 In this example problem, we are going to examine the torsional behavior of the J stiffener shown here, which comprises of a thin plate of 1 mm thickness along this semi-circular section and straight section connected to a base plate with a thickness of 2 mm. The other dimensions are shown here in this diagram. The problem asks us to calculate the maximum stress and angle of twist per unit length of the J-stringer profile for an internal torque of 5 Nm. Now the question also tells us we can give our answer as a magnitude, as the direction of the torque is not given. Basically, we're really interested in what is the magnitude of the angle of twist and the magnitude of the shear stress, not necessarily what direction it's occurring. The problem also gives us that the material has a shear modulus of 28 GPa. So let's look at this problem and look at how we can approach it. Well first, we have to take into consideration that the J-stiffener profile contains sections with different thicknesses. So we need to divide this profile into two thin rectangular plates with constant thicknesses that can act together. So we'll have section 1, which has a 1 mm thickness, plus section 2, which had our 2 mm thickness. Now because we are dividing it up, we have to be aware of superposition. So section 1 will carry a portion of the torque and section 2 will also carry a portion of the torque. But the total torque still has to equal our internal torque of 5 Nm. So that internal torque T is equal to T1 plus T2, and we'll call this equation I. Now when we analyze these separately, we also have to keep in mind of how they act together. So yes, the torques superimpose, but we also have a compatibility condition. Because these have to deform together, our angle of twist theta for the entire section must be the same as the angle of twist of section 1 and must be the same as the angle of twist of section 2. And we'll call this equation I. So this sets up the problem. Now let's start with looking at the angle of twist. We have a formula for the angle of twist of a thin rectangular plate. And that is the angle of twist per unit length, so d theta by dz, is equal to 3 times the torque in that plate divided by shear modulus times the width of the plate times the thickness of the plate cube. This was equation 3.6 from the lecture notes. So now what we can do is take this equation and substitute it into equation I. Now our compatibility equation says that the angle of twists are equal, and our equation is in terms of angle of twist per unit length. But if the angle of twists are the same and the loading is the same, then the angle of twist per unit length is also the same. So we can directly take this and say that 3t over gs t cubed for plate 1. So 3t1 divided by gs1 t1 cubed must equal 3t2 over gs2 t2 cubed. So this is just that the angle of twists are the same. We see immediately here that the threes will cancel out and our shear moduli will cancel out. And we could get t1 is equal to, if we take this up to the top, s1 t1 cubed over s2 t2 cubed, which I will write down here. And I'll substitute in s1, which is 35 millimeters is the length of the vertical section, plus 15 times pi is the perimeter of this semicircle. And this is multiplied by the thickness, which was 1 millimeter cubed. And we divide that by a width of 20 millimeters times a height of 2 millimeters to the power of 3. Now, you notice I've kept my units here in millimeters, where we normally convert to base units. But because I basically have a ratio of s t cubed over s t cubed, all of those units will cancel. This is a dimensionless term as long as I have consistent units. So I've left it in millimeters for that reason. If I take that expression, I can simplify that and get that t1 is equal to 0.513 t2. I can now substitute this result into equation i, so that I can get an expression of t2 in terms of our total torque of 5 Newton meters. So I get 5 Newton meters is equal to 0.513 t2 plus t2. So t2 becomes 5 divided by 1.513. So I will then get t2 is 3.3 Newton meters. Therefore t1 is 5 minus 3.3 or 1.7 Newton meters. Now we want to calculate the angle of twist, so we'll do it for section 2. So d theta 2 by dz is equal to 3 t2 divided by g s2 t2 cubed. I'll now do this, but now I have to convert everything to base units so that I can get a consistent set of units out. So 3 times 3.5 Newtons per meter divided by 28 times 10 to the 9 pascals times 0.02 meters is the width of the base plate times 0.002 meters is the thickness, the 2 millimeter thickness to the power of 3. If I simplify that I will get a value of 2.3 radians per meter which is 134 degrees per meter. Now the angle of twist is the same for section 1 and section 2. This was our compatibility equation and it's actually also equal to the angle of twist for the entire section. So the entire angle of twist d theta by dz is 134 degrees per meter. And this is 37 percent of a full revolution per meter. So that's half of the problem. We've calculated our angle of twist per unit length. Now we want to look at the maximum shear stress. And the maximum shear stress for a rectangular plate is given by equation 3.5 from our lecture notes where tau max is equal to 3 times the torque divided by s times the thickness squared. So we need to check both sections for this because we had a difference in the amount of torque but also a difference in the thickness. So we have to make sure that we check the maximum stress in both of those portions. So if we look at section 1, tau max in section 1 is 3 times the torque of 1.7 newtons per meter. And divided by our total length of the plate was 35 plus 15 pi times 10 to the minus 3 converting it to meters. And our thickness was 1 millimeter or 0.001 meters squared. So if we substitute all of this in, we get 62 megapascals. Now if we look at section 2, we have the higher torque but it's compensated by the higher thickness. But we still end up, despite the higher thickness and shorter width here, this combination with the higher torque results in a higher shear stress of 124 megapascals. So this is the maximum shear stress in the entire stiffener profile. And it's perhaps interesting to note that this is a very high shear stress. It's actually nearly the shear strength of a 2024 T3 aluminum alloy. So this shows that even for a very small torque of 5 newton meters, it's not a very high torque to be carried by a structure. But a small amount of torque can produce very, very high shear stresses in an open section. And this is why we don't use open sections as torque carrying structures, but sometimes we have newtons torques that can be carried by open sections if it's following the deformation of the rest of the structure. So sometimes we need to be able to check such profiles for their shear stress and deformation.