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Engineering science lesson 4
This document summarizes key concepts related to static engineering systems including: - Types of columns that can fail in true compression (short columns) or buckle before reaching full strength (long columns). - Types of beams classified by their support conditions. - Loads that can be applied to beams as concentrated loads or uniformly distributed loads. - Shear forces and bending moments that develop in beams due to applied loads, and sign conventions for positive and negative shear forces and moments. - Relationships for calculating shear forces and bending moments along beams subjected to different load types. - Examples of determining shear forces and bending moments at points along cantilever beams and beams with various concentrated and uniform loads
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Engineering science lesson 4
- 1. Chapter 1- Staticengineering systems 1.1 Simply supported beams 1.1.1 determination of shear force 1.1.2 bending moment and stress due to bending 1.1.3 radius of curvature in simply supported beams subjected to concentrated and uniformly distributed loads 1.1.4 eccentric loading of columns 1.1.5 stress distribution 1.1.6 middle third rule
- 2. Types of columns Dependingon the mode of failure, columns can be categorised in the following ways (a) a short column a column which will fail in true compression (b) a long column a column which buckles before full compressive strength is reached
- 3. Types of beams •Beams can be classified according to the manner in which they supported
- 6. Loads • Loads canbe applied on the beams in one of the following ways
- 7. Sheer force andbending moments • When a beam is loaded by forces or couples, internal stresses and strains are created. Consider a cantilever arrangement • It is convenient to reduce the resultant to a shear force, V, and a bending moment, M.
- 8. Sign conventions • Positiveshear forces always deform right hand face downward with respect to the left hand face. Positive shear stress acts clockwise while negative shear stress acts counter-clockwise • Positive bending moments always elongate the lower section of the beam. Positive moment compresses upper (sagging moments) whereas negative moment compresses lower (hogging moments)
- 9. Relationships for continuousloads • Consider the following beam segment with a uniformly distributed load with load intensity q. Note that distributed loads are positive when acting downward and negative when acting upward. • Summing forces vertically • Summing moments and discarding products of differentials because they are negligible compared to other terms
- 11. Relationships for concentratedloads • consider the following beam segment with a concentrated load, P. Again, concentrated loads are positive when acting downward and negative when acting upward. • Summing forces vertically • An abrupt change occurs in the shear force at a point where a concentrated load acts. As one moves from left to right through a point of load application, the shear force decreases by an amount equal to the magnitude of the downward load. • Summing the moments
- 13. Relationships for couples •Summing the moments
- 15. Sheer force andbending moment diagrams-Concentrated loads
- 16. Sheer force andbending moment diagrams-Uniform loads
- 17. Sheer force andbending moment diagrams-several concentrated loads
- 24. Example • Determine theshear force and bending moment at 1m and 4m from the right hand end of the beam shown in the figure. Neglect the weight of the beam. • A cantilever beam that is free at end A and fixed at end B is subjected to a distributed load of linearly varying intensity q. The maximum intensity of the load occurs at the fixed support and is equal to q0. Find the shear force V and bending moment M at distance x from the free end of the beam
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