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

Save this PDF as:

Size: px
Start display at page:

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

## Transcription

1 V. DEMENKO MECHNCS OF MTERLS LECTURE 6 Plne ending Deformtion. Digrms of nternl Forces (Continued) 1 Construction of ending Moment nd Shering Force Digrms for Two Supported ems n this mode of loding, first of ll, we should determine the support rections. Let us consider the method in following emple. Emple 1 nternl forces induced by concentrted force Given: F,, b. R.D.: Q( ), My( ). Let us find the support rections R nd R from the equtions of equilibrium of the bem: M = 0 R + b F = 0, (1) R = F + b, () M = 0 R ( + b) Fb = 0, (3) Fig. 1 R = Fb + b. (4) Let us check the support rections The result is true. Fb F F = 0, F + = 0. (5) + b + b Let us divide the bem into portions nd write the equtions of shering forces nd bending moments for n rbitrry cross-sections of two portions: - nd -. From the conditions of equilibrium of the left (-) or right (-) portion of the mentlly cut rod it follows tht for portion t distnce from support internl forces re described by the equtions.

2 V. DEMENKO MECHNCS OF MTERLS 015 Fb b = =, M ( ) R + 0 Q R Fb Fb = = = 0 = + b = = + b. (6) y Portion. n rbitrry section is situted t distnce within the limits F 0 b: Q ( ) = R = + b, (7) F Fb My ( ) = R= = 0 = + b = 0 = b + b. (8) Thus, in ech portion of the bem, the force Q is constnt in vlue nd is positive for portion nd negtive for portion. The moment depends linerly on ; it increses in portion from 0 to Fb ( + b) nd increses from ero to this vlue in portion. The digrms of Q ( ) nd M ( ) cn now be creted using results of these clcultions. y t should be noted tht the digrm of shering forces hs n brupt in the point of ppliction of specified eternl force, which is equl to the mgnitude of the force. R, F, For instnce, the Q ( ) digrm hs three brupts corresponding to the forces R. There re no ny brupt in the digrm of bending moments. Fig. Emple nternl forces induced by concentrted moment Given: M,, b. R.D.: Q (), M y (). From the equtions of equilibrium, the rections re since R = M R = + b, (9) M = 0; R + b M = 0, R = M, R = R. (10) + b Equtions of internl forces: Portion : 0 M Q ( ) = R = + b, (11)

3 Portion : 0 b V. DEMENKO MECHNCS OF MTERLS M M M ( ) = R = + b = = + b. (1) y 0 = = 0 M Q ( ) =+ R =+ + b, (13) M Mb M ( ) = R = + b = = + b y 0 = b = 0 (14) Note: point of n eternl moment ppliction, the bending moment digrm hs jump (brupt). t is equl to the mgnitude of the ctive moment pplied in this point. Emple 3 nternl forces induced by uniformly distributed lod Given: q, l. R.D.: Q( ), My( ). Let us determine the rections of supports nd from conditions of equilibrium: R Equtions of internl forces: n portion (0 l) : ql = R =. (15) Fig. 3 ql ql ql Q ( ) = q = = =, (16) 0 = l ql q ql My( ) = = 0 = = 0. (17) 8 = l = 0 = l

4 4 V. DEMENKO MECHNCS OF MTERLS 015 The shering force decreses linerly from ql to ql nd the bending moment vries long the bem length non-linerly nd ttins the mimum vlue in the bem midsection. Emple 4 nternl forces induced by linerly distributed lod Given: q m, l. R.D.: Q( ), My( ). Clcultion of the support rections: M = 0, ql 1 Rl l = 0, (18) 3 ql R =, (19) 6 M = 0, ql Rl l = 0, (0) 3 ql R =, (1) 3 Checking: Fig. 4 ql ql ql F = 0, + = 0. () 6 3 The equtions of Q( ) nd My( ) re s follows: q ql q ql ql Q ( ) = R + = 6 + l = l 6 = = l 3 ; (3) Etreme vlue of bending moment is clculted substituting into its eqution corresponding coordinte of the point. The lst one is determined equting to ero the derivtive of bending moment function which is described by sher force (check if you disgree): = 0 ql q e e 0 e l Q = + = =. (4) 6 l 3

5 V. DEMENKO MECHNCS OF MTERLS q ql q ql y 0 0 l l = l = e = 0 M = R = = = =. (5) Reltionships etween ending Moment M y, Shering Force Q nd Distributed Lod ntensity q() Let rod be fied in rbitrry mnner nd subjected to distributed lod of intensity q q( ) = in generl cse. There re other lods pplied to the bem. The direction dopted for q is considered positive (see Fig. 5). moments solte n element of length d from the rod nd pply to both its boundries the M y nd My dm y + nd lso shering forces Q nd Q + dq. Fig. 5 The lod q my be considered to be uniformly distributed within the infinitesimlly smll segment d (see Fig. 6). Equte to ero the sum of the projections of ll forces on verticl is : 0 Q + q d Q dq =. (6) Mking simplifictions, we obtin dq d ( ) = q. (7) Fig. 6 Equte to ero the sum of the moments with respect trnsverse is y (point ) d My + Q d+ q( ) d My dm y = 0. (8)

6 6 V. DEMENKO MECHNCS OF MTERLS 015 Mking simplifictions nd rejecting the smll quntity of higher order, we obtin dm y d ( ) ( ) = Q. (9) Thus the shering force in fct represents the derivtive of the bending moment with respect to the length of the rod. This conclusion ws proved in the emple mentioned bove. The derivtive of the shering force equls to the intensity of eternl distributed lod q. Emple 5 Generl cse of cntilever bem loding Given: = m, b = 1m, c = 3m, F = 50kN, q = 40 knm, M = 60kNm. R.D.: Q( ), My( )? Equtions of internl forces in n rbitrry cross-sections of the portions: 0 < < 0 50 Q =+ F q = = = = = 30kN, My ( ) =+ F q = 0 = = 0 = 0 knm. = Fig. 7 dm d e M y m? dm Q F q; d F = F qe = 0 e = = 1.5m. q = = Q ( ) = 0. e

7 V. DEMENKO MECHNCS OF MTERLS q e M = M 50( 1.5) ym y e = F e = = = knm. 0 < < b 30 Q =+ F q = kn, My ( ) = F( + ) q + = 0= 50 ( + 0) 40 1 () = 0 = 1= 10 knm. 0 < < b 30 Q = F q = kn, My ( ) = F( + b+ ) q + b+ M = 0= 10 = 1= 70kNm. Emple 6 Generl cse of two supported bem loding Fig. 8 Given: q = 0 knm, F = 50kN, M = 130 knm, = 3m, b= c= d = 1m. R.D.: Q( ), My( )? 1) Clcultion of rections R, M = 0 =+ M + F + b + + q R ( + b+ c+ d), R =+ 70 kn. R : The plus sign mens coincidence of R ctul direction with originlly selected.

8 8 V. DEMENKO MECHNCS OF MTERLS 015 M 0 M F c d q = =+ + + b+ c+ d + R + b+ c+ d, R =+ 40 kn. Checking: F = 0= R R + F + q = , i.e. the rections re determined correctly. ) Equtions of internl forces in n rbitrry cross-sections of corresponding portions: 0< < 1 70 Q = R = kn, 0 1 My =+ R = = 0 = = 70 knm. 0< < 1 70 Q = R = kn, 0 1 My =+ R d + M = 60 = = 10kNm. 0< < Q = R q = = 40 = = 0 knm, M y m? = = 0 e e Q R q e R 40 e = = = m, q 0 M ( e) = Re q =+ 40kNm. y V V V 0< < 1 0 Q = R q = kn, V My ( ) =+ R( + ) q + = 0= 30 = 1= 10 knm.

9 V. DEMENKO MECHNCS OF MTERLS Emple 7 Generl cse of two supported bem loding Given: q = 0 kn/m, F = 100kN, M = 60kNm, = m, b = 1m, c = 3m, d = 4 m. R.D.: Q( ), My( )? 1) Clcultion of rections R, R : orig M = 0 = R b+ c + F b+ c+ d c+ d qc ( + d) + b M + M q = = R (), 4= 190, R Fig. 9 R orig ct 190 = R = = kn. 4 The plus sign mens coincidence of direction with originlly selected. (b) M = 0=+ Fd + M M + R ctul c d orig + qc qd + R b+ c q + b+ c = 9 = 100( 4) orig orig 0 + R ( 4) 40( ) = R ( 4) 190, + R = 47.5 kn. The minus sign mens non-coincidence of Thus, originl direction of R ctul direction with originlly selected. R must be chnged on opposite. ct orig (c) Checking: F = 0 =+ R R + F + q qc ( + d) = 0, i.e. the rections re determined correctly.

10 10 V. DEMENKO MECHNCS OF MTERLS 015 ) Equtions of internl forces in n rbitrry cross-sections of corresponding portions: 0 < < = 0= 0 = 40 kn, My( ) q = 0 = Q q = = 0 < < b 87.5 Q = q R = kn, ctul = = 0 = 40 knm. My ( ) = q + Rctul+ M = 0=+ 80 = 1= 7.5 knm. 0 < < d = 0 = 4 Q =+ F q =+ 100kN =+ 0kN, My ( ) = F+ q = 0= 0 = 4 = 40kNm. V V 0 < < c = 0 = 3 V Q =+ F q d + R = 7.5kN = 87.5kN, V My ( ) = F( d + ) + R+ q( d + ) d + = 0= 40kN = 3 = 67.5kNm. 3 Construction of nternl Force Digrms for Stticlly Determinted Frmes y br system is ment ny structure consisting of rod-shped elements. f the elements of structure ct primrily in tension or compression, the br system is clled truss. f the elements of br system re primrily in bending or torsion, the system is clled frme. We will consider plne frmes. y stticlly determinte system is ment system for which ll the rections of the supports cn be determined by mens of equtions of equilibrium nd the internl forces t ny cross section cn lso be found by the method of sections.

11 V. DEMENKO MECHNCS OF MTERLS Given: F,, b. R.D.: N( ), Q( ), My( ) functions nd their grphs. Solution Equtions of internl forces in two portions re: 0 < < N ( ) = 0, 0 Q =, Fig y = = M = F = = F, 0 < < b =, Q ( ) = 0, N F Corresponding grphs re: M = F. y Fig. 11

12 1 V. DEMENKO MECHNCS OF MTERLS 015 Emple 9 Clcultion of internl forces in plne frme Given: q = 0 kn/m, l = 3m. orig v R = 15kN (ctul direction of R.D.: N( ), Q( ), My( ). t first let us determine the rections in supports nd : ) M = 0, R = 0, R =+ 15kN; b) F = 0, h R = qb =+ 60kN: c) M = 0, orig v R = 0, ct R v is opposite to originl). Let us write the equtions of the internl forces: N ( ) = 0 ; 15 N =+ R =+ kn; 60 N =+ R =+ kn; h 15 Q = R = kn; 0 Q = q 0 60 = = = = kn; l ct 15 Q =+ R = kn; v 0 My = R 0 45 = = = = knm; l q y = 3 = 0 M = R l =+ 45 = 45 knm; Fig. 1 ct 0 3 My = Rv 0 45 = = = = knm.

13 V. DEMENKO MECHNCS OF MTERLS Using this equtions the bending moment digrm nd the digrms of norml nd shering forces my be constructed: N ( ), kn Q ( ), kn My, knm Fig. 15 Checking the results, i.e. the equilibrium of the frme ngles. Fig. 16 Emple 10 Clcultion of internl forces in plne frme Given: q = 40 kn/m, F = 50kN, M = 40kNm, = m, b = 4m, c = 3m. R.D.: N( ), Q( ), My( )? 1) Clcultion of support rections: b orig orig () M = 0= M M + F q + qb + R 65 v b Rv = kn. The minus sign mens non-coincidence of R v ctul direction with originlly selected. Thus,

14 14 V. DEMENKO MECHNCS OF MTERLS 015 originl direction of R v must be chnged on opposite. orig ct (b) F = 0= F + R R + q + b R = 15 kn, i.e. ctul direction of R is upwrds. v (c) F = 0= R F R =+ 50 kn (right directed). h h ) Equtions of internl forces in n rbitrry cross-sections of the portions (see Fig. 17): 0 < < 0 N =, 0 Q =+ F q = = e F 50 = = = 1.5 m. q 40 Fig. 17 = 50 = 30 kn, = My( ) = F q = 0 = = 0 = =+ 0 knm, = 0 e e Q = F q =, 1 My( e) = Mym = Fe qe = knm. 0 < < c = = 15kN, Q ( ) F 50 N R 0 3 My = F = = 0 = = 150kNm. 0 < < c =+ =+ kn, = = 65kN, Q ( ) R 50 N R v 0 3 = = kn, My =+ Rh+ M = = 40 = =+ 190kNm. h

15 V V V 0 < < c V. DEMENKO MECHNCS OF MTERLS V =+ =+ 50 kn, 0 4 N R h Q = Rv + q = = 65 = = 95kN, V My ( ) = q + Rv+ M + Rhc = 0= 190 = 4 = 130kNm. V ct R 65 Q 0 v e = Rv + qe = e = = = 1.65 m, q 40 ct V V e ct ct y ( e) ym v e h ( 1.65) M = M = q + R + M + R c= = 8.4 knm. Fig. 18 Fig. 19 Fig. 0 3) Checking the results i.e. n equilibrium of the rods connection res (see Figs 1 ).

16 16 V. DEMENKO MECHNCS OF MTERLS 015 Fig. 1 Fig. 4 Construction of the Q () nd M y () Digrms for Curviliner ems Given: R, F. R.D: N( ), Q( ), My( ). The force F cn be resolved long the nd es into the components F sinα. F cosα nd Fig. 3 F = Fsinα, F = Fcosα, 0 N α = F = Fsinα 0 F 0, α = = α = π = α = π = 0 Q α =+ F = Fcos α F 0 F, α = = α = = π α = = π 0 My α = Fh = FRsinα 0 FR 0 α = = α = π = α = π =. Fig. 4

17 V. DEMENKO MECHNCS OF MTERLS Emple 11 Clcultion of internl forces in curviliner frme Given: M, F, R. R.D.: N( ), Q( ), My( ). Solution Equtions of internl forces re: 0 < α < π / N ( α ) = 0; 0 Q α = ; Fig. 5 My α = M. 0 < α < π / ( α) sinα 0 α = 0 α = π / N = P = = P, ( α) α α = 0 α = π / Q = Pcos = P = 0, y ( α) sinα α = 0 α = π / M = M + PR = M = M + PR. The grphs of internl forces re: Fig. 6

18 18 elements V. DEMENKO MECHNCS OF MTERLS 015 Emple 1 Clcultion of internl forces in plne frmes with curviliner Given: M = 40kHm, q = 10kN/m, = 1m. R.D.: N, Q, M y functions nd their grphs. Solution 1) Clcultion of support rections: F = 0 q R = 0 = 0 kn, R H H M F = 0 R M q = 0 = 30 kn, R rbitrry cross-sections of corresponding portions: 0 = = 30kN, Q ( ) = 0, 0 N R 0 α π 0 N = R cos = 30 = 0, α α = α = π α α = 0 α = π y V V F = 0 R R = 0 R = 30 kn. ) Equtions of internl forces in n M =. Q = R sin = 0 = 30 kn, ( α) y α= 0 α= π M = R 1 cos = 0 = R = 30 knm. 0 α π α α = 0 α = π N = R sin = 0 = 30 kn, y Fig. 7 α α = 0 α = π Q = R cos = 30 = 0,

19 V. DEMENKO MECHNCS OF MTERLS ( α) = 0 = M y = R 1+ sin M = 10 = 0 knm. α α π V V V 0 α V = = 30 kn, Qy ( ) RH q = 0 = N R V V q y АH = = 0 M = R = 0 = 0 knm. = = 0 = 0, Fig. 8 Glossry Stticlly determinte system (рус. статически опpеделимая система, укр. статично визначувана система) the system, rections nd internl forces of which could be determined solely from free-body digrms nd equtions of equilibrium. Truss (рус. ферма, укр. ферма) frme, generlly of steel, timber, concrete, or light lloy, built from members in tension nd compression.

### Chapter 5 Bending Moments and Shear Force Diagrams for Beams

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

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

Module Anlysis of Stticlly Indeterminte Structures by the Mtrix Force Method Version CE IIT, Khrgpur esson 8 The Force Method of Anlysis: Bems Version CE IIT, Khrgpur Instructionl Objectives After reding

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

E 160 Lb 2 Notes: Sher nd oment Digrms for ems Sher nd moment digrms re plots of how the internl bending moment nd sher vry long the length of the bem. Sign onvention for nd onsider the rbitrrily loded

### Module 1. Energy Methods in Structural Analysis

Module 1 Energy Methods in Structurl Anlysis Lesson 4 Theorem of Lest Work Instructionl Objectives After reding this lesson, the reder will be ble to: 1. Stte nd prove theorem of Lest Work.. Anlyse stticlly

### Calculus 2: Integration. Differentiation. Integration

Clculus 2: Integrtion The reverse process to differentition is known s integrtion. Differentition f() f () Integrtion As it is the opposite of finding the derivtive, the function obtined b integrtion is

### 1 Bending of a beam with a rectangular section

1 Bending of bem with rectngulr section x3 Episseur b M x 2 x x 1 2h M Figure 1 : Geometry of the bem nd pplied lod The bem in figure 1 hs rectngur section (thickness 2h, width b. The pplied lod is pure

### MECHANICS OF MATERIALS

9 The cgrw-hill Compnies, Inc. All rights reserved. Fifth SI Edition CHAPTER 5 ECHANICS OF ATERIALS Ferdinnd P. Beer E. Russell Johnston, Jr. John T. DeWolf Dvid F. zurek Lecture Notes: J. Wlt Oler Texs

### BME 207 Introduction to Biomechanics Spring 2018

April 6, 28 UNIVERSITY O RHODE ISAND Deprtment of Electricl, Computer nd Biomedicl Engineering BME 27 Introduction to Biomechnics Spring 28 Homework 8 Prolem 14.6 in the textook. In ddition to prts -e,

### Fundamental Theorem of Calculus

Fundmentl Theorem of Clculus Recll tht if f is nonnegtive nd continuous on [, ], then the re under its grph etween nd is the definite integrl A= f() d Now, for in the intervl [, ], let A() e the re under

### We partition C into n small arcs by forming a partition of [a, b] by picking s i as follows: a = s 0 < s 1 < < s n = b.

Mth 255 - Vector lculus II Notes 4.2 Pth nd Line Integrls We begin with discussion of pth integrls (the book clls them sclr line integrls). We will do this for function of two vribles, but these ides cn

### ( dg. ) 2 dt. + dt. dt j + dh. + dt. r(t) dt. Comparing this equation with the one listed above for the length of see that

Arc Length of Curves in Three Dimensionl Spce If the vector function r(t) f(t) i + g(t) j + h(t) k trces out the curve C s t vries, we cn mesure distnces long C using formul nerly identicl to one tht we

### 5.7 Improper Integrals

458 pplictions of definite integrls 5.7 Improper Integrls In Section 5.4, we computed the work required to lift pylod of mss m from the surfce of moon of mss nd rdius R to height H bove the surfce of the

### COLLEGE OF ENGINEERING AND TECHNOLOGY

COLLEGE OF ENGNEERNG ND TECHNOLOGY DEPRTMENT : Construction nd uilding Engineering COURSE : Structurl nlysis 2 COURSE NO : C 343 LECTURER : Dr. Mohmed SFN T. SSSTNT : Eng. Mostf Yossef, Eng. l-hussein

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

Elsticity nd Plsticity Stticlly indeterminte exmples - xil loded memers, rod in torsion, memers in ending Deprtment of Structurl Mechnics Fculty of Civil Engineering, VSB - Technicl University Ostrv 1

### DIRECT CURRENT CIRCUITS

DRECT CURRENT CUTS ELECTRC POWER Consider the circuit shown in the Figure where bttery is connected to resistor R. A positive chrge dq will gin potentil energy s it moves from point to point b through

### Section 14.3 Arc Length and Curvature

Section 4.3 Arc Length nd Curvture Clculus on Curves in Spce In this section, we ly the foundtions for describing the movement of n object in spce.. Vector Function Bsics In Clc, formul for rc length in

### Lecture 13 - Linking E, ϕ, and ρ

Lecture 13 - Linking E, ϕ, nd ρ A Puzzle... Inner-Surfce Chrge Density A positive point chrge q is locted off-center inside neutrl conducting sphericl shell. We know from Guss s lw tht the totl chrge on

### ES.182A Topic 32 Notes Jeremy Orloff

ES.8A Topic 3 Notes Jerem Orloff 3 Polr coordintes nd double integrls 3. Polr Coordintes (, ) = (r cos(θ), r sin(θ)) r θ Stndrd,, r, θ tringle Polr coordintes re just stndrd trigonometric reltions. In

### KINEMATICS OF RIGID BODIES

KINEMTICS OF RIGID ODIES Introduction In rigid body kinemtics, e use the reltionships governing the displcement, velocity nd ccelertion, but must lso ccount for the rottionl motion of the body. Description

### 13.4 Work done by Constant Forces

13.4 Work done by Constnt Forces We will begin our discussion of the concept of work by nlyzing the motion of n object in one dimension cted on by constnt forces. Let s consider the following exmple: push

### Here are the graphs of some power functions with negative index y (x) =ax n = a n is a positive integer, and a 6= 0acoe±cient.

BEE4 { Bsic Mthemtics for Economists BEE5 { Introduction to Mthemticl Economics Week 9, Lecture, Notes: Rtionl Functions, 26//2 Hint: The WEB site for the tetbook is worth look. Dieter Blkenborg Deprtment

### Review of Calculus, cont d

Jim Lmbers MAT 460 Fll Semester 2009-10 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some

### KEY CONCEPTS. satisfies the differential equation da. = 0. Note : If F (x) is any integral of f (x) then, x a

KEY CONCEPTS THINGS TO REMEMBER :. The re ounded y the curve y = f(), the -is nd the ordintes t = & = is given y, A = f () d = y d.. If the re is elow the is then A is negtive. The convention is to consider

### Conservation Law. Chapter Goal. 5.2 Theory

Chpter 5 Conservtion Lw 5.1 Gol Our long term gol is to understnd how mny mthemticl models re derived. We study how certin quntity chnges with time in given region (sptil domin). We first derive the very

### 99/105 Comparison of OrcaFlex with standard theoretical results

99/105 Comprison of OrcFlex ith stndrd theoreticl results 1. Introduction A number of stndrd theoreticl results from literture cn be modelled in OrcFlex. Such cses re, by virtue of being theoreticlly solvble,

### A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007

A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus

### Plates on elastic foundation

Pltes on elstic foundtion Circulr elstic plte, xil-symmetric lod, Winkler soil (fter Timoshenko & Woinowsky-Krieger (1959) - Chpter 8) Prepred by Enzo Mrtinelli Drft version ( April 016) Introduction Winkler

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

1 Ares Between Curves c 22 Donld Kreider nd Dwight Lhr We know tht if f is continuous nonnegtive function on the intervl [, b], then f(x) dx is the re under the grph of f nd bove the intervl. We re going

### Lecture 8. Newton s Laws. Applications of the Newton s Laws Problem-Solving Tactics. Physics 105; Fall Inertial Frames: T = mg

Lecture 8 Applictions of the ewton s Lws Problem-Solving ctics http://web.njit.edu/~sireno/ ewton s Lws I. If no net force ocects on body, then the body s velocity cnnot chnge. II. he net force on body

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

Things to Memorize: A Prtil List Jnury 27, 2017 Chpter 2 Vectors - Bsic Fcts A vector hs mgnitude (lso clled size/length/norm) nd direction. It does not hve fixed position, so the sme vector cn e moved

### Thomas Whitham Sixth Form

Thoms Whithm Sith Form Pure Mthemtics Unit C Alger Trigonometry Geometry Clculus Vectors Trigonometry Compound ngle formule sin sin cos cos Pge A B sin Acos B cos Asin B A B sin Acos B cos Asin B A B cos

### x 2 1 dx x 3 dx = ln(x) + 2e u du = 2e u + C = 2e x + C 2x dx = arcsin x + 1 x 1 x du = 2 u + C (t + 2) 50 dt x 2 4 dx

. Compute the following indefinite integrls: ) sin(5 + )d b) c) d e d d) + d Solutions: ) After substituting u 5 +, we get: sin(5 + )d sin(u)du cos(u) + C cos(5 + ) + C b) We hve: d d ln() + + C c) Substitute

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

REVIEW OF ALGEBRA Here we review the bsic rules nd procedures of lgebr tht you need to know in order to be successful in clculus. ARITHMETIC OPERATIONS The rel numbers hve the following properties: b b

### 5.2 Volumes: Disks and Washers

4 pplictions of definite integrls 5. Volumes: Disks nd Wshers In the previous section, we computed volumes of solids for which we could determine the re of cross-section or slice. In this section, we restrict

### Math 554 Integration

Mth 554 Integrtion Hndout #9 4/12/96 Defn. A collection of n + 1 distinct points of the intervl [, b] P := {x 0 = < x 1 < < x i 1 < x i < < b =: x n } is clled prtition of the intervl. In this cse, we

### Jackson 2.26 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell

Jckson 2.26 Homework Problem Solution Dr. Christopher S. Bird University of Msschusetts Lowell PROBLEM: The two-dimensionl region, ρ, φ β, is bounded by conducting surfces t φ =, ρ =, nd φ = β held t zero

### Mathematics Extension 2

00 HIGHER SCHOOL CERTIFICATE EXAMINATION Mthemtics Etension Generl Instructions Reding time 5 minutes Working time hours Write using blck or blue pen Bord-pproved clcultors my be used A tble of stndrd

### The Wave Equation I. MA 436 Kurt Bryan

1 Introduction The Wve Eqution I MA 436 Kurt Bryn Consider string stretching long the x xis, of indeterminte (or even infinite!) length. We wnt to derive n eqution which models the motion of the string

### 1B40 Practical Skills

B40 Prcticl Skills Comining uncertinties from severl quntities error propgtion We usully encounter situtions where the result of n experiment is given in terms of two (or more) quntities. We then need

### THREE-DIMENSIONAL KINEMATICS OF RIGID BODIES

THREE-DIMENSIONAL KINEMATICS OF RIGID BODIES 1. TRANSLATION Figure shows rigid body trnslting in three-dimensionl spce. Any two points in the body, such s A nd B, will move long prllel stright lines if

### The First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a).

The Fundmentl Theorems of Clculus Mth 4, Section 0, Spring 009 We now know enough bout definite integrls to give precise formultions of the Fundmentl Theorems of Clculus. We will lso look t some bsic emples

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

Physics 6C Solution of inhomogeneous ordinry differentil equtions using Green s functions Peter Young November 5, 29 Homogeneous Equtions We hve studied, especilly in long HW problem, second order liner

### Shear and torsion interaction of hollow core slabs

Competitive nd Sustinble Growth Contrct Nº G6RD-CT--6 Sher nd torsion interction of hollow core slbs HOLCOTORS Technicl Report, Rev. Anlyses of hollow core floors December The content of the present publiction

### Method of Localisation and Controlled Ejection of Swarms of Likely Charged Particles

Method of Loclistion nd Controlled Ejection of Swrms of Likely Chrged Prticles I. N. Tukev July 3, 17 Astrct This work considers Coulom forces cting on chrged point prticle locted etween the two coxil,

### Chapter 5 1. = on [ 1, 2] 1. Let gx ( ) e x. . The derivative of g is g ( x) e 1

Chpter 5. Let g ( e. on [, ]. The derivtive of g is g ( e ( Write the slope intercept form of the eqution of the tngent line to the grph of g t. (b Determine the -coordinte of ech criticl vlue of g. Show

### PHYS Summer Professor Caillault Homework Solutions. Chapter 2

PHYS 1111 - Summer 2007 - Professor Cillult Homework Solutions Chpter 2 5. Picture the Problem: The runner moves long the ovl trck. Strtegy: The distnce is the totl length of trvel, nd the displcement

### Math 8 Winter 2015 Applications of Integration

Mth 8 Winter 205 Applictions of Integrtion Here re few importnt pplictions of integrtion. The pplictions you my see on n exm in this course include only the Net Chnge Theorem (which is relly just the Fundmentl

### Linear Inequalities: Each of the following carries five marks each: 1. Solve the system of equations graphically.

Liner Inequlities: Ech of the following crries five mrks ech:. Solve the system of equtions grphiclly. x + 2y 8, 2x + y 8, x 0, y 0 Solution: Considerx + 2y 8.. () Drw the grph for x + 2y = 8 by line.it

### BRIEF NOTES ADDITIONAL MATHEMATICS FORM

BRIEF NOTES ADDITIONAL MATHEMATICS FORM CHAPTER : FUNCTION. : + is the object, + is the imge : + cn be written s () = +. To ind the imge or mens () = + = Imge or is. Find the object or 8 mens () = 8 wht

### Line Integrals. Partitioning the Curve. Estimating the Mass

Line Integrls Suppose we hve curve in the xy plne nd ssocite density δ(p ) = δ(x, y) t ech point on the curve. urves, of course, do not hve density or mss, but it my sometimes be convenient or useful to

### Chapters Five Notes SN AA U1C5

Chpters Five Notes SN AA U1C5 Nme Period Section 5-: Fctoring Qudrtic Epressions When you took lger, you lerned tht the first thing involved in fctoring is to mke sure to fctor out ny numers or vriles

### Properties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives

Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums - 1 Riemnn

### Exam CT3109 STRUCTURAL MECHANICS april 2011, 09:00 12:00 hours

Subfculty of Civil Engineering rk ech ge with your: Structurl echnics STUDENT NUBER : NAE : Em CT309 STRUCTURAL ECHANICS 4 ril 0, 09:00 :00 hours This em consists of 4 roblems. Use for ech roblem serte

### MAT137 Calculus! Lecture 20

officil website http://uoft.me/mat137 MAT137 Clculus! Lecture 20 Tody: 4.6 Concvity 4.7 Asypmtotes Net: 4.8 Curve Sketching 4.5 More Optimiztion Problems MVT Applictions Emple 1 Let f () = 3 27 20. 1 Find

### 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.

A. Limits - L Hopitl s Rule Wht you re finding: L Hopitl s Rule is used to find limits of the form f ( ) lim where lim f or lim f limg. c g = c limg( ) = c = c = c How to find it: Try nd find limits by

### Chapter 4 Contravariance, Covariance, and Spacetime Diagrams

Chpter 4 Contrvrince, Covrince, nd Spcetime Digrms 4. The Components of Vector in Skewed Coordintes We hve seen in Chpter 3; figure 3.9, tht in order to show inertil motion tht is consistent with the Lorentz

### Operations with Polynomials

38 Chpter P Prerequisites P.4 Opertions with Polynomils Wht you should lern: How to identify the leding coefficients nd degrees of polynomils How to dd nd subtrct polynomils How to multiply polynomils

### 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.

Problem 8.31 Sher flow in circulr section i 3 R θ s i 2 t Remove@"Globl` "D H remove ll symbols L The centroidl xes re locted t the center of the circle s shown bove. (1) Find bending stiffness: From symmetry,

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

Vectors Skill Achieved? Know tht sclr is quntity tht hs only size (no direction) Identify rel-life exmples of sclrs such s, temperture, mss, distnce, time, speed, energy nd electric chrge Know tht vector

### SECTION 9-4 Translation of Axes

9-4 Trnsltion of Aes 639 Rdiotelescope For the receiving ntenn shown in the figure, the common focus F is locted 120 feet bove the verte of the prbol, nd focus F (for the hperbol) is 20 feet bove the verte.

### State space systems analysis (continued) Stability. A. Definitions A system is said to be Asymptotically Stable (AS) when it satisfies

Stte spce systems nlysis (continued) Stbility A. Definitions A system is sid to be Asymptoticlly Stble (AS) when it stisfies ut () = 0, t > 0 lim xt () 0. t A system is AS if nd only if the impulse response

### Math 1431 Section M TH 4:00 PM 6:00 PM Susan Wheeler Office Hours: Wed 6:00 7:00 PM Online ***NOTE LABS ARE MON AND WED

Mth 43 Section 4839 M TH 4: PM 6: PM Susn Wheeler swheeler@mth.uh.edu Office Hours: Wed 6: 7: PM Online ***NOTE LABS ARE MON AND WED t :3 PM to 3: pm ONLINE Approimting the re under curve given the type

### 4. Calculus of Variations

4. Clculus of Vritions Introduction - Typicl Problems The clculus of vritions generlises the theory of mxim nd minim. Exmple (): Shortest distnce between two points. On given surfce (e.g. plne), nd the

### Abstract inner product spaces

WEEK 4 Abstrct inner product spces Definition An inner product spce is vector spce V over the rel field R equipped with rule for multiplying vectors, such tht the product of two vectors is sclr, nd the

### Chapter 6 Notes, Larson/Hostetler 3e

Contents 6. Antiderivtives nd the Rules of Integrtion.......................... 6. Are nd the Definite Integrl.................................. 6.. Are............................................ 6. Reimnn

### Integration Techniques

Integrtion Techniques. Integrtion of Trigonometric Functions Exmple. Evlute cos x. Recll tht cos x = cos x. Hence, cos x Exmple. Evlute = ( + cos x) = (x + sin x) + C = x + 4 sin x + C. cos 3 x. Let u

### CBE 291b - Computation And Optimization For Engineers

The University of Western Ontrio Fculty of Engineering Science Deprtment of Chemicl nd Biochemicl Engineering CBE 9b - Computtion And Optimiztion For Engineers Mtlb Project Introduction Prof. A. Jutn Jn

### Physics 121 Sample Common Exam 1 NOTE: ANSWERS ARE ON PAGE 8. Instructions:

Physics 121 Smple Common Exm 1 NOTE: ANSWERS ARE ON PAGE 8 Nme (Print): 4 Digit ID: Section: Instructions: Answer ll questions. uestions 1 through 16 re multiple choice questions worth 5 points ech. You

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

Limits Prtil Derivtives For single vrible function f (x), the limit lim x f (x) exists only if the right-hnd side limit equls to the left-hnd side limit, i.e., lim f (x) = lim f (x). x x + For two vribles

### APPROXIMATE INTEGRATION

APPROXIMATE INTEGRATION. Introduction We hve seen tht there re functions whose nti-derivtives cnnot be expressed in closed form. For these resons ny definite integrl involving these integrnds cnnot be

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

MA00 Clculus nd Bsic Liner Alger I Chpter Coordinte Geometr nd Conic Sections Review In the rectngulr/crtesin coordintes sstem, we descrie the loction of points using coordintes. P (, ) P(, ) O The distnce

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

. Eponentil nd rithmic functions.1 Eponentil Functions A function of the form f() =, > 0, 1 is clled n eponentil function. Its domin is the set of ll rel f ( 1) numbers. For n eponentil function f we hve.

### 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

Chpter 9: Green s functions for time-independent problems Introductory emples One-dimensionl het eqution Consider the one-dimensionl het eqution with boundry conditions nd initil condition We lredy know

### 0.1 THE REAL NUMBER LINE AND ORDER

6000_000.qd //0 :6 AM Pge 0-0- CHAPTER 0 A Preclculus Review 0. THE REAL NUMBER LINE AND ORDER Represent, clssify, nd order rel numers. Use inequlities to represent sets of rel numers. Solve inequlities.

### Solutions to Supplementary Problems

Solutions to Supplementry Problems Chpter 8 Solution 8.1 Step 1: Clculte the line of ction ( x ) of the totl weight ( W ).67 m W = 5 kn W 1 = 16 kn 3.5 m m W 3 = 144 kn Q 4m Figure 8.10 Tking moments bout

### Calculus - Activity 1 Rate of change of a function at a point.

Nme: Clss: p 77 Mths Helper Plus Resource Set. Copright 00 Bruce A. Vughn, Techers Choice Softwre Clculus - Activit Rte of chnge of function t point. ) Strt Mths Helper Plus, then lod the file: Clculus

### BEAM DIAGRAMS AND FORMULAS. Nomenclature

BEA DIAGAS AND FOULAS Nomencture E = moduus of esticity of stee t 9,000 ksi I = moment of inerti of em (in. 4 ) L = tot ength of em etween rection points (ft) m = mimum moment (kip-in.) = mimum moment

### Linear Motion. Kinematics Quantities

Liner Motion Physics 101 Eyres Kinemtics Quntities Time Instnt t Fundmentl Time Interl Defined Position x Fundmentl Displcement Defined Aerge Velocity g Defined Aerge Accelertion g Defined 1 Kinemtics

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

Unit #8 : The Integrl Gols: Determine how to clculte the re described by function. Define the definite integrl. Eplore the reltionship between the definite integrl nd re. Eplore wys to estimte the definite

### Columns and Stability

ARCH 331 Note Set 1. Su01n Columns nd Stilit Nottion: A = nme or re A36 = designtion o steel grde = nme or width C = smol or compression C c = column slenderness clssiiction constnt or steel column design

### Distributed Forces: Centroids and Centers of Gravity

Distriuted Forces: Centroids nd Centers of Grvit Introduction Center of Grvit of D Bod Centroids nd First Moments of Ares nd Lines Centroids of Common Shpes of Ares Centroids of Common Shpes of Lines Composite

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

Unit #8 : The Integrl Gols: Determine how to clculte the re described by function. Define the definite integrl. Eplore the reltionship between the definite integrl nd re. Eplore wys to estimte the definite

### 20 MATHEMATICS POLYNOMIALS

0 MATHEMATICS POLYNOMIALS.1 Introduction In Clss IX, you hve studied polynomils in one vrible nd their degrees. Recll tht if p(x) is polynomil in x, the highest power of x in p(x) is clled the degree of

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

Mth 1102: Clculus I (Mth/Sci mjors) MWF 3pm, Fulton Hll 230 Homework 2 solutions Plese write netly, nd show ll work. Cution: An nswer with no work is wrong! Do the following problems from Chpter III: 6,

### POLYPHASE CIRCUITS. Introduction:

POLYPHASE CIRCUITS Introduction: Three-phse systems re commonly used in genertion, trnsmission nd distribution of electric power. Power in three-phse system is constnt rther thn pulsting nd three-phse

### Overview of Calculus I

Overview of Clculus I Prof. Jim Swift Northern Arizon University There re three key concepts in clculus: The limit, the derivtive, nd the integrl. You need to understnd the definitions of these three things,

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

INTRODUCTION According to Newton s lw, prticle will ccelerte when it is subjected to unblnced forces. Kinetics is the study of the reltions between unblnced forces nd the resulting chnges in motion. The

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

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Deprtment of Physics Physics 8T Fll Term 4 In-Clss Problems nd 3: Projectile Motion Solutions We would like ech group to pply the problem solving strtegy with the

### Polynomials and Division Theory

Higher Checklist (Unit ) Higher Checklist (Unit ) Polynomils nd Division Theory Skill Achieved? Know tht polynomil (expression) is of the form: n x + n x n + n x n + + n x + x + 0 where the i R re the

### Multiple Integrals. Review of Single Integrals. Planar Area. Volume of Solid of Revolution

Multiple Integrls eview of Single Integrls eding Trim 7.1 eview Appliction of Integrls: Are 7. eview Appliction of Integrls: olumes 7.3 eview Appliction of Integrls: Lengths of Curves Assignment web pge

### Phys 7221, Fall 2006: Homework # 6

Phys 7221, Fll 2006: Homework # 6 Gbriel González October 29, 2006 Problem 3-7 In the lbortory system, the scttering ngle of the incident prticle is ϑ, nd tht of the initilly sttionry trget prticle, which

### Principles of Real Analysis I Fall VI. Riemann Integration

21-355 Principles of Rel Anlysis I Fll 2004 A. Definitions VI. Riemnn Integrtion Let, b R with < b be given. By prtition of [, b] we men finite set P [, b] with, b P. The set of ll prtitions of [, b] will

### MAT137 Calculus! Lecture 28

officil wesite http://uoft.me/mat137 MAT137 Clculus! Lecture 28 Tody: Antiderivtives Fundmentl Theorem of Clculus Net: More FTC (review v. 8.5-8.7) 5.7 Sustitution (v. 9.1-9.4) Properties of the Definite

### DYNAMIC EARTH PRESSURE SIMULATION BY SINGLE DEGREE OF FREEDOM SYSTEM

13 th World Conference on Erthque Engineering Vncouver, B.C., Cnd August 1-6, 2004 per No. 2663 DYNAMIC EARTH RESSURE SIMULATION BY SINGLE DEGREE OF FREEDOM SYSTEM Arsln GHAHRAMANI 1, Seyyed Ahmd ANVAR

### CONIC SECTIONS. Chapter 11

CONIC SECTIONS Chpter. Overview.. Sections of cone Let l e fied verticl line nd m e nother line intersecting it t fied point V nd inclined to it t n ngle α (Fig..). Fig.. Suppose we rotte the line m round

### Definite integral. Mathematics FRDIS MENDELU

Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová Brno 1 Motivtion - re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function defined on [, b]. Wht is the re of the

### Conducting Ellipsoid and Circular Disk

1 Problem Conducting Ellipsoid nd Circulr Disk Kirk T. McDonld Joseph Henry Lbortories, Princeton University, Princeton, NJ 08544 (September 1, 00) Show tht the surfce chrge density σ on conducting ellipsoid,