L 7Properties of Mechanical 01.docx

1、L 7Properties of Mechanical 01Properties of MaterialsII. Mechanical PropertiesThe mechanical properties are about the behaviour of materials when subject to forces. When a material is subject to external forces, then internal forces are set up in the material which oppose the external forces.When a

2、material is subject to external forces which stretch it then it is said to be in tension. TensileWhen a material is subject to forces which squeeze it then it is said to be in compression. CompressionIf a material is subject to forces which cause it to twist or one face to slide relative to an oppos

3、ite face then it is said to be shear. Shear forcesStress: The force per unit area.= = Unit for stress : 1 Pa = 1N/m2 , 1 MPa = 1N/mm2Q1. A bar of material with a cross-sectional area of 50 mm2 is subject to tensile forces of 100 N. What is the tensile stress? (Answer: 2 MPa )Q2. Calculate the stress

4、 produced in a 12.5 mm diameter aluminium alloy test piece when a mass of 3000Kg is applied axially along the bar. (Answer: 239 MPa )Q3. A 10mm diameter bar is clamped at one end and suspend vertically. Calculate the stress in the bar if a mass of 2000Kg is hung from the free end of the bar. (Answer

5、: 249.6 MPa )Strain: When a material is subject to tensile or compressive force, it changes in length.Strain = = = =Since strain is a ratio of two lengths it has no units. = 0.01 indicate that the change in length is 0.01 the original length.% of strain = 100Q4. A strip of material has a length of 5

6、0 mm. When it is subject to tensile forces it increases in length by 0.020mm. What is the strain? (Answer: 0.0004 )Q5. If Youngs Modulus of elasticity for this steel is 210,000 MPa, and the steel behaves elastically at the stress calculated in Q3, how much strain would be induced by the 2000Kg. (Ans

7、wer: 0.0012 )Q6. Compare the force required to produce a stress of 170MPa in a 25mm diameter bar and in a 50mm diameter bar. (Answer: i) 83.4 kN, ii) 333 kN)Youngs Modulus of Elasticity (E)E = = This applies only whilst the material behaves elastically i.e if the force is removed, the material will

8、return to its original dimensions. Youngs Modulus describes the elastic flexibility or stiffness of the material.Typical values for E are:Steel210,000 MPaRubber7 MPaAluminium70,000 MPaPlastic1,400 MPaCopper100,000 MPaWood13,500 MPaStrengthStrength refers to the ability of a material to withstand str

9、ess without failure.The strength of a material is the ability of it to resist the application of forces without breaking. The forces can be tensile, compressive or shear.The tensile stress the material can withstand without breaking, i.e Tensile strength = The compressive strength and shear strength

10、 are defined in a similar way.The unit of strength is the Pa. Tensile Strength Stress Limit of proportionality Upper Yield Stress Lower Yield Stress StrainStress strain graphQ7. Two rods 25mm in diameter of aluminium alloy must support a load of 222kN.(i) What is the stress?(ii) What is the strain?(

11、iii) If the aluminium rods are replaced by coppers and the strain can not exceed that in the aluminium, what diameter is required.HardnessAnother mechanical property that may be important to consider is hardness, which is a measure of a materials resistance to localised plastic deformation (e.g. a s

12、mall dent or a scratch).Callister, Jr. W. D., Hardness is a property that is often related to a broad range of mechanical and even physical properties. The engineering definition of hardness is “ a materials resistance to permanent indentation under static or dynamic forces” the tests used to determ

13、ine the hardness of a material are called indentation hardness tests.The indentation hardness is by far the most often used method for measuring hardness. There are three standardised indentation hardness tests:SEE BELOW The Brinell hardness test The Vickers hardness test The Rockwell hardness testT

14、he hardness of metals can be defined as resistance to plastic deformation and test methods are based on forcing an indenter into the surface under a known load.(i) The Brinell hardness testThis test utilises a steel ball as an indenter. The ball is pressed into the surface of a test specimen using f

15、orces ranging from 9.8 N to 29,400 N (mass 1 kg to 3000 kg). A hard steel ball indenter is used. Indenter diameters may be 1, 2, 5 or 10 mm.Brinell hardness is determined from the formula:Hardness Number, HB = Where F = applied force in Newtons, D = diameter of indenter (mm) and d = diameter of inde

16、ntation (impression, mm).To make a Brinell test, the surface of the specimen should be flat and reasonably well polished. Care is required that the hardness is not affected by the procedures used in preparing the specimen.Brinell hardness test results are always reported as:xHBy / z, e.g. 250HB10/30

17、00where, x = hardness number, y = indenter diameter in mm, z = mass used in kg.(ii) The Vickers hardness testThis test utilises a square based pyramidal diamond indenter in which the angle between the opposite faces is 136o. The force may be varied from 49 N to 1176 N by varying the mass from 5 to 1

18、20 kg. Varying the mass has no effect on hardness values in the Vickers test because Vickers hardness is independent of ratio.A square pyramidal diamond indenter, with an included angle = 136o, between opposite faces is forced into the surface of a metal under a constant load for 15 s.Hardness numbe

19、r, HV = Where F = indenting force in Newtons and d = mean length of impression diagonals (mm)Vickers hardness test results are always reported as:X HV y / z, ( e.g. 250HV 30)where, x = hardness number, y = mass used in kgExample 11.1 p212Indentation hardness tests are made on a sample of metal and t

20、he following results recorded:(1) using Vickers diamond test with 30 kg load the mean lengths of diagonals were: 1st impression 0.527 mm, 2nd impression 0.481 mm, 3rd impression 0.497 mm; HD = 1.854F/d2. For the first impression, HD = (1.854 30)/ 0.5272 = 200. For the second impression HD = 240, for

21、 the third HD = 225. The average of three impressions gives HD = 221.7.(2) using Brinell test with 10 mm ball and 3000 kg load the diameter of impression was 4.01 mm.Calculate the diamond and Brinell hardness numbers for the material and explain any variations in the results.HB = (2 3000) 10 (10 -)

22、= 228The small size of the indenter used in the Vickers test means that in an alloy an impression may be on a non- representative hard or soft spot. A number of determinations may have to be made to obtain a true average value. Using a 10 mm diameter ball indenter the Brinell test gives a result mor

23、e representative of the mean hardness of the material.Example 11.1The indenting loads normally used in connection with Vickers diamond tests are: 1kg, 2.5 kg, 5 kg, 10 kg, 20 kg, 30 kg and 50 kg. If determinations are most accurate when impression diagonals are approximately 0.5 mm, what indenting l

24、oads should be selected for the testing of:(i) aluminium samples with hardness of the order of HD = 20,(ii) brass samples with hardness of the order of HD = 60, and(iii) Steel samples with hardness of the order of HD = 200?Example 11.3Samples of pure copper in both the annealed and cold-worked condi

25、tions were subjected to Brinell hardness tests, using a 1 mm diameter ball indenter, with various loads. The test data are given below.(1) Calculate the Brinell hardness of the copper samples.(2) Complete a Meyer analysis and determine the Meyer constants.MaterialIndenting load (kg)Indentation diame

26、ter (mm)Annealed copper510150.3860.5400.636Cold worked copper1020300.3750.5270.632(iii) The Rockwell hardness testThExample 11.1(iii) Meyer hardness analysisThe Meyer relationship for the Brinell test is F = adn where F is the load (kgf), d is the diameter of the indentation and a and n are constant

27、s of the material and its condition. The a is related to the resistance to indenter penetration and n is the work hardening index. The relationship can be written:log F = log a + n log d Using a ball indenter of fixed diameter and a series of loads, the data recorded can be plotted as log F against

28、log d. A straight line should be obtained and the values of a and n determined.There are other types of hardness tests: Scratch hardness tests which utilise materials of known hardness to scratch unknown test pieces or specimens. Wear hardness which is a measurement of resistance to wear under speci

29、fic conditions. Rebound hardness which is measure as energy absorbed under impact loads.Toughness Toughness is a measure of the amount of energy required to cause failure(fracture) of a specimen. One measure of toughness is the area contained beneath the stress/strain curve for the material. Toughne

30、ss is measured in joules(J) where1 J = 1N.m =103 N.mmThe product of stress and strain isN.mm-2 mm. mm-1 = N.mm.mm-3, i.e toughness per unit volume.The more common methods for measuring the toughness of a material are the different types of notched bar impact test, such as the Izod test and Charpy te

31、st used for metal and for the impact testing of polymers.Most materials become more sensitive to notches as the temperature is reduced. In particular ferrous metals undergo a relatively sudden reduction in toughness at a temperature referred to as its transition temperature- the transition temperature can vary from almost +100oC to -100oC depending upon the chemical composition and the metallurgical structure of the steel.ElasticityElasticity refers to the ability of a material to deform without undergoing a permanent set or permanent