Fundamentals of Structural Analysis Episode 2 Part 7 potx

Other Topics by S. T. Mau 295 Example 7. Find the natural vibration period of a cantilever beam as shown. EI is constant and the mass is uniformly distributed with a density ρ per unit length of the beam. Assume there is no damping in the system. A cantilever beam with uniformly distributed mass. Solution. We shall limit ourselves to exploring the lateral vibration of the beam, although the beam can also have vibration in the axial direction. A rigorous analysis would consider the dynamic equilibrium of a typical element moving laterally. The resulting governing equation would be a partial differential equation with two independent variables, a spatial variable, x, and a time variable, t. The system would have infinite degrees of freedom because the spatial variable, x, is continuous and represents an infinite number of points along the beam. We shall pursue an approximate analysis by lumping the total mass of the beam at the tip of the beam. This results in a single degree of freedom (SDOF) system because we need to consider dynamic equilibrium only at the tip. Dynamic equilibrium of a distributed mass system and a lumped mass system. The dynamic equilibrium of this SDOF system is shown in the above figure. The dynamic equilibrium equation of the lumped mass is m 2 2 dt vd +kv =0 (6) where m= ρ L and k is the force per unit length of lateral deflection at the tip. We learn from beam analysis that the force at the tip of the beam needed to produce a unit tip deflection is 3EI/L 3 , thus k = 3EI/L 3 . L x v(x) ρ dx 2 2 dt vd V V+dV L v m 2 2 dt vd kv x Other Topics by S. T. Mau 296 An equivalent form of Eq. 3 is 2 2 dt vd + m k v =0 (6) The factor associated with v in the above equation is a positive quantity and can be replaced by ω 2 = m k (7) Then Eq. 6 can be put in the following form: 2 2 dt vd + ω 2 v =0 (6) The general solution to Eq. 6 is v = A Sin n ω t + B Cos n ω t, n=1, 2, 3… (5) The constants A and B are to be determined by the position and velocity at t=0. No matter what are the conditions, which are called initial conditions, the time variation of the lateral deflection at the tip is sinusoidal or harmonic with a frequency of n ω . The lowest frequency, ω , for n=1, is called the fundamental frequency of natural vibration. The other frequencies are frequencies of higher harmonics. The motion, plotted against time, is periodic with a period of T: T= ω π 2 (6) Harmonic motion with a period T. In the present case, if EI=24,000 kN-m 2 , L=6 m and ρ =100 kg/m, then k = 3EI/L 3 =333.33 kN/m, m= ρ L= 600kg, and ω 2 =k/m=0.555 (kN/m.kg)=555(1/sec 2 ). The fundamental v t T Other Topics by S. T. Mau 297 vibration frequency is ω =23.57 rad/sec, and the fundamental vibration period is T=0.266 sec. The inverse of T , denoted by f, is called the circular frequency: f = T 1 (6) which has the unit of circle per second (cps), which is often referred to as Hertz or Hz. In the present example, the beam has a circular frequency of 3.75 cps or 3.75 Hz. Interested readers are encouraged to study Structural Dynamics, in which undamped vibration, damped vibration, free vibration and forced vibration of SDOF system, multi- degree-of-freedom (MDOF) system and other interesting and useful subjects are explored. 299 Matrix Algebra Review 1. What is a Matrix? A matrix is a two-dimensional array of numbers or symbols that follows a set of operating rules. A matrix having m rows and n columns is called a matrix of order m-by-n and can be represented by a bold-faced letter with subscripts representing row and column numbers, e.g., A 3x7 . If m=1 or n=1, then the matrix is called a row matrix or a column matrix, respectively. If m=n, then the matrix is called a square matrix. If m=n=1, then the matrix is degenerated into a scalar. Each entry of the two dimensional array is called an element, which is often represented by a plain letter or a lower case letter with subscripts representing the locations of the row and column in the matrix. For example a 23 is the element in matrix A located at the second row and third column. Diagonal elements of a square matrix A can be represented by a ii . A matrix with all elements equal to zero is called a null matrix. A square matrix with all non-diagonal elements equal to zero is called a diagonal matrix. A diagonal matrix with all the diagonal elements equal to one is called a unit or identity matrix and is represented by I. A square matrix whose elements satisfy a ij =a ji is called a symmetric matrix. An identity matrix is also a symmetric matrix. A transpose of a matrix is another matrix with all the row and column elements interchanged: (a T ) ij =a ji . The order of a transpose of an m-by-n matrix is n-by-m. A symmetric matrix is one whose transpose is the same as the original matrix: A T = A. A skew matrix is a square matrix satisfying a ij = −a ji . The diagonal elements of a skew matrix are zero. Exercise 1. Fill in the blanks in the sentences below. A= ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 101 37 42 B= ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 1034 172 C= ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 843 451 312 D= ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ 7 5 2 E= [ ] 752 F= ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 800 050 002 G= ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 100 010 001 H= ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 000 000 000 K= ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ −− − 043 401 310 Matrix A is a ___-by___ matrix and matrix B is a ___-by___ matrix. Matrix A is the _____________ of matrix B and vice versa. Matrix Algebra Review by S. T. Mau 300 Matrices C and F are _________ matrices with ________ rows and ________ columns. Matrix D is a ________ matrix and matrix E is a ______ matrix; E is the __________ of D. Matrix G is an ________ matrix; matrix H is a ______ matrix; matrix K is a _______ matrix. In the above, there are _____ symmetric matrices and they are __________________. 2. Matrix Operating Rules Only matrices of the same order can be added to or subtracted from each other. The resulting matrix is of the same order with an element-to-element addition or subtraction from the original matrices. C+F = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 843 451 312 + ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 800 050 002 = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 1643 4101 314 C−F = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 843 451 312 − ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 800 050 002 = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 043 401 310 The following operations using matrices defined in the above are not admissible: A+B, B+C, D−E, and D−G. Multiplication of a matrix by a scalar results in a matrix of the same order with each element multiplied by the scalar. Multiplication of a matrix by another matrix is permissible only if the column number of the first matrix matches with the row number of the second matrix and the resulting matrix has the same row number as the first matrix and the same column number as the second matrix. In symbols, we can write B x D = Q and Q ij = ∑ = 3 1k kjik DB Using the numbers given above we have Q =B x D = BD = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 1034 172 ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ 7 5 2 = ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ++ ++ x701 3x5x24 1x7 7x5x22 = ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ 113 46 Matrix Algebra Review by S. T. Mau 301 P =Q x E = QE = ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ 113 46 [ ] 752 = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 791565226 32223092 We can verify numerically that P =QE = BDE= (BD)E= B(DE) We can also verify multiplying any matrix by an identity matrix of the right order will result in the same original matrix, thus the name identity matrix. The transpose operation can be used in combination with multiplication in the following way, which can be easily derived from the definition of the two operations. (AB) T =B T A T and (ABC) T =C T B T A T Exercise 2. Complete the following operations. E B = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 63 25 ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 1034 172 = DE = ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ 7 5 2 [ ] 752 = 3. Matrix Inversion and Solving Simultaneous Algebraic Equations A square matrix has a characteristic value called determinant. The mathematical definition of a determinant is difficult to express in symbols, but we can easily learn the way of computing the determinant of a matrix by the following examples. We shall use Det to represent the value of a determinant. For example, DetA means the determinant of matrix A. Det [5]=5 Det ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 63 25 = 5x Det [6] −3x Det [2] = 30 – 6 =24 Det ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 963 852 741 = 1x Det ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 96 85 −2x Det ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 96 74 + 3x Det ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 85 74 Matrix Algebra Review by S. T. Mau 302 =1x(−3)−2x(−6)+3x(−3)=0 A matrix with a zero determinant is called a singular matrix. A non-singular matrix A has an inverse matrix A -1 , which is defined by AA -1 =I We can verify that the two symmetric matrices at the left-hand-side (LHS) of the following equation are inverse to each other. ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − 812 141 211 ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − −− 113 13/43/10 33/103/31 = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 100 010 001 ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − −− 113 13/43/10 33/103/31 ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − 812 141 211 = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 100 010 001 This is because the transpose of an identity matrix is also an identity matrix and (AB)=I (AB) T =(B T A T )=(BA)=I T =I The above statement is true only for symmetric matrices. There are different algorithms for finding the inverse of a matrix. We shall introduce one that is directly linked to the solution of simultaneous equations. In fact, we shall see matrix inversion is an operation more involved than solving simultaneous equations. Thus, if solving simultaneous equation is our goal, we need not go through a matrix inversion first. Consider the following simultaneous equations for three unknowns. x 1 + x 2 + 2x 3 = 1 x 1 + 4x 2 − x 3 = 0 2x 1 − x 2 + 8x 3 = 0 The matrix representation of the above is ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − 812 141 211 ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ 3 2 1 x x x = ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ 0 0 1 Matrix Algebra Review by S. T. Mau 303 Imagine we have two additional sets of problems with three unknowns and the same coefficients in the LHS matrix but different right-hand-side (RHS) figures. ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − 812 141 211 ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ 3 2 1 x x x = ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ 0 1 0 and ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − 812 141 211 ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ 3 2 1 x x x = ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ 1 0 0 Since the solutions for the three problems are different, we should use different symbols for them. But, we can put all three problems in one single matrix equation below. ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − 812 141 211 ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 333231 232221 131211 xxx xxx xxx = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 100 010 001 Or, AX= I By definition, X is the inverse of A. The first column of X contains the solution to the first problem, and the second column contains the solution to the second problem, etc. To find X, we shall use a process called Gaussian Elimination, which has several variations. We shall present two variations. The Gaussian process uses each equation (row in the matrix equation) to combine with another equation in a linear way to reduce the equations to a form from which a solution can be obtained. (1) The first version. We shall begin by a forward elimination process, followed by a backward substitution process. The changes as the result of each elimination/substitution are reflected in the new content of the matrix equation. Forward Elimination. Row 1 is multiplied by (–1) and added to row 2 to replace row 2, and row 1 is multiplied by (−2) and added to row 3 to replace row 3, resulting in: ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − 430 330 211 ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 333231 232221 131211 xxx xxx xxx = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − 102 011 001 Row 2 is added to row 3 to replace row 3, resulting in: ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − 100 330 211 ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 333231 232221 131211 xxx xxx xxx = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − 113 011 001 Matrix Algebra Review by S. T. Mau 304 The forward elimination is completed and all elements below the diagonal line in A are zero. Backward Substitution. Row 3 is multiplied by (3) and added to row 2 to replace row 2, and row 3 is multiplied by (−2) and added to row 1 to replace row 1, resulting in: ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 100 030 011 ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 333231 232221 131211 xxx xxx xxx = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − −− 113 3410 227 Row 2 is multiplied by (−1/3) and added to row 1 to replace row 1, resulting in: ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 100 030 001 ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 333231 232221 131211 xxx xxx xxx = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − −− 113 3410 33/103/31 Normalization. Now that matrix A is reduced to a diagonal matrix, we further reduce it to an identity matrix by dividing each row by the diagonal element of each row, resulting in: ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 100 010 001 ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 333231 232221 131211 xxx xxx xxx = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − −− 113 13/43/10 33/103/31 Or, X= ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 333231 232221 131211 xxx xxx xxx = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − −− 113 13/43/10 33/103/31 Note that X is also symmetric. It can be derived that the inverse of a symmetric matrix is also symmetric. (2) The second version. We combine the forward and backward operations and the normalization together to reduce all off-diagonal terms to zero, one column at a time. We reproduce the original matrix equation below. ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − 812 141 211 ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 333231 232221 131211 xxx xxx xxx = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 100 010 001 Matrix Algebra Review by S. T. Mau 305 Starting with the first row, we normalize the diagonal element of the first row to one (in this case, it is already one) by dividing the first row by the vale of the diagonal element. Then we use the new first row to eliminate the first column elements in row 2 and row 3, resulting in ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − 430 330 211 ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 333231 232221 131211 xxx xxx xxx = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − 102 011 001 We repeat the same operation with the second row and the diagonal element of the second row to eliminate the second column elements in row 1 and row 3, resulting in ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − 100 110 301 ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 333231 232221 131211 xxx xxx xxx = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − − 113 03/13/1 03/13/4 The same process is done using the third row and the diagonal element of the third row, resulting in ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 100 010 001 ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 333231 232221 131211 xxx xxx xxx = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − −− 113 13/43/10 33/103/31 Or, X= ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 333231 232221 131211 xxx xxx xxx = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − −− 113 13/43/10 33/103/31 The same process can be used to find the solution for any given column on the RHS, without finding the inverse first. This is left to readers as an exercise. Exercise 3. Solve the following problem by the Gaussian Elimination method. ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − 812 141 211 ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ 3 2 1 x x x = ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ 1 6 3 Forward Elimination. Row 1 is multiplied by (–1) and added to row 2 to replace row 2, and row 1 is multiplied by (−2) and added to row 3 to replace row 3, resulting in: [...]... 0. 625 Node 8 9 10 11 12 13 0 0. 625 0 -0. 625 -0 .75 0 -0 .75 0 0 0.031 0 -0.031 -0. 023 -0. 023 Case (a): Nodal Solutions v(m) Node u(m) u(m) 1 2 3 4 0.011 0.011 0.011 0.011 -0.031 0 -0.031 Elongation (m) 0.000 0.011 0. 023 0.034 0.000 -0. 073 -0. 129 -0. 073 5 6 7 8 v(m) 0.045 0.045 0. 023 0.000 0.00 -0. 073 -0. 129 -0. 073 The reactions are: 0.5 MN upward at both node 1 and node 5 6 5 7 12 6 7 2 2 1 8 1 x 3 309 8 10... (MN) 1 2 3 4 5 6 7 0.563 0.563 0.1 87 0.1 87 -0.938 1.000 -0.3 12 Node 8 9 10 11 12 13 0 0.3 12 0 -0.3 12 -0. 375 -0. 375 0 0.016 0 -0.016 -0.011 -0.011 Case (b): Nodal Solutions v(m) Node u(m) u(m) 1 2 3 4 0.0 17 0.0 17 0.006 0.006 -0.041 0.040 -0.016 Elongation (m) 0.000 0.0 17 0.034 0.039 0.000 -0. 128 -0. 073 -0.041 5 6 7 8 v(m) 0.045 0.039 0. 028 0.0 17 0.00 -0.088 -0. 073 -0.041 The reactions are: 0 .75 MN upward... kN 2. 5 kN 4 kN 2 kN 3 kN 7. 5 kN 8 kN 0 kN 4.5 kN 4.5 kN 8 kN 6 kN 3 kN Solution to Problems by S T Mau (2- a) (2- b) (2- c) 4 kN 0 kN 5 kN −3 kN 3 kN 3 kN −5 kN 0 kN 4 kN 3 kN 5 kN 3 kN −6 kN 4 kN 3 kN (3-a) 2. 4 kN 1.8 kN (4-a) 5 kN 7. 8 kN −1.4 kN (4-b) −1. 12 0 kN 2 kN 2. 24 2. 24 2 kN 0 kN −1 kN 0 kN 0.5 kN 4.68 kN (4-c) 1 kN 0.5 kN 3 kN 1 kN 1 kN 5 −5.4 kN 5 6 .24 3 .24 3 .24 0.84 kN 2. 88 kN 4 kN 1. 12 kN... 4 kN 1. 12 kN 3 .2 kN 1 kN −1. 12 kN 5 kN 3.84 2. 4 4 kN (3-c) −4.8 kN −4 kN 5 kN −6 kN 3 kN (3-b) −3 kN −5 kN −5 kN 4 kN 4kN 5kN 4. 32 kN 2 kN 2. 24 −1. 12 −1 kN 1. 12 1 1 kN 1 kN 0.5 kN 1 kN 1.5 kN Problem 2 (1-b) Fa= −0.9 375 kN, Fb= 0.3 125 kN (1-a) Fa= −0. 625 kN, Fb= −0. 625 kN (2) Fa = −5. 37 kN, Fb = −5. 37 kN, Fc= −(Fa)y− (Fb)y= 4.80 kN (3) Fa= 2. 5 kN, Fb= 2. 5 kN, Fc = 0 (4-a) Fa=0, Fb= 12. 5 kN (4-b) Fa=0,... V 3.4 m -4.69 kN 17. 52 kN-m M M 14. 07 kN-m - 37. 5 kN-m (5) (6) 1 .25 kN 0 kN V V 6 .25 kN-m M M -3 .75 kN-m -10 kN-m (7) (8) V V -2. 5 kN M - 12 kN M 10 kN-m -36 kN-m (9) -15 kN-m 10 kN (10) 4 kN 4 kN 10 kN V V -10 kN 20 kN-m -2 kN -4 kN 4 kN-m 20 kN-m 2 kN-m M M -20 kN-m -4 kN-m 313 -4 kN-m Solution to Problems by S T Mau (11) 50 kN-m 10 kN 10 kN -10 kN M V T ( 12) 10 kN-m 2 kN V T M -2 kN (13) 10 kN -10... 0.5 ⎪ ⎪ 1. 62 ⎪ ⎬=⎨ ⎬ ⎨ ⎪− 1.0⎪ ⎪ − 1.5 ⎪ ⎪ 1.0 ⎪ ⎪ 0. 17 ⎪ ⎪ 0 ⎪ ⎪ 0.83 ⎪ ⎭ ⎩ ⎭ ⎩ (2) ⎡ 0.0 ⎢ 0.0 ⎢ ⎢ 0.0 ⎢ − 1.0 ⎢ 0.0 ⎢ ⎢ 0.0 ⎣ 04 Truss Analysis: Force Method, Part II Problem 5 (1) 0.0 526 mm to the right (2) There is no horizontal displacement at node 2 (3) Node 5 moves to the right by 0. 67 mm Problem 6 (1) The force in member 10 is 8.86/0.1 72 = 51.51 kN (2) The force in member 6 is 1. 62/ 0. 174 =9.31 kN... (2) matrix inversion, using the known inverse of the matrix at the LHS ⎡ 31 / 3 − 10 / 3 − 3⎤ ⎧ x1 ⎫ ⎧3⎫ ⎪ ⎪ ⎪ ⎪ ⎢− 10 / 3 4/3 1⎥ ⎨ x 2 ⎬ = ⎨6⎬ ⎥ ⎢ ⎢ −3 1 1⎥ ⎪ x3 ⎪ ⎪ 1⎪ ⎦⎩ ⎭ ⎩ ⎭ ⎣ 3 07 Matrix Algebra Review by S T Mau 308 Solution to Problems 02 Truss Analysis: Matrix Displacement Method Problem 1: k33= EAC2/L = 7 .2 MN/m, k34= EACS/L = 9.6 MN/m, k31= − EAC2/L = 7 .2 MN/m The effect of the change of. .. kN-m 2 kN -2 kN 2 kN T V 314 M Solution to Problems by S T Mau (15) 10 kN T M V -50 kN-m (16) -2 kN 2 kN T 10 kN-m 2 kN M V 06 Beam and Frame Analysis: Force Method, Part II Problem 3 (1) L /2EI L2 /8EI Shear(Rotation) diagram(LEFT) and Moment(Deflection) digram (RIGHT) (2) -L2 /8EI -L3 /24 EI -5L3 /48EI Shear(Rotation) diagram(LEFT) and Moment(Deflection) digram (RIGHT) (3) L/3EI -L/6EI 0.58L 0.064L2... reactions are: 0 .75 MN upward at node 1 and 0 .25 MN upward at node 5 6 5 7 12 6 7 2 2 1 13 8 1 1 x 3 1 3 11 10 9 1 y 8 4 5 4 1 Problem 4: Discuss the kinematic stability of each of the plane truss shown (1) Stable, (2) Unstable, (3) Stable, (4) Unstable, (5) Stable, (6) Unstable, (7) Stable, (8) Unstable, (9) Stable, (10) unstable 03 Truss Analysis: Force Method, Part I Problem 1 (1-a) (1-b) (1-c) 3 kN 5... Problem 2 The change of numbering for members results in a switch between (kG )2 and (kG)3, but has no effect on the resulting stiffness equation, because the nodal displacements are still the same The constrained stiffness equation is identical to Eq 16 Problem 3 Case (a): Member Solutions Elongation Member Force (m) (MN) Member Force (MN) 1 2 3 4 5 6 7 0. 375 0. 375 0. 375 0. 375 -0. 625 0 0. 625 Node 8 . Solutions Member Force (MN) Elongation (m) Member Force (MN) Elongation (m) 1 0. 375 0.011 8 0 0 2 0. 375 0.011 9 0. 625 0.031 3 0. 375 0.011 10 0 0 4 0. 375 0.011 11 -0. 625 -0.031 5 -0. 625 -0.031 12 -0 .75 0 -0. 023 6 0 0 13 -0 .75 0 -0. 023 7 0. 625 -0.031 Case (a):. 0.00 2 0.011 -0. 073 6 0.045 -0. 073 3 0. 023 -0. 129 7 0. 023 -0. 129 4 0.034 -0. 073 8 0.000 -0. 073 The reactions are: 0.5 MN upward at both node 1 and node 5. 1 2 3 4 5 6 7 8 12 1 3 1 4 1 56 7 89 1 10 11 12 13 x y Solution. kN 1 kN 0.5 kN 1.5 kN −1. 12 kN −1. 12 0 kN 2 kN 2. 24 −1 kN 0 kN 2. 24 0 kN 1 kN 1. 12 2 .24 −1. 12 −1 kN 1 kN 1 kN 1 4 kN 2 kN 1 kN Solution to Problems by S. T. Mau 3 123 12 Problem 4. (1) ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ −− −−− −−− − − 0.10.05.0 67. 00.00.0 0.00.05.0 67. 00.10.0 0.00.10.00.10.00.1 0.00.138.05.00.00.0 0.00.063.0 83.00.00.0 0.00.063.083.00.00.0 ⎪ ⎪ ⎪ ⎭ ⎪ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎨ ⎧ − 0 5.0 0.1 0 0 0 = ⎪ ⎪ ⎪ ⎭ ⎪ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎨ ⎧ − − − 5.0 5.0 5.0 88.0 63.0 63.0 (2) ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ −− −−− −−− − − 0.10.05.0 67. 00.00.0 0.00.05.0 67. 00.10.0 0.00.10.00.10.00.1 0.00.138.05.00.00.0 0.00.063.0 83.00.00.0 0.00.063.083.00.00.0 ⎪ ⎪ ⎪ ⎭ ⎪ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎨ ⎧ − 0 0.1 0.1 5.0 0 0 = ⎪ ⎪ ⎪ ⎭ ⎪ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎨ ⎧ − − − 83.0 17. 0 5.1 62. 1 04.1 21 .0 04