DOCSLIB.ORG

Introduction to Stiffness Analysis Stiffness Analysis Procedure

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Introduction to Stiffness Analysis Stiffness Analysis Procedure Introduction to Stiffness Analysis displacements. The number of unknowns in the stiffness method of The stiffness method of analysis is analysis is known as the degree of the basis of all commercial kinematic indeterminacy, which structural analysis programs. refers to the number of node/joint Focus of this chapter will be displacements that are unknown development of stiffness equations and are needed to describe the that only take into account bending displaced shape of the structure. deformations, i.e., ignore axial One major advantage of the member, a.k.a. slope-deflection stiffness method of analysis is that method. the kinematic degrees of freedom In the stiffness method of analysis, are well-defined. we write equilibrium equations in 1 2 terms of unknown joint (node) Definitions and Terminology Stiffness Analysis Procedure Positive Sign Convention: Counterclockwise moments and The steps to be followed in rotations along with transverse forces and displacements in the performing a stiffness analysis can positive y-axis direction. be summarized as: 1. Determine the needed displace- Fixed-End Forces: Forces at the “fixed” supports of the kinema- ment unknowns at the nodes/ tically determinate structure. joints and label them d1, d2, …, d in sequence where n = the Member-End Forces: Calculated n forces at the end of each element/ number of displacement member resulting from the unknowns or degrees of applied loading and deformation freedom. of the structure. 3 4 1 2. Modify the structure such that it fixed-end forces are vectorially is kinematically determinate or added at the nodes/joints to restrained, i.e., the identified produce the equivalent fixed-end displacements in step 1 all structure forces, which are equal zero. labeled Pfi for i = 1, 2, …, n later 3. Calculate the member fixed-end in the notes. forces in this kinematically 4. Introduce a unit displacement at restrained state at the nodes/ each displacement degree of joints of the restrained structure freedom identified in step 1 one due to the member applied at a time with all others equal to loads. Tables of member fixed- zero and without any loading on end forces due to member loads the structure, i.e., di = 1 with d1, for the kinematically restrained …, di-1, di+1, …, dn = 0 for i = 1, members are available later in 2, …, n. Sketch the displaced these notes. The member 5 6 structure for each of these cases. 5. Eliminate the error introduced in Determine the member-end forces step 3 to permit the displace- introduced as result of each unit ment at the nodes/joints. This is displacement for the kinematically accomplished by applying the restrained structure. These negative of the forces calculated member-end forces define the in step 3 and defines the kine- member-end stiffness coeffi- matically released structure. cients, i.e., forces per unit 6. Calculate the unknown node/ displacement. joint displacements. The member-end stiffness coefficients are vectorially added 7. Calculate the member-end at the nodes/joints to produce the forces. structure stiffness coefficients, which are labeled S for i = 1, 2, ij 7 8 …, n and j = 1, 2, …, n. 2 Illustration 1: Determine the degree of kinematic indeterminacy. The only To illustrate the stiffness method of unknown node/joint displacement analysis, we will first consider occurs at node B and it is a continuous beam structures. Start rotational displacement. Thus, the off by considering the two-span rotation at node B is labeled d1. beam shown in Figure 1. 2: Kinematically restrain the structure such that the displace- ments identified in step 1 equal zero. See Figure 2. Fig. 1 – Two-Span Continuous Beam Figure 2 – Kinematically Restrained Two-Span Beam of Figure 1 9 10 The heavy vertical line drawn through the horizontal roller support at B signifies that node B is “fixed” against displacement. Thus, the Figure 3 – “Fixed-End” Forces for the rotational displacement d = 0 for Kinematically Restrained Two-Span Beam of 1 Figure 1 the kinematically restrained Since span element (member) A-B structure of Figure 2. is not loaded, it will not produce any 3: Calculate the element/member fixed-end forces. However, fixed-end forces for the kinema- element (member) B-C is loaded tically restrained structure and and the “fixed-end” forces are vectorially add to obtain the fixed- labeled in Figure 3. They are end forces for the structure. simply the support reactions for the “fixed-fixed” beam. 11 12 3 Calculate the fixed-end forces for 4: Impose a unit displacement at the structure by vectorially adding each kinematic degree of freedom the member-end fixed-end forces. (DOF) to establish the structure Figure 4 – Joint Equilibrium at stiffness equations. the Kinematic Degree of Freedom for the Restrained Two-Span Beam of Figure 1 Figure 4 shows that qL2 M0 Pf1 B 12 Pf1 is drawn counterclockwise in Figure 4 since our sign convention is counterclockwise Figure 5 – Kinematically Restrained Two-Span moments are positive. Beam of Figure 1 Subjected to a Unit Displacement d = 1 13 1 14 Figure 5 shows the displaced shape The structure stiffness equations of the two-span beam for d1 = 1 as are expressed as well as the displaced shapes and [S] {d} = {P} – {Pf} member-end stiffness coefficients where [S] is the structure stiffness for the two elements comprising the matrix; {d} is the structure two-span beam of Figure 1. displacement vector; {P} is the Member-end stiffness coefficients applied structure concentrated are defined as the member-end force vector; and {Pf} is the forces resulting from the imposition structure fixed-end force vector of the single unit displacement for calculated in step 3. The applied the structure as shown in Figure 5. structure concentrated force vector Derivation of the member-end {P} lists the point forces for each stiffness coefficients (forces) shown structure displacement DOF. It in Figure 5 and others will be contains nonzero entries only at covered later in the notes. 15 16 4 the displacement DOF where a point force or moment is applied at Sij force at displacement DOF i the corresponding displacement due to a unit displacement at DOF. DOF j (i.e., dj = 1) with all other displacement DOF equal to The structure stiffness matrix zero (i.e., di = 0 for i = 1, …, j-1, coefficients are obtained by j+1, …, n). Stiffness coeffi- performing equilibrium at the nodes cients have units of force/ displacement (or moment/ for each structure DOF using the rotation). member-end stiffness coefficients. These structure stiffness matrix The structure stiffness coefficients are obtained by performing coefficients are designated as Sij and i = 1, 2, …, n and j = 1, 2, …, n. equilibrium calculations at the structure displacement degrees of freedom. 17 18 5: Eliminate the error introduced in For example structure: the kinematically restrained {d} = {d1} = d1 unknown structure: {P} = {0} = 0 [S] {d} = {P} – {Pf} {P } = {P } = qL2/12 f f1 6: Calculate the unknown structure Figure 6 – Equilibrium at Kinematic DOF 1 for displacements the Two-Span Beam of -1 Figure 1 {d} = [S] ({P} – {Pf}) For the example structure: Performing node equilibrium at 2 displacement DOF 1 gives (see d1 = L/8EI (-qL /12) Figure 6) gives = -qL3/96EI AB BC S11 = (4EI/L) + (4EI/L) 7: Calculate the member-end = 8EI/L 19 forces. 20 5 Vfb Mfb = Member Fixed-End {Qfb } Vfe Force Vector Mfe The member fixed-end forces are The beam member stiffness defined as equations can be written as Qfi = Qi in the kinematically determinate {Qbbbbb }4x1 [k ] 4x4 {u } 4x1 {Q f } 4x1 state due to member loading. Vb vb Mb = Member-End Force b {Qb } {ub } = Member-End Ve Vector ve Displacement Vector M e e 21 22 12 6L 12 6L Simplified Member End 22 Force Calculations EI 6L 4L 6L 2L [k ] bb 3 12 6L 12 6L 12EI 6EI L V(vv)()Vb be be fb LL32 6L 2L22 6L 4L (internal shear at beginning = Member Bending Stiffness node b) Matrix 6EI 2EI M(vv)(2)Mb be be fb L2 L The ij member stiffness coefficient (internal moment at beginning can be expressed mathematically node b) as 12EI 6EI V(vv)()Vebebefe32 kQij i LL u1j (internal shear at end node e) 6EI 2EI M(vv)(2)Mebebefe all other uk 0 (k j) L2 L (internal moment at end node e) 23 24 6 qL 6EI In order to apply equation (1) to the 2 16 AB L 2 calculation of element end forces, Vb 2EI qL 3 AB Mb L qL 48 compatibility between element {Q} V 6EI 96EI qL displacements u (i=1,2,3,4) and the e i 2 Me L 16 structure displacements dj (j=1,2, 4EI qL2 …, n) must be established. For the L 24 example beam: 6EI AB BC L2 e1b1d; d;allothers 0 BC Vb 4EI 1 AB 3 {Qfb } {0} BC Mb L qL qL L/6 {Q} BC T Ve 6EI96EI 2 1 {Qfb } qL/2 1L/61 L/6 M L2 L/6 e 2EI Since only one displacement is L T nonzero for each member, the 21qL qL22 27qL 5qL member end forces are 48 24 48 48 25 26 The shear force and bending Mathematical Expression of moment diagrams are given below Stiffness Superposition for the example structure. As stated in step 5, the structure stiffness equations are expressed as [S] {d} + {Pf} = {P} [S] structure stiffness matrix {d} structure displacement vector {Pf} structure fixed-end force vector {P} structure node/joint force vector 27 28 7 Referencing the previous figure on stiffness superposition, the first column of the figure represents the kinematically determinate state, i.e.
Load more

Recommended publications
  • Phys 140A Lecture 8–Elastic Strain, Compliance, and Stiffness
    Lecture 8 Elastic strains, compliance, and stiffness Review for exam Stress: force applied to a unit area Strain: deformation resulting from stress Purpose of today’s derivations: Generalize Hooke’s law to a 3D body that may be subjected to any arbitrary force. We imagine 3 orthogonal vectors 풙̂, 풚̂, 풛̂ embedded in a solid before we have deformed it. After we have deformed the solid, these vectors might be of different length and they might be pointing in different directions. We describe these deformed vectors 풙′, 풚′, 풛′ in the following way: ′ 풙 = (1 + 휖푥푥)풙̂ + 휖푥푦풚̂ + 휖푥푧풛̂ ′ 풚 = 휖푦푥풙̂ + (1 + 휖푦푦)풚̂ + 휖푦푧풛̂ ′ 풛 = 휖푧푥풙̂ + 휖푧푦풚̂ + (1 + 휖푧푧)풛̂ The components 휖훼훽 define the deformation. They are dimensionless and have values much smaller than 1 in most instances in solid state physics. How one ‘reads’ these vectors is (for example) 휖푥푥force along x, deformation along x; 휖푥푦shear xy-plane along x or y direction (depending which unit vector it is next to) The new axes have new lengths given by (for example): ′ ′ 2 2 2 풙 ∙ 풙 = 1 + 2휖푥푥 + 휖푥푥 + 휖푥푦 + 휖푥푧 Usually we are dealing with tiny deformations so 2nd order terms are often dropped. Deformation also changes the volume of a solid. Considering our unit cube, originally it had a volume of 1. After distortion, it has volume 1 + 휖푥푥 휖푥푦 휖푥푧 ′ ′ ′ ′ 푉 = 풙 ∙ 풚 × 풛 = | 휖푦푥 1 + 휖푦푦 휖푦푧 | ≈ 1 + 휖푥푥 + 휖푦푦 + 휖푧푧 휖푧푥 휖푧푦 1 + 휖푧푧 Dilation (훿) is given by: 푉 − 푉′ 훿 ≡ ≅ 휖 + 휖 + 휖 푉 푥푥 푦푦 푧푧 (2nd order terms have been dropped because we are working in a regime of small distortion which is almost always the appropriate one for solid state physics.
    [Show full text]
  • 19.1 Attitude Determination and Control Systems Scott R. Starin
    19.1 Attitude Determination and Control Systems Scott R. Starin, NASA Goddard Space Flight Center John Eterno, Southwest Research Institute In the year 1900, Galveston, Texas, was a bustling direct hit as Ike came ashore. Almost 200 people in the community of approximately 40,000 people. The Caribbean and the United States lost their lives; a former capital of the Republic of Texas remained a tragedy to be sure, but far less deadly than the 1900 trade center for the state and was one of the largest storm. This time, people were prepared, having cotton ports in the United States. On September 8 of received excellent warning from the GOES satellite that year, however, a powerful hurricane struck network. The Geostationary Operational Environmental Galveston island, tearing the Weather Bureau wind Satellites have been a continuous monitor of the gauge away as the winds exceeded 100 mph and world’s weather since 1975, and they have since been bringing a storm surge that flooded the entire city. The joined by other Earth-observing satellites. This weather worst natural disaster in United States’ history—even surveillance to which so many now owe their lives is today—the hurricane caused the deaths of between possible in part because of the ability to point 6000 and 8000 people. Critical in the events that led to accurately and steadily at the Earth below. The such a terrible loss of life was the lack of precise importance of accurately pointing spacecraft to our knowledge of the strength of the storm before it hit. daily lives is pervasive, yet somehow escapes the notice of most people.
    [Show full text]
  • Beam Element Stiffness Matrices
    Beam Element Stiffness Matrices CEE 421L. Matrix Structural Analysis Department of Civil and Environmental Engineering Duke University Henri P. Gavin Fall, 2014 Truss elements carry only axial forces. Beam elements carry shear forces and bending moments. Frame elements carry shear forces, bending moments, and axial forces. This document presents the development of beam element stiffness matrices in local coordinates. 1 A simply supported beam carrying end-moments Consider a simply supported beam resisting moments M1 and M2 applied at its ends. The flexibility relates the end rotations {θ1, θ2} to the end moments {M1,M2}: θ1 F11 F12 M1 = . θ2 F21 F22 M2 The flexibility coefficients, Fij, may be obtained from Castigliano’s 2nd Theo- ∗ rem, θi = ∂U (Mi)/∂Mi. First Column Second Column 2 CEE 421L. Matrix Structural Analysis – Duke University – Fall 2014 – H.P. Gavin The applied moments M1 and M2 are in equilibrium with the reactions forces V1 and V2; V1 = (M1 + M2)/L and V2 = −(M1 + M2)/L M + M x ! x V (x) = 1 2 M(x) = M − 1 + M L 1 L 2 L The total potential energy of a beam with these forces and moments is: 1Z L M 2 1Z L V 2 U = dx + dx 2 0 EI 2 0 G(A/α) By Castigliano’s Theorem, ∂U θ1 = ∂M1 ∂M(x) ∂V (x) Z L M(x) Z L V (x) = ∂M1 dx + ∂M1 dx 0 EI 0 G(A/α) x 2 x x Z L Z L Z L Z L L − 1 dx αdx L L − 1 dx αdx = + M1 + + M2 0 EI 0 GAL2 0 EI 0 GAL2 and ∂U θ2 = ∂M2 ∂M(x) ∂V (x) Z L M(x) Z L V (x) = ∂M2 dx + ∂M2 dx 0 EI 0 G(A/α) x x x 2 Z L Z L Z L Z L L L − 1 dx αdx L dx αdx = + M1 + + M2 0 EI 0 GAL2 0 EI 0 GAL2
    [Show full text]
  • Force/Deflection Relationships
    Introduction to Finite Element Dynamics ⎡ x x ⎤ 1. Element Stiffness Matrices u()x = ⎢1− ⎥ u = []C u (3) ⎣ L L ⎦ 1.1 Bar Element (iii) Derive Strain - Displacement Relationship Consider a bar element shown below. At the two by using Mechanics Theory2 ends of the bar, axially aligned forces [F1, F2] are The axial strain ε(x) is given by the following applied producing deflections [u ,u ]. What is the 1 2 relationship between applied force and deflection? du d ⎡ 1 1 ⎤ ε()x = = ()[]C u = []B u = ⎢− ⎥u (4) dx dx ⎣ L L ⎦ Deformed shape u2 u1 (iv) Derive Stress - Displacement Relationship by using Elasticity Theory The axial stresses σ (x) are given by the following assuming linear elasticity and homogeneity of F1 x F2 material throughout the bar element. Where E is Element Node the material modulus of elasticity. There are typically five key stages in the σ (x)(= Eε x)= E[]B u (5) 1 analysis . (v) Use principle of Virtual Work (i) Conjecture a displacement function There is equilibrium between WI , the internal The displacement function is usually an work done in deforming the bar, and WE , approximation, that is continuous and external work done by the movement of the differentiable, most usually a polynomial. u(x) is applied forces. The bar cross sectional area the axial deflections at any point x along the bar A = ∫∫dydz is assumed constant with respect to x. element. WI = ∫∫∫σ (x).ε (x)(dxdydz = ∫σ x).ε (x)dx∫∫dydz ⎡a1 ⎤ (6) u()x = a1 + a2 x = []1 x ⎢ ⎥ = [N] a (1) = A∫σ ()x .ε ()x dx ⎣a2 ⎦ ⎡F ⎤ W = []u u 1 = uT F (7) E 1 2 ⎢F ⎥ (ii) Express u()x in terms of nodal ⎣ 2 ⎦ displacements by using boundary conditions.
    [Show full text]
  • Deflection of Beams Introduction
    Deflection of Beams Introduction: In all practical engineering applications, when we use the different components, normally we have to operate them within the certain limits i.e. the constraints are placed on the performance and behavior of the components. For instance we say that the particular component is supposed to operate within this value of stress and the deflection of the component should not exceed beyond a particular value. In some problems the maximum stress however, may not be a strict or severe condition but there may be the deflection which is the more rigid condition under operation. It is obvious therefore to study the methods by which we can predict the deflection of members under lateral loads or transverse loads, since it is this form of loading which will generally produce the greatest deflection of beams. Assumption: The following assumptions are undertaken in order to derive a differential equation of elastic curve for the loaded beam 1. Stress is proportional to strain i.e. hooks law applies. Thus, the equation is valid only for beams that are not stressed beyond the elastic limit. 2. The curvature is always small. 3. Any deflection resulting from the shear deformation of the material or shear stresses is neglected. It can be shown that the deflections due to shear deformations are usually small and hence can be ignored. Consider a beam AB which is initially straight and horizontal when unloaded. If under the action of loads the beam deflect to a position A'B' under load or infact we say that the axis of the beam bends to a shape A'B'.
    [Show full text]
  • Mechanics of Materials Chapter 6 Deflection of Beams
    Mechanics of Materials Chapter 6 Deflection of Beams 6.1 Introduction Because the design of beams is frequently governed by rigidity rather than strength. For example, building codes specify limits on deflections as well as stresses. Excessive deflection of a beam not only is visually disturbing but also may cause damage to other parts of the building. For this reason, building codes limit the maximum deflection of a beam to about 1/360 th of its spans. A number of analytical methods are available for determining the deflections of beams. Their common basis is the differential equation that relates the deflection to the bending moment. The solution of this equation is complicated because the bending moment is usually a discontinuous function, so that the equations must be integrated in a piecewise fashion. Consider two such methods in this text: Method of double integration The primary advantage of the double- integration method is that it produces the equation for the deflection everywhere along the beams. Moment-area method The moment- area method is a semigraphical procedure that utilizes the properties of the area under the bending moment diagram. It is the quickest way to compute the deflection at a specific location if the bending moment diagram has a simple shape. The method of superposition, in which the applied loading is represented as a series of simple loads for which deflection formulas are available. Then the desired deflection is computed by adding the contributions of the component loads (principle of superposition). 6.2 Double- Integration Method Figure 6.1 (a) illustrates the bending deformation of a beam, the displacements and slopes are very small if the stresses are below the elastic limit.
    [Show full text]
  • Ch. 8 Deflections Due to Bending
    446.201A (Solid Mechanics) Professor Youn, Byeng Dong CH. 8 DEFLECTIONS DUE TO BENDING Ch. 8 Deflections due to bending 1 / 27 446.201A (Solid Mechanics) Professor Youn, Byeng Dong 8.1 Introduction i) We consider the deflections of slender members which transmit bending moments. ii) We shall treat statically indeterminate beams which require simultaneous consideration of all three of the steps (2.1) iii) We study mechanisms of plastic collapse for statically indeterminate beams. iv) The calculation of the deflections is very important way to analyze statically indeterminate beams and confirm whether the deflections exceed the maximum allowance or not. 8.2 The Moment – Curvature Relation ▶ From Ch.7 à When a symmetrical, linearly elastic beam element is subjected to pure bending, as shown in Fig. 8.1, the curvature of the neutral axis is related to the applied bending moment by the equation. ∆ = = = = (8.1) ∆→ ∆ For simplification, → ▶ Simplification i) When is not a constant, the effect on the overall deflection by the shear force can be ignored. ii) Assume that although M is not a constant the expressions defined from pure bending can be applied. Ch. 8 Deflections due to bending 2 / 27 446.201A (Solid Mechanics) Professor Youn, Byeng Dong ▶ Differential equations between the curvature and the deflection 1▷ The case of the large deflection The slope of the neutral axis in Fig. 8.2 (a) is = Next, differentiation with respect to arc length s gives = ( ) ∴ = → = (a) From Fig. 8.2 (b) () = () + () → = 1 + Ch. 8 Deflections due to bending 3 / 27 446.201A (Solid Mechanics) Professor Youn, Byeng Dong → = (b) (/) & = = (c) [(/)]/ If substitutng (b) and (c) into the (a), / = = = (8.2) [(/)]/ [()]/ ∴ = = [()]/ When the slope angle shown in Fig.
    [Show full text]
  • Fundamentals of Biomechanics Duane Knudson
    Fundamentals of Biomechanics Duane Knudson Fundamentals of Biomechanics Second Edition Duane Knudson Department of Kinesiology California State University at Chico First & Normal Street Chico, CA 95929-0330 USA [email protected] Library of Congress Control Number: 2007925371 ISBN 978-0-387-49311-4 e-ISBN 978-0-387-49312-1 Printed on acid-free paper. © 2007 Springer Science+Business Media, LLC All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Springer Science+Business Media, LLC, 233 Spring Street, New York, NY 10013, USA), except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use in this publication of trade names, trademarks, service marks and similar terms, even if they are not identified as such, is not to be taken as an expression of opinion as to whether or not they are subject to proprietary rights. 987654321 springer.com Contents Preface ix NINE FUNDAMENTALS OF BIOMECHANICS 29 Principles and Laws 29 Acknowledgments xi Nine Principles for Application of Biomechanics 30 QUALITATIVE ANALYSIS 35 PART I SUMMARY 36 INTRODUCTION REVIEW QUESTIONS 36 CHAPTER 1 KEY TERMS 37 INTRODUCTION TO BIOMECHANICS SUGGESTED READING 37 OF UMAN OVEMENT H M WEB LINKS 37 WHAT IS BIOMECHANICS?3 PART II WHY STUDY BIOMECHANICS?5 BIOLOGICAL/STRUCTURAL BASES
    [Show full text]
  • Chapter 10: Elasticity and Oscillations
    Chapter 10 Lecture Outline 1 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 10: Elasticity and Oscillations •Elastic Deformations •Hooke’s Law •Stress and Strain •Shear Deformations •Volume Deformations •Simple Harmonic Motion •The Pendulum •Damped Oscillations, Forced Oscillations, and Resonance 2 §10.1 Elastic Deformation of Solids A deformation is the change in size or shape of an object. An elastic object is one that returns to its original size and shape after contact forces have been removed. If the forces acting on the object are too large, the object can be permanently distorted. 3 §10.2 Hooke’s Law F F Apply a force to both ends of a long wire. These forces will stretch the wire from length L to L+L. 4 Define: L The fractional strain L change in length F Force per unit cross- stress A sectional area 5 Hooke’s Law (Fx) can be written in terms of stress and strain (stress strain). F L Y A L YA The spring constant k is now k L Y is called Young’s modulus and is a measure of an object’s stiffness. Hooke’s Law holds for an object to a point called the proportional limit. 6 Example (text problem 10.1): A steel beam is placed vertically in the basement of a building to keep the floor above from sagging. The load on the beam is 5.8104 N and the length of the beam is 2.5 m, and the cross-sectional area of the beam is 7.5103 m2.
    [Show full text]
  • Glossary of Materials Engineering Terminology
    Glossary of Materials Engineering Terminology Adapted from: Callister, W. D.; Rethwisch, D. G. Materials Science and Engineering: An Introduction, 8th ed.; John Wiley & Sons, Inc.: Hoboken, NJ, 2010. McCrum, N. G.; Buckley, C. P.; Bucknall, C. B. Principles of Polymer Engineering, 2nd ed.; Oxford University Press: New York, NY, 1997. Brittle fracture: fracture that occurs by rapid crack formation and propagation through the material, without any appreciable deformation prior to failure. Crazing: a common response of plastics to an applied load, typically involving the formation of an opaque banded region within transparent plastic; at the microscale, the craze region is a collection of nanoscale, stress-induced voids and load-bearing fibrils within the material’s structure; craze regions commonly occur at or near a propagating crack in the material. Ductile fracture: a mode of material failure that is accompanied by extensive permanent deformation of the material. Ductility: a measure of a material’s ability to undergo appreciable permanent deformation before fracture; ductile materials (including many metals and plastics) typically display a greater amount of strain or total elongation before fracture compared to non-ductile materials (such as most ceramics). Elastic modulus: a measure of a material’s stiffness; quantified as a ratio of stress to strain prior to the yield point and reported in units of Pascals (Pa); for a material deformed in tension, this is referred to as a Young’s modulus. Engineering strain: the change in gauge length of a specimen in the direction of the applied load divided by its original gauge length; strain is typically unit-less and frequently reported as a percentage.
    [Show full text]
  • On the Relationship Between Indenation Hardness and Modulus, and the Damage Resistance of Biological Materials
    bioRxiv preprint doi: https://doi.org/10.1101/107284; this version posted February 9, 2017. The copyright holder for this preprint (which was not certified by peer review) is the author/funder, who has granted bioRxiv a license to display the preprint in perpetuity. It is made available under aCC-BY 4.0 International license. On the relationship between indenation hardness and modulus, and the damage resistance of biological materials. David Labontea,∗, Anne-Kristin Lenzb, Michelle L. Oyena aThe NanoScience Centre, Department of Engineering, Cambridge, UK bUniversity of Applied Sciences, Bremen, Germany Abstract The remarkable mechanical performance of biological materials is based on intricate structure-function relation- ships. Nanoindentation has become the primary tool for characterising biological materials, as it allows to relate structural changes to variations in mechanical properties on small scales. However, the respective theoretical back- ground and associated interpretation of the parameters measured via indentation derives largely from research on ‘traditional’ engineering materials such as metals or ceramics. Here, we discuss the functional relevance of inden- tation hardness in biological materials by presenting a meta-analysis of its relationship with indentation modulus. Across seven orders of magnitude, indentation hardness was directly proportional to indentation modulus, illus- trating that hardness is not an independent material property. Using a lumped parameter model to deconvolute indentation hardness into components arising from reversible and irreversible deformation, we establish crite- ria which allow to interpret differences in indentation hardness across or within biological materials. The ratio between hardness and modulus arises as a key parameter, which is a proxy for the ratio between irreversible and reversible deformation during indentation, and the material’s yield strength.
    [Show full text]
  • How Beam Deflection Works?
    How Beam Deflection Works? Beams are in every home and in one sense are quite simple. The weight placed on them causes them to bend (deflect), and you just need to make sure they don’t deflect too much by choosing a large enough beam. As shown in the diagram below, a beam that is deflecting takes on a curved shape that you might mistakenly assume to be circular. So how do we figure out what shape the curvature takes in order to predict how much a beam will deflect under load? After all you cannot wait until the house is built to find out the beam deflects too much. A beam is generally considered to fail if it deflects more than 1 inch. Solution to How Deflection Works: The model for SIMPLY SUPPORTED beam deflection Initial conditions: y(0) = y(L) = 0 Simply supported is engineer-speak for it just sits on supports without any further attachment that might help with the deflection we are trying to avoid. The dashed curve represents the deflected beam of length L with left side at the point (0,0) and right at (L,0). We consider a random point on the curve P = (x,y) where the y-coordinate would be the deflection at a distance of x from the left end. Moment in engineering is a beams desire to rotate and is equal to force times distance M = Fd. If the beam is in equilibrium (not breaking) then the moment must be equal on each side of every point on the beam.
    [Show full text]