Question

2. For the pin-jointed truss shown in Figure Q2.1 applied at node 4. The Young's modulus E(GPa) is the same for the three truss vertical downward force P(kN) is a members. The cross sectional area of each of the truss members is indicated below and expressed in terms of a constant A. By using the stiffness method: (a) Compute the reduced stiffness matrix Kg [5 marks [10 marks (b) Calculate the global displacements of node 4 in terms of P, A and E (c) Calculate the axial forces in each of the truss members stating whether it is in tension or in compression in terms of P. [10 marks] 2 P (kN) (&ffz2e) G A 10 mm2 A (0428) A 2A-10 mm2 A, A 10 mm 60° 3 3 4m Figure Q2.1

Answer 1

P us m-3 2 2 Kese /_c CS s2 csS 3 CS AE -c -cs GO 2 C cs -s C S Foxes in members Pere A E 47 -5. 2

1 co-ordinates 2 joints 3 L1 4 m A1 10000 MM2 LET A-1 LET A 1 V 5.65 m A2 4 1 C 6.928 L2 20000 MM2 LET E-1 E 1 m 5 2 C 2.928 L3 8 m A3 10000 MM2 6 3 C C 7 6,928 m A1 0.01 M2 9 4 AND 1 element 1c 1 A2 0.02 M2 10 0 A3 0.01 M2 11 4 AND 2 element2 c -0.708 12 -0.708 S 13 4 AND 3 element 3c 0.5 -0.866 14 15 16 MEMBER STIFFNESS MATRIX 1 17 18 K1 0.0025 1 0 -1 0.0025 -0.0025 7 19 X AE 20 C -0.0025 C 1 -1 1 0.0025 C 21 0 0 22 23 4 7 7 24 K2 0.00354 0.50121 0.50121 -0.5012 -0.5012 0.00177 0.00177 -0.0018 -0.0018 25 0.50121 0.50121 -0.5012 -0.5012 0,00177 0,00177 -0.0018 -0.0018 X AE -0.0018 26 -0.5012 -0.5012 0.50121 0.50121 -0.0018 0.00177 0.00177 4 27 -0.5012 -0.5012 0.50121 0.50121 -0.0018 -0.0018 0.00177 0.00177 28 EEEE Nm un

7 5 6 7 8 30 K3 0,00125 0.25 0.433 0.25 -0.433 0,00031 0,00054 -0.0003 -0.0005 0.433 0.74996 0.00054 0,00094 -0.0005 AE -0.433 0.75 -0.0009 X -0.25 -0.433 0.433 -0.0003 -0.0005 0.00031 0.00054 6 0.25 -0.433 -0.75 0.433 0.74996 -0.0005 -0.0009 0,00054 0.00094 36 GLOBAL STIFFENESS MATRIX 1 6 7 0.0025 C C -0.0025 2 C 40 3 0.00177 0.00177 -0.0018 -0.0018 41 4 0.00177 0.00177 0 0 0.0018 -0.0018 AE X -0.0005 5 0.00031 0.00054 -0.0003 6 -0.0005 0.00054 0.00094 -0.0009 44 7 -0.0025 -0.0018 -0.0018 -0.0003 -0.0005 0.00459 0.00232 -0.0018 8 -0.0018 -0.0005 -0.0009 0.00232 0.00271 46 47 48 DISPLACEMENT MATRIX U1 F1 49 U2 F2 U3 F3 U4 F4 LET P 1 - U5 HERE, ONLY U7 AND U8 ARE NON ZERO F5 C U6 F6 U7 F7 0 U8 F8 1 56 N00 unD O OOO OOO 88853科的的印羽导5号等号导等令品528的%

H F K X 4 3 6 8 O0 25 UADUZD 2 3 00c021 ncoos4 0.0002 0.0005. 00954-0E9094- -0.0003 -0.0005 .0005 00005 0.00459 0.00232 -0b009 0.00232 6 -0.0025 -0.0018 -0.0018 -0.0018 -0.0018 7 U7 C 8 0 0 0.00271 U8 1 REDUCED STIFFNESS MATRIX AE X 0.00459 0.00232 U7 0 P 0.00232 0.00271 X U8 -1 P U7 -327.22 383.207 U8 327.22 648.187 X -1 327.217 -648.19 U7 X P/AE IN METERS U8 P/AE x

89 INTERNAL FORCES 90 1 -1 U7 91 MEMBER 1 0.0025 -1 1 X 1 X U8 92 U1 93 U2 94 95 0 X 96 0.0025 1 -1 327.217 P 97 -648.19 P 0 p 98 0 p 99 100 101 102 103 0.0025 -0.0025 0X 327.217 P 104 -648.19 P 0 p 0 p 105 106 107 108 FORCE IN MEMBER 1 kN 109 0.81804 P tension 110 111 elem 113 O O O

112 113 force in member 2 114 327.217 p 115 0.00354 -0.708 -0.708 0.70796 0.70796 X -648.19 p 116 0 p 117 0 p 118 119 120 327.217 p -648.19 p 121 0.0025 -0.0025 0.00251 0.00251 x 122 0 p 123 0 p 124 125 0.80437 p 126 P2 kN tension 127 128 129 FORCE IN MEMBER 3 130 327.217 p 131 0.08 -0.5 -0.866 0.5 0.866 X -648.19 p 132 0 p 0 p 133 134 135 327.217 p 136 -0.04 -0.0693 0.04 0.06928 X -648.19 p 137 138 0 p 0 p 139 140 141 P3 31.8177 P kN tension - 142 143