1) The document provides examples for determining internal forces in statically indeterminate truss structures. 2) It shows the process of finding the degree of indeterminacy, selecting a determinate system, calculating member forces from loads and redundant forces, and determining final member forces. 3) The final member forces are calculated using an equation that factors in the original forces, redundant forces, and a value for the "true redundant" determined from the forces.

1. 1
Eng. Abdulrahman Shaaban
Solved Examples for Consistent Deformation for trusses Structures
A) External determinate: Example (1)
Find the eternal forces at all members for the following truss.
Step 1: Determine the degree of indeterminacy
1) External determinacy:
Unknowns (U) = Reactions (R) = 4
Equations (E) = Statics equation + Special joint = 3
Number of indeterminacy (N) = 𝑈 − 𝐸 → 4 − 3 = 1
∴ it considered once statically indeterminate
2) Internal determinacy:
𝑈 = 𝑚𝑒𝑚𝑏𝑒𝑟𝑠 ( 𝑚) + 𝑅 = 9 + 4 = 13
𝐸 = 2 ∗ 𝐽𝑜𝑖𝑛𝑡 ( 𝐽) = 2 ∗ 6 = 12
N = 𝑈 − 𝐸 → 13 − 12 = 1 𝑏𝑢𝑡 𝑁 𝑓𝑟𝑜𝑚 𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑖𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑐𝑦 = 1 𝑎𝑙𝑠𝑜
so that it still once statically indeterminate
50 cm2
50 cm2
50 cm2
50 cm2
75 cm2
75 cm2
75 cm2
75 cm2
25 cm2

2. 2
Eng. Abdulrahman Shaaban
Step 2: Select proper statically determinate main system
Step 3: Coding members, calculate angles for inclined members and find reactions:
𝜃1 = tan−1(3 4⁄ ) = 36.870
Step 4: Find all internal forces at all members due to original loads
Hint: using joint method without details for each individual joint just find the force and but its value and its
type (use [+ve] for tension force and [-ve] for compression value) directly on the figure.
1 2
3
45
6
7 8
9
0
7.5 7.5
θ1
θ1 θ1
θ1
0 0
-7.5
-10-10
-7.5
12.5 12.5
-15
0
7.5 7.5
θ1
θ1 θ1
θ1

3. 3
Eng. Abdulrahman Shaaban
Step 5: Find all internal forces at all members due to applying unit redundant.
Step 6: Construct a table to summarize results and facilitation of calculation procedures.
Member L (cm) A (cm2
) N0 (t) N1 (t) N0N1*L/A N1
2
*L/A Nf
1 500 75 0 -1.25 0.00 10.42 -8.84
2 500 75 0 -1.25 0.00 10.42 -8.84
3 600 50 -7.5 +0.75 -67.50 6.75 -2.20
4 400 50 -10 +1 -80.00 8.00 -2.93
5 400 50 -10 +1 -80.00 8.00 -2.93
6 600 50 -7.5 +0.75 -67.50 6.75 -2.20
7 500 75 +12.5 -1.25 -104.17 10.42 3.66
8 500 75 +12.5 -1.25 -104.17 10.42 3.66
9 300 25 -15 0 0.00 0.00 -15.00
Σ -503.33 71.17
Step 7: Calculate the true value of redundant by the following eqn:
𝑥1 =
−𝑁0N1 ∗ L/A
𝑁1
2
∗ 𝐿/𝐴
= 7.07
Step 8: Calculate the final internal force at each member by the following eqn:
𝑁𝑓 = 𝑁0 + 𝑥1 ∗ 𝑁1
Step 9: Draw the final normal force diagram
-1.25
+0.75
+1+1
+0.75
-1.25 -1.25
0
1
0
0
θ1 θ1
1
-1.25
-8.84
-2.2
-2.93-2.93
-2.2
3.66 3.66
-15
7.07
7.5 7.5
7.07
-8.84

4. 4
Eng. Abdulrahman Shaaban
Example (2)
Step 1: Determine the degree of indeterminacy
1) External determinacy:
U = 4
E = 3
N = 𝑈 − 𝐸 → 4 − 3 = 1
∴ it considered once statically indeterminate
2) Internal determinacy:
𝑈 = 𝑚 + 𝑅 = 19 + 4 = 23
𝐸 = 2 ∗ 𝐽 = 2 ∗ 11 = 22
N = 𝑈 − 𝐸 → 23 − 22 = 1 𝑏𝑢𝑡 𝑁 𝑓𝑟𝑜𝑚 𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑖𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑐𝑦 = 1 𝑎𝑙𝑠𝑜
so that it still once statically indeterminate
Step 2: Select proper statically determinate main system
2A
2A 2A
2A
2A2A
3A

5. 5
Eng. Abdulrahman Shaaban
Step 3: Coding members, calculate angles for inclined members and find reactions:
𝜃1 = tan−1(6 2⁄ ) = 71.560
𝜃2 = tan−1(3 2⁄ ) = 56.310
Step 4: Find all internal forces at all members due to original loads
θ2
θ1
1 2
3
4
5
678
9
10 11 12 13
14 15 16 17 18 19
1212
0
1212
0
-6.33
-6.33
-6.33
-6.33
-7.21
-8-13.34-8
-7.21
+6 +10.67 +10.67 +6
+7.21
+4.81
-4.81 +4.81 -4.81 +7.21

6. 6
Eng. Abdulrahman Shaaban
Step 5: Find all internal forces at all members due to applying unit redundant.
Step 6: Construct a table to summarize results and facilitation of calculation procedures.
Member L (m) A N0 (t) N1 (t) N0N1*L/A N1
2
*L/A Nf
1 2√10 2 -6.33 +1.58 -31.63 7.89 -1.20
2 2√10 2 -6.33 -1.58 31.63 7.89 -11.46
3 2√10 2 -6.33 -1.58 31.63 7.89 -11.46
4 2√10 2 -6.33 +1.58 -31.63 7.89 -1.20
5 √13 1 -7.21 +1.80 -46.79 11.68 -1.37
6 4 1 -8 +2 -64.00 16.00 -1.51
7 4 3 -13.34 +2 -35.57 5.33 -6.85
8 4 1 -8 +2 -64.00 16.00 -1.51
9 √13 1 -7.21 +1.80 -46.79 11.68 -1.37
10 4 1 +6 -1.5 -36.00 9.00 1.13
11 4 2 +10.67 -3 -64.02 18.00 0.93
12 4 2 +10.67 -3 -64.02 18.00 0.93
13 4 1 +6 -1.5 -36.00 9.00 1.13
14 √13 1 +7.21 -1.80 -46.79 11.68 1.37
15 √13 1 -4.81 0 0.00 0.00 -4.81
16 √13 1 +4.81 0 0.00 0.00 4.81
17 √13 1 +4.81 0 0.00 0.00 4.81
18 √13 1 -4.81 0 0.00 0.00 -4.81
19 √13 1 +7.21 -1.80 -46.79 11.68 1.37
Σ -550.78 169.64
00
1 1
-1.58+1.58
-1.58 +1.58
+1.8
+2
+2+2
+1.8
-1.5
-1.8 0
0 0 0 -1.8
-3 -3 -1.5

7. 7
Eng. Abdulrahman Shaaban
Step 7: Calculate the true value of redundant by the following eqn:
𝑥1 =
−𝑁0N1 ∗ L/A
𝑁1
2
∗ 𝐿/𝐴
= 3.25
Step 8: Calculate the final internal force at each member by the following eqn:
𝑁𝑓 = 𝑁0 + 𝑥1 ∗ 𝑁1
Step 9: Draw the final normal force diagram
1212
3.25
-11.46
-1.2
-11.46
-1.2
-1.37
-1.51-6.85
-1.37
+1.13 +0.93 +0.93 +1.13
+4.81
-4.81+1.37 +4.81 -4.81 +1.37
3.25
-1.51

8. 8
Eng. Abdulrahman Shaaban
B) Internal determinate: Example (3)
Find the eternal forces at all members for the following truss.
Step 1: Determine the degree of indeterminacy
1) External determinacy:
U = 4
E = 3 + 1 = 4
N = 𝑈 − 𝐸 → 4 − 4 = 0
∴ it considered statically determinate
2) Internal determinacy:
𝑈 = 𝑚 + 𝑅 = 13 + 4 = 17
𝐸 = 2 ∗ 𝐽 = 2 ∗ 8 = 16
N = 𝑈 − 𝐸 → 17 − 16 = 1 so that it considered once statically indeterminate
Step 2: Select proper statically determinate main system

9. 9
Eng. Abdulrahman Shaaban
Step 3: Coding members, calculate angles for inclined members and find reactions:
𝜃1 = tan−1(3 4⁄ ) = 36.870
Step 4: Find all internal forces at all members due to original loads
Step 5: Find all internal forces at all members due to applying unit redundant.
26
64
64
0
1 2
3
456
7 8 11
θ1
26
64
64
0
-10
+8+29.34
-8
-26.67
+64
-43.33
-29.34
+16 +6
0
0
0
0
1
1
-0.6
0
00
0
0
-0.8
-0.8
-0.6
+1
9
10

10. 10
Eng. Abdulrahman Shaaban
Step 6: Construct a table to summarize results and facilitation of calculation procedures.
Member L (m) A N0 (t) N1 (t) N0N1*L/A N1
2
*L/A Nf
1 4 1 -29.34 0 0.00 0.00 -29.34
2 4 1 -8 -0.8 25.60 2.56 -19.17
3 5 1 -10 0 0.00 0.00 -10.00
4 4 1 +8 0 0.00 0.00 8.00
5 4 1 +29.34 -0.8 -93.89 2.56 18.17
6 4 1 +64 0 0.00 0.00 64.00
7 5 1 -43.33 0 0.00 0.00 -43.33
8 3 1 +16 -0.6 -28.80 1.08 7.62
9 5 1 -26.67 +1 -133.35 5.00 -12.71
10 5 1 0 +1 0.00 5.00 13.96
11 3 1 +6 -0.6 -10.80 1.08 -2.38
Σ -241.24 17.28
Step 7: Calculate the true value of redundant by the following eqn:
𝑥1 =
−𝑁0N1 ∗ L/A
𝑁1
2
∗ 𝐿/𝐴
= 13.96
Step 8: Calculate the final internal force at each member by the following eqn:
𝑁𝑓 = 𝑁0 + 𝑥1 ∗ 𝑁1
Step 9: Draw the final normal force diagram
64
64
26
-2.38
-10
+8+64
-43.33
-29.34
+18.17
-19.17
+7.62
-12.71
+13.96
0

11. 11
Eng. Abdulrahman Shaaban
Example (4)
Step 1: Determine the degree of indeterminacy
1) External determinacy:
U = 3
E = 3
N = 𝑈 − 𝐸 → 3 − 3 = 0
∴ it considered statically determinate
2) Internal determinacy:
𝑈 = 𝑚 + 𝑅 = 15 + 3 = 18
𝐸 = 2 ∗ 𝐽 = 2 ∗ 8 = 16
N = 𝑈 − 𝐸 → 18 − 16 = 2 so that it considered twice statically indeterminate
Step 2: Select proper statically determinate main system

12. 12
Eng. Abdulrahman Shaaban
Step 3: Coding members, calculate angles for inclined members and find reactions:
𝜃1 = tan−1(3 4⁄ ) = 36.870
Step 4: Find all internal forces at all members due to original loads.
Step 5: Find all internal forces at all members due to applying unit redundant.
9
1
9
6
2
3
4
5
6
8
9
12
13 14
15
7
10
11
θ1
9
+4.5
9
6
+4.5
0
0
-9
0
-7.5
0
+7.5
1
-0.8
-0.8
-0.6-0.6
+1
0
0
00 +1
1
-0.6 -0.6
-0.8
-0.8
0 0 0
0
N2
N1
N0

13. 13
Eng. Abdulrahman Shaaban
Step 6: Construct a table to summarize results and facilitation of calculation procedures.
Member L (m) A N0 (t) N1 (t) N2 (t) N0N1*L/A N0N2*L/A N1
2
*L/A N2
2
*L/A N1N2*L/A Nf
1 3 1 +4.5 -0.6 0 -8.1 0 1.08 0 0 6.36
2 3 1 +4.5 0 -0.6 0 -8.1 0 1.08 0 2.64
3 4 1 0 0 -0.8 0 0 0 2.56 0 -2.48
4 3 1 0 0 -0.6 0 0 0 1.08 0 -1.86
5 3 1 -9 -0.6 0 16.2 0 1.08 0 0 -7.14
6 4 1 0 -0.8 0 0 0 2.56 0 0 2.48
7 4 1 0 -0.8 -0.8 0 0 2.56 2.56 2.56 0.00
8 2.5 1 +7.5 +1 0 18.75 0 2.5 0 0 4.40
9 2.5 1 0 +1 0 0 0 2.5 0 0 -3.10
10 2.5 1 +7.5 +1 0 18.75 0 2.5 0 0 4.40
11 2.5 1 0 +1 0 0 0 2.5 0 0 -3.10
12 2.5 1 0 0 +1 0 0 0 2.5 0 3.10
13 2.5 1 -7.5 0 +1 0 -18.75 0 2.5 0 -4.40
14 2.5 1 0 0 +1 0 0 0 2.5 0 3.10
15 2.5 1 -7.5 0 +1 0 -18.75 0 2.5 0 -4.40
Σ 45.6 -45.6 17.28 17.28 2.56
Step 7: Calculate the true value of redundant by the following eqns:
𝛿10 + 𝑥1 𝛿11 + 𝑥2 𝛿12 = 0
𝛿20 + 𝑥1 𝛿21 + 𝑥2 𝛿22 = 0
𝛿10 = ∫
𝑁0 𝑁1 𝑙
𝐸𝐴
𝑑𝑙 𝛿20 = ∫
𝑁0 𝑁2 𝑙
𝐸𝐴
𝑑𝑙
𝛿11 = ∫
𝑁1 𝑁1 𝑙
𝐸𝐴
𝑑𝑙 𝛿22 = ∫
𝑁2 𝑁2 𝑙
𝐸𝐴
𝑑𝑙
𝛿12 = 𝛿21 = ∫
𝑁1 𝑁2 𝑙
𝐸𝐴
𝑑𝑙
45.6 + 𝑥1 ∗ 17.28 + 𝑥2 ∗ 2.56 = 0
−45.6 + 𝑥1 ∗ 2.56 + 𝑥2 ∗ 17.28 = 0
𝑥1 = −3.097 𝑥2 = 3.097
Step 8: Calculate the final internal force at each member by the following eqn:
𝑁𝑓 = 𝑁0 + 𝑥1 ∗ 𝑁1 + 𝑥2 ∗ 𝑁2

14. 14
Eng. Abdulrahman Shaaban
Step 9: Draw the final normal force diagram
9
+6.36
9
6
+2.64
-2.48
-1.86
-7.14
+2.48
+4.4
-3.1
-4.4
0
+3.1