STREMA BOOK

CHAIR UP!! Elevating Engineering Prowess Through In-Depth Scrutiny and Easier Approach An Instructional Material for CE-402 — STRENGTH OF MATERIALS Presented as a Final Project to The Faculty of College of Engineering Civil and Sanitary Engineering Department Alangilan, Batangas City Batangas State University The National Engineering University Gov. Pablo Borbon, Main Campus II Alangilan, Batangas City In Partial Fulfillment of the Requirements for the Degree of Bachelor of Science in Civil Engineering PREPARED AND SUBMITTED BY: GUALBERTO, IAN PAUL R. LEGASPI, CRISTINE JOY L. MALABANAN, RENZ MATHEW S. MAULION, ELOISA JANE B. ZAGALA, JAMES RYU C. SUBMITTED TO: ENGR. HILARION AQUINO III Instructor MAY, 2023



PREFACE CHAIR UP!! Elevating Engineering Prowess Through In-Depth Scrutiny and Easier Approach is an instructional material for Strength of Materials, made by students and for the students, or for anyone who would like to learn more about stresses, strains, deflection, etc.. It is particularly suitable as a textbook for Strength of Materials, the requisite course in the Engineering Mechanics series provided in the majority of university-level engineering programs. Engineering mechanics serves as both a foundation and a framework for the majority of engineering disciplines. Many issues in engineering, including civil, mechanical, aeronautical, and agricultural engineering, as well as engineering mechanics, are founded on statics and dynamics. Even in a field like electrical engineering, practice is essential. As a result, the sequence of engineering mechanics is crucial to the engineering curriculum. This series is not only required in and of itself, but engineering mechanics courses also assist to cement the student's mastery of other vital disciplines such as applied mathematics, physics, and graphics. Furthermore, these courses provide good opportunities to hone problem-solving skills. The major goal of studying engineering mechanics is to gain the ability to forecast the consequences of force, motion, stress, and deformations when performing engineering creative design duties. This capability necessitates more than a knowledge of physical and mathematical mechanics principles; it also necessitates the ability to visualize physical configurations in terms of real materials, actual constraints, and the practical constraints that govern the behavior of machines and structures. One of the key goals of a mechanics course is to assist students in developing the capacity to envision, which is so important in problem-solving. Indeed, building a meaningful mathematical model is frequently a more valuable experience than solving it. When the concepts and their limits are taught in the context of engineering application, maximum development is accomplished. This instructional material is the culmination of all of the authors’ knowledge and dedication for the sake of applying their expertise for the nation’s innovation. As we are future nation builders, we hope that this pursuit will be our stepping stone towards a sustainable and advanced future.

ACKNOWLEDGEMENT Our team expresses our deepest thanks and gratitude to the people who helped and shared their valuable assistance for the feasibility of this guidebook dedicated for learners. To Engr. Hilarion Aquino III, our instructor in Strength of Materials, for his guidance from the start, this guidebook would not be feasible without his guidance; To the chosen experts for accommodating the guidebook in this endeavor; To our understanding and supportive parents who gave endless support for doing the guidebook; Above all, to GOD ALMIGHTY for giving us the gift of everything. Without Him, this manual will not be possible. All of these we offer to You. Our wholehearted and sincerest thanks to all. I.P.R.G. C.J.J.L R.M.S.M E.J.B.M J.R.C.Z.

TABLE OF CONTENTS Topic Page Number OVERVIEW– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – 2 Chapter 1: SIMPLE STRESS– – – – – – – – – – – – – – – – – – – – – – – – – – – – 6 ➔ Exercises – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – 15 Chapter 2: DEFORMATION: HOOKE’S LAW, STATICALLY INDETERMINATE MEMBER– – – – – – – – – – – – – – – – – – – – –17 ➔ Exercises – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – 23 Chapter 3: THERMAL STRESSES, STATICALLY INDETERMINATE MEMBERS– – – – – – – – – – – – – – – – – – – – – – – – – – – 25 ➔ Exercises – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – 31 Chapter 4: TORSION – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – 33 ➔ Exercises – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – 38 Chapter 5: STRESSES ON THIN-WALLED CYLINDERS– – – – – – – – – – – – – – – – – – – – – – – – – – – –40 ➔ Exercises – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – 45 Chapter 6: SHEAR AND MOMENT IN BEAMS (INC. MOVING LOADS)– – – – – – – – – – – – – – – – – – – – – – – – – –47 ➔ Exercises – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – 79 Chapter 7: FLEXURAL STRESS – – – – – – – – – – – – – – – – – – – – – – – – – 81 ➔ Exercises – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – 85 Chapter 8: DEFLECTION IN BEAM – – – – – – – – – – – – – – – – – – – – – – – –87 ➔ Exercises – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – 102 GLOSSARY– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –103 ANSWER KEYS – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – 106 APPROVAL SHEET– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –122 REFERENCES – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – 123

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page | 2 strength of materials: OVERVIEW The mechanics of deformable bodies is concerned with the distribution of forces within bodies. The deformations generated by the internal forces cause stresses in the body, which can lead to failure of the materials. The basic principles in studying rigid body mechanics is a prerequisite for understanding courses like Strength of Materials or Mechanics of Materials. This serves as a starting point and, thus important in learning how to analyze internal stresses and its resulting deformation. After taking the Statics of Rigid Body, we must always remember the equilibrium of a deformable body. In stress analysis, the external loads are classified into two kinds: the surface forces and the body forces. Surface forces are forces that form when there is an interaction between two body’s surface that are in contact. The concepts included here are the area of contact, concentrated force, linear distributed force and object’s centroid. Body force, on the other hand, is the force that works on the volume of each mass and is unaffected by the material or environment around it. This force acts through the volume of the body. It is also important to note that linear distributed loadings result in a force with a magnitude equal to the area under the load diagram and a position that passes through the region's centroid. Figure 0.1. External loads in a Deformable Body We must also remember the concept of support reactions. This is used for two dimensional problems. Support produces a force in a particular direction on its attached member if it limits translation. Consequently, if rotation is prevented by the supports, then it produces a couple moment on that member. The different type of connection and its equal reaction on the member is seen in the figure below, Figure 2.

page | 3 Figure 0.2. Support Reactions More to this, the equations of equilibrium is significant in understanding complicated concepts in mechanics of materials. Equations of equilibrium must be satisfied in order to prevent a body from translating and rotating, and to achieve equilibrium state. To apply these equations, draw first a free-body diagram to understand and identify all the forces that are acting on the body. Figure 0. 3. Equation of Equilibrium The method of sections is used to calculate the internal resultant loadings acting on a sectioned body's surface. The internal resultant loadings present in a coplanar loadings consists of the normal force, shear force, torsional moment and bending moment. Normal Force (N) is the force that acts perpendicular to the area due to pushing and pulling of the external loads on the body. Shear Force (V) is the force that lies in the plane of the area. This is because of the external forces sliding over each other. Bending Moment (M) is the measure of the bending that occurs when a moment is applied. Lastly, Torsional Moment, also known as Torque (T), the measure of force that causes an object to twist or to rotate about an axis.

page | 4 Now that the key concepts are reviewed, we will now proceed onto leveling up. The principles we learned are applied continuously, thus, expanding the idea and its application in different situations. The focus of this book is to understand the concept of understanding the properties of materials when applied with loads and how the internal effect created in the body works. Strength of materials focuses on the internal effect of stress in a solid body that is subjected to an external loading. The change in shape or the deformation occurring when an external force acts on a body is of major interest. Strength of material is the resistance where the material of the body opposes this deformation. The study of the qualities of material things that enable them to resist the action of external forces, internal forces, and deformations caused by external forces is the principal objective of the subject strength of materials. The strength of materials depends on its mechanical properties like their elasticity, hardness and stiffness, and different materials combine these properties differently. This results in different abilities in the resistance of different forces. Engineers and other scientists use stress analysis information in the designing of structures like buildings and transportations. Stress analysis is significant in understanding the strength, weight and deformation of any materials. In theory, this course explains how to apply differential equations and solutions to investigate issues of various shapes and materials under various stress and strain situations generated by different situations.



page | 6 SIMPLE STRESSstrength of materials: chapter one — Introduction Stress is a measure of the magnitude of the forces that cause deformation. Stress, likewise referred to as unit strength, is expressed as a body's strength per unit area. The force is conveyed by the force on a member divided by its area, which can be expressed in psi or in N/mm2 or MPa. ������������������������������������ = ������������������������������������������ ������������������������������ ������������������������������������������������������ ������������������������ We can determine if there’s stress present when it meets the three criteria: the load, internal, and resisting area. When a body is affiliated with a load and is resisted by the area, which causes an internal effect on the body, then there’s stress present. Once it fails to satisfy the three criteria it can easily determine that the stress doesn't exist. Simple Stress According to Singer and Pytel in their book ‘Strength of Material’ Fourth Edition, stress can be said as simple stress if it is constant or uniform. The stress is uniformly distributed when the resultant of the applied forces passes through the centroid of the section or the body. Figure 1.1 Simple Stress

page | 7 Normal stress and Shear stress are the two classification of simple stress. The simple stress chart shown above demonstrates the two primary classifications of simple stress and the sub-classification of stress. A. NORMAL STRESS AND TENSILE STRESS Normal Stress In axially loaded elements, normal stress takes place when an applied force is perpendicular to the resisting area. Axial force, which can be tension or compression, is applied at the centroid of the member's cross-section. Normal stress includes tensile, compressive, and bearing stress. Members subjected to pure tension (or tensile force) are subjected to tensile stress, whereas members subjected to compressive force are subjected to compressive stress. Compressive force tends to shorten the member. Tension force, on the other hand, tends to lengthen the member. Bearing stress, on the other hand, is defined as the force exerted on a structure divided by its area. It is the tension that exists between the various bodies or the contact pressure between two separate bodies. Internal stress differs from compressive stress because it is caused by compressive forces. Tensile Stress Tensile stress happens whenever a load leads to the material stretching in the course of the load. It occurs when two antiparallel forces of identical magnitude act on an object along only one of its dimensions, immobilizing it. It is the resistance of an object to a force that has the ability to tear it apart. The shape of the cross-section has no impact on the strength of structures loaded under tension with similar cross-sectional areas. Tensile stress is also known as normal stress or tension. When the applied stress is much less than the object's tensile properties, the object completely or partially returns to its original form and size. As the tension attains tensile strength, the material is starting to stretch and fracture. The presented figure depicts one way of viewing a scenario involving tensile tension. When two people tug on a rope in opposite directions, the rope stretches as far as it can and at some point tears. The force acts on the

page | 8 axis when someone pulls on the rope. Tensile stress is an externally generated force that acts per unit area of the material and causes expansion, whereas tensile strength is the ability of a material to endure that stress until it breaks. Tensile stress is calculated as the greatest tension that an object can withstand without ripping, and it is usually expressed in N/mm2. Tensile stress is defined as the magnitude F of a force exerted along a rope divided by the cross-sectional area A of the rope in a perpendicular direction to the applied force. σ = ������ ������ Real-Life Application of Tensile Stress Tensile stress is synonymous with tensile forces or pulling. It is in charge of the elongation of the material along the axis of the applied load. Brittle materials tend to break before they reach their maximum strength, while ductile materials opt to bear the load. Tensile stress is applied to materials such as linking rods and staircase cables. The illustration portrays another instance of tensile stress in action. An arch bridge can be viewed in the image above, in which the bridge is anchored by an arch that gets linked by steel cables. For the purpose of preventing the bridge from collapsing, the arch supports the bridge's weight-bearing slab. The bridge's weight points downwards, while the steel cables connected to the bridge counteract this force, resulting in tensile stress. Compressive Stress Tensile stress is the exact opposite of compressive stress. A usual stress plane leads adjacent material sections to press toward each other. Compressive stress on bars, columns, and other structures causes shortening. Compressive stress is defined in the same way as tensile stress, but with negative numbers to show compression because delta L is the opposite. Compressive strength can be obtained by increasing the

page | 9 compressive stress. The materials will then be ductile or, in the case of brittle materials, will fracture. When brittle materials are subjected to compressive stress, the stored energy is abruptly released, resulting in the components shattering. When ductile materials are subjected to compressive stress, they compress but do not fail. The formula used in compressive stress is expressed as: σ = ������ ������ Where, F is the compressive force, A is the unit area and is the compressive stress. In a nutshell, when exposed to two equal and toward directions, compressive stress tends to shorten the material. The compressive stress acts normally on the area and pushes on it. Real Life Application of Compressive Stress Various kinds of matter respond differently to compression force. Brittle objects, for example, break when compressed, whereas elastic objects, when crushed, return to their original shape once the force is removed. The compression force is an inward force because it operates on the inside of the structure. Whenever mechanical or muscular pressure is applied to both ends of a spring, the spring's shape changes. When the pressure applied is inward, the spring opts for compression in size. In this scenario, a compression force is considered to be present. When an external force is exerted, the spring is stretched, resulting in a tension force.

page | 10 SAMPLE PROBLEM 1.1strength of materials: normal and tensile stresses — Given: ● F = 80 kN ● d = 6 cm ● h = 4 cm Required: ● Stress = ? Solution: σ = ������ ������ σ= 4������ π������2 σ= 4(−80 ������������) × ������������������2−������2 × (100 ������������)2 π(6 ������������)2 103 ������������ ������2 σ = 28. 29 ������������������

page | 11 B. BEARING STRESS AND SHEAR STRESS Bearing Stress When one solid body rests on another and transfers a load normal to it, the form of stress called “bearing stress” is developed at the surfaces in contact. Similar to direct compressive stress, the bearing stress, called σb, is a measure of the tendency for the applied force to crush the supporting member. Bearing stress is the contact pressure between the separate bodies. It differs from compressive stress, as it is an internal stress caused by compressive forces. Bearing Stress σ= ������������ ������ ������ ������ Bearing stress is the projected area of the contact surface. If plate thickness is “t” and rivet diameter is “d”, Then, ������ = ������������ ������ And so, bearing stress is σ = ������ , and it is computed in a manner similar to direct ������ ������ ������ ������ normal stresses where: Pb = compressive load Ab = area perpendicular to Pb Real Life Applications of Bearing Stress Some of the best applications and examples of bearing stress in real life are the soil pressure beneath a pier and the contact pressure between a rivet or bolt and the side of the hole it is in. If the bearing stress is large enough, it can locally crush the material, which can lead to more serious problems. To reduce bearing stresses, engineers sometimes employ bearing plates, the purpose of which is to distribute the contact forces over a larger area.

page | 12 Soil Bearing Capacity The bearing capacity of soil is perhaps the most important topic in soil engineering. Bearing capacity is the capacity of the soil to support loads applied to the ground. The bearing capacity of soil is the maximum average contact pressure between the foundation and the soil which should not produce shear failure in soil. In addition to providing a level platform for forms or masonry, footings spread out the weight of the house so the soil can carry the load. The load spreads out within the footing itself at about a 45-degree angle, and then spreads out in the soil at a steeper angle, more like 60-degrees from the horizontal. As the load under a footing spreads out, pressure on the soil diminishes. Soil directly under the footing takes the greatest load, and therefore should be thoroughly compacted. Soil Bearing Because the load spreads out, the pressure on the soil is greatest right beneath the footing. By the time we get down below the footing a distance equal to the footing’s width, the unit soil pressure has dropped by about half. Go down the same distance again, and the pressure has dropped by two-thirds. So, it's the soil right under the footing that is the most critical and also, typically, the most abused. Shear Stress Forces parallel to the area resisting the force cause shearing stress. It differs from tensile and compressive stresses, which are caused by forces perpendicular to the area on which they act. Shearing stress is also known as tangential stress. A shearing stress is an action that tends to slide on an adjacent section. τ= ������ ������ ������ where V is the resultant shearing force which passes through the centroid of the area Av being sheared.

page | 13 Single and Double Shear Single and Double Shear Since shearing stress is almost never evenly distributed across a segment, the shear stress formula must be regarded as yielding only the average shear stress. This has no bearing on the formula's effectiveness if we use an average shearing stress that accounts for the nonuniform distribution. Furthermore, when both the distance between the applied shearing pressures and the depth of the shearing are modest, the shearing stress distribution approaches homogeneity. Real Life Application of Shear Stress You'll find it amusing to learn that you're surrounded with shearing stress from the moment you wake up and get out of bed until you go back to sleep. It's because stress is present in practically every situation we encounter in our daily lives. So, in order to gain a better understanding, we'll look at some examples from real life. When you walk or run and your feet push the ground back to move forward, the force tends to deform a body by causing it to slip along a plane parallel to the applied stress. That's why it's shear stress.

page | 14 SAMPLE PROBLEM 1.2strength of materials: bearing and shear stresses — Given: ● d = 20 mm ● t = 25 mm ● shear strength = 350 MN/m2 Required: ● P=? Solution: V = τA P = 350[π(20)(25)] P = 549778.7 N P = 549.8 kN

page | 15 TEST YOUR KNOWLEDGEstrength of materials: exercises A 110-foot steel rod inside a lofty tower carries a 200-pound weight at its lower end. Determine the maximum normal stress in a circular rod with a diameter of 1/4 inch and weight of the rod. Answer: 4448.3 psi



page | 17 DEFORMATION:strength of materials: chapter two— HOOKE’S LAW, STATICALLY INDETERMINATE MEMBER Introduction Newton’s first law implies that an object oscillating back and forth is experiencing forces. Without force, the object would move in a straight line at a constant speed rather than oscillate. Consider, for example, plucking a plastic ruler to the left as shown in Figure 16.1.1. The deformation of the ruler creates a force in the opposite direction, known as a restoring force. Once released, the restoring force causes the ruler to move back toward its stable equilibrium position, where the net force on it is zero. However, by the time the ruler gets there, it gains momentum and continues to move to the right, producing the opposite deformation. It is then forced to the left, back through equilibrium, and the process is repeated until dissipative forces dampen the motion. These forces remove mechanical energy from the system, gradually reducing the motion until the ruler comes to rest. Figure 2.1: When displaced from its vertical equilibrium position, this plastic ruler oscillates back and forth because of the restoring force opposing displacement. When the ruler is on the left, there is a force to the right, and vice versa.

page | 18 The simplest oscillations occur when the restoring force is directly proportional to displacement. When stress and strain were covered in Newton’s Third Law of Motion, the name was given to this relationship between force and displacement was Hooke’s law: ������ = − ������������ Figure 2.2: (a) The plastic ruler has been released, and the restoring force is returning the ruler to its equilibrium position. (b) The net force is zero at the equilibrium position, but the ruler has momentum and continues to move to the right. (c) The restoring force is in the opposite direction. It stops the ruler and moves it back toward equilibrium again. (d) Now the ruler has momentum to the left. (e) In the absence of damping (caused by frictional forces), the ruler reaches its original position. From there, the motion will repeat itself. The force constant k is related to the rigidity (or stiffness) of a system—the larger the force constant, the greater the restoring force, and the stiffer the system. The units of k are newtons per meter (N/m). For example, k is directly related to Young’s modulus when we stretch a string. Figure 16.1.3 shows a graph of the absolute value of the restoring force versus the displacement for a system that can be described by Hooke’s law—a simple spring in this case. The slope of the graph equals the force constant k in newtons per meter. A common physics laboratory exercise is to measure restoring forces created by springs, determine if they follow Hooke’s law, and calculate their force constants if they do.

page | 19 Figure 2.3: (a) A graph of absolute value of the restoring force versus displacement is displayed. The fact that the graph is a straight line means that the system obeys Hooke’s law. The slope of the graph is the force constant k. (b) The data in the graph were generated by measuring the displacement of a spring from equilibrium while supporting various weights. The restoring force equals the weight supported, if the mass is stationary. SAMPLE PROBLEM 2.1strength of materials: deformation: hooke’s law, statically indeterminate member — Strategy: Consider the car to be in its equilibrium position x = 0 before the person gets in. The car then settles down 1.20 cm, which means it is displaced to a position x = −1.20×10−2 m. At that point, the springs supply a restoring force F equal to the person’s weight w=mg=(80.0kg)(9.80m/s2) = 784N. We take this force to be F in Hooke's law. Knowing F and x we can then solve the force constant k. Solution: 1. Solve Hooke’s law, F = −kx, for k: Figure 2.4: The mass of a car ������ = − ������ increases due to the ������ introduction of a passenger. This affects the displacement 2. Substitute known values and solve k: of the car on its suspension system. (credit: exfordy on ������ = − 784 ������ Flickr) −1.20 ������ 10−2 ������ ������ = 6. 53 ������ 104 ������/������

page | 20 Discussion: Note that F and x have opposite signs because they are in opposite directions—the restoring force is up, and the displacement is down. Also, note that the car would oscillate up and down when the person got in if it were not for damping (due to frictional forces) provided by shock absorbers. Bouncing cars are a sure sign of bad shock absorbers. Energy in Hooke’s Law of Deformation In order to produce a deformation, work must be done. That is, a force must be exerted through a distance, whether you pluck a guitar string or compress a car spring. If the only result is deformation, and no work goes into thermal, sound, or kinetic energy, then all the work is initially stored in the deformed object as some form of potential energy. The potential energy stored in a spring is ������������ = 1 ������������2. Here, we generalize the idea to elastic 2 ������������ potential energy for a deformation of any system that can be described by Hooke’s law. Hence, ������������ = 1 ������������2 2 ������������ where ������������ is the elastic potential energy stored in any deformed system that obeys ������������ Hooke’s law and has a displacement x from equilibrium and a force constant k. It is possible to find the work done in deforming a system in order to find the energy stored. This work is performed by an applied force ������ . The applied force is exactly ������������������ opposite to the restoring force (action-reaction), and so ������ = ������������. Figure 16.1.5 shows a ������������������ graph of the applied force versus deformation x for a system that can be described by Hooke’s law. Work done on the system is force multiplied by distance, which equals the area under the curve or 1 ������������2(Method A in the figure). Another way to determine the work is to 2 note that the force increases linearly from 0 to kx, so that the average force is 1 ������������, the 2 distance moved is x, and thus ������ = ������ ������ = [( 1 )������������(������) = 1 ������������2(Method B in the 2 2 ������������������ figure).

page | 21 Figure 2.5: A graph of applied force versus distance for the deformation of a system that can be described by Hooke’s law is displayed. The work done on the system equals the area under the graph or the area of the triangle, which is half its base multiplied by its height, or ������ = 1 ������������2. 2 SAMPLE PROBLEM 2.2strength of materials: deformation: hooke’s law, statically indeterminate member —

page | 22 Strategy for a: (a): The energy stored in the spring can be found directly from the elastic potential energy equation, because k and x are given. Entering the given values for k and x yields ������������ = 1 ������������2 = 1 (50. 0 ������/������)(0. 150 ������)2 2 2 ������������ = 0. 563 ������. ������ = 0.563 J Strategy for b: Figure 2.6: (a) In this Because there is no friction, the potential energy is converted entirely into kinetic energy. The expression for kinetic energy can be image of the gun, the solved for the projectile’s speed. spring is uncompressed before being cocked. (b) The spring has been Solution for b compressed a distance x, and the projectile is 1. Identify known quantities: in place. (c) When released, the spring ������������ = ������������ ������������ 1 ������������2 = 1 ������������2 = ������������ = 0. 563 ������ 2 2 converts elastic ������ ������������ ������������ potential energy ������������ 2. Solve for v: ������������ into kinetic energy. ������ = [ 2������������ = [ 2(0.563 ������) ] = 23. 7 (������/������)1/2 0.002 ������������ ]������������ ������ 3. Convert units: 23.7 m/s Discussion: (a) and (b): This projectile speed is impressive for a tranquilizer gun (more than 80km/h). The numbers in this problem seem reasonable. The force needed to compress the spring is small enough for an adult to manage, and the energy imparted to the dart is small enough to limit the damage it might do. Yet, the speed of the dart is great enough for it to travel an acceptable distance.

page | 23 Summary ● An oscillation is a back and forth motion of an object between two points of deformation. ● An oscillation may create a wave, which is a disturbance that propagates from where it was created. ● The simplest type of oscillations and waves are related to systems that can be described by Hooke’s law: ������ =− ������������ where F is the x restoring force, ������������ is the ������������ displacement from equilibrium or deformation, and ������������ = 1 ������������2. 2 ������������ TEST YOUR KNOWLEDGEstrength of materials: exercises A steel bar 50 mm in diameter and 2 m long is surrounded by a shell of a cast iron 5 mm thick. Compute the load that will compress the combined bar a total of 0.8 mm in the length of 2 m. For steel, E = 200 GPa, and for cast iron, E = 100 GPa. Answer: P=191.64 kN



page | 25 sTtrHenEgtRh oMf mAaLterSiaTlsR: cEhaSpStEerSt,hSrTeeA—TICALLY INDETERMINATE MEMBERS Introduction A body's proportions will change accordingly if its temperature is reduced or increased. The stresses that result from these changes in the body, however, are referred to as temperature stresses (Thermal Stresses) and their associated strains as temperature strains (Thermal Strains). There won't be any load or tension applied to the structure if thermal deformation is allowed to happen naturally. An internal tension is produced in some circumstances where temperature deformation is not allowed. Thermal stress is the name given to the internal stress produced. The term \"thermal stress\" and \"thermal strain\" refer to the tension and strain that a change in temperature causes in a body. When a body's temperature is elevated or dropped and it is not permitted to expand or contract naturally, thermal stress is caused in the body. When the body is free to grow or contract, no stress is put on it. The thermal stress is calculated as follows for a homogeneous rod positioned between unyielding supports as shown: Figure 3.1: homogeneous rod

page | 26 The sides of the object will alter if it is exposed to consistent temperature variations. If “α” is the coefficient of thermal expansion or contraction, the uniform thermal strain is given by: εt = α . (t2 – t1) or εt = α . Δt where, Δt = (t2 – t1) = Change in temperature. Take a bar of length L that is free to grow while being exposed to a uniform increase in temperature (t). The length of the bar then increased. Then the increase in the length of the bar is, εt . L = α . Δt . L Considered a body (for example a rod) which is heated to a specified temperature? Let: L = Original length of the rod, ● T = Rise in temperature, ● E = Young’s modulus, ● α = Coefficient of linear expansion and ● dL = Extension produced in the rod ● Due to the increase in the temperature, there is an extension produced in the rod. When the rod is allowed to expand freely, the extension produced in the rod is given by: Extension produced = α������������ ������������ = α������������ Figure 3.2: extension produced in the rod due to temperature increase

page | 27 This is depicted in the image above, where AB represents the rod's initial length and BB' represents the extension that the rod produced. Consider applying a compressive load P at the end BB'. As a result of this compressive load, the length of the rod decreases from (L + dL) to L. Compressive stress and strain is produced in the rod and is given by: ������������������������������������������������������������������ ������������������������������������ = ������������������������������������������������ ������������ ������������������������������ℎ = α������������ = α������������ = α������ ������������������������������������������������ ������������������������������ℎ ������ + α������������ ������ ������������������������������������ = ������������������������������������ (������) = α������������ Thrust or load on the rod is given by: ������������������������ ������������ ������ℎ������������������������ = ������������������������������������ (������) = α������������ (������) Thermal Stress and Strain Formulas When the rod is heated to a specific temperature and its ends are tightly fixed to the supports, the rod's expansion is restricted because both ends have been rigorously fixed. The rod has been put under stress and pressure. Thermal stresses and strains are the setting of stresses and strains within the rod. Thermal strain is given by: ������������������������������������ = ������������������������������������������������������ ������������������������������������������������������ = α������������ = α������ ������������������������������������������������ ������������������������������ℎ ������ Therefore: ������ = α������ Where: ● e = Thermal strain ● σ = Thermal stress Deformation due to temperature changes; δ = α������������ ������ Deformation due to equivalent axial stress:

page | 28 δ= ������������ = σ������ ������������ ������ ������ δ =δ ������ ������ α������������ = σ������ ������ Therefore: σ = α������������ Stress and Strain When the Supports Yield If the supports at the ends yield by an amount equal to the δ, then: ������������������������������������ ������������������������������������������������������ = ������������������������������������������������������ ������������������ ������������ ������������������������ ������������ ������������������������ − δ = α������������ − δ ������������������������������������ ������������������������������������ = ������������������������������������ ������������������������������������������������������ = α������������ − δ ������������������������������������������������ ������������������������������ℎ ������ ������������������������������������ ������������������������������������ = α������������ − δ (������) ������ or α������������ = ������ + σ������ ������ If the equilibrium equations are sufficient to calculate all the forces (including support reactions) that act on a body, these forces are said to be statically determinate. In statically determinate problems, the number of unknown forces is always equal to the number of independent equilibrium equations. If the number of unknown forces exceeds the number of independent equilibrium equations, the problem is said to be statically indeterminate. Static indeterminacy does not imply that the problem cannot be solved; it simply means that the solution cannot be obtained from the equilibrium equations alone. A statically indeterminate problem always has geometric restrictions imposed on its deformation. The mathematical expressions of these restrictions, known as the compatibility equations, provide us with the additional equations needed to solve the problem (the term

page | 29 compatibility refers to the geometric compatibility between deformation and imposed constraints). Because the source of the compatibility equations is deformation, these equations contain as unknowns either strains or elongations. We can, however, use Hooke’s law to express the deformation measures in terms of stresses or forces. The equations of equilibrium and compatibility can then be solved for the unknown forces. Procedure for Solving Statically Indeterminate Members 1. Draw the required free-body diagrams and derive the equations of equilibrium.Derive the compatibility equations. To visualize the restrictions on deformation, it is often helpful to draw a sketch that exaggerates the magnitudes of the deformations. 2. Use Hooke’s law to express the deformations (strains) in the compatibility equations in terms of forces (or stresses). 3. Solve the equilibrium and compatibility equations for the unknown forces. SAMPLE PROBLEM 3.1strength of materials: thermal stresses, statically indeterminate members — Given: ● cross-sectional area of 0.25������������2 ● tensile load at 70°F is 1200 lb ● α = (6. 5 × 10−6) in/(in·°F) ● E = 29 × 106 psi Required: ● What will be the stress at 0°F? What temperature will the stress be zero?

page | 30 Solution: 1. For the stress at 0°F: δ=δ +δ ������ ������������ σ������ = α������(������) + ������������ ������ ������������ σ = α������������ + ������ ������ σ = (6. 5 × 10−6)(29 × 106)(70) + 1200 0.25 σ = 17, 995 ������������������ 2. For the temperature that causes zero stress: δ =δ ������ ������������ α������(������) = ������������ ������������ α(������) = ������ ������������ (6. 5 × 10−6)(T - 70) = 1200 0.25(29 × 106) ������ = 95. 46 °������

page | 31 TEST YOUR KNOWLEDGEstrength of materials: exercises A bronze bar 3 m long with a cross sectional area of 320 mm2 is placed between two rigid walls as shown in figure below. At a temperature of -20°C, the gap Δ = 2.5 mm. Find the temperature at which the compressive stress in the bar will be 35 MPa. Use α = 18.0 × 10-6m/(m·°C) and E = 80 GPa. Answer: T = 50.6 °C



page | 33 TORSIONstrength of materials: chapter four— Introduction In order to transmit energy by rotation in workshops and factories, a turning force is always applied. This rotating force is applied to a pulley's rim, a keyed shaft, or any other suitable point that is spaced apart from the shaft's axis. Torque, also known as a turning moment or a twisting force, is the result of the turning force and the distance between the force's application point and the shaft's axis. Torsion is the twisting of the shaft as a result of this torsion. As shown in the picture, imagine a bar that is rigidly attached at one end and is twisted at the other end by a torque or twisting moment T equivalent to F d that is applied perpendicular to the axis of the bar. It is said that such a bar is in torsion. A structure is twisted by a torque. A torque generates a dispersion of stress over the cross section of the item as opposed to axial loads, which produce a uniform, or average, stress over the object's cross section. We'll keep things straightforward by concentrating on objects with a circular cross section, sometimes known as rods or shafts. The structure will twist along the long axis of the rod when a torque is applied, but its cross section will stay circular.

page | 34 Shafts are used for transmitting power and in that process, they are subjected to the following torques: ● A driving torque at the input end due to power transmitted. ● A resisting torque at the output end, exerted by the driven machinery. When a structural member is loaded by a coupling that causes rotation about its longitudinal axis, it twists, which is known as torsion. This applies the moment or couple in a vertical plane perpendicular to a member's longitudinal axis.Torques, or twisting couples of twisting moments, are the couples that cause a structural part to twist. A shaft is said to be in pure torsion when it is subjected to two equal and opposite pairs of twisting forces in a plane perpendicular to the shaft's longitudinal axis, or when the amplitude of the twisting moment is maintained throughout the shaft's length. Not every power transmission shaft is subjected to pure torsion, although some shaft components are. From this equation, we may deduce a few things right away. The first point may be self-evident: the greater the shear strain, which is again symbolized by the Greek letter gamma, the greater the angle of twist. Second, and this is a key distinction between structures that are stressed axially and those that are loaded with torque, the shear strain is not constant throughout the cross section. At the middle of the twisted rod, it is zero, and at the edge of the rod, it reaches a maximum value. Finally, the shear strain is smaller the longer the rod is. To understand what I'm referring to, picture the rod's cross section as a clock with simply an hour hand. The hour hand is at 12 o'clock when no torque is applied. The hour hand will rotate clockwise to a new position (let's say 2 o'clock) as the rod twists under a torque. The Greek letter phi is frequently used to represent the angle of twist, which is the angle between 2 and 12 o'clock. We may calculate the shear strain at any point along the cross section using this angle. Torsional Shearing Stress (τ) For a solid or hollow circular shaft subject to a twisting moment T, the torsional shearing stress τ at a distance ρ from the center of the shaft is τ= ������ρ and τ = ������������ ������ ������ ������������������ J stands for the second polar moment of the cross-sectional area in this equation. The phrase \"second moment of inertia\" is occasionally used to describe this, but let's avoid any

page | 35 confusion since that phrase already has a defined meaning with reference to the dynamic motion of objects. For solid cylindrical shaft: For hollow cylindrical shaft: ������ = π ������4 ������ = π (������4 − ������4) 32 32 τ= 16������ τ= 16������������ π������3 π(������4−������4) ������������������ ������������������ Angle of Twist The angle θ through which the bar length L will twist is: Θ= ������������ ������������ ������������������������������������������ ������������ where T is the torque in N·mm, L is the length of shaft in mm, G is shear modulus in MPa, J is the polar moment of inertia in mm4, D and d are diameter in mm, and r is the radius in mm. Power Transmitted by Shaft A shaft rotating with a constant angular velocity ω (in radians per second) is being acted by a twisting moment T. The power transmitted by the shaft is: ������ = ������ω = 2π������������ where T is the torque in N·m, f is the number of revolutions per second, and P is the power in watts.

page | 36 SAMPLE PROBLEM 4.1strength of materials: torsion — Given: ● solid aluminum shaft 2 in. in diameter ● G = 4 × 106 psi Required: ● maximum shearing stress in each segment and the angle of rotation of the free end Solution: τ= 16������ π������3 ������������������

page | 37 ★ For the 2-ft segment: τ= 16������ π������3 ������������������2 τ= 16(600)(12) π(2)3 ������������������2 τ������������������2 = 4583. 66 ������������������ ★ For the 3-ft segment: τ= 16������ π������3 ������������������3 τ= 16(800)(12) π(2)3 ������������������3 τ = 6111. 55 ������������������ ������������������3 ★ Angle of twist Θ= ������������ ������������ Θ= 1 Σ������������ ������������ Θ= 1 (600(2) + 800(3))(122) 1 π(24)( 4 × 106) 31 Θ = 0. 0825 ������������������ Θ = 4. 73°

page | 38 TEST YOUR KNOWLEDGEstrength of materials: exercises A steel propeller shaft is to transmit 4.5 MW at 3 Hz without exceeding a shearing stress of 50 MPa or twisting through more than 1° in a length of 26 diameters. Compute the proper diameter if G = 83 GPa. Answer: d = 352 mm



page | 40 strength of materials: chapter five— STRESSES ON THIN-WALLED CYLINDERS Introduction Few developments in engineering have had the same influence as the thin-walled pressure vessel. This extraordinary container has been a crucial component of industrial processes for decades due to its lightweight design and impressive strength. The thin-walled pressure vessel is a versatile and dependable solution for a variety of industrial applications, from storing compressed gasses to transferring high-pressure liquids. A pressure vessel is a pressurized container that is usually cylindrical or spherical in shape. (Please see the images below.) Figure 5.1. Pressure Vessels (a) Cylindrical Tank (b) Spherical Tank Bending stresses have little effect on the membrane-like shell of an ideal thin-walled pressure vessel over most of its surface. A sphere is regarded as the most structurally efficient shape for closed pressure vessels because internal pressure acts uniformly in all directions, resulting in uniform wall stresses. A cylindrical vessel, on the other hand, is slightly less efficient due to the fluctuation of wall stresses with direction, and end cap closing can dramatically affect the optimal

page | 41 membrane condition, demanding additional local reinforcements. However, the cylindrical shape may be easier to make and transport. Assumptions The key assumptions used here are wall thinness and geometric symmetries. These make obtaining average wall stresses analysis possible with simple free-body diagrams (FBD). Here is a more detailed list of assumptions: 1. Wall Thinness. The wall is assumed to be very thin compared to the other dimensions of the vessel. It should not exceed 1/10th of the diameter to ensure that the stresses are uniform across the wall. 2. Symmetries. In both vessels, we assumed that the stress (or force) are all the same in all directions. 3. Uniform Internal Pressure. When a pressure vessel has uniform and positive pressure inside, it is denoted by p. However, if the vessel is externally pressurized, like with atmospheric pressure, the pressure is called gage pressure and is found by subtracting the external pressure from the internal pressure. If the external pressure is higher than the internal one, as in submarine hulls, caution must be taken in applying stress formulas because wall buckling can cause instability and failure. (p=Δp=pi−po) 4. Ignoring End Effects. There are features that may affect symmetry assumptions such as supports and cylinder end caps, yet this may be ignored in basic design decisions such as picking up the thickness away from such regions. Cylindrical Vessels The cylinder has a thickness, t, and a diameter, d. This analysis is limited to \"Thin-Walled Pressure Vessels\". (Reminder: For a cylinder to qualify as \"thin walled\" the ratio of diameter to thickness (d/t) must be at least 10.)

page | 42 We can see that the forces induced by internal pressure (P) and circumferential stress (c) in the wall must be in balance if we cut this cylinder along its diameter. Circumferential stress, also known as tangential, girth, or hoop stress, acts perpendicular to the axis of a cylinder. The cylinder has a length L, an internal diameter d, and a thickness t. p represents the applied internal pressure. Consider a half-cylinder with a length of L that is sectioned along a diametric plane to study the equilibrium. The force owing to c acts on a 2tL area, and the resulting vertical pressure force is calculated using the projected horizontal area of DL. This gives, Equating forces: 2σ = ������ , σ������ = ������DL /2������L ������ Simplifying the equation to get the Circumferential Stress, σ������ = ������D / 2������ Furthermore, \"longitudinal stress\" refers to stress that acts parallel to a cylinder's long axis. ������ = ������ (π������)2 4 It is critical to remember that stress and pressure are measured in terms of force per unit area. To include stress and pressure in force equilibrium equations, multiply the stress or pressure by the area it acts on. The figures on the right depict the areas influenced by longitudinal stress and pressure. Once we've calculated the areas, we can use them to calculate the pressure and normal stress forces, as shown below. Equating forces: F = P , ������ (π������)2 = π������������σ 4 ������ As we simplify, the equation we get the longitudinal stress, σ= ������������ 4������ ������

page | 43 Spherical Vessels The theoretical ideal shape for a vessel that can withstand internal pressure is a sphere. As we isolate half of the sphere and its contents as a single free body, the stress in the vessel wall and the fluid pressure p operate on this free body. The free-body diagram gives the equilibrium condition σπ������������ = ������ π������2 , thus σ = ������������ 4 4������ Figure 5. (a) Spherical pressure vessel; (b) Cylindrical pressure vessel. Any portion that goes through the sphere's center produces the same outcome. The double curvature of the spherical vessel means that all stress directions surrounding the pressure point contribute to pressure resistance.