Strength of materials

STRENGTH OF MATERIALS TRAN MINH TU - University of Civil Engineering, Giai Phong Str. 55, Hai Ba Trung Dist. Hanoi, Vietnam S T R E N G T H O F M A T E R IA L S 1/10/2013 1 2• Ass. Prof. Tran Minh Tu, PhD. Eng. • Mon., Fri.: 12.15 -14.45 A.M. at Room 202 H1 • Office: 1. Floor – Lab. Building • E-mail: tpnt2002@yahoo.com • Tel: 04.8691 462 (off.). Handphone: 0912101173 • Course notes: • • Ref. : Ferdinand P. Beer, Jr. J. T. DeWolf, D. F. Mazurek. Mechanics of Mate

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rials. Mc Graw Hill . 2009 • Office hours: Tuesday 8:00-11:00 A.M. or by appointment GENERAL INFORMATION 1/10/2013 3 Sample reading list: Russell C. Hibbeler, Mechanics of Materials, 6/E (required text) Roy R. Craig, Jr (1996), Mechanics and Materials Bedford, Fowler & Liecht (2003), Statics and Mechanics and Materials Beer & Johnson, Mechanics and Materials James H. Gere, Mechanics and Materials, 6th edition 1/10/2013 4 Strength of Materials This course discusses stress calculation due to loads. Description includes stress-strain concept, tensile test, stress and strain due to axial loading, statically indeterminate case related to axial loading, introduction to plasticity and residual stress, stress and strain due to other loads, such as torsion, bending moment, and shear force, Mohr’s circle of stress, failure theory, deflection, statically indeterminate structures, and energy method. 1/10/2013 5 Princeton : Grading: Mid Term Exam - 20% Design Project - 30% Take Home Final Exam - 25% Problem set(s) - 15% Other (See Instructor) - 10% MIT: Grading Homework 25% Lab Assignments 30% Quiz 15% Case Study and Presentation 10% Final Exam 20% Class Attendance* Stanford: Grading Homework Assignments 25% Lab Reports 10% Midterm 25% Final Exam 40% Berkeley: Grading Weekly homework assignments(25%) Two Midterm Examinations(20%+20%) Final examination (35%). GENERAL INFORMATION 1 CHAPTER Introduction – Concept of Stress 1/10/2013 6 1.1. Introduction 1.2. Review of Static 1.3. Equilibrium of deformable body 1.4. Concept of Stress 1.5. Stress Under General Loadings 1.6. Strain 1.7. Types of loading 1.8. Assumptions 1.9. Principle Superposition Contents 1/10/2013 7 1/10/2013 8 1.1. Introduction - Consider a diving board as an example of a deformable body - From rigid-body equilibrium, given the weight of the diver, and the lengths L1, L2, we can determine the diving board support reactions at A and B. Questions of the following type can only be answered by employing the principles and procedures of strength of materials 1/10/2013 9 1.1. Introduction 1.1.1. Strength of materials • A branch of mechanics • It studies the relationship of – External loads applied to a deformable body, and – The intensity of internal forces acting within the body • Are used to compute the deformations of a body • Strength of Materials is a field of study that determines strength, stiffness, & stability 1.1. Introduction 1/10/2013 10 1/10/2013 11 1.1. Introduction Mechanics Rigid Body Mechanics Deformable Body Mechanics Strength of MaterialsStatic (Dynamic) Kinetic Kinematic Fluid Mechanics 1/10/2013 12 (2) Kinetic: ∑F = ma (1) Equilibrium ∑Fx = 0; ∑Fy = 0; ∑Fz = 0; ∑Mx= 0; ∑My= 0; ∑Mz= 0 Rigid Body • Static Deformable Body • Strength of Materials (1) Equilibrium ∑Fx = 0; ∑Fy = 0; ∑Fz = 0; ∑Mx= 0; ∑My= 0; ∑Mz= 0 (2) Stress – Strain relationship: s = Ee 1.1. Introduction 1/10/2013 13 1.1.2. Classification of Structural element 1.1. Introduction 1/10/2013 14 1.1. Introduction 1/10/2013 15 1.1. Introduction Classification of Structural elements Structural elements compose a structure and can be classified as by their forms (shapes and dimensions). • Three dimensional body 1/10/2013 16 • Plates and Shells 1.1. Introduction 1/10/2013 17 1.1. Introduction • Rods, Bars 1/10/2013 18 1.1. Introduction 1/10/2013 19 1.1. Introduction 1.1. Introduction 1.1.3. External loads • Surface forces – Area of contact – Concentrated force – Linear distributed force – Centroid C (or geometric center) • Body force (e.g., weight) 1/10/2013 20 1.1. Introduction 1.1.4. Support reactions • for 2D problems 1/10/2013 21 1.2. Review of Static 1.2.1. Structure Free-Body Diagram The first step towards solving an engineering problems is drawing the free body diagram of the element/structure considered • The structure is designed to support a 30 kN load • Perform a static analysis to determine the internal force in each structural member and the reaction forces at the supports • The structure consists of a boom and a rod joined by pins (zero moment connections) at the junctions and supports 1/10/2013 22 1.2. Review of Static • The structure is detached from the supports and the loads and reaction forces are indicated • Ay and Cy can not be determined from these equations • Conditions for static equilibrium: 1/10/2013 23     0 0.6 30 0.8m kN mC xM A   40kNxA  0x x xF A C   40kNx xC A     0 30 0kNy y yF A C     30kNy yA C   1.2. Review of Static • In addition to complete the structure, each component must satisfy the conditions for static equilibrium • Results: 40 40 30kN kN kNx x yA C C       0 0.8 0mB y yM A A     • Consider a free-body diagram for the boom: kN30yC substitute into the structure equilibrium equation 1/10/2013 24  0 0.8mB yM A   0yA  1.2.2. Component Free-Body Diagram 1.2. Review of Static x, y, z coordinates x, y plane 1/10/2013 25 1.2.3. Equilibrium of rigid body Internal forces: The forces set up within a body to balance the effect of the externally applied forces. 1.3. Equilibrium of deformable body 1.3.1. Method of Sections To obtain the internal loading acting on specific region – “cut” by imaginarily section through the region where internal loadings are to be determined. 1/10/2013 26 1.3. Equilibrium of deformable body Two parts of the body are separated => A free-body diagram of one of the parts is drawn. Internal loadings The Internal force is distributed on the area of section and represent effects of the material of the top part of the body acting on the adjacent material of the bottom part. The Resultant force FR and moment MR0 at any specific point O is shown in the fig. c. If a member is long and slender (rod, beam), the section considered is generally taken perpendicular to the longitudinal axis – is called a cross section 27 Define resultant force (FR) and moment (MRo) in 3D: • Normal force, N • Shear force, Q • Torsional moment or torque, T • Bending moment, M 1.3. Equilibrium of deformable body 1/10/2013 28 1.3. Equilibrium of deformable body 1/10/2013 29 - Coplanar Loadings. If the body is subjected to a coplanar system of forces (fig. a), then only normal-force, shear-force, and bending-moment components will exist at the section (fig. b). - If we use the x, y, z coordinate axes, as shown on the left segment. 0 ...zF N   0 ...yF Q   0 0 ...M M   1.3.2. Internal resultant loadings • For coplanar loadings: – Apply ∑ Fx = 0 to solve for N – Apply ∑ Fy = 0 to solve for Q – Apply ∑ MO = 0 to solve for M 1.3. Equilibrium of deformable body Procedure for Analysis • Method of sections 1. Choose a segment to analyze 2. Determine the Support Reactions 3. Draw a free-body diagram for whole body 4. Apply the equations of equilibrium 1/10/2013 30 The resultant internal loadings at a point located on the section of a body can be obtained by using the method of section. This requires the following steps. 1.3. Equilibrium of deformable body Procedure for analysis • Free-body diagram 1. Keep all external loadings in exact locations before “sectioning” 2. Indicate the unknown resultants, N, Q, M, and T at the section, normally at the centroid C of the sectioned area 3. Coplanar system of forces only include N, Q, and M 4. Establish x, y, z coordinate axes with origin at the centroid • Equations of equilibrium 1. Sum moments at section, about each coordinate axes where resultants act 2. This will eliminate unknown forces N and Q, with direct solution for M (and T) 3. Resultant force with negative value implies that the assumed direction in reality is opposite to that shown on the free-body diagram 1/10/2013 31 Example 1.1 1/10/2013 32 Example 1.1 1/10/2013 33 1.4. Concept of Stress • The main objective of the study of mechanics of materials is to provide the future engineer with the means of analyzing and designing various machines and load bearing structures. • Both the analysis and design of a given structure involve the determination of stresses and deformations. This chapter is devoted to the concept of stress. 1/10/2013 34 1.4. Concept of Stress Stress: The distribution of the internal force over the area on which it acts is expressed as the force intensity, that is force per unit area. There are two types of stress: normal stress and shear stress Consider the sectioned area subdivided in to small area such as DA Very small force DF acting on DA, will be replaced by three components 1/10/2013 35 1.4.1. Stress Analysis 1.4. Concept of Stress The sign convention for the normal stress is as follows: - A positive value for s indicates tensile stress, that is, the stress due to a force DF that pulls on the area on which it acts. - A negative value for s indicates compressive stress – push on 1.4.3. Shear stress • Intensity of force, or force per unit area, acting tangent to ΔA • Symbol used for the shear stress is t (tau) 1.4.2. Normal stress • Intensity of force, or force per unit area, acting normal to ΔA • Symbol used for the normal stress, is σ (sigma) 1/10/2013 36 1.4. Concept of Stress 1/10/2013 37 Units (SI system) • Newtons per square meter (N/m2) or a pascal (1 Pa = 1 N/m2) • kPa = 103 N/m2 (kilo-pascal) • MPa = 106 N/m2 (mega-pascal) • GPa = 109 N/m2 (giga-pascal) 1.4.3. Axial Loading: Normal Stress 1.4. Concept of Stress 1.4.4. Average shear stress: 1/10/2013 38 1.4. Concept of Stress A F A P avet Single Shear A F A P 2 ave t Double Shear 1/10/2013 39 1.4.5. Shearing Stress Examples 1.4. Concept of Stress • Bolts, rivets, and pins create stresses on the points of contact or bearing surfaces of the members they connect. dt P A P bs • Corresponding average force intensity is called the bearing stress, • The resultant of the force distribution on the surface is equal and opposite to the force exerted on the pin. 1/10/2013 40 1.4.6. Bearing Stress in Connections • The rod is in tension with an axial force of 50 kN. • The boom is in compression with an axial force of 40 kN and average normal stress of –26.7 MPa. • The minimum area sections at the boom ends are unstressed since the boom is in compression.    MPa167 m10300 1050 m10300mm25mm40mm20 26 3 , 26        N A P A endBCs • At the flattened rod ends, the smallest cross- sectional area occurs at the pin centerline, • At the rod center, the average normal stress in the circular cross-section (A = 314x10-6m2) is sBC = +159 MPa. 1.4. Concept of Stress 1/10/2013 41 1.4.7. Rod & Boom Normal Stresses 1.4. Concept of Stress • The cross-sectional area for pins at A, B, and C, 26 2 2 m10491 2 mm25         rA MPa102 m10491 N1050 26 3 ,     A P aveCt • The force on the pin at C is equal to the force exerted by the rod BC, • The pin at A is in double shear with a total force equal to the force exerted by the boom AB, MPa7.40 m10491 kN20 26,    A P aveAt 1/10/2013 42 1.4.8. Pin Shearing Stresses 1.4. Concept of Stress • Divide the pin at B into sections to determine the section with the largest shear force, (largest) kN25 kN15   G E P P MPa9.50 m10491 kN25 26,    A PG aveBt • Evaluate the corresponding average shearing stress, FBC = 50kN 1/10/2013 43 • To determine the bearing stress at A in the boom AB, we have t = 30 mm and d = 25 mm,    MPa3.53 mm25mm30 kN40  td P bs • To determine the bearing stress at A in the bracket, we have t = 2(25 mm) = 50 mm and d = 25 mm,    MPa0.32 mm25mm50 kN40  td P bs 1.4. Concept of Stress 1/10/2013 44 1.4.9. Pin Bearing Stresses 1.5. Stress Under General Loadings • A member subjected to a general combination of loads is cut into two segments by a plane passing through Q A V A V A F x z A xz x y A xy x A x D D  D D  D D  DD D limlim lim 00 0 tt s • The distribution of internal stress components may be defined as, 1/10/2013 45 1.5. Stress Under General Loadings 1/10/2013 46 • For equilibrium, an equal and opposite internal force and stress distribution must be exerted on the other segment of the member. • Stress components are defined for the planes cut parallel to the x, y and z axes. For equilibrium, equal and opposite stresses are exerted on the hidden planes. • It follows that only 6 components of stress are required to define the complete state of stress • The combination of forces generated by the stresses must satisfy the conditions for equilibrium: 0 0     zyx zyx MMM FFF     yxxy yxxyz aAaAM tt tt  DD 0 zyyzzyyz tttt  andsimilarly, • Consider the moments about the z axis: 1.5. Stress Under General Loadings 1/10/2013 47 1/10/2013 48 • Displacement is the relative movement of a point on a rod with respect to a point in space • Deformation is the relative movement of two points on a rod. L P d A B Point B displaces a distance d down. The deformation of the rod between points A and B is d. 1.6. Strain – Displacement and deformation 1.6. Strain 1.6.1. Deformation: When A force is applied to a body, it will tend to change the body’s shape and size - deformation 1.6.2. Strain: 1/10/2013 49 The strain is the geometrical expression of deformation caused by the action of stress on a physical body. The strain is calculated by first assuming a change between two body states: the beginning state and the final state Basically Strain are of three types 1. Longitudinal Strain: -It is defined as the increase in length (d) per unit original length (L) when deformed by the external force. Longitudinal Strain = d / L 2.Volumetric Strain:-It is defined as the change in volume (v) per unit original volume (V), when deformed by external force. Volume Strain = v / V 1/10/2013 50 - Normal Strain (longitudinal strain) 3. Shear Strain:-When the change takes place in the shape of the body, the strain is called a shear strain. 1.6. Strain - Shear Strain 1.7. Types of loading Mechanics of materials is a branch of applied mechanics that deals with the behaviour of solid bodies subjected to various types of loading Compression Tension (stretched) Bending Torsion (twisted) Shearing 1/10/2013 51 1/10/2013 52 • Continuous • Homogeneous • Isotropic • Perfectly elastic • Small deformation 1.8. Basic Assumptions 1/10/2013 53 • Continuous 1.8. Basic Assumptions 1/10/2013 54 • Homogeneous 1.8. Basic Assumptions 1/10/2013 55 • Isotropic Isotropic Anisotropic 1.8. Basic Assumptions 1/10/2013 56 1.8. Assumptions – Hooke’s Law • "Hooke's Law" is about stretching springs and wires. • Hooke's Law states:- the extension is proportional to the force • the spring will go back to its original length when the force is removed • so long as we don't exceed the elastic limit. Hooke’s Law 1/10/2013 57 s = E e t = G g 1.8. Assumptions – Hooke’s Law Young’s modulus Shear modulus 1.9. Principle Superposition Conditions must be satisfied: 1/10/2013 58 +1.9. Principle Superposition 1/10/2013 59 1/10/2013 60 1/10/2013 61 1/10/2013 62 THANK YOU FOR YOUR ATTENTION ! 1/10/2013 63

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