...Chapter 4 Shear Forces and Bending Moments 4.1 Introduction Consider a beam subjected to transverse loads as shown in figure, the deflections occur in the plane same as the loading plane, is called the plane of bending. In this chapter we discuss shear forces and bending moments in beams related to the loads. 4.2 Types of Beams, Loads, and Reactions Type of beams a. simply supported beam (simple beam) b. cantilever beam (fixed end beam) c. beam with an overhang 2 Type of loads a. concentrated load (single force) b. distributed load (measured by their intensity) : uniformly distributed load (uniform load) linearly varying load c. couple Reactions consider the loaded beam in figure equation of equilibrium in horizontal direction Fx = 0 HA - P1 cos = 0 HA = P1 cos MB = 0 - RA L + (P1 sin ) (L - a) + P2 (L - b) + q c2 / 2 = 0 (P1 sin ) (L - a) P2 (L - b) q c2 RA = CCCCCCC + CCCC + CC L L 2 L (P1 sin ) a P2 b q c2 RB = CCCCC + CC + CC L L 2 L for the cantilever beam Fx = 0 HA = 5 P3 / 13 12 P3 (q1 + q2) b Fy = 0 RA = CC + CCCCC 13 2 3 12 P3 q1b q1 b MA = 0 MA = CC + CC (L – 2b/3) + CC (L – b/3) 13 2 2 for the overhanging beam MB = 0 - RA L + P4 (L– a) + M1 = 0 MA = 0 - P4 a + RB L + M1 = 0 P4 (L– a) + M1 P4 a - M1 RA = CCCCCC RB = CCCC L L 4.3 Shear Forces and Bending Moments Consider a cantilever beam with a concentrated load P applied at the end A, at the cross section mn, the shear force and bending moment are found Fy =...
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...Mazurek Lecture Notes: Brock E. Barry U.S. Military Academy Pure Bending Copyright © 2015 McGraw-Hill Education. Permission required for reproduction or display. MECHANICS OF MATERIALS Pure Bending Seventh Edition Beer • Johnston • DeWolf • Mazurek Pure Bending: Prismatic members subjected to equal and opposite couples acting in the same longitudinal plane Fig. 4.2 (a) Free-body diagram of the barbell pictured in the chapter opening photo and (b) Free-body diagram of the center bar portion showing pure bending. Copyright © 2015 McGraw-Hill Education. Permission required for reproduction or display. 4-2 MECHANICS OF MATERIALS Other Loading Types Seventh Edition Beer • Johnston • DeWolf • Mazurek Fig. 4.3 (a) Free-body diagram of a clamp, (b) freebody diagram of the upper portion of the clamp. • Eccentric Loading: Axial loading which does not pass through section centroid produces internal forces equivalent to an axial force and a couple • Transverse Loading: Concentrated or distributed transverse load produces internal forces equivalent to a shear force and a couple • Principle of Superposition: The normal stress due to pure bending may be combined with the normal stress due to axial loading and shear stress due to shear loading to find the complete state of stress. 4-3 Fig. 4.4 (a) Cantilevered beam with end loading. (b) As portion AC shows, beam is not in pure bending. Copyright © 2015 McGraw-Hill Education. Permission required...
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...06 Solutions 46060_Part1 5/27/10 3:51 PM Page 329 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 6–1. Draw the shear and moment diagrams for the shaft. The bearings at A and B exert only vertical reactions on the shaft. A B 250 mm 800 mm 24 kN 6–2. Draw the shear and moment diagrams for the simply supported beam. M A 2m 2m 2 kN m 4 kN B 2m 329 06 Solutions 46060_Part1 5/27/10 3:51 PM Page 330 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 6–3. The engine crane is used to support the engine, which has a weight of 1200 lb. Draw the shear and moment diagrams of the boom ABC when it is in the horizontal position shown. a + ©MA = 0; 4 F (3) - 1200(8) = 0; 5 A -Ay + Ax 4 (4000) - 1200 = 0; 5 A 3 ft B 5 ft C FA = 4000 lb 4 ft + c ©Fy = 0; + ; ©Fx = 0; Ay = 2000 lb 3 (4000) = 0; 5 Ax = 2400 lb *6–4. Draw the shear and moment diagrams for the cantilever beam. 2 kN/m A 2m 6 kN m The free-body diagram of the beam’s right segment sectioned through an arbitrary point shown in Fig. a will be used to write...
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...mraChapter 9 9.1 Introduction Deflections of Beams in this chapter, we describe methods for determining the equation of the deflection curve of beams and finding deflection and slope at specific points along the axis of the beam 9.2 Differential Equations of the Deflection Curve consider a cantilever beam with a concentrated load acting upward at the free end the deflection in the y v is the displacement direction of the axis the angle of rotation (also called slope) is the angle between the x axis and the tangent to the deflection curve point m1 is located at distance x point m2 is located at distance x + dx slope at slope at m1 m2 is is +d denote O' the center of curvature and the radius of curvature, then d = ds is and the curvature 1 = 1 C = d C ds the sign convention is pictured in figure slope of the deflection curve dv C dx for = tan or ds j dx d C dx d C = dx = cos j 1 and d 2v CC dx2 = dv tan-1 C dx tan dv C dx j , then small = 1 C = 1 C = = if the materials of the beam is linear elastic = 1 C = M C EI [chapter 5] then the differential equation of the deflection curve is obtained d C dx d2v = CC dx2 = M C EI and v dV CC dx d 4v CC dx4 = -q q -C EI it can be integrated to find ∵ dM CC dx d 3v CC dx3 = V then V = C EI = 2 sign conventions for M, V and q are shown the above equations can be written in a simple form EIv" = M EIv"' = V EIv"" = -q this equations are valid only when...
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...forces. Stress is the lead to accurately describe and predict the elastic deformation of a body. Simple stress can be classified as normal stress, shear stress, and bearing stress. Normal stress develops when a force is applied perpendicular to the cross-sectional area of the material. If the force is going to pull the material, the stress is said to be tensile stress and compressive stress develops when the material is being compressed by two opposing forces. Shear stress is developed if the applied force is parallel to the resisting area. Example is the bolt that holds the tension rod in its anchor. Another condition of shearing is when we twist a bar along its longitudinal axis. This type of shearing is called torsion and covered in Chapter 3. Another type of simple stress is the bearing stress, it is the contact pressure between two bodies. Suspension bridges are good example of structures that carry these stresses. The weight of the vehicle is carried by the bridge deck and passes the force to the stringers (vertical cables), which in turn, supported by the main suspension cables. The suspension cables then transferred the force into bridge towers. Normal Stress Stress Stress is the expression of force applied to a unit area of surface. It is measured in psi (English unit) or in MPa (SI unit). Another unit of stress which is not commonly used is the dynes (cgs unit). Stress is the ratio of force over area. stress = force / area Simple Stresses There are three types of...
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...Experiment 3 Bending test – tensile strength Objective: 1. To investigate the relationship between load, span, width, height and deflection of a beam, placed on two bear affected by a concentrated load at the center. 2. To ascertain the coefficient of elasticity for steel, brass, aluminum and wood. Theory The stress-strain behavior of brittle materials (e.g. ceramic, low toughness composite material) is not usually ascertained by tensile tests as outline in Exp. 1. A more suitable transverse bending test is most frequently employed, in which a rod specimen either a circular or rectangular cross section is bent until fracture using a three- or four-point loading technique. The assessments are conducted according to ASTM Standard C 1161, “Standard Test Method for Flexural Strength of Advanced Ceramics at Ambient Temperature.” In this module, the apparatus has been design to enable students to carry out experiments on simply supported and cantilever beams in order to investigate:(a) the relationship between the deflections and the applied loads (b) the effect of variations in 1ength and cross sectional i.e. deflection per unit load. Simply supported beam with central point load For this arrangement, it can be shown that the deflection under the load i.e. maximum deflection Wl 3 ∆= 48 EI 15 where I = bd 3 12 ∆ l3 = W 4Ebd 3 ∴beam compliance Cantilever beam with end point load For this arrangement, it can be shown that the central deflection relative to the supports...
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...CE 422 REINFORCED CONCRETE DESIGN COURSE OUTLINE: I. Introduction to Reinforced Concrete II. Analysis and Design of Beams Singly Reinforced Beams Doubly Reinforced Beams T-Beams III. Shear and Diagonal Tension IV. Bonds, Development Length, Hooks and Splices Reinforcement V. Axially Loaded Columns VI. Eccentrically Loaded Columns (Short Columns Subjected to both Bending and Compression) VII. Long Columns VIII. Analysis and Design of Slabs One Way Slab Two Way Slab References: Any Reinforced Concrete Book Instructor: Engr. Emman Rey Quimson INTRODUCTION CONCRETE- a mixture of water, cement and gravel. REINFORCED CONCRETE MATERIALS: 1. Cement 2. Aggregates- comprises 65-75 % of the mixture Sand- aggregates passing the No. 4 sieve. Gravel- aggregates larger than the sand. 3. Water- comprises 15-20% of the mixture. 4. Steel Reinforcement- deformed bar -Grade 40 & Grade 60 are usual Grade of Reinforcement. -Grade 40- 40 000 psi -Grade 60- 60 000 psi TYPES OF STEEL REINFORCEMENT: 1. BILLET STEEL- newly made steel 2. RAIL STEEL-rerolled from old axle & rail. STRENGTH OF CONCRETE PROPORTIONS: CLASS | STRENGTH | AA | 4000 psi | A | 3000 psi | B | 2500 psi | C | 2000 psi | Unit weight of concrete-23.56 KN/m3 ; 2400 kg/ m3 Modulus of elasticity of concrete: @ 1500 kg/ m3 to 2500 kg/ m3 = Ec=Wc1.5 (0.043)f'c @normal weight (2400 kg/ m3)= Ec=4700 f'c LOADING...
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...AS 3600—2009 AS 3600—2009 Australian Standard® Concrete structures Accessed by NEWCREST MINING LIMITED on 14 Jul 2010 This Australian Standard® was prepared by Committee BD-002, Concrete Structures. It was approved on behalf of the Council of Standards Australia on 8 October 2009. This Standard was published on 23 December 2009. The following are represented on Committee BD-002: • • • • • • • • • • • • • • • • AUSTROADS Association of Consulting Engineers Australia Australian Building Codes Board Bureau of Steel Manufacturers of Australia Cement Concrete & Aggregates Australia—Cement Cement Concrete & Aggregates Australia—Concrete Concrete Institute of Australia Engineers Australia La Trobe University Master Builders Australia National Precast Concrete Association Australia Steel Reinforcement Institute of Australia University of Adelaide University of Melbourne University of New South Wales University of Western Sydney This Standard was issued in draft form for comment as DR 05252. Standards Australia wishes to acknowledge the participation of the expert individuals that contributed to the development of this Standard through their representation on the Committee and through the public comment period. Keeping Standards up-to-date Accessed by NEWCREST MINING LIMITED on 14 Jul 2010 Australian Standards® are living documents that reflect progress in science, technology and systems. To maintain their currency, all Standards are periodically reviewed, and...
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...break-junction technique, how can we judge the point when a single molecule is connecting two gold electrodes? Ans: A break junction is an electronic device which consists of two metal wires separated by a very thin gap, on the order of the inter-atomic spacing (less than a nanometer). This can be done by physically pulling the wires apart or through chemical etching or electromigration. As the wire breaks, the separation between the electrodes can be indirectly controlled by monitoring the electrical resistance of the junction. In this technique a metal wire is bent or pulled, often using a piezoelectric crystal to apply the necessary force. The bending or pulling causes the metal wire to break in a controlled manner since piezoelectric elongation can be controlled to a precision of angstroms or less. As the wire breaks, the separation between the electrodes can be indirectly controlled by monitoring the electrical current through the junction. Procedure: 1) Break junction method: * Fabricate a thin metal wire on top of flexible substrate (e.g.polymide) by optical or e-beam lithography, or glue the wire to substrate by epoxy. * Notch the wire with a sharp object (or e-beam lithography). * Bending the substrate to break the wire at notch point (flexible substrate: not broken). * Relax the bending to bring into two wire parts back to contact. Electromigration: * Fabricate a nanowire by e-beam lithography * Increase current passing...
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...Inertia Bending of a Connecting Rod Experiment 15th November 2015 Contents Abstract----------------------------------------------------------------------------------------------------2 * 1. Introduction-------------------------------------------------------------------------------------------2 1.1 Objective----------------------------------------------------------------------------------------------------2 1.2 Background------------------------------------------------------------------------------------------2 1.3 Theory-----------------------------------------------------------------------------------------------2 1.3.1 Bending moment-----------------------------------------------------------------------------2 1.3.2 The Right Angle Assumption--------------------------------------------------------------------------5 2. Apparatus----------------------------------------------------------------------------------------------6 * 3. Data-----------------------------------------------------------------------------------------------------7 * 4. Procedur-----------------------------------------------------------------------------------------------7 * 4.1 Recording-------------------------------------------------------------------------------------------7 * 4.2 Calibration------------------------------------------------------------------------------------------7 5. Results--------------------------------------------------...
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...Experiment 7: Deflection of beams (Effect of beam length and width) 1. OBJECTIVE The objective of this laboratory experiment is to find the relationship between the deflection (y) at the centre of a simply supported beam and the span, width. 2. MATERIALS - APPARATUS Steel Beams, Deflection measuring device, 500g weight 3. INTRODUCTORY INFORMATION The deflection of a beam, y, will depend on many factors such as: - • The applied load F (F=m•g). • The span L. • The width of the beam b, and its thickness h. Other factors such as position, method of loading, the material of which the beam is made will also influence the deflection. If we wish to find the relationship between y and one of the possible variables it is necessary to keep all the other possible variables constant throughout the experiment. 1. Length calculation In this experiment the same beam is used throughout and the centrally applied point load is kept constant. Thus keeping all possible variables other than the deflection y and the span L constant we may investigate the relationship between y and L. Let y[pic]Ln where n is to be found Then y = k•Ln where k is a constant Taking logarithms: log y = n log L + log k which is in the straight line form (y = mx + C). Thus plotting logy against log L will give a straight-line graph of slope “n” and “k” may be determined. 2. Width calculation In this experiment beams of the same material but of different width are...
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...* Lab Report: Reflection and Refraction Name: Sample Data and Answers DATA PROCEDURE A: Bouncing Light off a Flat Mirror FLAT MIRROR Angle of Incidence | Angle of Reflection | 24° | 23° | 45° | 45° | 63° | 63° | DATA PROCEDURE B: Bouncing Light off a Concave Mirror CONCAVE MIRROR Angle of Incidence | Angle of Reflection | 18° | 18.5° | 45° | 45° | 64° | 66° | DATA PROCEDURE C: Bouncing Light off a Convex Mirror CONVEX MIRROR Angle of Incidence | Angle of Reflection | 33° | 32° | 45° | 44° | 62° | 60° | DATA PROCEDURE D: The Bending of Light by Refraction REFRACTION: Calculated Index of Refraction, 1.33 INTERFACE 1 — From air into acrylic | INTERFACE 2 — From acrylic into air | Incident angle(in air) | Refracted angle(in acrylic) | Incident angle(in acrylic) | Refracted angle(in air) | 21° | 14° | 16° | 21° | 45° | 32° | 32° | 41° | 63° | 40° | 41° | 63° | PROCEDURE D: REFRACTION – Plot of versus from DataStudio, with the Linear Fit. DATA PROCEDURE E: Observing Dispersion DISPERSION At the Dispersion Interface(Light traveling from acrylic into air.) | Index of Refraction of Acrylic | Incident angle | Refracted angle(in air) | | | RED light | BLUE light | For RED light | For BLUE light | 20° | 26.3° | 27.9° | 1.30 | 1.37 | QUESTIONS PROCEDURES A, B, C: Bouncing Light off Mirrors * 1. According to the Law of Reflection, the incident angle and the reflection angle must be...
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... 1. When the deflections of a statically determinant beam are calculated using singularity functions, it is necessary to know the values of 2 boundary conditions. For each of the beams below, what are the boundary conditions? 2. Calculate the reactions and draw the SFD and BMD for the cantilever beams below. Using singularity functions, calculate the deflection at 2 metres, the deflection at the tip, and sketch the deflected shape. The cross section of the beam is 300 mm deep by 200 mm wide, and it is made of concrete with a Young’s modulus of 30,000 MPa. 4. Using singularity functions, derive (in terms of P) the equation for the upwards deflection at midspan of the beam below. EI = 10 x 106 Nm2 5. Determine the location and value of the maximum deflection for the beam below. How far from the centre is the point of maximum deflection (expressed as a percentage of the span length)? EI = 10 x 106 Nm2 6. Calculate the reactions and draw the shear force and bending moment diagrams for the beam below. EI = 10 x 106 Nm2 (Note that this has four reactions, so the 3 equations of equilibrium do not give sufficient information to solve the problem – it is statically indeterminate. The answer requires you to use the results from questions 4 and 5). 7. For the beam below, use singularity functions to determine an equation for the deflected shape, expressed in terms of x (measured from the left) and EI. If the beam is made of timber (E = 10,000 MPa) and the cross section...
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...msheikh@uow.edu.au Consultation time: Friday 3.00 -5.00 pm AGENDA FOR TODAY Topics covered weeks 1-6 Reinforced concrete (RC): an overview Properties of Concrete and Reinforcement Analysis and design of RC structures RC Design based on AS3600-2009 Critical Load Combinations Weeks 1-7 PART 1: DESIGN OF REINFORCED CONCRETE STRUCTURES Week 1: Introduction to Reinforced Concrete (RC) Week 2: Design of Beams- Serviceability Week 3: Design of Beams- Ultimate Strength Week 4: Design of Beams- Shear, Cracking, Detailing (In Class Quiz on Topics covered From Week 1 to Week 3) Week 5: Design of Slabs: One-Way slab Week 6: Design of Columns and Walls Week 7: MID-SESSION EXAM (Topics covered from weeks 1-6) Dr. Neaz Sheikh University of Wollongong Lecture Notes Based on Foster et al. (2010) 1 CIVL 311/CIVL 981 Autumn 2012 (Week 1) Weeks 8-13 PART 1: DESIGN OF STEEL STRUCTURES Week 8: Introduction to Structural Steel Design Week 9: Bending Strength of Stable Beams Week 10: Flexural-Torsional (Lateral) Buckling of Beams Week 11: Strength of Webs (In Class Quiz) Week 12: Axially Loaded Members Week 13: Connection Design CIVL 311 CO-REQUISITE ENGG 251: MECHANICS OF SOLIDS 5 ME H N S SOL DS NOTE: PRE-REQUISITE OF ENGG 251 ENGG 152: ENGINEERING MECHANICS Reference books SJ Foster, AE Kilpatrick and RF Warner “Reinforced Concrete Basics: Analysis and design of reinforced concrete structures”, Pearson Australia 2010 ISBN ISBN9781442538450 ...
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...Introduction to Finite Element Analysis (FEA) or Finite Element Method (FEM) Finite Element Analysis (FEA) or Finite Element Method (FEM) The Finite Element Analysis (FEA) is a numerical method for solving problems of engineering and mathematical physics. Useful for problems with complicated geometries, loadings, and material properties where analytical solutions can not be obtained. The Purpose of FEA Analytical Solution • • Stress analysis for trusses, beams, and other simple structures are carried out based on dramatic simplification and idealization: – mass concentrated at the center of gravity – beam simplified as a line segment (same cross-section) Design is based on the calculation results of the idealized structure & a large safety factor (1.5-3) given by experience. FEA • Design geometry is a lot more complex; and the accuracy requirement is a lot higher. We need – To understand the physical behaviors of a complex object (strength, heat transfer capability, fluid flow, etc.) – To predict the performance and behavior of the design; to calculate the safety margin; and to identify the weakness of the design accurately; and – To identify the optimal design with confidence Brief History Grew out of aerospace industry Post-WW II jets, missiles, space flight Need for light weight structures Required accurate stress analysis Paralleled growth of computers Common FEA Applications Mechanical/Aerospace/Civil/Automotive ...
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