Truss Deflections CE 331, Summer 2013 1 / 4 Given a truss with node and member numbers as indicated below: 4
4
5
6
3
7
2
1
1
5
2
3
Calculate the vertical deflection of Joint 2, 2V , due to the load(s) P. P
2V
Procedure: 1. Calculate the bar forces fi due to load(s) P. P
f4
f6
f3
f1
f7
f5
f2
Truss Deflections CE 331, Summer 2013 2 / 4 2. Calculate the member deformations i (elongation or shortening) due load(s) P. Use the formula for the axial deformation of a rod: f L
i  i i Ai E
3. Apply a unit load at the joint and direction of the desired deflection. Calculate the bar forces ui due to the unit load. These bar forces are unitless. u4
u6
u3
u7
u1
u5
u2
1
4. Multiply each member deformation i by the bar force due to the unit load ui and sum. 7
 2V    i u i i 1
Truss Deflections CE 331, Summer 2013 3 / 4 Example. Calculate the horizontal deflection at Joint 3 due to the 30 k load on the steel truss shown below. The bar forces due to the 30k load and various unit loads are provided to speed up the calculation process. For full credit, show detailed calculations for each number in the first row of your table. 14’
4’
9’
A = 1.5 in2 all members 30k
E = 29,000 ksi 4
4
5
8’
3
2
2
3
8’
1
1
Table 1. Bar lengths and forces due to loads shown in figures below. Bar Forces due to Loads in each Figure (tension = +’ve) Length, ft Member Fig 1, k
Fig 2
Fig 3
Fig 4
1 36.28
‐1.75
2.15
2.15
16.125
2 22.5
‐0.58
1.33
0.33
13.00
3 ‐42.15
2.03
‐1.16
‐1.16
24.08
4 ‐30.10
0.78
‐0.45
‐0.45
12.04
5 20.13
‐2.09
1.19
1.19
8.99
Figure 1
Figure 2
30 k
1
Figure 3
Figure 4
1
1
Truss Deflections CE 331, Summer 2013 E
A
4 / 4 29,000 ksi
1.5
in2
ft
L
k
fLoad
in
Member

U3_H
3_H
in
1
16.125
36.3
0.161
2.15
0.347
2
13
22.5
0.081
1.33
0.107
3
24.08
‐42.2
‐0.280
‐1.16
0.325
4
12.04
‐30.1
‐0.100
‐0.45
0.045
5
8.99
20.1
0.050
1.19
0.059
0.884
Sum:
   =
0.88
in
note: +'ve  means in the same direction as the unit load (See Figure 3)
Typical Calcs for Member 1
L1
= Length of Member 1 = 16.125 ft,calculated using dimensions provided in sketch
f1
i
= bar force in Member 1 due to load = 36.30 k T (from bar forces for Figure 1)
axial deformation of Member 1 due to actual loads = f x L / (A x E)
= (36.30 k) (16.13 ft x 12 in/ft) / ( 1.50 in2) (29,000 ksi) ) =
Ui3_H
0.161
= bar force in Member 1 due to a unit load at Joint 3 in the horizontal direction = 2.15
(from bar forces for Figure 3)
3_H
in
= deflection at Joint 3 in horizontal direction due to Member 1 deformation
=  x U3_H
= (0.161 in) (2.15) =
0.347
in