Solid Mechanics Module (EG1031) Lectures PDF

SCHOOL OF ENGINEERING Solid Mechanics Module (EG1031) Module Introduction Dr. Elsiddig Elmukashfi [email protected] Outline  What is Solid Mechanics?  Why do we study Solid Mechanics?  Examples and Applications  Solid Mechanics...

SCHOOL OF ENGINEERING Solid Mechanics Module (EG1031) Module Introduction Dr. Elsiddig Elmukashfi [email protected] Outline  What is Solid Mechanics?  Why do we study Solid Mechanics?  Examples and Applications  Solid Mechanics in nutshell  Solid Mechanics Module  Structural Elements  Next Lecture Lecture 1 - Introduction 2 Introduction What is Solid Mechanics? “Solid mechanics (also known as mechanics of solids or mechanics of materials) is the branch of continuum mechanics that studies the behaviour of solid materials, especially their deformation and failure under the action of forces, temperature changes, phase changes, and other external or internal agents.” Let’s break it down … Prediction of solid Different Different Deformation components Materials Conditions and Failure Lecture 1 - Introduction 3 Introduction Why do we study Solid Mechanics?  Solid mechanics plays a pivotal role in engineering, technology, and science, impacting various industries and improving our daily lives.  Engineering Applications: Solid mechanics principles are fundamental in designing structures, machines, and systems to ensure safety, reliability, and efficiency, which makes it is fundamental for civil, aerospace, nuclear, biomedical and mechanical engineering, for geology, and for many branches of physics and chemistry such as materials science  Material Optimisation: Understanding how materials deform and fail helps in selecting the right materials for specific applications.  Innovation: It drives innovation by enabling the development of advanced materials and structures.  Safety Assurance: Ensures the safety of critical infrastructure like bridges, buildings, and aerospace components. Lecture 1 - Introduction 4 Introduction Examples and Applications  Mechanical Engineering  Developing machinery, engines, and mechanical systems and optimising components for strength, durability, and efficiency. Analysis of ship Simulation of Crash Test Lecture 1 - Introduction 5 Introduction Examples and Applications  Aerospace Engineering  Designing aircraft and spacecraft structures and evaluating materials for extreme conditions and structural integrity. Analysis of Plane fuselage Lecture 1 - Introduction 6 Introduction Examples and Applications  Structural Engineering  Designing and analysing buildings, bridges, and infrastructure for safety and stability and ensuring load-bearing capacity and resistance to external forces. Analysis of the Terraced Tower Analysis of Bridge Lecture 1 - Introduction 7 Introduction Examples and Applications  Material Science  Investigating material properties, including elasticity, plasticity, and fracture mechanics and developing new materials with improved characteristics. Analysis of the Crack Growth in Analysis of Creep Failure Stainless Steel Lecture 1 - Introduction 8 Introduction Examples and Applications  Biomechanics  Studying the mechanics of biological tissues and prosthetic devices and advancing medical implants and orthopaedic solutions. Analysis of the knee Analysis of Spine Discs Lecture 1 - Introduction 9 Introduction Solid Mechanics in Nutshell Solid Mechanics Lecture 1 - Introduction 10 Introduction Solid Mechanics Module  Module Specifications  15 credits module.  22 Lectures (every Monday 9-11am except weeks 11 and 12).  11 Workshops includes: 2 Experimental laboratories (weeks 14-16 and 18-20). 9 Practical sessions (every Thursday 3-4pm from week 13).  The assessment: Practical assignments (20%). Examination (80%).  The re-assessment is only examination (100%). Lecture 1 - Introduction 11 Introduction Solid Mechanics Module  The intended Learning Outcomes (ILOs) 1. Articulate an understanding of the basic principles of solid mechanics and mechanics of materials with applications to mechanical and aerospace systems and mechanisms. 2. Demonstrate an analytical understanding of the different types of problems encountered in the design of mechanical and aerospace systems and mechanisms and an ability to identify and apply the theory required. 3. Interpret data and perform fundamental design calculations across the fields of material properties, structural mechanics, mechanics of machines. 4. Describe the main properties and characteristics of practical materials. Lecture 1 - Introduction 12 Introduction Solid Mechanics Module  Learning Materials  Lecture and practical sessions notes.  Recommended reference books: P.P. Benham, R.J. Crawford and C.G. Armstrong Mechanics of Engineering Materials - Longman Scientific & Technical. M.F. Ashby and D.R.H. Jones Engineering Materials 1 - Butterworth Heinemann.  And any online materials and/or books you like! Lecture 1 - Introduction 13 Introduction Solid Mechanics Module  Teaching Plan  The plan is to cover the basics of Solid Mechanics and materials.  Trusses and Frames.  Atoms, molecules and  Stress and strain. crystals  Beams.  Hooke's law.  Types of materials  Shafts.  Tresca’s criterion  Origin of elasticity.  2D solids.  Elongation of bars.  Origin of plasticity.  Bending of beams.  Origin of failure.  Twisting of shafts.  Failure. Structures Materials Response Lecture 1 - Introduction 14 Introduction Structure  A structure is something which transmits load from one place to another (usually the ground).  A structural element bear the primary loads and provide structural support and/or enhance stability, aesthetics, or functionality. Golden Gate Bridge Lecture 1 - Introduction 15 Introduction Force and Moment  A force is a vector quantity that can cause an object to change its velocity, i.e., to accelerate, unless counterbalanced by other forces. It can also cause the object to deform.  A moment is a vector quantity that quantifies the rotational effect of a force about a specific point or axis. 𝐹𝐹1 𝑎𝑎 = 𝐹𝐹2 𝑏𝑏 𝐹𝐹 𝐹𝐹 𝑏𝑏 𝐹𝐹1 𝑎𝑎 𝐹𝐹2 𝐹𝐹 = 𝑚𝑚𝑚𝑚 Forces Moment Lecture 1 - Introduction 16 Introduction How do structures carry loads?  Tension 𝐹𝐹 𝐹𝐹  Compression 𝐹𝐹 𝐹𝐹  Bending 𝑀𝑀 𝑀𝑀  Shear 𝐹𝐹 𝐹𝐹  Torsion 𝑇𝑇 𝑇𝑇 Lecture 1 - Introduction 17 Introduction Structural Elements  Trusses  They are structural systems composed of straight members (bars) connected at their ends by pin joints which are designed to carry axial loads only. Bars Little Belt Bridge The pin joint Lecture 1 - Introduction 18 Introduction Structural Elements  Struts  They structural components designed to primarily resist axial compression forces, pushing or bracing elements together to support and stabilise structures. Struts Struts Little Belt Bridge The pin joint Lecture 1 - Introduction 19 Introduction Structural Elements  Beams  The primarily carry loads applied perpendicular to their longitudinal axis such that they distribute the loads as bending moments and shear forces along their length. Beams Beams Concrete Building Wooden Building Lecture 1 - Introduction 20 Introduction Structural Elements  Columns  Columns are vertical structural elements designed to support compressive loads, primarily acting along their longitudinal axis. Columns Columns Concrete Building Wooden Building Lecture 1 - Introduction 21 Introduction Structural Elements  Cables  They are slender, flexible, and tension-only structural elements designed to resist axial loads, primarily in tension. Wembley Stadium Tower Bridge Lecture 1 - Introduction 22 Introduction Structural Elements  Ties  They are structural elements that primarily resist axial tension forces, keeping two or more components together by pulling them in tension. London Eye Ties Lecture 1 - Introduction 23 Introduction Structural Elements  Arches  They are curved structural elements that support loads primarily through axial compression. Arches in Babylon Roman aqueduct (France) Lecture 1 - Introduction 24 Introduction Structural Elements  Shafts  They are cylindrical or rod-like mechanical components that transmit rotational motion and torque from one part of a machine to another. Axel Shafts Gear Box Car Axle Lecture 1 - Introduction 25  Introduction Structural Elements  Plates  They are two-dimensional structural elements with a relatively small thickness compared to their other dimensions, they can carry loads primarily through bending and flexural deformation. Slab Cabin floor Airplane Fuselage Building Skelton Lecture 1 - Introduction 26 Introduction Structural Elements  Shells  They are three-dimensional structural elements that have a curved geometry (sphere or cylinder), they can carry loads through membrane action (tension or compression across the surface) and bending. Cylindrical shell Spherical shell Airplane Fuselage Spherical Tank Lecture 1 - Introduction 27 Introduction Next Lecture – Introduction to Statics  We are going to cover the following topics: Static equilibrium Free body diagram Types of support The concept of external forces The concept of internal forces Statical determinacy requirements Lecture 1 - Introduction 28  SCHOOL OF ENGINEERING Solid Mechanics Module (EG1031) Introduction to Statics Dr. Elsiddig Elmukashfi [email protected] Outline  Introduction  Static equilibrium  Types of Load  Types of Support  Statical Determinacy  Next Lecture Lecture 2 - Introduction to Statics 2 Introduction The intended Learning Outcomes (ILOs) 1. Have a good grasp of the principles of equilibrium, which are fundamental to the understanding of all mechanical problems. 2. Be familiar with the concepts of force, moment and reaction and the importance of sign conventions. 3. Understand how applied loads and support conditions are idealised in structural analysis. 4. Be able to draw free body diagrams and calculate forces in simple planar structures. 5. Know the difference between internal and external forces. 6. Be able to classify a general structure into statically determinate, statically indeterminate, or mechanisms. Lecture 2 - Introduction to Statics 3 Static Equilibrium  Equilibrium is a fundamental concept in mechanics and is crucial for analysing structures and systems.  Equilibrium may be thought of as a special case of Newton’s Second Law in the case where there are no accelerations, i.e. the problem is static.  In one dimension, Newton’s Second Law reads: 𝐹𝐹 = 𝑚𝑚𝑚𝑚 = 0 if no acceleration  Consider the following example: 𝐹𝐹2 𝑚𝑚 𝐹𝐹1 + 𝐹𝐹3 𝐹𝐹 = 𝐹𝐹1 − 𝐹𝐹2 − 𝐹𝐹3 = 𝑚𝑚 0 = 0 Lecture 2 - Introduction to Statics 4 Static Equilibrium  In two dimensions, there are two independent direct equilibrium equations. 𝐹𝐹1 𝐹𝐹2 𝐹𝐹𝑦𝑦 = 𝐹𝐹 sin 𝜃𝜃 𝐹𝐹𝑥𝑥 = 0 𝐹𝐹 𝑦𝑦 𝜃𝜃 𝐹𝐹𝑦𝑦 = 0 𝐹𝐹𝑥𝑥 = 𝐹𝐹 cos 𝜃𝜃 𝑧𝑧 𝑥𝑥 𝐹𝐹3 𝐹𝐹4  The polygon of forces is a graphical method used to find the resultant of multiple forces. 𝐹𝐹1 𝐹𝐹4  The resultant force is a single force that represents the combined effect of multiple concurrent forces acting on an object which is zero in the case of equilibrium. 𝐹𝐹3 𝐹𝐹2 𝑅𝑅𝑥𝑥 = 𝐹𝐹𝑥𝑥 , 𝑅𝑅𝑦𝑦 = 𝐹𝐹𝑦𝑦 , 𝑅𝑅 = 𝑅𝑅𝑥𝑥2 + 𝑅𝑅𝑦𝑦2. Lecture 2 - Introduction to Statics 5 Static Equilibrium  In three dimensions, there are three independent direct equilibrium equations. 𝐹𝐹1 𝐹𝐹2 𝐹𝐹𝑥𝑥 = 0 𝐹𝐹𝑧𝑧 = 𝐹𝐹 cos 𝜃𝜃𝑧𝑧 𝜃𝜃𝑧𝑧 𝐹𝐹 𝐹𝐹𝑦𝑦 = 0 𝑧𝑧 𝐹𝐹𝑦𝑦 = 𝐹𝐹 cos 𝜃𝜃𝑦𝑦 𝜃𝜃𝑥𝑥 𝜃𝜃𝑦𝑦 𝑦𝑦 𝐹𝐹3 𝐹𝐹𝑥𝑥 = 𝐹𝐹 cos 𝜃𝜃𝑥𝑥 𝐹𝐹𝑧𝑧 = 0 𝐹𝐹4 𝑥𝑥  Similarly, resultant force can be calculated and it should equal to zero. 𝑅𝑅𝑥𝑥 = 𝐹𝐹𝑥𝑥 , 𝑅𝑅𝑦𝑦 = 𝐹𝐹𝑦𝑦 , 𝑅𝑅𝑧𝑧 = 𝐹𝐹𝑧𝑧 , 𝑅𝑅 = 𝑅𝑅𝑥𝑥2 + 𝑅𝑅𝑦𝑦2 + 𝑅𝑅𝑧𝑧2. Lecture 2 - Introduction to Statics 6 Static Equilibrium  The rotational motion is also expressed using Newton’s Second Law using angular acceleration. 𝑀𝑀 = 𝐼𝐼𝑎𝑎𝜃𝜃 = 0 if no acceleration 𝑜𝑜  where 𝐼𝐼 is the moment of inertia (refer to dynamics books). 𝐹𝐹  The moment is defined as the force times its lever arm: 𝑟𝑟 𝑀𝑀 = 𝐹𝐹𝐹𝐹  In two dimensions, there is one 𝐹𝐹1 independent moment equilibrium equation. 𝑟𝑟3 𝑜𝑜 𝑟𝑟4 𝐹𝐹2 𝑟𝑟1 𝑟𝑟2 𝑀𝑀 = 𝑀𝑀1 + 𝑀𝑀2 + 𝑀𝑀3 + 𝑀𝑀4 = 𝑚𝑚 0 = 0 𝑦𝑦 ⇒ 𝐹𝐹1 𝑟𝑟1 + 𝐹𝐹2 𝑟𝑟2 + 𝐹𝐹3 𝑟𝑟3 + 𝐹𝐹4 𝑟𝑟4 = 0 𝑧𝑧 𝑥𝑥 𝐹𝐹3 𝐹𝐹4 Lecture 2 - Introduction to Statics 7 Static Equilibrium  The moment can be expressed in vector form as 𝑜𝑜 𝐌𝐌 = 𝐅𝐅 × 𝐫𝐫 𝐅𝐅 𝐫𝐫  where 𝐫𝐫 is a vector from the point 𝑂𝑂 about which moments are taken to any point along the line of action of 𝐅𝐅. Bold face indicates vector and ‘×’ is a cross product (refer to vector algebra books). 𝐹𝐹𝑧𝑧  The vector representation of a force in three 𝐤𝐤 𝐅𝐅 dimensions can be expressed in terms of the unit vectors 𝐢𝐢, 𝐣𝐣 and 𝐤𝐤 in 𝑥𝑥, 𝑦𝑦 and 𝑧𝑧-directions, respectively. 𝐹𝐹𝑦𝑦 𝐣𝐣 𝐢𝐢 𝐅𝐅 = 𝐹𝐹𝑥𝑥 𝐢𝐢 + 𝐹𝐹𝑦𝑦 𝐣𝐣 +𝐹𝐹𝑧𝑧 𝐤𝐤, 𝐅𝐅 = 𝐹𝐹𝑥𝑥2 + 𝐹𝐹𝑦𝑦2 + 𝐹𝐹𝑧𝑧2 𝐹𝐹𝑥𝑥  Similarly, the position vector 𝐫𝐫 and moment 𝐌𝐌. 𝐫𝐫 = 𝑟𝑟𝑥𝑥 𝐢𝐢 + 𝑟𝑟𝑦𝑦 𝐣𝐣 +𝑟𝑟𝑧𝑧 𝐤𝐤, 𝐫𝐫 = 𝑟𝑟𝑥𝑥2 + 𝑟𝑟𝑦𝑦2 + 𝑟𝑟𝑧𝑧2 𝐌𝐌 = 𝑀𝑀𝑥𝑥 𝐢𝐢 + 𝑀𝑀𝑦𝑦 𝐣𝐣 +𝑀𝑀𝑧𝑧 𝐤𝐤, 𝐌𝐌 = 𝑀𝑀𝑥𝑥2 + 𝑀𝑀𝑦𝑦2 + 𝑀𝑀𝑧𝑧2 Lecture 2 - Introduction to Statics 8 Static Equilibrium  In three dimensions, there are three 𝐅𝐅1 independent moment equilibrium equations. 𝐅𝐅2 𝐫𝐫1 𝑜𝑜 𝐫𝐫2 𝑧𝑧 𝑀𝑀𝑥𝑥 = 0, 𝑀𝑀𝑦𝑦 = 0, 𝑀𝑀𝑧𝑧 = 0. 𝐫𝐫𝟒𝟒 𝐫𝐫3  The three equations are the components of 𝑦𝑦 𝐅𝐅3 the moment vector 𝐅𝐅4 𝑥𝑥 𝐌𝐌 = 𝐅𝐅1 × 𝐫𝐫1 +𝐅𝐅2 × 𝐫𝐫2 +𝐅𝐅3 × 𝐫𝐫3 +𝐅𝐅4 × 𝐫𝐫4 = 𝟎𝟎  It should be noted that the zero is a vector. 𝟎𝟎 = 0𝐢𝐢 + 0𝐣𝐣 + 0𝐤𝐤 Lecture 2 - Introduction to Statics 9 Static Equilibrium Example 1  Consider two forces act on a point 𝑂𝑂. One 4N has magnitude 4 N and acts in the 𝑥𝑥-direction. The other has magnitude 3 N and acts at 30° 3N 𝑦𝑦 to the vertical as shown in the figure. Calculate the magnitude and direction of the 30° force that should be applied so that the object 𝑧𝑧 𝑥𝑥 is in equilibrium  We can use Polygon of forces approach or vector approach. Use the vector approach: 𝐅𝐅2 𝐅𝐅1 = 3 cos 60° 𝐢𝐢 + 3 sin 60° 𝐣𝐣 𝑦𝑦 𝐅𝐅1 60° 𝐅𝐅2 = −4 𝐢𝐢  The resultant force is 𝑧𝑧 𝑥𝑥 𝐑𝐑 = 𝐅𝐅1 + 𝐅𝐅2 = −2.5 𝐢𝐢 + 2.6 𝐣𝐣 Lecture 2 - Introduction to Statics 10 Static Equilibrium  Thus, to achieve equilibrium, we need to apply a force equal in magnitude and opposite in direction to the resultant. 𝐅𝐅 = −𝐑𝐑 = 2.5 𝐢𝐢 − 2.6 𝐣𝐣  The magnitude of the forces is 𝐅𝐅 = 2.52 + 2.62 = 3.6 N 2.6 𝜃𝜃 = atan = 43.8° 2.5 𝐅𝐅 𝐅𝐅2 𝐅𝐅 𝜃𝜃 2.6 𝑦𝑦 𝜃𝜃 𝐅𝐅1 2.5 𝑥𝑥 𝐑𝐑 𝑧𝑧 30° Lecture 2 - Introduction to Statics 11 Static Equilibrium Example 2  Consider the structure in the figure. The bar 𝐴𝐴𝐴𝐴 is rigid and carries a block of a 60° mass 𝑚𝑚 using a rigid string. Also, the bar is connected to walls using rigid strings. 𝐴𝐴 𝐵𝐵 Find the forces in the strings. Hint: the 𝑚𝑚 weight of the bar is negligible. 𝐿𝐿/3 2𝐿𝐿/3  The Free Body Diagram (FBD)  It is a visual representation used in 𝑇𝑇2 engineering and physics to isolate one 𝑇𝑇3 part of a structure and consider equilibrium of all the forces acting on 60° that part. 𝑇𝑇1 𝐴𝐴 𝐵𝐵  We have three unknowns 𝑇𝑇1 , 𝑇𝑇2 and 𝑇𝑇3 so we need three equations. 𝑚𝑚𝑚𝑚 Lecture 2 - Introduction to Statics 12 Static Equilibrium 𝑇𝑇2 𝑇𝑇3 60° 𝑇𝑇3 sin 60° 𝑦𝑦 𝑇𝑇1 𝐴𝐴 𝐵𝐵 𝑇𝑇3 cos 60° 𝐿𝐿/3 2𝐿𝐿/3 𝑧𝑧 𝑥𝑥 𝑚𝑚𝑚𝑚  The horizontal equilibrium: 𝐹𝐹𝑥𝑥 = 𝑇𝑇3 cos 60° − 𝑇𝑇1 = 0 (1)  The vertical equilibrium: 𝐹𝐹𝑦𝑦 = 𝑇𝑇2 + 𝑇𝑇3 sin 60° − 𝑚𝑚𝑚𝑚 = 0 (2)  The moment equilibrium about 𝐴𝐴: 𝐿𝐿 𝑀𝑀𝑧𝑧 = 𝑇𝑇3 sin 60° 𝐿𝐿 − 𝑚𝑚𝑚𝑚 =0 (3) 3 Lecture 2 - Introduction to Statics 13 Static Equilibrium  Solving Eq.(3) yields: 2 𝑇𝑇3 = 𝑚𝑚𝑚𝑚 = 0.38𝑚𝑚𝑚𝑚 3 3  It follows from Eqs.(1) and (2): 1 𝑇𝑇1 = 𝑚𝑚𝑚𝑚 = 0.19𝑚𝑚𝑚𝑚 3 3  and 2 𝑇𝑇2 = 𝑚𝑚𝑚𝑚 = 0.67𝑚𝑚𝑚𝑚 3  Thus, we can directly determine the forces if the number of equations equals the number of unknown forces. Lecture 2 - Introduction to Statics 14 Types of Load  Structures are designed to sustain various types of loads and possible combinations of loads that could act on them during their lifetime.  Structural loads are external forces and can be broadly classified into four groups: 1. Dead Loads  They are structural loads of a constant magnitude over time. They include the self- weight of structural members, such as walls, ceilings, beams, columns, etc. 2. Live Loads  They are moveable or temporarily attached to a structure. They include the loads on a building created by the storage of furniture and equipment, occupancy (people), and impact. 3. Impact Loads  They are sudden or rapid loads applied on a structure over a relatively short period of time compared with other structural loads. Examples of impact loads are loads from moving vehicles, vibrating machinery, or dropped weights Lecture 2 - Introduction to Statics 15 Types of Load 4. Environmental Loads  They are loads is exerted by environment on structures. Some of these loads are: Rain loads: They are loads due to the accumulated mass of water on a rooftop during a rainstorm or major precipitation. Wind loads: They are pressures exacted on structures by wind flow. Wind forces have been the cause of many structural failures in history in coastal regions. Snow loads: In some geographic regions, the force exerted by accumulated snow on buildings’ roofs can be quite enormous, and it can lead to structural failure. Seismic loads: The ground motion caused by seismic forces in many geographic regions of the world can be quite significant and often damages structures. Hydrostatic and Earth Pressures: Retaining structures must be designed against overturning and sliding caused by hydrostatic and earth pressures to ensure the stability of their bases and walls. Lecture 2 - Introduction to Statics 16 Types of Load Modelling of Load  Loads are commonly idealised as one of the following types:  Point Load 𝐹𝐹  Point Moment 𝑀𝑀 𝐿𝐿/2  Distributed Load 𝑤𝑤𝑤𝑤 𝑤𝑤 𝐿𝐿  Triangular distributed Load 𝑤𝑤𝑤𝑤/2 𝑤𝑤 𝐿𝐿 𝐿𝐿/3 Lecture 2 - Introduction to Statics 17 Types of Support Degrees of Freedom (DOFs)  They refer to the number of independent ways a mechanical system can move or be displaced.  DOFs typically includes translational (movement in 𝑥𝑥, 𝑦𝑦, 𝑧𝑧) and rotational (rotation about 𝑥𝑥, 𝑦𝑦, 𝑧𝑧) freedoms. 𝑧𝑧 𝑢𝑢𝑧𝑧 𝑦𝑦 𝑢𝑢𝑥𝑥 𝑥𝑥 𝑢𝑢𝑦𝑦 Lecture 2 - Introduction to Statics 18 Types of Support Degrees of Freedom (DOFs)  They refer to the number of independent ways a mechanical system can move or be displaced.  DOFs typically includes translational (movement in 𝑥𝑥, 𝑦𝑦, 𝑧𝑧) and rotational (rotation about 𝑥𝑥, 𝑦𝑦, 𝑧𝑧) freedoms. 𝜃𝜃𝑧𝑧 𝑧𝑧 𝑦𝑦 𝜃𝜃𝑦𝑦 𝑥𝑥 𝜃𝜃𝑥𝑥 Lecture 2 - Introduction to Statics 19 Types of Support  In two dimensions, there are three DOFs, two translational and one rotational. 𝑢𝑢𝑦𝑦 𝜃𝜃𝑧𝑧 𝑦𝑦 𝑢𝑢𝑥𝑥 𝑧𝑧 𝑥𝑥 Lecture 2 - Introduction to Statics 20 Types of Support  Structural support refers to elements within a structure that prevent movement or some of the DOFs.  Reaction forces are the forces at the supports generated by the constraint.  Types of supports: 𝑦𝑦 1. Roller 𝑢𝑢𝑦𝑦 = 0 𝑥𝑥 𝑉𝑉 2. Pinned 𝑦𝑦 𝑢𝑢𝑥𝑥 = 𝑢𝑢𝑦𝑦 = 0 𝑥𝑥 𝐻𝐻 𝑉𝑉 3. Fixed or built-in 𝑦𝑦 𝑥𝑥 𝐻𝐻 𝑢𝑢𝑥𝑥 = 𝑢𝑢𝑦𝑦 = 𝜃𝜃𝑧𝑧 = 0 𝑀𝑀 𝑉𝑉 Lecture 2 - Introduction to Statics 21 Internal Forces  The loads acting on a structure are external forces.  The internal forces are generated by external forces and act inside the structural members.  The internal forces can most easily be visualised by making an imaginary cut through the structure.  When cutting a structural member and revelling the internal forces, they should be in equilibrium with the external forces. cut 𝑉𝑉 𝐻𝐻 𝐻𝐻 𝐻𝐻 𝑀𝑀 𝑉𝑉 𝑀𝑀 𝑉𝑉 𝑀𝑀  The internal forces depend on the external forces and the structural element. Lecture 2 - Introduction to Statics 22 Internal Forces Example 3  The figure shows Galileo's illustration of a cantilever beam that is fixed in one edge and unsupported at the other. If the beam is 2 m long and has mass per unit length of 7.5 kg/m, and a rock E has a mass 50 kg. Calculate the reactions at the wall. Also determine the internal forces at distance 𝑥𝑥 from the wall.  Let’s model the problem and figure out the FBD A 𝑤𝑤 A D 𝐻𝐻 D 7.5 kg/m C 𝑦𝑦 B C B E 𝑀𝑀𝑅𝑅 𝑉𝑉 𝑚𝑚𝑚𝑚 50 kg 𝑧𝑧 𝑥𝑥 Lecture 2 - Introduction to Statics 23 Internal Forces  The second step is to obtain the reaction forces. 𝐿𝐿/2 𝑤𝑤𝑤𝑤  The horizontal equilibrium: A 𝐻𝐻 D 𝐹𝐹𝑥𝑥 = 𝐻𝐻 = 0 C 𝑦𝑦 B 𝑀𝑀R 𝑚𝑚𝑚𝑚  The vertical equilibrium: 𝑉𝑉 𝐿𝐿 𝑧𝑧 𝑥𝑥 𝐹𝐹𝑦𝑦 = 𝑉𝑉 − 𝑤𝑤𝑤𝑤 − 𝑚𝑚𝑚𝑚 = 0 ⟹ 𝑉𝑉 = 𝑤𝑤𝑤𝑤 + 𝑚𝑚𝑚𝑚  The moment equilibrium about the left edge: 𝐿𝐿 𝑀𝑀𝑧𝑧 = −𝑀𝑀R − 𝑤𝑤𝑤𝑤 − 𝑚𝑚𝑚𝑚 𝐿𝐿 = 0 2 𝑤𝑤𝐿𝐿2 ⟹ 𝑀𝑀R = − − 𝑚𝑚𝑚𝑚𝑚𝑚 2 Lecture 2 - Introduction to Statics 24  Internal Forces  The third step is to obtain the internal forces. 𝑥𝑥/2 𝑤𝑤𝑤𝑤 A 𝑤𝑤 𝑆𝑆 A 𝑆𝑆 𝐻𝐻 𝐹𝐹 𝐻𝐻 𝐹𝐹 𝑦𝑦 B B 𝑀𝑀R 𝑀𝑀 𝑀𝑀R 𝑀𝑀 𝑉𝑉 𝑉𝑉 𝑥𝑥 𝑥𝑥 𝑥𝑥 𝑧𝑧  The horizontal equilibrium: 𝐹𝐹𝑥𝑥 = 𝐹𝐹 − 𝐻𝐻 = 0 ⟹ 𝐹𝐹 = 0  The vertical equilibrium: 𝐹𝐹𝑦𝑦 = 𝑉𝑉 − 𝑤𝑤𝑤𝑤 − 𝑆𝑆 = 0 ⟹ 𝑆𝑆 = 𝑤𝑤 𝐿𝐿 − 𝑥𝑥 + 𝑚𝑚𝑚𝑚  The moment equilibrium about the cut: 𝑥𝑥 𝑤𝑤𝑥𝑥 2 𝑀𝑀𝑧𝑧 = −𝑀𝑀R − 𝑉𝑉 𝑥𝑥 + 𝑤𝑤𝑤𝑤 + 𝑀𝑀 = 0 ⟹ 𝑀𝑀 = 𝑀𝑀R + 𝑉𝑉 𝑥𝑥 − 2 2 Lecture 2 - Introduction to Statics 25 Statical Determinacy  It is a condition in structural analysis where the number of equations available to solve for unknown forces (𝑁𝑁Eq ) is equal to the number of unknowns (𝑁𝑁Un ). 𝑁𝑁Eq = 𝑁𝑁Un  We call such structure is statically determinate.  Structures violate this equality are called statically indeterminate.  Structures with 𝑁𝑁Eq < 𝑁𝑁Un , are over constraint and the number of additional unknowns is called statically indeterminate with a degree of redundancy, i.e. equal to 𝑁𝑁Un − 𝑁𝑁Eq.  Structures with 𝑁𝑁Eq > 𝑁𝑁Un , are mechanisms as they have infinite number of solutions for the unknown forces.  This analysis yields different formula for different types of structures and boundary conditions. Lecture 2 - Introduction to Statics 26 Statical Determinacy Example 4  Consider example 3, Galileo’s illustration of a cantilever, and determine whether the cantilever is statically determinant or not. A 𝑤𝑤 A D 𝐻𝐻 D 7.5 kg/m C 𝑦𝑦 B C B E 𝑀𝑀𝑅𝑅 𝑉𝑉 𝑚𝑚𝑚𝑚 50 kg 𝑧𝑧 𝑥𝑥  The number of unknowns 𝑁𝑁Un = 3  The number of equations 𝑁𝑁Eq = 3  Thus, the system is statically determinant. Lecture 2 - Introduction to Statics 27 Next Lecture Analysis of pin jointed frames  We are going to cover the following topics: Pin jointed frames External forces in pin jointed frames Internal forces in pin jointed frames using sectioning method. Internal forces in pin jointed frames using matrix method. Lecture 2 - Introduction to Statics 28  SCHOOL OF ENGINEERING Solid Mechanics Module (EG1031) Introduction to Frames Dr. Elsiddig Elmukashfi [email protected] Outline  Introduction  Statical Determinacy  Analysis of Frames Method of Joints (Resolution at joints) Method of section  Next Lecture  Some Problems Lecture 3 - Introduction to Frames 2 Introduction The intended Learning Outcomes (ILOs) 1. Be able to draw free body diagrams and calculate forces in simple planar assemblies. 2. Be able to classify frameworks as statically determinate, statically indeterminate, or mechanisms. 3. Appreciate the importance of sign conventions. 4. Be able to determine forces in statically determinate pin-jointed frames using method of joints. 5. Understand how simple pin-jointed frame problems can be solved computationally using method of section. Lecture 3 - Introduction to Frames 3 Introduction  A pin-jointed structure, is also known as a truss, transmit no moment, and provide no resistance to rotation of the bars.  Pin-jointed frames offer versatility and stability, making them prevalent in structural engineering for their load-bearing capabilities and ease of analysis. Eiffel Tower Power Transmission Tower Formula 1 Car Lecture 3 - Introduction to Frames 4 Introduction What makes a frame to be a truss? 1. The structural members (bars) are joined together at their ends with pins or hinges, allowing for rotational movement. 𝑇𝑇1 𝑇𝑇1 𝑂𝑂 𝑂𝑂 𝑂𝑂 𝑂𝑂 𝑇𝑇2 𝑇𝑇2 𝑇𝑇3 𝑇𝑇3 Case 1 Case 2 In case 1, the moment equilibrium about 𝑂𝑂 is not satisfied due to the contribution of 𝑇𝑇2. In case 2, the moment equilibrium about 𝑂𝑂 is satisfied. Lecture 3 - Introduction to Frames 5 Introduction 2. The external loads are applied only at the joints, not on the bars. 3. The weight of the structural members (bars) is negligible. Consider the following structure: 𝐴𝐴 𝑇𝑇3 𝐴𝐴 𝑇𝑇2 𝑇𝑇8 𝑇𝑇7 𝐵𝐵 𝐵𝐵 𝑤𝑤 𝑇𝑇4 𝑇𝑇6 𝑇𝑇1 𝑚𝑚𝑚𝑚 𝑇𝑇5 𝑚𝑚 Let’s examine the internal forces 𝑇𝑇3 𝐴𝐴 𝑇𝑇2 𝑤𝑤𝑤𝑤 𝑇𝑇4 𝑇𝑇1 𝑚𝑚𝑚𝑚 Lecture 3 - Introduction to Frames 6 Introduction Rigid-jointed frames  They are structural frameworks composed of interconnected members joined together at their ends with rigid joints which do not allow for rotational movement.  The structural members in rigid-jointed frames carry bending moment and shear force in addition to axial force which are transferred across the joints.  Frames can take external loads on the frame members and self-weight of each member. Pin joint Rigid joint Lecture 3 - Introduction to Frames 7 Introduction Dimensionality  Frames can be classified based on the dimensionality: 1. Two-dimensional (Planar) Frames: Frames are confined to a two-dimensional plane where their members lie in the same plane and are interconnected. They are commonly used in planar trusses and plane frames. Analysis typically involves forces in the 𝑥𝑥 and 𝑦𝑦-directions. 2. Three-dimensional (Space) Frames: Frames existing in three-dimensional space where their members extend in various spatial directions, not limited to a single plane. They are often encountered in complex architectural structures, space frames, and industrial frameworks. Analysis involves forces and moments in multiple directions (𝑥𝑥, 𝑦𝑦, 𝑧𝑧) and rotational components.  Classifying frames based on dimensionality helps us determine the appropriate analysis and design methods for different structural systems. Lecture 3 - Introduction to Frames 8 Introduction Dimensionality  Examples of types of frames: Planar Frame Space Frame Lecture 3 - Introduction to Frames 9 Introduction Free Body Diagram for Pin-jointed Frames  Components of a FBD for Pin-Jointed Frames: 1. Depict the pin joint as a point where members meet. 2. Show all external forces and reactions acting on the joint or section. 3. Represent members as straight lines, indicating their orientation and unknown forces. 4. Label forces with appropriate notations (e.g., tension, compression). 𝐴𝐴 𝑇𝑇3 𝑇𝑇2 𝑇𝑇8 𝑇𝑇7 𝐵𝐵 𝐴𝐴 𝐵𝐵 𝑇𝑇4 𝑇𝑇6 𝑇𝑇1 𝑇𝑇5 Lecture 3 - Introduction to Frames 10 Statical Determinacy  Consider two-dimensional planar pin-jointed frames.  How many unknowns do we have? Number of reactions, 𝑟𝑟, (externally). The force in each bar, 𝑏𝑏, (internally).  How many equations can we set up to solve for this? Horizontal and vertical equilibrium at every pin, i.e., i.e. twice the number of pins.  So for a structure to be statically determinate the number of reactions, plus the number of bars must be exactly twice the number of pins 𝑏𝑏 + 𝑟𝑟 = 2𝑗𝑗  where 𝑗𝑗 is the number of pin joints, 𝑏𝑏 + 𝑟𝑟 > 2𝑗𝑗 for statically indeterminate and 𝑏𝑏 + 𝑟𝑟 < 2𝑗𝑗 for mechanism.  To be pedantic, this is a necessary but not sufficient condition (see Example 2)!. Lecture 3 - Introduction to Frames 11 Statical Determinacy Example 1  Consider the three pin-jointed frames. Examine whether they are statically determinate, indeterminate or mechanism. 𝑏𝑏 = 4 𝑏𝑏 = 5 𝑏𝑏 =6 𝑟𝑟 = 3 𝑟𝑟 = 3 𝑟𝑟 = 3 2𝑗𝑗 = 8 2𝑗𝑗 = 8 2𝑗𝑗 = 8 𝑏𝑏 + 𝑟𝑟 − 2𝑗𝑗 = −1 𝑏𝑏 + 𝑟𝑟 − 2𝑗𝑗 = 0 𝑏𝑏 + 𝑟𝑟 − 2𝑗𝑗 = 1 Mechanism Statically determinate Statically indeterminate Lecture 3 - Introduction to Frames 12 Statical Determinacy Example 2  Consider the following pin-jointed frame. Examine whether it is statically determinate, indeterminate or mechanism. If the frame is not statically determinate, suggest a method to achieve determinacy. Local mechanism 𝑏𝑏 = 8, 𝑟𝑟 = 4, 2𝑗𝑗 = 12 ⇒ 𝑏𝑏 + 𝑟𝑟 − 2𝑗𝑗 = 0  The formula suggests that the structure is statically determinate, BUT it is a mechanism! Lecture 3 - Introduction to Frames 13 Statical Determinacy  What should we do? Add one bar to stabilise the four-sided part of the frame. Remove one of the constrained degrees of freedom. 𝑏𝑏 = 9, 𝑟𝑟 = 3, 2𝑗𝑗 = 12 ⇒ 𝑏𝑏 + 𝑟𝑟 − 2𝑗𝑗 = 0 Lecture 3 - Introduction to Frames 14 Analysis of Frames  There are four methods that can be used to obtain the reaction and internal forces. 1. Method of Joints (Resolution at joints): It analyses forces within individual joints of the frame and applies equilibrium equations to solve for unknown member forces. 2. Method of sectioning: It adopts cutting the frame into sections to isolate specific members (i.e., members of interest). Then, applies equilibrium equations to solve for member forces in the selected sections. 3. Graphical Method (Bow's Notation): It represents the frame as a graphical diagram of the forces. Then using polygon of forces to determine the unknown forces. 4. Matrix Method (Stiffness Method): It applies matrix algebra to analyse member forces and displacements which is commonly used for more complex structures and in finite element analysis. Lecture 3 - Introduction to Frames 15 Analysis of Frames Method of Joints (Resolution at joints)  It is an analytical technique used to determine internal forces within individual joints in pin-jointed frames and trusses  It is based on the principle of static equilibrium, treating each joint as a separate free-body diagram.  The Method of Joints is particularly suitable for simple structures with relatively few members and joints.  It provides a systematic approach to analyse pin-jointed frames, helping engineers determine member forces efficiently.  Sign convention: In this course we define all tensile forces as positive and compressive forces as negative. (+) (−) Positive force Negative force Lecture 3 - Introduction to Frames 16 Analysis of Frames  The method of joints steps are: 1. Isolate a joint of interest. 2. Draw the forces acting on the joint, including applied loads and reactions. 𝑇𝑇4 𝑇𝑇3 𝐴𝐴 𝑦𝑦 𝑇𝑇5 𝑇𝑇2 𝑧𝑧 𝑥𝑥 𝑇𝑇1 3. Apply force equilibrium equations to solve for unknown member forces. 𝐹𝐹𝑥𝑥 = 0, 𝐹𝐹𝑦𝑦 = 0. 4. Repeat for other joints until all forces are determined. Lecture 3 - Introduction to Frames 17 Analysis of Frames Example 3  Consider the pin-jointed frame in the figure. All members are 5 m long and forces are in kN. Find the reaction forces in A and E, then determine the forces in all bars. 5 10 B D 𝑦𝑦 All angles are 60° 𝐻𝐻A A C E 𝑧𝑧 𝑥𝑥 10 𝑉𝑉A 𝑉𝑉E 15  The first step is to check the statical determinacy: 𝑏𝑏 = 7, 𝑟𝑟 = 3, 2𝑗𝑗 = 10 ⇒ 𝑏𝑏 + 𝑟𝑟 − 2𝑗𝑗 = 0 Lecture 3 - Introduction to Frames 18 Analysis of Frames  The second step is to determine the reaction forces:  The horizontal equilibrium: 𝐹𝐹𝑥𝑥 = 𝐻𝐻A − 10 = 0 ⟹ 𝐻𝐻A = 10 kN  The vertical equilibrium: 𝐹𝐹𝑦𝑦 = 𝑉𝑉A + 𝑉𝑉E − 30 = 0  The moment equilibrium about 𝐴𝐴: 𝑀𝑀𝑧𝑧 = 5 × 2.5 + 15 × 5 + 10 × 7.5 − 𝑉𝑉E × 10 = 0 ⟹ 𝑉𝑉E = 16.25 kN  It follows from the vertical equilibrium that: ⟹ 𝑉𝑉A = 13.75 kN Lecture 3 - Introduction to Frames 19 Analysis of Frames  The third step is to isolate pin joints and analyse them.  Joint A: 𝑇𝑇AB  The horizontal equilibrium: 𝐻𝐻A A 60° 𝑇𝑇AC 𝐹𝐹𝑥𝑥 = 𝑇𝑇AC + 𝑇𝑇AB cos 60° + 10 = 0  The vertical equilibrium: 𝑉𝑉A 𝐹𝐹𝑦𝑦 = 𝑇𝑇AB sin 60° + 13.75 = 0 ⟹ 𝑇𝑇AB = −15.88 kN  It follows from the horizontal equilibrium that: ⟹ 𝑇𝑇AC = −2.06 kN  Joint B: 5 𝑇𝑇BD  The horizontal equilibrium: B 60° 𝐹𝐹𝑥𝑥 = 𝑇𝑇BD + 𝑇𝑇BC cos 60° − 𝑇𝑇AB cos 60° = 0 𝑇𝑇AB 60° 𝑇𝑇BC Lecture 3 - Introduction to Frames 20 Analysis of Frames  The vertical equilibrium: 𝐹𝐹𝑦𝑦 = −5 − 𝑇𝑇BC sin 60° − 𝑇𝑇AB sin 60° = 0  Solving the equations yield: ⟹ 𝑇𝑇BC = 10.1 kN and 𝑇𝑇BD = 13.0 kN  Joint C: 𝑇𝑇BC 𝑇𝑇CD  The horizontal equilibrium: 60° 𝐹𝐹𝑥𝑥 = 𝑇𝑇CE + 𝑇𝑇CD cos 60° − 𝑇𝑇BC cos 60° 𝑇𝑇AC 60° 60° 𝑇𝑇CE −𝑇𝑇AC − 10 = 0 C 10 15  The vertical equilibrium: 𝐹𝐹𝑦𝑦 = 𝑇𝑇CD sin 60° + 𝑇𝑇BC sin 60° − 15 = 0 ⟹ 𝑇𝑇CD = 7.22 kN Lecture 3 - Introduction to Frames 21 Analysis of Frames  It follows from the horizontal equilibrium that: ⟹ 𝑇𝑇CE = 9.38 kN  Joint D: 10  The horizontal equilibrium: 𝑇𝑇BD D 𝐹𝐹𝑥𝑥 = −𝑇𝑇BD − 𝑇𝑇CD cos 60° + 𝑇𝑇DE cos 60° = 0 60° 60° 𝑇𝑇DE ⟹ 𝑇𝑇DE = −18.76 kN 𝑇𝑇CD  We now have forces calculated in all members.  We have not used horizontal equilibrium at D or equilibrium at E — we can (and should) use these to check our answers. Lecture 3 - Introduction to Frames 22 Analysis of Frames Method of sections  It is an analytical technique used to determine internal forces within specific members of pin-jointed frames and trusses, unlike the Method of Joints, it focuses on isolating and analysing entire sections of the structure.  The method of sections steps are: 1. Select a section of the frame that includes the member(s) of interest. 2. Cut through the selected section, creating a free-body diagram of the part of the structure being analysed. 3. Include external loads and reactions acting on the section. 4. Apply equilibrium equations and moment equations to solve for unknown forces in the member. 5. Repeat for other sections if necessary. Lecture 3 - Introduction to Frames 23 Analysis of Frames Example 4  Consider the frame in Example 3. Determine the forces in members BD and CE only. 5 10 B D 𝑦𝑦 All angles are 60° 𝐻𝐻A A C E 𝑧𝑧 𝑥𝑥 10 𝑉𝑉A 𝑉𝑉E 15  From previous solution: 𝐻𝐻A = 10 kN, 𝑉𝑉A = 13.75 kN, and 𝑉𝑉E = 16.25 kN Lecture 3 - Introduction to Frames 24  Analysis of Frames  The first step is to select the section.  Cut the structure through BD, CD and CE 5 10 5 𝑇𝑇BD B D B D 𝑇𝑇CD 𝐻𝐻A A C 𝐻𝐻A A C E 𝑦𝑦 𝑇𝑇CE 10 10 𝑉𝑉A 𝑉𝑉E 𝑉𝑉A 15 15 𝑧𝑧 𝑥𝑥  The moment equilibrium about C: 𝑀𝑀𝑧𝑧 = −𝑇𝑇BD × 5 sin 60° − 𝑉𝑉A × 5 + 5 × 2.5 = 0 ⟹ 𝑇𝑇BD = 13.0 kN Lecture 3 - Introduction to Frames 25 Analysis of Frames  The moment equilibrium about D: 𝑀𝑀𝑧𝑧 = −10 × 5 sin 60° + 10 × 5 sin 60° − 𝑉𝑉A × 7.5 + 5 × 5 +15 × 2.5 + 𝑇𝑇CE × 5 sin 60° = 0 ⟹ 𝑇𝑇CE = 9.38 kN  Thus, using the three equilibrium equations, we can solve for a maximum of unknowns.  Therefore, occasionally you may need to use sections which cut more than three members  You will then have to make a second cut (or use some other information) to calculate some of the forces. Lecture 3 - Introduction to Frames 26 Next Lecture Introduction to Mechanics of Materials  We are going to cover the following topics: The concept of stress The concept of strain Materials behaviour in one-dimensional loading (i.e., uniaxial loading). Elastic-plastic behaviour in one-dimension. Lecture 3 - Introduction to Frames 27 Some Problems Problem 1  For both pin-jointed frames in the Figure below, state whether it is statically determinate, statically indeterminate or a mechanism. If the frame is statically indeterminate give number of redundancies, and if it is a mechanism give the number of degrees of freedom of movement. (a) (b)  Answers: a) Statically determinate, no local mechanism. b) Mechanism with one degree of freedom. Lecture 3 - Introduction to Frames 28 Some Problems Problem 2  For the pin-jointed truss shown in the Figure below: a) Find the reactions at A and E. b) Determine the forces in all the bars. H I J 2m 2m F G 𝑦𝑦 A B C D E 40 kN 40 kN 𝑧𝑧 𝑥𝑥 3m 3m 3m 3m  Answers: a) 𝑉𝑉A = 50 kN, 𝐻𝐻A = 0, 𝑉𝑉E = 30 kN. b) 𝑇𝑇AB = 𝑇𝑇BC = 37.5 kN, 𝑇𝑇CD = 𝑇𝑇𝐷𝐷𝐷𝐷 = 22.5 kN. Lecture 3 - Introduction to Frames 29  SCHOOL OF ENGINEERING Solid Mechanics Module (EG1031) Introduction to Mechanics of Materials Dr. Elsiddig Elmukashfi [email protected] Outline  Introduction  The Stress  The Strain  The Uniaxial Loading  The Stress-Strain Relationship  The Linear Elasticity  The Plastic Deformation  Next Lecture Lecture 4 - Introduction to Mechanics of Materials 2 Introduction The intended Learning Outcomes (ILOs) 1. Have a good grasp of the key concepts in materials mechanics and appreciate the importance of the field. 2. Understand the concept of stress and the different types of stresses 3. Understand the concept of strain and the different types of strains. 4. Appreciate the importance of sign conventions for stress and strain. 5. Be familiar with the elastic and plastic deformation and their materials properties. 6. Understand how what is a uniaxial deformation state and how to solve problems for stress and strain. Lecture 4 - Introduction to Mechanics of Materials 3 Introduction Mechanical Behaviour of Materials  Materials Mechanics is a branch of engineering that focuses on understanding how materials respond to various applied forces and loads.  It explores the mechanical behaviour of materials, including how they deform, transmit forces, and withstand external conditions.  Key Concepts: 1. Stress: Internal resistance by the body to the external load applied to it. 2. Strain: Measure of material deformation under external load. 3. Constitutive behaviour: models describe the material responses to different mechanical and/or thermal. 4. Material Properties: properties that reflect the relationship between its response (deformation or failure) to an applied load or force. Important mechanical properties are strength, hardness, ductility and stiffness. Lecture 4 - Introduction to Mechanics of Materials 4  Introduction Spring Experiment  Consider the spring in the Figure.  In the first step, we load the spring using a mass 𝑚𝑚 and record the deflection to be 𝑥𝑥. 𝑥𝑥  In the second step, we load the same 2𝑥𝑥 𝑚𝑚 spring using a mass 2𝑚𝑚 and record the deflection to be 2𝑥𝑥. 2𝑚𝑚  How do we relate the load to deflection? 𝐹𝐹1 = 𝑚𝑚𝑚𝑚 𝐹𝐹 ∝ 𝑥𝑥 ⇒ 𝐹𝐹 = 𝑘𝑘𝑘𝑘 𝐹𝐹1 𝐹𝐹2 = 2𝑚𝑚𝑚𝑚  Where 𝑘𝑘 is the spring constant (stiffness). 𝑚𝑚  Thus, we can relate the internal force to the deflection using a linear relationship with a 𝐹𝐹1 parameter (i.e., this is a constitutive law). Lecture 4 - Introduction to Mechanics of Materials 5 Introduction The Problem  The problem is to determine the deformation and failure of a structure! Given Determine Determine Prediction the External Internal the Failure Forces Forces Deformation  Example: Analysing a truss for the deformation and failure. B D 𝑦𝑦 A C E 𝑧𝑧 𝑥𝑥 𝐹𝐹 Lecture 4 - Introduction to Mechanics of Materials 6 The Stress  Stress is a fundamental concept in solid mechanics and it represents the internal resistance of a material to the external forces.  Stress is denoted by 𝜎𝜎 and defined as the force, 𝐹𝐹, per unit area, 𝐴𝐴: 𝐹𝐹 𝜎𝜎 = 𝐴𝐴 𝐴𝐴 𝐹𝐹 𝐹𝐹 𝐹𝐹 𝜎𝜎  The dimension of the stress is 𝐹𝐹 𝑀𝑀 = 𝐿𝐿2 𝐿𝐿𝑇𝑇 2  The units of the stress in the SI units (N, m, kg, s) is N/m2 (Pa).  The sign of the stress is positive when it is tension and negative when it is compression. Lecture 4 - Introduction to Mechanics of Materials 7 The Stress  There are two types of stress that a structure can experience: Normal Stress: is the stress generated by a force acts perpendicular (or "normal") to the surface of an object. It is usually denoted by the Greek letter Sigma 𝜎𝜎. Shear Stress: is the stress generated by a force acts parallel to the surface of an object. It is usually denoted by the Greek letter 𝜏𝜏. 𝐹𝐹 𝐹𝐹 𝜎𝜎 = 𝜏𝜏 = 𝐴𝐴 𝐴𝐴 𝐴𝐴 𝐹𝐹 𝐴𝐴 𝐹𝐹 𝐹𝐹 𝐹𝐹 Normal stress Shear stress Lecture 4 - Introduction to Mechanics of Materials 8 The Strain  Strain is a key concept in solid mechanics and it quantifies the deformation of a material under the influence of external forces and.  Strain is denoted by 𝜀𝜀 and defined as a dimensionless quantity (has no unit), typically expressed as a ratio or percentage change in length. Δ𝐿𝐿 𝜀𝜀 = 𝐿𝐿0  where Δ𝐿𝐿 is the change in length and 𝐿𝐿0 is the original length. Unloaded Rod 𝐿𝐿0 Uniaxially Loaded 𝐹𝐹 𝐹𝐹 Rod 𝐿𝐿0 Δ𝐿𝐿  The sign of the strain is positive when it is increase in length and negative when it is decrease in length. Lecture 4 - Introduction to Mechanics of Materials 9 The Strain  There are two types of strain that a structure can experience: Normal Strain: is the elongation of an object in response to a normal stress (i.e., perpendicular to a surface), and is denoted by the Greek letter epsilon 𝜀𝜀. Shear Strain: is the deformation of an object is response to a shear stress (i.e. parallel to a surface), and is denoted by the Greek letter gamma 𝛾𝛾. Δ𝐿𝐿 Δ𝐿𝐿 𝜀𝜀 = 𝛾𝛾 ≈ tan 𝛾𝛾 = 𝐿𝐿0 ℎ0 ℎ0 𝐿𝐿0 Δ𝐿𝐿 𝐹𝐹 𝐹𝐹 𝐹𝐹 𝛾𝛾 𝐿𝐿0 Δ𝐿𝐿 𝐹𝐹 Normal strain Shear strain Lecture 4 - Introduction to Mechanics of Materials 10 The Strain  The stain due to mechanical loading is called mechanical strain.  There are other sources of strain that are caused by environment, e.g., temperature, moisture, etc.  The thermal strain results from temperature changes in a material such that as a material is heated or cooled, its atoms or molecules expand or contract, leading to a change in size..  The thermal strain is calculated using the material's coefficient of thermal expansion. Δ𝐿𝐿 𝛼𝛼𝐿𝐿0 Δ𝑇𝑇 Δ𝐿𝐿 = 𝛼𝛼𝐿𝐿0 Δ𝑇𝑇 ⟹ 𝜀𝜀th = = = 𝛼𝛼Δ𝑇𝑇 𝐿𝐿0 𝐿𝐿0  where 𝛼𝛼 is thermal expansion coefficient and its units is 1/℃.  The total strain of a material is the sum of the mechanical strain 𝜀𝜀m and the thermal strain 𝜀𝜀th : 𝜀𝜀 = 𝜀𝜀m + 𝜀𝜀th Lecture 4 - Introduction to Mechanics of Materials 11 Uniaxial Loading  Uniaxial loading is a type of loading in where a material or structure experiences forces or loads in a single predominant direction.  It is characterised by the application of forces along one primary axis, causing either tension (stretching) or compression (shortening) in that direction. 𝑥𝑥 𝐿𝐿0 Unloaded bar 𝐹𝐹 𝐹𝐹 𝐹𝐹 𝐹𝐹 𝐿𝐿0 Δ𝐿𝐿 Δ𝐿𝐿 𝐿𝐿0 Tensile Load Compressive Load Lecture 4 - Introduction to Mechanics of Materials 12 Uniaxial Loading  The stress and strain are defined along 𝑥𝑥-direction, respectively, as 𝐹𝐹 Δ𝐿𝐿 𝜎𝜎𝑥𝑥 = and 𝜀𝜀𝑥𝑥 = 𝐴𝐴 𝐿𝐿0  The deformation also takes place along other directions (lateral directions) such that we need to define the strain components in other directions. Lateral strain = −𝜈𝜈 × Direct strain  This property of a material is known as Poisson's ratio, and it is denoted by the Greek letter nu 𝜈𝜈. 𝜀𝜀𝑦𝑦 = 𝜀𝜀𝑧𝑧 = −𝜈𝜈𝜀𝜀𝑥𝑥 𝑧𝑧 𝜀𝜀𝑦𝑦 = 𝜀𝜀𝑧𝑧 𝜎𝜎𝑥𝑥 𝑥𝑥 𝜎𝜎𝑥𝑥 𝑦𝑦 𝐿𝐿0 Δ𝐿𝐿 Lecture 4 - Introduction to Mechanics of Materials 13 Stress-Strain Relationship  The stress-strain relationship is a fundamental concept in mechanics of materials and be used to compare materials, regardless of the specimen geometry.  It describes how materials respond to applied loads by relating the stress (internal resistance) to the strain (deformation).  The stress-strain relationship depends on the material system, e.g., rubber materials exhibit pure elastic behaviour and metals show elastic-plastic behaviour.  Uniaxial Tensile Test is a standard experiment in materials testing to determine the stress-strain behaviour.  It involves applying an axial tensile load to a specimen (usually a cylindrical shape) along a single axis such that the test provides valuable data on a material's stress- strain behaviour.  During the test, the specimen undergoes deformation, typically elongation, and reaches the final failure as it experiences tension. Lecture 4 - Introduction to Mechanics of Materials 14 Stress-Strain Relationship  The figure below, shows a typical testing machine where a dog-bone specimen is tested and a typical load-displacement curve of mild steel.  The experiment is performed by applying the uniaxial load monotonically until failure. 𝐹𝐹 N 𝐹𝐹, Δ𝐿𝐿/2 3000 2000 1000 1 2 3 4 Δ𝐿𝐿 mm 𝐹𝐹, Δ𝐿𝐿/2 Testing Machine Load-displacement curve Lecture 4 - Introduction to Mechanics of Materials 15 Stress-Strain Relationship  The figure below, shows the stress-strain relationship calculated from the load- displacement data.  The stress-strain relationship show the elastic modulus 𝐸𝐸, elastic limit 𝜎𝜎y , the tensile strength 𝜎𝜎TS , and the strain to failure 𝜀𝜀f (ductility limit). 𝜎𝜎 MPa Tensile Strength (onset of necking) 600 Initial yield point 𝜎𝜎TS 400 𝜎𝜎y 200 Ductility limit 𝐸𝐸 Elastic modulus 𝜀𝜀 % 5 10 15 𝜀𝜀f 20 Stress-strain behaviour Lecture 4 - Introduction to Mechanics of Materials 16 The Linear Elasticity  The linear elasticity describes the behaviour of materials within their elastic limit, where deformation is directly proportional to the applied stress.  In the linear elastic region, materials return to their original shape when the load is removed.  Mathematically, it is represented by Hooke's Law.  One dimensional Hooke's Law is expressed as: For the normal stress-strain: 𝜎𝜎 = 𝐸𝐸𝐸𝐸 For the shear stress-strain: 𝜏𝜏 = 𝐺𝐺𝐺𝐺  where 𝐸𝐸 is elastic modulus (it is also called Young’s modulus) and 𝐺𝐺 is the shear modulus.  The elastic an shear moduli are material parameters and they have a unit of stress. Lecture 4 - Introduction to Mechanics of Materials 17 The Linear Elasticity  The elastic (Young’s) modulus varies between different materials.  The figure is taken from Ashby and Jones Engineering Materials - 1 book which illustrates the wide range of the elastic modulus.  The figure shows the different groups of materials (we will learn about that in the next lecture!). Bar chart of 𝐸𝐸-modulus data Lecture 4 - Introduction to Mechanics of Materials 18 The Linear Elasticity  Numerical data for Young’s modulus: Lecture 4 - Introduction to Mechanics of Materials 19 The Linear Elasticity  Numerical data for Young’s modulus (Cont’d): Lecture 4 - Introduction to Mechanics of Materials 20 The Plastic Deformation  Plastic deformation occurs when a material undergoes permanent, non- reversible deformation beyond its elastic limit (yield strength).  Plastic deformation involves the rearrangement of atomic or molecular structures such that the material does not return to its original shape after unloading.  The total strain can be decomposed 𝜎𝜎 MPa into elastic and plastic contributions 600 𝜀𝜀 = 𝜀𝜀el + 𝜀𝜀pl 𝜎𝜎y  where 𝜀𝜀el and 𝜀𝜀pl are the elastic and 400 𝜎𝜎y0 plastic contributions, respectively. 200  The yield strength is a function of 𝐸𝐸 𝐸𝐸 the material history and the yield condition is defined by: 𝜀𝜀 % 5 10 15 20 𝑓𝑓 𝜎𝜎 = |𝜎𝜎| − 𝜎𝜎𝑦𝑦 ≥ 0 𝜀𝜀pl 𝜀𝜀el Stress-strain behaviour Lecture 4 - Introduction to Mechanics of Materials 21 The Plastic Deformation  The plastic behaviour is essential to understand the material yielding, ductility, and failure.  The figure is taken from Ashby and Jones Engineering Materials - 1 book which illustrates the wide range of the yield strength.  The figure shows the different groups of materials (we will learn about that in the next lecture!). Bar chart of Yield Strength data Lecture 4 - Introduction to Mechanics of Materials 22 The Plastic Deformation  Numerical data for yield strength, tensile strength and strain to failure: Lecture 4 - Introduction to Mechanics of Materials 23 The Plastic Deformation  Numerical data for yield strength, tensile strength and strain to failure (Cont’d): Lecture 4 - Introduction to Mechanics of Materials 24 The Plastic Deformation  Numerical data for yield strength, tensile strength and strain to failure (Cont’d):  Note: Bracketed 𝜎𝜎TS data for brittle materials refer to the modulus of rupture 𝜎𝜎r (we will discuss it in the future). Lecture 4 - Introduction to Mechanics of Materials 25 Examples Example 1  Consider the bar in the figure. The bar material has the properties 𝐸𝐸,𝜎𝜎𝑦𝑦 and 𝛼𝛼. The edges of the bar are rigidly clamped. The bar is then uniformly 𝑥𝑥 heated up to a temperature 𝑇𝑇 from initial temperature 𝑇𝑇0. Find the stress in the bar and the temperature at which the bar will start to yield.  The total strain is 𝜀𝜀 = 𝜀𝜀m + 𝜀𝜀th = 0  where 𝜎𝜎 𝜀𝜀m = and 𝜀𝜀th = 𝛼𝛼Δ𝑇𝑇 𝐸𝐸  and Δ𝑇𝑇 = 𝑇𝑇 − 𝑇𝑇0.  Therefore, 𝜎𝜎 + 𝛼𝛼Δ𝑇𝑇 = 0 ⟹ 𝜎𝜎 = −𝐸𝐸𝛼𝛼Δ𝑇𝑇 𝐸𝐸  The yield condition is |𝜎𝜎| ≥ 𝜎𝜎y : 𝜎𝜎y 𝜎𝜎y = −𝐸𝐸𝛼𝛼Δ𝑇𝑇y ⟹ Δ𝑇𝑇y = 𝐸𝐸𝐸𝐸 Lecture 4 - Introduction to Mechanics of Materials 26 Next Lecture Introduction to Materials  We are going to cover the following topics: Classes of materials and microstructure (i.e., Crystalline and amorphous materials). Relation between elastic properties and interatomic bonding and structure (i.e., interatomic potential, crystal’s structure). Engineering properties and selection of materials. Lecture 4 - Introduction to Mechanics of Materials 27  SCHOOL OF ENGINEERING Solid Mechanics Module (EG1031) Introduction to Materials Dr. Elsiddig Elmukashfi [email protected] Outline  Introduction  Materials  Atomic Bonding Atomic Structure Interatomic bond Types of Bonds  Next Lecture Lecture 4 - Introduction to Mechanics of Materials 2 Introduction The intended Learning Outcomes (ILOs) 1. Have a good grasp of the matter structure. 2. Understand the concept of interatomic potential. 3. Understand the concept of interatomic force. 4. Be familiar with the different types of interatomic bonds. 5. Understand how the character of interatomic bond affects the properties. Lecture 5 - Introduction to Materials 3 Introduction  Materials have been shaping human history since the dawn of civilisation.  Historically, the development and advancement of societies have been intimately tied to the members’ ability to produce and manipulate materials to fill their needs.  Civilizations have been designated by the level of their materials development (Stone Age, Bronze Age, Iron Age). Stone age weapon Bronze age weapon Iron age weapon  Form very limited number of materials, in the early days of human civilisation, materials utilisation has become totally a selection process, thanks to the understanding of properties and manufacturing processes. Lecture 5 - Introduction to Materials 4 Introduction  Everything that is designed and made by people is made of some material or combination of materials.  The central question for a design engineer is: How do I choose the best material for a given application?  A natural approach to this question is to attempt to match the attributes or properties of the material to the needs or requirements of the engineering design problem. Materials Properties Design Requirements Density Function Strength Size Stiffness Shape Weight Melting Temperature … Durability  This turns out not to be straightforward as there is rarely an exact correspondence between properties and requirements of the engineering design problem. Lecture 5 - Introduction to Materials 5 Materials  Materials are categorised into four primary classes based on their atomic or molecular structure: Metals Composites Ceramics Polymers Lecture 5 - Introduction to Materials 6 Materials  Any individual material has many properties or attributes.  There are common classes of property that the designer must consider. Mechanical properties Physical properties Transport properties  Modulus  Density  Optical properties  Yield and tensile (refractive index, strength.  Melting temperature transparency, …)  Hardness.  Thermal expansivity  Thermal properties (conductivity,  Fracture toughness  Latent heat of fusion diffusivity, …)  Fatigue strength  Electrical properties  Heat capacity  Creep strength (conductivity/,  … permittivity,…)  …  …  There is no agreement on common classes of property that the designer must consider. Lecture 5 - Introduction to Materials 7 Atomic Bonding Atomic Structure  Materials are made up of atoms, consisting of positively charged protons and uncharged neutrons in a central nucleus, surrounded by negatively charged electrons.  the electrons are considered to exhibit both wave-like and particle-like characteristics Electrons such that their position is considered to be - cloud - the probability of being at various locations - - around the nucleus. - - - - Nucleus  Almost all the mass is in the nucleus, and a +16 typical nucleus has size about 10−14 m. But - - - the atoms in a typical solid are spaced - - - about 10−10 m apart. -  Hence solids consist almost entirely of - empty space! The silicon atom  The attraction between electrons and nuclei pulls atoms together. Lecture 5 - Introduction to Materials 8 Atomic Bonding Interatomic bond  There are two forces involved when two atoms are in proximity.  The forces are the attraction between oppositely charged electrons and nuclei pulls atoms together (attractive), and the repulsion between the similarly charged electrons keeps them apart (repulsive).  The figure shows electron sharing in the ionic bond between a sodium atom and a chlorine atom in the sodium chloride. - Na+ - Cl− - - - - - - - - - - - - +11 - +17 - - - - - Attraction - - - - - - - - - Sodium atom Chlorine atom Lecture 5 - Introduction to Materials 9 Atomic Bonding  The relationship between atoms can be modelled approximately by assuming that the potential energy 𝑈𝑈 of the pair of atoms at distance 𝑟𝑟 consists of two power law components: 𝐴𝐴 𝐵𝐵 𝑈𝑈 = − + 𝑟𝑟 𝑚𝑚 𝑟𝑟 𝑛𝑛  Appropriate choice of constants 𝐴𝐴, 𝐵𝐵, 𝑚𝑚, and 𝑛𝑛 can give a good representation of bonds between various types of atom.  The question is what does the potential energy represent?  Consider that work is done (energy expended) - 𝐹𝐹, 𝛿𝛿𝛿𝛿 𝐹𝐹, 𝛿𝛿𝛿𝛿 - - - - - - - when a force moves through a distance in the - - - + - - - - - + - - - - - - - - - - direction of the force - - 𝑟𝑟 𝛿𝛿𝑈𝑈 = 𝐹𝐹𝐹𝐹𝐹𝐹  Thus, by considering infinitesimally small changes in position d𝑈𝑈