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Statics and Matrix Algebra Problems with Solutions, Exams of Engineering Analysis
Two problems that involve statics and matrix algebra. The first problem requires determining the reactions at a fixed point, identifying zero force members, and calculating the forces in certain members of a truss system using the method of joints and the method of sections. The second problem involves writing a system of equations in matrix form, finding the determinant of a matrix using cofactor expansion, determining the inverse of a matrix using row operations, and solving the system of equations using the inverse of the matrix. Intermediate work is shown for each step.
Typology: Exams
2018/2019
1 / 8
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(A)
FBD (1.5 point) Σ 𝐹𝑥 = 0 : (0.5 point) 15 - Ax = 0, Ax = 15 kN (0.5 point) Σ 𝐹𝑦 = 0 : (0.5 point) Ay – 10 – 10 – 10 – 5 = 0, Ay = 35 kN (0.5 point) Σ 𝑀 = 0 : (0.5 point) MA – 103 – 106 – 109 – 512 – 15*10 = 0, MA = 390 kN.m (1 point) (B) Zero force members: GF (4 points), [Also GF by inspection and KJ from part C is correct (2 points each)] (- 0.5 points for each wrong member. This cannot add up to less than -4)
(C)
Node K FBD (1 point) Σ 𝐹𝑥 = 0 : (0.5 point) FKI = 15 kN (0.5 point) Σ 𝐹𝑦 = 0 : (0.5 point) FKJ = 0 kN (0.5 point) Node J FBD (1 point) Σ 𝐹𝑥 = 0 : (0.5 point) 0.8321 * FIJ + 0.9864 * FHJ = 0 kN (0.5 point) Σ 𝐹𝑦 = 0 : (0.5 point) 0.5547 * FIJ – 0.1644 * FHJ – 5 = 0 kN (0.5 point) FIJ = 7.21 kN (0.5 point), FHJ = -6.08 kN (0.5 point)
Problem #4 (25 Points) Consider the system of equations: 2x+5y-2z = - x+3y+9z = 33 4x+10y+z = - (A) Write the system of equations in matrix form, AX = B (3 pts) (B) Find the determinant of matrix A using cofactor expansion along column #2. (6 pts) (C) Determine A-1, the inverse of A, using only row operations. (10 pts) (D) Using A-1, solve the system of equations for the variables x, y, and z. (6 pts) Note: To receive full credit, all intermediate work should be shown. Solution (A) AX=B
൩ (1 point each matrix/vector) (B) C 12 = (-1)1+2^ [(11)-(94)] = 35 (1.5 point) C 22 = (-1)2+2^ [(21)-(-24)] = 10 (1.5 point) C 32 = (-1)3+2^ [(29)-(-21)] = -20 (1.5 point) det (A) = 535+310+10*(-20) = 5 (1.5 point) (C)
൩ R 1 *0.5 (1 point)
൩ R 2 -R 1 (1 point)
൩ R 3 -4R 1 (1 point)
൩ R 2 *2 (1 point)
൩ R 1 -2.5R 2 (1 point)
൩ R 3 /5 (1 point)
൩ R 2 -20*R 3 (1 point)
൩ R 1 +51*R 3 (1 point)
൩ (1 point) A-1^ =
൩ (1 point) (D) X= A-1^ B (1 point) ቈ
൩ (2 point) x = 3 (1 point) y = -2 (1 point) z = 4 (1 point)