Chapter 5 Bending Moments and Shear Force Diagrams for Beams

Similar documents
Module 2. Analysis of Statically Indeterminate Structures by the Matrix Force Method. Version 2 CE IIT, Kharagpur

V. DEMENKO MECHANICS OF MATERIALS LECTURE 6 Plane Bending Deformation. Diagrams of Internal Forces (Continued)

CE 160 Lab 2 Notes: Shear and Moment Diagrams for Beams

Module 1. Energy Methods in Structural Analysis

MECHANICS OF MATERIALS

1 Bending of a beam with a rectangular section

13.4 Work done by Constant Forces

DIRECT CURRENT CIRCUITS

BME 207 Introduction to Biomechanics Spring 2018

COLLEGE OF ENGINEERING AND TECHNOLOGY

approaches as n becomes larger and larger. Since e > 1, the graph of the natural exponential function is as below

Kirchhoff and Mindlin Plates

20 MATHEMATICS POLYNOMIALS

and that at t = 0 the object is at position 5. Find the position of the object at t = 2.

Shear and torsion interaction of hollow core slabs

AQA Further Pure 2. Hyperbolic Functions. Section 2: The inverse hyperbolic functions

Chapter 4 Contravariance, Covariance, and Spacetime Diagrams

4 VECTORS. 4.0 Introduction. Objectives. Activity 1

Solutions to Supplementary Problems

Higher Checklist (Unit 3) Higher Checklist (Unit 3) Vectors

The Wave Equation I. MA 436 Kurt Bryan

PROPERTIES OF AREAS In general, and for an irregular shape, the definition of the centroid at position ( x, y) is given by

Chapter 6 Notes, Larson/Hostetler 3e

Appendix 3, Rises and runs, slopes and sums: tools from calculus

5.2 Volumes: Disks and Washers

Math 1102: Calculus I (Math/Sci majors) MWF 3pm, Fulton Hall 230 Homework 2 solutions

Physics 116C Solution of inhomogeneous ordinary differential equations using Green s functions

Chapter 1 - Functions and Variables

1.2. Linear Variable Coefficient Equations. y + b "! = a y + b " Remark: The case b = 0 and a non-constant can be solved with the same idea as above.

Equations of Motion. Figure 1.1.1: a differential element under the action of surface and body forces

Plates on elastic foundation

A B= ( ) because from A to B is 3 right, 2 down.

In-Class Problems 2 and 3: Projectile Motion Solutions. In-Class Problem 2: Throwing a Stone Down a Hill

I1 = I2 I1 = I2 + I3 I1 + I2 = I3 + I4 I 3

SOLUTIONS TO CONCEPTS CHAPTER

Goals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite

ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER /2019

Statically indeterminate examples - axial loaded members, rod in torsion, members in bending

DERIVATIVES NOTES HARRIS MATH CAMP Introduction

6.5 Plate Problems in Rectangular Coordinates

Partial Derivatives. Limits. For a single variable function f (x), the limit lim

Solution Manual. for. Fracture Mechanics. C.T. Sun and Z.-H. Jin

The Algebra (al-jabr) of Matrices

ARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac

i 3 i 2 Problem 8.31 Shear flow in circular section The centroidal axes are located at the center of the circle as shown above.

10 Deflections due to Bending

Before we can begin Ch. 3 on Radicals, we need to be familiar with perfect squares, cubes, etc. Try and do as many as you can without a calculator!!!

Chapter 0. What is the Lebesgue integral about?

7. Indefinite Integrals

A. Limits - L Hopital s Rule. x c. x c. f x. g x. x c 0 6 = 1 6. D. -1 E. nonexistent. ln ( x 1 ) 1 x 2 1. ( x 2 1) 2. 2x x 1.

We know that if f is a continuous nonnegative function on the interval [a, b], then b

Dynamics: Newton s Laws of Motion

NUMERICAL INTEGRATION. The inverse process to differentiation in calculus is integration. Mathematically, integration is represented by.

MORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.)

Lecture 13 - Linking E, ϕ, and ρ

10.5. ; 43. The points of intersection of the cardioid r 1 sin and. ; Graph the curve and find its length. CONIC SECTIONS

u t = k 2 u x 2 (1) a n sin nπx sin 2 L e k(nπ/l) t f(x) = sin nπx f(x) sin nπx dx (6) 2 L f(x 0 ) sin nπx 0 2 L sin nπx 0 nπx

Chapters 4 & 5 Integrals & Applications

Columns and Stability

Explain shortly the meaning of the following eight words in relation to shells structures.

SUMMER KNOWHOW STUDY AND LEARNING CENTRE

MASTER CLASS PROGRAM UNIT 4 SPECIALIST MATHEMATICS WEEK 11 WRITTEN EXAMINATION 2 SOLUTIONS SECTION 1 MULTIPLE CHOICE QUESTIONS

INTRODUCTION. The three general approaches to the solution of kinetics problems are:

DEFINITION OF ASSOCIATIVE OR DIRECT PRODUCT AND ROTATION OF VECTORS

9.4. The Vector Product. Introduction. Prerequisites. Learning Outcomes

Sections 1.3, 7.1, and 9.2: Properties of Exponents and Radical Notation

A wire. 100 kg. Fig. 1.1

16 Newton s Laws #3: Components, Friction, Ramps, Pulleys, and Strings

Things to Memorize: A Partial List. January 27, 2017

Lesson Notes: Week 40-Vectors

A. Limits - L Hopital s Rule ( ) How to find it: Try and find limits by traditional methods (plugging in). If you get 0 0 or!!, apply C.! 1 6 C.

On the diagram below the displacement is represented by the directed line segment OA.

Review of Calculus, cont d

Improper Integrals, and Differential Equations

Unit #9 : Definite Integral Properties; Fundamental Theorem of Calculus

Higher Maths. Self Check Booklet. visit for a wealth of free online maths resources at all levels from S1 to S6

99/105 Comparison of OrcaFlex with standard theoretical results

P 1 (x 1, y 1 ) is given by,.

Families of Solutions to Bernoulli ODEs

Plate Theory. Section 11: PLATE BENDING ELEMENTS

fractions Let s Learn to

Ch AP Problems

Plate Theory. Section 13: PLATE BENDING ELEMENTS

4. Calculus of Variations

Phys101 Lecture 4,5 Dynamics: Newton s Laws of Motion

Set up Invariable Axiom of Force Equilibrium and Solve Problems about Transformation of Force and Gravitational Mass

3.4 Numerical integration

Math 113 Exam 2 Practice

Week 10: Line Integrals

4.4 Areas, Integrals and Antiderivatives

Chapter 1: Logarithmic functions and indices

In Section 5.3 we considered initial value problems for the linear second order equation. y.a/ C ˇy 0.a/ D k 1 (13.1.4)

Section 7.2 Velocity. Solution

1 Line Integrals in Plane.

Lecture 1. Functional series. Pointwise and uniform convergence.

Applications of Bernoulli s theorem. Lecture - 7

1. a) Describe the principle characteristics and uses of the following types of PV cell: Single Crystal Silicon Poly Crystal Silicon

Sample pages. 9:04 Equations with grouping symbols

PHYS Summer Professor Caillault Homework Solutions. Chapter 2

Thomas Whitham Sixth Form

Transcription:

Chpter 5 ending Moments nd Sher Force Digrms for ems n ddition to illy loded brs/rods (e.g. truss) nd torsionl shfts, the structurl members my eperience some lods perpendiculr to the is of the bem nd will cse only sher nd bending in the bem. The current chpter together with Chpters 6 to 8 will focus on such n issue. 5. SHER FORCE ND ENDNG MOMENTS DGRMS FOR EMS Sher Force Digrm (SFD)indictes how force pplied perpendiculr to the is (i.e. prllel to cross section) of bem is trnsmitted long the length of tht bem. ending Moment Digrm (MD) will show how the pplied lods to bem crete moment vrition long the length of the bem. These digrms re used to determine the norml nd sher stresses s well s deflection nd slopes in the following chpters. 5. EM SGN CONVENTON (S&4 th :56-57; 5 th :56-57) t ny point long its length, bem cn trnsmit bending moment M() nd sher force V(). f loded bem is cut, the definitions of positive distributed lod, sher force nd positive bending moment re s Fig. 5. below: ositive internl bending moment ositive distributed lod ositive internl sher force Fig. 5. em sher force nd bending moment sign convention Where distributed lod cts downwrd on the bem; internl sher force cuses clockwise rottion of the bem segment on which it cts; nd the internl moment cuses compression in the top fibers of the segment, or to bend the segment so tht it holds wter. 5. RETONSH ETWEEN EM ODNGS (S&4 th :64-68; 5 th :64-68) bem (Fig. 5.) is loded with verticl forces F i, bending moments M i nd distributed lods w(). F F w() F..D. of element d w() M() M() dm() d d M M V() d V() dv() d d d Fig. 5. Trnsversely loded bem nd free body digrm of element d ecture Notes of Mechnics of Solids, Chpter 5

ook t the FD of n elementl length d of the bove loded bem (Fig. 5.). s it hs n infinitesiml length, the distributed lod cn be considered s Uniformly Distributed od (UD) with constnt mgnitude w() over the differentil length d. t is now necessry to equte the equilibrium of the element. Strting with verticl equilibrium dv ( ) ( ) ( ) ( ) F y = = V w d V d = d (5.) Dividing by d in the limit s d, dv ( ) = w( ) d (5.) Tking moments bout the right hnd edge of the element: d dm ( ) ( ) ( ) ( ) ( ) M R.H. Edge = = M V d w d M d = d (5.3) Dividing by d in the limit s d, dm ( ) = V ( ) d (5.4) Eqs. (5.) & (5.3) re importnt when we hve found one nd wnt to determine the others. 5.3 ENDNG MOMENT ND SHER FORCE EQUTONS ntroductionry Emple - Simply Supported em y using the free body digrm technique, the bending moment nd sher force distributions cn be clculted long the length of the bem. et s tke simply supported bem, Fig. (5.3), s n emple to shown the solutions: F..D. (globl equilibrium) F..D. (method of section -) M() o V() R Y =(-/) R Y =/ R Y =(-/) Fig. 5.3 FD of bem cut before force Step : Cut bem just before the force (i.e. Section -), nd drw free body digrm including the unknown sher force nd bending moment s in Fig. 5.3. Tke moments bout the right hnd end (O): M o = = M ( ) = M ( ) = (5.5) To determine the sher force, use Eq. (5.4), giving tht: dm ( ) ( ) V = = (5.6) d To verify Eq. (5.6), equte verticl equilibrium: ecture Notes of Mechnics of Solids, Chpter 5

F y = V ( ) = V ( ) = which is the sme eqution s (5.6). These then, re the equtions for the bending moment nd sher force vrition in the rnge of. To find out the rest of the bending moment nd sher force distributions, it is necessry to now crry out similr nlysis, but cutting the bem just before the end (Section -). Step : Cut bem just before the right hnd end (RHE) F..D. (Section -) M() o V() R Y =( -/) Fig. 5.4 FD of bem cut before the right hnd end Equte moments bout the right side: M = = ( ) M ( ) = giving: M ( ) = ( ) = (5.7) nd using Eq. (5.4), the sher force eqution is : dm ( ) ( ) V = = (5.8) d These epressions for the bending moment nd sher force cn now be plotted ginst to produce the Sher Force nd ending Moment Digrms s Fig. 5.5: ecture Notes of Mechnics of Solids, Chpter 5 3

oding Digrm (-/) / V() (-/) Sher Force Digrm -/ M() (-/) ending Moment Digrm Fig. 5.5 Sher Force nd ending Moment Digrms for simply supported bem Mculey's Nottion (4 th :59-599; 5 th :59-599) The two sets of equtions for V() nd M(), Eqs. (5.5), (5.6), (5.7) nd (5.8), cn be condensed to just one set of equtions if we use specil type of nottion clled Mculey's Nottion. The bove equtions would look like this (to be derived in Emple 5.) V ( ) = (5.9) M ( ) = (5.) Where the nottion hs the following mening: n = ( ) n for < for ( n ) (5.) when differentiting: n n = n for n for n = for n = (5.) Remrks To derive the bending moment eqution by using Mculey's nottion, you my need to do the following: ) Determine the ground rections from globl equilibrium; ) Cut the bem just before the right hnd end; ecture Notes of Mechnics of Solids, Chpter 5 4

3) Equte the cut FD to equilibrium bout the right hnd end; 4) ll length terms in the bending moment/sher force equtions MUST be written using Mculey's nottion; 5) lwys indicte the powers, even if they re or. Emple 5.: s in the introductory emple, determine the sher force nd bending moment equtions nd plot them for simply-supported bem s in the introductory emple. Step : Determine the ground rections; F..D. (globl equilibrium) We hve R Y =(-/) R Y = (-/) nd R Y = / R Y =/ Step : Drw FD of bem cut just before the RHS (Section -). F..D. (Section -) o R Y =( -/) V() M() Step 3: Equilibrium for FD of bem cut just before the RHS (Section -). Tke moments bout RHS: M O = = M ( ) = M ( ) = nd differentiting w.r.t. '', s Eq. (5.4), gives the sher force eqution s: dm ( ) ( ) V = = d Step 4: lotting the Sher Force nd ending Moment Digrms ecture Notes of Mechnics of Solids, Chpter 5 5

ccording to M() nd V() to depict the digrms, ook t the equtions segment by segment When M ( ) = ( ) = nd V ( ) = ( ) = To plot this segment in the digrm, firstly look t the boundry points s =,M ( ) = nd =,M ( ) = ( / ). Drw two points nd then connect them becuse the eqution gives line. ikewise, one cn plot Sher Force Digrm in this region. When M ( ) = ( ) ( ) = = nd V ( ) = ( ) ( ) = = Remrks: lese drw globl FD of the bem firstly nd follow by its Ser Force nd ending Moment Digrms. The reson for doing this is tht when you get sufficient eperience, you my be ble to directly plot the Sher Force Digrm by observing the eternl forces s well s plot ending Moment Digrms by observing the Sher Force Digrm. Nevertheless you MUST still work out nd indicte the loctions nd vlues (including or ve) t ll turning points in the digrms in detil. t is lso interesting to note tht concentrted forces (e.g. rection forces nd eternl forces) correspond to inclined line in MD nd horizontl line in SFD. oding Digrm (-/) / V() (-/) Sher Force Digrm -/ M() (-/) ending Moment Digrm Emple 5.: Determine the sher force nd bending moment equtions nd plot them for simply-supported bem loded with UD. ecture Notes of Mechnics of Solids, Chpter 5 6

Step : Determine the ground rections; From globl equilibrium the ground rection forces cn be found to be both equl to w/ s, w F..D. (globl equilibrium) w/ w/ M = = R Y ( w) = R Y = w / ( upwrds) Fy = = RY RY w = R Y = w / ( upwrds) Step : Drw FD of bem cut just before the RHS (Section -). F..D. (Section -) w R w / o M() V() R Y =w/ Step 3: Equilibrium for FD of bem cut just before the RHS (Section -). s fr s V() nd M() re concerned the UD cn be temporrily replced by its resultnt R w (=w) pplied t the centroid of the UD distribution in the moment equilibrium eqution. So if we tke moments bout the RHS of the bem we get: O = = R Y Rw M = M w w M ( ) = nd differentiting w.r.t. '', s Eq. (5.4), gives : dm ( ) ( ) w V = = w d Step 4: lotting the Sher Force nd ending Moment Digrms ccording to M() nd V() to depict the digrms ( ) = ( w / ) ( w ) M ( ) ecture Notes of Mechnics of Solids, Chpter 5 7

w oding Digrm w/ V() w/ w/ Sher Force Digrm -w/ M() w /8 rbol ending Moment Digrm t is worth pointing out tht one should not completely replce such UD by its corresponding resultnt concentrted force R w (=w) in the beginning of the solution. There is significnt difference of the Sher Force nd ending Moment Digrms between concentrted force (Emple 5.) nd UD (Emple 5.). t is lso interesting to note tht the UD corresponds to n inclined line in the Sher Force Digrm nd qudrtic curve (prbol) in the ending Moment Digrms. ecture Notes of Mechnics of Solids, Chpter 5 8

Emple 5.: Determine the sher force nd bending moment equtions nd plot them for bem loded with UD between nd nd two concentrted forces t C nd E. F..D. (globl equilibrium) R Y UD=w=kN/m kn m 5m 5m m R DY C D kn E Step : Determine the ground rections; M = = 5 5 RDY 3 = R DY = 3. 5kN Fy = = R Y RDY = R Y = 7. 5kN Step : Drw FD of bem cut just before the RHS (Section -). Note: The only problem with Mculey's Nottion is tht it does not work when UD stops. t however does work for UD which strts nywhere long bem nd continues to the end. The problem cn be corrected by pplying UD of equl mgnitude but opposite sense where the first UD ends. F..D. (Section -) nd ppliction of equivlent UD w=kn/m kn o M() 7.5kN 3.5kN V() Step 3: Equilibrium for FD of bem cut just before the RHS (Section -). Tke moments bout RHS: M O = = 7. 5 5 3. 5 M ( ) = 7. 5 5 3. 5 nd differentiting w.r.t. '', s Eq. (5.4), gives the sher force eqution s: dm V ( ) = = 7. 5 5 3. 5 d ( ) M ( ) ecture Notes of Mechnics of Solids, Chpter 5 9

Step 4: lotting the Sher Force nd ending Moment Digrms oding Digrm UD=w=kN/m 7.5kN kn m 5m 5m m 3.5kN C D kn E Sher Force Digrm V() kn 7.5 -.5 -.5 ending Moment Digrm M() knm qudrtic 8.5 5.5 - gin, the UD segment corresponds to n inclined line in SFD nd qudrtic curve in MD. Emple 5.3: Determine the sher force nd bending moment equtions nd plot them for cntilever bem loded with moment M = 4kNm nd force F= kn. Globl F..D. M M =4kNm kn C R Y 4m.5m Step : Determine the ground rections; The cntilever bem is fully clmped in the left hnd end s shown. The ground rection for this point should hve rection force R Y nd rection moment M.. So the globl equilibrium is given s F y = = R Y = R Y = kn (- downwrds) M = = M 4 5. 5 = M = 95kNm (- clockwise) ecture Notes of Mechnics of Solids, Chpter 5

Step : Drw FD of bem cut just before the RHS (Section -). F..D. (Section -) 95kNm M =4kNm O M() kn 4m V() Step 3: Equilibrium for FD of bem cut just before the RHS (Section -). Tke moments bout RHS: O 4 M = = 95 4 ( ) M ( ) M = 95 4 4 nd differentiting w.r.t. '', s Eq. (5.4), gives the sher force eqution s: dm ( ) ( ) V = = = d Step 4: lotting the Sher Force nd ending Moment Digrms 95kNm oding Digrm M =4kNm kn C kn Sher Force Digrm V() kn - M() knm 95 ending Moment Digrm 5 55 t is interesting to observe tht due to concentrted bending moments M (rection moment) nd M (eternl moment), there is respectively sudden lep nd drop in the ending Moment Digrm. n ddition, such concentrted moments do not ffect the Sher Force Digrm (Note tht the drop t point (the left end) in SFD is due to the concentrted rection force R Y. n fct, both M nd M do not pper in sher force eqution V() t ll). ecture Notes of Mechnics of Solids, Chpter 5