Truss Example
Truss Example
Truss Example
Figure Q1 shows a pin-jointed truss with loads 22.7 kN and 45 kN applied at joint A. In
addition to the load shown, member AB was found fabricated 5 mm too short.
a) Form the global stiffness relations corresponding to the unconstrained degrees of freedom
and solve for the non-zero degrees of freedom.
b) Analyse for the axial forces in members 1 and 3 if, the structure is still acted upon by the
22.7 kN load (without the 45 kN load) and member AB was considered cut to the right
length (no fabrication error).
Given A = 322 mm2 for all members in the pin-jointed truss and E = 200 GPa
6m 45 kN
1
22.7 kN
A
6m
2
3
B C
9m 9m
Figure Q1
Learning Outcomes:
This test is designed to access student ability’s on:
able to identify and number the unconstrained degree of freedoms of the indeterminate
truss and able to determine the non-zero degrees of freedom.
able to establish the element stiffness matrices of all members in global coordinate system.
able to assemble the structure stiffness equation
able to determine the axial forces for the given members using the element stiffness
matrices and force vectors.
ECS468 (INDETERMINATE STRUCTURES)
4
Solution:
6m 45 kN 2
1
1
22.7 kN
A
6m
6 2 8
3
B 5 C 7
9m 9m
Figure Q1
a) Form the global stiffness relations corresponding to the unconstrained degrees of freedom
and solve for the non-zero degrees of freedom. (CO2-PO3)
Du D1 = ? Qu Q3 = ?
D2 = ? Q4 = ?
Q5 = ?
Dk D3 = 0 Q6 = ?
D4 = 0 Q7 = ?
D5 = 0 Q8 = ?
D6 = 0
D7 = 0 Qk Q1 = 22.7
D8 = 0 Q2 = - 45
ECS468 (INDETERMINATE STRUCTURES)
C x2 CxC y C x2 CxC y
AE C x C y C y2 CxC y C y2
k
L C x2 CxC y C x2 CxC y
C x C y C y2 CxC y C y2
Member 1, L = 10.82 m
xF xN 90 y F y N 6 12
C x cos x 0.8318, CY sin x 0.5545
L 10.82 L 10.82
3 4 1 2
0.6919 0.4612 0.6919 0.4612 3
0.4612 0.3075
AE 0.4612 0.3075 4
K1
10.82 0.6919 0.4612 0.6919 0.4612 1
0.4612 0.3075 0.4612 0.3075 2
Member 2, L = 10.82 m
xF x N 9 0 yF y N 60
Cx cos x 0.8318, CY sin x 0.5545
L 10.82 L 10.82
5 6 1 2
0.6919 0.4612 0.6919 0.4612 5
AE 0.4612 0.3075 0.4612 0.3075 6
K2
10.82 0.6919 0.4612 0.6919 0.4612 1
0.4612 0.3075 0.4612 0.3075 2
ECS468 (INDETERMINATE STRUCTURES)
Member 3, L = 10.82 m
xF xN 9 18 yF y N 6 0
Cx cos x 0.8318, CY sin x 0.5545
L 10.82 L 10.82
7 8 1 2
0.6919 0.4612 0.6919 0.4612 7
AE 0.4612 0.3075 0.4612 0.3075 8
K3
10.82 0.6919 0.4612 0.6919 0.4612 1
0.4612 0.3075 0.4612 0.3075 2
ECS468 (INDETERMINATE STRUCTURES)
K = K1 + K2 + K3
1 2 3 4 5 6 7 8
Q5 0 0.8318
Q 0 AE 0 . 005 0 . 5545
6
Q1 0 10.82 0.8318
2 0
Q 0 .5545
Q5 0 0.0003844
Q6 0 AE 0 . 0002562
Q1 0 0.0003844
Q
2 0
0 . 0002562
ECS468 (INDETERMINATE STRUCTURES)
Q = KD + Q0
1 2 3 4 5 6 7 8
22.7 2.0757 0.4612 0.6919 0.4612 0.6919 0.4612 0.6919 0.4612 D1 0.0003844 1
45 0.4612 0.9225 0.4612 0.3075 0.4612 0.3075 0.4612 0.3075 D 0.0002562 2
2
Q3 0.6919 0.4612 0.6919 0.4612 0 0 0 0 0 0 3
Q4 AE 0.4612 0.3075 0.4612 0.3075 0 0 0 0 0 0 4
Q 10 . 82 0 AE
0.6919 0.4612 0 0 0.6919 0.4612 0 0 0.0003844
5
5
Q6 0.4612 0.3075 0 0 0.4612 0.3075 0 0 0 0.0002562 6
Q7 0.6919 0.4612 0 0 0 0 0.6919 0.4612 0 0 7
D1 2.987 x103 m
D2 0.0127m
c) Analyse for the axial forces in members 1 and 3 if, the structure is still acted upon by the
22.7 kN load (without the 45 kN load) and member AB was considered cut to the right
length (no fabrication error). (CO2-PO3)
D1 2.067 x10 3 m
3
D 2 1 .032 x10 m
ECS468 (INDETERMINATE STRUCTURES)
L DFX
DFY
Member 1
0
0.8318 0.5545 0.8318 0.5545
64400 0
q1 3
10.82 2.067 x10
3
1.032 x10
q1 6.827 kN T
Member 3
0
0.8318 0.5545 0.8318 0.5545
64400 0
q3 3
10.82 2.067 x10
3
1.032 x10
q3 6.827kN C