Statics – Truss Problem I. Statics We are going to start our discussion with something very familiar. We are going to look at a simple statically determinate truss. The type of structure you analyzed in Statics. Trusses are characterized by linear elements (beams) that are pinned together at their ends. The pinned connection offers no resistance to rotation so there is no moment around the joint. Consider the frame shown below: The members in the structure are numbered with circles around them. We call these members elements. The points where the elements join are called nodes. In the problem above there are 5 elements and 4 nodes. We can draw free body diagrams at the nodes. E -F5F5R3R2-F4F4F1-F1F2R1-F2F3F31 2 4 3In the free body diagrams above, we are using the variables F for the forces in the elements. The subscripts on these forces refer to the element number. The variables R represent the reactions. Their subscripts are assigned arbitrarily. We know that the sum of the forces at each node must equal the externally applied force at that point. There are two equations for each point, the horizontal equation and the vertical equation. We can write the equations as: F1F2F3F4F5R1R2R3External1H - F1 = -E 1V - F5 = 0 2H F1 +.707F3 = 0 2V - F2-.707F3 = 0 3H + F4 = 0 3V F2 + R1 = 0 4H -.707F3 + R3= 0 4V +.707F3 + F5 + R2 = 0 The row above the equations lists all of the unknowns. The equations are written to keep the unknown terms in their respective columns. The column on the left list the equations needed to solve the forces in the system. The number in this column refers to the node, the H for horizontal forces, and V for vertical forces. Empty cells in the equation table indicate the force or reaction specified in the column does not apply to that particular node. These empty cells could be filled with zeros. The coefficients and sign of the coefficients for the element forces can be readily determined by looking at the direction of the element with respect to the node we are analyzing. This is shown in the diagram below. Ø ElementNode Horizontal Reference )sin()cos(θθ==tCoefficienVerticaltCoefficienHorizontalThere are several things to notice about the equations above: 1. There are 8 equations and 8 unknowns. This system of equations can be solved for the forces in the elements and the reactions. 2. In this particular problem, there are 5 elements and one unknown force per element. There are also 3 unknown reactions. This gives us a total of 8unknowns in the problem. There are 2 equations for each node, a vertical equation and a horizontal equation which gives us a total of 8 equations. We can solve this system since we have as many equations as we have unknowns. In general, problems of this type must satisfy the equation shown below if they are solvable. 2 * Nodes = Elements + Reactions 3. If we add another element as shown at the right, there will be 8 equations and 9 unknowns. The problem will no longer be statically determinate and cannot be solved using the technique we are discussing. 4. In this type of problem, there must be at least 3 reactions. Two reactions are required to eliminate the X and Y translation and another reaction is required to eliminate any rotation of the object. The basic premise is that the object cannot move. It is in static equilibrium. 123 4External Force E Element 5 12346II Matrix Notation The equations above can be rewritten in matrix notation as shown below. (2.1) ⎪⎪⎪⎪⎪⎩⎪⎪⎪⎪⎪⎨⎧⎪⎪⎪⎪⎪⎭⎪⎪⎪⎪⎪⎬⎫=⎪⎪⎪⎪⎪⎩⎪⎪⎪⎪⎪⎨⎧⎪⎪⎪⎪⎪⎭⎪⎪⎪⎪⎪⎬⎫×⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣⎡−−−−−000000001010707.0010000707.00001000100000100000000707.1000000707.010001000000000001132154321ERRRFFFFF External Forces Or as: []}{}{ EFM=× (2.2) Matrix based upon shape of the structure Forces and ReactionsThis structure is common to many types of engineering problems. The left-hand-side is completely dependent upon the geometry and the right-hand-side upon the driving forces. If the geometry does not change, we can examine many load cases without changing the left-hand-side of the equation. This can lead to solution efficiencies we will discuss later. III Matrix Algebra Review 3.0 Matrix Multiplication Matrix multiplication is a relatively simple operation where the rows of the first matrix are multiplied times the columns of the second matrix. The results are shown below. ⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡++++++++++++++++++=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡×⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡irhogliqhngkiphmgjfreodlfqendkfpemdjcrboalcqbnakcpbmajrqponmlkjihgfedcba (2.3) In matrix notation we can write: [][][]CBA=× (2.4) which is an equivalent statement. Note that in general, matrix multiplication is not commutative so that: [][][]CAB≠× . (2.5) You can easily prove this by multiplying matrix [B] times [A]. The product will not be equal to [C]. For the two matrices to be equal, each term of the two matrices must be equal. 3.1 Matrix Transpose A matrix transpose is created by swapping the rows and columns of a matrix. This is shown below. ⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡ifchebgdaihgfedcbaT (2.6) The superscript T indicates the transpose operation.3.2 Identity Matrix The identity matrix is a special matrix composed on 1s on the diagonal and 0s everywhere else. ⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡100010001 (2.7) The identity matrix has a property similar to the scalar number 1. Any matrix multiplied by the identity matrix is equal to itself. This is shown below. [I] is the identity matrix and [M] is any other matrix. [] [ ][]MMI=× (2.8) or [][][]MIM=× (2.9) 3.3 Matrix Inverse If the matrix [B] is the inverse of [A] then: [][][]IAB=× (2.10) Another way of stating this is: [] [][]IAA =×−1 (2.11) The -1 superscript indicates the inverse of a matrix. The inverse of a matrix cannot be easily created with simple row column operations as could the transpose of a matrix but it does have important uses. In solving problems that can be represented mathematically as shown in equation (2.2) []}{}{ EFM=× (2.2) We can multiply both sides by the inverse of [M] [] [][]}{}{11EMFMM