Advanced Methods of Structural Analysis

236 7 The Force Method

columns 2 and 3, respectively. For computation of unit displacementı 11 , the entries
of the column 5 should be summated. Similar procedure should be performed for
computation of loaded displacement1P(column 4).

Ta b l e 7. 5 Calculation of internal forces in the members of a statically indeterminate truss

Members EAl NN 1 NP^0 NN 1 NP^0 EAl NN 12 EAl NN 1 X 1 ND

NP^0 CNN 1 X 1

NN 1 Nl

EA

0 1 2 3 4 5 6 7 8

Vertical

members

1–7 3 0 120 0 0 0 120 0

2–6 3 0 0 0 0 0 0 0

3–5 3 0 120 0 0 0 120 0

Lower

chord

0–7 4 0.667 160 426:88 1.774 96:288 63:71 169:98
7–6 4 0.667 160 426:88 1.774 96:288 63:71 169:98
6–5 4 0.667 160 426:88 1.774 96:288 63:71 169:98
5–4 4 0.667 160 426:88 1.774 96:288 63:71 169:98
Upper chord 1–2 4 1.333 160 853:12 7.107 192:43 32:43 172:92
2–3 4 1.333 160 853:12 7.107 192:43 32:43 172:92
Diagonals 0–1 5 0.833 200  833 3.469 120:25 79:75 332:16
1–6 5 0.833 0 0 3.469 120:25 120:25 500:84
6–3 5 0.833 0 0 3.469 120:25 120:25 500:84
3–4 5 0.833 200  833 3.469 120:25 79:75 332:14
Factor 1=EA 1=EA 1=EA 1=EA

1PD
5; 079:76
EA
I ı 11 D
35:186
EA
The primary unknown isX 1 D1P=ı 11 D144:36kN. Column 6 contains the
computation of the first term of (7.8). Computation of final internal force in each
element of the truss is provided according to formula (7.8) and shown in column 7.

Reaction of supports Knowing the internal forceswe can calculate the reactionR
of any support. Free-body diagrams of joints 0 and 6 are presented in Fig.7.13e.
Equilibrium equations for joint 0 is
X
YD 0 W R 0 79:75sin ̨D^0 !R 0 D47:85kN:

It is obvious thatR 4 DR 0 D47:85kN. Equilibrium equation for joint 6

X

YD 2 120:25sin ̨CR 6 D 0

leads to the following reaction at the intermediate support of the truss:
R 6 D 2 120:250:6D144:3kN. Pay attention, this result has been obtained
early as the primary unknownX 1.

Static verificationFor truss in whole the equilibrium equations
X
YD 2 47:85C144:3 2  120 D 240  240 D0;
X
XD 0 W H 0 D0: