§3.2 Cables, bars and springs Topics for this section: Why do we assume cables and ropes are inextensible, and bars are rigid? Springs: what are they, how are they modeled, and what are they good for? Introduction to statically indeterminate problems.

Why do we assume cables and ropes are inextensible, and bars are rigid? 1st reason: Theses assumptions are usually accurate, because cables, ropes and bars are often very stiff, meaning they deform little when they are loaded. 2nd reason: With these assumptions, the geometry of the structure in the equilibrium configuration is usually known. Consider the following example.....

Example 1: Consider a structure that has known geometry before loads are applied. If a 100 N vertical force is applied at point A, determine the forces supported by members AB and AC. Equilibrium equations: ∑F x = 0: F AB cos  + F AC cos  = 0,(1) ∑F y = 0: F AB sin  – F AC sin  – 100 N = 0.(2)

Remarks: 1) The angles  and  in Eqs. (1) and (2) must be those in the equilibrium configuration of the structure. 2) Question: is  = 30° and  = 45°, or are these angles unknown? In other words, do Eqs. (1) and (2) contain two unknowns or four unknowns? 3) If  = 30° and  = 45° before loads are applied, then after loads are applied, because all materials are deformable, these angles will usually be different (and unknown) – Mechanics of Materials is needed to account for this. 4) If members are undeformable, then  and  remain unchanged after the force is applied, and the unknowns F AB and F AC may be determined using only the equilibrium equations. ∑F x = 0: F AB cos  + F AC cos  = 0,(1) ∑F y = 0: F AB sin  – F AC sin  – 100 N = 0.(2)

Take-away message: In statics, we generally assume cables and ropes are inextensible, and bars are rigid. With these assumptions, the geometry of a structure typically does not change when forces are applied, and hence, the geometry of the structure in its equilibrium configuration is generally known. Under these circumstances, the equations of equilibrium are sufficient to determine all unknowns, and furthermore, this solution is exact. Sometime, it is necessary or desirable to model the deformability of a structure. Some options include: Springs: these are simple structural elements that deform when they are subjected to forces... more on this very soon! Mechanics of Materials: this subject introduces more general models for how materials and structural components deform.

Springs A spring is a structural member that deforms when it is subjected to a force. Cables, ropes, bars, and other members are often modeled as springs. Complicate structures are sometimes modeled as springs. Materials are modeled as springs (at least over a range of conditions). © Flat Earth Images RF

A spring deforms when it is subjected to a force. A linear spring has force that is proportional to stretch. An elastic spring returns to its original geometry when the force is removed. Spring law

F s = force supported by spring. F s > 0 tension. F s < 0 compression.  = stretch of spring.  = 0 is the unstretched position.  = final length  initial length. = L – L 0  > 0 extension.  < 0 contraction. k = spring stiffness (≥ 0 always). dimensions: (force/length)... or... (force/displacement) Sign conventions used in spring law

Alternative sign conventions Sometimes, it may be more convenient to use other sign conventions when writing the spring law. For example, let F s be positive in compression, and  be positive in contraction. Then,

Example 2: A model for an electrical switch is shown. The spring has 7 mm unstretched length. Contact at point D occurs when  = 40°. Determine the force F needed to cause the switch to make contact at point D. Model the collar at A as a particle, and assume bar CD is frictionless.

Example 3: Determine the stiffness and initial length for the spring of a pogo stick so that the pogo stick begins to compress when 50 lb is applied, and fully compresses when 200 lb is applied. © Siede Preis/Getty Images RF © Rubberball/Getty Images RF

FBD: Draw the FBDs if the force applied by the user is W.

A statically determinate structure has as many unknowns as equilibrium equations — generally, all of the unknowns can be determined. A statically indeterminate structure has as more unknowns than equilibrium equations — not all (usually none!) of the unknowns can be determined. Physical explanation: a statically indeterminate structure has more load paths than are required to provide equilibrium. Statically indeterminate structures... a brief introduction

Example 4: Which of the structures is statically determinate and which is statically indeterminate?