Consider the truss made of 2 bars with lengths 1 m, orientated in the structure shown in Figure 1. The two bars have the same cross-sectional areas A = 5.10-4 m² and modulus of elasticity E = 2-10¹¹ N/m². A force of F1,x = 10 kN is applied in the x-direction at node 1. 45 O 2 = 90° 2 1 1m 3 1m → F₁x = 10KN Figure 1: Truss assembly formed of two bars. Nodes and bars are numbered, with bars numbered with underlines. a) Using the numbering system of nodes and bars shown in Figure 1, state the 2 local stiffness matrices for each bar and combine these into a global stiffness system for all nodal displacements of the truss. Use the matrix system given in Eq. 1 below relating the local displacements of a single bar to the local forces applied to its nodes. Ui fix fiy AE Vi U j fjx L fjy Vj Here C= cos(0) and S = sin(0), where is the angle of orientation of the bar. CS -C² -CS -CS -S² C2 C² CS S² -C² -CS -CS -S² CS CS S² (1)

Elements Of Electromagnetics
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Consider the truss made of 2 bars with lengths 1 m, orientated in the structure shown in Figure 1. The
two bars have the same cross-sectional areas A = 5·10-4 m² and modulus of elasticity E = 2·10¹¹ N/m².
10 kN is applied in the x-direction at node 1.
A force of F1,x
=
850
P
2
=
90°
2
AE
L
1
1m
3
1m
F1,x = 10KN
Figure 1: Truss assembly formed of two bars. Nodes and bars are numbered, with bars numbered
with underlines.
X
a) Using the numbering system of nodes and bars shown in Figure 1, state the 2 local stiffness matrices
for each bar and combine these into a global stiffness system for all nodal displacements of the
truss. Use the matrix system given in Eq. 1 below relating the local displacements of a single bar
to the local forces applied to its nodes.
fix
fiy
fjx
fjy
Here C = cos(0) and S = sin(0), where is the angle of orientation of the bar.
CS -C² -CS
C²
CS S² -CS -S²
-C² -CS C²
-CS -S² CS
CS
S²
R
Wi
Vi
U j
(1)
Transcribed Image Text:Consider the truss made of 2 bars with lengths 1 m, orientated in the structure shown in Figure 1. The two bars have the same cross-sectional areas A = 5·10-4 m² and modulus of elasticity E = 2·10¹¹ N/m². 10 kN is applied in the x-direction at node 1. A force of F1,x = 850 P 2 = 90° 2 AE L 1 1m 3 1m F1,x = 10KN Figure 1: Truss assembly formed of two bars. Nodes and bars are numbered, with bars numbered with underlines. X a) Using the numbering system of nodes and bars shown in Figure 1, state the 2 local stiffness matrices for each bar and combine these into a global stiffness system for all nodal displacements of the truss. Use the matrix system given in Eq. 1 below relating the local displacements of a single bar to the local forces applied to its nodes. fix fiy fjx fjy Here C = cos(0) and S = sin(0), where is the angle of orientation of the bar. CS -C² -CS C² CS S² -CS -S² -C² -CS C² -CS -S² CS CS S² R Wi Vi U j (1)
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