Chapter 6 Trusses and frames

A different way to discover a truss is by means of subtraction. Imagine your first initial design for a bridge is to make it from one huge chunk of solid steel. This would be wildly heavy and expensive. So you could cut holes out of the chunk here and there, greatly diminishing the weight and amount of material used, but not much reducing the strength. Between these holes you would see other heavy regions of metal from which you might cut more holes leading to a more savings of weight at not much cost in strength. In fact, the reduced weight in the middle decreases the load on the outer parts of the structure possibly making the whole structure stronger rather than weaker^††margin: Advanced aside: There are three ways that having more material in a structure can make it weaker: 1) the extra material adds to the gravitational load, as for the imagined bridge here, 2) Added material can be wedged in, causing the structure to fight itself (so called locked-in or internal stresses), and 3) material in the wrong place can cause stress concentrations and thus weak spots. . Eventually you would find yourself with very holey swiss cheese, a structure that looks something like a collection of bars attached from end to end in vaguely triangular patterns; like a microscopic picture of spongy bone (fig. 6.2):

As opposed to a solid block, a truss

• •

  Uses less material;

• •

  Puts less gravity load on other parts of the structure;

• •

  Leaves space for other things of interest (e.g., cars, cables, wires, people).

Real trusses are usually not made by removing material from a solid but by joining bars of steel, wood, or bamboo with welds, bolts, rivets, nails, screws, glue, or lashings. Once you are aware, you will notice trusses making up bridges, radio towers, and large-scale construction equipment. Early airplanes were flying trusses (fig. 6.3).

Bamboo trusses have been used as scaffoldings for millennia. Birds have had bones whose internal structure is truss-like since they were dinosaurs.

Trusses are practical, sturdy, and light structures.

But trusses are also prominent near the front of elementary mechanics books because

• •

  They are perhaps the easiest example of a complex mechanical system that a student can analyze;

• •

  They illustrate a variety of more general structural mechanics issues;

• •

  They help build intuition about structures that are not really trusses; the engineering mind can see an underlying conceptual truss where there is no physical truss.

What is a truss?

The sturdiness of most trusses comes from the inextensibility of the bars, not the resistance to rotation at the joints (as in the sticks and rubber-band examples at the start of this section). To make the analysis simpler, the relatively small resistance to rotation in the joints (even welded joints) is totally neglected in truss analysis. Thus the interaction of the bars with their neighbors is by forces, with no couples; each bar has one net force acting on each end. So:

An ideal truss is an assembly of two-force members.

Or, if you like, an ideal truss is a collection of bars connected at their ends with frictionless pins. Loads are only applied at the pins. In engineering analysis, the word ‘truss’ refers to an ideal truss even though the object of interest might have, say, welded joint connections.

Had we assumed the presence of welding equipment in your study room, the opening paragraph of this section would have described the welding of metal bars instead of the attachment of pencils with rubber bands. Even with welded-together steel you would have found that the triangles would be much more rigid than the V or square.

An ideal truss is a collection of bars connected at frictionless joints at which are applied loads as shown in fig. 6.5b (the load at a joint can be $\stackrel{\rightharpoonup }{\mathrm{𝟎}}$ and thus not show on either the sketch of the truss or the free-body diagram of the truss). A truss is held in place with supports which are idealized in 2D as either being fixed pins (as for joint E in fig. 6.5a) or as a pin on a roller (as for joint G in fig. 6.5a). Reaction forces, the forces on the truss at the supports, show on a FBD of the whole truss (fig. 6.5b) and also on a FBD of any joint at a support. Each bar is a two-force body (fig. 6.5c), with the same magnitude of tension pulling away from each end. A joint can be cut free with a conceptual chain saw, fooling each bar stub with the bar tension, as in the free-body diagram of a joint in fig. 6.5d.

The bar tensions can be negative. A bar with a tension of, say, $T=-5000\text{N}$ is said to be in compression. A tension of $-5000\text{N}$ is a compression of $5000\text{N}$.

In elementary truss analysis you are given the design of a truss and the loads applied to it. Your goal is to ‘solve the truss’ which means you are to find the reaction forces and the tensions in the bars (sometimes called the ‘bar forces’). As an engineer, this allows you to determine the needed strengths for the bars.

Trusses are always analyzed by the same basic method used in all of mechanics, the ‘method of free-body diagrams’.

• •

  Free-body diagrams are drawn of the whole truss and of various parts of the truss.

• •

  The equilibrium equations are applied to each free-body diagram, and

• •

  The resulting equations are solved for the unknown bar forces and reactions.

The ‘method of free-body diagrams’ is classically subdivided into two sub-methods.

• •

  In the method of joints you draw free-body diagrams of every joint and apply the force-balance equations to each free-body diagram.

• •

  In the method of sections you draw a free-body diagrams of one or more parts of the structure, each of which includes 2 or more joints and apply force and moment balance to the part or parts.

These two methods can be used separately or in conjunction.

In the rest of this chapter we cover the method of joints, the method of sections, computer solution using the method of joints, and miscellaneous advanced truss topics.

The elementary truss analysis you are about to learn is straightforward and fun. You will learn it without difficulty. However, the analysis of trusses at a more advanced level is mysteriously deep and has occupied great minds from the mid-nineteenth century (e.g., Maxwell and Cauchy) to the present (see, e.g., box 6.5 on page 6.5).

Let’s start with an example.

Consider this planar model of the arm of a construction derrick (see fig.6.7). Assume $F$ and $d$ are known. This truss has joints A-S (skipping ‘F’ to avoid confusion with the load). As is common in truss analysis, we totally neglect the force of gravity on the truss elements

.

The goal is to find the tensions in the bars (the so-called ‘bar forces’).

The method of joints is a subset of the more general method of free-body diagrams. Free-body diagrams are drawn of the joints. Here is the method-of-joints recipe:

• •

  Draw a free-body diagram of the whole structure and write 3 independent equilibrium equations (6 in 3D) and solve for unknown reactions if you can. This step is technically superfluous, but is so-often a time-saver that it’s best to just do it.

• •

  Draw free-body diagrams of all $n$ joints, 18 such in the example above.

• •

  For each joint free-body diagram you write the force-balance equations, each of which can be broken down into 2 scalar equations (3 in 3D).

• •

  Solve the $2n$ joint equations ($3n$ in 3D) for the unknown bar forces and reactions. In the example above this is $18\ast 2=36$ equations for 33 unknown tensions and 3 unknown reactions (which you may have found from the FBD of the whole structure, but need not have).

Solving 36 simultaneous equations is generally only feasible with a computer, which is one way to go about things. However, for simple triangulated structures, like the one in fig. 6.7, you can find a sequence of joints for which hand solution is easy. If you solve the equilibrium equations as you go, there are at most two unknown bar forces at each joint. By this means, the joint force-balance equations can be solved, even for some complex structures, without computers.

From the free-body diagram of the whole structure (fig. 6.7) we find that

Note, we picked a sign convention for the graphical representation of forces on the Free-Body Diagram (see pages 1.1 and 2.64) and let the algebra possibly generate negative numbers: at S the support pushes on the arm with a force of $-8F_{Ay}$ which is pulling (if $F_{Ay}>0$)

Note that for tension the order of subscripts is not meaningful. The tension $T_{BC}$ is the same scalar as the tension $T_{CB}$. $T_{BC}=T_{CB}$ is the amount of pulling on joint B and also the amount of pulling on joint C. That the two force vectors are negatives of each other is accounted for by the definition of tension as pulling. This unimportance of the order of subscripts is in contrast with the case of position vectors where ${\stackrel{\rightharpoonup }{𝒓}}_{\mathrm{BC}}$ is the position vector from B to C (also called ${\stackrel{\rightharpoonup }{𝒓}}_{\mathrm{C}/\mathrm{B}}$). For position vectors ${\stackrel{\rightharpoonup }{𝒓}}_{\mathrm{BC}}={\stackrel{\rightharpoonup }{𝒓}}_{\mathrm{C}/\mathrm{B}}=-{\stackrel{\rightharpoonup }{𝒓}}_{\mathrm{B}/\mathrm{C}}=-{\stackrel{\rightharpoonup }{𝒓}}_{\mathrm{CB}}$. Summarizing, the subscript order has meaning for ${\stackrel{\rightharpoonup }{𝒓}}_{\mathrm{AB}}$ but not for $T_{AB}$.

In the solve-by-hand method of joints, we first find a joint with at most 2 bars connected. Then we work our way through the structure, one joint at a time, picking joints with at most 2 unknown bar tensions. For each joint we will use

Typically many bars in a truss are parallel to the $x$ or $y$ axis so we often fall into the routine of immediately reducing the above vector equilibrium equation to the component equations

For the truss in fig. 6.7

• •

  Joint B has only two bars connected (see fig. 6.8). Force balance using FBD 6.8 tells us at a glance that

  $$\sum F_x=0\Rightarrow T_{\mathrm{DB}}=0\text{and}\sum F_y=0\Rightarrow T_{\mathrm{AB}}=0$$

• •

  Now you can draw a free-body diagram of joint A where there are only two unknown tensions (since we just found $T_{\mathrm{AB}}$), namely $T_{\mathrm{AD}}$ and $T_{\mathrm{AC}}$. Force balance gives two scalar equations

  	${\displaystyle \sum F_x}=0$	$\Rightarrow$	$F_{Ax}-T_{AC}-\sqrt{2}{T}_{AD}/2=0$
  	${\displaystyle \sum F_y}=0$	$\Rightarrow$	$-F_{Ay}+\underset{0}{\underbrace{T_{AB}}}+\sqrt{2}{T}_{AD}/2=0$

  which you can solve to find $T_{AD}=\sqrt{2}{F}_{Ay}$ and $T_{AC}=F_{Ax}-F_{Ay}$.

• •

  Next is joint C. Force balance for joint C will tell you $T_{\mathrm{CD}}$ and $T_{\mathrm{CE}}$.

• •

  Then you can work your way through the alphabet of joints. Using the bar tensions you have already found you can find, one at a time, joints with only two unknown tensions.

That’s it for the method of joints for simple structures.

Just by looking at joint B and thinking about the free-body diagram you could probably pick out that bars DB and AB must be zero-force members.

Here we explain the unnecessary but useful trick of recognizing such zero-force members even before systematically using the method of joints. Zero-force members are bars with $T=0$, like bars AB, BD and CD in the truss of Fig. 6.7. The basic idea is this:

If there is any direction for which only one bar contributes a force on a joint, then that bar is a zero-force member.

In particular:

• •

  At any joint where

  • –

    there are no loads, and

  • –

    where there are only two unknown non-parallel bar forces, and

  • –

    where all known bar-tensions are zero,

  then the two new bar tensions are both zero (e.g., joint B in fig. 6.8).

• •

  At any joint where all bars but one are in the same direction as the applied load (if any), the one bar is a zero-force member (see joints C, G, H, K, L, O, and P in fig. 6.7).

In the truss of fig. 6.7 bars AB, BD, CD, EG, IH, JK, ML, NO, and PQ are all zero-force members. Sometimes it is useful to keep track of the zero-force members by marking them with a zero (see fig. 6.9).

Most elementary texts, like this one, start with structures that yield easily to the method of joints. These are structures where you can totally solve the equilibrium equations for the joints, one joint at a time; each new joint introduces two scalar equations and two unknown bar-tensions.

For more complex trusses, this straightforward approach can fail a few ways:

• •

  Some structures are not designed in a straightforward triangulated manner and cannot be solved 2 equations at a time. Although the method of joints may still yield a solution, it may require simultaneous solution of all of the equilibrium equations^††margin: In the language of linear algebra: simple structures yield equilibrium equations that are naturally in upper triangular form. More complicated structures do not yield an upper-triangular form. .

• •

  Many structures cannot be solved (that is, the bar tensions can’t be found) by using the laws of statics alone. Such are appropriately called ‘statically indeterminate’ (e.g., see page 7.1).

For this first truss section we only consider structures that are statically determinate and easily solved. See sec. 6.5 for a detailed discussion of static determinacy.

Trusses can carry big loads with little use of material and can look nice (See fig. 6.11). They are used in many structures. Why don’t engineers use trusses for all structural designs? Here are some reasons to consider not using a truss:

• •

  Trusses are relatively difficult to build, involving many small parts^††margin: If an Indian says to you ”Go and count the rivets in the Howrah bridge.” she means go away and do something that will take a very long time. The bridge has many rivets (and bars and joints). and thus requiring much time and effort to assemble.

• •

  Trusses can be sensitive to damage when forces are not applied at the anticipated joints. They are especially sensitive to loads on the middle of the bars.

• •

  Trusses inevitably depend on the tension strength in some bars. Some common building materials (e.g., concrete, stone, and clay) crack easily when pulled.

• •

  Trusses often have little or no redundancy, so failure in one part can lead to total structural failure.

• •

  The triangulation that trusses require can use space that is needed for other purposes (e.g., doorways, rooms).

• •

  Trusses tend to be stiff, and sometimes more flexibility is desirable (e.g., diving boards, car suspensions).

• •

  In some places, some people consider trusses unaesthetic. (e.g., the Washington Monument is not supposed to look like the Eiffel Tower).

Nonetheless, for situations where you want a stiff, light structure that can carry known loads at pre-defined points, a truss is often the best design choice.

After you have mastered the elementary 2D truss analysis of the previous section, you might wonder

• •

  Do the ideas generalize to 3D? Yes, with only minor elaboration.

• •

  Does at least one of the methods presented always work? Yes, if you just look at the homework problems for elementary truss analysis. And yes again for many practical structures. But some trusses cannot be analyzed by the simple methods. In this section we classify trusses into types. One type, statically determinate trusses, can be analyzed by simple statics methods; other trusses require study of deformations as well as statics.

The concepts for 3D trusses are basically the same as for 2D trusses with these differences;

• •

  In the method of joints, each joint is associated with 3 scalar force-balance equations instead of 2;

• •

  In the method of sections, and in the free-body diagram of the whole structure one has 6 scalar equations instead of 3;

• •

  To hold the structure in place takes at least 6 reaction components instead of 3;

• •

  The rule-of-thumb check for static determinacy of a grounded structure in 3D is $b+r=3j$ instead of the 2D relation $b+r=2j$ (See sec. 6.5 on 6.5 for discussion of these formulas);

• •

  The rule of thumb for rigidity for a floating truss in 3D is $b+6=3j$ instead of the 2D relation $b+3=2j$.

(Recall: $b$= number of bars; $r$ = number of reaction components; & $j$ = number of joints.) There are various ways to think about the number six in the counts above. Assuming the structure is more than a point, six is the number of ways a rigid structure can move in three-dimensional space (three translations and three rotations), six is the number of equilibrium equations for the whole structure (one 3D vector moment, and one 3D vector force), and six is the number of constraints needed to hold a structure in place.

A tripod is the simplest rigid 3D structure. With four joints ($j=4$), three bars ($b=3$), and nine unknown reaction components ($r=3\times 3=9$), it exactly satisfies the equation $3j=b+r$, a check for determinacy of rigidity of 3D structures.

A tripod is the 3D equivalent of the two-bar truss shown in fig. 6.111a on page 6.111.

The simplest rigid floating structure in 3D is a tetrahedron. With four joints ($j=4$) and six bars ($b=6$) it exactly satisfies the equation $3j=b+6$ which is a check for determinacy of rigidity of floating 3D structures.

A tetrahedron is thus, in some sense, the 3D equivalent of a triangle in 2D.

Any closed polyhedron, with each face a triangle of rods, is a rigid structure. This includes a tetrahedron (above), an octahedron, a cube with a diagonal on each face, an icosahedron, and Buckminster Fuller’s geodesic domes.

Well, so Cauchy thought. It turns out that there are some strange non-convex polyhedra that are not rigid. But, for practical purposes, if you see triangles all around the outside of a structure you can assume it’s rigid.

The truss shown in the figure carries a load $F=10\mathrm{kN}$ at joint D. The truss is designed with nine rods, six of which (the inclined ones) have the same length $d=2\mathrm{m}$. Rods BC, EC, DE and BD form a square.

1. 1.

   Find the support reactions at joints A and F.

2. 2.

   Find the tensions in rods BD and BC.

1. 1.

   Support reactions: To find the support reactions at A and F, we draw the free-body diagram of the entire truss (see fig. 6.15). We are given that $d=2\mathrm{m}$ and that $\mathrm{\angle }\mathrm{ABD}=\mathrm{\angle }\mathrm{DEF}=\pi /2$. Therefore, $\mathrm{\ell }=\sqrt{2}d=2\sqrt{2}\mathrm{m}$.

   The scalar force-balance equation in $x$-direction readily gives $R_{A_x}=0$. The scalar moment-balance equation about point A gives

   $$2\mathrm{\ell }{R}_F-\mathrm{\ell }F=0\Rightarrow R_F=\frac{F}{2}=5\mathrm{kN}.$$

   Now, from the scalar force balance in the $y$-direction, we have

   $$R_{A_y}+R_F-F=0\Rightarrow R_{A_y}=F-R_F=5\mathrm{kN}.$$

   Answer: $R_{A_x}=0,R_{A_y}=5\mathrm{kN},R_F=5\mathrm{kN}$

2. 2.

   Tensions in BD and BC: We can find the tensions in rods BC and BD by analyzing the equilibrium of joint B. As you can see, joint B has three unknown forces acting on it, namely the tensions of rods AB, BC and BD. Since the joint equilibrium equations (only two scalar equations) can only solve for two unknowns, we need to start at joint A, determine $T_{AB}$ first and then move on to joint B.

   The free-body diagrams of the joints A and B are shown in fig. 6.16. Let us first consider the equilibrium of joint A. From the scalar force-balance equations, we have

   	${\displaystyle \sum F_y}=0$	$\Rightarrow$	$R_{A_y}+T_{AB}\sin \theta =0$
   		$\Rightarrow$	$T_{AB}=-R_{A_y}/\sin \theta =-5\mathrm{kN}/(1/\sqrt{2})=-7\mathrm{kN}.$
   	${\displaystyle \sum F_x}=0$	$\Rightarrow$	$T_{AB}\cos \theta +T_{AD}=0$
   		$\Rightarrow$	$T_{AD}=-T_{AB}\cos \theta =-7\mathrm{kN}(1/\sqrt{2})=5\mathrm{kN}.$

   Filename:sfig8-4-1a

   Now, we analyze joint B. From the geometry of forces, it is clear that writing scalar force-balance equations in the $x^{\prime }$ and $y^{\prime }$ directions will be advantageous. For example, the force balance in the $x^{\prime }$ direction immediately gives $T_{BD}=0$. The force balance in the $y^{\prime }$ direction gives

   $$-T_{AB}+T_{BC}=0\Rightarrow T_{BC}=T_{AB}=-7\mathrm{kN}.$$

   Answer: $T_{BC}=-7\mathrm{kN},T_{BD}=0$

   Note that it is easy to spot bar BD as a zero-force member since it is perpendicular to rods AB and BC.

For the truss tower shown in the figure, assume all horizontal and vertical rods to be 1$\mathrm{m}$long and rods numbered 16 and 18 to be 0.5$\mathrm{m}$long. Given that the horizontal load on the truss $F=500\text{N}$, find the tension in rod 15.

The free-body diagrams of the three joints are shown in fig. 6.18. Let us first consider the equilibrium of joint K. A simple inspection (or force balance in the $y$-direction) shows that bar 18 is a zero-force member. The force balance in the horizontal direction then immediately gives $T_{19}=F=500\text{N}$. Thus,

Next, we consider the equilibrium of joint I. Since $T_{19}$ is already known, there are only two unknown forces, $T_{14}$ and $T_{17}$ at this joint. The force balance in the horizontal direction gives

Now we proceed to joint J. Note that we used only one scalar equation (force balance in the $x$-direction) at joint I, since we do not need $T_{14}$. Similarly, to find $T_{15}$, we only need the force balance in the horizontal direction at joint J:

Note: We did not have to find support reactions first in order to proceed to other joints as in the previous sample. As long as you can find a sequence of joints with just two unknown forces at each joint, up to the force that you need to determine, you can easily find the force with hand calculations.

The truss shown in the figure is made up of five horizontal and six inclined rods. All inclined rods are 1$\mathrm{m}$ long and at right angles to each other. The truss carries two vertical loads, $F_1=4\mathrm{kN}$ and $F_2=1\mathrm{kN}$ as shown. Find the tensions in rods CE, DE, and DF.

To find tensions in rods CE, DE and DF, we can either use joints C and D, or joints E and F. However, for either set we need to start from other joints since there are more than two unknown forces at each joint. Let us start from joint G and work our way through joints F and E. To start at joint G, however, we first need to determine the support reaction G.

The free-body diagram of the entire truss is shown in fig. 6.20 where we have numbered the rods for convenience. The scalar moment-balance equation about point A in the $z$-direction gives

The force-balance equations give

Now, we are ready to proceed from joint G. The free-body diagrams of joints G, F, and E are shown in fig. 6.21.

The truss shown in the figure has four horizontal bays, each of length 1 $\mathrm{m}$. The top bars make ${20}^{\circ }$ angle with the horizontal. The truss carries two loads of 40  kN and 20  kN as shown. Find the forces in each bar. In particular, find the bars that carry the maximum tensile and compressive forces.

So, we go joint by joint, draw the free-body diagram of each joint and write the equilibrium equations. After we get all the equations, we can solve them on a computer. All joint equations are just force equilibrium equations, i.e., ${\displaystyle \sum \stackrel{\rightharpoonup }{𝐅}}=\stackrel{\rightharpoonup }{\mathrm{𝟎}}.$

• •

  Joint A:

  $$(A_x+T_1+T_{10}\cos {\alpha }_1)\hat{\mathit{ı}}+(A_y+T_{11}+T_{10}\sin {\alpha }_1)\hat{\mathit{ȷ}}=\stackrel{\rightharpoonup }{\mathrm{𝟎}}.$$

  (6.1)

• •

  Joint B:

  $$(-T_1+T_2+T_8\cos {\alpha }_2)\hat{\mathit{ı}}+(T_9+T_8\sin {\alpha }_2)\hat{\mathit{ȷ}}=\stackrel{\rightharpoonup }{\mathrm{𝟎}}.$$

  (6.2)

• •

  Joint C:

  $$(-T_2+T_3+T_6\cos {\alpha }_3)\hat{\mathit{ı}}+(T_7+T_6\sin {\alpha }_3)\hat{\mathit{ȷ}}=F_1\hat{\mathit{ȷ}}.$$

  (6.3)

• •

  Joint D:

  $$(T_4-T_3)\hat{\mathit{ı}}+T_5\hat{\mathit{ȷ}}=\stackrel{\rightharpoonup }{\mathrm{𝟎}}.$$

  (6.4)

• •

  Joint E:

  $$(-T_4-T_{15}\cos \theta )\hat{\mathit{ı}}+T_{15}\sin \theta \hat{\mathit{ȷ}}=F_2\hat{\mathit{ȷ}}.$$

  (6.5)

• •

  Joint F:

  $$(-T_6\cos {\alpha }_3+(T_{15}-T_{14})\cos \theta )\hat{\mathit{ı}}+(-T_6\sin {\alpha }_3+(T_{14}-T_{15})\sin \theta -T_5)\hat{\mathit{ȷ}}=\stackrel{\rightharpoonup }{\mathrm{𝟎}}.$$

  (6.6)

• •

  Joint G:

  $$(-T_8\cos {\alpha }_2+(T_{14}-T_{13})\cos \theta )\hat{\mathit{ı}}+((T_{13}-T_{14})\sin \theta -T_8\sin {\alpha }_2-T_7)\hat{\mathit{ȷ}}=\stackrel{\rightharpoonup }{\mathrm{𝟎}}.$$

  (6.7)

• •

  Joint H:

  $$(-T_{10}\cos {\alpha }_1+(T_{13}-T_{12})\cos \theta )\hat{\mathit{ı}}+((T_{12}-T_{13})\sin \theta -T_{10}\sin {\alpha }_1-T_9)\hat{\mathit{ȷ}}=\stackrel{\rightharpoonup }{\mathrm{𝟎}}.$$

  (6.8)

• •

  Joint I:

  $$(I_x+T_{12}\cos \theta )\hat{\mathit{ı}}+(-T_{11}-T_{12}\sin \theta )\hat{\mathit{ȷ}}=\stackrel{\rightharpoonup }{\mathrm{𝟎}}.$$

  (6.9)

Dotting each equation from (6.1) to (6.9) with ˆ ı and ˆ ȷ , we get the required 18 equations. We need to define all the angles that appear in these equations (${\alpha }_1,{\alpha }_2,{\alpha }_3$, and $\theta$) before we are ready to solve the equations on a computer.

Now, we are ready for a computer solution. You can enter the 18 equations in matrix form or as your favorite software package requires and get the solution by solving for the unknowns. Here is a pseudocode to set up and solve the matrix equation. Let us order the unknown forces in the form

so that $x_1$ to $x_{15}$ = $T_1$ to $T_{15}$; $x_{16}=A_x$, $x_{17}=A_y$, and $x_{18}=I_x$.

Entering and solving full matrix equation:

theta = pi/9                      % specify theta in radians
alpha1 = atan(3*tan(theta)        % calculate alpha1
alpha2 = atan(2*tan(theta))       % calculate alpha2 from arctan
alpha3 = theta                    % calculate alpha3 from arctan
C = cos(theta), S = sin(theta)    % compute all sines and cosines
C1 = cos(alpha1), S1 = sin(alpha1)
C2 = .. ..

F1 = 20;                           % input given external loads
F2 = 40;

A = [1 0 0 0 0 0 0 0 0 C1 0 0 0 0 0 1 0 0     % enter matrix A row-wise
     0 0 0 0 0 0 0 0 0 S1 1 0 0 0 0 0 1 0
     :
     :
     0 0 0 0 0 0 0 0 0 0 -1 -S 0 0 0 0 0 0]
b = [0 0 0 0 0 F1 0 0 0 F2 0 0 0 0 0 0 0 0]’  % enter column vector b

solve A*x = b for x

The solution obtained from the computer is

A simple 3-D truss: The 3-D truss shown in the figure has 12 bars and 6 joints. Nine of the 12 bars that are either horizontal or vertical have length $\mathrm{\ell }=1\mathrm{m}$. The truss is supported at A on a ball and socket joint, at B on a linear roller, and at C on a planar roller (all three supports are on the ground). The loads on the truss are ${\stackrel{\rightharpoonup }{𝑭}}_1=-50\text{N}\widehat{𝒌},{\stackrel{\rightharpoonup }{𝑭}}_2=-60\text{N}\widehat{𝒌}$ , and ${\stackrel{\rightharpoonup }{𝑭}}_3=30\text{N}\hat{\mathit{ȷ}}$. Find all support reactions and the tension in bar BC.

The free-body diagram of the entire structure is shown in fig. 6.26. Let the support reactions at A, B, and C be ${\stackrel{\rightharpoonup }{𝑹}}_A=R_{A_x}\hat{\mathit{ı}}+R_{A_y}\hat{\mathit{ȷ}}+R_{A_z}\widehat{𝒌},{\stackrel{\rightharpoonup }{𝑹}}_B=R_{B_x}\hat{\mathit{ı}}+R_{B_z}\widehat{𝒌},\text{and}{\stackrel{\rightharpoonup }{𝑹}}_C=R_C\widehat{𝒌}$. Then the moment balance about point A, ${\displaystyle \sum {\stackrel{\rightharpoonup }{𝑴}}_{\mathrm{A}}}=\stackrel{\rightharpoonup }{\mathrm{𝟎}}$, gives

Note that ${\stackrel{\rightharpoonup }{𝑭}}_1$ passes through A and, therefore, produces no moment about A. Now we compute each term in the equation above.

Substituting these products in eqn. (6.10), and dotting the resulting equation with $\hat{\mathit{ȷ}},\widehat{𝒌},\text{and}\hat{\mathit{ı}}$, respectively, we get

Thus, $\stackrel{\rightharpoonup }{𝑩}=R_{B_x}\hat{\mathit{ı}}+R_{B_z}\widehat{𝒌}=-15\sqrt{3}\text{N}\hat{\mathit{ı}}+30\text{N}\widehat{𝒌}$ and ${\stackrel{\rightharpoonup }{𝑹}}_C=\stackrel{\rightharpoonup }{\mathrm{𝟎}}$. Now from the force balance, ${\displaystyle \sum \stackrel{\rightharpoonup }{𝑭}}=\stackrel{\rightharpoonup }{\mathrm{𝟎}}$, we find ${\stackrel{\rightharpoonup }{𝑹}}_A$ as

To find the force in bar BC, we draw a free-body diagram of joint B (which connects BC) as shown in fig. 6.27. Now, writing the force balance for the joint in the $x$-direction, i.e., $[{\displaystyle \sum \stackrel{\rightharpoonup }{𝐅}}=\stackrel{\rightharpoonup }{\mathrm{𝟎}}]\cdot \hat{\mathbf{ı}}$, gives

Thus, the force in bar BC is $T_{BC}=30\text{N}$ (tensile force).

A 3-D truss solved on the computer: The 3-D truss shown in the figure is fabricated with 12 bars. Bars 1–5 are of length $\mathrm{\ell }=1\mathrm{m}$, bars 6–9 have length $\mathrm{\ell }/\sqrt{2}(\approx 0.71\mathrm{m})$, and bars 10–12 are cut to size to fit between the joints they connect. The truss is supported at A on a ball and socket, at B on a linear roller, and at C on a planar roller. A load $F=2\mathrm{kN}$ is applied at D as shown. Write all equations required to solve for all bar forces and support reactions and solve the equations using a computer.

• •

  Joint A:

  $$T_1\hat{\mathit{ı}}+\frac{T_6}{\sqrt{2}}(\hat{\mathit{ı}}+\widehat{𝒌})+\frac{T_{10}}{\sqrt{6}}(\hat{\mathit{ı}}+2\hat{\mathit{ȷ}}+\widehat{𝒌})+T_4\hat{\mathit{ȷ}}+R_{A_x}\hat{\mathit{ı}}+R_{A_y}\hat{\mathit{ȷ}}+R_{A_z}\widehat{𝒌}=\stackrel{\rightharpoonup }{\mathrm{𝟎}}.$$

• •

  Joint B:

  $$-T_1\hat{\mathit{ı}}+\frac{T_7}{\sqrt{2}}(-\hat{\mathit{ı}}+\widehat{𝒌})+T_2\hat{\mathit{ȷ}}+\frac{T_{12}}{\sqrt{2}}(-\hat{\mathit{ı}}+\hat{\mathit{ȷ}})+R_{B_y}\hat{\mathit{ȷ}}+R_{B_z}\widehat{𝒌}=\stackrel{\rightharpoonup }{\mathrm{𝟎}}.$$

• •

  Joint C:

  $$-\frac{T_6}{\sqrt{2}}(\hat{\mathit{ı}}+\widehat{𝒌})-\frac{T_7}{\sqrt{2}}(-\hat{\mathit{ı}}+\widehat{𝒌})+T_5\hat{\mathit{ȷ}}+\frac{T_{11}}{\sqrt{6}}(\hat{\mathit{ı}}+2\hat{\mathit{ȷ}}-\widehat{𝒌})=\stackrel{\rightharpoonup }{\mathrm{𝟎}}.$$

• •

  Joint D:

  $$-T_2\hat{\mathit{ȷ}}-\frac{T_{11}}{\sqrt{6}}(\hat{\mathit{ı}}+2\hat{\mathit{ȷ}}-\widehat{𝒌})-T_3\hat{\mathit{ı}}+\frac{T_9}{\sqrt{2}}(-\hat{\mathit{ı}}+\widehat{𝒌})-F\widehat{𝒌}=\stackrel{\rightharpoonup }{\mathrm{𝟎}}.$$

• •

  Joint E:

  $$-T_4\hat{\mathit{ȷ}}+\frac{T_{12}}{\sqrt{2}}(\hat{\mathit{ı}}-\hat{\mathit{ȷ}})+T_3\hat{\mathit{ı}}+\frac{T_8}{\sqrt{2}}(\hat{\mathit{ı}}+\widehat{𝒌})+R_{E_z}\widehat{𝒌}=\stackrel{\rightharpoonup }{\mathrm{𝟎}}.$$

• •

  Joint F:

  $$-T_5\hat{\mathit{ȷ}}-\frac{T_8}{\sqrt{2}}(\hat{\mathit{ı}}+\widehat{𝒌})-\frac{T_{10}}{\sqrt{6}}(\hat{\mathit{ı}}+2\hat{\mathit{ȷ}}+\widehat{𝒌})-\frac{T_9}{\sqrt{2}}(-\hat{\mathit{ı}}+\widehat{𝒌})=\stackrel{\rightharpoonup }{\mathrm{𝟎}}.$$

Now we can separate out 3 scalar equations from each of the joint vector equations by dotting them with $\hat{\mathit{ı}},\hat{\mathit{ȷ}}$, and $\widehat{𝒌}$.

Thus, we have 18 required equations for the 18 unknowns. Before we go to the computer, we need to do just one more little thing. We need to order the unknowns in some way in a one-dimensional array. So, let

Thus $x_1=R_{A_x},x_2=R_{A_y},\mathrm{\dots },x_7=T_1,x_8=T_2,\mathrm{\dots },x_{18}=T_{12}$. Now we are ready to go to the computer, feed these equations, and get the solution. We enter each equation as part of a matrix [A] and a vector {b} such that [A]{x} = {b}. Here is the pseudocode:

sq2i = 1/sqrt(2)            % define a constant
sq6i = 1/sqrt(6)            % define another constant
F = 2                       % specify given load
A(1,[1 7 12 16]) = [1 1 sq2i sq6i]
A(2,[2 10 16]) = [1 1 2*sq6i]
.
.
A(18,[14 15 16]) = [sq2i sq2i sq6i]
b(12,1) = F
form A and b setting all other entries to zero
solve A*x = b for x

The solution obtained from the computer is the one-dimensional array x which after decoding according to our numbering scheme gives the following answer.

The method of joints can solve any solvable truss. So why learn a different method? There are two basic reasons.

1. 1.

   Sometimes one only wants to know a little and the method of joints is cumbersome.

   Example: Difficulty in finding just one bar tension.

   Say you are interested only in $T_{\mathrm{KM}}$ in the truss of fig. 6.7 on page 6.7. With the method of joints we could find $T_{\mathrm{KM}}$ by working through the joints one at a time. To get to joint K we would have to draw free-body diagrams of at least 8 other joints first. And for each we would have to solve two simultaneous equations.

2. 2.

   Sometimes the method of joints doesn’t best reveal basic structural ideas.

   Example: Difficulty in understanding trends.

   Again look at the truss of fig. 6.7. With the method of joints we would find, after all the algebra, that all the bars on the bottom (AC, CE, EH, HJ, JL, LN, NP, PR) have compression (negative tension) and that each bar has more compression than the one to its right. Similarly the top is all tension with the tension increasing with the bars more to the left. Are these trends just a consequence of lots of algebra?

The method of sections provides a shortcut, particularly for elementary textbook-like problems. And the method of sections can explain some structural trends.

Say you are just trying to find one bar tension, for example $T_{\mathrm{KM}}$ in the truss of fig. 6.7. For simplicity we limit our attention to 2D structures.

For a given truss and given bar tension of interest there is no guarantee that the recipe applies. You can always find a section cut through the bar of interest, but there may be too-many unknowns in the free-body diagrams of both of the resulting sub-structures.

Because 2D statics of finite objects gives three scalar equations we can generally find all three unknown bar tensions from a section cut that goes through 3 bars.

Look at the free-body diagram from a section cut in fig. 6.47. Moment balance about point J (about an axis through J in the $z$ direction) gives: