Static equilibrium requires a balance of forces and a balance of moments.

Similar documents
7.2.1 Basic relations for Torsion of Circular Members

See the solution to Prob Ans. Since. (2E t + 2E c )ch - a. (s max ) t. (s max ) c = 2E c. 2E c. (s max ) c = 3M bh 2E t + 2E c. 2E t. h c.

BENDING OF BEAM. Compressed layer. Elongated. layer. Un-strained. layer. NA= Neutral Axis. Compression. Unchanged. Elongation. Two Dimensional View

Physics 201 Lecture 18

ME 210 Applied Mathematics for Mechanical Engineers

Physics Tutorial V1 2D Vectors

b) (5) What average force magnitude was applied by the students working together?

06 - ROTATIONAL MOTION Page 1 ( Answers at the end of all questions )

Physics 2A Chapter 10 - Moment of Inertia Fall 2018

From Newton to Einstein. Mid-Term Test, 12a.m. Thur. 13 th Nov Duration: 50 minutes. There are 20 marks in Section A and 30 in Section B.

Physics C Rotational Motion Name: ANSWER KEY_ AP Review Packet

Physics 4A Chapter 8: Dynamics II Motion in a Plane

Lecture 5. Torsion. Module 1. Deformation Pattern in Pure Torsion In Circular Cylinder. IDeALab. Prof. Y.Y.KIM. Solid Mechanics

Stress, Cauchy s equation and the Navier-Stokes equations

Chapter 7-8 Rotational Motion

Circular Motion & Torque Test Review. The period is the amount of time it takes for an object to travel around a circular path once.

Physics 111 Lecture 12. SJ 8th Ed.: Chap Static Equilibrium. Overview - Equilibrium Defined. Static Equilibrium Examples

Rigid Body Dynamics 2. CSE169: Computer Animation Instructor: Steve Rotenberg UCSD, Winter 2018

STRENGTH OF MATERIALS 140AU0402 UNIT 3: BEAMS - LOADS AND STRESSES

Physics 1114: Unit 5 Hand-out Homework (Answers)

Mechanics Topic D (Rotation) - 1 David Apsley

TRANSILVANIA UNIVERSITY OF BRASOV MECHANICAL ENGINEERING FACULTY DEPARTMENT OF MECHANICAL ENGINEERING ONLY FOR STUDENTS

- 5 - TEST 1R. This is the repeat version of TEST 1, which was held during Session.

To Feel a Force Chapter 7 Static equilibrium - torque and friction

Chapter 12. Kinetics of Particles: Newton s Second Law

AH Mechanics Checklist (Unit 2) AH Mechanics Checklist (Unit 2) Circular Motion

Physics 107 TUTORIAL ASSIGNMENT #8

2 Governing Equations

MAGNETIC FIELD INTRODUCTION

Magnetic Dipoles Challenge Problem Solutions

Math Notes on Kepler s first law 1. r(t) kp(t)

Rotational Motion: Statics and Dynamics

Section 26 The Laws of Rotational Motion

7.2. Coulomb s Law. The Electric Force

Rectilinea Motion. A foce P is applied to the initially stationay cat. Detemine the velocity and displacement at time t=5 s fo each of the foce histoi

Physics NYB problem set 5 solution

SAMPLE LABORATORY SESSION FOR JAVA MODULE B. Calculations for Sample Cross-Section 2

c) (6) Assuming the tires do not skid, what coefficient of static friction between tires and pavement is needed?

PHYS 1114, Lecture 21, March 6 Contents:

Chapter 4: The laws of motion. Newton s first law

Chapter 4. Newton s Laws of Motion

3.6 Applied Optimization

Chapter 2: Basic Physics and Math Supplements

Physics Fall Mechanics, Thermodynamics, Waves, Fluids. Lecture 6: motion in two and three dimensions III. Slide 6-1

Electrostatics (Electric Charges and Field) #2 2010

Construction Figure 10.1: Jaw clutches

OSCILLATIONS AND GRAVITATION

Principles of Physics I

The Laws of Motion ( ) N SOLUTIONS TO PROBLEMS ! F = ( 6.00) 2 + ( 15.0) 2 N = 16.2 N. Section 4.4. Newton s Second Law The Particle Under a Net Force

LINEAR PLATE BENDING

Physics 111 Lecture 5 Circular Motion

VECTOR MECHANICS FOR ENGINEERS: STATICS

( ) ( ) Review of Force. Review of Force. r = =... Example 1. What is the dot product for F r. Solution: Example 2 ( )

e.g: If A = i 2 j + k then find A. A = Ax 2 + Ay 2 + Az 2 = ( 2) = 6

5. Pressure Vessels and

ESCI 342 Atmospheric Dynamics I Lesson 3 Fundamental Forces II

Sections and Chapter 10

MCV4U Final Exam Review. 1. Consider the function f (x) Find: f) lim. a) lim. c) lim. d) lim. 3. Consider the function: 4. Evaluate. lim. 5. Evaluate.

Lab 10: Newton s Second Law in Rotation

Physics 2020, Spring 2005 Lab 5 page 1 of 8. Lab 5. Magnetism

PH 221-3A Fall EQUILIBRIUM and. Lectures Chapter 12 (Halliday/Resnick/Walker, Fundamentals of Physics 8 th edition)

Right-handed screw dislocation in an isotropic solid

6.4 Period and Frequency for Uniform Circular Motion

Rotational Motion. Every quantity that we have studied with translational motion has a rotational counterpart

PHYS Summer Professor Caillault Homework Solutions. Chapter 5

When a mass moves because of a force, we can define several types of problem.

Math 2263 Solutions for Spring 2003 Final Exam

Chapter 13 Gravitation

Objectives: After finishing this unit you should be able to:

21 MAGNETIC FORCES AND MAGNETIC FIELDS

Physics. Rotational Motion.

Written as per the revised syllabus prescribed by the Maharashtra State Board of Secondary and Higher Secondary Education, Pune.

Physics 2212 GH Quiz #2 Solutions Spring 2016

1) Consider an object of a parabolic shape with rotational symmetry z

AST 121S: The origin and evolution of the Universe. Introduction to Mathematical Handout 1

Section 8.2 Polar Coordinates

KEPLER S LAWS OF PLANETARY MOTION

Conservation of Angular Momentum

AP * PHYSICS B. Circular Motion, Gravity, & Orbits. Teacher Packet

Chap 5. Circular Motion: Gravitation

b) (5) What is the magnitude of the force on the 6.0-kg block due to the contact with the 12.0-kg block?

Force can be exerted by direct contact between bodies: Contact Force.

KEPLER S LAWS AND PLANETARY ORBITS

EFFECTS OF FRINGING FIELDS ON SINGLE PARTICLE DYNAMICS. M. Bassetti and C. Biscari INFN-LNF, CP 13, Frascati (RM), Italy

, and the curve BC is symmetrical. Find also the horizontal force in x-direction on one side of the body. h C

Understanding the Concepts

Math 209 Assignment 9 Solutions

Article : 8 Article : 8 Stress Field. and. Singularity Problem

Phys 201A. Homework 6 Solutions. F A and F r. B. According to Newton s second law, ( ) ( )2. j = ( 6.0 m / s 2 )ˆ i ( 10.4m / s 2 )ˆ j.

Physics 11 Chapter 4: Forces and Newton s Laws of Motion. Problem Solving

Ch. 4: FOC 9, 13, 16, 18. Problems 20, 24, 38, 48, 77, 83 & 115;

Chapter 5 Force and Motion

Chapter 1: Introduction to Polar Coordinates

EF 152 Exam #1, Spring, 2011 Page 1 of 5

PS113 Chapter 5 Dynamics of Uniform Circular Motion

Easy. r p 2 f : r p 2i. r p 1i. r p 1 f. m blood g kg. P8.2 (a) The momentum is p = mv, so v = p/m and the kinetic energy is

Dynamics of Rotational Motion

2. Electrostatics. Dr. Rakhesh Singh Kshetrimayum 8/11/ Electromagnetic Field Theory by R. S. Kshetrimayum

Physics 207 Lecture 5. Lecture 5

2. Plane Elasticity Problems

Transcription:

Static Equilibium Static equilibium equies a balance of foces and a balance of moments. ΣF 0 ΣF 0 ΣF 0 ΣM 0 ΣM 0 ΣM 0 Eample 1: painte stands on a ladde that leans against the wall of a house at an angle of 0 o. ssume that the painte is a midheight of the ladde. Calculate the minimum coefficient of fiction fo static equilibium.

0 ) 0 1 1 1 Σ Σ P m m P F P P P P F peson ladde μ μ μ μ μ μ 1 ) 1 ) 1 g m m P g m m P p l p l 0 0) ) 0) 0) 1 Sin L g m m P LCos LSin P p l 0.1763 μ Static equilibium: Sum of vetical and hoiontal foces must be eo. Taking the moment about point O and setting it equals to eo:

Load nalsis 3-D Newton Fist Law: bod at est tend to emain at est and a bod in motion at constant velocit will tend to maintain that velocit unless acted upon b an etenal foce. Newton Second Law: The time ate of change of momentum of a bod is equal to the magnitude of applied foce and acts in the diection of the foce. Newton Thid Law: when two paticles inteact, a pai of equal and opposite eaction foces will eist at thei contact point. This foce pai will have the same magnitude and act along the same diection line but have opposite sense.

Newton s second law can be witten fo a igid bod in two foms, one fo linea foces and one fo moments o toques M G H G ma F & Σ Σ FFoce; mmass, a acceleation M G moment about the cente of gavit G H & time ate of change of the moment o the angula momentum about the CG ma F ma F ma F Σ Σ Σ k j i H G ˆ ˆ ˆ ω ω ω M M M ω ω α ω ω α ω ω α ) ) ) Σ Σ Σ f the, and aes ae chosen to coincide with the pincipal aes of inetia of the bod. Whee, and ae the pincipal centoidal mass moments of inetia second moments of mass). n 3-D Eule s Equations

Fo -D: Whee α is the angula acceleation ΣF ma ΣF ma ΣM Moment of a Foce about an ais α

3-D Equilibium Eample: Two tansmission belts pass ove sheaves welded to an ale suppoted b beaings at B and D. The adius at.5 and at C. The ale otates at constant speed. Find T and the eaction foces at B, D. ssume that beaing at D eets no aial thust and neglect weights of sheaves and ale.

8in 6in 6in

B C D 33.75lb 67.5lb 33.75lb 33.75lb 33.75lb 0.5lb-in

B C D 4lb 70lb 8lb 4lb 8lb 336lb-in

Definitions Centoid of ea Cente of Gavit of an aea): Point, ) that defines the geometic cente of the aea Fist Moment of an ea with espect to the - ais Fist Moment of an ea with espect to the - ais Q Q τ VQ b

Eample : Calculate the cente of gavit of the ectangle: a)without a hole b)with a hole of dimensions c and d a) Without a hole 1 1 a b a b a a b b a b a b n i i n i i

b) With a hole d c b a d c c e b a a d c b a d c d f b a b n i i n i i ) ) 1 1 Second Moment o Moment of netia of an ea O J ρ Rectangula moments of inetia Pola moments of inetia

Radius of Gation O O O J Eample 3: Find the moment of inetia of the cicula aea about the and aes, the pola moment of inetia and the adius of gation about the and aes. φ φ Cos Sin Cos ea φ φ φ Cos 4 4 4 4 4 4 J Cos Sin π π π φ φ φ π π

Paallel-is Theoem ' d Eample 4: ) 4 ' 4 ' 4 16 5 4 4 4 π. π π d π

σ M

Mass Moment of netia t is the poduct of the element s mass and the squae of the element s distance fom the ais. m ) m a m ) m a m ) m a

Load Classification and Sign Convention Classification with espect to the method of application: a)nomal tensile b)nomal compessive c)shea d)bending e)tosion f) Combined

υ > 0 υ deflection The sign convention that will be used hee is as follows:

Distibuted loads, Shea Foce and Bending Moment in Beams

4 4 3 3 E -q q V E V V M E M E M υ υ υ υ

Successive ntegation Method V q V ) V ) q ) 600 C V ) 600 C Fo 0 < < 1 V ) 600 C1 Fo 0 C 650 1 V ) 600 650 Fo 1m then V 1) 350N 1 1 0 < < 1 650N /3m 1050N

N V then m Fo V C V Fo C V Fo 1050 ) 150 600 ) 150 150 500 350 1) 1 600 ) 1 < < 550 1) 1 0 0 0) 0 650 100 ) 650 600 ) ) 650 600 ) 1 0 3 3 3 < < M then Fo C M then Fo C M V M V Fo

0 ) 500 550 1) 1 150 100 ) 150 600 ) ) 150 600 ) 1 4 4 3 < < M then Fo C M then Fo C M V M V Fo M Ma 550N-m

Nomal Stess and Stain: Whee F: foce, nomal to σ F the coss-sectional aea, 0 : oiginal coss-sectional aea 0 1Pa 1 N.m - ; 1MPa 10 6 Pa; 1GPa10 9 Pa Nomal stain o tensile stain o ial stain ε l l l 0 0 Δl l 0

Hooke s Law When stains ae small, most of mateials ae linea elastic. Young s modulus Nomal: Δl F o lo E σ Εε σ Spings: the sping ate k F Δl o E l o E ε

Tosion Loading esulting fom the twist of a shaft. l Z θ θ γ θ, l G G G θ θ γ τ θ θ,, G Shea Modulus of Elasticit Shea stain Twist Moment o Toque l G T θ τ θ θ,, J ea Pola Moment of netia J G l T l J G T θ θ o J T τ θ, Thus: J T o Ma τ ngula sping ate: l G J T k a θ

Maimum Stesses in Beams We can obtain the nomal and shea stesses fom fleue and shea fomulas σ M τ VQ b Whee: σ is the nomal stess acting on the coss section, M is the bending moment, is the distance fom the neutal ais and is the moment of inetia of the coss sectional aea with espect to the neutal ais. τ is the shea stess at an point in the coss section, V is the shea foce, Q is the fist moment of the coss sectional aea outside of the point in the coss section whee the stess is being found, and b is the width of the coss section.

The nomal stesses obtained fom the fleue fomula have thei maimum values at the fathest distance fom the neutal ais. The nomal stesses ae calculated at the coss section of maimum bending moment. The shea stess obtained fom the shea fomula usuall have thei highest value at the neutal ais. The shea stesses ae calculated at the coss section of maimum shea foce. n most cicumstances, these ae the onl stesses that ae needed fo design puposes. Howeve to obtain a moe complete pictue of the stesses, we will need to detemine the pincipal stesses and maimum shea stesses at vaious points in the beam.

Beams of Rectangula Coss Section Conside a simple ectangula beam below, and a coss section to the left of the load. Points and E ae at the top and bottom of the beam. Point C is in the midheight of the beam and points B and D ae in between. f Hooke s law applies linea elasticit), the nomal and shea stesses at each of these five points can be eadil calculated fom the fleue and shea fomulas. ll the elements of vetical and hoiontal faces, ae in plane stess, because thee is no stesses acting pependicula to the plane of the figue.

Points and E elements ae in uniaial compessive and tensile stesses espectivel. Point C neutal ais) element is in pue shea. Points B and D elements have both nomal and shea stesses. Stesses in a beam of ectangula coss section: a) simple beam with points, B, C, D, and E on the side of the beam; b) nomal and shea stesses acting on stess elements at points, B, C, D, and E; c) pincipal stess; and d) maimum shea stesses.

We ma use eithe the tansfomation equations of plane stess o the Moh s cicle to find the stesses at each point along the height of the beam o to descibe how the pincipal stesses changes as we go fom to top to the bottom of the beam. B investigating the stesses at man coss sections of the beam, we can detemine how the pincipal stesses va thoughout the beam. Stess tajecto: Gives the diections of the pincipal stesses. Stess Contous: Cuves connecting points of equal pincipal stess Pincipal-stess tajectoies fo beams of ectangula coss section: a) cantileve beam, and b) simple beam. Solid lines epesent tensile pincipal stesses and dashed lines epesent compessive pincipal stesses.)

Eample 8-3: simple beam B with a span length L 6ft suppots a concentated load P 10800lb acting a distance c ft fom the ighthand suppot see figue below). The beam is made of steel and has a ectangula coss section width bin and height h 6in). nvestigate the pincipal stesses and maimum shea stesses at coss section mn located at a distance 9in fom the end of the beam. Conside onl the in-plane stesses) Point in) 3 B C 1 D 0 E -1 F - G -3

Solution The eaction of the beam at suppot is R P/3 3600lb, and theefoe the bending moment and shea foce at the section mn ae M R 3600lb)9in) 3400lb-in V R 3600lb Nomal stess on coss section mn bh ) 3400 lb in ) 3 in ) 6 in ) M M 1 σ X 900 3 1 Whee has units in inches and σ has units in psi. The stesses calculated ae positive when in tension. Note that a positive value of uppe half of the beam) gives a negative stess, as epected. Shea stesses on coss section mn The shea stesses ae given b the shea fomula τ in which the fist moment Q fo a ectangula coss section is VQ b

) ) 3 3 4 6 4 1 the shea fomula becomes thus, 4 h bh V h b b bh V b VQ τ h b h h b Q 4 h b b Q h/ V Ma 3 τ

The shea stesses τ acting on the face of the stess element ae positive upwads, wheeas the actual shea stesses τ act downwad. Theefoe VQ 6V h τ 3 b bh 4 Substituting the numeical values into this equation gives τ XY 6 3600 lb ) in ) 6in ) n which has units of inches and τ has units of psi Note: The maimum shea stess that occus at the neutal ais in a ectangula section can be simplified b using the following equation: τ Ma 3V 6in ) 50 ) 4 3 9

Calculation of stesses on coss section mn We divide the height of the beam into si equal intevals and label the coesponding points fom to G. Point in) σ psi) τ psi) 3-700 0 B -1800-50 C 1-900 -400 D 0 0-450 E -1 900-400 F - 1800-50 G -3 700 0 The nomal stesses va lineal fom a compessive stess of -700psi at the top of the beam point ) to a tensile stess of 700psi at the bottom of the beam point G). The shea stesses have a paabolic distibution with a maimum stess at the neutal ais point D).

Pincipal Stesses and Maimum Shea Stesses The pincipal stesses and maimum shea stesses at each of the seven points though G ma be detemined fom the following equations σ σ σ σ σ 1, ± σ σ τ MX τ ) τ )

Since thee is no nomal stess in the diection, this equation simplifies to: σ τ MX τ ) σ σ σ 1, ± τ ) Thus, b substituting the values of σ and τ, we can calculate the pincipal stesses σ 1 and σ and the maimum shea stess τ ma. Point in) σ psi) τ psi) σ 1 psi) σ psi) τ ma psi) 3-700 0 0-700 1350 B -1800-50 34-1834 934 C 1-900 -400 15-105 60 D 0 0-450 450-450 450 E -1 900-400 105-15 60 F - 1800-50 1834-34 934 G -3 700 0 700 0 1350

The lagest tensile stess anwhee in the beam is the nomal stess at the bottom of the beam at the coss section of maimum bending moment σ tens ) ma 14400psi). The lagest shea stess occus to the ight of the load P V R B 700lb). Theefoe, the lagest value that occus at the neutal ais is τ ) ma 900psi The lagest shea stess anwhee in the beam occus at 45 o planes at eithe the top o bottom τ ) ma 14400 / 700psi