Engineering & Technology · Statics, Stress & Deformation
Trusses, frames, beams and idealised structural elements
A triangle of three bars can carry a downward force at its top joint. A single horizontal beam carries a similar load very differently. The outline alone does not tell us the internal action: connections and load locations decide which idealisation is credible.
Use Forces, Energy & Simple Systems and Material Structure & Properties first. Resolve horizontal and vertical components and use simple ratios and powers. The chapter defines each new mechanics symbol before applying it; 3–4–5 triangles avoid a trigonometry prerequisite in the worked truss. All dimensions, loads, material parameters, limits and records are constructed teaching inputs, not equipment specifications, permissible loads or testing instructions.
Back to the chapterAfter reading, you can
Recognise when a two-force member model is appropriate and follow a load through joints.
Choose an element for the relationship
A structural element is a simplified representation of how part of a structure carries load. A tie primarily carries axial tension along its length. A strut carries axial compression. A beam can carry transverse loading through internal shear and bending moment. A frame combines connected members whose joint interactions may include moments. The same physical piece can need different models under different loading.
For a two-force member in equilibrium, its only two external forces are equal, opposite and collinear. If they pull its ends apart, call the action tension; if they push inward, compression. The straight slender bar model usually places that line along the member. Loads distributed along it, appreciable self-weight or a joint couple would invalidate the pure two-force assumption.
A truss uses a restricted joint model
An ideal planar truss connects straight members with frictionless pins and applies external loads only at joints. Neglect member weight here. Each member can then carry a single axial force, and each joint supplies two component force-balance equations. First solve the external reactions of the whole truss, then isolate a joint to reveal its member forces.
Assuming every member force initially pulls away from a joint is a bookkeeping choice. A negative answer means the member pushes the joint in the opposite direction. Keep that sense when drawing the adjacent joint: the member’s force on each end joint is opposite. Summing the internal forces of the complete structure cancels paired interactions.
A solved force is the beginning of structural reasoning
The figure’s two sloping members meet at a top joint. Their horizontal components cancel by symmetry while their upward components support the load. At a bottom joint, the diagonal’s horizontal component must be balanced by the base member. This makes an otherwise hidden tension path visible.
A frame with rigid joints cannot generally be solved by declaring all pieces two-force members. Connections transfer turning effects, and member bending matters. Even an appropriate truss model leaves later checks: deformation, compressive buckling, joint eccentricity and additional load cases. A zero force for one load case is not proof that a member is unnecessary.
Use a 3–4–5 triangle to resolve joint forces
The original truss has a 6 m base and 4 m height, so each half has horizontal run 3 m and diagonal length 5 m. A 100 N apex load gives 50 N upward reaction at each support. If C is the magnitude of compression in either slope, top-joint vertical balance gives 2C(4/5)=100. Thus C=62.5 N.
At the left base joint, the compressed slope pushes downward and left. Its vertical component is 50 N, balanced by the support. Its horizontal component is 62.5×3/5=37.5 N left. The base therefore pulls right with 37.5 N: tension. These component balances explain the forces without assigning any strength or permissible load to the bars.
Try the reasoning
Why does loading a bar halfway along its length threaten the two-force idealisation?
Show the answer and reasoning
It introduces another external force and generally bending, so the end forces need not be simply collinear.
Are the slopes in tension in this example?
Show the answer and reasoning
No. Their compressive forces push upward on the top joint, providing the needed support.
Does a correct axial-force solution prove the compressed slope remains straight?
Show the answer and reasoning
No. Buckling and deformation require additional geometry, material and boundary information.
What to carry forward
Identify the load path and joint model before assigning axial, shear or bending behaviour to a member.
References
Teaching resources and research supporting this lesson’s explanations and examples.
Daniel Baker and William Haynes · Engineering Statics2 sources