SAMPLE OF THE STUDY MATERIAL PART OF CHAPTER 1 STRESS AND STRAIN

1.1 Stress & Strain SAMPLE OF THE STUDY MATERIAL PART OF CHAPTER 1 STRESS AND STRAIN Stress is the internal resistance offered by the body per unit area. Stress is represented as force per unit area. Typical units of stress are N/m, ksi and MPa. There are two primary types of stresses: normal stress and shear stress. Normal stress,, is calculated when the force is normal to the surface area; whereas the shear stress,, is calculated when the force is parallel to the surface area. P P normal _ to _ area A parallel _ to _ area A Linear strain (normal strain, longitudinal strain, aial strain),, is a change in length per unit length. Linear strain has no units. Shear strain,, is an angular deformation resulting from shear stress. Shear strain may be presented in units of radians, percent, or no units at all. L parallel _ to _ area Height tan [ in radians] A composite bar consists of an aluminum section rigidly fastened between a bronze section and a steel section as shown in Fig. 1.1. Aial loads are applied at the positions indicated. Determine the stress in each section. Page : 1

Bronze A = 1. in. Aluminum A = 1. 8 in. Steel A = 1. 6 in. 4000 lb 7000 lb 9000 lb 000 lb 1.3 ft 1.6 ft 1.7 ft Solution Figure 1.1 To calculate the stresses, we must first determine the aial load in each section. The appropriate free-body diagrams are shown in Fig. 1. below from which we determine P = 4000 lb (tension), and P = 7000 lb (compression). 4000 lb P br 4000 lb 9000 lb P al 4000 lb 9000 lb 000 lb P st Figure 1. The stresses in each section are σ = = 3330 psi (tension) σ =.. σ = σ =.... = 780 psi (compression) = 4380 psi (compression) Note that neither the lengths of the sections nor the materials from which the sections are made affect the calculations of the stresses. Page :

1. Hooke s Law: Aial and Shearing Deformations Hooke s law is a simple mathematical relationship between elastic stress and strain: stress is proportional to strain. For normal stress, the constant of proportionality is the modulus of elasticity (Young s Modulus), E. E The deformation,, of an aially loaded member of original length L can be derived from Hooke s law. Tension loading is considered to be positive, compressive loading is negative. The sign of the deformation will be the same as the sign of the loading. L L E This epression for aial deformation assumes that the linear strain is proportional to the normal stress and that the cross-sectional area is constant. E When an aial member has distinct sections differing in cross-sectional area or composition, superposition is used to calculate the total deformation as the sum of individual deformations. PL AE P PL AE When one of the variables (e.g., A), varies continuously along the length, PdL AE P L AE dl AE The new length of the member including the deformation is given by The algebraic deformation must be observed. L f L Hooke s law may also be applied to a plane element in pure shear. For such an element, the shear stress is linearly related to the shear strain, by the shear modulus (also known as the modulus of rigidity), G. G The relationship between shearing deformation, s and applied shearing force, V is then epressed by VL s AG Page : 3

If a tension test bar is found to taper uniformly from (D a) to (D + a) diameter, prove that the error involved in using the mean diameter to calculate the Young s Modulus is Let the two diameters be (D + a) and (D a) as shown in Fig. 1.3. Let E be the Young s Modulus of elasticity. Let the etension of the member be δ. P (D + a) (D - a) P L Figure 1.3 Then, δ = E = ()() ( ) If the mean diameter D is adopted, let E be the computed Young s modulus. Then δ = E = Hence, percentage error in computing Young s modulus = = = 100 = 10a D percent. 100 100 Page : 4

A weight of 45 kn is hanging from three wires of equal length as shown in Fig. 1.4. The middle one is of steel and the two other wires are of copper. If the cross section of each wire is 3 sq. mm, find the load shared by each wire. Take E = 07 N/ mm and E = 14.N/mm. Cu Steel Cu 45 kn Figure 1.4 Let P = load carried by the copper wire P = load carried by the steel wire f = stress in copper wire f = stress in steel wire Then, P + P + P = 45000 N or P + P = 45000 Also, strain in steel wire = strain in copper wire Hence = or = =. = 0.6 But, 45000 = P + P = (f a) + (f a) = 3 (f + f ) = 3( 0.6 f + f ) Page : 5

f = = 63.5 N/mm. P = f a = 63.5 3 = 0.447 kn P =. = 1.77 kn Compute the total elongation caused by an aial load of 100 kn applied to a flat bar 0 mm thick, tapering from a width of 10 mm to 40 mm in a length of 10 m as shown in Fig. 1.5. Assume E = 00 GPa.[1 Pa = 1N/m ] 0 mm 0 mm thick m 60 mm P = 100 kn y n d P = 100 kn 10 m Figure 1.5 Consider a differential length for which the cross-sectional area is constant. Then the total elongation is the sum of these infinitesimal elongations. At section m-n, the half width y (mm) at a distance (m) from the left end is found from geometry to be = or y = (4 + 0) mm And the area at that section is A = 0(y) = (160 + 800)mm At section m n, in a differential length d, the elongation is given by δ = dδ = = 0.500 d 160 + 800 ()( )( ) Page : 6

from which the total elongation is d δ = 0.500 160 + 800 = 0.500 160 [ln(160 + 800] = (3.13 10 ) ln = 3.44 10 m = 3.44 mm A compound tube is made by shrinking a thin steel tube on a thin brass tube. A and A b are crosssectional areas of steel and brass tubes respectively, E and E are the corresponding values of the Young s Modulus. Show that for any tensile load, the etension of the compound tube is equal to that of a single tube of the same length and total cross-sectional area but having a Young s Modulus of. = = strain (i) Where P and P are stresses in steel and brass tubes respectively. P A + P A = P. (ii) where P total load on the compound tube. From equations (i) and (ii);. A P + P A = P P A A = P P =.. (iii) Etension of the compound tube = dl = etension of steel or brass tube dl =. l.. (iv) From equations (iii) and (iv); dl = =.. (v) Let E be the Young s modulus of a tube of area (A + A ) carrying the same load and undergoing the same etension dl = ( ). (vi) Page : 7

equation (v) = (vi) = ( ) or E = The diameters of the brass and steel segments of the aially loaded bar shown in Fig. 1.6 are 30 mm and 1 mm respectively. The diameter of the hollow section of the brass segment is 0 mm. Determine: (i) The displacement of the free end; (ii) The maimum normal stress in the steel and brass Take E = 10 GN/m and E = 105 GN/m. In the Figure shown below, A = π 4 (1) = 36π mm = 36π 10 m 10 kn 1 mm 5 KN 30 mm 0 mm A Steel B Brass C D 0.15 m Figure 1.6 0. m 0.15 m (A ) = π 4 (30) = 5π mm = 5π 10 m (A ) = π 4 [(30) (0) ] = 15π mm = 15π 10 m (i) The maimum normal stress in steel and brass: 10 10 σ = 36π 10 10 MN/m = 88. 4 MN/m (σ ) = 5 10 MN 10 = 7. 07 MN/m 5π 10 m Page : 8

(σ ) = 5 10 15π 10 10 MN/m = 1. 73 MN/m (ii) The displacement of the free end: The displacement of the free end δl = (δl ) + (δl ) + (δl ) 88.4 0.15 = 10 10 10 + 7.07 0. 1.73 0.15 σl 105 10 + 10 105 10 δl = 10 E = 6.316 10 + 1.347 10 + 1.515 10 = 9.178 10 m or 0.09178 mm. A beam AB hinged at A is loaded at B as shown in Fig. 1.7. It is supported from the roof by a.4 cm long vertical steel bar CD which is 3.5 cm square for the first 1.8 m length and.5 cm square for the remaining length. Before the load is applied, the beam hangs horizontally. Take E = 10 GPa Determine: (i) The maimum stress in the steel bar CD; Roof (ii) The total elongation of the bar. Given: A =.5.5 = 6.5 cm = 6.5 10 m ; A = 3.5 3.5 = 1.5 10 m ; l = 1.8m l = 0.6 m E = 10 GPa. Let P = The pull in the bar CD. Then, taking moments about A, we get P 0.6 = 60 0.9 P = 90 kn 3.5 cm.5 cm B C 1 E D 1.8 cm 0.6 cm A (i) The maimum stress in the steel bar CD, σ ma : 0.3 m 60 kn 0.6 m The stress shall be maimum in the portion DE of the steel bar CD. σ = P 90 10 MN = 10 A 6.5 10 m = 144 MN/m Figure 1.7 Page : 9

(ii) The total elongation of the bar CD, δl = δl + δl = Pl A E + Pl A E = P E l + l A A 90 10 = 10 10 1.8 1.5 10 + 0.6 6.5 10 = 0.48 10 (1469.38 + 960) = 0.001039 m 1. 04 mm 1.3 Stress-Strain Diagram Actual rupture strength Stress Ultimate strength σ = P A Yield point Rupture strength Elastic limit Proportional limit 0 Figure 1.8 Strain ε = δ L Proportional Limit: It is the point on the stress strain curve up to which stress is proportional to strain. Elastic Limit: It is the point on the stress strain curve up to which material will return to its original shape when unloaded. Page : 10

Yield Point: It is the point on the stress strain curve at which there is an appreciable elongation or yielding of the material without any corresponding increase of load; indeed the load actually may decrease while the yielding occurs. Ultimate Strength: It is the highest ordinate on the stress strain curve. Rupture Strength: It is the stress at failure 1.4 Poisson s Ratio: Biaial and Triaial Deformations Poisson s ratio,, is a constant that relates the lateral strain to the aial strain for aially loaded members. lateral Theoretically, Poisson s ratio could vary from zero to 0.5, but typical values are 0.33 for aluminum and 0.3 for steel and maimum value of 0.5 for rubber. Poisson s ratio permits us to etend Hooke s law of uniaial stress to the case of biaial stress. Thus if an element is subjected simultaneously to tensile stresses in and y direction, the strain in the direction due to tensile stress is /E. Simultaneously the tensile stress y will produce lateral contraction in the direction of the amount y /E, so the resultant unit deformation or strain in the direction will be y E E Similarly, the total strain in the y direction is y y E E Hooke s law can be further etended for three-dimensional stress-strain relationships and written in terms of the three elastic constants, E, G, and. The following equations can be used to find the strains caused due to simultaneous action of triaial tensile stresses: 1 y z E 1 y y z E aial Page : 11

z 1 E z y y G yz yz G z z G y For an elastic isotropic material, the modulus of elasticity E, shear modulus G, and Poisson s ratio are related by E G 1 E G 1 The bulk modulus (K) describes volumetric elasticity, or the tendency of an object's volume to deform when under pressure; it is defined as volumetric stress over volumetric strain, and is the inverse of compressibility. The bulk modulus is an etension of Young's modulus to three dimensions. For an elastic, isotropic material, the modulus of elasticity E, bulk modulus K, and Poisson s ratio are related by E 3K 1 A thin spherical shell 1.5 m diameter with its wall of 1.5 cm thickness is filled with a fluid at atmospheric pressure. What intensity of pressure will be developed in it if 160 cu.cm more of fluid is pumped into it? Also, calculate the hoop stress at that pressure and the increase in diameter. Take E = 00 GPa and = 0.33. At atmospheric pressure of fluid in the shell, there will not be any increase in its volume since the outside pressure too is atmospheric. But, when 160 cu. cm of fluid is admitted into it forcibly, the sphere shall have to increase its volume by 160 cu. cm. Page : 1

increase in volume = 160 cu. cm. V = (4 / 3) πr = (4 / 3) π 75 cu. cm. e = 160 V e = 3e e = = increase in diameter = = 0.00453 cm 1.5 Thermal stresses = ( / ) Temperature causes bodies to epand or contract. Change in length due to increase in temperature can be epressed as ΔL = L.α.t Where, L is the length, α (/ o C) is the coefficient of linear epansion, and t ( o C) is the temperature change. From the above equation thermal strain can be epressed as: ϵ = = αt If a temperature deformation is permitted to occur freely no load or the stress will be induced in the structure. But in some cases it is not possible to permit these temperature deformations, which results in creation of internal forces that resist them. The stresses caused by these internal forces are known as thermal stresses. When the temperature deformation is prevented, thermal stress developed due to temperature change can be given as: σ = E.α.t A copper rod of 15 mm diameter, 800 mm long is heated through 50 C. (a) What is its etension when free to epand? (b) Suppose the epansion is prevented by gripping it at both ends, find the stress, its nature and the force applied by the grips when one grip yields back by 0.5 mm. α = 18.5 10 per C E = 15 kn/mm Page : 13

(a) Cross-sectional area of the rod = 15 = sq.mm Etension when the rod is free to epand = αtl = 18.5 10 50 800 = 0.74 mm (b) When the grip yields by 0.5 mm, etension prevented = = αtl - 0.5 = 0.4 mm Temperature strain =. Temperature stress =. 15 10 = 37.5 N/mm (comp.) Force applied = 37.5 = 667 N A steel band or a ring is shrunk on a tank of 1 metre diameter by raising the temperature of the ring through 60 C. Assuming the tank to be rigid, what should be the original inside diameter of the ring before heating? Also, calculate the circumferential stress in the ring when it cools back to the normal temperature on the tank. α = 10 per C and E = 00 kn/mm Let d mm be the original diameter at normal temperature and D mm, after being heated through 60 C. D should be, of course, equal to the diameter of the tank for slipping the ring on to it. Circumference of the ring after heating = π D mm. Circumference of the ring at normal temperature = π d mm The ring after having been slipped on the tank cannot contract to π d and it cools down resulting in tensile stress in it. Contraction prevented = π(d d) Temperature strain = = α T or ( ) from which d = 999.4 mm = α T = 10 60 Page : 14

circumferential temperature stress or stress due to prevention of contraction of the ring = α TE = 10 60 10 = 1 N/mm A steel rod.5 m long is secured between two walls. If the load on the rod is zero at 0 C, compute the stress when the temperature drops to 0 C. The cross-sectional area of the rod is 100 mm, α = 11.7 µm/(m. C), and E = 00 GPa. Solve, assuming (a) that the walls are rigid and (b) that the walls spring together a total distance of 0.500 mm as the temperature drops. Part a. Imagine the rod is disconnected from the right wall. Temperature deformations can then freely occur. A temperature drop causes the contraction represented by δ in Fig. 1.9. To reattach the rod to the wall, it will evidently require a pull P to produce the load deformation δ. From the sketch of deformations, we see that δ = δ, or, in equivalent terms α( T)L = = from which we have σ = Eα( T) = (00 10 )(11.7 10 )(40) = 93.6 10 N/m = 93.6 MPa Ans. δ T P δ P δ T P Figure 1.9 Part b. When the walls spring together, Fig. 1.9 shows that the free temperature contraction is equal to the sum of the load deformation and the yield of the walls. δ P Yield Page : 15

Hence δ = δ + yield Replacing the deformations by equivalent terms, we obtain αl( T) = + yield or (11.7 10 )(.5)(40) = from which we obtain σ = 53.6 MPa. σ(.) + (0.5 10 ) A composite bar made up of aluminium and steel is firmly held between two unyielding supports as shown in Fig. 1.10. A 10 cm 15 cm B 00 kn Aluminum bar Steel bar C 10 cm 15 cm Figure 1.10 An aial load of 00 kn is applied at B at 50. Find the stresses in each material when the temperature is 100. Take E for aluminium and steel as 70 GN/m and 10 GN/m respectively. Coefficient of epansion for aluminium and steel are 4 10 per and 11.8 10 per respectively. Given: A = 10 cm = 10 10 m ; A = 15 cm = 15 10 m ; l = 10 cm = 0.1 m; l = 15 cm = 0.15 m; t = 50, t = 100 ; Load, P = 00 kn E = 70 GN/m ; E = 10 GN/m ; α = 4 10 per, α = 11.8 10 per. Stresses; σ al ; σ s ; Out of 00 kn load(p) applied at B, let P kn be taken up by AB and (0 - P ) kn by BC. Since the supports are rigid, Page : 16

Elongation of AB = contraction of BC. P l A E = P l A E = (P P )l A E (P 10 ) 0.1 10 10 70 10 = (00 P ) 10 0.15 15 10 10 10 P (00 P ) = 10 0.15 10 10 70 10 P = 66.67 0.333 P 10 0.1 15 10 = 0.333 10 10 P = 50 kn Stress in aluminium, (σ ) = = = 50 MN/m (Tensile) Stress in steel, (σ ) = = = () = 100 MN/m (Compressive) These are the stresses in the two materials (Aluminium and steel) at 50. Now let the temperature be raised to 100. In order to determine the stresses due to rise of temperature, assume that the support at C is removed and epansion is allowed free. Rise of temperature = t t = 100 50 = 50 Epansion of AB = l. α. t = 0.1 4 10 50 = 10 10 m... (i) Epansion of BC = l. α. t = 0.15 11.8 10 50 = 88.5 10 m. Let a load be applied at C which causes a total contraction equal to the total epansion and let C be attached to rigid supports. If this load causes stress σ N/m in BC, its value must be 15 10 σ and hence stress in AB must be σ = 1.5 σ N/m Total contraction caused by the load = σ 0.15 150 0.1 + (ii) 10 10 70 10 From eqns. (i) ans (ii), we have σ 0.15 1.5σ 0.1 + 10 10 70 10 = 10 10 + 88.5 10 = 08.5 10 or 0.15σ 10 + 0.15σ 70 = 10 08.5 10 = 08500 0.6 σ = 10 08500 = 43785000 Page : 17

σ = 7.97 10 N/m or 7.97 MN/m (Compressive) At 100, Stress in aluminium, σ = - (σ ) + 1.5 7.97 = 50 + 1.5 7.97 = 59.45 MN/m (comp.) Stress in steel, σ = (σ ) + 7.97 = 100 + 7.97 = 17.97 MN/m (comp.) 1.6 Thin-Walled Pressure Vessels Cylindrical shells Figure 1.11 F 0 : ( t) p( r) 0 Hoop stress or circumferential stress = pr/t = pd/t z 1 Page : 18

y t σ da r z P da Longitudinal stress = pr/t = pd/4t Figure 1.1 F 0 : ( rt) p( r ) 0 y σ σ t z σ σ r Figure 1.13 Page : 19

Spherical shells σ da t C r P da Figure 1.14 F 0 : ( rt) p( r ) 0 pr Hoop stress = longitudinal stress = 1 t σ σ σ σ = σ Figure 1.15 Page : 0

A cylindrical shell 900 mm long, 150 mm internal diameter, having a thickness of metal as 8 mm, is filled with a fluid at atmospheric pressure. If an additional 0000 mm of fluid is pumped into the cylinder, find (i) the pressure eerted by the fluid on the cylinder and (ii) the hoop stress induced. Take E = 00 kn/mm and = 0.3 Let the internal pressure be p. N/mm. Hoop stress = f = = = 9.375p N/mm Longitudinal stress = f = = 4.6875p N/ mm Circumferential strain = e = f = (9.375 0.3 4.6875) = 7.96875 Longitudinal strain = e = f = (4.6875 0.3 9.375) = 17.815 Increase in volume = δv = e V = 0000mm 17.815 (i) p = 14.1 N/mm (150) 900 = 0000 (ii) f =. = = 13.4 N/mm 1.7 Mohr s Circle Mohr's circle gives us a graphic tool by which, we can compare the different stress transformation states of a stress cube to a circle. Each different stress combination is described by a point around the circumference of the circle. Compare the stress cube to a circle created using the circle offset y a ave y and R y Page : 1

y σ y τ y σ τ y τ y σ -face coordinate: (, y ) y-face coordinates: ( y, y ) τ y σ y Figure 1.16 -τ y (σ, -τ y ) B R y y σ +τ σ ave A (σ, τ y ) R y y Figure 1.17 Notes: +τ (meaning counterclockwise around the cube) is downward - τ (meaning clockwise around the cube) is up on the ais A rotation angle of θ on the stress cube shows up as θ on the circle diagram and rotates in the same direction. The largest and smallest values of σ are the principle stresses, σ 1 and σ. Page :

The largest shear stress, τ ma is equal to the radius of the circle, R. The center of the circle is located at the value of the average stress, σ ave If σ 1 = σ in magnitude and direction (nature) the Mohr circle will reduce into a point and no shear stress will be developed. If the plane contain only shear and no normal stress (pure shear), then origin and centre of the circle will coincide and maimum and minimum principal stress equal and opposite. σ 1 = + τ, σ = -τ The summation of normal stresses on any two mutually perpendicular planes remains constant. σ + σ y = σ 1 + σ 1.8 Applications: Thin-Walled Pressure Vessels Cylindrical shells: Hoop stress or circumferential stress = σ = Longitudinal stress = σ = y σ t σ z σ σ r Figure 1.18 Page : 3

Figure 1.19 τ = pd 4t τ ( ) = σ = pd 8t Spherical shells: Hoop stress = longitudinal stress = σ = σ = σ σ σ σ = σ Figure 1.0 Page : 4

Figure 1.1 τ = σ = pd 8t At a certain point a material is subjected to the following strains: ε = 400 10 ; ε = 00 10 ; γ = 350 10. Determine the magnitudes of the principal strains, the directions of the principal strain aes and the strain on an ais inclined at 30 clockwise to the -ais. Page : 5

Mohr s strain circle is as shown in Figure 1.. ε y ε 1 y θ 1 = 30 30 ε 100 0 100 00 300 400 500 θ 1 60 60 ε = 100 ε ε θ 00 ε θ = 100 + 1 γ ε 1 = 500 By measurement: ε = 500 10 ε 100 10 Figure 1. θ = = 30 θ = 90 + 30 = 10 ε = 00 10 the angles being measured counterclockwise from the direction of ε. Page : 6

Draw Mohr s circle for a -dimensional stress field subjected to (a) pure shear (b) pure biaial tension, (c) pure uniaial tension and (d) pure uniaial compression. Mohr s circles for -dimensional stress field subjected to pure shear, pure biaial tension, pure uniaial compression and pure uniaial tension are shown in Fig. 1.3(a) to 1.3(d). τ τ σ n 0 σ n σ σ σ 1 σ 1 τ (a) (b) 0 σ n 0 σ σ 1 (c) Figure 1.3 σ (d) A thin cylinder of 100 mm internal diameter and 5 mm thickness is subjected to an internal pressure of 10 MPa and a torque of 000 Nm. Calculate the magnitudes of the principal stresses. Given: d = 100 mm = 0.1 m; t = 5 mm = 0.005 m; D = d + t = 0.1 + 0.0050.11 m, p = 10 MPa, 10 10 N/m ; T = 000 Nm. Page : 7

Principal stresses, σ 1, σ : Longitudinal stress, σ = σ = =. = 50 10 N/m = 50 MN/m. Circumferential stress, σ = σ = =. = 100 MN/m. To find the shear stress, using the relation, = τ, we have τ = τ = TR Ip = T R π 3 (D d ) = Principal stresses are calculated as follows: σ, σ = σ + σ = 50 + 100 ± σ σ + τ 50 100 ± 000 (0.05 + 0.005) π 3 (0.11 0.1 ) + (4.14) = 75 ± 34.75 = 109.75 and 40.5 MN/m Hence, σ (Major principal stress) = 109.75 MN/m ; (Ans.) σ (minor principal stress) = 40.5 MN/m (Ans.) = 4.14 MN/m A solid shaft of diameter 30 mm is fied at one end. It is subject to a tensile force of 10 kn and a torque of 60 Nm. At a point on the surface of the shaft, determine the principle stresses and the maimum shear stress. Given: D = 30 mm = 0.03 m; P = 10 kn; T = 60 Nm Principal stresses (σ 1, σ ) and maimum shear stress (τ ma ): Tensile stress, σ = σ =. = 14.15 10 N/m or 14.15 MN/m Page : 8

τ σ σ τ Figure 1.4 As per torsion equation, = Shear stress, I = =. = (.) = 11.3 10 N/m or 11.3 MN/m The principal stresses are calculated by using the relations: σ, σ = σ + σ ± σ σ + τ Here σ = 14.15 MN/m, σ = 0; τ = τ = 11.3 MN/m σ, σ = 14.15 ± 14.15 + (11.3) = 7.07 ± 13.35 = 0.45 MN/m, 6.75 MN/m. Hence, major principal stress, σ 1 = 0. 45 MN/m (tensile) (Ans.) Minor principal stress, σ = 6.75 MN/m (compressive (Ans.) Maimum shear stress, τ = σ σ =.(.) = 13.35 mm/m (Ans.) A thin cylinder with closed ends has an internal diameter of 50 mm and a wall thickness of.5 mm. It is subjected to an aial pull of 10 kn and a torque of 500 Nm while under an internal pressure of 6 MN/m. (i) Determine the principal stresses in the tube and the maimum shear stress. (ii) Represent the stress configuration on a square element taken in the load direction with direction and magnitude indicated; (schematic) Page : 9

Given: d = 50 mm = 0.05 m D = d + t = 50 +.5 = 55 mm = 0.055 m; Aial pull, P = 10kN; T = 500 Nm; p = 6MN/m (i) Principal stress (σ 1, σ ) in the tube and the maimum shear stress (τ ma ): σ = pd 4t + P πdt = 6 10 0.05 4.5 10 + 10 10 π 0.05.5 10 σ = pd t = 6 10 0.05 = 60 10.5 10 Principal stresses are given by the relations: σ, σ = σ σ = 30 10 + 5.5 10 = 55.5 10 N/m ± σ σ + τ We know that, = where J = (D d ) = [(0.055) (0.05) ] =.848 10 m 500.848 10 = τ (0.055/) or, τ = 500 (0.055/).848 10 = 48.8 10 N/m Now, substituting the various values in eqn. (i), we have σ, σ = 55.5 10 + 60 10 = (55.5 + 60) 10 ± 55.5 10 60 10 + (48.8 10 ) ± 4.84 10 + 330.96 10 = 57.75 10 ± 48.33 10 = 106.08 MN/m, 9.4 MN/m Hence, principal stress are: σ 1 = 106.08 MN/m ; Maimum shear stress, τ = σ σ =.. σ = 9.4 MN/m Ans. = 48.33 MN/m Ans. Page : 30

(ii) Stress configuration on a square element: pd t Y pd 4t + P πdt Square element pd 4t + P πdt X Pd t Figure 1.5 Page : 31