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ENGR20003 Engineering Materials
Elementary 2D stress and strain and composite actions (Part 3) Massoud Sofi Dept. of Infrastructure Engineering Room
Learning objectives • Concept of stress and strain • Plane stress and strain • Mohr’s circle for plane stress • Composite materials • Material failure • Be able to describe ductile yielding, strain hardening and the effects of unloading/reloading of a ductile material beyond the yield limit. • Be able to describe how improvement in ductility results in better load-sharing behaviour in structures. • Be able to explain why a ductile material has better resistance to impact; and identify when ductility is particularly important in practice. By the end of this lecture, students should : Learning objectives • Definitions of strength and ductility; yielding by shear dislocation of crystal lattice; work hardening • Ductility and load sharing behaviour • Ductility and resistance to impact and dynamic actions Content Definition of strengths and Ductility F Yield Strength (Fy) Design Strength (Fd) Ultimate Strength (Fu) uy y u =Ductility Ratio Overstrength Ratio d u F F = Shear dislocation of crystal lattice Tensile force leading to shear stress and deformation Poisson Effect A piece of metal is made up of a large number of crystal lattices in small irregular blocks separated by grain boundaries a crystal grain Shear dislocation of crystal lattice Small stresses distort crystal Larger stresses cause slippages Shear dislocation of crystal lattice However, slip could propagate along the crystal lattice Shear dislocation of crystal lattice Work hardening A piece of metal is made up of a large number of crystal lattices in small irregular blocks separated by grain boundaries a crystal grain Crystal lattices typically have random orientation When shear stress is applied, slips only occur in some crystal lattices. s e Yielding commences when slip occurs Work hardening More crystal lattices slip as Stress is increased. s e Further yielding Work hardening Sequence depends on orientation. Which crystal lattice would slip next ? s e Further yielding Work hardening Work hardening by unloading and reloading s e Unloading and reloading On unloading and reloading : material is stronger and less ductile new ductility range new yield strength Ductile versus Brittle behaviour F Ductile Behaviour Brittle Behaviour F Examples : mild steel, reinforced concrete (with good seismic detailing) Examples : glass and ceramics, aluminium reinforced concrete with poor detailing • Definitions of strength and ductility; yielding by shear dislocation of crystal lattice; work hardening • Ductility and load sharing behaviour • Ductility and resistance to impact and dynamic actions Content Hangers of different lengths ( ) AE W 2 2 = ( ) AE W = 2 Stresses and strains on the two hangers are the same = = 2 2 e es E= A = sectional area; E = Young’s modulus w w 2 Load sharing between the hangers 2 Stresses and strains on the two hangers are now very different 2 e s e s 2w With both hangers made of brittle material 2 e s e s 2w With both hangers made of ductile material What will happen to the stress in each hanger if the load is increased further ? The shorter hanger is now spared from failure Load sharing between the hangers Example 2 W Two hangers are made up of grey cast iron with ultimate tensile strength of 275 MPa. The Young’s modulus value of cast iron is 170000 MPa. Both hangers have the same cross sectional area of 100 mm2. a) Determine the weight W that can be supported by the hangers. b) Determine the force that is acting on each hanger before failure. c) Determine the weight W that can be supported by the hangers if the hangers are ductile (assume that the ultimate strength is now the yielding strength of the material). Load sharing between the hangers A1 = A2 E1 = E2 l1 = 2l2 For compatibility, 1 = 2 As e = / l e1 = 1/2 e2 Example 2 W 1 2 = Cable 2 will fail first at 2 = 275 170000 = 0.0016 At that time, the strain in Cable 1 1 = 0.0008 Load sharing between the hangers Example 2 W 1 2 = 1 + 2 a) The weight that can be supported by the hangers: = 170000 × 100 0.0016 + 0.0008 = 40800 N = 40.8 kN b) The force acting on each hanger before failure: 1 = 1 = 170000 × 100 × 0.0008 = 13.6 kN 2 = 2 = 170000 × 100 × 0.0016 = 27.2 kN c) When the hangers are ductile both hangers can reach their yielding strength = 2 = 2 × 275 × 100 = 55 kN = = Load sharing between the hangers Another load sharing example = 2 e e s e s With both ties made of brittle material Weight blocks =e Upper tie Lower tie es e s With both ties now made of ductile material The cantilever has extra capacity to take more weight blocks Another load sharing example Load sharing with a cable With ties made of brittle material Collapse !! The structure would have been able to take more loads had the ties been made ductile. Are you able to explain this by stress- strain diagrams ? Reinforced concrete without ductile detailing V P Examples of reinforced concrete with ductile detailing Column bent during an earthquake • Definitions of strength and ductility; yielding by shear dislocation of crystal lattice; work hardening • Ductility and load sharing behaviour • Ductility and resistance to impact and dynamic actions Content Meaning of “strong” and “weak” Strong Weak Meaning of “strong” and “weak” F F Meaning of “strong” and “weak” Mass of hammer = M Velocity of hammer = V K.E. = ½ M V2 Area < ½ M V2 F F Area = ½ M V2 Meaning of “strong” and “weak” F F Area = ½ M V2 Hard to push the nail in by hand --- wood appears to be very strong Nail can be easily driven in by hammer -- wood is not all that strong Meaning of “strong” and “weak” Example Two hangers are made up of grey cast iron with ultimate tensile strength of 275 MPa. The Young’s modulus value of cast iron is 170000 MPa. Both hangers have the same cross sectional area of 100 mm2. The hangers are supporting a rigid plate. a) Determine the mass M that can be supported by the hangers, if the mass is to be dropped from 1 m height. The weight of the plate can be ignored. b) If the hangers are now made up of a ductile material . The yield strength is 275 MPa, the Young’s modulus value is 170000 MPa, ductility is 2. Determine the mass that can be supported by the hangers if it is to be dropped from 1 m height. M 1 m 1.5 m Ductility and resistance to impact actions Example M 1 m 2 m Potential energy = strain energy M g h = area under the curve of F- relationship a) = 275 × 100 = 27500 N At ultimate condition = ∆ ∆ = = 27500 2000 100 × 170000 = 3.2mm F- relationship for the two hangers F u = 3.2 mm 2Fu = 55000 N Strain energy = ½ 55000 3.2 = 88000 Nmm = 88 Nm Potential energy = strain energy M g h = 88 M = 88/(9.81 x 1)= 9 kg Ductility and resistance to impact actions Example M 1 m 2 m Potential energy = strain energy M g h = area under the curve of F- relationship a) = 27500 N Hangers are ductile ∆ = = 27500 2000 100 × 170000 = 3.2mm F- relationship for the two hangers F y = 3.2 mm 2Fy = 55000 N Strain energy = ½ 55000 3.2 + 3.2 x 55000 = 264000 Nmm = 264 Nm Potential energy = strain energy M g h = 264 M = 264/(9.81 x 1)= 27 kg Ductility is 2, u = 2 x 3.2 =6.4 mm u = 6.4 mm Ductility and resistance to impact actions Time to think • What is really meant by being “strong” in the context of load and impact. • When is strength important and when is ductility important. • More examples … Bucket gradually filled up with sand Point of failure Strength is important ---- ductility not important F Dropping buckets of sand Both the ductility and strength are important F Another example when ductility is important Yet another example when ductility is important ….. in an earthquake Kinetic energy is ½ MV2 Strength and Ductility (toughness) F Toughness is a measure of the strain energy absorption capacity up to the point of fracture high strength but low ductility (low toughness) lower strength but higher ductility (higher toughness) ie. better impact resistance End of Lecture on Strength and Ductility Now look at the subject objectives and refresh how much you have learnt from the lecture End of Lecture