Question 1 Bar 1 1 m Horo C gog(0) 90° Bar 2 = 10 kN in (A) 1m Figure 1: Truss assembly formed of two bars Figure 1 shows a truss structure consisting of two bars of 1 m in length, cross-sectional area A = 5∙10−4 m² and modulus of elasticity E = 200GPa. Nodes 1 and 2 are held fixed, and a force of 10 kN is applied in the x-direction at node 3. 45° a) Form the local and global stiffness matrices, apply the boundary conditions and find the displace- ments of the nodes for the given force. Use the following formula relating the local displacements of a single bar to the local forces applied to its nodes. 2 fix C² CS -C²-CS] Ui AE fiy CS S² 图 []&&&式] -CS -S² Vi fjx L - C² -CS C2 CS U j fjy -CS -S² CS S2 Vj X joro A in the har (1)

Elements Of Electromagnetics
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Question 1
Bar 1
1m
fix
fiy
fjæ
90⁰
fjy
Here C = cos(0) and S
Bar 2
10 kN
=
1m
45°
2
Figure 1: Truss assembly formed of two bars
=
Figure 1 shows a truss structure consisting of two bars of 1 m in length, cross-sectional area A
5.10 4 m² and modulus of elasticity E 200GPa. Nodes 1 and 2 are held fixed, and a force of
10 kN is applied in the x-direction at node 3.
=
y
a) Form the local and global stiffness matrices, apply the boundary conditions and find the displace-
ments of the nodes for the given force. Use the following formula relating the local displacements
of a single bar to the local forces applied to its nodes.
X
Vi
C²
CS -C² -CS Wi
AE CS S² -CS -S²
- C² -CS C² CS
S²
L
U j
-CS -S² CS
sin (0), where is the angle of orientation of the bar.
(1)
Transcribed Image Text:Question 1 Bar 1 1m fix fiy fjæ 90⁰ fjy Here C = cos(0) and S Bar 2 10 kN = 1m 45° 2 Figure 1: Truss assembly formed of two bars = Figure 1 shows a truss structure consisting of two bars of 1 m in length, cross-sectional area A 5.10 4 m² and modulus of elasticity E 200GPa. Nodes 1 and 2 are held fixed, and a force of 10 kN is applied in the x-direction at node 3. = y a) Form the local and global stiffness matrices, apply the boundary conditions and find the displace- ments of the nodes for the given force. Use the following formula relating the local displacements of a single bar to the local forces applied to its nodes. X Vi C² CS -C² -CS Wi AE CS S² -CS -S² - C² -CS C² CS S² L U j -CS -S² CS sin (0), where is the angle of orientation of the bar. (1)
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