Chapter 15: Visualizing Stress and Strain

Transcription

1 hapter 15: Visualizing Stress and Strain Measuring Stress We can see deformation, if it's large enough, but we cannot see stress, and in most cases we cannot measure it directl. Instead, we can use a strain gauge to measure the strain on the surface of an object that is being elasticall deformed, and use Young's Modulus to estimate the stress. The earliest strain gauges in the 1930s were made of fine wire. If ou glue a ver thin wire along the bottom surface of a beam, then load the beam in bending, the wire will stretch and become thinner. The thinner the wire becomes, the higher its electrical resistance. measuring this resistance ou can determine strain, then calculate stress as E, provided the object is being ε strained elasticall. Modern strain gauges use the same principal, but are made of thin metal foil instead of wire. The foil is embedded in a pad that is eas to glue to the part being tested. The longer the conductor, the more sensitive the gauge. With a foil strain gauge, the conductor loops back on itself man times. The total wire length is the loop length times twice the number of loops; in this cartoon, the total length is 14 times the loop length. loop length Stress at the ase of a Short lock onsider the short block in hapter 1, Eample #4. point load is offset from both neutral aes. The total stress at the four corners is psi, psi, psi, and -1.5 psi. 3 ft. W 1 kips 0.5 ft. 1.5 ft. D 4 ft..5 ft. In plan view, point has a tensile stress, while points,, and D have compressive stresses psi Plan view D -1.5 psi D We can calculate the stress at an point along the edges b interpolating from the corner stresses, because the stress varies linearl from one corner to the net. Likewise we can calculate the stresses within the area of the block b interpolating from the edge stresses..5 ft. Point load location psi psi 4 ft. 1

2 One wa to visualize these stresses is with a grid pattern in plan view, identifing the stress at each location, or node. In the diagram, the nodes are at half-foot intervals D lternativel, we can calculate contour lines of stress (a.k.a. isostress lines, coloring the spaces between the lines. The pattern shows that the pink areas are in tension, while the blue areas are in compression. The picture helps us see that the isostress lines are diagonals, and there is a zero stress line stretching from the edge to the D edge psi D Mohr's ircle Normal and shear stresses in uniaial tension Take a tensile bar, cut it in half transversel to the ais, and glue the bar back together. If ou pull on the bar along the -ais, a normal stress P develops in the cut adhesive. rea is the cut surface area of the bar (the cross-sectional area. There is no shear stress, because the adhesive is not loaded in shear. onsider a little square piece of adhesive: it has a normal stress acting on the left and right faces, but no normal stress on the top or bottom faces, and no shear stresses. P

3 ut a similar tensile bar in half along its -ais, and glue the bar back together. If ou load the bar as shown, a shear stress P develops in the adhesive. gain, area is the cut surface area of the bar. There is no normal stress, because the adhesive is onl loaded in shear. little square piece of adhesive has a shear stress acting on the top and bottom faces, in the direction. The square becomes a parallelogram as it distorts. We could call this shear stress, because it acts along the direction. Think about how the little square of adhesive distorts relative to other little squares of adhesive on its left and right: its neighbors tr to prevent the piece from parallelogramming, so there are shear stresses in the direction too. The stress required to produce the parallelogram is the same along each edge, so ; we'll use a single term. positive shear stress is one that distorts the square clockwise relative to its base; a negative shear stress distorts the square counterclockwise. ut another bar at an angle to the ais, and glue it back together. If ou pull on the bar, the adhesive is loaded in normal tension and in shear. The two segments of the bar want to pull awa from each other and slide along each other. Let's define two new aes, ' and ', perpendicular and parallel to the cut surface. We have a shear stress acting parallel to the cut surface, and a tensile stress ' acting perpendicular to the cut surface, in the ' direction. We don't have to actuall cut the bar and use an adhesive; instead, we can imagine the stresses that act in various directions within the material. Draw a square that represents the piece of adhesive, and rotate it in different directions. We can calculate the stresses in the ' and ' directions b using equations from Statics: sum of the forces in an particular direction equals zero. P P cut cut Positive shear stress Negative shear stress ' ' '' ' First, draw the square element. In the general case, there could be a vertical normal stress, so include this stress in the sketch. θ cut 3

4 Net, cut the square along a diagonal in the ' direction, so the normal stress acting on this surface is ' and the shear stress acting on this surface is ''. The new '-' coordinate aes are tilted at angle θ counterclockwise from the original - coordinate aes. θ '' ' ' ' The square has a depth (the thickness of the stressed part, so each edge is actuall an area in three dimensions. Let the area of the cut surface be area a. Using trigonometr, the area of the left surface is a cos θ and the area of the bottom surface is asin θ. rea a cosθ θ rea a rea a sinθ The force acting on each face is equal to the stress times the area. We have 6 forces acting on the triangular bod. a cosθ θ ' a ' a a cosθ a sinθ a sinθ Look at the components of the forces acting in the ' direction. If ou add them up, the sum is zero. ' a acos θ+ asin θ+ a cos θsin θ Notice that area a cancels, so we can write ' cos θ+ sin θ+ cos θsin θ There are some trig identities that will help simplif the equation into something useful: cos 1+cos θ θ, sin θ 1 cosθ, sin θ cosθ sin θ, and cosθcos θ sin θ. a cosθ sinθ a cosθ cosθ a sinθ cosθ θ ' a ' a a sinθ sinθ 4

5 Now we get ' 1+cos θ 1 cos θ + + sin θ + cos θ+ + cos θ + sin θ + + cos θ+ sin θ Following the same procedure, we can calculate the sum of the forces in the ' direction. ' ' a a cos θsin θ+ a sinθ cos θ asin θ+ a cos θ a cosθ cosθ a cosθ sinθ θ ' a ' a rea a cancels. Using the trig identities, we can simplif the equation: ' ' ( + cos θsin θ+ (cos θ sin θ sin θ ( + cos θ ( sin θ+ cos θ a sinθ sinθ a sinθ cosθ We can calculate the stress in the ' direction b taking angle θ+90. Using free-bod diagrams and trig identities, we get: ' + cos θ sin θ ' ' ' θ+90 '' Now we have three equations that are functions of,, and, and θ. Let's pick values of normal and shear stresses, and see how ', ', and '' var b plotting stress vs. angle θ. The graph at the right shows ' as a function of angle θ if we pick 1ksi, 0.5 ksi, and 1. ksi. The graph is a sine wave, with peaks, valles, and zero values. Picking different values of,, and will change the amplitude and vertical position of the sine wave, but not the period. Stress (ksi ' Notice that the maimum values of ' occur at θ40 and 0, the minimum values occur at θ130 and 310, and ' 0 at θ100, 155, 83, and 385. ngle θ (degrees 5

6 This graph shows ' as a function of angle θ for the same set of,, and..0 ' The maimum values of ' occur at θ130 and 310, the minimum values occur at θ40 and 0, and ' 0 at θ13, 65, 193, and 35. This graph of ' is shifted 90 to the right of the graph of ' ; as a result, ou can find the angles for the maimum values of ' b adding 90 to the angles for the maimum values of '. Stress (ksi ngle θ (degrees This graph shows '' as a function of angle θ for the same set of,, and..0 The maimum values of '' occur at θ175 and 355, shifted 45 from the maimum values of. The minimum values of '' occur at θ85 and 65, and ' ' 0 at θ39, 19, 19, and 309. Stress (ksi '' ngle θ (degrees We can see how the three stresses interrelate b plotting them together. There are normal stresses for which the shear stress is zero, and there are shear stresses for which the normal stress is zero. The values and angles depend on the input conditions (,, and. Let's take another look at the equations for ' and '' : ' + + cos θ+ sin θ ' ' ( sin θ+ cosθ Stress (ksi ' '' ' ngle θ (degrees Rewrite the ' equation, putting two terms on the left: ' + cos θ+ sin θ Square both sides of the ' and '' equations: ( ' + ( cos θ+( sin θ cos θ+ sin θ ' ' ( sin θ ( sin θcos θ+ cos θ Net, add these two equations together: ( ' + + ' ' ( cos θ+( sin θcos θ+ sin θ + ( sin θ ( sin θ cosθ+ cos θ 6

7 We can eliminate two terms because ( sin θcos θ ( sin θ cos θ0. Rewriting the equation, we have: ( ' + + ' ' ( (cos θ+sin θ+ (sin θ+cos θ Using the trig identit cos θ+sin θ1, the equation simplifies to ( ' + + ' ' ( +. The term + is an average of two stress values, so let's call it avg. Now the equation is even simpler: ( ' avg + ' ' ( +. If we define the right side of the equation as R ( + then ( ' avg + ' ' R. This is the equation of a circle with a radius R and a center at coordinates ( avg,0. In 1880, Otto Mohr developed this method for visualizing stresses in two-dimensional and three-dimensional objects. His graphical method is also used for diagramming strains. We can use Mohr's ircle to determine the value and direction of maimum stresses within a loaded part. Follow this 6-step process for drawing Mohr's circle for the stresses acting at a point within a stressed object. For best results, use a compass and straightedge on graph paper. In this eample, use the same stresses as before: 1ksi, 0.5 ksi, and 1. ksi. Step 1 Draw two aes. Label the horizontal ais and the vertical ais. Step Mark the center of the circle at coordinates ( avg,0 and label it for center. In this case, avg + 1 ksi+0.5 ksi 0.75ksi. 7

8 Step 3 Mark point (, on the graph; in this case, (1ksi, 1. ksi. Label this point. (, - Step 4 Placing our compass point at the center of the circle, draw a circle passing through this point. The radius of Mohr's circle is: (, R ( + 1 ksi 0.5 ksi ( +(1. ksi 1.6 ksi Notice the circle also passes through point (,, which is (0.5 ksi,1. ksi. Mark this point and label it. line passing through points and is the diameter of Mohr's circle. R (, - Step 5 The maimum normal stress occurs at point D; we call it 1. The minimum normal stress occurs at point E; we call it. The maimum shear stress occurs at point F; we call this ma.these three stresses are called Principal Stresses. You can measure these values off the graph, or ou can calculate them: 1 avg +R0.75ksi+1.6 ksi1.976ksi avg R0.75 ksi 1.6 ksi ksi E (, F R ma D 1 (, - ma R1.6ksi Step 6 Starting on line segment -, measure the angle counterclockwise about point to line segment -D. This angle is θ. tan θ ( / (1. ksi 1ksi 0.5 ksi 4.8 θtan θ This is the angle of the principal stress 1. The angle of principal stress is θ ( θ - E (, F R ma D θ 1 (, - 8

9 Going back to the original stress element, we had 1ksi, 0.5 ksi, and 1. ksi. If we rotate the element counterclockwise 39.1, in this orientation there are no shear stresses, and the principal stresses are ksi tension and ksi compression. The maimum shear stress ma 1.6 ksi is oriented 45 from the maimum normal stress, at θ ksi 1. ksi 1 ksi ' ' ksi θ ksi Eample #1 n aluminum rod with a diameter of 0.5 inches is pulled with a load of 000 lb. as shown. alculate the applied stresses,, and. Use Mohr's circle to find the principal stresses 1 and, angle θ, and the maimum shear stress ma at point. Solution The applied normal stress is in the direction; there is no applied shear stress. 0 P 4 P lb. kip π d 10. ksi π(0.5in lb. 0 P avg + 0 ksi+10.ksi 5.09 ksi ma 5.1 ksi F R ( + ( 0 ksi 10.ksi +(0 ksi 5.09 ksi 1 avg +R5.09 ksi+5.09 ksi10. ksi avg R5.09 ksi 5.09 ksi0ksi ma R5.09ksi ngle θ0, b inspection. 1 (, - ( avg, 0 (, (0, 0 (5.1 ksi, 0 (10. ksi, 0 9

10 Eample # The rod in Eample #1 is pulled as before, and also twisted with a torque of 45 ft. lb. Solution Once again the applied normal stress is in the direction, but the torque creates an applied shear stress. 0 T P P 4 P lb. kip π d π(0.5in lb. 10. ksi as before T c J where c d and J π d 4 Substituting, T d 3 16 T 4 π d π d ft lb. 3 π(0.5 in. 3 1in. ft. kip.0 ksi 10 3 lb. avg + R ( + 0 ksi+10.ksi 5.09 ksi ( 0 ksi 10.ksi +(.0 ksi.6ksi 1 avg +R5.09 ksi+.6 ksi7.1 ksi avg R5.09 ksi.6ksi 16.9ksi ma R.6 ksi ma.6 ksi F (, (10. ksi,.0 ksi E D θ ksi ksi R -77 (, - (0 ksi, -.0 ksi tan θ (.0ksi 0ksi 10. ksi 4.3 θtan θ ; We can also use Mohr's circle to find out the stress in a plane at an angle α. Rotate counterclockwise through angle α from line segment - to a new line segment '- to find the shear and normal stresses acting in this direction. 10

11 Eample #3 Two 3 cm diameter wooden dowels are glued together on a 30 diagonal as shown. tensile load of 1.5 kn is applied. What are the normal and shear stresses acting on the glue joint? P α 30 Solution The applied normal stress is in the direction; there is no applied shear stress. P 4 P kn (100cm MPa m π d π(3cm m 10 3 kn.1mpa 0 0 avg +.1MPa+0 MPa.1 MPa ma.1 MPa F ' (3. MPa, 1.8 MPa R α60 1 (, - ( avg, 0 (, (0, 0 (.1 MPa, 0 (4. MPa, 0 R ( +.1MPa 0 MPa ( +(0 MPa.1MPa 1 avg +R.1MPa+.1MPa4.4 MPa avg R.1 MPa.1MPa0 MPa ma R.1 MPa ngle θ0, b inspection. Rotate counterclockwise through angle α from line segment - to a new line segment '-. The normal stress at point ' is avg +Rcos α.1mpa+.1 MPa cos( MPa. The shear stress at point ' is R sin α.1mpa sin( MPa. Now ou can compare these numbers with the tensile strength and shear strength of the adhesive. For eample, if the shear strength of the adhesive is.5 MPa, then select a smaller angle α that keeps the shear stress below.5 MPa. Theories of Failure Design engineers don't prevent failure, the manage it b designing products and structures so that when the fail due to old age, abuse, or ordinar service, nobod is injured. How do we define failure? It depends on the product or structure. bridge deck could be described as a failure if it deflects too far, even though it does not break. shear pin in a snowblower fails if if does not break when the blades are overloaded b ice, or does break under ordinar snow load. In Strength of Materials, elastic formulas do not work in the plastic zone, and usuall we do not want plastic deformation during service, therefore we can define failure as ielding for most products or structures. 11

12 William Rankine, a 19 th centur Scottish engineer, developed the Maimum Normal Stress Theor. He considered materials having the same strength in tension and compression (such as steel, not cast iron. He proposed a failure stress criterion f P ma where f is the failure stress and Pma is the maimum principal stress. If we plot principal stresses 1 and on and coordinates, the part is safe if the combined stress point is within the square. - f f Safe - f f Not safe 1 harles-ugustin de oulomb, a 19 th centur French phsicist, and Henri Tresca, a 19 th centur French engineer, developed the Maimum Shear Stress Theor (also known as the Tresca Theor, which defines failure as f ma ma min. This theor is more conservative than the Maimum Normal Stress Theor, as it cuts off the upper left and lower right corners of the square safe zone. - f f Safe f Not safe 1 - f In metals, we have a linear elastic curve on the stressstrain diagram. The area under the curve equals the strain energ per unit volume, u ε. Richard Edler von Mises was a Jewish mathematician who escaped ustria when the Nazis rose to power, and later joined the facult at Harvard. He developed the most widel-used failure criterion. From the energ per unit volume equation, he derived: f where 1 and are the principal stresses. The von Mises or Maimum Distortion Energ Theor produces an elliptical safe zone. Notice on all three stress graphs, the safe envelopes all pass through the same four points on the two aes, marked in red. number of other failure theories appl to special cases, such as the Maimum Strain Theor of arré de Saint-Venant, and the Internal Friction Theor. However, von Mises is used widel in finite element programs because it fits more data than other methods. ε area ε Safe Not safe 1 Ke Equations The average normal stress (center of Mohr's circle is avg +, and the radius is R ( + The principal stresses are 1 avg +R and avg R. The maimum shear stress ma R Find the angle of principal stresses with tan θ [ ( ] θtan 1 [ ( ] θ [θ] and 90 + [θ] 1

13 Problems for hapter 15: Visualizing Stress and Strain 1. Using the results of Homework Problem 1-5, identif the location of the zero-stress line.. t what angle does the maimum shear stress act in Eample #1 of this chapter? 3. Draw Mohr's circle for point on the surface of a clindrical pressure tank. The wall thickness is 3/16 in., the diameter is 30 in., and the internal gas pressure is 400 psi. Note: is the longitudinal stress, is the hoop stress. alculate the principal stresses and the maimum shear stress. 4. The tank in problem #3 is welded from three pieces: the ends are made of two hemispherical caps, and the bod is made b welding a strip of steel in a heli at a 45 angle (it looks like a paper towel roll. Use Mohr's circle to find the shear and normal stresses acting on the 45 weld. 5. Use Mohr's circle to find an algebraic epression for the maimum shear stress in a spherical pressure vessel, in terms of pressure, inside diameter, and wall thickness. 6. Which is better, using a graphical method for evaluating stresses, or using equations? Eplain our answer. 13