CH1_LECTURE1.2-simple stress and strain.pdf

ROYAL UNIVERSITY OF BHUTAN 1 1 1 COLLEGE OF SCIENCE AND TECHNOLOGY PHUNTSHOLING: BHUTAN TSM 202:STRENGTH OF MATERIAL Lecture 1.2 Simple Stresses and Strain 1...

ROYAL UNIVERSITY OF BHUTAN 1 1 1 COLLEGE OF SCIENCE AND TECHNOLOGY PHUNTSHOLING: BHUTAN TSM 202:STRENGTH OF MATERIAL Lecture 1.2 Simple Stresses and Strain 1 3/09/2020 : Nimesh Chettri 8/7/2022  ANALYSIS OF BARS OF VARYING SECTIONS A bar of different lengths and of different diameters (and hence of different cross-sectional areas) is shown in Figure below. Let this bar is subjected to an axial load P. 2 Though, each section is subjected to the same axial load P, yet the stresses, strains and change in lengths will be different. The total change in length will be obtained by adding the changes in length of individual section. Let p= Axial load acting on the bar, L1 =Length of section 1, A1 =Cross-sectional area of section 1, L2, A2 =Length and cross-sectional area of section 2, L3 A3 = Length and cross-sectional area of section 3, and 8/7/2022 E=Young's modulus for the bar  𝐿𝑜𝑎𝑑 𝑃 Then stress for the section 1, σ1 = 𝐴𝑟𝑒𝑎 𝑠𝑒𝑐𝑡𝑖𝑜𝑛1 = 𝐴 1 𝑃 𝑃 Similarly stresses for the section 2 and section 3 are given as, σ2 = & σ3 = 3 𝐴2 𝐴3 σ1 𝑃 The strains in different sections are given by ε1 = =𝐴𝐸 𝐸 1 σ2 𝑃 σ3 𝑃 Similarly strains for the section 2 and section 3 are given as, ε2 = = 𝐴 𝐸 & ε3 = =𝐴𝐸 𝐸 2 𝐸 3 Change in length of section 1 dL But strain in section 1 = i.e. ε1 = L 1 Length of section 1 1 where dLl = change in length of section 1. 𝑃L Change in length of section 1, dLl = ε1 L1 , Therefore dLl = 𝐴 𝐸l 1 Similarly changes in length of section 2 and of section 3 are obtained as Change in length of section 2, 𝑃L 𝑃L dL2 = ε2 L2 , Therefore dL2 = 𝐴 𝐸2 and change in length of section 3, dL3 = ε3 L3 , Therefore dL3 = 𝐴 𝐸3 2 3 Total change in the length of the bar, 𝑷L 𝑷L2 𝑷L 𝑷 Ll L2 L3 dL= dL1 +dL2 + dL3 = 𝑨 𝑬l + + 𝑨 𝑬3 or + + ---------1 𝟏 𝑨𝟐 𝑬 𝟑 𝑬 𝑨𝟏 𝑨𝟐 𝑨𝟑 Above Equation is used when the Young's modulus of different sections is same. If the Young's modulus of different sections is different, then total change in length of the bar is given by, Ll L2 L3 8/7/2022 dL= P + + ---------2 𝑨𝟏𝑬𝟏 𝑨𝟐𝑬𝟐 𝑨𝟑𝑬𝟑  Solved example: An axial pull of 35000 N is acting on a bar consisting of three lengths as shown in Figure. If the Young's modulus = 2.1 X 105 N/mm2 determine: (i) stresses in each section and (ii) total extension of the bar. 4 Sol: Given: Axial pull, P = 35000 N Length of section 1, L1 = 20 cm = 200 mm Dia. of section 1, Dl = 2 cm = 30 mm Area of section 1, A1 = π/4(202) = 100 π mm2 Length of section 2, L2 = 25 cm = 250 mm Dia. of section 2, D2 = 3 cm = 30 mm Area of section 2, A2 = π/4(302) = 225 π mm2 Length of section 3, L3 = 22 cm = 220 mm Dia. of section 3, D3 = 5 cm = 50 mm Area of section 3, A3 = π/4(502) = 625 π mm2 8/7/2022 Young’s Modulus E= 2.1 X 105 N/mm2  (i) Stresses in each section 𝐴𝑥𝑖𝑎 𝐿𝑜𝑎𝑑 35000 5 Stress in section 1, σ1 = = = 111.406 N/mm2 𝐴𝑟𝑒𝑎 𝑠𝑒𝑐𝑡𝑖𝑜𝑛1 100 π 𝐿𝑜𝑎𝑑 35000 Stress in section 2, σ2 = = = 49.514 N/mm2 𝐴𝑟𝑒𝑎 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 2 225 π 𝐿𝑜𝑎𝑑 35000 Stress in section 3, σ3 = = = 17.825 N/mm2 𝐴𝑟𝑒𝑎 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 3 625 π (ii) Total extension of the bar 𝑷 Ll L2 L3 dL= + + 𝑬 𝑨𝟏 𝑨𝟐 𝑨𝟑 35000 200 250 220 dL= + + 2.1 X 105 100 π 225 π 625 π dL=0.183 mm 8/7/2022  ANALYSIS OF UNIFORMLY TAPERING CIRCULAR ROD A bar uniformly tapering from a diameter D1 at one end to a diameter D2 at the other and 6 is shown in the slide here. Let P =Axial tensile load on the bar, L = Total length of the bar, E =Young's modulus. Consider a small element of length dx of the bar at a distance x from the left end. Let the diameter of the bar be Dx at a distance x from the left end. 𝐷1 −𝐷2 𝐷1 −𝐷2 Dx=D1- x =D1- kx where k = 𝐿 𝐿 Area of cross-section of the bar at a distance x from the left end π π Ax = Dx2 = (D1 - kx)2 4 4 𝐿𝑜𝑎𝑑 𝑃 Now the stress at a distance x from the left end is given by, σx = =π Ax (Dx − kx)2 4 4𝑃 Therefore σx = 8/7/2022 π(D1 − kx)2  The strain εx in the small element of length dx is obtained by, 7 𝑆𝑡𝑟𝑒𝑠𝑠 σ 4𝑃 εx = E = E1 = ------------------------1 πE(D1 − kx)2 4𝑃 Extension of the small elemental length dx = Strain. dx = εx. dx i.e..dx πE(D1 − kx)2 Total extension of the bar is obtained by integrating the above equation between the limits 0 and L, Total extension, 𝐿 4𝑃 4𝑃 𝐿 dL= ‫׬‬0. 𝑑𝑥 = πE ‫׬‬0 (D1 − kx)−2. 𝑑𝑥 πE(D1 − kx)2 4𝑃 𝐿 (D1 − kx)−2 (−k).𝑑𝑥 = πE ‫׬‬0 multiplying and dividing by (–k) (−𝑘) 4𝑃 1 𝐿 4𝑃 1 𝟏 𝐷1 −𝐷2 = = substituting k= in the above equation, we πEk (D1 − kx) 0 πEk (D1 − kL) 𝑫𝟏 𝐿 get total extension. 4𝑃 1 𝟏 𝟒𝑷𝑳 i.e. = πE𝑫 𝑫 if the rod is of uniform diameter the D1 = D2 = D πE 𝐷 1 −𝐷2 𝐿 (D1 − 𝐷 1 −𝐷2 𝐿 L) 𝑫𝟏 𝟏 𝟐 𝟒𝑷𝑳 Therefore total extension, dL= πE𝑫𝟐 8/7/2022  Solved example: Find the modulus of elasticity for a rod, which tapers uniformly 8 from 30 mm to 15 mm diameter in a length of 350 mm. The rod is subjected to an axial load of 5.5 kN and extension of the rod is 0.025 mm. Sol: Given: Larger diameter, D1 = 30 mm Smaller diameter, D2 = 15 mm Length of rod, L = 350 mm Axial load, P = 5.5 kN = 5500 N Extension, dL = 0.025 mm Using equation derived before, we get 4𝑃𝐿 dL= πE𝐷1𝐷2 4𝑃𝐿 4 𝑥 5000 𝑥 350 E= = πdL𝐷1𝐷2 π x 30 x 15 x 0.025 E= 2.178 x 105 N/mm2 8/7/2022  ANALYSIS OF UNIFORMLY TAPERING RECTANGULAR ROD A bar of constant thickness and uniformly tapering in width 'from one end to the other end is shown in 9 Figure below: Let P = Axial load on the bar L= Length of bar a = Width at bigger end b =Width at smaller end E = Young's modulus t = Thickness of bar Consider any section X-X at a distance x from the bigger end. Width of the bar at the section x-x 𝑎−𝑏 𝑥 𝑎−𝑏 i.e. a - = a – kx -----k = 𝐿 𝐿 Thickness of bar at section X-X = t Area of the section X-X = Width x thickness i.e = (a-kx)t 8/7/2022  𝐿𝑜𝑎𝑑 𝑃 Stress n the section X-X = = Area (a−kx)t 10 𝑆𝑡𝑟𝑒𝑠𝑠 Extension of the small elemental length dx = Strain x Length dx = x dx E 𝑃 [ ] (a−kx)t 𝑃 = x dx = dx E E(a−kx)t Total extension of the bar is obtained by integrating the above equation between the limits 0 and L. 𝐿 𝑃 𝑃 𝐿 𝑑𝑥 𝑃 Total extension, dL = ‫׬‬0 E(a−kx)t dx = ‫׬‬0 (a−kx) = - [log 𝑒 𝑎 − 𝑘𝑙 − log 𝑒 Et Et a] 𝑃 = [log 𝑒 𝑎 − log 𝑒 (𝑎 − 𝑘𝐿)] Etk 𝑃 𝑎 = [log 𝑒 ( )] Etk 𝑎−𝑘 𝑙 𝑷𝑳 𝒂 = 𝒍𝒐𝒈𝒆 ( ) Et(a−b) 𝒃 8/7/2022  Solved example: A rectangular bar made of steel is 2.8 m long and 15 mm thick. The rod is subjected to an axial tensile load of 40 kN. The width of the rod varies from 75 mm at 11 one end to 30 mm at the other. Find the extension of the rod if E = 2 x 105 Nlmm2. Sol: Length,= L = 2.8 m = 2800 mm Thickness, = t= 15mm Axial load,= P = 40 kN = 40,000 N Width at bigger end, a = 75 mm Width at smaller end, b = 30 mm Value of E = 2 x105 N/mm2 Let dL = Extension of the rod. 𝑃𝐿 𝑎 40000 𝑥 2800 75 Therefore, = log 𝑒 ( ) = = log 𝑒 ( ) Et(a−b) 𝑏 2x 105 x 15 (75−30) 30 = 0.8296 x 0.9163 = 0.76 mm 8/7/2022  ANALYSIS OF BARS OF COMPOSITE SECTIONS A member, made up of two or more other members of equal lengths but of different materials rigidly integrated with each other and behaving as one unit for extension or compression when subjected to 12 an axial tensile or compressive loads, is called a composite bar. For the composite bar the following two points are Important : 1. The extension or compression in each bar is equal. Hence deformation per unit length i.e., strain in each member is equal. 2. The total external load on the composite member is equal to the sum of the loads carried by each different material. Figure in this slide shows a composite bar made up of two different materials. Let P = Total load on the composite member, L = Length of composite member and also length of members of different materials, Al = Area of cross-section of member 1, A2 =Area ·of cross-section of member 2, E1 =Young's Modulus of member 1, E2 =Young's Modulus of member 2, Pl = Load shared by member 1, P2 = Load shared by member 2, σ1 = Stress induced in member 1, and σ 2 = Stress induced in member 2. 8/7/2022  Therefore, total load on the composite member is equal to the sum of the load carried by-the two members is P=P1 +P2 --------1 13 𝐿𝑜𝑎𝑑 𝑐𝑎𝑟𝑟𝑖𝑒𝑑 𝑏𝑦 𝑏𝑎𝑟 1 The stress in member 1 = Therefore, P1= σ1 A1 -----------2 Area of cross section of member 1 Similarly load in member 2, P2 = σ2 A2 ----------3 Substituting in equation 1 the total load P = σ1 A 1 + σ1 A2 -------------4 Since the ends of the two members are rigidly connected, each member will change in length by the same amount. Also the length of each member is same and hence the ratio of change in length to the original length (i.e, strain) will be same for each member. 𝑆𝑡𝑟𝑒𝑠𝑠 𝑖𝑛 𝑚𝑒𝑚𝑏𝑎𝑟 1 Strain in member 1 = = σ1 /E1 Young′s modulus of member 1 And strain in member 2 = σ2 /E2 We know that strain in member 1 = strain in member 2 i.e σ1 /E1 = σ2 /E2 --------5 From equations 3 and 4 the stresses σ1 and σ2 can be determined. By substituting the values of σ1 and σ2 in equations 2 and 3, the load carried by different materials may be computed. Note: The ratio of the modulus of elasticity of two different material is know and Modular ratio i.e E1/E2 8/7/2022  Solved Example: A steel rod of 3 cm diameter is enclosed centrally in a hollow copper tube of external diameter 5 cm and internal diameter of 4 cm. The composite bar is then subjected to 14 an axial pull of 45000 N. If the length of each bar is equal to 15 cm, determine :(i) The stresses in the rod and tube, and (ii) Load carried by each bar. Take E for steel = 2.1 x l05 N/mm2 and for copper = 1.1 x l05 N/mm2 Sol: Dia. of steel rod = 3 cm = 30 rom 𝜋 Area of steel rod As = 4 (30)2 = 706.86 mm2 Externa dia. Of copper tube = 50 mm Internal dia. Of copper tube = 40 mm 𝜋 Area of copper tube = 4 (502 - 402) = 706.86 mm2 Axial load P= 45000 N Length of each bar L = 15 cm Young’s Modulus for steel Es = 2.1 x 105 N/mm2 Young’s Modulus for copper Ec = 1.1 x 105 N/mm2 8/7/2022  (i) The stresses in the rod and tube 15 Let σs = stress in steel, Ps = Load carried by steel rod, σc = Stress in copper , Pc = load carried by copper tube We know that strain in steel = strain in copper i.e. σs /Es = σc /Ec ---------1 Therefore σs = Es/Ec x σc = 1.90 σc We also know that, Load = stress x area and total load (P) = load on steel + load on copper i.e. P = σs As + σc Ac P = 1.90 σs x 706.86 + σc x 706.86 = 45000 σc = 21.90 N/mm2 , substituting value of σs in equation 1 we get σs = 41.77 N/mm2 (ii) Load carried by each bar Load = stress x area Ps = σs As = 41.77 x 706.86 = 29.6 kN Pc = 45 – 29.6 = 15.5 kN 8/7/2022  Thermal (Temperature ) Stresses All materials expand when temperature rises 16 and contract when temperature falls. The change in length due to change in temperature is found to be directly proportional to length (L) of the member and the change in temperature. Therefore, lets consider ‘t’ as change in temperature and ‘L’ as length of the member. Hence, change in length (dL)is given as follows; dL∝ 𝒕. 𝑳, Let ′𝛼’ which is constant of proportionality, Then we get, dL= 𝜶. t.L, constant of proportionality ′𝛼 ’ is called coefficient of thermal expansion and is defined as change in unit length of the material due to unit change in temperature. 8/7/2022  If the ends of the body are fixed to rigid supports, so that its expansion is prevented then compressive stress and strain will be set up in 17 the rod. These stresses and strains are known as thermal stresses and thermal strain. Extension prevented dL 𝛼.t.L Thermal strain 𝜀𝑡= = = 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ 𝐿 𝐿 = 𝜶. t And Thermal Stress σt = Thermal strain x E = 𝛂. t.E If free expansion is prevented partly (𝛼. t.L − 𝛿) as show in the figure, then stress and strain corresponds to amount of free expansion prevented. Therefore Shortening (strain) caused by obstruction is given as, dL = 𝛼.t.L - 𝛿 or PL PL = 𝛼.t.L - 𝛿 --------------------------dL = 𝐴𝐸 𝐴𝐸 8/7/2022  Solved Example: A rod is 2 m long at a temperature of 10° C. Find the expansion 18 of the rod, when the temperature is raised to 80° C. If this expansion is prevented, find the stress induced in the material of the rod. Take E = 1011 Nlm2 and 𝛼 = 0.000012 per degree centigrade. Length of rod, L=2 m=200 cm Initial temperature, T1= 10° C Final temperature, T2= 80° C Rise in temperature, T = T2 - Tl = 80° - 10° = 70° C Young’s Modulus, E= 1011 N/m2 Coefficient of linear expansion, 𝛼 = 0.000012 1. Expansion of rod = 𝛼.t.L = 0.000012 x 70 x 200 = 0.168 cm 2. Thermal stress= σt = 𝛼. t.E = 0.000012 x 70 x 1011 = 84 N/mm2 8/7/2022  Complimentary Shear Stress and Shear Strain 19 Consider first figure (2 D plane with unit thickness subjected to shear stresses), the force acting on face AD and BC = F = 𝜏𝑥𝑦 𝑥 𝐴𝐷 = 𝜏𝑥𝑦 𝑥 𝐵𝐶-----------1 Moment due to force, M1= 𝜏𝑥𝑦 𝑥 𝐴𝐷 x AB= 𝜏𝑥𝑦 𝑥 𝐵𝐶 x CD To maintain the equilibrium of the system, there must be restoring couple whose moment must be equal to this couple Therefore, the force acting on face DC and AB = F = 𝜏𝑦𝑥 𝑥 𝐷𝐶 = 𝜏𝑦𝑥 𝑥 𝐴𝐵-----------2 Moment due to force, M2= 𝜏𝑦𝑥 𝑥 𝐴𝐵𝑥 𝐴𝐷 = 𝜏𝑦𝑥 𝑥 𝐶𝐷 x BC , M1 =M2 i.e. 𝜏𝑥𝑦 𝑥 𝐴𝐷 x AB = 𝜏𝑦𝑥 𝑥 𝐴𝐵𝑥 𝐴𝐷 Therefore, 𝛕𝐱𝐲 = 𝛕𝐲𝐱 8/7/2022  Strain Energy When a body is subjected to externa forces, the body will in turn generates resisting force which varies 20 proportional to the applied force. The figure in the slide shows variation of resisting force with deformation. In the process of proportional increase of resisting force with deformation, the work is done by this force which is stored as energy within the body. On the removal of the external force this stored energy springs back the material to the original position. This energy stored in a body due to straining of the body is called strain energy. Let us consider a bar of length L, cross sectional are A and subjected to axial load P.. Let resistance developed be R. When deformation is 0, R=0. When deformation is dL = 𝜀𝐿, 𝑅 = 𝑃 Therefore, Work done by the resisting force = Average resistance x dL 0+𝑃 𝑃 = x 𝜀𝐿 = 𝜀𝐿 2 2 𝜎𝐴 𝜎 𝜀𝐿 = 𝜀𝑉 -------AL =V We know that strain energy = work done by internal force R 2 2 𝜎 1 SE= 𝜀𝑉 = Stress x 𝑆𝑡𝑟𝑎𝑖𝑛 𝑥 𝑣𝑜𝑙𝑢𝑚𝑒 2 2 𝜎𝜎 SE= 𝑉 -------Strain = 𝜎 /E 2𝐸 𝟏 𝝈𝟐 SE= 𝑽 𝟐 𝑬 8/7/2022 21 THANK YOU 8/7/2022