DOCSLIB.ORG

Beam Deflection by Integration.Pptx

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Beam Deflection by Integration.Pptx 14 January 2011 Mechanics of Materials CIVL 3322 / MECH 3322 Deflection of Beams The Elastic Curve ¢ The deflection of a beam must often be limited in order to provide integrity and stability of a structure or machine, or ¢ To prevent any attached brittle materials from cracking 2 Beam Deflection by Integration 1 14 January 2011 The Elastic Curve ¢ Deflections at specific points on a beam must be determined in order to analyze a statically indeterminate system. 3 Beam Deflection by Integration The Elastic Curve ¢ The curve that is formed by the plotting the position of the centroid of the beam along the longitudal axis is known as the elastic curve. 4 Beam Deflection by Integration 2 14 January 2011 The Elastic Curve ¢ At different types of supports, information that is used in developing the elastic curve are provided l Supports which resist a force, such as a pin, restrict displacement l Supports which resist a moment, such as a fixed end support, resist displacement and rotation or slope 5 Beam Deflection by Integration The Elastic Curve 6 Beam Deflection by Integration 3 14 January 2011 The Elastic Curve 7 Beam Deflection by Integration The Elastic Curve ¢ We can derive an expression for the curvature of the elastic curve at any point where ρ is the radius of curvature of the elastic curve at the point in question 1 M = ρ EI 8 Beam Deflection by Integration 4 14 January 2011 The Elastic Curve ¢ If you make the assumption to deflections are very small and that the slope of the elastic curve at any point is very small, the curvature can be approximated at any point by 2 v is the deflection of d v M the elastic curve = dx2 EI 9 Beam Deflection by Integration The Elastic Curve ¢ We can rearrange terms d 2v EI = M dx2 10 Beam Deflection by Integration 5 14 January 2011 The Elastic Curve ¢ Differentiate both sides with respect to x d ⎛ d 2v⎞ dM EI = = V(x) dx ⎝⎜ dx2 ⎠⎟ dx 11 Beam Deflection by Integration The Elastic Curve ¢ And again differentiate both sides with respect to x d 2 ⎛ d 2v⎞ dV EI = = w(x) dx2 ⎝⎜ dx2 ⎠⎟ dx 12 Beam Deflection by Integration 6 14 January 2011 The Elastic Curve d 4v ¢ So there are EI 4 = w(x) three paths to dx finding v d 3v EI = V(x) dx3 d 2v EI = M (x) dx2 13 Beam Deflection by Integration Boundary Conditions 14 Beam Deflection by Integration 7 14 January 2011 Boundary Conditions 15 Beam Deflection by Integration Boundary Conditions 16 Beam Deflection by Integration 8 14 January 2011 Boundary Conditions 17 Beam Deflection by Integration Boundary Conditions 18 Beam Deflection by Integration 9 14 January 2011 Boundary Conditions 19 Beam Deflection by Integration Combining Load Conditions 20 Beam Deflection by Integration 10 14 January 2011 Cantilever Example ¢ Given a cantilevered beam with a fixed end support at the right end and a load P applied at the left end of the beam. ¢ The beam has a length of L. 21 Beam Deflection by Integration Cantilever Example ¢ If we define x as the distance to the right from the applied load P, then the moment function at any distance x is given as M (x) = −Px 22 Beam Deflection by Integration 11 14 January 2011 Cantilever Example ¢ Since we have a function for M along the beam we can use the expression relating the moment and the deflection M (x) = −Px d 2v EI = M(x) dx2 d 2v EI 2 = −Px 23 Beam Deflection by Integration dx Cantilever Example ¢ Isolating the variables and integrating ⎛ d 2v⎞ EI = −Px ⎝⎜ dx2 ⎠⎟ ⎛ dv⎞ Px2 EI = − + C ⎝⎜ dx ⎠⎟ 2 1 24 Beam Deflection by Integration 12 14 January 2011 Cantilever Example ¢ Integrating again ⎛ dv⎞ Px2 EI = − + C ⎝⎜ dx⎠⎟ 2 1 Px3 EIv = − + C x + C 6 1 2 25 Beam Deflection by Integration Cantilever Example ¢ To be able to define the function for v, we need to evaluate C1 and C2 ⎛ dv⎞ Px2 EI = − + C ⎝⎜ dx⎠⎟ 2 1 Px3 EIv = − + C x + C 6 1 2 26 Beam Deflection by Integration 13 14 January 2011 Cantilever Example ¢ The right end of the beam is supported by a fixed end support therefore the slope of the deflection curve is 0 and the deflection is 0 ⎛ dv⎞ Px2 EI = − + C ⎝⎜ dx⎠⎟ 2 1 Px3 EIv = − + C x + C 6 1 2 27 Beam Deflection by Integration Cantilever Example ¢ In terms of boundary conditions this means ⎛ dv⎞ Px2 EI = − + C dv ⎝⎜ dx⎠⎟ 2 1 x = L : = 0 dx Px3 EIv = − + C x + C x = L :v = 0 6 1 2 28 Beam Deflection by Integration 14 14 January 2011 Cantilever Example ¢ Evaluating the expressions at the boundary conditions ⎛ dv⎞ Px2 dv EI ⎜ ⎟ = − + C1 x = L : = 0 ⎝ dx ⎠ 2 dx 2 2 PL PL x = L :v = 0 EI (0) = − + C1 ⇒ C1 = 2 2 Px3 PL2 EIv = − + x + C 6 2 2 PL3 PL2 PL3 EI (0) = − + L + C2 ⇒ C2 = − 6 2 3 29 Beam Deflection by Integration Cantilever Example ¢ So the expression for the slope (θ) and the deflection (v) are given by 1 ⎛ Px2 PL2 ⎞ dv θ = ⎜ − + ⎟ x = L : = 0 EI ⎝ 2 2 ⎠ dx P θ = L2 − x2 x = L :v = 0 2EI ( ) 1 ⎛ Px3 PL2 PL3 ⎞ v = − + x − EI ⎝⎜ 6 2 3 ⎠⎟ P v = −x3 + 3xL2 − 2L3 6EI ( ) 30 Beam Deflection by Integration 15 14 January 2011 Combining Load Conditions 31 Beam Deflection by Integration Example 1 32 Beam Deflection by Integration 16 14 January 2011 Example 1 33 Beam Deflection by Integration Homework ¢ Show a plot of the shear, bending moment, slope, and deflection curves identifying the maximum, minimum, and zero points for each curve. Use separate plots for each function. ¢ Show the mathematical expression(s) for each function. ¢ Problem P10.4 ¢ Problem P10.8 ¢ Problem P10.11 34 Beam Deflection by Integration 17 .
Load more

Recommended publications
  • 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.
    [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]
  • 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]
  • Chapter 4 Deflection and Stiffness
    Chapter 4 Deflection and Stiffness Faculty of Engineering Mechanical Dept. Chapter Outline Shigley’s Mechanical Engineering Design Spring Rate Elasticity – property of a material that enables it to regain its original configuration after deformation Spring – a mechanical element that exerts a force when deformed Nonlinear Nonlinear Linear spring stiffening spring softening spring Fig. 4–1 Shigley’s Mechanical Engineering Design Spring Rate Relation between force and deflection, F = F(y) Spring rate For linear springs, k is constant, called spring constant Shigley’s Mechanical Engineering Design Tension, Compression, and Torsion Total extension or contraction of a uniform bar in tension or compression Spring constant, with k = F/d Shigley’s Mechanical Engineering Design Tension, Compression, and Torsion Angular deflection (in radians) of a uniform solid or hollow round bar subjected to a twisting moment T Converting to degrees, and including J = pd4/32 for round solid Torsional spring constant for round bar Shigley’s Mechanical Engineering Design Deflection Due to Bending Curvature of beam subjected to bending moment M From mathematics, curvature of plane curve Slope of beam at any point x along the length If the slope is very small, the denominator of Eq. (4-9) approaches unity. Combining Eqs. (4-8) and (4-9), for beams with small slopes, Shigley’s Mechanical Engineering Design Deflection Due to Bending Recall Eqs. (3-3) and (3-4) Successively differentiating Shigley’s Mechanical Engineering Design Deflection Due to Bending 0.5 m (4-10) (4-11) (4-12) (4-13) (4-14) Fig. 4–2 Shigley’s Mechanical Engineering Design Example 4-1 Fig.
    [Show full text]
  • Bending Analysis of Simply Supported and Clamped Circular Plate P
    SSRG International Journal of Civil Engineering (SSRG-IJCE) Volume 2 Issue 5–May 2015 Bending Analysis of Simply Supported and Clamped Circular Plate P. S. Gujar 1, K. B. Ladhane 2 1Department of Civil Engineering, Pravara Rural Engineering College, Loni,Ahmednagar-413713(University of Pune), Maharashtra, India. 2Associate Professor, Department of Civil Engineering, Pravara Rural Engineering College, Loni,Ahmednagar- 413713(University of Pune), Maharashtra, India. ABSTRACT : The aim of study is static bending appropriate plate theory. The stresses in the plate can analysis of an isotropic circular plate using be calculated from these deflections. Once the analytical method i.e. Classical Plate Theory and stresses are known, failure theories can be used to Finite Element software ANSYS. Circular plate determine whether a plate will fail under a given analysis is done in cylindrical coordinate system by load [1]. using Classical Plate Theory. The axisymmetric Vanam et al.[2] done static analysis of an bending of circular plate is considered in the present isotropic rectangular plate using finite element study. The diameter of circular plate, material analysis (FEA).The aim of study was static bending properties like modulus of elasticity (E), poissons analysis of an isotropic rectangular plate with ratio (µ) and intensity of loading is assumed at the various boundary conditions and various types of initial stage of research work. Both simply supported load applications. In this study finite element and clamped boundary conditions subjected to analysis has been carried out for an isotropic uniformly distributed load and center concentrated / rectangular plate by considering the master element point load have been considered in the present as a four noded quadrilateral element.
    [Show full text]
  • Explicit Analysis of Large Transformation of a Timoshenko Beam : Post-Buckling Solution, Bifurcation and Catastrophes Marwan Hariz, Loïc Le Marrec, Jean Lerbet
    Explicit analysis of large transformation of a Timoshenko beam : Post-buckling solution, bifurcation and catastrophes Marwan Hariz, Loïc Le Marrec, Jean Lerbet To cite this version: Marwan Hariz, Loïc Le Marrec, Jean Lerbet. Explicit analysis of large transformation of a Timoshenko beam : Post-buckling solution, bifurcation and catastrophes. Acta Mechanica, Springer Verlag, 2021, 10.1007/s00707-021-02993-8. hal-03306350 HAL Id: hal-03306350 https://hal.archives-ouvertes.fr/hal-03306350 Submitted on 29 Jul 2021 HAL is a multi-disciplinary open access L’archive ouverte pluridisciplinaire HAL, est archive for the deposit and dissemination of sci- destinée au dépôt et à la diffusion de documents entific research documents, whether they are pub- scientifiques de niveau recherche, publiés ou non, lished or not. The documents may come from émanant des établissements d’enseignement et de teaching and research institutions in France or recherche français ou étrangers, des laboratoires abroad, or from public or private research centers. publics ou privés. Explicit analysis of large transformation of a Timoshenko beam : Post-buckling solution, bifurcation and catastrophes Marwan Hariza,1, Lo¨ıc Le Marreca,2, Jean Lerbetb,3 aUniv Rennes, CNRS, IRMAR - UMR 6625, F-35000 Rennes, France bUniversit´eParis-Saclay, CNRS, Univ Evry, Laboratoire de Math´ematiques et Mod´elisation d’Evry, 91037, Evry-Courcouronnes, France Abstract This paper exposes full analytical solutions of a plane, quasi-static but large transformation of a Timoshenko beam. The problem isfirst re-formulated in the form of a Cauchy initial value problem where load (force and moment) is prescribed at one-end and kinematics (translation, rotation) at the other one.
    [Show full text]
  • 7.4 the Elementary Beam Theory
    Section 7.4 7.4 The Elementary Beam Theory In this section, problems involving long and slender beams are addressed. As with pressure vessels, the geometry of the beam, and the specific type of loading which will be considered, allows for approximations to be made to the full three-dimensional linear elastic stress-strain relations. The beam theory is used in the design and analysis of a wide range of structures, from buildings to bridges to the load-bearing bones of the human body. 7.4.1 The Beam The term beam has a very specific meaning in engineering mechanics: it is a component that is designed to support transverse loads, that is, loads that act perpendicular to the longitudinal axis of the beam, Fig. 7.4.1. The beam supports the load by bending only. Other mechanisms, for example twisting of the beam, are not allowed for in this theory. applied force applied pressure cross section roller support pin support Figure 7.4.1: A supported beam loaded by a force and a distribution of pressure It is convenient to show a two-dimensional cross-section of the three-dimensional beam together with the beam cross section, as in Fig. 7.4.1. The beam can be supported in various ways, for example by roller supports or pin supports (see section 2.3.3). The cross section of this beam happens to be rectangular but it can be any of many possible shapes. It will assumed that the beam has a longitudinal plane of symmetry, with the cross section symmetric about this plane, as shown in Fig.
    [Show full text]
  • Teaching Engineering Reasoning Using a Beam Deflection Lab
    AC 2010-2146: TEACHING ENGINEERING REASONING USING A BEAM DEFLECTION LAB Natasha Smith, University of Southern Indiana Page 15.1173.1 Page © American Society for Engineering Education, 2010 Teaching Engineering Reasoning using a Beam Deflection Lab Abstract Well crafted laboratories reinforce theoretical concepts presented in class, but also sharpen students’ technical reasoning skills and provide practice in technical communication. This paper presents an introductory mechanics laboratory on beam deflection, suitable for freshmen engineering courses or as an opening week experiment for Strengths of Materials. The lab consists of 4 distinct experiments, each requiring students to measure maximum deflection of a simply supported beam. In each experiment, a single variable is adjusted (first load, then span, then cross-sectional geometry, and finally material). Although the procedure is likely similar to laboratories typically conducted in Mechanics courses after simple beam theory has been presented, the author suggests alternative objectives and student requirements targeted toward a less experienced audience. The primary goal of the exercise is to model engineering reasoning and begin a discussion on what makes for sound analysis. In conjunction with this goal, a standard for technical reporting also begins to form. To fulfill a secondary purpose, a simple addendum to the experiment is useful to show students the differences between the questions addressed by Statics (i.e. forces on rigid bodies) and those they will face in Strengths of Materials. The final objective is to demonstrate how stiffness is affected by both geometry and material, highlighting implications for design. The lab has been conducted in various forms in both an introductory design course and as a first lab in Strengths of Materials.
    [Show full text]
  • Minimizing Deflection and Bending Moment in a Beam with End Supports
    MINIMIZING DEFLECTION AND BENDING MOMENT IN A BEAM WITH END SUPPORTS Samir V. Amiouny John J. Bartholdi, III John H. Vande Vate School of Industrial and Systems Engineering Georgia Institute of Technology Atlanta, Georgia 30332, USA December 3, 1991; revised April 3, 2003 Abstract We give heuristics to sequence blocks on a beam, like books on a bookshelf, to minimize simultaneously the maximum deflection and the maximum bending moment of the beam. For a beam with simple supports at the ends, one heuristic places the blocks so that the maximum deflection √ is no more than 16/9 3 ≈ 1.027 times the theoretical minimum and the maximum bending moment is within 4 times the minimum. Another heuristic allows maximum deflection up to 2.054 times the theoretical minimum but restricts the maximum bending moment to within 2 times the minimum. Similar results hold for beams with fixed supports at the ends. Key words: combinatorial mechanics, heuristics, sequencing, beam, de- flection, bending moment 1 1 Introduction The limiting factor in the design of beams loaded by weights is often either the permissible deflection [6] or the bending moment [8]. For given loads, the positions at which they are placed determine the deflection and bending moment along the beam; therefore deflection and bending moment can be controlled to some extent by judicious sequencing of the loads. Unfortunately, as we shall show, it can be computationally difficult to determine the best sequence of loads on the beam; however we give fast heuristics that position the loads so that both the maximum deflection and the maximum bending moment are guaranteed to be not “too much” larger than the minimum possible.
    [Show full text]