Mechanical Properties of Metals PDF

Chapter 2 Mechanical Properties of metals Dr.H.Elhendawi 1 1 Dr.H.Elhendawi 2 2  What will happen if …? Building a bridge of concrete and aluminum? Dr.H....

Chapter 2 Mechanical Properties of metals Dr.H.Elhendawi 1 1 Dr.H.Elhendawi 2 2  What will happen if …? Building a bridge of concrete and aluminum? Dr.H.Elhendawi 3 3  Objectives Demonstrate your understanding of 1 mechanical properties, elasticity, stress, strain, Hook’s Law, elastic limit, ,ultimate strength and elastic moduli. Write and apply formulas for calculating Elasticity Modulus. 2 3 Solve problems involving each of the parameters in the above objectives. Dr.H.Elhendawi 4 4 I- Concepts Dr.H.Elhendawi 5 5  1- Mechanical Properties of Solid 1-The material properties that determine its behavior under applying external force (loads).)‫خواص المادة التى تحدد سلوكها عند تعرضها لقوى خارجية (األحمال‬ 2-They are always related to elastic and plastic behavior of material..‫ترتبط دائما بالسلوك المرن و اللدونه للمواد‬ 3- The mechanical properties of materials such as elasticity , plasticity , elastic modulus, yield strength, fracture strength , ductility and hardness. ‫الخصائص الميكانيكيةة للمةواد م ةل المرونةه و اللدونةة و معامةل المرونةه اجهةاد ااجعةان و اجهةاد الكسةر و‬.‫ااستطالة و الصالبة‬ Dr.H.Elhendawi 6 6  2- Elasticity  Elasticity is the ability of an object or material to resume its normal shape after being stretched or compressed..‫ المرونة قدرة الجسم او المادة على استعاده شكلها الطبيعى بعد الشد أو الضغط‬ Dr.H.Elhendawi 7 7 When you hear the word "stress," what do you usually think of ? No, this is stress! Dr.H.Elhendawi 8 8  3- Stress  Stress is the investigation of the internal resistance of the body, the nature of forces set up within a body to balance the effect of the externally applied forces (load). ‫أى طبيعةةة القةةوى الموجةةودة داخةةل الجسةةم‬,‫ االجه ا هةةو دراسةةة المقاومةةة الداخليةةة للجسةةم‬. )‫لموازنة تا يرالقوى الخارجية المطبقة عليه (الحمل‬ 1 9 Dr.H.Elhendawi 0  4- Strain  Strain is a measure of the degree of deformation due external force..‫ االنفع ل هو مقياس مدى التغير الحادث نتيجة تأ يرالقوة الخارجية‬ 1 10 Dr.H.Elhendawi 1  Normal Force (1) Normal Stress (𝝈): The Force (F) per unit area (𝑨𝒐 ). 𝑭.‫القوة المؤثرة على وحدة المساحات من العينه تحت االختبار‬:‫االجهاد‬ 𝝈= 𝑨𝒐 Where: F applied force ‫القوة المطبقة‬ 𝑨𝒐 original cross section area ‫مساحة مقطع االسطوانة‬ 𝑨𝒐 = 𝝅𝒓𝟐 ‫مساحة الدائرة‬ 𝑨𝒐 = 𝑳𝟐 ‫مساحة المكعب‬ 𝑵 𝑰𝒃 𝒅𝒚𝒏 Unit: = 𝑷𝒂 𝑶𝒓 = 𝒑𝒔𝒊 , 𝑶𝒓 = 𝒃𝒂𝒓𝒚𝒆 (Pressure unit) 𝒎𝟐 𝒊𝒏𝟐 𝒄𝒎𝟐 How to reduce stress ? 1 11 Dr.H.Elhendawi 2  Normal Strain Normal Strain (𝜺): a measure of the degree of deformation due external force.‫ يقيس مدى التغير الحادث نتيجة تأثيرالقوة الخارجية‬:‫االنفعال‬ 𝑳𝒇 −𝑳𝒐 𝜺= 𝑳𝒐 1 Dr.H.Elhendawi 12 3 Different types of loads ‫أنواع الحمل المختلفة‬ 1 13 Dr.H.Elhendawi 4  ‫‪Parallel Force‬‬ ‫قوة موازية (قوة القص)‪(1) Parallel Force (Shear Force):‬‬ ‫تحريك السطح الحر من الجسم ‪.‬‬ ‫قوتان متعاكسان فى االتجاه على مستوين متوازيين‬ ‫‪‬‬ ‫يحدث قص أو قطع فى الجسم ‪.‬‬ ‫قوتان متعاكسان فى االتجاه على مستوين متماسين‬ ‫‪‬‬ ‫‪Shear Stress (𝛕):‬‬ ‫𝑭‬ ‫=𝛕‬ ‫𝒐𝑨‬ ‫‪Where:‬‬ ‫ ‬ ‫‪F‬‬ ‫‪applied parallel force‬‬ ‫ ‬ ‫‪𝑨𝒐 original tangential cross section area‬‬ ‫‪1‬‬ ‫‪14‬‬ ‫‪Dr.H.Elhendawi‬‬ ‫‪5‬‬ Shear Strain (𝜸): ∆𝒍 𝜸 = 𝒕𝒂𝒏𝜽 = 𝒍𝒐 Where: ∆𝒍 linear displacement 𝒍𝒐 height Dr.H.Elhendawi 1 15 6  1 16 Dr.H.Elhendawi 7  II-Stress& Strain relationship (Hook’s Law) If a force (F) is exerted on an object, the length of it will change by ∆𝐿. Experimentally ∆𝐿 is proportional to the force exerted on the object. 𝑭 = 𝒌 ∆𝑳 Where F: the force pulling on the object(N) ‫قوة السحب على الجسم‬ ∆𝐋 : the change in the length of the object (m). k : proportionality constant This Law is valid for most solid from iron to bone. If the forces are great enough, the object will break, or fracture. 1 17 Dr.H.Elhendawi 8  Hook’s Law  Stress is proportional to strain and the proportionality constant is called the elastic modulus and it depends on the material being deformed as well as on the nature of the deformation. ‫ يتناسب اإلجهاد تناسبا طرديا مع اإلنفعال و ثابت التناسب تسمى معامل المرونة ويعتمد معامل المرونة‬.)‫على نوع المادة وطبيعة التغير الحادث فى المادة(اتجاة القوة المؤثرة على الجسم‬ 1 18 Dr.H.Elhendawi 9 Stress-Strain Curve 192 Dr.H.Elhendawi 0  Stress& Strain Curve Elastic Region Plastic Region 1. Nonpermanent 2. Permanent ‫ تغير غير ائم (يعو الجسم لشكله األصلى‬ ‫ ائم (ال يعو الجسم لشكلة بزوال‬.) ‫بزوال االجه‬ ) ‫االجه‬ 2. At relativity low stresses. 2. At relativity high stresses.. ‫ يح ث عن االجه ات الصغيرة نسبي‬ ‫ يح ث عن االجه ات الكبيرة‬. ‫نسبي‬ 3. Stress is directed proportional 3. Stress and strain are with strain (Obeys Hook’s Law). not proportional (don’t ‫ يتن سب االجه طر ي مع االنفع ل (يخضع‬ obeys Hook’s Law). )‫لق نون هوك‬ ‫ ال يتن سب االجه مع االنفع ل(ال‬.)‫يخضع لق نون هوك‬ Dr.H.Elhendawi 2 20 1  III- Elastic Region Properties  From Hook’s Law the proportionality constant (k) is called modulus of elasticity. [OA] ‫من قانون هوك ابت التناسب يسمى معامل المرونة‬  There are three types of deformations and their elastic moduli are 1.Young’s Modulus 2.Shear Modulus 3.Bulk Modulus.‫ يوجد ثالث من معامالت المرونة وتحدد على حسب نوع فى االجهاد‬  The elastic moduli depend on the type of stress and strain..‫ معامل المرونة يعتمد على نوع ااجهاد و اانفعال‬  Elastic modulus measures the resistance if the materials to elastic strain which called stiffness..‫ يقيس معامل المرونة مقاومة المواد لالجهاد المرن و الذى يسمى الصالبة‬ 2 21 Dr.H.Elhendawi 2  1- Elasticity Moduli 1.Young’s Modulus 2.Shear Modulus 3.Bulk Modulus  Elasticity in length  Elasticity in shape  Elasticity in volume 𝑉𝑜𝑙𝑢𝑚𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 𝑁𝑜𝑟𝑚𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠 𝝈 𝑩= 𝒀= = 𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 𝑉𝑜𝑙𝑢𝑚𝑒 𝑠𝑡𝑟𝑎𝑖𝑛 𝑁𝑜𝑟𝑚𝑎𝑙𝑠𝑡𝑟𝑎𝑖𝑛 𝜺 𝑮= 𝐹 Τ𝑨𝒐 ∆𝑷 𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑎𝑖𝑛 = ∆𝑉 = = 𝑭/𝑨𝒐 − ൗ𝑉 − ∆𝑉ൗ𝑉 𝛕 F/Ao F/Ao 0 0 ∆𝑳/𝑳𝒐 = = = ∆𝑃 γ ∆L/Lo tan θ = −𝑉0 ∆𝑉 2 22 Dr.H.Elhendawi 3 ‫‪ Proportional Limit (OA):‬‬ ‫حد التناسب‬ ‫‪It is the end point of linear relationship between stress and strain.‬‬ ‫‪ ‬بعض المواد اتكون العالقة بين ااجهاد واانفعال فى منطقة المرونة عالقة خطية م ل الحديد الزهر وبالتالى‬ ‫ايمكن تحديد معامل المرونه لها ‪.‬‬ ‫‪ ‬معامل المرونة هو مدى مقاومة المادة لالنفعال المرن ويسمى ‪. stiffness‬‬ ‫‪ ‬فكلما زاد معامل المرونة قل اانفعال المرن ‪.‬‬ ‫‪ Elastic Limit (B) :‬‬ ‫ح المرونة‬ ‫‪The deformations are completely recovered when the load is removed.‬‬ ‫‪Dr.H.Elhendawi‬‬ ‫‪2‬‬ ‫‪23‬‬ ‫‪4‬‬  2- Poisson Ratio  The ratio of the lateral and axial (longitudinal) strain..‫ النسبة بين االنفعال المستعرض الى االنفعال الطولى‬  In elastic region εx = εy , εx ∝ εz εx εz ⱱ= − = − Dimensionless εy εy https://www.w3schools.blog/poissons-ratio Relation between Elastic Moduli Y = E = 2G(1 + ⱱ) https://www.geeksforgeeks.org/poissons-ratio/ 2 24 Dr.H.Elhendawi 5 Table: Yong’s modulus ,Shear’s modulus, Bulk’s modulus and Poisson ratio at room temperature. Materials Yong’s modulus Shear’s modulus Bulk’s modulus Poisson ratio (𝟏𝟎𝟏𝟎 𝑵/𝒎𝟐 ) (𝟏𝟎𝟏𝟎 𝑵/𝒎𝟐 ) (𝟏𝟎𝟏𝟎 𝑵/𝒎𝟐 ) Aluminum 7 2.5 7 0.33 Brass 9.1 3.5 6.1 0.35 Glass 6.5 –7.8 2.6 –3.2 5 - 5.5 0.18 - 0.3 Copper 11 4.2 14 0.35 Steel 20 8.4 6 0.27 Tungsten 35 14 20 0.28 Water - - 0.21 2 25 Dr.H.Elhendawi 6  IV- Plastic Region Properties  Yield point or elastic limit point (B) (σy): : Is the stress corresponding to the yielding point..‫ نقطة االزعان أو حد المرونة المنطقة الفاصلة بين منطقة المرونة و اللدونة‬  Is the stress at which plastic deformation begins (measure the resistance to plastic deformation). ‫ هﻮ اﻹجه الﺬﻱ يﺒﺪﺃ بعﺪﻩ االنفع ل اللﺪن و يقﻴﺲ مﺪﻱ مق ومة الﻤ ة لﻼنفع ل‬ ‫ فى بعض الموا يصعب تح ي هذﻩ النقطة ولهذا يتم رسم خط موازى للعﻼقة الخطية‬.0.002 ‫يب ﺃ من انفع ل قيمته‬ 2 26 Dr.H.Elhendawi 7  IV- Plastic Region Properties  Elastic limit or yield strength (σy): ‫اجهاد االذعان‬ 1. The stress corresponding to the yielding point. 2. the stress at which plastic deformation begins. 3. There are lower and upper in same materials. 2 27 Dr.H.Elhendawi 8  IV- Plastic Region Properties  Tensile strength (TS): ‫اجهاد االختناق‬ 1. It is the stress at the maximum point on the stress-strain curve. ‫أعلى قيمة لالجهاد على المنحنى‬ 2. It is the maximum stress that can be sustained by a structure in tension and if maintained fracture will result..‫أقصى اجهاد تتحمله المادة واجا استمر لفترة طويلة تنكسر المادة‬  Fracture strength (B):.‫اجهاد الجى يحدث عنده انهيار للمادة و تنكسر‬ 2 28 Dr.H.Elhendawi 9  V-Ductile and Brittle materials Ductile materials : ‫مواد مطيلية‬ 1. Materials that exhibit large plastic deformation before fracture..‫ المواد التى تظهر انفعال لدن كبير قبل ان يحدى له كسر‬ 2. Cable of absorbing energy or shock.‫ قادرة على امتصاص الصدمات والطاقة‬ 3. If they become over loaded, they will exhibit large deformation before failing. ‫ عند زيادة ااحمال تظهر انفعال لدن كبير قبل الكسر‬ Dr.H.Elhendawi 3 29 0  V- Ductile and Brittle materials Brittle materials : ‫مواد قاصفة‬ 1. Materials that exhibit large little or no plastic deformation before fracture..‫ المواد التى تظهر انفعال لدن صغير او يظهربصورة قليلة قبل ان يحدث له كسر م ل الزجاج او الحديد الزهر‬ 2. Percent elongation(EL) : is the fracture strain expressed as a percent. 𝑙𝑓 − 𝑙𝑜 %𝐸𝐿 = 𝑥100 𝑙𝑜 330 Dr.H.Elhendawi 1 Brittle material Ductile material Plastic material 3 31 Dr.H.Elhendawi 2  (VI) Factors affecting on behavior of material 1- Chemical components.  Steel has brittle behavior when it contains a high carbon, and it is ductile when the carbon content is reduced. 2- Temperature.  At low temperatures materials become harder and more brittle  At high temperatures materials become softer and more ductile. 3 32 Dr.H.Elhendawi 3  ‫)‪(VII) Working Stress (σw‬‬ ‫‪ It is the stress which used in design calculation to restrict the applied load to‬‬ ‫‪one that is less than the material can fully support.‬‬ ‫‪ ‬هﻮ اﻹجه الﺬﻱ يﺴﺘﺨﺪم فﻲ حﺴ ب ت تﺼﻤﻴﻢ الﻤﻨﺸﺂت وﺃجﺰاء اﻵالت حﺘﻲ نﻀﻤﻦ ﺃن اﻹجه ات الﻮاقعة علﻴه ﺃقﻞ مﻦ ﺃقﺼﻲ ﺇجه‬ ‫تﺘﺤﻤله فﻲ مﻨﻄقة الﻤﺮونة وب لﺘ لﻲ ﺇذا حﺪث ﺃى تﺠ وز مف جﺊ لﻦ نﺪخﻞ مﺒ شﺮة ﺇلى مﻨﻄقة اللﺪونة وب لﺘ لى يﺘﺤقﻖ األم ن فﻲ الﺘﺼﻤﻴﻢ‪.‬‬ ‫𝒚𝝈‬ ‫= 𝒘𝝈‬ ‫𝑭 ‪𝑺.‬‬ ‫‪2- Safety Factor (S.F): is a ratio of the yielding stress (𝝈𝒚 ) divided by the allowed‬‬ ‫‪stress(𝝈𝒘 ).‬‬ ‫‪ ‬مع مل االم ن هو النسبة بين اجه االذع ن واجه التشغيل‪.‬‬ ‫‪3‬‬ ‫‪33‬‬ ‫‪Dr.H.Elhendawi‬‬ ‫‪4‬‬  The reasons for using the working stress 1- The design load may be different from the actual load.. ‫حمل التصميم قد يكون مختلف عند الحمل لحقيقى‬ 2- Avoid error inexact measurements due to errors in fabrication.. ‫تجنب خطأ القياسات غير الدقيقة بسبب أخطاء التصنيع‬ 3- Some materials, such as wood, concrete can show high variability in mechanical properties. ‫بعض المواد م ل الخشب والخرسانة من الممكن أن يحدث تغير في خواصها الميكانيكية بعد فترة من‬.‫الزمن‬ 3 34 Dr.H.Elhendawi 5  Chapter Laws 1. 𝑭┴𝑨 𝑭 ∆𝑳 𝝈 𝑭𝑳𝒐 𝝈= , 𝜺= ,𝑬 = 𝒀 = = 𝑨𝒐 𝑳𝒐 𝜺 𝑨∆𝑳 2. F//A 𝑭 ∆𝒙 𝝉 𝑭𝑳 𝝉= ,𝜸 = 𝜽 = ,𝑮 = = 𝑨𝒐 𝑳 𝜸 𝑨∆𝒙 εx εy 3. ⱱ = − εy =− εy 4. 𝐘 = 𝐄 = 2G(1 + ⱱ) 𝝈𝒚 5. 𝝈𝒘 = 𝑺.𝑭 3 35 Dr.H.Elhendawi 6 Ex(1) A pendulum consists of a big sphere of mass m = 30 kg hung from the end of a steel wire that has a length L = 15 m, a cross-sectional area A = 9 × 10-6 m2 and Young’s modulus Y = 200 × 109 N/ m2. Find the tensile stress on the wire and the increase in its length. Solution 𝑚 = 30 𝑘𝑔, 𝐿𝑜 = 15𝑚 , 𝐴 = 9𝑥106 𝑚2 , Y = 200 × 109 N/ m2 Tensile stress = 𝐴𝐹 =𝑚𝑔 = (30𝑥9.8) = 3.27𝑥10 7 𝑁 𝐴 (9𝑥10−6 ) 𝑚2 𝐹 Τ𝐴 𝑌= ∆𝐿ൗ 𝐿0 The increase in its length (∆𝐿). 𝐹/𝐴 3.27𝑥107 −3 𝑚 ∆𝐿 = 𝐿0 = 15 = 2.45𝑥10 𝑌 200 × 109 Dr.H.Elhendawi 3 36 7 Ex(2) A piece of copper originally 305 mm long is pulled in tension with a stress of 276 MPa. if the deformation is entirely elastic, what will be the resultant elongation? Solution 𝐿𝑜 = 305𝑥10−3 𝑚, 𝑃 = 276𝑥106 𝑃𝑎 The increase in its length (∆𝐿). 𝑃 276𝑀𝑃𝑎 ∆𝐿 = 𝐿0 = 305𝑚𝑚 = 0.76 𝑚𝑚 𝑌 11× 10 𝑀𝑃𝑎 4 3 37 Dr.H.Elhendawi 8 Ex(3) A telephone wire 120 m long and 2.2 mm in diameter is stretched by a force of 380N. What is the longitudinal stress? if the length after stretching is 120.10m, what is the longitudinal strain? Determine young's modulus for the wire. Solution The cross−sectional area of the wire is 𝜋𝐷2 𝜋(2.2𝑥10−3 𝑚)2 𝐴0 = = = 3.8𝑥10−6 𝑚2 4 4 𝐹 380𝑁 𝑆𝑡𝑟𝑒𝑠𝑠 = = = 100 𝑀𝑁/𝑚2 =100 MPa 𝐴𝑜 3.8𝑥10−6 𝑚2 ∆𝐿 0.1 𝑚 𝑆𝑡𝑟𝑒𝑠𝑠 𝑆𝑡𝑟𝑎𝑖𝑛 = = = 8.3𝑥10−4 𝑚 𝐸= = 120𝑥109 𝑃𝑎 𝐴𝑜 120 𝑚 𝑆𝑡𝑟𝑎𝑖𝑛 3 38 Dr.H.Elhendawi 9 Ex(4) The upper surface of a cube of gelatin ,5.0 cm on a side , is displaced 0.64 cm by a tangential force. If the shear modulus of the gelatin is 940 Pa, what is the magnitude of the tangential force? Solution 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 𝜏 𝐹 Τ𝐴 𝐹 Τ𝐴 Shear modulus (G) = 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑎𝑖𝑛 = 𝛾 = ∆𝐿 ൗ𝐿0 = 𝑡𝑎𝑛𝜃 ∆𝐿 0.64𝑐𝑚 𝛾= = = 0.128 𝐿𝑜 5𝑐𝑚 𝜏 = 𝐺𝛾 = (940𝑃𝑎)(0.128) = 120.32𝑃𝑎 𝐹 = 𝜏𝐴 = 120.32 0.05𝑚 2 = 0.3 𝑁 4 39 Dr.H.Elhendawi 0 Ex(5) sphere is initially surrounded by air, and the air pressure A solid brass exerted on it is 105 N/ m2. The sphere is lowered into the ocean to a depth where the pressure is 2x107 N/ m2 The volume of the sphere in air is 0.5 m3. By how much does this volume change once the sphere is submerged? Solution ∆𝑃 = 𝑃𝑜𝑐𝑒𝑎𝑛 − 𝑃𝑎𝑖𝑟 ∆𝑃 𝐵 = −𝑉0 ∆𝑉 ∆𝑃 ∆𝑉 = −𝑉0 𝐵 𝑁 𝑁 2𝑥107 2 − 105 2 ∆𝑉 = − 0.5 𝑚3 𝑚 𝑁 𝑚 = −1.6𝑥10−4 𝑚3 6.1𝑥108 2 𝑚 4 40 Dr.H.Elhendawi 1 Ex(6)Aluminum specimen has a diameter of 𝒅𝒐 =25mm and gauge length of Lo=250mm.If a force of 165kN elongates the gauge length 1.2mm,determine the modulus of elasticity. Also, determine by how much the force causes the diameter of the specimen to contract. Takeⱱ=0.33. Solution F 165x10−3 8 σ= = = 3.36x10 Pa A 3.14(12.5x10−3 )2 ∆L 1.2 ε= = = 4.8x10−3 Lo 250 𝜎 3.36𝑥108 𝑃𝑎 𝐸=𝑌= = = 70 𝐺𝑃𝑎 𝜀 4.8𝑥10−3 𝜀𝑙𝑎𝑡𝑒𝑟𝑎𝑙 = −ⱱ𝜀𝑎𝑥𝑖𝑎𝑙 =-(0.33)(4.8x10−3 )=1.584x10−3 ∆𝑑 = 𝜀𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝑑𝑜 =(-1.584x10−3 )(25mm)=0.0396 mm 4 41 Dr.H.Elhendawi 2  4 42 Dr.H.Elhendawi 3 Recourses https://www.sciencebuddies.org/science-fair-projects/references/stress-strain-strength-an-introduction- to-materials-science https://www.rgpvonline.com/answer/basic-mechanical-engineering/28.html https://m.blog.naver.com/k_wooram/221656514972 Serway, Raymond A., and John W. Jewett. Physics for scientists and engineers. Cengage learning, 2018. 4 43 Dr.H.Elhendawi 4  4 45 Dr.H.Elhendawi 5