Mechanics of Materials Engr Lecture 10 Axial Deformation #1

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Mechanics of Materials Engr 350 - Lecture 10 Axial Deformation #1

Axial Deformation Mechanics of materials approach: real structural elements are analyzed as idealized models with simplified loads Solutions are therefore approximations Solutions are generally satisfactory (within 10% of actual) Some reasons these approximations work is: Allowable Stress Design (ASD) – Applying Factor of Safety Load and Resistance Factor Design – Statistical analysis of strength and loads Saint-Venant’s Principle

Saint-Venant’s principle States that the stress, strain, and displacement fields due to two different but statically equivalent force distributions on parts of a body far away from the loading points are approximately the same In other words….localized maximum stresses diminish rapidly as the distance from the point of loading increases Consider Fig. 5.1 and 5.2 in Philpot In future courses, consider application sites

Deformations of axially loaded bars Consider a bar loaded by a force, F Recall: 1) Normal strain: 2) Hooke’s law: 3) Normal stress: 𝜀 𝑎𝑥𝑖𝑎𝑙 = 𝛿 𝐿 F F 𝜎 𝑎𝑥𝑖𝑎𝑙 =𝐸∗ 𝜀 𝑎𝑥𝑖𝑎𝑙 𝜎 𝑎𝑥𝑖𝑎𝑙 = 𝐹 𝐴 Combine these to get the FLEA equation 𝛿= 𝐹∗𝐿 𝐸∗𝐴

Deformations of axially loaded bars cont… Same constraints as Hooke’s Law (linear-elastic region only!) Fine point: The previously derived equation is only for homogeneous, prismatic member loaded by forces at the end (Prismatic member means the combination of E*I is constant) If multiple members are used, deformations are additive 𝛿= 𝑖 𝐹 𝑖 ∗ 𝐿 𝑖 𝐸 𝑖 ∗ 𝐴 𝑖 Sign convention 𝛿 is + if the member elongates 𝛿 is - if the member compressed or contracted

Non-prismatic members What about cases where F, or E*I are not constant? Consider a post such as a solid tapered concrete footing A long slender hanging rod/wire here the mass cannot be neglected In this case, each increment of length is subjected to either a different force and/or has a different cross-sectional area Let the length of each section become infinitesimally small 𝑑𝛿= 𝐹 (𝐿) 𝐸∗𝐴 (𝐿) 𝑑𝑥 𝛿= 0 𝐿 𝑑𝛿 = 0 𝐿 𝐹 (𝐿) 𝐸∗𝐴 (𝐿) 𝑑𝑥

Practice Problem (MM Module M5.3)

Example 5.2

Example 5.2