Chapter three. Laith Batarseh

Transcription

1 Next Previous 3/23/2014 Chapter three Laith Batarseh Home End Tensile test Tensile test is a test performed on material to define its strength to tensile axial loads. The machine used is called the tensile tester or the tensile test machine. The main objective from the tensile test is to relate between the generated stress inside the material (due to the application of external loads or forces) and the strain (i.e. deformation) results from the application of the external force. 1

2 Tensile machine and specimen Tensile machine Tensile specimen Tensile test procedures 1:- chose a suitable specimen and put two gauge points on the specimen 2:- measure the initial diameter (Do) and initial length (Lo) of the specimen 3:- fix the specimen on the machine 4:- start to apply a tensile force (P) on the specimen for certain consecutive periods of time 5:- measure the deflection in length (δl = L L o ) for each value of P 6:- calculate the stress (σ) using the following formula: 7:- calculate the strain (ε) using the following formula: P A L Lo 8:- continuo until the material break down (fracture point) 9:- finally, draw σ Vs. ε to have stress-strain diagram 2 ; Ao Do o 4 2

3 Stress-strain diagram of ductile materials Ductile material is a material needs large strain to reach the break point. Mild steel, aluminum, brass and other metals are classified as ductile materials. The stress strain diagram for such materials is shown in the figure. Stress-strain diagram of ductile materials We can divide the stress strain diagrams into four regions: Elastic region. Yield region. Strain hardening region Necking region 3

4 Elastic region. We can conclude about this region:- the relation between the stress and the strain is linear if the material is unloaded, the material go back to its initial condition (i.e. zero strain). This region starts from the origin (0,0) to elastic limit. There is other value of stress is in our interest: the proportional limit. The proportional limit is the value of stress where the linear proportional region end. This dose not mean that the elastic region end at this value but it will continue until the elastic limit but not in linear fashion. In many cases, its difficult to find the elastic limit so the yield strength (which is the beginning of the next region) is taken instead. Yield region. We can conclude about this region:- This region starts from the yield stress (σ y ). From this region the material enters the plastic region which means if the load is gone the material will not return back to its original condition and there will be residual strain (i.e. permanent deformation) in the material. As its illustrated from stress strain diagram, in this region no additional stress is needed to continuo the strain (i.e. constant stress, incensing strain ). In the previous stress diagram, this region was enlarge to explain its significant rule in our analysis. 4

5 Strain hardening region. We can conclude about this region:- In this region, the material struggles to strain. The relation between the stress and the strain is none liner proportional relation. The material fight the loading until it reach its maximum strength (σ u ) which is called the ultimate strength or the ultimate tensile strength (UTS). Necking region. We can conclude about this region:- when the material exceeds its maximum strength, the bonding between the material crystal starts to break and a necking (reduction in the area) starts to appear in it. this is the start of the of the end of the material and the necking continuo until there is no enough area to handle the stress and the fracture occurs. Fractured specimen 5

6 Tensile test Practical considerations True stress strain diagram Due to the reduction occurred in the material in the period of the tensile test, there is a value of stress is different from the one we assumed previously (P/A o ). assume we have the ability to measure the area through the test ( call it A), then a true value of stress can be determined (σ T = P/A). If σ T is used instead of σ in drawing a stress strain diagram, this diagram is called true stress strain diagram. In many cases and because as engineers we design in the elastic region, we use the first type of stress strain diagram which is called engineering stress diagram to define the material properties. You can observe that both true and engineering stress strain diagrams share the same elastic region. Realistic stress strain diagram for mild steel 6

7 Ductile and brittle materials As said before, ductile material is a material needs large strain to reach the break point. On the contrary, brittle material is a material can not hold a large to reach fracture. Glass is one of the famous examples on such materials. There is an important note about some types of ductile materials which is:- there are some materials that have a stress-strain diagrams where the yield strength can not be determined easily. For such cases, the offset method is used. This method is shown in the figure for the aluminum alloy. Observe the strain value chosen for the aluminum alloy (i.e. ε = =0.0%) also observe the line drawn from this value parallel to the elastic line Percents of elongation and reduction of area For ductile material Percent of Percent of elongation L f L L reduction of area o o A o A A o f 7

8 Hooke s Law Hooke s law is a mathematical formula represents the linear relation between the stress and the strain in the elastic region. Mathematically:- σ = E.ε The proportional constant (E) is called modulus of elasticity or Young s modulus of elasticity. In addition, E represents the mathematical slope of the elastic line. So, after drawing the stress strain diagram, to find the modulus of elasticity, find the slope of the elastic line. By mathematics: Where: σ p is the stress at the proportional limit ε p is the strain at the proportional limit p E p Plastic deformation As said before, when a load exceed the elastic region ( or enter the plastic region) and then be released ( unloaded ), the material will remain with plastic deformation or plastic strain. To find the value of this stain, we use the offset method as illustrated in the next figure As you can see, we exceed the elastic region when the stress become 600 MPa while the yield strength is 450 MPa. Observe the line drawn from 600 MPa stress parallel to the elastic line 8

9 Strain energy There are some concepts related to the strain and the energy given to (or taken from), u, the material when the load is applied (or released). These concepts are: modulus of resilience and modulus of toughness. Modulus of resilience is the strain energy density when the material is loaded in the elastic region. Simply, it is the area under the elastic line. Or: 2 1 p ur 1 p p 2 2 E Modulus of toughness is the strain energy given to or taken from the material through the whole loading period (i.e. until fracture). Mathematically, it is the area under the stress strain diagram. Such modulus is important when design a member that may suddenly loaded over the elastic region. EXAMPLE 1 The stress strain diagram for an aluminum alloy that is used for making aircraft parts is shown in Fig If a specimen of this material is stressed to 600 MPa, determine the permanent strain that remains in the specimen when the load is released. Also, find the modulus of resilience both before and after the load application. 9

10 EXAMPLE 1 (cont) Solutions When the specimen is subjected to the load, the strain is approximately mm/mm. The slope of line OA is the modulus of elasticity, From triangle CBD, 450 E 75.0 GPa BD E CD CD CD 0.008mm/mm EXAMPLE 1 (cont) Solutions This strain represents the amount of recovered elastic strain. The permanent strain is OC mm/mm (Ans) Computing the modulus of resilience, u r 1.35MJ/m (Ans) 3 u MJ/m (Ans) r initial final pl pl pl pl Note that the SI system of units is measured in joules, where 1 J = 1 N m. 3 10

11 11

12 Poisson s ratio When the specimen or the material is subjected to a tensile load, the elongation occurred will results also as a reduction in the area (or the specimen diameter). This reduction is also strain however, it is in the lateral direction. This process is illustrated in the figure. The relation between the axial strain and the lateral strain by the Poisson s ratio (ν) which is given as Lat D / Do L / L long D D D L L L o o o 12

13 EXAMPLE 2 A bar made of A-36 steel has the dimensions shown in Fig If an axial force of P = 80kN is applied to the bar, determine the change in its length and the change in the dimensions of its cross section after applying the load. The material behaves elastically. Copyright 2011 Pearson Education South Asia Pte Ltd EXAMPLE 2 (cont) Solutions The normal stress in the bar is z P A Pa From the table for A-36 steel, E st = 200 GPa z z E st mm/mm The axial elongation of the bar is therefore m (Ans) z z L z Copyright 2011 Pearson Education South Asia Pte Ltd 13

14 EXAMPLE 2 (cont) Solutions The contraction strains in both the x and y directions are v x y st z m/m The changes in the dimensions of the cross section are L x L y x y x y m (Ans) m (Ans) Copyright 2011 Pearson Education South Asia Pte Ltd Shear stress strain diagram The shear stress strain diagram is drawn where the shear stress is the vertical axis and the shear strain is the horizontal axis as shown in the figure As in the stress strain diagram, there is elastic region in the shear stress strain diagram and the Hook s law is applied in this region : τ = G.γ Where: G is the shear modulus there is relation between G an E and is given as: E G

15 EXAMPLE 3 A specimen of titanium alloy is tested in torsion and the shear stress strain diagram is shown in Fig. 3 25a. Determine the shear modulus G, the proportional limit, and the ultimate shear stress. Also, determine the maximum distance d that the top of a block of this material, shown in Fig. 3 25b, could be displaced horizontally if the material behaves elastically when acted upon by a shear force V. What is the magnitude of V necessary to cause this displacement? Copyright 2011 Pearson Education South Asia Pte Ltd EXAMPLE 3 (cont) Solutions By inspection, the graph ceases to be linear at point A. Thus, the proportional limit is pl This value represents the maximum shear stress, point B. Thus the ultimate stress is 504 MPa (Ans) u 360MPa (Ans) Since the angle is small, the top of the will be displaced horizontally by tan d 50mm 0.008rad d 0.4 mm Copyright 2011 Pearson Education South Asia Pte Ltd 15

16 EXAMPLE 3 (cont) Solutions The shear force V needed to cause the displacement is avg V V ; 360MPa V 2700kN (Ans) A Copyright 2011 Pearson Education South Asia Pte Ltd 16