# NORMAL STRESS. The simplest form of stress is normal stress/direct stress, which is the stress perpendicular to the surface on which it acts.

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1 NORMAL STRESS The simplest form of stress is normal stress/direct stress, which is the stress perpendicular to the surface on which it acts. σ = force/area = P/A where σ = the normal stress P = the centric axial load A = the area of the section The normal stress is usually expressed in pascals (Pa), where one pascal is equal to one newton per square metre, that is, 1 Pa = 1 N/m 2. A pascal is a very small unit of stress, so one can usually expect to see stresses expressed in kpa or MPa. Page 5-1

2 Example 1 A steel bar of rectangular cross-section, 3 cm by 2 cm, carries an axial load of 30 kn. Estimate the average tensile stress over a normal cross-section of the bar. Solution The area of a normal cross-section of the bar is A = 0.03*0.02 = 0.6 x 10-3 m 2 The average tensile stress over this cross-section is then σ = P/A = 30 x 10 3 / 0.6 x 10-3 = 50 x 10 6 N/m 2 = 50 MPa Page 5-2

3 HOOKE S LAW For many materials, the lower end of the stress-strain curve is a straight line. This behaviour was recognised by Robert Hooke and stated as Hooke s Law. Hooke s Law: For an elastic body, stress is proportional to strain. σ = E ε The constant E is called the elastic modulus, modulus of elasticity, or Young s modulus. E is equal to the slope of the stress-strain curve. E = stress/strain = σ / ε Since strain is dimensionless, E has the same units as stress, e.g. Pa, MPa. The value of E for a given material is a constant. Materials with a high modulus of elasticity have a high resistance to elastic deformation, and are said to be stiff. Page 5-3

4 NORMAL STRAIN A member carrying a tensile load will stretch. The stretch is usually called deformation, and the symbol for deformation is δ. The strain is dimensionless. ε = δ / L where ε = the normal strain δ = the normal deformation L = the original length of the member before deformation From Hooke s law, σ = E ε When the stress and strain are caused by axial loads, we have P/A = E* ( δ / L ) δ = PL/AE Page 5-4

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6 TENSILE TEST The mechanical properties of materials used in engineering are determined by tests performed on small specimens of the material. The tests are conducted in materials-testing laboratories equipped with testing machines. A tensile test machine is shown as below:- A stress-strain diagram for a structural steel in tension is shown in the following figure. Page 5-6

7 There are several points of interest, which can be identified on the curves as follows: 1. Proportional limit The maximum stress for which stress is proportional to strain. That is, stress at point A. 2. Elastic limit Maximum stress that can be applied to a material without producing a permanent plastic deformation or permanent set when the load is removed. That is, stress at point B. 3. Yield point Stress for which the strain increases without an increase in stress. That is, the horizontal portion of the curve BC. 4. Ultimate strength Maximum stress material can support up to failure. That is, the stress at point D. At this point the test piece begins, visibly, to neck. The material in the test piece in the region of the neck as almost perfectly plastic at this stage and from thence, onwards to fracture, there is a reduction in nominal stress. 5. Breaking strength Stress in the material based on original cross-sectional area at the time it breaks. It is also called fracture or rupture strength. That is, the stress at point E. Page 5-7

8 Stress-strain curves for other materials Different materials have different stress-strain curves. The following stress-strain curves for aluminum alloy and concrete. Aluminum alloy It is a ductile material, which does not have a yield point. A line drawn parallel to the linear portion of the stress-strain curve from a strain of (i.e. 0.2%) intersects the stress-strain curve. The intersection point is defined as a yield point. Concrete It is a brittle material. Page 5-8

9 POISSON S RATIO When a load is applied along the axis of a bar, axial strain is produced. At the same time, a lateral (perpendicular to the axis) strain is also produced. If the axial force is in tension, the length of the bar increases and the cross-section contracts or decreases. That is, a positive axial stress produces a positive axial strain and a negative lateral strain. For a negative axial stress, the axial strain is negative and the lateral strain is positive. The ratio of lateral strain to axial strain is called Poisson s ratio. It is constant for a given material provided that the material is not stressed above the proportional limit, is homogeneous, and has the same physical properties in all directions. ν = lateral strain / axial strain = - lateral strain / axial strain The negative sign ensures that Poisson s ratio is a positive number. The value of Poisson s ratio, ν, varies from 0.25 to 0.35 for different metals. For concrete, it may be as low as ν = 0.1 and for rubber as high as ν = 0.5. Page 5-9

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11 Example 2 An aluminum alloy sample is tested in tension. When the stress is 150 MPa the normal strain is 2.1 x 10-3 m/m. Calculate the modulus of elasticity for this alloy. Solution The modulus of elasticity: E = σ / ε = 150 x 10 6 /2.1 x 10-3 = x 10 9 Pa = GPa Example 3 A 2m long round bar of polystyrene plastic with a diameter of 25 mm carries a 5 kn tensile load. If the modulus of elasticity of the polystyrene is 3.1 GPa, calculate the longitudinal deformation in the bar. Solution The deformation, δ = εl = σl/e = PL/AE = 5000*2/[(π*25 2 /4) x 10-6 *3.0 x 10 9 ) = 6.79 x 10-3 m = 6.79 mm Page 5-11

12 Example 4 Accurate experimental measurements in a compression test of a 200 mm long square sample with a 50 by 50 mm cross section give a longitudinal deformation of 0.1 mm and a transverse deformation of mm. Determine Poisson s ratio for the material. Solution To calculate Poisson s ratio, it is necessary to first determine both the longitudinal and transverse strains: ε l ε t = δ l / L = -0.1/200 = m/m = δ t / L = 0.008/50 = m/m Now Poisson s ratio may be calculated: ν = ε t /ε l = / = 0.32 Page 5-12

13 Example 5 A steel bar has the dimensions shown in the following figure. If an axial force of P = 80 kn 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. Take E st = 200 GPa and ν st = 0.3. The material behaves elastically. Solution The normal stress in the bar is, σ Z = 80 x 10 3 / (0.1*0.05) = 16 x 10 6 Pa = 16 MPa Thus the strain in the z direction is, ε Z = σ Z / E st = 16 / 200 x 10 3 = 80 x 10-6 The axial elongation of the bar is therefore, L Z = ε Z L Z = 80 x 10-6 *1500 = 120 x 10-6 m = 120 µm Page 5-13

14 The contraction strains in both x and y directions are, ε x = ε y = -ν st ε Z = -0.3* 80 x 10-6 = -24 µm/m Thus the changes in the dimensions of the cross section are, L x = ε x L x = -24 x 10-6 *0.1 = -2.4 µm L Y = ε Y L Y = -24 x 10-6 * 0.05 =-1.2 µm Example 6 A steel rod is loaded as shown in the following figure. Determine the deformation of the steel rod. Page 5-14

15 Solution We divide the rod into the three-component parts as indicated in the following figures. To find the internal forces P 1, P 2 and P 3, we must cut sections through each of the component parts, drawing each time the free-body diagram of the portion of rod located to the right of the section. Expressing that each of the free bodies is en equilibrium, we obtain successively. P 1 = 400 kn P 2 = -100 kn = 200 kn P 3 P = il δ i i A E i i = 400 x 10 3 *0.3 /(600 x 10-6 * 200 x 10 9 ) + (-100 x 10 3 *0.3 /(600 x 10-6 * 200 x 10 9 ) +200 x 10 3 *0.4 /(200 x 10-6 * 200 x 10 9 ) = 2.75x 10-3 m = 2.75 mm Page 5-15

16 ALLOWABLE STRESS The allowable stress is the maximum stress that is considered safe for a material to support under certain loading conditions. The stress may be used to design load-supporting members of structures and machines. Allowable stress values are determined by tests and from experience gained from the performance of previous designs under service conditions. Allowable stress is also sometimes called the working or design stress. FACTOR OF SAFETY The factor of safety is defined as the ratio of some load that represents the strength for the member to the allowable load for the member. That is, Factor of safety, F. S. = ultimate load for the member allowable load for the member For tension member, where the load is equal to stress multiplied by area, the ratio of the loads is identical to the ratio of stresses. Accordingly, for a tension member a factor of safety that is based on the ultimate stress is equal to the ratio of the ultimate stress to the allowable stress. Thus, F.S. = σ u / σ a Values of the factor of safety used to design members depend on many factors. Among these are the nature of the loads, variation in material properties, types of failures, uncertainly in analysis and the environment to which the member is exposed. Factors of safety range in value from over 1 to 20 with values between 3 and 15 common. Page 5-16

17 Example 7 A hollow cylinder is to be designed to support a compressive load of 650 kn. The allowable compressive stress σ a = 69.2 MPa. Compute the outer diameter of the cylinder if the wall thickness is 50 mm. Solution Solving for the required area, we have A req = P/σ a = 650 x 10 3 /69.2 = 9393 mm 2 Considering a hollow cylinder, the area is equal to A req = π * (d o 2 d i 2 ) /4 where d o = outer diameter of the hollow cylinder d i = inner diameter of the hollow cylinder Therefore, A req = π * (d o 2 d i 2 ) /4 = 9339 mm 2 (1) Given that the wall thickness is 50 mm, d o d i = 100 mm (2) By solving the above equations 1 and 2, d o = mm Page 5-17

18 STRESS CONCENTRATIONS The formula σ = P/A for the stress in an axially loaded bar is based on a uniform stress distribution over the cross-sectional area of the bar. If the loads are applied through two rigid plates at the two ends of the bar, this assumption of uniform stress distribution is reasonable. But if the load is concentrated at a point, very high local stresses can appear three are stress will not be uniformly distributed. The high local stresses are known as stress concentrations. Page 5-18

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20 SAINT-VENANT S PRINCIPLE At a point directly under the load, the maximum stress may be several times the average stress σ ave = P/A. As we move away from the point of load application, the maximum stress drops rapidly. At a distance b (b =the width of the bar) away from the end of the bar, the stress distribution becomes nearly uniform. The above observation illustrates Saint-Venant s principle, which states that: At a reasonable distance away from the loaded region, the effects of local stress concentration become unimportant. The reasonable distance may be taken to be the largest dimension of the loaded region (in the above case, it is the width of the bar). Saint-Venant s principle applied not only to axial loads but also to practically any type of load. The practical meaning of this important principle is that stresses can be calculated by the usual mechanics of materials formulas (such as σ = P/A), except in the immediate neighbourhood of loaded regions or changes in geometry. Page 5-20

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22 PROBLEMS 1. A 10 m long tie rod stretches 4.7 mm. Calculate the strain in the tie rod. Ans. ε = x 10-3 m / m 2. A 600 mm long concrete sample has a deformation of 1.70 mm when it fails under an axial compressive load. Determine the average normal strain in the sample at failure. Ans. ε = m/m 3. The strain in a 4 m long steel tension member was found to he 450 x 10-6 m/m. Calculate the total deformation in the member. Ans. δ = 1.80 mm 4. A member subject to an axial load of 25 kn compression has a strain of m/m. If the original length of the member was 0.40 m, determine the deformation of the member. Ans. δ = mm 5. A strip of bronze 5 by 20 mm carries an axial tensile load of 30 kn. In a length of 200 mm, there is a deformation of mm. Calculate the modulus of elasticity for the bronze. Ans. E = 117 GPa 6. A 50 m steel surveyor s chain has a length of m when an axial tensile load is applied. Find the percent deformation in the chain. Ans. % deformation = % 7. Determine the modulus of elasticity for steel piano wire if it has a strain of 2.62 x 10-3 m/m when the stress in it is 550 MPa. Ans. E = 210 GPa Page 5-22

23 8. If a medium-strength concrete has a modulus of elasticity of 20 GPa, find the strain in the concrete when the stress is -8.0 MPa. Ans. ε = x 10-3 m/m 9. A sample of platinum was tested in tension. The longitudinal strain was found to be x 10-3 m/m and the transverse strain was x 10-3 m/m when the normal stress was 120 MPa. Determine Poisson's ratio for platinum. Ans. ν = When the shear stress in invar was found to be 75 MPa, the corresponding shear strain was 1.33 x 10-3 m/m. Determine the shear modulus of elasticity for invar. Ans. G = 56.4 GPa 11. Determine the shear modulus of elasticity for zinc if E = 110 GPa and ν = [Note: E = 2 G(1+ ν ) ] Ans. G = 44.0 GPa Page 5-23

24 PROBLEMS 1. A tie bar is 2 m in length, has a circular cross-section of 19 mm diameter and carries a longitudinal load of 35 kn. Calculate the stress in the bar and the change in length (E = 200 kn/mm 2 ). Ans. ( N/mm 2, 1.23 mm) 2. A steel tie is 1.4 m long, has a cross-sectional area of 110 mm 2 and carries a tensile load of 10.5 kn. If the value of Young's Modulus of Elasticity (E) is 200 kn/mm 2 and Poisson's ratio ν = 0.3, calculate: (i) the direct tensile stress, (ii) the longitudinal strain, (iii) the lateral strain and (iv) the change in length. Ans. (i) N/mm 2, (ii) 0.48 x 10-3, (iii) x 10-3, (iv) 0.67 mm) 3. A hollow cylindrical steel tube with an outer diameter of 300 mm is to be used as a column to carry a vertical load of 2000 kn. If the direct stress in the steel is not to exceed 120 N/mm 2, calculate: (i) the thickness of metal required in the wall of the tube (Hint: calculate the internal diameter) and (ii) the change in external diameter under load. Assume that E = 200 kn/mm 2 and ν = 0.3. Ans. (i) mm, (ii) mm) 4. A steel tie bar 1.1 m long and 50 mm diameter is subject to a tensile stress of 120 N/mm 2. Determine: (i) the extension (ii) the change in lateral dimension and (iii) the change in volume. Assume that E = 200 kn/mm 2 and Poisson's ratio ν =0.3. Ans. (i) 0.66 mm, (ii) mm, (iii) 518 mm 3 ) 5. A square bar, 10 by 10 mm, is tested in tension. At a load of 5 kn the deformation in a 50 mm length is mm, and at a load of 25 kn the deformation is mm. If the stress-strain diagram is a straight line between these two points, calculate the modulus of elasticity for this material. Ans: E = 108 GPa 6. A concrete cylinder with a diameter of 150 mm is tested in compression. At a load of 88.0 kn the deformation in 200 mm is mm, and at a load of kn the deformation is mm. Calculate the modulus of elasticity if the stress-strain diagram is assumed to be straight between these two points. Page 5-24

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