Deflection Of Simply Supported Beam With Udl

Brijendra Gupta. Chapter 9 deflections of beams chapter 9 beams on elastic foundations where is the maximum deflection in a simply supported load quora solved where on the beam below. Beam Deflection Equations are easy to apply and allow engineers to make simple and quick calculations for deflection. I was able to determine the Shear Force Diagram, but currently I'm struggling with the Bending moment diagram. The beam is supported at each end, and the load is distributed along its length. The bending moment at the two ends of the simply supported beam and at the free end of a cantilever will be zero. Deflection and Slope of Beams •As load is applied on a beam, it deflects. simply supported beam? When a simply supported beam is subjected to a moving udl longer than the span, the absolute maximum bending moment occurs when the whole span is loaded. The center reaction Q4. Load Calculation For Beam. I believe that changing formulation to MPC is the solution. 044) /64 = 1. 2 Differential Equations of the Deflection Curve consider a cantilever beam with a concentrated load acting upward at the free end the deflection v is the displacement in the y direction the angle of rotation of the axis. ! The beam has a length of L. Design StatusVertical shear Moment Buckling Deflection PASS PASS PASS PASS. Cantilever beam. M is lesser and hence lighter materials of construction can be used it resist the bending moment. 5) Option to calculate singly supported, fixed and cantilevered beams. 8 × 10 6 psi, υ 12 = 0. Depending on the load applied, it undergoes shearing and bending. Its mode of deflection is primarily by bending. a simply supported beam spanning say 4m is loaded with a UDL of say 18kn/m. suggested that the finite element method give less deflection as compared with the results obtained using beam theory for any specific location along span length. Hence, the fundamental equation in finding deflections is: 2 2 x x d y M dx EI In which the subscripts show that both M and EI are functions of x and so may change along the length of the beam. Whereas the deflection of simply supported composite beam with partial shear connection is 24. Because the beam is pinned to its support, the beam cannot experience deflection at the left-hand support. 1457\times 10^{-3} m^{4}, {/eq}use the principle of superposition. Deflection Chart of Beam at various Condition. approximate solutions. Fixed-Fixed Beam with UDL is a beam type that has fixed supports on both ends and transverse uniformly distributed load. 5kN/m2, Permanent: 0. (b) Deflection at where the point load is acting, and state its direction. Simply-Supported Beams Figure: A simply-supported beam. Fig:1 Formulas for Design of Simply Supported Beam having. A simply support by the original beam is usually a good choice, but sometimes another point is more convenient. Area Moment Method. A simply supported beam is a type of beam that has pinned support at one end and roller support at the other end. 30, and t k = 0. 2 Problems of Practices on Deflection of Beam of the section for the simply supported beam is J i n the left half and 0. It gives the diagrammatic representation of bending moments at different points of beam. Question: The A-36 steel simply-supported beam is subjected to the concentrated and distributed loads as shown. BEAM DIAGRAMS AND FORMULAS Table 3-23 (continued) Shears, Moments and Deflections 13. Tutorial from Simply Supported beam and UDL load. 3 away from A and another 5. ) MAXIMUM SLOPE AND DEFLECTION OF SIMPLY SUPPORTED BEAMS Loading condition Maximum slope Deflection ( y) Max. Tutorial (Hindi) Slope and Deflection of Beams: GATE (Mechanical) 15 lessons • 3 h 21 m. If the beam carries a U. Max Deflection Of A Simply Supported Beam Posted on September 27, 2019 by Sandra Design of simply supported beam decrease deflection at the cantilever deflection of a simply supported beam puter aided deflection and slope determine the maximum deflection. distance from the centre of the shaft to the point of application of the force For the simply supported beam PQ shown, Wire Body/Line Body/Beam: Supported - Line , a simply supported vertex is not realistic and leads to Boundary Condition Application. Simply Supported Beam With Udl. An explanation of the variables:. Deflection Of Simply Supported Beam With Uniformly Varying Load Posted on May 4, 2020 by Sandra Uniformly varying load 37 review bending moment and shear force moment area theorems civil er uniformly varying load 37 review deflection on a cantilever beam. Beam Formulas. Deflection function under concentrated loads P and distributed load q can be easily obtained respectively as fallows: )0( )2( 24 )0() 16 1 12 1 ( 323 23 LxxLxxL EI qx y LxxLx EI P y p p dd dd (1) When calculating the deflection caused by the secondary moment, the upper and lower T-beam can be combined into a small one regarded as fixed-end of. Objectives. The beam consists of a solid rectangular part with a depth 3 times the width (D = 3B). Simply-Supported Beams Figure: A simply-supported beam. 9 away from a on a 6m beam. Simply-Supported Beams. 2018/2019. 5 J in right half. A simply supported beam is the most simple arrangement of the structure. 5 m Structural width, b 0. For a simply-supported beam, we use the following boundary conditions: w(0)=0. Thanks all. Substitute x for L/2 gives: EIy max = = = Thus y max = but total load on the beam = W = wL. This calculator uses standard formulae for slope and deflection. The beam is supported at each end, and the load is distributed along its length. Simply supported beam due to point load 2. For all individual loads and all load combinations the global deformation of the beam member nodes and the support nodes are transferred into beam member deflection results. The beam consists of a solid rectangular part with a depth 3 times the width (D = 3B). The following page shows the line, shear force and bending moment diagrams for this beam. Calculate bending moments, shear, slope and deflection of a simply supported and guided beam at any point along its length when it is subject to a UDL (uniformly distributed load). In this example we take a beam with the UDL of 20 kN/m applied to the centre of the beam as shown. General shapes are rectangular sections, I beams, wide flange beams and C channels. One of the most powerful functions is using it as a beam deflection calculator (or beam displacement calculator). and so the equation for maximum deflection on a simply supported beam with a total UDL = W is : y max = HOME. FOR CONCENTRATING LOAD- (End reaction X L1 ) ( where L1 is the distance between end reaction and that point where the load or external force is acted ) 2. so i was given a problem by my mechanical principles teacher to work out the max vertical deflection of a simply supported beam with two point loads one a distance of 5. Assume simple beam theory is applicable for the simply supported beam shown. 5 J in right half. Let R 1 & R 2 be the reactions then,. Will automatically generate results curves and has the ability to find values at a specific location. ) and uniformly varying loads (u. (AU Nov/Dec 2016) It will occur at centre 3 c 5 WL y 384 EI =-UNIT - 3 DEFLECTION. Slope And Deflection Of Simply Supported Beam With Udl. Note that the maximum stress quoted is a positive number, and corresponds to the largest stress magnitude in the beam. Simply supported beam moment i. 2 inches under a live load only deflection limit of L/360. A uniform load of 2. In multiple span continuous beams, will load in one span produce stress the other spans? A2. Use (a) A36 steel and (b) steel with f y = 345 MPa and permissible live load deflection of span. A I-meter-long, simply supported copper beam (E = 117 GPa) carries uniformly distributed load q. A simply supported beam with a uniformly distributed load. This video shows how you can calculate deflection and slope of a cantilever beam with a point load at the free end using the double integration method. Moment Area Method For Deflection Of Beam | Part 4 | Simply Supported Beam Example 1 | Midpoint Load Part 3 | Multiple Point Loads and UDL on different beams Simply Supported Beam Example. Calculate the width & depyh of the beam if permissible bending stress is 7N/mm2 and central deflection is not exceed 1cm. a) represents a beam subject to a uniformly distributed load (udl) of magnitude w, across its length, l. A I-meter-long, simply supported copper beam (E = 117 GPa) carries uniformly distributed load q. This tool lets you to predict the deflection and stress of Beam. Deflection & Slope - Cantilever Beam with a Point Load at the Free End Deflection & Slope - Cantilever Beam with a Point Load at the Free End. Area Moment Method. It gives the diagrammatic representation of bending moments at different points of beam. A simply supported beam of length l carries a concentrated load W at distances of a and b from the two ends. Single simply supported unprotected steel beam subjected to nominal gravity loads has been considered for the investigation. Example 1. Moment Area Method For Deflection Of Beam | Part 4 | Simply Supported Beam Example 1 | Midpoint Load Part 3 | Multiple Point Loads and UDL on different beams Simply Supported Beam Example. M Equation is: w W1 b c 3 b c a w b 6 c l b c + Note that Macaulay terms are integrated with respect to, for example, (x -a) and they must be ignored when negative. M max max = wl 2 8 6. The beam is considered to. Metric round tube beam deflection calculator. (i) Derive the expression for strain energy due to bending. the member with simply supported ends. On completion of this tutorial you should be able to solve the slope and deflection of the following types of beams. Chapter 9 deflections of beams chapter 9 beams on elastic foundations where is the maximum deflection in a simply supported load quora solved where on the beam below. 5 J in right half. Design & Mfg. Beam bending is analyzed with the Euler-Bernoulli beam equation. Shear Force Diagram (kN) 0. If {eq}I_{z} = 0. 30, and t k = 0. Macaulay's method is to be used. Calculation Reference Roark's Formulas for Stress and Strain. If you're unsure about what deflection actually is, click here for a deflection definition Below is a concise beam deflection table that shows how to calculate the maximum deflection in a beam. We will also touch on the selection of a Concrete Beam. If AB = 5m and BC = 7m Span AB is loaded with udl 40kN/m and span BC with udl 30kN/m. The number 0. (1) Figure shows a simply supported beam of span 5m carrying two point loads. Simply supported beam moment i. when =0 15 x The plate deflection Δp is Δp= x δ 16 Where, δ is the primitive beam deflection of the x-x strip. Hi! Does anyone know how to calculate the maximum bending moment and deflection in a simply supported beam with a UDL acting across its entire length and a point load acting at some distance other than the centre of the span? It's for the design of a timber trimming beam in a loft conversion. Beam formulas with shear and mom calculator for ers slope and deflection cantilever propped cantilever with udl k6nqmkeedz4w beam formulas with shear and mom fixed both ends beam udl. 9 kN/m is applied to the beam. Structural Beam Deflection, Stress, Bending Equations and calculator for a Beam supported Both Ends Overhanging Supports Symmetrically, Uniform Load. CALCULATING BENDING MOMENT OF SIMPLY SUPPORTED BEAM 1. Simply supported beam due to self - weight 3. E = 200 kN/mm2, I = 16 x 10 8 mm4. It carries a uniformly distributed load of 10 KN/m as shown in figure. Cantilever Beams Part 1 – Beam Stiffness (continued) The next step would be to solve for the stress distribution in the beam generated by the given deflection. This can be used to observe the calculated deflection of a simply supported beam or of a cantilever beam. I want to complete our discussion of beam deflections by looking at another method of solution, the method of superposition. Obviously software could handle this question much more readily, I am going to assume you are expected to do this by hand. (These assume that the beam is uniform, i. Simply-Supported Beams Figure: A simply-supported beam. •The deflection can be observed and measured directly. Calculation Reference Roark's Formulas for Stress and Strain. The slope (q) & deflection (y) at any spanwise location can be derived. If you're unsure about what deflection actually is, click here for a deflection definition Below is a concise beam deflection table that shows how to calculate the maximum deflection in a beam. Skip navigation Sign in. The following image illustrates a simply supported beam. , determination of deflections of simple beams of simple portal. (b) Deflection at where the point load is acting, and state its direction. Structural Beam Deflection, Stress, Bending Equations and calculator for a Beam supported Both Ends Overhanging Supports Symmetrically, Uniform Load. Deflection of RC (Rectangular Beam) Simply Supported With UDL Elastic modulus, E 7. Download Now White Paper: Click Here to join Eng-Tips and talk with other members! Have calculation deflection at x for a Partial UDL from first principles. A method is for efficiently determining a very large number of terms in the series. Use strain energy method. ) and uniformly varying loads (u. Calculate the factored design loads (without self-weight). Simply supported beam with point force in the middle. If all the diagrams can be fitted on a single plot, do so. Tkinter GUI based python program can solve for shear, moment, slope, and deflection of a simply supported beam with cantilevers at either end. Beam Simply Supported at Ends - Concentrated load P at the center 2 1216 Pl E I (2 ) 2 2 3 Px l l for 0yx x 12 4 2 EI 3 max Pl 48 E I x 7. Aug 11, 2019 - Explore alii9131's board "Bending moment" on Pinterest. This calculator is for finding the slope and deflection at a section of simply supported beam subjected to uniformly distributed load (UDL) on a portion of span. If {eq}I_{z} = 0. Many structures can be approximated as a straight beam or as a collection of straight beams. A beam of uniform rectangular section 200 mm 4+3. Beam equations for Resultant Forces, Shear Forces, Bending Moments and Deflection can be found for each beam case shown. 1 Deflection curve of a beam. The deflection and stress levels predictions are required for beam design for a given shape, load, boundary and materials. Deflection Simply Supported Beam Udl Posted on April 30, 2020 by Sandra Deflection of a simply supported beam a simply supported beam acb has point fixed beam with udl ering max bending moment in a cantilever deflection of cantilever beam. Simply Supported Beam - With UDL. Academic year. The total load on beam is the UDL multiplied by the length of the beam, i. Total force on beam being wl. suggested that the finite element method give less deflection as compared with the results obtained using beam theory for any specific location along span length. Given the simply supported beam under a UDL (as shown in the figure below), determine ∆ C. Because the beam is pinned to its support, the beam cannot experience deflection at the left-hand support. The calculator produced a report suitable for building regulation approval which shows the bending, shear and deflection for the beam are all within safe limits. Find out: a) The flexural stiffness b) The dimensions of the section. Hi! Does anyone know how to calculate the maximum bending moment and deflection in a simply supported beam with a UDL acting across its entire length and a point load acting at some distance other than the centre of the span? It's for the design of a timber trimming beam in a loft conversion. The equations given here are for homogenous, linearly elastic materials, and where the rotations of a beam are small. Let say for an example, If we want to calculate the Deflection for Simply Supported Beam having 1m [1000mm] with acting load of 1000N. Question: The A-36 steel simply-supported beam is subjected to the concentrated and distributed loads as shown. A simply supported beam of span 8 m carries a udl of 4 kN/m over the entire span and two point loads of 2 kN at 2 m from each support. Table of Contents. Deflection of RC (Rectangular Beam) Simply Supported With UDL Elastic modulus, E 7. Moment Area Method For Deflection Of Beam | Part 4 | Simply Supported Beam Example 1 | Midpoint Load Part 3 | Multiple Point Loads and UDL on different beams Simply Supported Beam Example. Some of the allowed Deflections are shown in the example. Deflection Of Simply Supported Beam With Uniformly Varying Load Posted on May 4, 2020 by Sandra Uniformly varying load 37 review bending moment and shear force moment area theorems civil er uniformly varying load 37 review deflection on a cantilever beam. Stiffness of a simply supported rc beam simply supported beam with udl beams fixed at both ends continuous beam calculator clearcalcs calculate the maximum deflectionHow To Calculate The Maximum Deflection Of A. Calculate the factored design loads (without self-weight). In this video derive an expression for deflection of beam with udl load solve by double integration method. Simply supported beam with point force in the middle. If {eq}I_{z} = 0. LO2 : To determine deflection for simply supported steel beams TASK 2 (A) Determine deflections in simply supported steel beams with point loads and a uniformly distributed load provided in Task 1 (a) to 1 (e). CHAP 4 FINITE ELEMENT ANALYSIS OF BEAMS AND FRAMES 2 INTRODUCTION SIMPLY SUPPORTED BEAM • Assumed deflection curve • Strain energy L ˛ ˛ 4 2 0 3 2 4 EI pL UV C C L ˛ ˛ 4 4 00 35 24 0 2 dEIpL pL CC dC L EI ˛ ˛˛ 13 EXAMPLE - SIMPLY SUPPORTED BEAM cont. Max Deflection Of A Simply Supported Beam Posted on September 27, 2019 by Sandra Design of simply supported beam decrease deflection at the cantilever deflection of a simply supported beam puter aided deflection and slope determine the maximum deflection. Hi there, I need to calculate the safe UDL for a beam 457x152 UB 52 simply supported at both ends with a span of both 6m and 12m (2no. If all the diagrams can be fitted on a single plot, do so. The beam dimensions are b = 1. 3 Of The Lecture Notes, It Is Shown That The Deflection Of A Simply Supported Beam With A UDL Is Given By Nu = W/24E I (z^4 - 2Lz^3 + L^3 Z). [A/M-15] The deflection at the MID SPAN is given by: y B = WL3/ 3EI y B = (25000 x 30003) / (3 x 2. State the location of maximum shear force in a simple beam with any kind of loading. A simply supported beam with a point load at the middle. Classical beam theory may be very difficult to apply to more complicated structures such as bridges and, therefore, the methods of numerical integration of curvatures along with harmonic analysis using curvatures are investigated. Here we display a specific beam loading case. Write the maximum value of deflection for a simply supported beam of constant EI, span L carrying central concentrated load W. The beam is now cantilevered from this support. , slope and deflection. 044) /64 = 1. Beam Simply Supported at Ends – Concentrated load P at the center 2 1216 Pl E I (2 ) 2 2 3 Px l l for 0yx x 12 4 2 EI 3 max Pl 48 E I x 7. Let say for an example, If we want to calculate the Deflection for Simply Supported Beam having 1m [1000mm] with acting load of 1000N. Moment Area Method For Deflection Of Beam | Part 5 | Simply Supported Beam Example 2 | UDL. Find: (a) What is the Maximum deflection ratio of beam 1 to beam 2? Solution:( ) beam1 = ( ) beam2 = x = =. 1-2 Simply-Supported Composite Beams Edition 2. In this study, Navier's solution for the analysis of simply supported rectangular plates is extended to consider rigid internal supports. Let us consider a deflection of a simply supported beam which is subjected to a concentrated load W acting at a distance 'a' from the left end. 30, and t k = 0. How To Draw Shear Force Bending Moment Diagram. Deflection of Beam: Deflection is defined as the vertical displacement of a point on a loaded beam. The beams are assumed to be exposed to temperature. This may be reduced slightly to account for beneficial effects of compression near the support, in accordance with code provisions. GTU MIMP 827,319 views. Use (a) A36 steel and (b) steel with f y = 345 MPa and permissible live load deflection of span. A uniform distributed load of 1000 N/m is applied to the lower horizontal members in the vertical downward direction. Its vertical reaction at right support will be equal to? assume that a beam is an end span of a girder which is continuous over several support and is integral with its supports. 6): (a) the simply supported beam, (b) the overhanging beam, and (c) the cantileoer beam. Its mode of deflection is primarily by bending. Beam forces calculation, uniformly distributed load, concentrated loads, beam deflection, metric units, online spreadsheet. The maximum deflection is measured as 1. Calculation Reference Roark's Formulas for Stress and Strain. Note that the maximum stress quoted is a positive number, and corresponds to the largest stress magnitude in the beam. A steel pipe 50mm internal diameter and wall thickness 2mm is simply supported on a span of 6m. and L = 12. Simply supported beam due to point load and UDL. i XEI , (e) Loading Upward loading positive Fig. You will need to determine the moment of inertia of the cross section and the distance from. R1 x 6 = 1000×3 + (200×3)3/2 = 3600. • The three simple‐beam bending moment diagrams thus obtained are shown in Figure. Draw a BMD for each loading (including the support reactions of the original beam. Slope and Deflection Calculator for Simply Supported Beam with uniform load on full span. This post gives a solved design example of a laterally restrained beam […]. Use of Macaulay's technique is very convenient for cases of discontinuous and/or discrete loading. IT carries a UDL of 9kN/m run over the centre length. 3 Simply-supported beam carrying a uniformly distributed load A beam of uniform flexural stiffness EI and span L is simply-supported at its ends, Figure 13. If AB = 5m and BC = 7m Span AB is loaded with udl 40kN/m and span BC with udl 30kN/m. ML diagram (simple beam bending moment diagram 3 MBA bending moment diagram due to external loads) A 1 B 2 15 MAB Bending Moment Diagram. The deflection at center of beam and various failure patterns are studied. 1 Prediction of girder deflection using classical beam theory for a simply supported girder subjected to a. A simply supported beam is the most simple arrangement of the structure. Cold-rolled RHS, SHS and channel. The centre of gravity of the UDL is at ½ of 8. Line Diagram. The "BeamAnal" calculates Shear Force, Bending Moment and Deflection at 31 positions along the member length. Find the deflection under the point load in terms of EI. (A) Derive. Al – Raheimy College of Engineering, Babylon University Abstract This paper presents a theoretical investigation of the free transverse vibrations of a uniform beam that is simply supported at both ends subjected to a static axial force. In Eurocode 0, characteristic loads can be factored down for serviceability calculations. D2 is the deflection on the outboard end of the overhang. E=1x104 N/mm2. Computer Aided Deflection And Slope Yses Of Beams. Bending Analysis of Simply Supported and Clamped Circular plate - Free download as PDF File (. Use strain energy method. Calculate the height h of the beam if the maximum bending stress is 90 MPa and the modulus of elasticity is 200 GPa. Beam Supported at Both Ends - Uniform Continuous Distributed Load. You can find comprehensive tables in references such as Gere, Lindeburg, and Shigley. 4 10 1 384) 5 (9 384 4 7 4 4 Deflection is in downward direction. Classical beam theory may be very difficult to apply to more complicated structures such as bridges and, therefore, the methods of numerical integration of curvatures along with harmonic analysis using curvatures are investigated. 9 away from a on a 6m beam. The beams were simply supported at the ends and subjected to a concentrated load applied at the mid-span. After leaning this unit, you must be able to. Find plastic moment and BMD at collapse. 30, and t k = 0. A simply supported beam cannot have any translational displacements at its support points, but no restriction is placed on rotations at the supports. Therefore the general deflection equation for a simply supported beam with a UDL. 2 Problems of Practices on Deflection of Beam of the section for the simply supported beam is J i n the left half and 0. Simply Supported Beam With Overhang On One Side Exle. Deflection is (with a simple centerloaded beam) is PL^3/48EI The various deflections are as follows: (i) for a simply supported beam with point load (center)=PL^3/48EI (ii) // // // UDL= 5PL^4. 9 kN/m is applied to the beam. known solution for the simply supported plate with the uniform loading giving the deflection function for the strip case is combined with that for a solution of deflection function which shows the effects due to the edges W =Ws +We. 21 Beam Deflection by Integration ! 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. LO2 : To determine deflection for simply supported steel beams TASK 2 (A) Determine deflections in simply supported steel beams with point loads and a uniformly distributed load provided in Task 1 (a) to 1 (e). Formula: Hollow Square Tube Deflection = (f x l x l x l) / (t x y x (((s - k 4) 4 /12) - ((s - (2. 9 with a = L qL4 ( C)1 = CCC b4E Ib the deflection ( C)2 due to a force T acting on C is obtained use conjugate beam method TL2 TL L 2L ( C)2 = M = CCC L + CC C C b3E Ib b EIb 2 3 2TL3 = CCC b3E Ib the elongation of the cable is Th ( C)3 = CC EcAc compatibility equation. b) shear force diagram shows the regions of maximum shear, for this beam these correlate to the reaction forces. 1457\times 10^{-3} m^{4}, {/eq}use the principle of superposition. Academic year. Fixed and Simply Supported Ends UDL Along Entire Length. 6 - Cantilevers “Rules of Thumb” the width (b) of a rectangular beam should be between 1/3 and 2/3 of the effective. MU = wu L. Solution: The equation we work with is ∆=, Where M o = bending moment distribution due to actual or real loading M 1 = bending moment distribution due to virtual or unit loading E = modulus of elasticity of the material of beam I = moment of inertia of beam section L = beam span ∆ c = deflection at point C. There are other issues too. In the example, the Maximum Deflection allowed is controlled the Code. SS beam subjected to point and UDL - Example. 6): (a) the simply supported beam, (b) the overhanging beam, and (c) the cantileoer beam. The problem with standard formulae, is that you have to create then anyway! To be accurate, that formula would require that the first load is S/2 from the support at each end, where s = L/n. Beam Supported at Both Ends - Uniform Continuous Distributed Load. 25 MNm2 and using Castigliano’s theorem, determine the deflection at the centre of the beam. Simply Supported Beam Deflection from Loading Function Example Course Description This course builds on the concept of force and moment equilibrium learnt from first year engineering mechanic and physics courses and focuses on the internal actions and deformations experienced by simple structural members under loading. Beams are assumed to be simply supported. The simplest form of this equation is as follows:. Therefore the general deflection equation for a simply supported beam with a UDL. 1 simply supported beam subjected to point load. This calculator is for finding the slope and deflection at a section of simply supported beam subjected to uniformly distributed load (UDL) on a portion of span. The series is given by 422 2. This defection is ¼ times the deflection of a simply supported beam. Slope And Deflection Of Simply Supported Beam With Udl. This section covers shear force and bending moment in beams, shear and moment diagrams, stresses in beams, and a table of common beam deflection formulas. is EIy = Maximum deflection occurs at mid-span when x = L/2. 3 Simply-supported beam carrying a uniformly distributed load A beam of uniform flexural stiffness EI and span L is simply-supported at its ends, Figure 13. In a simply supported beam with udl or symmetrically located point loads, both the maximum moment and max deflection are at the center of the span and the minimum moments and minimum deflections. Its mode of deflection is primarily by bending. Simply Supported Beam - With UDL More Beams. In order to calculate reaction R1, take moment at point C. (According to Newton’s third law there is for every action an equal and opposite reaction. Integrated into each beam case is a calculator that can be used to determine the maximum displacements, slopes, moments, stresses, and shear forces for this beam problem. UNIT - III 1. •The deflection can be observed and measured directly. Deflection can be calculated by standard formula will only give the deflection of common beam configurations and load cases at discrete locations , or by methods such as virtual. The supported end of the beam may be built into masonry or it may be a projection from a simply supported beam. A I-meter-long, simply supported copper beam ( E = 117 GPa) carries uniformly distributed load q. Fig:1 Formulas for Design of Simply Supported Beam having. at r = 0 and it is given by Deflection (w) max = 4 5 64 1 qa D P P. This beam deflection calculator is designed to calculate the deflection of a simply supported cantilever with a single load point at one end. The centre of gravity of the UDL is at ½ of 8. Simply Supported Beam - With UDL More Beams. 00 m = 40 kN. Calculate the height h of the beam if the maximum bending stress is 90 MPa and the modulus of elasticity is 200 GPa. the point loads are both 15. Because the beam is pinned to its support, the beam cannot experience deflection at the left-hand support. A beam of length of 6 m is simply supported at its ends. Moment Equation For Cantilever Beam With Distributed Load. This free online calculator is developed to provide a software tool for calculation of deflection and slope at any section of simply supported beam (without overhangs) subjected to point load, uniformly distributed load, varying load and applied moments on the span or on the supports. Deflection at the centre: A simply supported beam of span 6 m is subjected to Udl of 24 kN/m for a length of 2 m from left support. 353 mm came from beam theory - the equation for maximum deflection of a simply-supported beam subjected to a UDL taken this website. Its vertical reaction at right support will be equal to? assume that a beam is an end span of a girder which is continuous over several support and is integral with its supports. (i) Derive the expression for strain energy due to bending. A cantilever beam with a uniformly distributed load. The double integration method is a powerful tool in solving deflection and slope of a beam at any point because we will be able to get the Simply supported Beams: Deflection Of Beams Mechanism Of Solid. Thus D = 0. 6kN/m2 Width of load perpendicular to beam, or height of load supported by beam: 3. 3 SOLUTIONS FOR BEAM-COLUMNS (DEFLECTION PROBLEM). Tutorial (Hindi) Slope and Deflection of Beams: GATE (Mechanical) 15 lessons • 3 h 21 m. We have already seen terminologies and various terms used in deflection of beam with the help of recent posts and now we will be interested here to calculate the deflection and slope of a simply supported beam carrying a point load at the midpoint of the beam with the help of this post. 2 Differential Equations of the Deflection Curve consider a cantilever beam with a concentrated load acting upward at the free end the deflection v is the displacement in the y direction the angle of rotation of the axis. Figure 1: Typical cantilever beam studied. In a simply supported beam with udl or symmetrically located point loads, both the maximum moment and max deflection are at the center of the span and the minimum moments and minimum deflections. The beam is supported at each end, and the load is distributed along its length. A simply supported beam is the most simple arrangement of the structure. Max Deflection Of A Simply Supported Beam Posted on September 27, 2019 by Sandra Design of simply supported beam decrease deflection at the cantilever deflection of a simply supported beam puter aided deflection and slope determine the maximum deflection. Deflection Of Simply Supported Beam With Uniformly Varying Load Posted on May 4, 2020 by Sandra Uniformly varying load 37 review bending moment and shear force moment area theorems civil er uniformly varying load 37 review deflection on a cantilever beam. In this example we take a beam with the UDL of 20 kN/m applied to the centre of the beam as shown. The bending moment at the two ends of the simply supported beam and at the free end of a cantilever will be zero. Slope And Deflection Of Simply Supported Beam With Uvl February 23, 2019 - by Arfan - Leave a Comment Beam deflection formulas cantilever beam with uniformly varying load cantilever beams moments and deflections work wordpress ii b tec. Deflection at supports in a simply supported beam is maximum b. Example - Beam with Uniform Load, English Units. Will automatically generate results curves and has the ability to find values at a specific location. I was to prepare the Shear force diagram and bending moment diagram for simply supported beam with UDL acting throughout the beam and two Point Loads anywhere on the beam. A simply supported beam cannot have any translational displacements at its support points, but no restriction is placed on rotations at the supports. M Equation is: w W1 b c 3 b c a w b 6 c l b c + Note that Macaulay terms are integrated with respect to, for example, (x -a) and they must be ignored when negative. Simply Supported Beam Distributed Load Formulas:. 6 and an allowable stress of 175 N/mm. We have already seen terminologies and various terms used in deflection of beam with the help of recent posts and now we will be interested here to calculate the deflection and slope of a simply supported beam carrying a point load at the midpoint of the beam with the help of this post. of simply supported beams are obtained. Hi! Does anyone know how to calculate the maximum bending moment and deflection in a simply supported beam with a UDL acting across its entire length and a point load acting at some distance other than the centre of the span? It's for the design of a timber trimming beam in a loft conversion. Torsion reinforcement. zero deflection on the sides given by x = 0 and x = a, representing simple supports. I can't seem to find a deflection formula for partial UDL's for a simply supported beam. Christian Otto Mohr The length of a conjugate beam is always equal to the length of the actual beam. 6 - Cantilevers “Rules of Thumb” the width (b) of a rectangular beam should be between 1/3 and 2/3 of the effective. Assumption: Plane sections remain plane, the strain distribution will be as shown. Moment Area Method For Deflection Of Beam | Part 3 | Cantilever Beam Example 3 | UDL & Moment Part 11 | Overhang Beam | UVL | Question 4 Part 7 | Multiple Loads on Simply Supported Beam. This calculator uses standard formulae to determine the values of slope and deflection at the required section. Because the beam is pinned to its support, the beam cannot experience deflection at the left-hand support. Solution in the images below, you can also refer these sites to get the deflection values of simply supported beam for different loading conditions by putting the desired values. BEAMS: STATICALLY INDETERMINATE (9. Find i) UDL it may carry if the bending stress is not to exceed 100N/mm 2. For all individual loads and all load combinations the global deformation of the beam member nodes and the support nodes are transferred into beam member deflection results. 4 LATERALLY SUPPORTED BEAMS. 2c Shear Force Diagram for Simply Supported Beam with UDL b) Bending Moment Diagram for simply supported beam with UDL The generated bending moment diagram using Matlab is as shown below. Good question, Thanks For The A2A, OK fine, now why we consider larger side as a depth? Let Us take one example Assume you have one Steel scale which have depth is d and breadth is b. (5)If the beam carries a U. In this study, Navier's solution for the analysis of simply supported rectangular plates is extended to consider rigid internal supports. Let us consider a deflection of a simply supported beam which is subjected to a concentrated load W acting at a distance 'a' from the left end. Just like a pivot, the wall is capable of exerting an upwards reaction force R 1 on the beam. The loads are symmetric about the centre of the beam. Solution: The equation we work with is ∆=, Where M o = bending moment distribution due to actual or real loading M 1. The beam is a steel wideflange section with E 628 10 psi and an allowable bending stress of 17,500 psi in both tension and compression. Fig:1 Formulas for Design of Simply Supported Beam having. Macaulay's method (the double integration method) is a technique used in structural analysis to determine the deflection of Euler-Bernoulli beams. A simply supported beam of uniform flexural rigidity EI and span l, carries two. 3 Solution for a Beam with Fixed Axial Displacements The problem is solved under the assumption of simply-supported end condition, and the line load is distributed accordingly to the cosine function. Fig 1 shows a simply-supported beam AB subjected to a partial UDL 3 kN/m and a point load 50 kN. Case 3:-When simply supported beam is subjected to a single concentrated load at mid-span. For this reason, the analysis of stresses and deflections in a beam is an important and useful topic. the deflection ( C)1 due the uniform load can be found from example 9. In Eurocode 0, characteristic loads can be factored down for serviceability calculations. A simply support by the original beam is usually a good choice, but sometimes another point is more convenient. Chapter 9 deflections of beams chapter 9 beams on elastic foundations where is the maximum deflection in a simply supported load quora solved where on the beam below. Slope And Deflection Of Simply Supported Beam With Uvl February 23, 2019 - by Arfan - Leave a Comment Beam deflection formulas cantilever beam with uniformly varying load cantilever beams moments and deflections work wordpress ii b tec. Simply supported beam with a point load at the centre and with two or more point loads - SFD and BMD in case of simply supported beam with a central point load - Example. The Navier solution for the rectangular plate simply supported o n all sides and under a uniformly distributed load, , as shown in Figure 1, is presented in Chapter 5 of Timoshenk o’s text. Point of contraflexure = 0 (iii) Three Hinged Parabolic Arch Having Abutments at Different Levels. The beam consists of a solid rectangular part with a depth 3 times the width (D = 3B). FOR CONCENTRATING LOAD- (End reaction X L1 ) ( where L1 is the distance between end reaction and that point where the load or external force is acted ) 2. As shown in figure below. and a uniformly distributed live load of 550 lbs/ft. Given that the flexural rigidity EI = 100,000 kNm2 for the entire beam, determine: (a) The rotation at Support A, and state its sense of rotation. Based on the type of deflection there are many beam deflection formulas given below, w = uniform load (force/length units) V = shear. When lateral deflection of the compression flange of a beam is prevented by providing effective lateral support (restraint), the beam is said to be laterally supported. [A/M-15] The deflection at the MID SPAN is given by: y B = WL3/ 3EI y B = (25000 x 30003) / (3 x 2. , determination of deflections of simple beams of simple portal. A simply supported beam of uniform flexural rigidity EI and span l, carries two. Simply supported beam moment i. 5x10-4 m 4 and E = 1x10 7 kN/m 2 (i) The fixed end moment for the beam carrying udl: M A = M B = 12 2 WL = KNm x 75. 1 x 105 x 108) y B = 10. 7KN and the UDL is 1. Continuous Beam Three Span With Udl. Deflection JAE, that sounds like the condition he's looking for. distance from the centre of the shaft to the point of application of the force For the simply supported beam PQ shown, Wire Body/Line Body/Beam: Supported - Line , a simply supported vertex is not realistic and leads to Boundary Condition Application. 6 × 10 6 psi, G 12 = 0. A steel beam (178 x 102 x 19 UB S275) 3m long was selected. In practice however, the force may be spread over a small area, although the dimensions of this area should be substantially smaller than the beam span length. Moment Area Method For Deflection Of Beam | Part 3 | Cantilever Beam Example 3 | UDL & Moment Part 11 | Overhang Beam | UVL | Question 4 Part 7 | Multiple Loads on Simply Supported Beam. Use strain energy method. 3 Solution for a Beam with Fixed Axial Displacements The problem is solved under the assumption of simply-supported end condition, and the line load is distributed accordingly to the cosine function. For completeness and BCO submission you should also considered shear forces and deflection of the beam. Take moment about point C, for reaction R1 $$\sum M_{c}\space = 0$$. The centre of gravity of the UDL is at ½ of 8. Question: The A-36 steel simply-supported beam is subjected to the concentrated and distributed loads as shown. 4 ENES 220 ©Assakkaf Statically Determinate Beam When the equations of equilibrium are sufficient to determine the forces and stresses in a structural beam, we say that this beam is statically determinate Statically Indeterminate Beams LECTURE 18. RE: Beam Formulas for Multiple Point Loads. CHAPTER 2 : DEFLECTION (MACAULAY METHOD) PART 1 by Saffuan Wan Ahmad Let us again consider a simply supported beam AB of length L and carrying concentrated load P at mid span,C as shown below. The beam is supported at each end, and the load is distributed along its length. Solution in the images below, you can also refer these sites to get the deflection values of simply supported beam for different loading conditions by putting the desired values. 1 Simply supported circular plate subjected to uniformly distributed load The deflection, moment and transverse shear are to be finite at the center of the plate (r = 0). I was able to determine the Shear Force Diagram, but currently I'm struggling with the Bending moment diagram. The Young’s Modulus of the beam is 30 x 10^6 Psi. ! The beam has a length of L. shear force and bending moment diagram for simply supported beam with udl - Duration: 19:45. 2d Bending Moment Diagram for Simply Supported. Draw the positive. The tables below give equations for the deflection, slope, shear, and moment along straight beams for different end conditions and loadings. Calculate the flexural stiffness of a simply supported beam which will limit the deflection to 2 mm at the middle. The beam dimensions are b = 1. 01 radians at the ends. The beam is supported at each end, and the load is distributed along its length. (b) Deflection at where the point load is acting, and state its direction. In the above image, I need to find the deflection at point C and D of the simple supported beam. The simplest form of this equation is as follows:. Skip navigation Sign in. Example 1. These programs are of high quality but the approach, and hence the output, is too complex for a quick solution of continuous beam problems. Remember the actual value of v is still intact in the deflection as it is part of the flexural rigidity of the plate. Uniform Load on full span. The calculator produced a report suitable for building regulation approval which shows the bending, shear and deflection for the beam are all within safe limits. In a coil spring, the stress is distributed evenly along the length of the coil. Let say for an example, If we want to calculate the Deflection for Simply Supported Beam having 1m [1000mm] with acting load of 1000N. By taking into account the stiffness of the composite joints, the deflection is therefore reduced by roughly 50%. Another method of determining the slopes and deflections in beams is the area-moment method, which. The load on the conjugate beam is the M/EI diagram of the loads on the actual beam. (B) Explain how deflection in beams affects structural stability. The deflection of the beam towards a particular direction when force is applied on it is called Beam deflection. You can find comprehensive tables in references such as Gere, Lindeburg, and Shigley. 3 away from A and another 5. The Euler-Bernoulli equation describes a relationship between beam deflection and applied external forces. 5 kN/m × 8. In this post i will briefly explain different types of beams. The ClearCalcs beam calculator allows the user to input the geometry and loading of a beam for analysis in a few simple steps. FOR CONCENTRATING LOAD- (End reaction X L1 ) ( where L1 is the distance between end reaction and that point where the load or external force is acted ) 2. shear force and bending moment diagram for simply supported beam with udl - Duration: 19:45. Could anyone please tel me how to calculate the Bending Stress, Compressive stress and the Deflection for any beam (UB 203x102x23) subjected to Uniform Distributed Load of 8063 N/m over a 8 m long beam (simply supported beam). Deflection & Slope - Cantilever Beam with a Point Load at the Free End Deflection & Slope - Cantilever Beam with a Point Load at the Free End. Engineers adopt deflection limits which suit the nature of the building. Good question, Thanks For The A2A, OK fine, now why we consider larger side as a depth? Let Us take one example Assume you have one Steel scale which have depth is d and breadth is b. The calculator produced a report suitable for building regulation approval which shows the bending, shear and deflection for the beam are all within safe limits. This square tubing deflection calculator calculates tube deflection for square based on length of beam, tube size, decimal gauge and load. D1 is the maximum deflection midspan between supports. Where y is the deflection at the point, and x is the distance of the point along the beam. 9 with a = L qL4 ( C)1 = CCC b4E Ib the deflection ( C)2 due to a force T acting on C is obtained use conjugate beam method TL2 TL L 2L ( C)2 = M = CCC L + CC C C b3E Ib b EIb 2 3 2TL3 = CCC b3E Ib the elongation of the cable is Th ( C)3 = CC EcAc compatibility equation. Use strain energy method. ALL calculators require a Premium Membership. Deflection of Cantilever Beam. All structural/civil formula books, etc seem to neglect this information. This can be used to observe the calculated deflection of a simply supported beam or of a cantilever beam. General shapes are rectangular sections, I beams, wide flange beams and C channels. Remember the actual value of v is still intact in the deflection as it is part of the flexural rigidity of the plate. The beam is also pinned at the right-hand support. EI is constant. Problem 691 Determine the midspan deflection for the beam shown in Fig. For a simply sagging supported structure with distributed mass - or load due to gravitational force - can be estimated as. What is the absolute maximum bending moment due to a moving udl longer than the span of a simply supported beam? 11. deflection curve of beams and finding deflection and slope at specific points along the axis of the beam 9. Chapter 9 deflections of beams chapter 9 beams on elastic foundations where is the maximum deflection in a simply supported load quora solved where on the beam below. Let say for an example, If we want to calculate the Deflection for Simply Supported Beam having 1m [1000mm] with acting load of 1000N. Deflection of RC (Rectangular Beam) Simply Supported With UDL Elastic modulus, E 7. The loads are symmetric about the centre of the beam. 01 radians at the ends. This calculator uses standard formulae to determine the values of slope and deflection at the required section. Introduction Universal beam sections are normally employed in buildings to carry load. 2018/2019. 353 mm came from beam theory - the equation for maximum deflection of a simply-supported beam subjected to a UDL taken this website. The ends A and C are hinged supports and B is a continuous support. (Hint: Apply Case No. Differential Equation of the Deflection Curve of Beam: When a beam is subjected to pure bending couple M, it is bent into a circular arc the radius of curvature. As shown in figure below. It carries a UDL of 9 kN/m over the entire length. The beams used for frame work are selected on the basis of deflection, amongst other factors. For information on beam deflection, see our reference on. Formula: Hollow Square Tube Deflection = (f x l x l x l) / (t x y x (((s - k 4) 4 /12) - ((s - (2. STATIC ANALYSIS OF A SIMPLY SUPPORTED BEAM WITH UNIFORMLY DISTRIBUTED LOAD Figure 1 All dimensions are in mm Objective: To find the deflection, stress, strain, shear force and bending moment diagram of simply supported beam with uniformly distributed load as shown in Figure 1. This beam deflection calculator is designed to calculate the deflection of a simply supported cantilever with a single load point at one end. A simply supported beam is a type of beam that has pinned support at one end and roller support at the other end. Beam bending is analyzed with the Euler-Bernoulli beam equation. (ii) Derive the expression for strain energy due to. The center reaction Q4. y of a simply supported beam under uniformly distributed load (Figure 1) is given by EI qx L x dx d y 2 ( ) 2 2 − = (3) where. 1 Simply supported circular plate subjected to uniformly distributed load The deflection, moment and transverse shear are to be finite at the center of the plate (r = 0). In multiple span continuous beams, will load in one span produce stress the other spans? A2. I did not have large deflection turned on. Find expressions for the total strain energy of the beam and the deflection under load. Now we have the deflection for a simply supported beam with UDL acting on the whole length. Sign conventions for load, S. Any help would be much appreciated! Thanks. INTRODUCTION (Fig. In this study, Navier's solution for the analysis of simply supported rectangular plates is extended to consider rigid internal supports. 0GPa Tributary width for loading, tw 6. 4 Analysis of Singly Reinforced Beam Bending Consider a simply supported singly reinforced rectangular beam Load causes - deflection (downwards) - bottom of the beam will be in tension while top in compression. From the geometry of the figure, (6. The beam is now cantilevered from this support. 6 and an allowable stress of 175 N/mm. The Navier solution for the rectangular plate simply supported o n all sides and under a uniformly distributed load, , as shown in Figure 1, is presented in Chapter 5 of Timoshenk o’s text. deflection (YId Cantilever with concentrated load Wat end WL2 2EI W 6E1 - ~2~3 - 3 ~2~ + x3~ WL3 3EI. Shear Force diagram for Simply supported Beam with three segments, when third segment start and end is reversed from other 2, then the in-correct shear force diagram generated by Staad pro, Please clarify. Christian Otto Mohr The length of a conjugate beam is always equal to the length of the actual beam. 7KN and the UDL is 1. I believe that changing formulation to MPC is the solution. The centre of gravity of the UDL is at ½ of 8. 4 Analysis of Singly Reinforced Beam Bending Consider a simply supported singly reinforced rectangular beam Load causes - deflection (downwards) - bottom of the beam will be in tension while top in compression. Based on the supports of the beam, Following are some of the classification of the beam. The equivalent UDL method is a useful tool for estimating the deflection in a simply supported beam with a complex loading. These programs are of high quality but the approach, and hence the output, is too complex for a quick solution of continuous beam problems. Chapter-5 Deflection of Beam Page- 7 (ix) A simply supported beam with a continuously distributed load the intensity of which at any point 'x' along the beam is x sin x ww L ⎛⎞π = ⎜⎟ ⎝⎠ (i) A Cantilever beam with point load at the free end. The calculator produced a report suitable for building regulation approval which shows the bending, shear and deflection for the beam are all within safe limits. Here we display a specific beam loading case. Monash University. Deflection Of Simply Supported Beam With Uniformly Varying Load Posted on May 4, 2020 by Sandra Uniformly varying load 37 review bending moment and shear force moment area theorems civil er uniformly varying load 37 review deflection on a cantilever beam. A simply supported beam cannot have any translational displacements at its support points, but no restriction is placed on rotations at the supports. The force is concentrated in a single point, located in the middle of the beam. when the beam is loaded with a 14KNload at each one third span point, it failed. helping to develop a coherent picture of the deflection behaviour. Deflection of RC (Rectangular Beam) Simply Supported With UDL Elastic modulus, E 7. Beam Deflection Equations are easy to apply and allow engineers to make simple and quick calculations for deflection. Markings for. Deflection Simply Supported Beam Udl Posted on April 30, 2020 by Sandra Deflection of a simply supported beam a simply supported beam acb has point fixed beam with udl ering max bending moment in a cantilever deflection of cantilever beam. The modulus of elasticity is 205GPa. Beam Deflection Ronni Kaptanband Co. Good question, Thanks For The A2A, OK fine, now why we consider larger side as a depth? Let Us take one example Assume you have one Steel scale which have depth is d and breadth is b. Simple Support Beam with Overhanging. Aug 11, 2019 - Explore alii9131's board "Bending moment" on Pinterest. 1 Classification based on supports; 2 Simply Supported Beam : U. See more ideas about Bending moment, Civil engineering construction, Civil engineering design. 044) /64 = 1. Lesson 15 of 15 • 2 upvotes • 12:56 mins. Chapter 9 deflections of beams chapter 9 beams on elastic foundations where is the maximum deflection in a simply supported load quora solved where on the beam below. The shear force diagram indicates the shear force withstood by the beam section along the length of the beam. Total force on beam being wl. The beam is laterally supported. Slope and Deflection for sim. The dead load does not include the self-weight of the beam. Differential Equation of the Deflection Curve of Beam: When a beam is subjected to pure bending couple M, it is bent into a circular arc the radius of curvature. Skip navigation Sign in. 9 away from a on a 6m beam. Beams - Fixed at One End and Supported at the Other - Continuous and Point Loads.
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# Double Angle Formulas: Finding $\tan 2\theta$

I am trying to find $$\tan 2\theta$$ where $$sin \theta = \frac{5}{13}$$ and $$\theta$$ is in Quadrant One.

According to my textbook, $$\tan 2\theta = \frac{120}{119}$$, but I get $$\frac{-10}{13}$$ instead.

The Identity I am using:

$$\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^{2}\theta}$$

My Process:

Since $$y = 5\;$$ and $$r = 13,\; x = 12.$$

Apply Tangent Double Angle Formula: $$\frac{2(\frac{5}{12})}{1 - (\frac{5}{12})^2}$$

$$\frac{\frac{10}{12}}{1 - \frac{25}{12}} \cdot \frac{12}{12}$$

$$\frac{10}{12-25}$$

$$\frac{-10}{13}$$

What am I doing wrong?

• $(5/12)^2$ is not $25/12$. It is $25/144$. – Nick Nov 22 '18 at 0:12
• Erm, that's probably my issue then. – LuminousNutria Nov 22 '18 at 0:12

Alternatively:

$$\sin\theta = 5/13$$, implies $$\cos\theta = \sqrt{1-(5/13)^2} = 12/13$$.

$$\sin 2\theta = 2\sin \theta \cos \theta = 120/169$$.

$$\cos 2\theta = 2 \cos^2 \theta-1 = 1 - 2 \sin^2 \theta = 119/169$$.

This gives $$\tan 2\theta = \sin 2\theta/ \cos 2 \theta = 120/119$$.

I made a mistake solving the problem. $$(\frac{5}{12})^2 \neq \frac{25}{12}$$.

Actually, $$(\frac{5}{12})^2 = \frac{25}{144}$$.

Taking that into account:

$$\frac{\frac{10}{12}}{1 - \frac{25}{144}} \cdot \frac{12}{12}$$

$$\frac{10}{12-\frac{300}{144}}$$

$$\frac{10}{12 - \frac{25}{12}} \cdot \frac{12}{12}$$

$$\frac{120}{144-25}$$

$$\frac{120}{119}$$

Therefore, $$\tan 2\theta = \frac{120}{119}$$


$\sigma$-algebra generated by a subset

I'm new to this concept of a $\sigma$-algebra generated by a collection of subsets.

Let $\Omega = \{a, b, c, d\}$ and \begin{align} &\mathcal{F}_1 = \{\Omega, \emptyset, \{a\}\} \\ &\mathcal{F}_2 = \{\Omega, \emptyset, \{a\}, \{b, c, d\}\}\text{.} \end{align} I wish to show that $$\sigma\langle \mathcal{F}_1 \rangle = \bigcap_{\mathcal{F} \in \mathcal{I}(\mathcal{F}_1)}\mathcal{F} = \mathcal{F}_2$$ where $\mathcal{I}(\mathcal{F}_1) = \{\mathcal{F}: \mathcal{F}_1 \subset \mathcal{F} \text{ and }\mathcal{F} \text{ a }\sigma\text{-algebra on }\Omega \}$.

I know immediately from this that any $\sigma$-algebra $\mathcal{F}$ must, at the very least, contain elements of $\mathcal{F}_1$: $$\{\Omega, \emptyset, \{a\}\}$$ but other than literally listing every possible collection of subsets of $\Omega$ and checking which are $\sigma$-algebras, and then intersecting such sets, I don't see what's an efficient way to do this.

Is my way of thinking correctly or is there a quicker way?

• Complete the collection of subsets to be closed under the conditions for an sigma algebra, i.e., closed under the operations of countable union or intersection and complementary. This is the minimal sigma algebra that contain the collection. – Masacroso Feb 22 '16 at 19:26
• @Masacroso Ah, so that's why it's called the "smallest" $\sigma$-algebra generated by a set. Thank you! – Clarinetist Feb 22 '16 at 19:29

Yes. but one quick shortcut is that the complements of the sets are in the sigma algebra as well, so you can easily see $\{b,c,d\}$ must be in $\sigma\langle\mathcal{F}_1\rangle$.

• And we know that this $\{b, c, d\}$ set with everything else in $\mathcal{F}_1$ must be the "smallest" $\sigma$-algebra generated by $\mathcal{F}_1$. Thank you! ... now, the question is, how do I prove this rigorously? I don't see how else to do this other than finding every possible $\mathcal{F}$ and then intersecting them. – Clarinetist Feb 22 '16 at 19:31
• @Clarinetist, no, it's the smallest also because it is the minimum extension of the current set to a sigma-algebra. So suffices to find one, and show that none of its subsets are sigma algebras. – gt6989b Feb 22 '16 at 19:34
• Very good and clear explanation. Thank you! – Clarinetist Feb 22 '16 at 19:35

Here are some more general statements that imply the desired equation. We assume the following without proof.

1. The intersection of $\sigma$-algebras is again a $\sigma$-algebra.
2. A set of the form $\lbrace \emptyset, \Omega, A, A^c \rbrace$ is a $\sigma$-algebra.

Let $\mathscr{G}$ (particular case $\mathcal{F}_1$) be any set of subsets of $\Omega$. $\mathscr{A}:=\bigcap_{\mathcal{F} \in \mathcal{I}(\mathscr{G})}\mathcal{F}$ is nonempty because the powerset $\mathscr{P}(\Omega)$ has $\mathscr{P}(\Omega) \in \mathcal{I}(\mathscr{G})$. It follows that $\mathscr{G} \subset \mathscr{A}$ and that, because of assumption 1, $\mathscr{A}$ is a $\sigma$-algebra. If $\mathscr{A}'$ is another $\sigma$-algebra with $\mathscr{G} \subset \mathscr{A}'$, then, because $\mathscr{A}'$ is one of the sets over which the intersection is taken, we have $\mathscr{A} \subset \mathscr{A}'$. So $\mathscr{A}$ satisfies the definition of the smallest $\sigma$-algebra generated by $\mathscr{G}$, i.e. $\sigma(\mathscr{G}) = \mathscr{A}$.

We conclude $\sigma(\mathcal{F}_1) = \bigcap_{\mathcal{F} \in \mathcal{I}(\mathscr{\mathcal{F}_1})} \mathcal{F}$.

We now show that $\sigma(\mathcal{F}_1) = \mathcal{F}_2$. Note that $A \in \mathcal{F}_1$ implies $A \in \sigma(\mathcal{F}_1)$ and $\lbrace a \rbrace \in \sigma(\mathcal{F}_1)$ implies $\lbrace b,c,d \rbrace = \lbrace a \rbrace^c \in \sigma(\mathcal{F}_1)$, so that $\mathcal{F}_2 \subset \sigma(\mathcal{F}_1)$. By assumption 2, $\mathcal{F}_2$ is a $\sigma$-algebra. We obtain $\sigma(\mathcal{F}_1) = \mathcal{F}_2$.

Note

It holds in general that $\sigma(\lbrace A \rbrace) = \sigma(\lbrace \emptyset, \Omega, A \rbrace) = \lbrace \emptyset, \Omega, A, A^c \rbrace$, we could have proved this instead of proving this for our particular case in the last step.


# Show that limit does not exist (two variables)

I just started looking into multiple variable calculus and limits involving them. I'm not amazing at limits either.

I want to answer this question:

Show that the following limit does not exist

$$\lim_{(x,y)\to (0,0)}\frac{x^2y^2}{x^2y^2+(x-y)^2}$$

So, my working:

$$\lim_{x\to 0}\frac{x^2y^2}{x^2y^2+(x-y)^2}=\lim_{x\to 0}\frac{0}{y^2}=0$$

and

$$\lim_{y\to 0}\frac{x^2y^2}{x^2y^2+(x-y)^2}=\lim_{y\to 0}\frac{0}{x^2}=0$$

I wasn't sure what to do after this as they're both 0 but using the fact it can approach from any direction, I tried substituing $$y=x$$, not sure if that's correct - or my working.

So, let $$y=x$$, then

$$\lim_{(x,y)\to (0,0)}\frac{x^2y^2}{x^2x^2+(x-y)^2}=\lim_{x\to 0}\frac{x^4}{x^4+(x-x)^2}=1$$

Therefore limit doesn't exist.

Is this somewhat correct? What's the best way to answer a question like this?

Also, when showing that this limit does not exist, do I need to find different values of limits for both $${x\to 0}$$ AND $${y\to 0}$$? Or is one enough, e.g. if I just find two different values for two limits for $${x\to 0}$$ without using $${y\to 0}$$ in my calculation at all, is that fine?

Thanks!

What you have done is correct. The limit exists only if the value of the limit along every direction that leads to $$(0,0)$$ is same.

So when you calculate $$\lim_{x\to 0}\frac{x^2y^2}{x^2y^2+(x-y)^2}$$ you are calculating limit along the line $$x=0$$.

Similarly,
$$\lim_{y\to 0}\frac{x^2y^2}{x^2y^2+(x-y)^2}$$ is limit along line $$y=0$$.

And the last limit you calculated is along line $$y=x$$.

So to answer your question, yes it would have been perfectly acceptable if you did not calculate limit along $$y=0$$. Just showing two examples where the limit comes out to be different along different directions is enough to show limit does not exist.

• "The limit exists only if the value of the limit along every direction ... is same" A word of caution: that's a necessary, but not a sufficient condition. It can happen that all directional limits give the same value, and the limit does not exist. – leonbloy Feb 6 at 14:35
• @leonbloy Do you have an example of that case? – Shufflepants Feb 6 at 15:48
• @leonbloy I don't know of any way by which we can prove all directional limits exist and have same value, except by using epsilon-delta definition. Though I have very limited knowledge of multi-variable calculus. So it would be great if you could give an example where all directional limits give the same value, and the limit does not exist. – Swapnil Rustagi Feb 6 at 15:53
• If all directional limits give the same value, then the function might not be continuous at that point; it would need to be defined at that point, with a value equal to all the limits. But I think if all possible directional limits exist and are equal, then the function can be said to have that limit. (Unless anyone has a counterexample?) – gidds Feb 6 at 16:28
• @Shufflepants $f(x,y)=xy^2/(x^2+y^4)$ See here ocw.umb.edu/mathematics/math-240-calculus-iii-spring-2010/… (from page 19) – leonbloy Feb 6 at 17:11

Your reasoning is ok. If the limit existed, you would obtain the same value for every directional limit, which was not the case.

• How can I improve my reasoning? – Mandingo Feb 6 at 11:48
• No much room for improvement here... It is pretty standard. Maybe you could have just used directional limits instead of starting with the iterated limits, but in the end it is quite the same. – PierreCarre Feb 6 at 11:54

What you did is completely correct, but perhaps you can ease it a little bit as follows:

For $$\;x=0\;,\;\;y\to0\;$$ the limit clearly is $$\;0\;$$, whereas for $$\;x=y\;$$ and $$\;x\to0\;$$ we get

$$\frac{x^4}{x^4+(x-x)^2}=1\xrightarrow[x=y\to0]{}1$$

Thus the limit depends on the path chosen $$\;\implies\;$$ the limit doesn't exist...and this is the relevant argument: going on different paths yields different limits.

You did right.

A possible improvement is to compute the limit along an arbitrary line: along $$y=mx$$ you have to compute $$\lim_{x\to0}\frac{x^2(mx)^2}{x^2(mx)^2+(x-mx)^2}= \lim_{x\to0}\frac{m^2x^4}{m^2x^4+x^2(1-m)^2}= \lim_{x\to0}\frac{m^2}{m^2+(1-m)^2x^{-2}}= \begin{cases} 0 & m\ne 1 \\[4px] 1 & m=1 \end{cases}$$ Doing this way may immediately show how to solve the business. Of course, if you find that all limits along lines are equal, you cannot conclude that the limit exists; instead, you should try along other curves, if you suspect the limit doesn't exist.

What you did is correct, but note that after proving that the limit is $$0$$ when you take $$y=0$$, the fact that it is also $$0$$ when you take $$x=0$$ is irrelevant. What matters here is that going to $$(0,0)$$ through two different directions leads you to two distinct limits. Therefore, there is no (global) limit at $$(0,0)$$.


# Is $\mathbb R^2$ a field?

I'm new to this very interesting world of mathematics, and I'm trying to learn some linear algebra from Khan academy.

In the world of vector spaces and fields, I keep coming across the definition of $\mathbb R^2$ as a vector space ontop of the field $\mathbb R$.

This makes me think, Why can't $\mathbb R^2$ be a field of its own? Would that make $\mathbb R^2$ a field and a vector space?

Thanks

-
A field has multiplication. How would you define multiplication on $\mathbb R^2$ so that it is a field? (There is a way to do so, but it isn't "obvious" until you realize that the resulting field is the complex numbers...) – Thomas Andrews Dec 4 '12 at 16:07
$\mathbb{R}^2$ can be a field but with multiplication defined as follows: $(a,b)(c,d) = (ac - bd, ad + bc)$. Indeed, this is one way of defining the complex numbers. – Rankeya Dec 4 '12 at 16:08
But, if you want to try to do this for $\mathbb{R}^n$, for $n \geq 3$ such that $\mathbb{R}$ is naturally embedded in $\mathbb{R}^n$ as a subfield, then it is not possible to do so, and this is a harder fact to prove. – Rankeya Dec 4 '12 at 16:11
ok this just answered my follow-up question. Is there any proof of that being true? – vondip Dec 4 '12 at 16:11
Vondip - Perhaps this is at a slight tangent, but a significant difference between R and C is that R is an ordered field and C is not. e.g. 5 is larger than 3, but which is "larger", 4 + 7i or 6 + 5i ? (Answer: well, defining how "large" or "the length" a complex number is is not as obvious as for the reals. In fact, there are many different ways of defining the length of a complex number). Just something to think about. – Adam Rubinson Dec 4 '12 at 16:37

If you define:

$$(a,b)+(x,y):=(a+x,b+y)$$

$$(a,b)\cdot (x,y):=(ax-by,ay+bx)$$

then the set $\,\Bbb R^2=\Bbb R\times\Bbb R\,$ turns into a field, and a rather well known and important one. Can you identify it?

-
Spoiler: take a look at one of the comments given. $\^\smile\^$ – FrenzY DT. Dec 4 '12 at 16:10
complex numbers indeed! fantastic how it all connects! Would that make R^2 a field and a vector space? – vondip Dec 4 '12 at 16:11
Yes, it would, because addition in $\mathbb{R}^2$, as @DonAntonio defines it, is component wise (the usual way). Remember, the vector space structure depends only on the underlying abelian group. – Rankeya Dec 4 '12 at 16:15
What do you mean "vector space"? Any field is a vector space over any subfield, so the field $\,\Bbb R^2\cong\Bbb C\,$ is a vector field ove an infinite number of subfields, say $\,\Bbb C\,,\,\Bbb R\,,\,\Bbb Q\,,\,\Bbb Q(i)\ldots\,$ , etc. – DonAntonio Dec 4 '12 at 16:15
So would that mean that I any vector space could be defined when F -being the field, as : F^n ? – vondip Dec 4 '12 at 16:17

It is important to understand that a set on its own has no algebraic structure. By defining operators on $\mathbb{R}^2$ you could turn it into (almost) anything you like.

The natural operators on $\mathbb{R}^2$, namely $(x, y) + (a, b) \mapsto (x+a, y+b)$ and $(x, y) \cdot (a, b) \mapsto (x\cdot a, y\cdot b)$ do not define a field as $(0, 1)$ has no multiplicative inverse.

-

Usually in mathematics one defines these structures as tuples

A field is a triple $(K,+,\cdot)$ such that $K$ is a set and [...] and $\cdot:K \times K \rightarrow K$

A Vectorspace is a triple $(V,+,\cdot)$ such that $V$ is a set and [...] and $\cdot: K \times V \rightarrow V$

So your question is meaningless: A set (say $\mathbb R^2$) cannot be a field or a vectorspace or a group or anything - only if you add some additional structure (most of the time operations) you can ask this question.

For example $\mathbb R^2$ can be the set used in the definition of a field, as well as the underlying set used in the definition of a vectorspace. And we are happy, the addition operation $$+:\mathbb R^2 \times \mathbb R^2 \rightarrow \mathbb R^2$$ is the same, and the multiplicaton for "the" vectorspace structure $$\mathbb R \times \mathbb R^2 \rightarrow \mathbb R^2$$ is "compatible" with the multiplication for "the" field structure $$\mathbb R^2 \times \mathbb R^2 \rightarrow \mathbb R^2$$

-

Adding to the above answer. With the usual exterior multiplication of the $\mathbb{R}-\{0\}$ as a ring with the natural addition and multiplication you can not make a field out of $\mathbb{R}^{2}-(0,0)$ \ But there may exist other products such as the one in the answers which can make a field out of ${\mathbb{R}\times\mathbb{R}}-\{ 0\}$ \ According to one of the theorems of Field theory every field is an Integral domain. So by considering : ${\mathbb{R}\times\mathbb{R}}-\{ 0\}$ With the following natural product: $(A,B)*(C,D)=(AB,CD)$ We see that $(1,0)*(0,1)=(0,0)$ Which means that $\mathbb{R}^{2}$is not an integral domain and hence not a field.

-
Why are you considering $\mathbb{R}^*\times\mathbb{R}^*$? – Tobias Kildetoft Feb 23 '15 at 10:16
Bcoz he is asking that if $\mathbb{R}^{2}$is a field ... And I think he meant with the exterior multiplication and exterior addition. Otherwise it is obvious we can consider it as field aince it is isomorphism to $\mathbb{C}$ and hence a field – Romel Feb 23 '15 at 10:21
But what does that have to do with this? You are removing way more elements than $0$ when you consider this (in fact, it is clear that the usual multiplication does turn this into a group). – Tobias Kildetoft Feb 23 '15 at 10:22
And since the natural multiplication is defined on $\mathbb{R}$ We have to consider $\mathbb{R}^{*}$ as the set which the natural multiplication works on – Romel Feb 23 '15 at 10:24
No, to be a field we would need $(\mathbb{R}\times\mathbb{R})\setminus \{0\}$ to be a group, not $\mathbb{R}^*\times\mathbb{R}^*$. – Tobias Kildetoft Feb 23 '15 at 10:25


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# Each term of a certain sequence is calculated by adding a particular

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30 Jul 2018, 00:45
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Each term of a certain sequence is calculated by adding a particular constant to the previous term. The 2nd term of this sequence is 27 and the 5th term is 84. What is the 1st term of this sequence?

(A) 20
(B) 15
(C) 13
(D) 12
(E) 8

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30 Jul 2018, 10:41
Bunuel wrote:
Each term of a certain sequence is calculated by adding a particular constant to the previous term. The 2nd term of this sequence is 27 and the 5th term is 84. What is the 1st term of this sequence?

(A) 20
(B) 15
(C) 13
(D) 12
(E) 8

let the constant be $$d$$ and the first term be $$a$$

so 2nd term $$= a+d=27$$ and 5th term will be $$a+4d=84$$. Subtract the two equations to get

$$3d=57 =>d=19$$

Hence $$a=27-19=8$$

Option $$E$$
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Updated on: 31 Jul 2018, 17:11
Bunuel wrote:
Each term of a certain sequence is calculated by adding a particular constant to the previous term. The 2nd term of this sequence is 27 and the 5th term is 84. What is the 1st term of this sequence?

(A) 20
(B) 15
(C) 13
(D) 12
(E) 8

let d=difference between terms
84-27=57=3d
d=19
27-19=8=1st term
E

Originally posted by gracie on 30 Jul 2018, 18:15.
Last edited by gracie on 31 Jul 2018, 17:11, edited 1 time in total.
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30 Jul 2018, 19:34
Bunuel wrote:
Each term of a certain sequence is calculated by adding a particular constant to the previous term. The 2nd term of this sequence is 27 and the 5th term is 84. What is the 1st term of this sequence?

(A) 20
(B) 15
(C) 13
(D) 12
(E) 8

Let the first term be 'a' and the constant being added be d.

So, 2nd term becomes -
$$a_2$$ = a + d = 27 ------ (1)

and the 5th term becomes -
$$a_5$$ = a + d + d + d + d = 84
Hence,
$$a_5$$ = a + 4d = 84 ------ (2)

Now multiplying equation (1) by 4 to eliminate d

hence (1) becomes -

$$a_2$$ * 4 =>
4a + 4d = 108 ---- (3)

Subtracting equation (2) from (3)

4a + 4d - a - 4d = 108 - 84

3a = 24

a = 8

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01 Aug 2018, 16:36
Bunuel wrote:
Each term of a certain sequence is calculated by adding a particular constant to the previous term. The 2nd term of this sequence is 27 and the 5th term is 84. What is the 1st term of this sequence?

(A) 20
(B) 15
(C) 13
(D) 12
(E) 8

Let n = the constant added to each term to get the next term.

We have:

2nd term = 27

3rd term = 27 + n

4th term = 27 + 2n

5th term = 27 + 3n

Since we are given that the 5th term is 84, we have:

27 + 3n = 84

3n = 57

n = 19

Since the second term was given as 27, we know the first term is 27 - n. So the first term is 27 - 19 = 8.

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# Iterated Integrals - setting up limits of integration

1. Sep 30, 2008

### mirandasatterley

1. The problem statement, all variables and given/known data

Find the volume of the region under the graph of f(x,y) = x+y and above the region y2≤x, 0≤x≤9

3. The attempt at a solution

From these equations, x will be integrated from 0-9, but i'm not sure about y.

My thinking is that y will be intgrated from 0-3 because y2≤x and the smallest value of x is 0, and the square root of 0 is 0, so that is the smallest y, and the largest x value possible is 9 and the positive quare root of 9 is 3, so this is the largest value of y,
So I would integrate:

∫0-9∫0-3 (x+y)dydx.

Is this correct or am I missing something, where the limits of integration also involve using f(x,y)?

2. Sep 30, 2008

### HallsofIvy

Staff Emeritus
No, that is not correct. The region $a\le x\le b$, $c\le y\le d$, with a, b, c, d numbers is always a rectangle and the figure here is not a rectangle. But the bounds do NOT involve f(x,y)- that is a "z" value and goes inside the integral as you have it.

Always draw a picture for problems like this. y^2= x is a parabola, of course, "on its side". The line x= 9 is a vertical line crossing the parabola at (9,3) and at (9, -3). Yes, you can integrate with x going from 0 to 9. On your picture, mark an arbitrary "x" by marking a point on the x-axis between 0 and 9. Now draw a vertical line from one boundary to the other. The y bounds, for that x, are y values of those endpoints: $(x, -\sqrt{x})$, and $(x, -\sqrt{x})$. Your integral is
$$\int{x=0}^9\int_{y=-\sqrt{x}}^{\sqrt{x}} f(x)dy dx= \int{x=0}^9\int_{y=-\sqrt{x}}^{\sqrt{x}}(x+ y) dy dx$$

Of course, like any double integral, you can reverse the order of integration. If you look at your picture you will see that y ranges, overall, from -3 to 3. Draw a horizontal line across the parabola, representing an arbitrary value of y in that range. It will have left endpoint on the parabola: x= y^2, and right endpoint on the vertical line x= 9. Those will now be the limits of integration for this order:
$$\int_{y= -3}^3\int_{x= y^2}^9 f(x,y)dx dy= \int_{y= -3}^3\int_{x= y^2}^9(x+ y)dx dy$$

Try it both ways. You should get the same answer.

3. Sep 30, 2008

### mirandasatterley

In a similar situation where i have to switch the order of integration from
∫0-3∫y2-9 f(x,y) dxdy to dydx,

Is this also a parabola on it's side, with intercepts through (3,9) and (0,9), meaning that I would be setting up the new limits of integration, fro the portion of the parabola above y=0?

4. Sep 30, 2008

### mirandasatterley

So, ∫0-9∫Square root of x -3 f(x,y) dydx


# Inequality (-3/x) < 3

1. Aug 23, 2013

### Qube

1. The problem statement, all variables and given/known data

(-3/x) < 3

2. Relevant equations

Dividing/multiplying an inequality causes the inequality sign to change.

3. The attempt at a solution

I keep getting the wrong solution. I tried two methods. I cannot get the textbook solution (x < -1)

Method one:

-3 < 3x
-1 < x

Method two:

Divide both sides by negative one, resulting in:

3/x > -3
3 > -3x
-1 < x

2. Aug 23, 2013

### Staff: Mentor

Here you have multiplied by x, which could be positive or negative (or zero). Since you don't know the sign of x, you need two cases - one where x is assumed to be positive and the other where x is assumed to be negative.
Again, you've multiplied by x, whose sign you don't know. Same comments as above apply here.

3. Aug 23, 2013

### Qube

All right! Now I have -1 < x if x is positive and -1 > x if x is negative. How do I determine which is correct? Is the time for trial-and-error?

4. Aug 23, 2013

### Staff: Mentor

The book's answer is wrong or at least incomplete.

If x > 0, you get x > - 1. This means that x > 0 AND x > -1. Together, these mean that x > 0.
If x < 0, you get x < -1. This means that x < 0 AND x < -1. Together, these mean that x < -1.

If you look at the graph of y = -3/x, you'll see that it has two parts. For the curve on the left, -3/x < 3 when x < -1. For the curve on the right, -3/x < 3 for any x > 0.

5. Aug 23, 2013

### Qube

So I guess I'll have to do a bit of trial and error to determine the exact intervals? As in, x > -1 is technically correct, but the best answer is that x is actually greater than 0 or x > 0?

How do you recommend solving these problems? I'm having massive trouble.

6. Aug 23, 2013

### verty

Just remember the rule, NEVER multiply or divide by x or an expression containing x, it could be negative. Read Mark44's reply again, the answer is: x < -1 OR x > 0.

Example: $\frac{x+2}{x-3} > 0$. If $x-3 > 0, x+2 > 0$, and if $x-3 < 0, x+2 < 0$. Can you give the answer for this example?

7. Aug 23, 2013

### rock.freak667

Not to detract from what the others are saying, if you multiply by x, you are multiplying by an unknown value which can be +ve or -ve and will thus change the sign of the inequality (i.e. < might become >). So what you can do to prevent this is multiply instead by x2 as this will always be positive.

Then you can solve it as you would using normal algebraic means. You will just need to be careful when doing this as you might pick up an extra solution. But simply analyzing your solution set for the problem would lead you in the correct path.

8. Aug 23, 2013

### Ray Vickson

Turn your inequality into one of the form
$$\frac{1}{x} < a$$
or
$$\frac{1}{x} > a$$
I will let you figure out which inequality applies, and what is the value of $a$.

Anyway, once you have a simple inequality in $1/x$, you can envision the graph $y = 1/x$, and then figure out what the x-region must be for the corresponding inequality in y.

Note: this suggested method is safe; it never has you multiplying or dividing by an x of unknown sign, at least, not until the very end.

9. Aug 24, 2013

### vanhees71

I think it's time to give the solution to clarify the issue:

You just have to do a case differentiation. First you can simplyfy the inequality by divding through $-3$:
$$-\frac{3}{x} < 3 \Leftrightarrow \frac{1}{x}>-1.$$
Now for $x>0$ the inequality is always fulfilled and for $x<0$ you get by multiplying with $(-x)>0$
$$-1>x \; \Leftrightarrow \; x<-1$$.
Thus the inequality is fulfilled for either $x>0$ or $x<-1$.

10. Aug 24, 2013

You can also try
\begin{align*} \frac{-3}{x} & < 3 \tag{Given} \\ \frac{-3}{x} \cdot x^2 & < 3x^2 \\ -3x & < 3^2 \\ 0 & < 3x^2 + 3x \tag{Now factor to obtain}\\ 0 & < 3x\left(x + 1\right) \end{align*}

If you set up a sign table (or graph $3x(x+1)$ you'll find that the inequality is solved
for numbers $x < -1$ or $x > 0$ as stated in other places. In short, it is not that one or the other of these two inequalities gives the solution set, is that the solution set consists of any number that satisfies either the first or the second of these two.

11. Aug 24, 2013

### Infrared

I think this problem is being made much harder than it really is.

$\frac{-3}{x}<3 \\ \frac{-3}{x}-3<0 \\ \frac{1}{x}+1>0 \\ \frac{x+1}{x}>0$

So $x>0$ or $x<-1$

12. Aug 24, 2013

### Ray Vickson

You are doing exactly what I suggested the OP do, but he/she never reported back.

13. Aug 25, 2013


# The definitions of limit infimum and limit supremum

I have begun reading Rosenthal's "A First Look to Rigorous Probability Theory" and in order to reinforce my calculus background I am studying through the appendix section. Here he defined the limit of a sequence of real numbers as $\lim_{n \to \infty}x_n=x$ if for each $0 <\epsilon$ there is a natural number $N$ such that it is $|x-x_n| < \epsilon$ for $n > N$.

Then the limit infimum and supremum are defined as $\liminf_n x_n = \lim_{n \to \infty} \inf_{k\geq n} x_k$ and $\limsup_n x_n = \lim_{n \to \infty} \sup_{k\geq n} x_k$.

1. My first question is about the interpretation of the limit infimum and limit supremum definitions. It looks to me like they are defined as the limits of sequences as well. For example, for the infimum, one can define a sequence $a_n = \inf_{k \geq n}x_k$ where $x_n$ was already a sequence of real numbers. Then the limit infimum is the limit of the sequence $a_n$ , $\lim_{n \to \infty}a_n$, so the infimums of the subsequences $x_k$ where $k \geq n$ are converging if the limit exists. It is the similar for the supremum. Is my interpretation correct here?

2. It is said the limit infimum and supremum do always exist, though they can be infinite. I did not understand this; why it is so?

3. In the book it is stated that the limit $\lim_{n \to \infty} x_n$ exists if and only if it is $\liminf_n x_n = \limsup_n x_n$. I could not prove this to myself. How can it be shown that this statement is true?

• 1. ok, 2. they are monotone sequences. – enzotib Jul 23 '14 at 7:28
• For 3. In general the limit infimum is the least value to which a subsequence of ${a_n}$ can converge and the limit supremum is the greater value to wihch a subsequence of ${a_n}$ can converge – Dimitri Jul 23 '14 at 7:44

These are answers to the secondary questions in the comments after my answer to the original question; I cannot put the answers in comments, because the size of a comment is limited.

The second question. Let $(x_n)$ be a sequence taking values in the set $\{0,1,2,\ldots,9\}$, and $a:=\liminf_n x_n$. If $0$ appears infinitely many times in the sequence, then $a=0$. If $0$ appears only finitely many times, but $1$ appears infinitely often, then $a=1$. And so on. This is a slick answer. It is entirely another matter to answer the question with a definite value for a particular given sequence, such as the sequence of digits after the comma in the decimal expansion of $\pi$. Well, if you can prove that the digit $0$ appears infinitely many times in the decimal expansion of $\pi$, then you have $a=0$, and so on; in this case it takes a serious theoretical effort to determine the $\liminf$.

As for the third question, let $(x_n)$ be a sequence of real numbers such that $a:=\liminf_n x_n$ is a real number (that is, $-\infty<a<\infty$). Since the sequence of infima $a_n:=\inf_{k\geq n} x_n$ is increasing (meaning that $m\leq n$ implies $a_m\leq a_n$ -- I hate "nondecreasing"), the limit $a=\lim_{n\to\infty} a_n$ is actually the supremum: $a=\sup_n a_n$.
$\quad$Let $a_1\leq a$ and $a'<a_1$; then $a'<a$. We claim that there exists $m$ such that $x_n\geq a'$ for every $n\geq m$: since $a=\sup_n a_n$, there exists $m$ such that $a_m\geq a'$; but then $x_n\geq a_m\geq a'$ for every $n\geq m$.
$\quad$Now let $a_1>a$, and set $a':=\tfrac{1}{2}(a+a_1)$; we have $a<a'<a_1$. In this case we claim that given any $m$ there exists $n\geq m$ such that $x_n<a'$: since $a_m=\inf_{k\geq m} x_k\leq a<a'$, there exists $n\geq m$ such that $x_n<a'$.
$\quad$We have proved that $a$ is the largest of all real numbers $a_1$ with the property that for every $a'<a_1$ there are only finitely many $n$ such that $x_n<a'$.

1. Yes, in the book you are reading, $\liminf_n x_n$ and $\limsup_n x_n$ are defined as limits of sequences. Note that the sequence $a_n=\inf_{k\geq n} x_k$ is increasing (which means that $m\leq n$ implies $a_m\leq a_n$), while the sequence $b_n=\sup_{k\geq n} x_k$ is decreasing.

2. The $\liminf$ and $\limsup$ are taken in the extended real line $\overline{\mathbb{R}}=\{-\infty\}\cup\mathbb{R}\cup\{\infty\}$. As an ordered set, $\overline{\mathbb{R}}$ is obtained from the ordered set $\mathbb{R}$ by adding to it the bottom (the least) element $-\infty$ and the top (the greatest) element $\infty$. As a topological space, $\overline{\mathbb{R}}$ is homeomorphic to any closed interval $[a,b]$ (with $a<b$) in $\mathbb{R}$, and is therefore compact. For example, $x\mapsto(2/\pi)\arctan x$ is a homeomorphism $\overline{\mathbb{R}}\to[-1,1]$ (where it is understood that ${-}\infty\mapsto-1$ and $\infty\mapsto1$), and at the same time it is an isomorphism of (totally) ordered sets. All this means that the limits, $\liminf$'s, and $\limsup$'s in the extended real line are just `infinitely stretched out' versions of the limits, $\liminf$'s, and $\limsup$'s in a closed interval. Since in a closed interval all increasing/decreasing sequences converge, so then do all increasing/decreasing sequences in the extended real line; in particular, $\liminf_n x_n$ and $\limsup_n x_n$ always exist. When you are thinking about the extended real line, you can imagine it as a closed interval, you can' go astray with that. If you have to prove some assertion about sequences in the extended real line, formulated in terms of the ordering and the topology (which is, after all, also determined by the ordering), then it suffices to prove the assertion for sequences in a closed interval.

3. You can assume that $x_n$ is a sequence in a closed interval; this will make your life easier, because $\liminf_n x_n$ and $\limsup_n x_n$ are real numbers (belonging to the interval). Now prove the following: $a=\liminf_n x_n$ is the largest real number with the property that for every real number $a'<a$ there are only finitely many $n\in\mathbb{N}$ such that $x_n\leq a'$, and the analogous assertion for $b=\limsup_n x_n$ (which you will formulate yourself --- just turn everything upside-down). With these characterizations of $\liminf_n x_n$ and $\limsup_n x_n$ under the belt, you will easily prove that $\lim_{n\to\infty} x_n$ exists iff $\liminf_n x_n=\limsup_n x_n$.

Remark. You did not ask this, but since I have mentioned imagining the extended real line as a closed interval$\ldots$ $~$You can actually see $\overline{\mathbb{R}}$ as the closed interval $[-1,1]$, and transfer anything that's happening in $\overline{\mathbb{R}}$ to $[-1,1]$. Let $f(y)=\tan((\pi/2)y)$ for $y\in[-1,1]$, so that $f^{-1}(x)=(2/\pi)\arctan(x)$ for $x\in\overline{\mathbb{R}}$. Now define, for all $y,z\in[-1,1]$, \begin{aligned} \mathit{sum}(y,z)~ &:= f^{-1}(f(y)+f(z))~,\\ \mathit{prod}(y,z)~ &:= f^{-1}(f(y)f(z))~. \end{aligned} Actually the operations $\mathit{sum}$ and $\mathit{prod}$ are not defined everywhere on $[-1,1]\times[-1,1]$; for example, $\mathit{plus}$ is not defined at the two points $(1,-1)$ and $(-1,1)$ that correspond to the risque limits of the form $\infty-\infty$. Write the definitions of $\mathit{sum}$ and $\mathit{prod}$ in Mathematica (or a similar program), and then draw their $3\text{D}$ diagrams, placing on all three axes the labels $-\infty$, $-1$, $0$, $1$, $\infty$ at positions, respectively, $-1$, $-1/2$, $0$, $1/2$, $1$. You will be able to observe in the diagrams the behavior of the two operations all the way to the infinity (in either direction); at the points where the operations are not defined you will clearly see the discontinuities (at these points the diagrams contain vertical lines --- more precisely, the closures of the diagrams contain vertical lines).

• A quick question: What exactly do we mean by a sequence in a closed interval? Do we mean that all values $x_n$ are mapped to values in an interval $[a,b]$ such that it is $a \leq x_n \leq b$? (a and b are real numbers) – Ufuk Can Bicici Jul 23 '14 at 9:42
• Yes, we mean exactly that. – chizhek Jul 23 '14 at 9:58
• I have another question, it may be weird but still it bugged me. Let's assume that the sequence $x_n$ takes the values of $\pi$'s digit values after the comma. So each $x_n$ takes a value in the set $(0,1,2,3,4,5,6,7,8,9)$. If we think of the sequence $a_n = \inf_{k \geq n} x_k$ this sequence is clearly nondecreasing and bounded in $[0,9]$, but how can we know the value of limit $\liminf_n x_n$? Since we don't know exactly the values of $x_n$ for very large $n$s, we can be never sure about the value for $a_n$ for these values. – Ufuk Can Bicici Jul 23 '14 at 12:36
• Continued: The infimum can be zero, one or any value, we cannot observe this exactly for this almost random sequence $x_n$ at large $n$s. So, what can we say about the value $\liminf_n x_n$ for such sequences? – Ufuk Can Bicici Jul 23 '14 at 12:39
• For the third question, unfortunately I have failed to find a rigorous proof for that if $a$ is the largest real number such that for every $a' < a$ there is finitely many $x_n < a'$ then $a=\liminf_n x_n$. But intuitively it is clear that this is valid: If we have a plot of a bounded sequence $x_n$ and visualize a vertical line $y=a'$ then there will be finitely many $x_n$ points under that line and as soon as we increase the line over $a$, the number of $x_n$ points will be infinitely many! But I frustratingly fail to express that visualization as a mathematically valid statement... – Ufuk Can Bicici Jul 23 '14 at 15:00


# Smallest number with specific number of divisors

Is there a general method for finding smallest number of specific number of divisors? I am doing "Higher Algebra by Barnard JM Child" and came across a question that "find the smallest number with 24 divisors", that's how I tried to solve it, alert me if I am wrong:

Since $24$ can not have more than $4$ prime factors, the number can not have more than 4 prime factors.

As a single number it is :$2$^${23}$

as product of two numbers: $2^5*3^3$,$2^{11}*3$,$2^7*3^2$, since $5+3<7+2<11+1$, so $2^5*3^3$ is the min of these numbers

as product of three numbers :$2^5*3*5$,$2^3*3^2*5$ since $3+2+1<5+1+1$, hence $2^3*3^2*5$ is the lesser of two

as product of 4 numbers $2^2*3*5*7$

$k=\min(2^{23},2^5*3^3,2^3*3^2*5,2^2*3*5*7)$

=$2^3*3^2*5=360$

The above method seems to be fishy and laborious, is there a general approach to find the smallest number with specific number of divisors?

• For multicharacter exponents, put them in braces. So 2^{11} to get $2^{11}$ Also, \min gets it set as a function. – Ross Millikan Sep 17 '13 at 15:51
• oeis.org/A005179 – barrycarter Sep 7 '17 at 19:17

This approach is not fishy at all, but can be laborious. Note that you can't use $5+3 \lt 7+2$ to conclude that $2^5*3^3 \lt 2^7*3^2$, although it is true, you have to take the ratio and use $3 \lt 2^2$. For example, if you were looking for $36$ factors, one comparison would be $2^8*3^3$ versus $2^5*3^5$. Despite the fact that $5+5 \lt 3+8, 2^5*3^5=7776 \gt 6912=2^8*3^3$. You can take logs and compare $5 \log 2 + 3 \log 3$ with $7 \log 2 + 2 \log 3$ to get it right.

I don't know an easier way. Your example shows the failure of a greedy algorithm. Factor the desired number of factors, here as $3*2^3$. Starting from the largest factor, find the cheapest way to get that many. So start with $2^2$, which has $3$ factors. Now you need to double it. You can either multiply by a new prime, clearly $3$, or increase the exponent of $2$ to $5$. Since the first has a factor $3$ and the second $8$, we choose $2^2*3$ Now we want to double again, and our choices are $2^3, 3^2, \text{ or } 5$ and we take $5$, giving $2^2*3*5$ One more doubling comes from $7$, and we come up with $2^2*3*5*7=420$, not the best.

Take any number $N=p_1^{m_1}...p_k^{m_k}$, where $p_i$ are its prime divisors, and compute how many divisors it has. Each divisor would be a product of the same primes in varying powers, $D=p_1^{d_1}...p_k^{d_k}$, where $0\leq d_i \leq m_i$. Different divisors have different collections of powers, so the number of divisors will be $(m_1+1)(m_2+1)...(m_k+1)$.

Now let's find the smallest $N$ such that $(m_1+1)(m_2+1)...(m_k+1)=24$. There are only few possibilities to break 24 into a product of decreasing numbers: $3*2*2*2=4*3*2=6*4=6*2*2=8*3=12*2=24$. The corresponding candidates with the smallest primes chosen for the smallest powers would be these ones:

$2^{3-1}*3^{2-1}*5^{2-1}*7^{2-1}=2^2*3*5*7=420$

$2^{4-1}*3^{3-1}*5^{2-1}=2^3*3^2*5=360$

$2^5*3^3=2592$

$2^5*3*5=480$

$2^{11}*3=6144$

$2^{23}=8388608$

Then just pick the smallest one, which happens to be 360.

• @ Michael That's what I tried at first,this algorithm becomes laborious if we are dealing with such numbers whose number of divisors has many prime factors,so I want a generalized method in order to avoid comparing numbers>>> – Tom Lynd Sep 17 '13 at 17:23

A005179 lists some resources on the problem. In particular:

Grost, M. (1968). The Smallest Number with a Given Number of Divisors. The American Mathematical Monthly, 75(7), 725-729. doi:10.2307/2315183

From the introduction:

Given $h = q_1 q_2 \dots q_n$, with primes $q_1 \le q_2 \le \dots \le q_n$, let $A(h)$ be the smallest number with $h$ divisors.

In many cases, $$A(h) = 2^{q_1-1} 3^{q_2-1} \dots p_n^{q_n-1}$$

The primary objective of the paper is to determine the exceptions.

We call these numbers ordinary, from R. Brown, The minimal number with a given number of divisors, Journal of Number Theory 116 (2006) 150-158. In this paper Brown shows almost all $A(h)$ are ordinary, in particular "We show here that all square-free numbers are ordinary and that the set of ordinary numbers has natural density one."