple – we can calculate forces in members by creating an imaginary cut through the structure, drawing the forces acting on the ends of the cut members, and then conducting a force and moment balance for one of the two parts of the structure.

 

The only minor drawback of this trick is that you have to compute the reaction forces acting on the structure first (by conducting a force and moment balance for the whole thing.

 

Let’s work through an example to show how this works.

 

Example 1: Calculate the internal forces in members CD and CG of the structure shown below.

We’ll use the method of sections to do this.  The first step is to calculate reaction forces at A and E, using the standard procedure.  Here’s a free body diagram for the full structure (we followed the usual rules to draw the reaction forces at the two joints)

 

Here’s a force and moment balance table

 

It’s easy to solve the equilibrium equations to see that

 

Now, we section the structure, by creating an imaginary cut through several members.  The cut must split the structure into two completely separate parts.  It’s a good idea to cut as few members as possible.  It’s also a good idea to cut through members that we are interested in.  The choice shown below is good.

 

Finally, we perform a force and moment balance for either of the two parts of the structure.  It won’t make any difference which half you use – but it makes sense to choose the simplest one.  In this case it’s best to use the portion DEFG on the right.

 

Here’s the force balance table (note that member CG is at 45 degrees to the vertical, from the geometry of the structure).  The member tensions are taken to act at the joints when writing down position vectors (remember that we can always translate a force parallel to itself without changing its moment)

 

 

 

 

 

 

 

This leaves us with three equations for three unknown member forces (remember that we already calculated  )

These can easily be solved to see that

Presto!  We have the member forces we were asked for.  Using the earlier method, we would have had to set up 20 equation in 20 unknown forces, and then solved them.

 

 

 

9.8 Internal forces in a general solid shaft

 

Let’s now work out the forces that act inside a stiff shaft.  Assume that the ends of the shaft are loaded in some way (eg through some kind of joint).  Then, create an imaginary cut through some representative cross-section, and ask what kind of forces and moments are required to resist relative motion of the two neighboring segments of the shaft.

 

 

 

We know that:

 

(i)                  The shaft resists extension (relative motion of the adjacent faces in the i direction) – this requires a force  on the surface shown (and an equal and opposite force on the adjacent face);

(ii)                The shaft resists shearing (relative motion in the j and k directions) which requires forces  on the surface shown;

(iii)               The shaft resists twisting (relative rotation about the i axis).  This requires a moment  on the surface shown;

(iv)              The shaft resists bending (relative rotation about the j and k axes).  This requires moments  on the surface shown.

 

Thus, in general, any internal surface inside a stiff solid object is subject to three components of force and three components of moment. (These are actually statically equivalent resultants of pressures and shear tractions acting on the internal surfaces).

 

So, internal forces in a stiff member are just like the reactions associated with a clamped joint.

 

Forces in solid shafts and beams will be discussed in more detail in EN31.