1 ROIHAN MASFIZAIZAN MOHD AZAM
Acknowledgement i Alhamdulillah, first and foremost, praises and thanks to the God, the Almighty, for His showers of blessing throughout my writing to complete the e-book successfully. We cannot express enough thanks to the Director, Deputy of Director and Head of Mechanical Engineering Department Polytechnic Banting Selangor for their continued support and encouragement. We offer our sincere appreciation for the learning opportunities provided by them. Our completion of this e-book could not have accomplished without the support our colleagues. The ideas, comments and advices are very useful and its have been made our e-book content is better. All of that support in times to complete this e-book will not be forgotten. Finally, we would also like to thank our parents, family members and friends who helped a lot in finalizing this e-book within the limited time frame. Thank you.
ii This e-book written is contain main topic and subtopic which relates to Chapter 1 of syllabus DJJ30103 Strength of Material for polytechnics students. The content discussed is related to basic knowledge, concepts, examples and calculation or problem solving. This e-book also explained the definition and formulas used in this topic. It comes with solved examples, quizzes and final year question to strengthen student understanding. For quizzes the scheme answer is showed to student for enable to compare their answer. This e-book hopefully will further aid students in understanding the chapter 1 through explicit explanations or through solved examples and illustrations. Abstract
Declarations iii All Right Reserved. No part of this publication may be reproduced or transmitted in any form or by any electronic or mechanical including photocopy, recording, or any information storage and retrieval system without permission from Politeknik Banting Selangor and Bahagian Instruksional dan Pembelajaran Digital JPPKK. AUTHORS Roihan Binti Romli Masfizaizan Binti Manaf Mohd Azam Bin Mohd Daud EDITOR Mohd Azam Bin Mohd Daud PUBLISHED BY Politeknik Banting Selangor Persiaran Ilmu, Jalan Sultan Abdul Samad, 42700 Banting, Selangor Darul Ehsan. TELEPHONE : 03-3120 1657 / 03-3120 1625 FAX : 03-3120 1706 website : https://pbs.mypolycc.edu.my/ e-mel : [email protected]
iv Table Of Contents Acknowledgement i Abstract ii Declaration iii Table of Contents iv 1.0 Introduction 01 1.1 Classification of Load 01 1.2 Effect of load 03 1.3 Stress 03 1.4 Strain 04 1.5 Young’s Modulus 04 1.6 Hooke’s Law 09 1.6.1 Hooke Law Statement of Springs 09 1.6.2 Hooke Law Statement Relating Stress & Strain 10 1.6.3 Stress-Strain Diagram 11 1.7 Shear Stress and Shear Strain 12 1.7.1 Difference Single Shear And Double Shear 14 1.7.2 Modulus of Rigidity 14 1.8 Factor of Safety 15 1.9 Poisson Ratio 16 Tutorial 20 Past Semester Final Year Exam Question 22 Scheme 24 References 35
Dynamic load A force is any action that tends to maintain or change the motion of a body or distort it. A force acting on an object causes the object to change its shape or size. When there is an interaction between two objects, they exert forces on each other, these forces are equal in magnitude but opposite in direction. 1.0 Introduction Static load 1.1 Classification Of Load 1 2 A load is a force that acts on A load is a force that acts on a material. Loads can be categorized into static loads, dynamic loads, impact loads and fatigue loads as shown in figure 1.1. A static load is a load that does not change. For example, a building as shown in Figure 1. The area of this building has to bear a constant load as long as the building is not moved or demolished. A dynamic load is a load that is constantly changing. For example, vehicles crossing the road. Figure 2 shows a train passing through a rail where the rail has to bear a variable load every time the train crosses the track. (Image Sources : 1zoom.me) Figure 1: Static load Figure 2: Dynamic load 1
Alternating load 3 Impact load 4 It is a load that acts immediately. For example, when a knock or impact of a hammer is applied to a nail (Figure 3). As the hammer hits the nail, the load will act instantly on the nail. Alternating load is a load that changes over time. For example, when a load is suspended on a spring (Figure 4), the spring will try to prevent elongation from occurring due to the applied load. Figure 3: Impact load (Image Sources : Julian Cassell’s DIY Blog) Figure 4: Alternating load (Image Sources : girlwalls.blogspot.com) 2
1.2 Effect of Load Figure 5: Effect of force (Image Sources: Linear Motion Tips) 1.3 Stress Stress is the measure of an external forces acting over the cross-sectional area of any object or body (Figure 6). Its is simply a ratio of the applied force to the cross-sectional area of the material. The external force acting on a body opposed by an internal force causes deformation to occur. Figure 5 show the effect produced by internal forces. A F = Unit : 2 @ Pascal (Pa) F Figure 6: Force applied to the bar 3
1.4 Strain Strain is the deformation or displacement of material that results from an applied stress (Figure 7). When you pull on a tension rod, you can see the rod physically increase in length. It is defines the ratio of change in length divided by original length. 1.5 Young’s Modulus Young’s Modulus, E or the modulus of elasticity is a mechanical property that measures the tensile or compressive stiffness of materials when force is applied. Symbol: Unit: 2 @ Pa Formula: ′ = = ɛ = Original length, L Extension, x = Strain can also be expressed as a percentage Unit : Dimension-less Figure 7: Deformation on bar 4
A mild steel bar 2.05 m long contracts by 0.24 mm when tensile load is applied to it. Determine the strain and the percentage strain. EXAMPLE 1.2 SOLUTION: ɛ = = 0.24 × 10−3 2.05 = 1.171 × 10−4 Percentage strain = 1.171 × 10−4 × 100% = 0.0117% Original length, Lo = 2.05 Elongation, x =0.24 × 10−3 EXAMPLE 1.1 A bar with a cross sectional area of 150 mm2 has a tensile force of 17 kN applied to it. Determine the stress in the bar. Cross-sectional area, A = 150 × 10−62 Force, F = 17 × 103 σ = 17 × 10 3 150 × 10 −6 = 113.33 × 106 NΤm 2 = 133.33 MPa SOLUTION: = 5
EXAMPLE 1.4 A rod has cross-sectional area of 2030 cm2 an elongation 1.28 mm when 206 kN of tensile load applied. The length of rod 5.5 m. Calculate: a. Tensile stress b. Tensile strain c. Young modulus SOLUTION: Area, A = 2030 × 10−4 2 Elongation, x = 1.28 × 10−3 Force, F = 206 × 103 Length, Lo = 5.5 EXAMPLE 1.3 An aluminum bar carries a tensile load of 23 kN. Determine the diameter of the bar if the stress is 10 MPa. SOLUTION: = = = 23 ×103 10×106 = 2.3 × 10−32 2.310−3 = 2 4 d = 0.054m Force, F = 23 × 103 Stress, σ = 10 × 106 6
SOLUTION: ) = = 206 ×103 2030×10−4 = 1.01 × 106 2 = 1.01MPa b) = = 1.28 × 10−3 5.5 = 2.33 × 10−4 c) = = 1.01×106 2.33×10−4 = 4.33 × 109 2 = 4.33 EXAMPLE 1.5 A 6.07 m long alloy wire is loaded with 123 kN. If the stress in the wire is 74 MN/m2 and Young’s Modulus of alloy is 208 GN/m2 , calculate: a) Strain. b) Elongation c) Diameter of the alloy Length, Lo = 6.07 Force, F = 123 123 × 103 Stress, σ = 74 × 106 2 Young Modulus, E = 208 × 109 2 a) = = 74 × 106 208 × 109 = 3.56 × 10−4 b) = = (3.56 × 10−4 )(6.07) = 2.16 × 10−3 c) = = 123×103 74×106 = 1.66 × 10−32 1.667 × 10−3= 2 4 = 0.046 7
Test your understanding by answering the Quiz. Click button below: 8
1.6 Hooke’s Law Hooke’s law states that the force (F) required to extend or compress a spring by some distance (x) varies proportionately with respect to that distance (Figure 8). 1.6.1 Hooke Law Statement of Springs Hooke's Law is a principle of physics related to the properties of a material. First statement relates to the elongation of spring subject to the application of force. Second statement is known as the law of elasticity and relates the relation of stress and strain of materials. Figure 8: The force is proportional to the extension (Image Sources: www.springsfast.com) 9
Hooke’s law states that the stress of a material is proportional to its strain within the elastic limit of the material 1.6.2 Hooke Law Statement Relating Stress & Strain = Hooke’s law for Springs is expressed as: = F - Force (N) x - Extension or compression of the spring (m) k - Spring constant (N/m) σ - Stress (Pascal) E - Modulus of elasticity or Young’s modulus (Pascal) ε - Strain (dimensionless) 10
Figure 9: Stress Strain Diagram (Image Sources: www.scienceabs.com) Watch the video to better understand the Tensile Test and Stress - Strain graph. A stress strain diagram or stress strain curve is used to illustrate the relationship between a material’s stress and strain. A stress strain curve can be constructed from data obtained in any mechanical test where load is applied to a material and continuous measurements of stress and strain are made simultaneously. Suppose that a metal specimen is placed in a Universal Testing Machine. Since the axial load increases gradually in increments, the total elongation over the gauge length is measured at each load increment and this is continued until failure of the specimen occurs (Figure 9). 1.6.3 Stress-Strain Diagram 11
1.7 Shear Stress And Shear Strain Let's consider a rivet used to connect two plates as shown in the Figure 10. When two equal forces act on two plates in opposite directions. Then the shear force acts on the rivet and the shear stress will be induced on the cross section of the rivet. Figure 10: Shear Force (Image Sources: extrudesign.com) Based on Figure 7, several new properties of materials can obtain as following: ❖ Yield Stress, = ❖ Maximum Tensile Stress, = ❖ Percentage Elongation = ℎ × 100% ❖ Reduction = − × 100% 12
A type of stress that acts coplanar with cross section of material. 1 2 3 Shear strain is the tangent of the angle, and is equal to the length of deformation at its maximum divided by the perpendicular length in the plane of force application Can be calculated by taking the ratio of force per unit area Can be calculated by taking the ratio of displacement / length = = Where: τ - Shear stress F - Force applied A - Area of cross-section Unit of Shear stress is /2 Where: - Shear strain - Displacement - Length Shear strain has no unit Shear stress measure Tangential force quantity. Shear strain measure Angular changes of body. 13
Load is applied on only one plane Load is applied on two planes More shear at each failure plane Less shear at each failure plane 1.7.1 Difference The Single Shear and Double Shear 1.7.2 Modulus of Rigidity Modulus of Rigidity, (Shear Modulus) is the coefficient of elasticity for a shearing force. It is defined as: “The ratio of shear stress to the displacement per unit sample length (shear strain)“ Can be calculated by shear stress, over shear strain, = = / / = 14
1.8 Factor of Safety The factor of safety is the load-carrying capacity of the system in excess of what the system actually supports. Bridges, buildings, safety equipment, and fall protection all start with a factor of safety. The safety factor determines how much stronger the system is than it needs to be. The safety factor is the backbone of all safety structures and equipment and comes from the engineer. In the planning phase of all safety structures and equipment, engineers determine the overload required for any object to remain safe in the event of an emergency. A factor of safety increases the safety of people and reduces the risk of failure of a product. Working Stress - When designing machine parts, it is desirable to keep the stress lower than the maximum or ultimate stress at which failure of the material takes place. It is also known as safe or allowable stress. Ultimate Stress - The stress, which attains its maximum value is known as ultimate stress. It is defined as the largest stress obtained by dividing the largest value of the load reached in a test to the original cross-sectional area of the test piece. Factor of Safety is defined as the ratio of the maximum stress to the working stress. () = 15
Figure 11: Poisson Ratio (Image Sources: whatispiping.com) 1.9 Poisson Ratio Poisson's ratio of materials is a very important parameter in materials science and engineering mechanics. When a force is applied to a bar it deforms (elongates or compresses) in the axial (longitudinal) direction. At the same time, deformation is observed in the transverse (width) direction as well. Poisson's ratio relates these changes in the transverse direction and the axial direction. Poisson's ratio is defined as the ratio of transverse strain to axial strain under the influence of the same force (Figure 11). It is material property and remains constant. 16
Figure 12: Poisson Ratio on bar (Image Sources: whatispiping.com) Figure 12 shows A tensile force (F) is applied in a bar of diameter do and length lo . With the action of this force F, the bar elongates and final length in l. Also, the diameter reduces and the final diameter is d. 17
SOLUTION: EXAMPLE 1.6 A piece of aluminium wire is 5 m long and has a cross-sectional area of 100 mm2 . it is subjected to increasing loads, the extension being recoded for each load applied. The result are: Draw the load/extension graph and determine the modulus of elasticity for the material of the wire. a) = = = ℎ = (1.4 × 106 ) 5 100 × 10−6 = 70 G ℎ = = = 2.95 × 103 2.1 × 10−3 = 1.4 × 106 2 Load (kN) 0 1.10 2.95 4.71 7.00 8 .00 Extension (mm) 0 0.8 2.2 3.4 5.0 6.5 Figure 13 : A graph of Load versus Extension 18
EXAMPLE 1.7 A tensile test is performed on a brass specimen 10 mm in diameter using a gauge length of 50 mm. when the tensile test load, F reaches a value of 20 kN, the distance between the gauge marks has increased by 0.122 mm. Calculate: a) Modulus of elasticity of the brass b) Poisson’s ratio if the diameter of gauge decreases by 0.0083 mm SOLUTION: Length, Lo = 50 × 10−3 Diameter, D = 10 × 10−3 Force, F = 20 × 103 Extension, x = 0.122 × 10−3 Calculate Modulus of Elasticity & Poisson Ratio? = 4 10 × 10−3 2 = 7.85 × 10−5 2 = = 20 × 103 7.85 × 10−3 = 254.65 × 106 = = 0.122 × 10−3 50 × 10−3 = 2.44 × 10−3 = = 254.65 × 106 2.44 × 10−3 = 104.36 = ∆ = 0.0083 × 10−3 10 × 10−3 = 8.3 × 10−4 = = 8.3 × 10−4 2.4 × 10−3 = 0.346 a) b) 19
TUTORIAL (Image courtesy of dozr.com) 1. State the type of force in the situation according to the picture below: (Image courtesy of shutterstock.com) (Image courtesy of sawsonskates.com) (Image courtesy of boingboing.net) a c d b 20
2. Define in engineering terms below: a. Normal stress b. Normal strain 3. A tube of outside diameter 60 mm and inside diameter 30 mm is subjected to a tensile load of 70 kN. Determine the stress in the tube. 4. A load with 56 mm diameter and 455 mm length has been subjected to an axial load of 322 kN. The final length is 461 mm. Calculate the stress and the strain of the bar. 5. The ratio is called ______________. 6. A steel bar has diameter of 20 mm was loaded with tensile load. Determine load, F and the Poisson ratio if the final diameter is 19.04 mm. Given stiffness of alloy, E is 70 GPa and strain 4 × 10−3 . 7. State two reasons why tensile testing is important in engineering field. 8. A compression test was applied to a 28 mm diameter and 100 mm length aluminium specimen using Universal Testing Machine. When the load applied was 90kN, the shortening observed was 0.092mm. Calculate the modulus of elasticity. 21
PAST SEMESTER FINAL EXAM QUESTIONS a. Compare the Modulus Of Elasticity and Modulus Of Rigidity (4 marks) b. A tensile test has been done on steel rod. The result stress versus strain in tensile test was shown in figure 1a. Fill a name of point in box. (6 marks) Figure 1a a. State THREE (3) effects on the material under load (6 marks) b. Plot graph stress verses strain for mild steel after testing and plot upper and lower yield point. Explain the condition of the specimen in between yield point. (8 marks) 22 SESSION 2 2021/2022 SESSION JUNE 2019
a. List THREE (3) types of forces and illustrate them using suitable figure. (6 marks) b. A rod with 4.5m length and cross-sectional area of 1050mm2 c. elongates by 6.56mm when 65kN tensile force is applied on both sides. Calculate: I. Tensile stress in the rod. (2 marks) II. Strain in the rod. (2 marks) III. Young Modulus of the rod. (2 marks) IV. The safety factor if the maximum stress is 332 MN/m2 . (2 marks) SESSION JUNE 2018 a. Define the terms as below and state its unit. i. Stress (2 marks) ii. Strain (2 marks) iii. Safety factor (2 marks) b. A copper wire 4 m long was act by 100 kN of tensile load. If the stress applied is 60 MN/m2 and given ECopper = 112 GN/m2 , calculate: i. The strain in the copper (2 marks) ii. The elongation of copper (2 marks) iii. The diameter of copper (4 marks) 23 SESSION DECEMBER 2018
SCHEME 24
1) The following are the types of load except: a) Impact load b) Alternating load c) Dynamic load d) Bending load 2) State the effect of load on the given situation: ▪ Remove the water from the sponge a) Compression b) Tension c) Torsion d) Shear ▪ Slice the bread using a knife a) Compression b) Tension c) Torsion d) Shear ▪ Sitting on the bed mattress a) Compression b) Tension c) Torsion d) Shear ▪ The rope that pulls the car being towed a) Compression b) Tension c) Torsion d) Shear 3) The unit of strain is: (a) Pascals (b) Metres (c) Dimension-less (d) Newtons 4) The unit of stiffness is: (a) Newtons (b) Pascals (c) Newtons per metre (d) Dimension-less 5) The unit of Young’s modulus of elasticity is: (a) Pascals (b) metres (c) dimension-less (d) newtons 6) A wire is stretched 3 mm by a force of 150 N. Assuming the elastic limit is not exceeded, the force that will stretch the wire 5 mm is: (a) 150 N (b) 250 N (c) 90 N (d) 450 N Scheme – Quiz 25
7) 100mm2 equal to m2 a. 100x10-3m b. 100x10-6m2 c. 100x10-4m2 d. 100x10-3m2 8) A rectangular bar having a cross-sectional area of 80 mm2 has a tensile force of 20 kN applied to it. Determine the stress in the bar. a. 250MPa b. 255MPa c.200MPa d. 252MPa 9) A metal bar 5.0 m long extends by 0.05 mm when a tensile load is applied to it. The percentage strain is: a. 0.1 b. 0.01 c. 0.001 d. 0.0001 Scheme – Quiz 26
Scheme – Tutorial 1) a. Tensile b. torsion c. Shear d. Compression 2) a. Normal stress - When the external force acts perpendicular to the cross-sectional area of any object or a body. b. Normal strain - Deformation of a material body under the action of applied forces. Given : Outside diameter 60 mm Inside diameter 30 mm Tensile load of 70 kN Determine the stress in the tube. A= 4 2 − 2 A= 4 60 × 10−3 2 − 300 × 10−3 2 = 2.12 × 10−3 2 = = 70 × 103 2.12 × 10−3 = 33 × 106 Given: Diameter,d = 56 × 10−3 Length,L = 455 Load, F = 322 × 103 Final length = 461 × 10−3 Calculate stress and strain. Stress: A = 4 56 × 10−3 2 = 2.46 × 10−3 2 = = 322 × 103 2.46 × 10−3 = 130.89 × 106 3) 4) 27
= 461 × 10−3 − 455 × 10−3 = 6 × 10−3 = = 6 × 10−3 455 × 10−3 = 0.0131 Strain: 5) The ratio is called Young’s Modulus. Given: Diameter, d = 20 × 10−3 Final Diameter, = 19.04 × 10−3 = 70 × 109 Strain, = 4 × 10−3 Determine force, F and Poisson Ratio. Force: A= 4 20 × 10−3 2 = 3.14 × 10−4 2 E = = = 70 × 109 4 × 10−3 = 280 = = 280 × 106 3.14 × 10−4 = 87.92 Poisson Ratio: = ∆ = 20 − 19.04 20 = 0.048 Poisson Ratio,ν = = 0.048 4×10−3 = 12 6) Scheme – Tutorial 28
Given: Diameter,d =28 × 10−3 Length,L = 100 × 10−3 Load, F = 90 × 103 Elongation,x = 0.092 × 10−3 Determine Modulus of Elasticity? A = 4 28 × 10−3 2 = 6.16 × 10−4 2 Modulus of Elasticity: = = 90 × 103 6.16 × 10−4 = 146.1 × 106 E = σ ε = 146.1×106 9.2×10−4 = 158.80 GPa = = 0.092 × 10−3 100 × 10−3 = 9.2 × 10−4 8) 7) Two reasons why tensile testing is important in engineering field. • To determine the maximum strength of materials. • To know the yield nature of the material. Scheme – Tutorial 29
Scheme – Past Semester Final Year Questions SESSION 2 2021/2022 a. Compare the Modulus Of Elasticity and Modulus Of Rigidity. ❑ Modulus of elasticity is used to calculate the deformation of an object when a deforming force acts at right angles to a surface of the object. ❑ Modulus of rigidity is used to calculate deformations when a deforming force acts parallel to the surface of an object b. The result stress versus strain in tensile test. Fill a name of point: A – Maximum point B – Fracture point C – Upper yield point D – Young Modulus E – Elastic region F – Plastic region 30
a. THREE (3) effects on the material under load: i. Elongation ii. Shortening iii. Shear Or any else, reasonable. SESSION JUNE 2019 b. Plot graph stress verses strain for mild steel after testing and plot upper and lower yield point. Explain the condition of the specimen in between yield point. ❑ Material still longer ever there is no increment load of tensile load ❑ Cross sectional area of material decreasing ❑ Material goes to plastic range Choose ONE or else is reasonable Scheme – Past Semester Final Year Questions 31
a. THREE (3) types of forces and illustrate using suitable figure. SESSION DECEMBER 2018 Tensile force Compression force Shear force Given: Length, L = 4.5 Cross-sectional area, A = 1050 × 10−62 Elongation, x = 6.56 × 10−3 Tensile force, F = 65 × 103 Maximum stress = 332 × 106 Determine tensile stress, strain, Young’s Modulus and Safety Factor. b. = = 65 × 103 1050 × 10−6 = 61.90 × 106 = = 6.56 × 10−3 4.5 = 1.46 × 10−3 i. ii. Scheme – Past Semester Final Year Questions 32
= = 61.9 × 106 1.46 × 10−3 = 42.40 × 109 = = 332 × 106 61.9 × 106 = 5.36 iii. iv. SESSION JUNE 2018 a) Define: i. Stress is the measure of an external forces acting over the cross-sectional area of any object or body. Unit - N/m2 . ii. Strain is the deformation or displacement of material that results from an applied stress. Unit – dimension less. iii. Factor of Safety is defined as the ratio of the maximum stress to the working stress. Unit – dimension less Scheme – Past Sem Final Year Questions 33
b) A copper wire 4 m long was act by 100 kN of tensile load. If the stress applied is 60 MN/m2 and given E copper = 112 GN/m2 , calculate: i. The strain in the copper = = = 60 × 106 112 × 109 = 5.36 × 10−4 ii. The elongation of copper = = = (5.36 × 10−4 )(4) = 2.14 × 10−3 iii. The diameter of copper = = = 100 × 103 60106 = 1.67 × 10−3 = 2 4 2= 4 1.67 × 10−3 = 0.046 Scheme – Past Semester Final Year Questions 34
35 References 1. Ferdinand P. Beer, E. Russell Johnston, Jr., John T. DeWolf, David F. Mazurek (2015). Mechanics of Materials (7th ed.). 2 Penn Plaza, New York: McGraw-Hill Education. 2. Roihan, Tamil M., Siti.H., Ros. S., Siti.M., Marliyana, Norliza,(2019). Strength of Materials, (3rd Ed.). Politeknik Nilai. 3. Young's modulus (2022). Retrieved from https://en.wikipedia.org/wiki/Young%27s_modulus 4. Danielle Collins (2019). Mechanical properties of materials: Stress and strain. Retrieved from https://www.linearmotiontips.com/mechanical-properties-ofmaterials-stress-and-strain/ 5. Popular lifehacks (2021). What is the purpose of stress strain diagram? Retrieved from https://teacherscollegesj.org/whatis-the-purpose-of-stress-strain-diagram/ 6. Romel Verterra (2022). Stress-Strain Diagram. Retrieved from https://mathalino.com/reviewer/mechanics-and-strength-ofmaterials/stress-strain-diagram 7. Sundar Dannana (2018). What is Shear Stress and Shear Strain? Retrieved from https://extrudesign.com/shear-stressand-shear-strain/