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ES2501: Statics/Unit 20-1 : Internal Forces in Beams

ES2501: Statics/Unit 20-1 : Internal Forces in Beams. Introduction:. Beam: Slender structural member subjected to lateral loading. In contrast to:. bar:. Subjected to axial loading. shaft:. Internal forces in a beam :. Subjected to torque.

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ES2501: Statics/Unit 20-1 : Internal Forces in Beams

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  1. ES2501: Statics/Unit 20-1: Internal Forces in Beams Introduction: Beam:Slender structural member subjected to lateral loading In contrast to: bar: Subjected to axial loading shaft: Internal forces in a beam: Subjected to torque Three types of internal forces in a beam: N ---- Normal force (to prevent relative axial motion at C of two parts) V ---- Shear force (to prevent relative lateral sliding at C of two parts) M ---- Bending moment (to prevent relative rotation at C of two parts) Cross-section

  2. ES2501: Statics/Unit 20-2: Internal Forces in Beams Introduction: Sign conventions: Conventional “+” direction Nature of Internal forces: Distributed shear stresses Distributed normal stresses Internal forces are in fact normal and shear stresses distributed over a cross-section area. N, V, and M are simplified models as resultant force in the axial direction, the resultant force in the lateral direction, and the resultant moment about the so-called neutral axism respectively.

  3. ES2501: Statics/Unit 20-3: Internal Forces in Beams Example 1: Find the internal forces of a beam, subjected to a point load, at a specified location Find internal forces on cross-sections at C and D Support Reactions: Internal forces at C: Note: these are reactions that do not exactly the conventional signs of internal forces

  4. ES2501: Statics/Unit 20-4: Internal Forces in Beams Example 1 (con’d): Find internal forces on cross-sections at C and D Note: the problem can also been equivalently solved by equilibriun of the BD part The same results but more convenient Internal forces at D: Note: The normal force is zero in general for the case where a load acts perpendicularly to the axis of a straight beam

  5. ES2501: Statics/Unit 20-5: Internal Forces in Beams Example 2: Find internal forces on cross-sections at C and D Distributed load Support Reactions: Internal forces at C:

  6. ES2501: Statics/Unit 20-6: Internal Forces in Beams Example 2 (con’d): Find internal forces on cross-sections at C and D Distributed load Statically equivalent load Internal forces at D: Or, equivalently by equilibrium of BD: The same results

  7. ES2501: Statics/Unit 20-7: Internal Forces in Beams Comments: N-diagram V-diagram M-diagram Find internal forces at any position Graphic representation Finding internal forces in a beamis the first step of a complete beam analysis Both need M(x) first

  8. ES2501: Statics/Unit 20-8: Internal Forces in Beams Comments: Superposition principle Problem II: by distributed load only Problem I: by P only By Beam analysis or results in Handbook Reactions Internal forces Superposition Total solution

  9. Comments: ES2501: Statics/Unit 20-9: Internal Forces in Beams Statiacally inderminate problems: # of static equilibrium equations =3 # of unknowns =6 Properties of symmetry may help: By symmetry.

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