External Work. When a force F undergoes a displacement dx in the same direction i as the force, the work done is

Similar documents
Module 4 : Deflection of Structures Lecture 4 : Strain Energy Method

14. *14.8 CASTIGLIANO S THEOREM

Chapter 2: Deflections of Structures

Mechanical Properties of Materials

Consider an elastic spring as shown in the Fig.2.4. When the spring is slowly

2 marks Questions and Answers

Mechanical Engineering Ph.D. Preliminary Qualifying Examination Solid Mechanics February 25, 2002

CHAPTER 5 Statically Determinate Plane Trusses

This procedure covers the determination of the moment of inertia about the neutral axis.

CHAPTER 5 Statically Determinate Plane Trusses TYPES OF ROOF TRUSS

Stress Analysis Lecture 3 ME 276 Spring Dr./ Ahmed Mohamed Nagib Elmekawy

D : SOLID MECHANICS. Q. 1 Q. 9 carry one mark each. Q.1 Find the force (in kn) in the member BH of the truss shown.

2. Rigid bar ABC supports a weight of W = 50 kn. Bar ABC is pinned at A and supported at B by rod (1). What is the axial force in rod (1)?


Using the finite element method of structural analysis, determine displacements at nodes 1 and 2.

Chapter 5 Elastic Strain, Deflection, and Stability 1. Elastic Stress-Strain Relationship

Indeterminate Analysis Force Method 1

Mechanics of Materials MENG 270 Fall 2003 Exam 3 Time allowed: 90min. Q.1(a) Q.1 (b) Q.2 Q.3 Q.4 Total

KINGS COLLEGE OF ENGINEERING DEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK. Subject code/name: ME2254/STRENGTH OF MATERIALS Year/Sem:II / IV

2. Determine the deflection at C of the beam given in fig below. Use principal of virtual work. W L/2 B A L C

SSC-JE MAINS ONLINE TEST SERIES / CIVIL ENGINEERING SOM + TOS

Symmetric Bending of Beams

Method of Virtual Work Frame Deflection Example Steven Vukazich San Jose State University

INTRODUCTION TO STRAIN

JUT!SI I I I TO BE RETURNED AT THE END OF EXAMINATION. THIS PAPER MUST NOT BE REMOVED FROM THE EXAM CENTRE. SURNAME: FIRST NAME: STUDENT NUMBER:

Chapter 11. Displacement Method of Analysis Slope Deflection Method

D : SOLID MECHANICS. Q. 1 Q. 9 carry one mark each.

Interstate 35W Bridge Collapse in Minnesota (2007) AP Photo/Pioneer Press, Brandi Jade Thomas

Question 1. Ignore bottom surface. Solution: Design variables: X = (R, H) Objective function: maximize volume, πr 2 H OR Minimize, f(x) = πr 2 H

FIXED BEAMS IN BENDING

5. What is the moment of inertia about the x - x axis of the rectangular beam shown?

k 21 k 22 k 23 k 24 k 31 k 32 k 33 k 34 k 41 k 42 k 43 k 44

ME 323 Examination #2 April 11, 2018

Mechanics of Materials II. Chapter III. A review of the fundamental formulation of stress, strain, and deflection

UNIT-I STRESS, STRAIN. 1. A Member A B C D is subjected to loading as shown in fig determine the total elongation. Take E= 2 x10 5 N/mm 2

Stress Analysis Lecture 4 ME 276 Spring Dr./ Ahmed Mohamed Nagib Elmekawy

TORSION INCLUDING WARPING OF OPEN SECTIONS (I, C, Z, T AND L SHAPES)

BEAM DEFLECTION THE ELASTIC CURVE

QUESTION BANK DEPARTMENT: CIVIL SEMESTER: III SUBJECT CODE: CE2201 SUBJECT NAME: MECHANICS OF SOLIDS UNIT 1- STRESS AND STRAIN PART A

BEAMS AND PLATES ANALYSIS

Final Exam Ship Structures Page 1 MEMORIAL UNIVERSITY OF NEWFOUNDLAND. Engineering Ship Structures

Method of Consistent Deformation

M. Vable Mechanics of Materials: Chapter 5. Torsion of Shafts

Module 3. Analysis of Statically Indeterminate Structures by the Displacement Method

Name :. Roll No. :... Invigilator s Signature :.. CS/B.TECH (CE-NEW)/SEM-3/CE-301/ SOLID MECHANICS

Comb resonator design (2)

Torsion of Shafts Learning objectives

and F NAME: ME rd Sample Final Exam PROBLEM 1 (25 points) Prob. 1 questions are all or nothing. PROBLEM 1A. (5 points)

2012 MECHANICS OF SOLIDS

Level 7 Postgraduate Diploma in Engineering Computational mechanics using finite element method

The University of Melbourne Engineering Mechanics

Finite Element Analysis Prof. Dr. B. N. Rao Department of Civil Engineering Indian Institute of Technology, Madras. Module - 01 Lecture - 13

Strength of Materials II (Mechanics of Materials) (SI Units) Dr. Ashraf Alfeehan

STATICALLY INDETERMINATE STRUCTURES

STRUCTURAL SURFACES & FLOOR GRILLAGES

Mechanics of Materials Primer

UNIVERSITY OF SASKATCHEWAN ME MECHANICS OF MATERIALS I FINAL EXAM DECEMBER 13, 2008 Professor A. Dolovich

Module 2. Analysis of Statically Indeterminate Structures by the Matrix Force Method

1-1 Locate the centroid of the plane area shown. 1-2 Determine the location of centroid of the composite area shown.

Review of Strain Energy Methods and Introduction to Stiffness Matrix Methods of Structural Analysis

Outline. Organization. Stresses in Beams

8-5 Conjugate-Beam method. 8-5 Conjugate-Beam method. 8-5 Conjugate-Beam method. 8-5 Conjugate-Beam method

R13. II B. Tech I Semester Regular Examinations, Jan MECHANICS OF SOLIDS (Com. to ME, AME, AE, MTE) PART-A

PES Institute of Technology

QUESTION BANK SEMESTER: III SUBJECT NAME: MECHANICS OF SOLIDS

Module 3. Analysis of Statically Indeterminate Structures by the Displacement Method

ME Final Exam. PROBLEM NO. 4 Part A (2 points max.) M (x) y. z (neutral axis) beam cross-sec+on. 20 kip ft. 0.2 ft. 10 ft. 0.1 ft.

Name (Print) ME Mechanics of Materials Exam # 1 Date: October 5, 2016 Time: 8:00 10:00 PM

Review Lecture. AE1108-II: Aerospace Mechanics of Materials. Dr. Calvin Rans Dr. Sofia Teixeira De Freitas

ME C85/CE C30 Fall, Introduction to Solid Mechanics ME C85/CE C30. Final Exam. Fall, 2013

DESIGN OF BEAMS AND SHAFTS

BE Semester- I ( ) Question Bank (MECHANICS OF SOLIDS)

MECHANICS OF MATERIALS

PDDC 1 st Semester Civil Engineering Department Assignments of Mechanics of Solids [ ] Introduction, Fundamentals of Statics

= 50 ksi. The maximum beam deflection Δ max is not = R B. = 30 kips. Notes for Strength of Materials, ET 200

Chapter 6: Cross-Sectional Properties of Structural Members

[7] Torsion. [7.1] Torsion. [7.2] Statically Indeterminate Torsion. [7] Torsion Page 1 of 21

Example 3.7 Consider the undeformed configuration of a solid as shown in Figure 3.60.

Theory of Structures

[5] Stress and Strain

UNIT 1 STRESS STRAIN AND DEFORMATION OF SOLIDS, STATES OF STRESS 1. Define stress. When an external force acts on a body, it undergoes deformation.

The science of elasticity

Shafts: Torsion of Circular Shafts Reading: Crandall, Dahl and Lardner 6.2, 6.3

NAME: Given Formulae: Law of Cosines: Law of Sines:

IDE 110 Mechanics of Materials Spring 2006 Final Examination FOR GRADING ONLY

ME325 EXAM I (Sample)

Advanced Structural Analysis EGF Section Properties and Bending

MECE 3321 MECHANICS OF SOLIDS CHAPTER 3

UNIT-I Introduction & Plane Stress and Plane Strain Analysis

SECTION A. 8 kn/m. C 3 m 3m

ME 323 FINAL EXAM FALL SEMESTER :00 PM 9:00 PM Dec. 16, 2010

Comb Resonator Design (2)

Chapter 8 Supplement: Deflection in Beams Double Integration Method

Solution: The strain in the bar is: ANS: E =6.37 GPa Poison s ration for the material is:

EMA 3702 Mechanics & Materials Science (Mechanics of Materials) Chapter 3 Torsion

3 2 6 Solve the initial value problem u ( t) 3. a- If A has eigenvalues λ =, λ = 1 and corresponding eigenvectors 1

MECHANICS OF MATERIALS

Chapter 4-b Axially Loaded Members

ε t increases from the compressioncontrolled Figure 9.15: Adjusted interaction diagram

Chapter 5 Structural Elements: The truss & beam elements

Transcription:

Structure Analysis I Chapter 9

Deflection Energy Method

External Work Energy Method When a force F undergoes a displacement dx in the same direction i as the force, the work done is du e = F dx If the total displacement is x the work become U e x = F dx 0 1 U e = 2 P The force applied gradually

External Work Energy Method If P is already applied to the bar and that another force F`i is now applied The work done by P when the bar undergoes the further deflection ` is then U e ' = P '

The work of a moment is defined by the product of the magnitude of the moment M and the angle dθ then du e = M dθ If the total angle of rotation is θ the work become: θ U e 0 = M dθ U e 2 1 = Mθ The moment applied gradually

Energy Method Strain Energy Axial Force σ = Eε σ = ε = N A L = NL AE P = N U = 1 P 2 2 N L U i = 2AE N=internal normal force in a truss member caused by the real load L = length of member A = cross-sectional area of a member E = modulus of elasticity of a member

Energy Method Strain Energy Bending dθ = U = 1 2 M dx EI Mθ du U i i = L = 0 2 M dx 2EI 2 M dx 2EI

Principle of Virtual Work P = Work of Etrnl External Loads u δ Work of Internal Loads Virtual Load 1. = u. dl Real displacement

Method of Virtual Work: Trusses External Loading 1. = u. dl 1. = n. NL AE 1 = external virtual unit load acting on the truss joint in the stated t direction of n = internal virtual normal force in a truss member caused by the external virtual unit load = external joint displacement caused by the real load on the truss N = internal normal force in a truss member caused by the real load L = length of member A = cross-sectional area of a member E = modulus of elasticity of a member

Method of Virtual Work: Trusses Temperature 1. = n α T L 1 = external virtual unit load acting on the truss joint in the stated direction of n = internal virtual normal force in a truss member caused by the external virtual unit load = external joint displacement caused by the temperature change. α = coefficient of thermal expansion of member T = change in temperature of member L = length of member

Method of Virtual Work: Trusses Fabrication Errors and Camber 1. = n L 1 = external virtual unit load acting on the truss joint in the stated direction of n = internal virtual normal force in a truss member caused by the external virtual unit load = external joint displacement caused by fabrication errors L = difference in length of the member from its intended size as caused by a fabrication error.

Example 1 The cross sectional area of each member of the truss show, is A = 400mm2 & E = 200GPa. a) Determine the vertical displacement of joint C if a 4-kN force is applied to the truss at C

Solution 1. = n. NL AE A virtual force of 1 kn is applied at C in the vertical direction

Member n (KN) N (KN) L (m) nnl AB 0.667 2 8 10.67 AC 0.833 2.5 5 10.41 CB 0.833 2.5 5 10.41 Sum 10.67 nnl AE 10.67 AE 10.67 ( kn 1. = = = 6 2 ( m ) 6 ( kn / m = 0.000133 m = 0.133 mm 2 400 10 200 10 ) )

Example 2 Text book Example 8-14 Determine vertical displacement at C A = 0.5 in 2 E=29 (10) 3 ksi

Example 3

Method of Virtual Work: Beam = 1. u. dl M d θ = dx EI L M 1.. = m EI dx 0 1 = t l i t l it l d ti th b i th t t d di ti f 1 = external virtual unit load acting on the beam in the stated direction of m = internal virtual moment in a beam caused by the external virtual unit load = external joint displacement caused by the real load on the truss M = internal moment in a beam caused by the real load L = length of beam I = moment of inertia of cross-sectional E = modulus of elasticity of the beam

Method of Virtual Work: Beam Similarly the rotation angle at any point on the beam can be determine, a unit couple moment is applied at the point and the corresponding internal moment have to be determine m θ L θ 1( KN. m). θ = 0 m M EI dx

Example 4 Determine the displacement at point B of a steel beam E = 200 Gpa, I = 500(10 6 ) mm 4

Solution x dx x dx x x dx M m L 6 6 ) 6 ( ) 1 ( 1. 10 4 10 3 10 2 = = = = m EI EI EI dx EI m 0 15 ) 15(10 ) 15(10 4 1. 3 3 0 0 0 0 m EI 0.15 ) )(10 500(10 ) 200(10 ) ( ) ( 12 6 6 = = =

Another Solution Real Load Virtual Load B B = = 200(10 0.15m 6 2000 ) 500(10 6 )(10 12 ) 7.5

Example 5

Example 6 Determine the Slope θ and displacement at point B of a steel p p p beam E = 200 Gpa, I = 60(10 6 ) mm 4

Solution Virtual Load

Real Load Real Load 3 3 ) 3 ( 1) ( ) 3 ( (0) 1 10 2 10 10 5 = = + = = L x x dx dx x dx x dx M m θ d 0094 0 ) 3(5 ) 3(10 2 1. 2 2 5 5 5 0 0 = = + = = EI EI EI EI dx EI m θ θ θ 0.0094 rad ) 60(10 ) 200(10 2 ) ( ) ( 6 6 = = θ B

Another Solution Real Load Virtual Load θ = B 112 112 = 1 = EI EI 112 = 0.0094 200(10 9 ) 60 10 6 rad

Example 7

Example 8

Example 9

Example 9b Determine the horizontal deflection at A

Solution Real Load Virtual Load

A = 100 500 1250 0 + 2.5 = EI EI EI 1250 = 0.031031 m = 9 6 200(10 ) 200 10

Virtual Strain Energy Caused by: Axial Load U = U n nnl AE n = internal virtual axial load caused by the external virtual unit load N = internal normal force in the member caused by the real load L = length of member A = cross-sectional area of the member E = modulus of elasticity of the material

Virtual Strain Energy dy = γ dx γ = τ / G dy = τ / G dx ( ) Caused by: Shear τ = K ( V / A) V dy = K dx GA du = νdy = νk V / A dx S ( ) L νvν V US = K dx GA 0

Virtual Strain Energy Caused by: Shear L νv U S = K dx GA 0 ν= internal virtual shear in the member caused by the external virtual unit load V = internal shear in the member caused by the real load G= shear modulus of elasticity for the material A = cross-sectional area of the member K = form factor for the cross-sectional area K=1.2 for rectangular cross sections K=10/9 for circular cross sections K=1.0 for wide-flange and I-beams where A is the area of the web.

Virtual Strain Energy γ = cd θ / dx γ = τ / G Tc τ = J τ Caused by: Torsion γ τ T d θ = dx = dx = dx c Gc GJ tt du t = tdθ = dx GJ U t = ttl GJ

Virtual Strain Energy Caused by: Torsion U t = ttl GJ t= internal virtual torque caused by the external virtual unit load T= internal shear in the member caused by the real load G= shear modulus of elasticity for the material J = polar moment of inertia of the cross-sectional L = member length

Virtual Strain Energy Caused by: Temperature d θ = U temp α T m c L 0 dx mα α T = c m dx

Virtual Strain Energy Caused by: Temperature U temp L = 0 m α T c m dx m= internal virtual moment in the beam caused by the external virtual unit load or unit moment α = coefficient of thermal expansion Tm = change in temperature between the mean temperature and the temperature at the top or the bottom of the beam c = mid-depth of the beam

Example 10

Example 11