Cantilever Beams Part 1 - Beam Stiffness
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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. -
DEFINITIONS Beams and Stringers (B&S) Beams and Stringers Are
DEFINITIONS Beams and Stringers (B&S) Beams and stringers are primary longitudinal support members, usually rectangular pieces that are 5.0 or more in. thick, with a depth more than 2.0 in. greater than the thickness. B&S are graded primarily for use as beams, with loads applied to the narrow face. Bent. A type of pier consisting of two or more columns or column-like components connected at their top ends by a cap, strut, or other component holding them in their correct positions. Camber. The convex curvature of a beam, typically used in glulam beams. Cantilever. A horizontal member fixed at one end and free at the other. Cap. A sawn lumber or glulam component placed horizontally on an abutment or pier to distribute the live load and dead load of the superstructure. Clear Span. Inside distance between the faces of support. Connector. Synonym for fastener. Crib. A structure consisting of a foundation grillage and a framework providing compartments that are filled with gravel, stones, or other material satisfactory for supporting the structure to be placed thereon. Check. A lengthwise separation of the wood that usually extends across the rings of annual growth and commonly results from stresses set up in wood during seasoning. Creep. Time dependent deformation of a wood member under sustained load. Dead Load. The structure’s self weight. Decay. The decomposition of wood substance by fungi. Some people refer to it as “rot”. Decking. A subcategory of dimension lumber, graded primarily for use with the wide face placed flatwise. Delamination. The separation of layers in laminated wood or plywood because of failure of the adhesive, either within the adhesive itself or at the interface between the adhesive and the adhered. -
Beam Structures and Internal Forces
ENDS 231 Note Set 13 S2008abn Beam Structures and Internal Forces • BEAMS - Important type of structural members (floors, bridges, roofs) - Usually long, straight and rectangular - Have loads that are usually perpendicular applied at points along the length Internal Forces 2 • Internal forces are those that hold the parts of the member together for equilibrium - Truss members: F A B F F A F′ F′ B F - For any member: T´ F = internal axial force (perpendicular to cut across section) V = internal shear force T´ (parallel to cut across section) T M = internal bending moment V Support Conditions & Loading V • Most often loads are perpendicular to the beam and cause only internal shear forces and bending moments M • Knowing the internal forces and moments is necessary when R designing beam size & shape to resist those loads • Types of loads - Concentrated – single load, single moment - Distributed – loading spread over a distance, uniform or non-uniform. 1 ENDS 231 Note Set 13 S2008abn • Types of supports - Statically determinate: simply supported, cantilever, overhang L (number of unknowns < number of equilibrium equations) Propped - Statically indeterminate: continuous, fixed-roller, fixed-fixed (number of unknowns < number of equilibrium equations) L Sign Conventions for Internal Shear and Bending Moment Restrained (different from statics and truss members!) V When ∑Fy **excluding V** on the left hand side (LHS) section is positive, V will direct down and is considered POSITIVE. M When ∑M **excluding M** about the cut on the left hand side (LHS) section causes a smile which could hold water (curl upward), M will be counter clockwise (+) and is considered POSITIVE. -
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. -
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 -
A Simple Beam Test: Motivating High School Teachers to Develop Pre-Engineering Curricula
Session 2326 A Simple Beam Test: Motivating High School Teachers to Develop Pre-Engineering Curricula Eric E. Matsumoto, John R. Johnston, E. Edward Dammel, S.K. Ramesh California State University, Sacramento Abstract The College of Engineering and Computer Science at California State University, Sacramento has developed a daylong workshop for high school teachers interested in developing and teaching pre-engineering curricula. Recent workshop participants from nine high schools performed “hands-on” laboratory experiments that can be implemented at the high school level to introduce basic engineering principles and technology and to inspire students to study engineering. This paper describes one experiment that introduces fundamental structural engineering concepts through a simple beam test. A load is applied at the center of a beam using weights, and the resulting midspan deflection is measured. The elastic stiffness of the beam is determined and compared to published values for various beam materials and cross sectional shapes. Beams can also be tested to failure. This simple and inexpensive experiment provides a useful springboard for discussion of important engineering topics such as elastic and inelastic behavior, influence of materials and structural shapes, stiffness, strength, and failure modes. Background engineering concepts are also introduced to help high school teachers understand and implement the experiment. Participants rated the workshop highly and several teachers have already implemented workshop experiments in pre-engineering curricula. I. Introduction The College of Engineering and Computer Science at California State University, Sacramento has developed an active outreach program to attract students to the College and promote engineering education. In partnership with the Sacramento Engineering and Technology Regional Consortium1 (SETRC), the College has developed a daylong workshop for high school teachers interested in developing and teaching pre-engineering curricula. -
I-Beam Cantilever Racks Meet the Latest Addition to Our Quick Ship Line
48 HOUR QUICK SHIP Maximize storage and improve accessibility I-Beam cantilever racks Meet the latest addition to our Quick Ship line. Popular for their space-saving design, I-Beam cantilever racks can allow accessibility from both sides, allowing for faster load and unload times. Their robust construction reduces fork truck damage. Quick Ship I-beam cantilever racks offer: • 4‘ arm length, with 4” vertical adjustability • Freestanding heights of 12’ and 16’ • Structural steel construction with a 50,000 psi minimum yield • Heavy arm connector plate • Bolted base-to-column connection I-Beam Cantilever Racks can be built in either single- or double-sided configurations. How to design your cantilever rack systems 1. Determine the number and spacing of support arms. 1a The capacity of each 4’ arm is 2,600#, so you will need to make sure that you 1b use enough arms to accommodate your load. In addition, you can test for deflection by using wood blocks on the floor under the load. 1c Use enough arms under a load to prevent deflection of the load. Deflection causes undesirable side pressure on the arms. If you do not detect any deflection with two wood blocks, you may use two support arms. Note: Product should overhang the end of the rack by 1/2 of the upright centerline distance. If you notice deflection, try three supports. Add supports as necessary until deflection is eliminated. Loading without overhang is incorrect. I-Beam cantilever racks WWW.STEELKING.COM 2. Determine if Quick Ship I-Beam arm length is appropriate for your load. -
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. -
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. -
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'. -
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. -
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.