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Published byNigel Lambert Modified over 9 years ago
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Chapter 5 Analysis and Design of Beams for Bending
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5.1 Introduction -- Dealing with beams of different materials:
steel, aluminum, wood, etc. -- Loading: transverse loads Concentrated loads Distributed loads
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-- Supports Simply supported Cantilever Beam Overhanging
Continuous Fixed Beam
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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:
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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
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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!
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5.2 Shear and Bending-Moment Diagrams
Determining of V and M at selected points of the beam
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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
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5.3 Relations among Load, Shear
and Bending Moment Relations between Load and Shear Hence, (5.5)
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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’)
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Relations between Shear and Bending Moment
or (5.7)
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MD – MC = area under shear curve between points C and D
(5.7) MD – MC = area under shear curve between points C and D
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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)
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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.
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5.5 Using Singularity Functions to Determine Shear and Bending Moment in a Beam
Beam Constitutive Equations
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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 “+”
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Rules for Singularity Functions
(5.15) (5.14)
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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
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Rule #3:
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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.
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(5.11) (5.12) After integration, and observing that We obtain as before (5.13)
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(5.16) and (5.17)
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Example 5.05
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Sample Problem 5.9
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5.6 Nonprismatic Beams
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Load and Resistance Factor Design
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Example 5.03
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