Analysis of Trusses, Frames and Machines

Introduction

Structures and mechanical devices are assemblies of members that transmit forces from applied loads to supports or between connected components. A roof truss carries gravity loads through axial tension and compression; a crane frame transfers loads through combinations of axial force, shear, and bending; and a machine mechanism transmits motion and force through pinned links. Engineering mechanics provides systematic methods for determining these external reactions and internal actions.

This chapter treats planar systems under static loading. The central idealizations are that bodies are rigid, connections are represented by appropriate force constraints, and equilibrium governs the analysis. These idealizations are sufficiently accurate for preliminary design and for many structures whose deformations are small.

Learning Objectives

After studying this chapter, you should be able to:

• distinguish trusses, frames, and machines;
• identify support reactions and assess static determinacy;
• locate zero-force members in planar trusses;
• analyze trusses by the method of joints and the method of sections;
• determine internal axial force, shear force, and bending moment in frames;
• analyze machines and linkages as connected rigid bodies;
• construct basic axial-force, shear-force, and bending-moment diagrams;
• recognize modeling assumptions, limitations, and professional safety responsibilities.

1. Structural Models and Terminology

A two-force member is acted upon by forces at only two points. For equilibrium, the forces must be equal, opposite, and collinear; therefore the member carries only axial tension or compression. An ideal truss is composed of straight two-force members connected by frictionless pins, with external loads applied at joints. Each member is consequently a two-force member.

A frame contains at least one multiforce member. Loads may act between joints, and members can carry axial force, shear, and bending moment. A machine is an assembly designed to transmit or modify motion and force. Its members may be two-force links, but the assembly usually has input and output forces or couples and is analyzed member by member.

Important terms include:

• Joint or node: connection point between members.
• Panel: portion of a truss between adjacent joints.
• Tension: axial force pulling a member apart.
• Compression: axial force pushing a member together.
• Internal force: force exposed when a body is imagined to be cut.
• Free-body diagram (FBD): isolated body showing all external forces and couples.
• Static determinacy: condition in which equilibrium equations alone are sufficient to determine unknown reactions and member forces.

For a planar rigid body, the independent equilibrium equations are

Here, F_x and F_y are force components in newtons (N), and M_O is the moment about point O in newton-metres (N·m). These equations apply to a body in planar static equilibrium, with a consistent sign convention.

A moment of a force about O is

where r is the distance from O to the force application point, F is force magnitude, and θ is the angle between the position vector and force. Equivalently, moment equals force times perpendicular distance. The expression assumes a rigid body and uses a chosen positive rotational sense.

2. External Reactions and Determinacy

A pin support supplies two reaction components, normally A_x and A_y. A roller supplies one reaction normal to its supporting surface. A fixed support supplies two force components and a reaction moment in a planar problem. The reactions must be drawn on the FBD before member forces are considered.

For a pin-connected planar truss with m members, j joints, and r external reaction components, a common determinacy relation is

If the geometry is stable and the relation holds, the truss is generally statically determinate. If m + r > 2j, it is externally or internally indeterminate; if m + r < 2j, it is unstable or deficient. The count is only a necessary condition: a geometrically unstable arrangement can satisfy the equation, so stability must also be checked.

For an ideal planar truss, each joint provides two scalar equations. The entire truss must first satisfy global equilibrium. Reactions should be solved before internal member forces unless a carefully selected section permits otherwise.

3. Zero-Force Members

Zero-force members carry no axial force for the particular loading case, although they may be essential for stability, construction, or a different loading case.

Two standard rules apply to an unloaded joint:

1. If only two non-collinear members meet at the joint, both members are zero-force members.
2. If three members meet, two are collinear, and there is no external load or support reaction at the joint, the non-collinear member is a zero-force member.

These rules follow directly from joint equilibrium. In the second case, the non-collinear member is the only member capable of balancing a force component perpendicular to the collinear pair, so its force must be zero. The rules must not be applied at a joint carrying a load, a support reaction, or a relevant applied couple. After removing an identified zero-force member, the rules may be applied again to the reduced idealized joint arrangement.

Zero-force members are not necessarily useless. They may prevent a mechanism, brace the structure against lateral loads, reduce slenderness, or provide alternate load paths.

4. Method of Joints

The method of joints applies equilibrium to each pin joint separately. It is most efficient when forces in many or all members are required.

Procedure

1. Draw the complete truss FBD and calculate support reactions.
2. Identify zero-force members.
3. Select a joint with no more than two unknown member forces.
4. Assume unknown member forces act away from the joint; this assumes tension.
5. Apply ∑F_x = 0 and ∑F_y = 0.
6. Move to adjacent joints and continue.
7. Interpret a positive result as tension under the assumed direction and a negative result as compression.
8. Check the final results using equilibrium of the entire truss or an unused joint.

Worked Example 1: Simple triangular truss

A triangular truss has joints A(0, 0), B(4, 0), and C(2, 3), in metres. A is a pin and B is a roller. A downward load of 10 kN acts at C. Determine the support reactions and member forces.

Reactions. Taking moments about A,

Vertical equilibrium gives

There is no horizontal applied load, so A_x = 0.

Joint C. Each inclined member has length √(2² + 3²) = √13 m. Its direction components are horizontal 2/√13 and vertical 3/√13. Assume member forces F_AC and F_BC act away from C. Horizontal equilibrium gives

Vertical equilibrium gives

Therefore,

The negative sign means both inclined members are in compression, with magnitude 6.01 kN.

Joint A. The horizontal member force is zero because A_x = 0 and the horizontal component of the inclined-member force is balanced by the horizontal member. Using the known inclined force,

Thus AB is in tension, while AC and BC are in compression. The results satisfy joint B and global equilibrium.

5. Method of Sections

The method of sections is efficient when forces in only a few members are needed. An imaginary cut is passed through the truss, and one portion is isolated. The cut should pass through no more than three unknown member forces in a planar determinate truss, because the isolated portion provides only three independent equilibrium equations.

Procedure

1. Calculate external reactions.
2. Pass a section through the target members.
3. Select the simpler side of the cut.
4. Draw its FBD, showing cut-member forces as unknown axial forces.
5. Apply force and moment equilibrium.
6. Use moments about the intersection of two unknown-member lines of action whenever possible; this eliminates those two unknowns.

The method relies on the fact that each cut member transmits an axial force along its own axis. It is not directly valid for a frame member that also carries shear and bending moment unless the complete internal actions are included.

Worked Example 2: Section analysis

Suppose a section through a determinate truss cuts three members, and on the left portion the known external forces produce a 24 kN·m counterclockwise moment about the intersection of the lines of action of two cut-member forces. The third cut member has a perpendicular distance of 0.8 m from that point. Taking the third-member force as tension and counterclockwise positive,

The member is therefore in compression at 30 kN. This calculation is valid only if the stated external moment includes all forces on the isolated portion and the third member’s line of action is correctly measured.

6. Frames and Internal Actions

Frames differ from ideal trusses because their members may be loaded between joints and may transmit moments through rigid connections. Analyze a frame by first isolating the entire frame to determine external reactions, then isolating individual members or subassemblies to determine internal forces.

At a cut through a planar member, the internal actions are conventionally represented by:

• N, axial force, in newtons;
• V, shear force, in newtons;
• M, bending moment, in newton-metres.

The sign convention must be stated. A common convention considers positive axial force as tension, positive shear as upward on the left face of a cut segment, and positive bending moment as sagging. Other conventions are acceptable if used consistently.

At an internal pin, the bending moment is zero because the pin cannot transmit a couple. The force components at the pin are equal and opposite on the two connected bodies. A two-force frame member has no internal shear or bending in the ideal model, but a multiforce member generally does.

Worked Example 3: Internal actions in a loaded beam segment

A simply supported beam carries a 6 kN point load at x = 2 m from the left support and spans 5 m. The left and right reactions are found from equilibrium:

Then

For a section between the left support and the load, 0 ≤ x < 2 m,

For a section to the right of the load, 2 < x ≤ 5 m,

At x = 2 m, the bending moment is 7.2 kN·m from either expression. The shear changes abruptly by 6 kN at the point load, while the moment remains continuous. The moment is zero at both simple supports, as required for this idealization.

7. Axial-Force, Shear-Force, and Bending-Moment Diagrams

Diagrams are graphical representations of internal actions as functions of position. They support design decisions by showing where a member experiences its largest demand.

Useful relationships for a beam with distributed load intensity w(x) are

where V is shear in N, M is moment in N·m, x is distance in m, and w is load per unit length in N/m. The signs depend on the adopted convention. Thus:

• where distributed load is zero, shear is constant;
• where shear is constant, moment varies linearly;
• where shear is linear, moment is parabolic;
• a concentrated force causes a jump in shear;
• a concentrated applied couple causes a jump in moment;
• an internal maximum or minimum bending moment commonly occurs where V = 0, provided the point lies within the member.

For an axial-force diagram, positive values indicate tension under the chosen convention and negative values compression. For a truss, the axial force is constant within each ideal member. For a frame, axial force may vary along a member when distributed or transverse loads act.

Best practices for diagrams

1. Establish a clear coordinate and sign convention.
2. Mark supports, point loads, applied couples, and member discontinuities.
3. Divide the member into intervals at every load or connection change.
4. Derive expressions from an FBD rather than sketching from memory.
5. Label values at supports, load points, and extrema.
6. Check boundary conditions and jumps.
7. Use units on axes and numerical values.

8. Machines and Connected Bodies

A machine is analyzed by applying equilibrium to the entire assembly and then to selected links. The first FBD may determine support reactions, while subsequent FBDs determine the force transmitted through a pin or the input force required to balance an output load.

For a pin-connected link carrying forces at two pin centers and no other load, the link is a two-force member. If a third force or couple acts on it, it becomes a three-force or multiforce member. For a three-force member in equilibrium, the three lines of action must be concurrent or parallel. This concurrency condition can greatly simplify machine analysis.

Internal pin forces should not be treated as external forces when the complete machine is isolated; they occur in equal and opposite pairs and cancel. They must be included when one link is isolated.

Illustrative machine analysis

Consider a hand-operated lever pinned to a frame at O. An effort force P acts at distance a from O, and an output resistance R acts at distance b on the opposite side. If the forces are perpendicular to the lever and the lever is static,

The relation assumes negligible lever weight, frictionless pin action, and perpendicular force lines. If the effort is inclined, use the perpendicular moment arm or write the cross-product moment. The result demonstrates mechanical advantage: increasing a relative to b reduces the required effort, but usually increases travel at the effort point.

9. Detailed Applied Case: Preliminary Analysis of a Roof Truss

Background. An illustrative design team is evaluating a small steel roof truss for a workshop. The truss is idealized as a pin-connected planar assembly. Roof loads are transferred to the panel joints, and the supports consist of a pin at one end and a roller at the other.

Problem. The team must identify support reactions, determine whether any members are zero-force members under symmetrical gravity loading, estimate critical member forces, and decide whether a detailed design review is required.

Analysis. The engineer first prepares the global FBD and checks that the applied vertical loads and geometry produce a stable arrangement. The pin and roller reactions are calculated from the three global equilibrium equations. At an unloaded joint near the gable, two non-collinear members meet; both are identified as zero-force members for that load case. However, the engineer records that wind uplift or asymmetric snow loading could activate them.

A method-of-joints analysis is then started at a support joint with two unknown member forces. Tension is assumed for each unknown. Negative computed values identify compression members. Because only selected forces near midspan are needed for connection sizing, a section is also taken through three members. A moment equation about the intersection of two cut-member lines of action directly gives the third member force. Results are checked against equilibrium of the opposite portion.

Approach and solution. The member-force schedule is reported with member identification, magnitude, sign, and interpretation. Units are kN, and the assumptions are stated: pin joints, joint loading, negligible self-weight or explicitly included self-weight, small deformation, and static loading. Members in compression are flagged for buckling assessment; members in tension are flagged for connection and net-section checks.

Outcome. The equilibrium analysis supplies force demands but does not by itself certify the truss. The design team must verify material strength, slenderness, connection capacity, fatigue where relevant, corrosion protection, fire requirements, construction tolerances, load combinations, and the governing building code. A zero-force result is not permission to omit a member without a stability and alternate-load-path review.

Learning points. Idealized analysis is a model, not a complete design. Correct load paths, support assumptions, sign interpretation, independent checks, and transparent documentation are essential. Safety-critical decisions require qualified engineering judgment and applicable regulations.

10. Limitations, Technology, and Professional Practice

The methods in this chapter assume static loading, idealized supports, rigid-body equilibrium, and sufficiently small deformation. Real structures may exhibit joint friction, connection eccentricity, member self-weight, fabrication imperfections, elastic deformation, dynamic effects, residual stress, and material or geometric nonlinearity. Truss members that appear pin-connected may develop secondary bending because of rigid connections or out-of-plane restraint.

Computer-aided structural-analysis software can automate equilibrium equations, display force diagrams, and perform parametric studies. It cannot correct an incorrect model. A sound workflow is to create a hand-calculated simplified model first, compare software reactions and selected member forces, inspect units and boundary conditions, and investigate any discrepancy. AI-assisted tools may help classify members or check arithmetic, but generated results require verification, traceable assumptions, and human responsibility. Safety, licensing, confidentiality, and code compliance remain professional obligations.

A good engineering report includes the structural idealization, FBDs, coordinate system, sign convention, equations, units, reaction checks, member-force tables, diagrams, assumptions, load cases, software version if used, and limitations. Never conceal a negative result by changing signs or omit a potentially governing load case merely because it produces inconvenient forces.

Chapter Summary

Planar trusses are idealized as pin-connected two-force members, so each member carries axial tension or compression. Global equilibrium determines support reactions, while joint equilibrium determines member forces one joint at a time. Sections provide an efficient alternative when only selected forces are required. Zero-force-member rules follow from equilibrium at unloaded joints, but zero force applies only to the specified load case and does not imply structural irrelevance.

Frames and machines contain multiforce members and must be analyzed through carefully selected free-body diagrams. Internal actions include axial force, shear, and bending moment. Axial-, shear-, and moment diagrams are constructed from equilibrium and from the relationships between distributed load, shear, and moment. Reliable practice requires consistent signs, dimensional checks, independent verification, awareness of idealization limits, and compliance with safety codes.

Key Terms

Truss: Pin-connected assembly of ideal two-force members.

Frame: Structural assembly containing one or more multiforce members.

Machine: Connected mechanism that transmits or modifies force and motion.

Joint: Point at which members are connected.

Two-force member: Member acted upon by two equal, opposite, collinear forces.

Free-body diagram: Isolated-body representation showing external actions.

Support reaction: Force or couple exerted by a support.

Static determinacy: Condition in which equilibrium equations determine all unknowns.

Zero-force member: Member carrying zero force for a specified loading case.

Method of joints: Truss method applying two equilibrium equations at each joint.

Method of sections: Method using an imaginary cut through selected truss members.

Axial force: Internal force along a member’s longitudinal axis.

Shear force: Internal force transverse to a member’s longitudinal axis.

Bending moment: Internal couple resisting flexural rotation.

Tension: Axial action that tends to elongate a member.

Compression: Axial action that tends to shorten a member.

Static equilibrium: State satisfying zero resultant force and zero resultant moment.

Load path: Route by which applied loads are transferred to supports.

Review Questions

Short Answer Questions

1. What distinguishes an ideal truss member from a typical frame member?
2. State the three independent planar equilibrium equations.
3. Under what conditions is a non-collinear member at an unloaded three-member joint a zero-force member?
4. Why is the method of sections useful in truss analysis?
5. What physical action causes a jump in a shear-force diagram?

Long Answer Questions

1. Explain the complete method-of-joints procedure, including assumptions, sign interpretation, and checks.
2. Compare trusses, frames, and machines in terms of member forces, connections, and free-body diagrams.
3. Derive the support reactions and internal-force functions for a simply supported beam with one point load.
4. Explain how a cut-member moment equation can isolate one unknown force in a truss.
5. Discuss the limitations of idealized equilibrium analysis in real structural design.

Analytical / Application Questions

1. A triangular truss has a 12 kN downward load at its apex and symmetric supports. Explain how you would determine reactions and identify whether either inclined member is in tension or compression.
2. A section through a truss cuts three members, and one cut-member line passes through the chosen moment center. Describe how you would select the moment center and solve the remaining cut-member force.
3. A lever has an effort arm of 0.6 m and a resistance arm of 0.15 m. If the resistance is 800 N and both forces are perpendicular to the lever, determine the required effort and state the assumptions behind the calculation.

Multiple Choice Questions with Answers

Question 1

For a planar rigid body in static equilibrium, which set of equations must be satisfied?

A. ∑F_x = 0 only

B. ∑F_y = 0 and ∑M = 0 only

C. ∑F_x = 0, ∑F_y = 0, and ∑M = 0

D. ∑M = 0 only

Correct Answer: C. ∑F_x = 0, ∑F_y = 0, and ∑M = 0

Question 2

An ideal truss member is generally modeled as carrying which type of internal action?

A. Axial force only

B. Bending moment only

C. Shear force only

D. Axial force, shear, and bending independently

Correct Answer: A. Axial force only

Question 3

At an unloaded joint with two non-collinear members, the member forces are

A. equal and tensile

B. equal and compressive

C. both zero

D. opposite and nonzero

Correct Answer: C. both zero

Question 4

A planar truss has 9 members, 6 joints, and 3 external reaction components. What does m + r = 2j indicate?

A. 9 + 3 = 2(6), so the count is consistent with static determinacy

B. 9 + 3 > 2(6), so the truss must be unstable

C. 9 + 3 < 2(6), so the truss must be determinate

D. The truss has no reactions

Correct Answer: A. 9 + 3 = 2(6), so the count is consistent with static determinacy

Question 5

In the method of sections, why is a cut through no more than three unknown member forces usually selected for a planar determinate truss?

A. A truss has only three members

B. A planar isolated portion provides three independent equilibrium equations

C. Each member has three forces

D. More than three members cannot be cut physically

Correct Answer: B. A planar isolated portion provides three independent equilibrium equations

Question 6

A negative member force obtained after assuming tension indicates

A. zero force

B. compression

C. a moment reaction

D. an unstable support

Correct Answer: B. compression

Question 7

For a beam with the stated sign convention, a concentrated transverse force generally causes what feature in the shear diagram?

A. A jump

B. A parabolic segment

C. No change

D. A continuous maximum only

Correct Answer: A. A jump

Question 8

For a simply supported beam with no applied end couples, the bending moment at an ideal simple support is generally

A. equal to the reaction force

B. zero

C. maximum

D. infinite

Correct Answer: B. zero

Question 9

A lever has effort arm 0.6 m, resistance arm 0.15 m, and resistance 800 N. With perpendicular forces, the effort is

A. 100 N

B. 200 N

C. 800 N

D. 3,200 N

Correct Answer: B. 200 N

Question 10

Which statement about a zero-force member is correct?

A. It is unnecessary under every possible load

B. It cannot contribute to stability

C. It carries zero force for the specified load case but may be important under other loads

D. It always carries a bending moment

Correct Answer: C. It carries zero force for the specified load case but may be important under other loads