Chapter 5 Analysis and Design of Beams for Bending 

5.1 Introduction -- Dealing with beams of different materials: steel, aluminum, wood, etc. -- Loading: transverse loads  Concentrated loads  Distributed loads

-- Supports  Simply supported  Cantilever Beam  Overhanging  Continuous  Fixed Beam

A. Statically Determinate Beams -- Problems can be solved using Equations of Equilibrium B. Statically Indeterminate Beams -- Problems cannot be solved using Eq. of Equilibrium -- Must rely on additional deformation equations to solve the problems. FBDs are sometimes necessary:

FBDs are necessary tools to determine the internal (1) shear force V – create internal shear stress; and (2) Bending moment M – create normal stress From Ch 4: (5.1) (5.2) Where I = moment of inertia y = distance from the N. Surface c = max distance

 Recalling, elastic section modulus, S = I/c, hence (5.3) hence For a rectangular cross-section beam, (5.4) From Eq. (5.3), max occurs at Mmax  It is necessary to plot the V and M diagrams along the length of a beam.  to know where Vmax or Mmax occurs!

5.2 Shear and Bending-Moment Diagrams Determining of V and M at selected points of the beam

Sign Conventions The shear is positive (+) when external forces acting on the beam tend to shear off the beam at the point indicated in fig 5.7b 2. The bending moment is positive (+) when the external forces acting on the beam tend to bend the beam at the point indicated in fig 5.7c  Moment

5.3 Relations among Load, Shear and Bending Moment Relations between Load and Shear Hence, (5.5)

VD – VC = area under load curve between C and D Integrating Eq. (5.5) between points C and D (5.6) VD – VC = area under load curve between C and D (5.6’) 1 (5.5’)

Relations between Shear and Bending Moment or  (5.7)

MD – MC = area under shear curve between points C and D (5.7) MD – MC = area under shear curve between points C and D

5.4 Design of Prismatic Beams for Bending -- Design of a beam is controlled by |Mmax| (5.1’,5.3’) Hence, the min allowable value of section modulus is: (5.9)

Question: Where to cut? What are the rules? Answer: whenever there is a discontinuity in the loading conditions, there must be a cut. Reminder: The equations obtained through each cut are only valid to that particular section, not to the entire beam.

5.5 Using Singularity Functions to Determine Shear and Bending Moment in a Beam Beam Constitutive Equations

Notes: Sign Conventions: In this set of equations, +y is going upward and +x is going to the right. 2. Everything going downward is “_”, and upward is +. There is no exception. 3. There no necessity of changing sign for an y integration or derivation.   Sign Conventions: 1. Force going in the +y direction is “+” 2. Moment CW is “+”

Rules for Singularity Functions (5.15) (5.14)

Rule #2: Distributed load w(x) is zero order: e.g. wo<x-a>o 2. Pointed load P(x) is (-1) order: e.g. P<x-a>-1 3. Moment M is (-2) order: e.g. Mo<x-a>-2

Rule #3:

Rule #4: Set up w = w(x) first, by including all forces, from the left to the right of the beam. Integrating w once to obtain V, w/o adding any constants. Integrating V to obtain M, w/o adding any constants. Integrating M to obtain EI , adding an integration constant C1. Integrating EI to obtain EIy, adding another constant C2. 6. Using two boundary conditions to solve for C1 and C2.

(5.11) (5.12) After integration, and observing that We obtain as before (5.13)

(5.16) and (5.17)

Example 5.05

Sample Problem 5.9

5.6 Nonprismatic Beams

Load and Resistance Factor Design

Example 5.03