Compound Bars and Thermal Stresses (EGS 21107) 06/10/2024 PDF

Introduction to Mechanics of Materials - 06/10/2024 EGS 21107 Compound bars EGS 21107: INTRODUCTION TO MECHANICS OF MATERIALS 1 About the course Course code: EGS 21603 Course title: Intr...

Introduction to Mechanics of Materials - 06/10/2024 EGS 21107 Compound bars EGS 21107: INTRODUCTION TO MECHANICS OF MATERIALS 1 About the course Course code: EGS 21603 Course title: Introduction to Mechanics of Materials Credit hours: 2 Course webpage: https://lms.uofk.edu/course/view.php?id=6074 Instructor: Dr. Amged Osman Abdelatif 2 Dr. Amged O. Abdelatif 1 Introduction to Mechanics of Materials - 06/10/2024 EGS 21107 Compound bars Any tensile or compressive members which consists of two or more bars or tubes in parallel, usually of different materials is called a compound bars (or composite bars). The main idea of analysis is to apply both equilibrium and compatibility condition for solution. 3 Example 1 For the figure, the system is subjected to load P. Find the load taken by the rod and the tube. For the rod take 𝐸 = 𝐸1, 𝐴 = 𝐴1 and, for the tube take 𝐸 = 𝐸2, 𝐴 = 𝐴2. 4 Dr. Amged O. Abdelatif 2 Introduction to Mechanics of Materials - 06/10/2024 EGS 21107 Solution Equilibrium equation: 𝑊1 + 𝑊2 = 𝑃 Compatibility equation: 𝜀1 = 𝜀2 𝑊1 𝑊2 𝑊1.𝐴2 𝐸2 ∴ = >>>> 𝑊2 = 𝐴1 𝐸1 𝐴2 𝐸2 𝐴1 𝐸1 𝑊1.𝐴2 𝐸2 ∴ 𝑊1 + = 𝑃 >>>> 𝐴1 𝐸1 𝑃.𝐴1 𝐸1 𝑃.𝐴2 𝐸2 ∴ 𝑊1 = & ∴ 𝑊2 = 𝐴1 𝐸1 +𝐴2 𝐸2 𝐴1 𝐸1 +𝐴2 𝐸2 5 Example 2 A concrete column, 50 cm square, is reinforced with four steel rods, each 2.5 cm in diameter, embedded in the concrete near the corners of the square. If Young's modulus for steel is 200 GPa and that for concrete is 14 GPa, estimate the compressive stresses in the steel and concrete when the total thrust on the column is 1 MN. 6 Dr. Amged O. Abdelatif 3 Introduction to Mechanics of Materials - 06/10/2024 EGS 21107 Solution 𝜋 Area of steel: 𝐴𝑠 = 4 × 252 = 1960 𝑚𝑚2 4 Area of concrete: 𝐴𝑐 = 500 × 500 − 1960 = 248000 𝑚𝑚2 Equilibrium Equation: 𝑊𝑐 + 𝑊𝑠 = 106 𝑁 𝑊𝑐 𝑊𝑠 Compatibility Equation: 𝜀𝑐 = 𝜀𝑠 >>> = 𝐴𝑐 𝐸𝑐 𝐴𝑠 𝐸𝑠 106 106 This gives: 𝑊𝑐 = 𝐴 𝐸 >>>> 𝜎𝑐 = 1960×200×103 = 3.62 𝑀𝑃𝑎 1+ 𝑠 𝑠 248000+ 𝐴𝑐 𝐸𝑐 14×103 106 106 𝑊𝑠 = 𝐴 𝐸 >>>> 𝜎𝑠 = 248000×14×103 = 51.76 𝑀𝑃𝑎 1+ 𝑐 𝑐 1960+ 𝐴𝑠 𝐸𝑠 200×103 7 Quiz: Calculate strain on the rod. 8 Dr. Amged O. Abdelatif 4 Introduction to Mechanics of Materials - 06/10/2024 EGS 21107 Thermal stresses EGS 21107: INTRODUCTION TO MECHANICS OF MATERIALS 9 10 Dr. Amged O. Abdelatif 5 Introduction to Mechanics of Materials - 06/10/2024 EGS 21107 Thermal stresses Generally, if the temperature increases, the body will expand, whereas if the temperature decreases, it will contract. 𝛿𝑇 = 𝐿𝛼 ∆𝑇 Where: 𝛼 is the expansion coefficient (1/F, 1/C, 1/K) ∆𝑇 the algebraic change in temperature 𝐿 the original length of the member 𝛿𝑇 the algebraic change in the length 11 Thermal stresses ∵ 𝛿𝑇 = 𝐿𝛼 ∆𝑇 Then the thermal strain is: 𝛿𝑇 𝜀𝑇 = = 𝛼 ∆𝑇 𝐿 This gives stress as: 𝜎𝑇 = 𝜀𝑇 × 𝐸 12 Dr. Amged O. Abdelatif 6 Introduction to Mechanics of Materials - 06/10/2024 EGS 21107 Example 1 13 Solution 14 Dr. Amged O. Abdelatif 7 Introduction to Mechanics of Materials - 06/10/2024 EGS 21107 Example 2 A steel tube of 2.4 cm external diameter and 1.8 cm internal diameter encloses a copper rod of 1.5 cm diameter to which it is rigidly joined at each end. If at a temperature of 10o C there is no longitudinal Stress, calculate the Stresses in the rod and tube when the temperature is raised to 200o C. 𝐸𝑠 = 210,000 𝑀𝑃𝑎 steel 𝐸𝑐 = 100,000 𝑀𝑃𝑎 copper 𝛼𝑠 = 11 × 10−6 /𝑜 𝐶 𝛼𝑐 = 18 × 10−6 /𝑜 𝐶 15 Solution 16 Dr. Amged O. Abdelatif 8 Introduction to Mechanics of Materials - 06/10/2024 EGS 21107 Explain why bricks fall Brick wall Steel rod Window 17 Homework 4 Solve the following problems: Text Book: Strength of Materials – G. H. Ryder – 3rd Edition Page: 22 Problems: 8, 9 Text Book: Mechanics of Materials – R. C. Hibbeler– 8th Edition Page: 146 & 157 Problems: 4-42, 4-84, 18 18 Dr. Amged O. Abdelatif 9