COMPLEX STRESS TUTORIAL 3 COMPLEX STRESS AND STRAIN

Transcription

1 COMPLX STRSS TUTORIAL COMPLX STRSS AND STRAIN This tutorial is not part of the decel unit mechanical Principles but covers elements of the following sllabi. o Parts of the ngineering Council eam subject C05 ngineering Science. o Parts of the ngineering Council eam subject D09 Mechanics of Solids. You should judge our progress b completing the self assessment eercises. These ma be sent for marking at a cost (see home page). On completion of this tutorial ou should be able to do the following. plain a comple stress situation. Derive formulae for comple stress situations. Analse and solve stresses in a comple stress situation. Solve problems using a graphical method (Mohr s Circle) plain the use of strain gauge rosettes to determine principal strains and stresses. Appl the theor to combined bending and torsion problems. D.J.DUNN

2 . COMPLX STRSS Materials in a stressed component often have direct and shear stresses acting in two or more directions at the same time. This is a comple stress situation. The engineer must then find the maimum stress in the material. We will onl consider stresses in two dimensions, and. The analsis leads on to a useful tool for solving comple stress problems called Mohr's' Circle of Stress.. DRIVATION OF QUATIONS Consider a rectangular part of the material. Stress acts on the plane and acts on the plane. The shear stress acting on the plane on which acts is and act on the plane on which acts. The shear stresses are complementar and so must have opposite rotation. We will take clockwise shear to be positive and anti-clockwise as negative. Suppose we cut the material in half diagonall at angle θ as shown and replace the internal stresses in the material with applied stresses σ θ and τ θ. In this case we will do it for the bottom half. The dimensions are, and t as shown. Figure Now turn the stresses into force. If the material is t m thick normal to the paper then the areas are t and t on the edges and t /sin or t /cos on the sloping plane. The forces due to the direct stresses and shear stresses are stress area and as shown. Figure Figure D.J.DUNN

3 The material is in equilibrium so all the forces and moments on the plane must add up to zero. We now resolve these forces perpendicular and parallel to the plane. To make it easier the forces are labelled a, b, c, d, e, f, g, and h. Figure 4 a = t sin. c = t cos. e = t sin g = t cos b = t cos d = t sin f = t cos h = t sin All the forces normal to the plane must add up to (t /sin ). Balancing we have a + c + e + g = (t /sin) All the forces parallel to the plane must add up to (t /sin ) - f + h+ b - d = (t /sin ) Making the substitutions and conducting algebraic process will ield the following results. t t sin t cos t sin t cos sin sin sin cos sin sin cos cos sin sin sin cos tan tan cos sin sin cos cos sin sin cos Perhaps the point should have been made earlier that the shear stress on both planes are equal (read up complementar stress) so we denote them both = = cos sin...(.) Repeating the process for the shear stress we get sin cos...(.) D.J.DUNN

4 Consider the case where =5 MPa, =5 MPa and =00 MPa 75 50cos 00sin 50sin 00cos If the value of and are plotted against the resulting graphs are as shown below. Figure 5 In this eample the maimum value of σ θ is about 90 MPa. The plane with this stress is at an angle of about o. The maimum shear stress is about MPa on a plane at angle 77 o. These general results are the same what ever the values of the applied stresses. The graphs show that σ θ has a maimum and minimum value and a mean value not usuall zero. These are called the PRINCIPAL STRSSS. The principal stresses occur on planes 90 o apart. These planes are called the PRINCIPAL PLANS. The shear stress τ θ has an equal maimum and minimum value with a mean of zero. The ma and min values are on planes 90 o apart and 45 o from the principal planes. This is of interest because brittle materials fail on these planes. For eample, if a brittle material is broken in a tensile test, the fracture occurs on a plane at 45 o to the direction of pull indicating that the fail in shear. Further it can be seen that the principal planes have no shear stress so this is a definition of a principle plane. τ θ = 0 when σ θ = on the principal planes where it is maimum or minimum. There are several theories about wh a material fails usuall. The principle stresses and maimum shear stress are used in those theories. D.J.DUNN 4

5 . DTRMINING TH PRINCIPAL STRSSS AND PLANS You could plot the graph and determine the values of interest as shown in the previous section but is convenient to calculate them directl. The stresses are a maimum or minimum on the principal planes so we ma find these using ma and min theor (the gradient of the function is zero at a ma or min point). First differentiate the function and equate to zero. d sin cos 0 d cos sin tan...(.) There are two solutions to this equation giving answers less than 60 o and the differ b 90 o. From this the angle of the principal plane ma be found. If this angle is substituted into equation (.) and algebraic manipulation conducted the stress values are then the principal stresses and are found to be given as ma 4...(.4) 4 min...(.5) Repeating the process for equation (.), we get ma min 4 4..(.6) (.7) WORKD XAMPL No. Using the formulae, find the principal stresses for the case shown below and the position of the principal plane. Figure 6 D.J.DUNN 5

6 SOLUTION τ 50 tanθ σ σ = -7.6o or 08.4o = -5.8o or 54.o Putting this into equations. and. we have cos sin o (00 00) cos( 7.6 ) 50sin( 7.6 σ 08.MPa 08.MPa θ Using the other angle (00 00) cos(08.4 σθ -8.MPa -8.MPa sin cos o 00 00sin cos 7.6 From equations.6 and.7 we have o o ) 50sin( (This is as epected) o o ) ) ma = (/)( - ) = 58. MPa min = -(/)( - ) = -58. MPa It is easier to use equations.4 and.5 as follows. ma min MPa MPa The plot confirms these results. D.J.DUNN 6

7 .. MOHR'S CIRCL OF STRSS Although we can solve these problems easil these das with computer programmes or calculators, it is still interesting to stud this method of solution. Mohr found a wa to represent equations. and. graphicall. The following rules should be used.. Draw point 'O' at a suitable position (which is possible to see with eperience). Measure and along the horizontal ais using a suitable scale and mark A and B.. Find the centre (M) half wa between the marked points. 4. Draw up at A and down at B if it is positive or opposite if negative (as in this eample). 5. Draw the circle centre M and radius MC. It should also pass through D. 6. Draw the diagonal CD which should pass through M. 6. Measure O and this is. Measure OF and this is. The angle is shown. ma is the radius of the circle. Figure 7 Consider the triangle. The sides are as shown. Figure 8 D.J.DUNN 7

8 Appling trigonometr we find the same results as before so proving the geometr represents the equations. tan (.) ma 4...(.4) min 4...(.5) WORKD XAMPL No. Show that the solution in worked eample No. ma be obtained b constructing the circle of stress. SOLUTION Following the rules for construction figure 8 is obtained. Figure 9 D.J.DUNN 8

9 WORKD XAMPL No. A material has direct stresses of 0 MPa tensile and 80 MPa compressive acting on mutuall perpendicular planes. There is no shear stress on these planes. Draw Mohrs' circle of stress and determine the stresses on a plane 0 o to the plane of the larger stress. SOLUTION Since there is no shear stress, and are the principal stresses and are at the edge of the circle. Select the origin '0' and plot +0 to the right and -80 to the left. Draw the circle to encompass these points and mark the centre of the circle. Using a protractor draw the required plane at 40 o to the horizontal (assumed anticlockwise) through the middle of the circle. Draw the verticals. Scale the direct stress and shear stress. These ma be checked with the formulae if confirmation is required. Figure 0 D.J.DUNN 9

10 SLF ASSSSMNT XRCIS No. a) Figure 0 shows an element of material with direct stresses on the and planes with no shear stress on those planes. Show b balancing the forces on the triangular element that the direct and shear stress on the plane at angle anti-clockwise of the plane is given b cos sin Show how Mohr s circle of stress represents this equation. Figure b) An elastic material is subjected to two mutuall perpendicular stresses 80MPa tensile and 40 MPa compressive. Determine the direct and shear stresses acting on a plane 0o to the plane on which the 80 MPa stress acts. (Hint for solution) The derivation is the same as in the notes but with no shear stress, i.e. the stresses shown on fig.9 are the principal stresses. (50 MPa and 5 MPa) D.J.DUNN 0

11 . Define the terms principal stress and principal plane. A piece of elastic material has direct stresses of 80 MPa tensile and 40 MPa compressive on two mutuall perpendicular planes. A clockwise shear stress acts on the plane with the 80 MPa stress and an equal and opposite complementar shear stress acts on the other plane. The maimum principal stress in the material is 00 MPa tensile. Construct Mohrs' circle of stress and determine the following. i. The shear stress on the planes. (5 MPa) ii. The maimum shear stress. (80 MPa) iii. The minimum principal stress. (-60 MPa) iv. The position of the principal planes. (0o C.W. of 80 MPa direction) (Hint for solution. Make the larger stress act on the plane and the other on plane. Mark off, and. This is enough data to draw the circle.) D.J.DUNN

12 D.J.DUNN. COMPLX STRAIN. PRINCIPAL STRAINS In an stress sstem there are mutuall perpendicular planes on which onl direct stress acts and there is no shear stress. These are the principal planes and the stresses are the principal stresses. These are designates, and. The corresponding strains are the principal strains, and. Figure The strain in each direction is given b Most often we onl stud dimensional sstems in which case = 0 so )...(....(.) It is more practical to measure strains and convert them into stresses as follows. From equation. + = Substituting for in equation we have Similarl we can show These formulas should be used for converting principal strains into principal stresses.

13 SLF ASSSSMNT XRCIS No.. The principal strains acting on a steel component are and 6. Determine the principal stresses. = 05 GPa = 0. (Ans..47 MPa and.79 MPa). The principal strains acting on a steel component are -00 and 60. Determine the principal stresses. = 05 GPa = 0. (Ans. 9. MPa and -.4 MPa). MOHR'S CIRCL OF STRAIN Two dimensional strains ma be analsed in much the same wa as two dimensional stresses and the circle of strain is a graphic construction ver similar to the circle of stress. Strain is measured with electrical strain gauges (not covered here) and a tpical single gauge is shown in the picture. First consider how the equations for the strain on an plane are derived. Figure Consider a rectangle A,B,C,D which is stretched to A',B',C',D under the action of two principal stresses. The diagonal rotates an angle from the original direction. The plane under stud is this diagonal at angle to the horizontal. Figure The principal strains are = B/AB so B = AB and = B'/CB so B' = CB BH is nearl the same length as FG. FB' = FG + GB' = B cos + B' sin = AB cos + CB sin The strain acting on the plane at angle is = FB'/DB = cos + sin = ½( + ) + ½( - ) cos...(.) D.J.DUNN

14 BF is nearl the same length as HG so BF = H - G = B sin - B' cos BF = AB sin + CB cos = BF/DB = cossin - cossin = ½( - )sin...(.4) quations. and.4 ma be compared to the equations for dimensional stress which were = ½( + ) + ½{( - )}cos...(.5) = ½{( - )}sin...(.6) It follows that a graphical construction ma be made in the same wa for strain as for stress. There is one complication. Comparing equations 4 and 6, is the shear stress but is not the shear strain. It can be shown that the rotation of the diagonal (in radians) is in fact half the shear strain on it () and negative so equation 4 becomes -½ = ½( - )sin...(.7) These equations ma be used to determine the strains on two mutuall perpendicular planes and b using the appropriate angle. Alternativel, the ma be solved b constructing the circle of strain. D.J.DUNN 4

15 . CONSTRUCTION Figure 5. Draw point 'O' at a suitable position (which is possible to see with eperience). Measure and along the horizontal ais using a suitable scale.. Find the centre half wa between the marked points. 4. Draw the circle. 5. Draw in the required plane at the double angle. 6. Measure and and. SLF ASSSSMNT XRCIS No.. The principal strains in a material are 500 and 00. Calculate the direct strain and shear strain on a plane 0 o anti clockwise of the first principal strain. (Answers 450 and -7..). Repeat the problem b constructing the circle of strain.. The principal strains in a material are 600 and -00. Determine the direct strain and shear strain on a plane.5 o clockwise of the first principal strain. (Answers 480 and -566.) D.J.DUNN 5

16 . STRAIN GAUG ROSTTS It is possible to measure strain but not stress. Strain gauges are small surface mounting devices which, when connected to suitable electronic equipment, enable strain to be measured directl. In order to construct a circle of strain without knowing the principal strains, we might epect to use the strains on two mutuall perpendicular plains ( and ) and the accompaning shear strain. Unfortunatel, it is not possible to measure shear strain so we need three measurements of direct strain in order to construct a circle. The three strain gauges are convenientl manufactured on one surface mounting strip and this is called a strain gauge rosette. There are two common forms. One has the gauges at 45 o to each other and the other has them at 60 o to each other. The method for drawing the circle of strain is different for each. We shall label the gauges A, B and C in an anti clockwise direction as shown.. CONSTRUCTION OF TH 45 o CIRCL Figure 6 We shall use a numerical eample to eplain the construction of the circle. Suppose the three strains are A =700 B = 00 C = 00 Figure 7. Choose a suitable origin O. Scale off horizontal distances from O for A, B and C and mark them as A, B and C.. Mark the centre of the circle M half wa between A and C. 4. Construct vertical lines through A, B and C 5. Measure distance BM 6. Draw lines A A' and C C' equal in length to BM 7. Draw circle centre M and radius M A' = M C' 8. Draw B B' Scaling off the values we find = 74, =58 and the angle = 0 o The first principal plane is hence 5 o clockwise of plane A. D.J.DUNN 6

17 SLF ASSSSMNT XRCIS No.4. The results from a 45 o strain gauge rosette are A = 00 B = -00 C = 00 Draw the strain circle and deduce the principal strains. Determine the position of the first principal plane. Go on to convert these into principal stresses given = 05 GPa and = 0.9. (Answers = 60 = -0 =8. MPa The first principal plane is 49 o clockwise of plane A) =6 MPa D.J.DUNN 7

18 . CONSTRUCTION OF A 60 o ROSTT Figure 8. Select a suitable origin O. Scale of A, B and C to represent the three strains A, B and C. Calculate OM = (A + B + C)/ 4. Draw the inner circle radius MA 5. Draw the triangle (60 o each corner) 6. Draw outer circle passing through B' and C'. 7. Make sure that the planes A, B and C are in an anti clockwise direction because ou can obtain an upside down version for clockwise directions. 8. Scale off principal strains. SLF ASSSSMNT XRCIS No.5. The results from a 60 o strain gauge rosette are A = 700 B = 00 C = -00 Draw the strain circle and deduce the principal strains. Determine the position of the first principal plane. Go on to convert these into principal stresses given = 05 GPa and = 0.5. (Answers = 7 = -98 =49.4 MPa =-.6 MPa The first principal plane is o anti clockwise of plane A) D.J.DUNN 8

19 4. COMBIND BNDING, TORSION AND AXIAL LOADING. When a material is subjected to a combination of direct stress, bending and torsion at the same time, we have a comple stress situation. A good eample of this is a propeller shaft in which torsion is produced. If in addition there is some misalignment of the bearings, the shaft will bend as it rotates. If a snap shot is taken, one side of the shaft will be in tension and one in compression. The shear stress direction depends upon the direction of the torque being transmitted. Figure 9 The bending and shear stresses on their own are a maimum on the surface but the will combine to produce even larger stresses. The maimum stress in the material is the principal stress and this ma be found with the formulae or b constructing Mohr s circle. In earlier work it was shown that the maimum direct and shear stress was given b the following formula. σ σ σ σ 4τ σ σ 4τ σma σ τma If there is onl one direct stress in the aial direction and an accompaning shear stress (assumed positive), then putting = 0 we have the following. σ σ 4τ σ 4τ σma τma If the aial stress is onl due to bending, then = B. From the bending and torsion equations we have formula for B and as follows. M MD M TR TD 6T σb and τ I I J J πd πd Substituting for and we get the following result. 6 6 σma M T M and τ ma T M πd πd It was shown earlier that the angle of the principal plane could be found from the following formula. tan. Putting = and = 0 this becomes tan. If there is onl bending stress and torsion we ma substitute for B and as before, we get the following formula. T tanθ M If the direct stress is due to bending and an additional aial load, (e.g. due to the propeller pushing or pulling), the direct stresses should be added together first to find as the are in the same direction. You could draw Mohr s circle to solve these problems or use the appropriate formulae. D.J.DUNN 9

20 WORKD XAMPL No.4 A propeller shaft has a bending stress of 7 MPa on the surface. Torsion produces a shear stress of 5 MPa on same point of the surface. The propeller pushes and puts a compressive stress of MPa in the shaft. Determine the following. The principal stresses on the surface. The position of the principal plane. SOLUTION Since we have stress values, the problem is best solved b drawing Mohr s circle. At the point considered we have two a direct stresses and a shear stress. The total direct stress is 7 - = 5 MPa. Let this be and let the shear stress be positive on this plane. will be zero and the shear stress will be negative on the plane. Construction of the circle ields principle stresses of 8.09 and -.09 MPa. The principle plane is.7 o clockwise of the plane. Figure 0 D.J.DUNN 0

21 Check the answers b using the formulae. σ ma σ σ σ 4τ 4τ τ ma 5 tan MPa MPa o WORKD XAMPL No.5 A solid circular shaft 00 mm diameter is subjected to a bending moment of 00 Nm and a Torque of 400 Nm. Calculate the maimum direct stress and shear stress in the shaft. SOLUTION As there is no other aial load we can use the simpler formulae. σ σ τ τ ma ma ma ma tanθ 6 πd 6 πd T M 4.07 MPa T.55 MPa M T M M τ π(0.) ma θ 5. o 6 π(0.) θ 6.6 o D.J.DUNN

22 SLF ASSSSMNT XRCIS No.6. A shaft is subjected to bending and torsion such that the bending stress on the surface is 80 MPa and the shear stress is 0 MPa. Determine the maimum stress and the direction of the plane on which it occurs relative to the ais of the shaft. (66.5 MPa 5.8 o clockwise of the ais). A solid circular shaft 0 mm diameter is subjected to a torque of 000 Nm and a bending moment of 8000 Nm. Calculate the maimum direct and shear stress and the angle of the principal plane. (47.9 MPa, 4. MPa and 7 o ). A solid circular shaft 60 mm diameter transmits a torque of 50 Nm and has a bending moment of 9 Nm. Calculate the maimum direct and shear stress and the angle of the principal plane. (.4 MPa,. MPa and 9.9 o ) 4. A solid circular shaft is 90 mm diameter. It transmits 8 knm of torque and in addition there is a tensile aial force of 8 kn and a bending moment of knm. Calculate the following. i. The maimum stress due to bending alone. (4.9 MPa) ii. The shear stress due to torsion alone. (55.9 MPa) iii. The stress due to the aial force alone. (.6 MPa) iv. The maimum direct stress. (8.5 MPa) v. The maimum shear stress. (59.9 MPa) vi. The position of the principal plane. (4.4 o relative to the ais) D.J.DUNN