Mohr's Circle: Stress Transformation Explained
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Stress transformation without memorised trigonometry. We start with a small square under a known stress state, cut it at an angle, and watch the normal and shear stress on that cut change as the angle turns. Plotting the pair as the cut rotates draws a circle on its own, and three lines of algebra prove it is exactly a circle, handing over its centre and its radius. From there the circle is read rather than calculated: the principal stresses and the angle where they act, the maximum shear and why it always sits forty five degrees away from the principal directions, and a six step recipe for drawing the circle for any new stress state from scratch.
Here is a small square of material, cut out of something bigger and magnified. Nothing about it is moving. But the material all around it is pushing and dragging on every face, and those pushes and drags are what we mean by a stress state. On the two vertical faces, the material pulls straight outward, perpendicular to the face. That is a normal stress. We call it sigma x, and here it is eighty megapascals of tension. On the two horizontal faces there is a second normal stress, sigma y, twenty megapascals. Same idea, turned ninety degrees. And there is a third thing going on. The material also drags along the faces, parallel to them rather than perpendicular. That is shear stress, tau x y, thirty megapascals. It has to appear on all four faces at once, or the little square would spin. Three numbers describe this state completely. Eighty, twenty, thirty. Hold on to them, because every single number in the rest of this lecture comes out of exactly those three. Now the question that actually matters. Those three numbers belong to the faces I happened to draw. Slice the square some other way, and the material pushes and drags on that new face by different amounts. Here is that cut. I will name it by the direction of its normal, the line pointing straight out of the cut face, at an angle theta measured round from the x direction. The stress on that face splits into the same two pieces as before. A normal stress along the arrow, sigma theta. And a shear stress running along the face itself, tau theta. Turn the cut, and both of them change. Watch the two arrows as theta grows. Bring it back the other way and they change again. Neither of these numbers is a property of the material on its own. Both of them depend entirely on where you decided to cut. So here is where the equations come in. These two hand you sigma theta and tau theta for any angle you like, and you have almost certainly met them before. Written out, they look like trigonometry to memorise. Let us take them apart instead. The first piece is simply the average of the two normal stresses. Fifty megapascals for us. There is no theta anywhere in it, so it is exactly the same on every cut you could possibly make. The second piece is half the difference of the two normal stresses, thirty for us, multiplied by the cosine of two theta. It swings between plus thirty and minus thirty as the cut turns. The third piece is tau x y, thirty again, multiplied by the sine of two theta. It is zero when the cut sits on the x face, and it is largest at forty five degrees. The shear equation is built from the same two amplitudes, swapped over, with one sign flipped. That is the whole of the difference between them. And now the thing everybody wonders about. Why two theta, and not just theta? The reason is geometry rather than algebra, and it is worth thirty seconds. A cut at theta, and a cut at theta plus one hundred eighty degrees, are the same plane. Watch the cut go all the way round, and it lands exactly where it started. So everything has to repeat after half a turn of the cut. The only way to make a cosine repeat after half a turn is to double the angle inside it. That is where the two comes from. It is not a fudge factor. So that is the shape of the answer. A constant, plus a cosine, plus a sine, everything running in two theta. Plot sigma theta against tau theta as the cut turns, and that shape is going to draw something specific. In a moment it draws a circle.
Same square, same three numbers. This time I am going to keep score. Every time I turn the cut, I get a pair of numbers out of it, and I am going to plot that pair as a single point. Horizontal axis, the normal stress on the cut. Vertical axis, the shear stress on the cut. Both in megapascals. This plane has nothing to do with where anything sits in space. It is a scoreboard. Start with the cut we already know, the one lying on the x face, theta equals zero. Sigma is eighty and tau is thirty, so the point goes here. Call it X. Now turn the cut round to twenty degrees. The normal stress has climbed to ninety two, and look at the green arrow. The shear has very nearly gone. Push on out to forty five. On the way it passes twenty two and a half, where the shear is exactly zero for an instant, and then it comes back pointing the other way. Minus thirty. At sixty seven and a half, sigma is back to fifty, which is the average we started from, and the shear has reached its most negative value, minus forty two point four. Ninety degrees puts the cut on the y face. Twenty across, minus thirty up. That point is Y, and notice it is straight across the plot from X. Now let it run all the way to one hundred eighty degrees, and the path closes on itself. It is a circle. Nobody drew a circle. The two equations drew it. Two things about that trip are worth writing down. The cut turned through one hundred eighty degrees, and the point went round exactly once. That is the two theta, showing up as geometry instead of algebra. And the point ran clockwise while the cut turned counterclockwise. Some books draw the shear axis pointing down so that the two rotations agree. Nothing else changes if they do. I am leaving it the honest way up. So let us prove it, because the proof is three lines long and it hands us the centre and the radius for free. Give the pieces short names. C for the average, a for half the difference, b for the shear. Move C across to the left in the first equation. Then sigma minus C is a cos two theta plus b sin two theta, and the shear equation is already in that form. Now square them both and add. The cross terms come with opposite signs and cancel each other. Cosine squared plus sine squared is one. Everything collapses. What is left is sigma minus C, squared, plus tau squared, equals a squared plus b squared. That is the equation of a circle, and it is not an approximation to one. The centre sits at sigma equals C, on the horizontal axis, always. The radius is the square root of a squared plus b squared. For our numbers that is root eighteen hundred, forty two point four megapascals. And that is Mohr's circle. It is not a memory aid and it is not a drawing trick. It is the actual set of every stress pair this material can show you, on every cut you could ever make through it. So let us read it.
There is the circle, standing on its own, with its centre at fifty and its radius forty two point four. Before we read anything off it, one more piece of algebra, and it is the piece that makes the whole thing easy. A cosine and a sine of the same angle, added together, always make a single cosine of that angle, shifted along. The amplitude of that single wave is R, and the shift is what we are going to call two theta p. So both transformation equations collapse into these. Sigma is the centre plus R times a cosine. Tau is minus R times a sine. One amplitude, one angle, and that angle counts from the principal direction. Look at what that says. As the cut turns, sigma and tau are the horizontal and the vertical coordinates of one point going round a circle of radius R, centred at C. Those two lines are the circle, written out. Now we read. The rightmost point of the circle is the largest normal stress this material will ever feel, on any cut whatsoever. The leftmost point is the smallest. Those two are the principal stresses. Centre plus radius, and centre minus radius. Ninety two point four, and seven point six megapascals. That is the entire calculation, and you can do it in your head. But notice something else about those two points. Both of them are sitting on the horizontal axis, so the shear on those two cuts is exactly zero. That is the real definition of a principal plane. Not biggest. No shear. The circle makes those two statements into the same statement, because the only two points on it with zero shear are the two ends of the horizontal diameter, and those are the two extremes of sigma. So which cut do you have to make to get them? Read that off the circle too. Here is X again, our starting point, the x face. It sits thirty across and thirty up from the centre. Thirty and thirty means the radius out to X leans at forty five degrees. To get from X round to sigma one, you swing that radius clockwise through forty five degrees. In symbols, the tangent of two theta p is b over a, which is twice the shear over the difference of the normal stresses. Thirty over thirty is one, so two theta p is forty five degrees. And angles on the circle are always two theta. So the physical cut only has to turn half as far. Theta p is twenty two and a half degrees. One quick check before we move on. Add the two principal stresses together. Ninety two point four, plus seven point six, is one hundred exactly. Now add the two you started with. Eighty plus twenty is also one hundred. That is not luck. The sum of the two normal stresses is the same on every cut, because it is twice the centre, and turning the cut never moves the centre. It is the cheapest check you will ever run on your own arithmetic. Here are those two angles side by side, because keeping them straight is most of the skill. On the left, forty five degrees, swept at the centre of the circle. On the right, the same square rotated by twenty two and a half. And on that rotated square, the faces carry ninety two point four and seven point six, straight pushes with nothing dragging along them at all. No shear arrows anywhere. That is what principal means. So the ends of the circle gave us the biggest and smallest push, and the angle we swept to reach them gave us the cut they act on. Now let us go to the top of the circle instead.
The two ends of the circle gave us the extremes of normal stress. Now go to the other extreme. The top of the circle and the bottom of the circle are the two cuts with the most shear this material will ever show you. How much shear? Read the height. The centre sits on the axis, so the highest point of the circle is exactly one radius up. The maximum shear is the radius. Forty two point four megapascals. And that is not a coincidence of our numbers. The centre of a Mohr circle always lies on the horizontal axis, so the top is always exactly R above it. Maximum shear equals the radius, for every stress state there is. Now read the other coordinate while you are up there. The top point sits directly above the centre, so the normal stress on that cut is fifty. The average. That is worth saying out loud, because people get it wrong. The planes of maximum shear are not free of normal stress. They carry the average of sigma x and sigma y, whatever that average happens to be. So where is that cut? Straight up from the centre is a quarter turn round from sigma one. Ninety degrees on the circle. And angles on the circle are two theta, so halve it. Forty five degrees on the material. Theta s is twenty two and a half plus forty five, which is sixty seven and a half degrees. There is the picture. On the left, the right angle at the centre of the circle. On the right, the same square turned by sixty seven and a half degrees, which is forty five degrees on from the principal orientation. Its faces are dragged hard, forty two point four each way, and they are also pushed by the average, fifty. Shear and normal stress together, which is exactly what the top of the circle told us. And here is the fact to carry out of the room. The planes of maximum shear always sit forty five degrees from the principal planes. Always. For every stress state, because the top of a circle is always a quarter turn from its right hand end. Which explains something you have already seen in a lab. Pull a mild steel bar in simple tension. Sigma x is the tension, sigma y is zero, and there is no shear on those faces at all. So the circle runs through the origin, and its centre and its radius are both half the tension. The highest point is up here, half the tension worth of shear, at a quarter turn round the circle. A quarter turn on the circle, half of that on the bar. Forty five degrees. Which is precisely the angle those shear bands run at when a ductile bar yields, and precisely why they run at all.
So here is the whole thing as a procedure. Six steps. Hand somebody a new stress state and they can do this on the back of an envelope, and I want you leaving able to do it without looking anything up. Step one. Draw your axes. Normal stress running across, shear stress running up. Pick a scale that fits your numbers and stick to it, because the circle only looks like a circle if both axes match. Step two. Plot the x face as a single point. Sigma x across, tau x y up. Eighty and thirty. That one is X. Step three. Plot the y face. Sigma y across, but minus tau x y up. Twenty and minus thirty. That one is Y. The sign flip on that second one is not a rule to memorise. It is exactly what the shear equation gives you at ninety degrees, and it is what makes X and Y land on opposite ends of one diameter. Step four. Join X to Y with a straight line. Where it crosses the horizontal axis is the centre, and that crossing is always the average of the two normal stresses. Fifty for us. Step five. The radius is the distance from the centre out to X. Thirty across, thirty up, so root eighteen hundred. Forty two point four. Step six. Put your compass on the centre, open it out to X, and draw. That is the circle, and the four numbers you came for are already sitting on it. Right hand end, ninety two point four. Left hand end, seven point six. Top of the circle, forty two point four of shear, at a normal stress of fifty. And the angle from X round to the right hand end is two theta p. Six steps, and not one of them is a formula you have to trust. Every one of them is something you can see. One more use, and then you have everything. Suppose somebody now asks you for the normal and shear stress on a cut at sixty degrees. You do not go back to the transformation equations. You already have the answer. Start at X, because X is theta equals zero. Turn through twice sixty, so one hundred twenty degrees, and turn clockwise, because that is the way the point runs. You land here. Call it Q. Now just read the coordinates. Sigma is sixty one megapascals, and tau is minus forty one. That minus sign is information, not a nuisance. It tells you which way along the face the material is being dragged. That is the point of the circle. Once you have drawn it, every cut through that material is already answered. You are not calculating any more. You are reading. So, four things to carry out of the room. The centre is the average of the two normal stresses, and turning the cut never moves it. The radius is the square root of half the difference squared, plus the shear squared. That is just the length of the line from the centre out to X, so it is Pythagoras and nothing more. The principal stresses are the centre plus and minus the radius, and they act on the cut you reach by turning through half the angle you measure on the circle. And the maximum shear is the radius itself, sitting forty five degrees away from the principal planes, on faces that also carry the average normal stress. None of that was memorised. All of it came from watching one pair of numbers move as a cut turned. Draw the circle first, and the trigonometry never comes back.
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