Simple Stresses and Strains PDF

# Simple Stresses and Strains ## **Page 1** **Solution:** For the type of given loading, a change in the length of wire is given by; $\frac{8}{\pi} \frac{\Delta L}{E}$ **For the steel wire,** $\frac{P \times 2000}{4(\frac{3}{2})^2 \times (2 \times 10^5)} = 0.001415 P$ **Weight suspended P =...

# Simple Stresses and Strains ## **Page 1** **Solution:** For the type of given loading, a change in the length of wire is given by; $\frac{8}{\pi} \frac{\Delta L}{E}$ **For the steel wire,** $\frac{P \times 2000}{4(\frac{3}{2})^2 \times (2 \times 10^5)} = 0.001415 P$ **Weight suspended P =** $0.75 \times 0.001415 P$ = 530 N **For the brass wire,** $4.65 = \frac{530 \times 2500}{π(2)^2 x E_b}$ **Ep =** 421974 **Modulus of elasticity for the brass wire,** $E_b = \frac{421974}{4.65} = 90747 N/mm^2$ ## 21.4. STRESS-STRAIN DIAGRAM - **Stress-strain curve** is a graphical plot of stress-versus strain. - These quantities are experimentally obtained by subjecting a metallic bar of uniform cross-section to a gradually increasing tensile load till failure of the bar occurs. - The test is conducted in a tensile testing machine on a test specimen having the appearance/configuration as shown in Fig. 21.4. - The specimen has collars provided at both the ends for gripping it firmly in the fixtures of the machine. - The central portion of the test specimen is somewhat smaller than the end regions, and this central section constitutes the gauge length over which elongations are measured. - An extensiometer (dial gauge) is used to measure very small changes in length. - After that vernier scale on the machine is used to measure extension. - Load and extension are simultaneously recorded till the specimen breaks. - **Stress** is calculated by dividing the load by the original cross-sectional area of the test specimen. - **Strain** is calculated by dividing the extension of a given length (gauge length) by the original unstrained length. - That is: $\epsilon = \frac{\delta l}{l}$ and stress $\sigma = \frac{P}{A}$ - **Stress-strain curve** is then plotted by having obtained numerous pairs of values of stress and strain; stress plots as ordinate and strain as abscissa on a graph taking suitable scale. ## **Page 2** **Figure 21.5 shows the typical behaviour of stress-strain curve for mild specimen and its salient features are :** * **Proportional limit:** Stress is a linear function of strain and the material obeys Hook's law. * This proportionality extends upto point A and this point is called proportional limit or limit of proportionality. * 0-A is a straight line portion of the curve-and its slope represents the value of modulus of elasticity. * **Elastic limit:** Beyond proportional limit, stress and strain depart from straight line relationship. * However, the material remains elastic upto state point B. * The word elastic implies that the stress developed in the material is such that there is no residual or permanent deformation when the load is removed. * Upto to this point, the deformation is reversible or recoverable. * Stress at B is called the elastic limit stress, this represents the maximum unit stress to which a material can be subjected and is still able to return to its original form upon removal of load. * **Yield point:** Beyond elastic limit, the material shows considerable strain even though there is no increase in load or stress. * This strain is not fully recoverable, i.e., there is no tendency of the atoms to return to their original positions. * The behaviour of the material is inelastic and the onset of plastic deformation is called yielding of the material-Yielding pertains to the region C-D and there is drop in load at the point D. * The point C is called the upper yield point and point D is the lower yield point. * The difference between the upper and lower yield point is small and the quoted yield stress is usually the lower value. * **Ultimate strength or tensile strength:** After yielding has taken place, the material becomes strain hardened (strength of the specimen increases) and an increase in load is required to take the material to its maximum stress at point E. * Strain in this portion is about 100 times than that of the portion from 0 to D. * Point E represents the maximum ordinate of the curve and the stress at this point is known either as ultimate stress or the tensile stress of the material. * **Breaking strength:** In the portion EF, there is falling off the load (stress) from the maximum until fracture takes place at F. * The point F is referred to as the fracture or breaking point and the corresponding stress is called the breaking stress. ## Page 3 **The apparent fall in stress from E to F may be attributed to the fact that stress calculations are made on the basis of original cross-sectional area. In fact elongation of the specimen is accompanied by reduction in cross-sectional area and this reduction becomes significant near the ultimate stress. In case stress calculations are based on actual area, the curve would be seen to rise until fracture occurs. For mild steel, the test piece breaks making a cup and cone type fracture; the two pieces can be joined together to find out the diameter (actual area) at the neck under the specimen breaks.** **For many ductile materials other than mild steel, e.g., aluminium, copper etc. no definite yield point is obtained. For such materials, the strain-strain curve plots as shown in Fig. 21.6.** **For brittle materials, like cast iron, no appreciable deformation is obtained and the failure occurs without yielding (Fig. 21.7).** * **Proof stress:** Quite often it is desired to determine the stress at which a specified permanent extension takes place in a tensile test. * The extension specified is usually 0.1, 0.2 or 0.5 percent of gauge length. Such a stress is known as proof stress. * For determining 0.2% proof stress from stress-strain curve, a point G representing 0.2%, i.e., 0.002 is marked on the strain axis. * A line GH is then drawn parallel to initial slope line OA. * The stress at the point where this line cuts the curve is 0.2% proof stress.. * **Working stress and safety factor:** Working stress is the allowable stress for design purpose. * During design of an element, it is to be kept in mind that actual stress developed in the element does not exceed the working stress. * Frequently such a stress is determined by dividing either the yield stress or the ultimate stress by a number termed the safety factor. * The safety factor accounts for: * Internal flows in the material, * Stress concentration, * Uncertainities about the magnitude and nature to which the machine element is subjected, etc. * The value of safety factor depends upon the judgement and experience of the designer and is usually governed by : * Type of loading, * Reliability of material and, manufacturing processes such as castings and forgings, * Extent of damage which will be caused if the machine element fails. ## Page 4 **The allowable or working stress used in design calculations is taken as** **Working stress =** $\frac{ultimate stress}{safety factor}$ **Material Classification :** Materials are commonly classified as : * **(i) Homogeneous and isotropic material:** A homogeneous material implies that the elastic properties such as modulus of elasticity and Poisson's ratio of the material are same everywhere in the material system. * Isotropic means that these properties are not directional characteristics, i.e., an isotropic material has same elastic properties in all directions at any one point of the body. * **(ii) Rigid and linearly elastic material:** A rigid material is one which has no strain regardless of the applied stress. A linearly elastic material is one in which the strain is proportional to the stress. * **(iii) Plastic material and rigid-plastic material:** For a plastic material, there is definite stress at which plastic deformation starts. A rigid-plastic material is one in which elastic and time-dependent deformations are neglected. The deformation remains even after release of stress (load). * **(iv) Ductile and brittle material:** A material which can undergo 'large permanent' deformation in tension, i.e., it can be drawn into wires is termed as ductile. A material which can be only slightly deformed without rupture is termed as brittle. **Ductility of a material is measured by the percentage elongation of the specimen or the percentage reduction in cross-sectional area of the specimen when failure occurs. If I is the original length and l' is the final length, then** % increase in length = $\frac {l'-l}{l}$ x 100 **The length l' is measured by putting together two portions of the fractured specimen. Likewise if A is the original area of cross-section and A' is the minimum cross-sectional area at fracture, then** % age reduction in area = $\frac {A- A'}{A}$ × 100 **A brittle material like cast iron or concrete has very little elongation and very little reduction in cross-sectional area. A ductile material like steel or aluminium has large reduction in area and increase in elongation. An arbitrary percentage elongation of 5% is frequently taken as the dividing line between these two classes of material.** **Example 21.6. The following data was recorded during tensile test made on a standard tensile test specimen :** ## Page 5 **Original diameter and gauge length = 15 mm and 60 mm; minimum diameter at fracture = 10 mm; distance between gauge points at fracture = 75 mm; load at yield point and at fracture = 40 kN and 45 kN; maximum load that specimen could take = 70 kN.** **Make calculations for (a) yield strength, ultimate tensile strength and breaking strength (b) percentage elongation and percentage reduction in area after fracture (c) nominal and true stress and fracture.** **Solution:** - **Original area Ap=** $\frac{\pi}{4}$ (15)2 = 176.625 mm² - **Final area A, =** $\frac{\pi}{4}$ (10)2 = 78.5 mm² (at fracture) **a) Yield strength = ** $\frac{40 x 10^3}{176.625}$ V= 226.47 N/mm² = 226.47 × 10^6 N/m² = 226.47 MPa **Ultimate tensile strength =** $\frac{70 x 10^3}{176.625}$ = 396.32 N/mm² = 396.32 × 10^6 N/m² = 396.32 MPa **Breaking strength =** $\frac{45 × 10^3}{ 176.625}$ = 254.78 N/mm² = 254.78 × 10^6 N/m² = 254.78 MPa **It may be noted that calculations for all the strengths are made on the basis of original area at the test section** **b) Percentage elongation =** $\frac{75-60}{60}$ × 100 = 25% **Percentage reduction in area =** $\frac{176.625-78.5}{176.625}$ × 100 = 55.56% **c) Nominal stress =** $\frac{load at fracture}{original area}$ = $\frac{45 × 10^3}{176.625}$ = 254.78 N/mm² **The nominal strength and the breaking strength are synonymous** **True stress =** $\frac{load at fracture}{final areá}$ = $\frac{45 x 10^3}{78.5}$ = 573.25 ## **21.5. EXTENSION OF A TAPERED BAR** **Consider a circular bar that tapers uniformly from diameter d₁ at the bigger end to diameter d₂ at the small end, and subjected to axial tensile load P (Fig. 21.11).** **Let attention be focused on an elementary strip of length dx at distance x from the bigger end. Diameter of the elementary strip,** $dx = d₁ - (d₁ - d₂) \frac{x}{l}$ = $d₁ − k x$ where k = $(d₁ – d₂)/l$