I want to conclude our discussion of structural analysis with a discussion of static determinacy. And here we will only look at equilibrium of trusses and beams, which we've discussed already somewhat in static section 3A and 3B. So first applying this to simple trusses. The rigid truss, of course, will not collapse under the application of a load. So in this case, the first example ABCD, if I apply a load to this truss at the point B, the members AB and BC and CD Will all rotate, and the truss will collapse. In other words, it's not rigid. On the other hand, the simple triangular truss here, ABC, is rigid, because if I apply loads of those pins, it will remain rigid and will not collapse, at least under the application of a moderate load. We construct a simple truss by starting with a triangle and then successively adding two members and one connection. So, in this case, we can start off with the truss ABC, then we add a new node D and two members BD and CD and we continue on this way and it remains a simple truss. This can apply to much more complex situations, for example if we start with a triangular truss EBD, then we add the node point C and two members BC and CD. Next I can add the node A and two new members AB and BC. Next, the node F and new members DF and CD, and finally, the node G and the members GF and AG, etc. So, this remains a simple truss and we can build up trusses of arbitrary complexity in this way, even though they remain in the definition of a simple truss. Now this means that there is a simple relationship between the number of members m and the number of joints j. Where the relationship is the number of members m is equal two times the number of joints minus three. Now our conditions of equilibrium at the points or joints provide two j equations because at each joint or node we have two equilibrium equations. Sum of the forces in the x direction is zero and sum in the y direction is equal to zero. There's no moment equation because there is no moment about the pins. So generally speaking, if we have m, unknown member forces. In other words we have m members with an unknown force in them and we have r, unknown reactions or supports. And for a simple truss, we just showed that 2j is equal to m plus three, therefore we can solve it for m member forces and three reaction forces at the supports from statics. And equilibrium of the entire truss provides three additional equilibrium equations which are not independent of the pin equations. So, this example here, we can look at in this case we have a simply supported truss. We have a pin support here. So we have a vertical force and a horizontal force here. And a roller support at D. Therefore we have only one component of force there. So in that case we have three unknown reactions. So generally speaking then we can solve a trust if m plus r is equal to 2j it's statically determinant. So this leads to a general consideration of determinancy that if the number of unknowns is equal to m plus 3, the number of members plus three unknown reactions. And, in this case, the number of members is equal to nine. One, two, three, four, five, six, seven, eight, nine, members. We have six joints. One, two, three, four, five, six and three reactions. two reactions at A and one reaction at the roller support D. Therefore, in this case 2j is equal to 12, m plus 3 is equal to 12. Therefore, m plus r is equal to 2j. And we're okay. We can solve it. It's statically determinant. So this leads to a general classification that if m plus r is equal to 2j for a simple truss then it is statically determinate, we can solve it from statics alone. On the other hand, if m plus r is greater than 2j, then we have too many unknowns and it's statically indeterminate. We cannot solve it from statics alone. If m plus r is less than 2j then there are not enough member forces, the truss is not rigid, it's unstable. And for the statically indeterminate here, this quantity m plus r minus 2j, is known as the degree of static indeterminacy. And this is the corresponding section from the reference handbook, which states the same thing. In addition they also include in here a plain frame and the difference is that a plain frame has one or two built in or cantilever ends. In other words at least one member is a multi-force member. But I don't think we'll need to worry about that here. So, some examples. The first one, a simple triangle ABC, in this case we have three members. So M is equal to three. And if we have three reactions then we have 3 plus 3 is equal to 6, is equal to two times the number the joints, three joints, 2j, therefore, if that is simply supported that is statically determined it's okay. This case, the second one here ABCD. We have in this case we have four members, m is equal to four, and again if it's simply supported, we have three reaction forces. So m plus r is equal to seven, which is less than 2j. We have four joints, so 2j is equal to eight. So therefore this situation is unstable. So this will collapse inwards like this under an applied load. How about the bottom one? In this case, m plus r, we have nine members if you count them up there. And if it's simply supported with three reaction forces, We have m plus r, is equal to 9 plus 3, is equal to 12. And we have 6 joints, so that's equal to 2j, so that one is also okay. It's statically determinant. We can solve it. This concept also applies to beams, which I won't go to any detail but we can also have statically indeterminate beams and this is called a propped cantilever. So we have a ridged cantilever support at A at the left hand end and we have a roller support now at B of the right hand end. So in this case we have the number of members is one, AB the cantilever beam, but now we have four reactions. We have a horizontal component of force, HA at the support. A vertical reaction, RA and also because it's a rigid cantilever, we have a moment at A. And also we have a vertical component, a force B at the right hand side. Therefore we have four unknowns here. So m plus r is equal to one plus four is equal to five, but 2j we have two joints at A and B so 2j = 2 x 2 = 4 so therefore m plus r is greather than 2j, in other words it's statically indeterminate. Which we knew already. We have four unknowns but only three equations therefore we cannot solve this by statics alone. And the quantity m plus r minus 2j is equal to 1 plus 4 minus 4 is equal to 1. So then we would say this beam is statically indeterminate to the first degree. So therefore we cannot solve this by statics, in other words the equilibrium equations alone. We need additional equations and in this case you can either do it by solving the differential equations of the deflection curve, those differential equations that we had earlier. But this is only possible for the simplest of problems. Or we can also use superposition whereby we deploy compatibility and relationships between force and displacement, similar to we did in this module on mechanics of materials, section IIb. So in this case we might proceed something like this. We might remove the support of the right hand side, and we say in this case that we choose RB as the static redundant role support, and we remove it there, and then we term this a released structure. Or alternatively, we can remove the moment of the cantilever support, as the static redundant. We remove that there, and then again, we have a so-called released structure, which we can solve by statics. So this is just an overview of the idea of determinacy and indeterminacy in trusses and beams and this concludes our discussion of structural analysis.
