Mechanics of Materials Spring 2024 Dr. Ibrahim Nabil Eldesouky Faculty of Engineering German International University (GIU) 7 February 2024 Dr. Ibrahim Eldesouky 1 Objective • The objective of the course is to introduce the Mechanical Engineering students to the analysis and design of various load bearing structures. • Students will learn how to analyze the stresses and deformations of various structural components subjected to axial loading, torsion and bending as well as its application to the design of simple structures. 7 February 2024 Dr. Ibrahim Eldesouky 2 Course content 1. Introduction to the concept of stress. 2. Axial Loading. 3. Torsion. 4. Pure Bending. 5. Analysis and Design of Beams for Bending. 6. Shearing Stresses in Beams. 7. Transformations of Stress and Strain 8. Principle Stresses Under a Given Loading Different types of loading 7 February 2024 Dr. Ibrahim Eldesouky 3 Reference 7 February 2024 Dr. Ibrahim Eldesouky 4 Course information Lecturer & TA availability Grading • Dr. Ibrahim Eldesouky • Assignments: 10% • Project: 20% • Midterm Exam: 30% • Final Exam: 40% • Office: S1.310 • Office hours: Sunday 3rd slot, or by appointment • Email: ibrahim.eldesouky@giu-uni.de • Eng. Salma Emam Contact hours • Office: S1.211 • Credit points: 4 ECTS • Lectures: • Email: salma.emam@giu-uni.de • Eng. Yomna Nouh • Email: yomna.nouh@giu-uni.de 7 February 2024 • Wednesday: 4th slot (weekly) • Tutorials: once per week (check your schedule) Dr. Ibrahim Eldesouky 5 Chapter 1 – Concept of Stress 7 February 2024 Dr. Ibrahim Eldesouky 6 Review of Statics • The structure is designed to support a 30 kN load • The structure consists of a bar and rod joined by pins at the junctions and supports • Perform a static analysis to determine the force in each structural member and the reaction forces at the supports 7 February 2024 Dr. Ibrahim Eldesouky 7 7 February 2024 Dr. Ibrahim Eldesouky 8 Reactions 7 February 2024 Dr. Ibrahim Eldesouky 9 Structure Free-Body Diagram • Structure is detached from supports and the loads and reaction forces are indicated • Conditions for static equilibrium: π = 0 = π΄ 0.6m − 30kN 0.8m π¨π = ππkN πΉ = 0 =π΄ +πΆ πͺπ = −π¨π = −ππkN πΉ = 0 = π΄ + πΆ − 30kN = 0 π¨π + πͺπ = ππkN • Ay and Cy can not be determined from these equations 7 February 2024 Dr. Ibrahim Eldesouky 10 Component Free-Body Diagram • In addition to the complete structure, each component must satisfy the conditions for static equilibrium • Consider a free-body diagram for the bar: ο₯ M B ο½ 0 ο½ ο Ay ο¨0.8 m ο© Ay ο½ 0 substitute into the structure equilibrium equation C y ο½ 30 kN • Results: π¨πΏ = ππkN → πͺπ = ππkN ← πͺπ = ππkN ↑ Reaction forces are directed along bar and rod 7 February 2024 Dr. Ibrahim Eldesouky 11 Method of Joints • The bar and rod are 2-force members, i.e., the members are subjected to only two forces which are applied at member ends • For equilibrium, the forces must be parallel to an axis between the force application points, equal in magnitude, and in opposite directions • Joints must satisfy the conditions for static equilibrium which may be expressed in the form of a force triangle: π¨π© 7 February 2024 Dr. Ibrahim Eldesouky π©πͺ 12 Stress Analysis Can the structure safely support the 30 kN load? • From a statics analysis FAB = 40 kN (compression) FBC = 50 kN (tension) • At any section through member BC, the internal force is 50 kN with a force intensity or stress of dBC = 20 mm P 50 ο΄103 N ο³ BC ο½ ο½ ο½ 159 MPa A 314 ο΄10-6 m 2 • From the material properties for steel, the allowable stress is ο³ all ο½ 165 MPa • Conclusion: the strength of member BC is adequate 7 February 2024 Dr. Ibrahim Eldesouky 13 Design • Design of new structures requires selection of appropriate materials and component dimensions to meet performance requirements • For reasons based on cost, weight, availability, etc., the choice is made to construct the rod from aluminum ο¨ο³all= 100 MPa). What is an appropriate choice for the rod diameter? P ο³ all ο½ A Aο½ d2 Aο½ο° 4 dο½ 4A ο° ο½ ο¨ P ο³ all ο½ 50 ο΄103 N 100 ο΄106 Pa 4 500 ο΄10ο 6 m 2 ο° ο½ 500 ο΄10ο 6 m 2 ο© ο½ 2.52 ο΄10ο2 m ο½ 25.2 mm • An aluminum rod 26 mm or more in diameter is adequate 7 February 2024 Dr. Ibrahim Eldesouky 14 Centric & Eccentric Loading • A uniform distribution of stress in a section infers that the line of action for the resultant of the internal forces passes through the centroid of the section. • A uniform distribution of stress is only possible if the concentrated loads on the end sections of two-force members are applied at the section centroids. This is referred to as centric loading. • If a two-force member is eccentrically loaded, then the resultant of the stress distribution in a section must yield an axial force and a moment. • The stress distributions in eccentrically loaded members cannot be uniform or symmetric. 7 February 2024 Dr. Ibrahim Eldesouky 15 Axial Loading: Normal Stress • The resultant of the internal forces for an axially loaded member is normal to a section cut perpendicular to the member axis. • The force intensity on that section is defined as the normal stress. π = lim → ΔπΉ Δπ΄ ππππ = π· π¨ • The normal stress at a particular point may not be equal to the average stress but the resultant of the stress distribution must satisfy P ο½ ο³ ave A ο½ ο² dF ο½ ο² ο³ dA A • The detailed distribution of stress is statically indeterminate, i.e., can not be found from statics alone. 7 February 2024 Dr. Ibrahim Eldesouky 16 Shearing Stress • Forces P and P’ are applied transversely to the member AB. • Corresponding internal forces act in the plane of section C and are called shearing forces. • The resultant of the internal shear force distribution is defined as the shear of the section and is equal to the load P. • The corresponding average shear stress is, πave = π· π¨ • Shear stress distribution varies from zero at the member surfaces to maximum values that may be much larger than the average value. • The shear stress distribution cannot be assumed to be uniform. 7 February 2024 Dr. Ibrahim Eldesouky 17 Shearing Stress Examples Single Shear ο΄ ave ο½ 7 February 2024 Double Shear P F ο½ A A ο΄ ave ο½ Dr. Ibrahim Eldesouky P F ο½ A 2A 18 Bearing Stress in Connections • Bolts, rivets, and pins create stresses on the points of contact or bearing surfaces of the members they connect. • The resultant of the force distribution on the surface is equal and opposite to the force exerted on the pin. • Corresponding average force intensity is called the bearing stress, π 7 February 2024 Dr. Ibrahim Eldesouky 19 Stress Analysis & Design Example • Determine the stresses in the members and connections of the structure shown. • From a statics analysis: FAB = 40 kN (compression) FBC = 50 kN (tension) • Must consider the maximum normal stresses in AB and BC, and the shearing stress and bearing stress at each pinned connection 7 February 2024 Dr. Ibrahim Eldesouky 20 Rod & Bar Normal Stresses • The rod is in tension with an axial force of 50 kN. • At the rod center, the average normal stress in the circular cross-section (A = 314 x 10-6 m2) is ο³BC = +159 MPa. • At the flattened rod ends, the smallest crosssectional area occurs at the pin centerline, A ο½ ο¨20 mm ο©ο¨40 mm ο 25 mm ο© ο½ 300 ο΄10ο 6 m 2 P 50 ο΄103 N ο³ BC ,end ο½ ο½ ο½ 167 MPa A 300 ο΄10ο 6 m 2 • The bar is in compression with an axial force of 40 kN and average normal stress of –26.7 MPa. • The minimum area sections at the bar ends are unstressed since the bar is in compression. 7 February 2024 Dr. Ibrahim Eldesouky 21 Pin Shearing Stresses • The cross-sectional area for pins at A, B, and C, 2 ο¦ 25 mm οΆ ο6 2 A ο½ ο°r ο½ ο°ο§ ο· ο½ 491ο΄ 10 m ο¨ 2 οΈ 2 • The force on the pin at C is equal to the force exerted by the rod BC, P 50 ο΄103 N ο΄ C , ave ο½ ο½ ο½ 102 MPa ο 6 2 A 491ο΄10 m • The pin at A is in double shear with a total force equal to the force exerted by the boom AB, ο΄ A, ave ο½ 7 February 2024 Dr. Ibrahim Eldesouky P 20 kN ο½ ο½ 40.7 MPa ο 6 2 A 491ο΄10 m 22 Pin Bearing Stresses • To determine the bearing stress at A in the bar AB, we have t = 30 mm and d = 25 mm, ο³b ο½ P 40 kN ο½ ο½ 53.3 MPa td ο¨30 mm ο©ο¨25 mm ο© • To determine the bearing stress at A in the bracket, we have t = 2(25 mm) = 50 mm and d = 25 mm, ο³b ο½ 7 February 2024 P 40 kN ο½ ο½ 32.0 MPa td ο¨50 mm ο©ο¨25 mm ο© Dr. Ibrahim Eldesouky 23 Pin Shearing Stresses • Divide the pin at B into sections to determine the section with the largest shear force, PE ο½ 15 kN 50 kN PG ο½ 25 kN (largest) • Evaluate the corresponding average shearing stress, ο΄ B, ave ο½ 7 February 2024 PG 25 kN ο½ ο½ 50.9 MPa A 491ο΄10ο 6 m 2 Dr. Ibrahim Eldesouky 24 Units 7 February 2024 Dr. Ibrahim Eldesouky 25 Any questions? 7 February 2024 Dr. Ibrahim Eldesouky 26
