1. Theory of Structures(1) Lecture No. 1

8. Roller Hinged Fixed Link

10. Y B X A Y A Solution of Example 1 1- ∑ X = 0 XA – 5 = 0 XA = 5t 2- ∑MB = 0 YA × 6 – 2 × 6 × 3 + (4 × 3)/2 × 1 = 0 YA= 5t 3- ∑ Y = 0 YA + YB = 2 × 6+ (4 × 3)/2=18 YB = 13t

11. Solution of Example 2 1- ∑ X = 0 X A = 3t 2- ∑ M B = 0 Y A × 4 + 8 + 4 × 1 – 3 × 1= 0 Y A = -2.25t ↑ Y A =2.25t ↓ 3- ∑ y = 0 Y A + 2× 2 = Y B 2.25+ 4 = Y B Y B = 6.25 t ↑ Y A X A Y B

12. Solution of Example 3 1- ∑ X = 0 XB – 4 = 0 XB = 4t 2- ∑ MA = 0 10YB – 4 × 12 – 2 × 9 – 2 × 5× 2.5 – 4 × 3= 0 YB = 10.3 t 3- ∑ Y = 0 YA + YB = 4 + 2 + 10 = 16 YA = 5.7 t Y B Y A X B

13. Solution of Example 4 1- ∑ X = 0 XA – 4 = 0 XA = 4t 2- ∑ Y = 0 YA – 1 × 6 – 5 – 5 – 2.5 = 0 YA = 18.5 t 3- ∑ M / A = 0 MA – 5 × 2 – 5 × 4 – 2.5 ×6 - 1×6×3 = 0 M A = 63 t.m

14. Solution of Example 5 R 1 = 1/2 × 4 × 8 = 16 t R 2 = 0.5 × 4 = 2 t 1- ∑ X = 0 X A – R 2 = 0 X A = 2t 2- ∑ M A = 0 11 Y B + 2 × 13 – R 2 × 2 – 2 × 1 – R 1 × 7= 0 Y B = 8.36 t 3- ∑ Y = 0 Y A + Y B + 2 – 2 – 16 = 0 YA= 7.64t

15. Solution of Example 6 1- ∑ X = 0 X B – 20 = 0 X B = 20t 2- ∑ M A = 0 6 Y B + 20 × 3 – 10 × 6 – 20 × 6 = 0  Y B = 20t 3- ∑Y = 0 Y A + Y B = 20 + 30 + 10 = 60  Y A = 40t

16. Solution of Example 7 ∑ MO = 0 3 × 0 + 8 × 1 + 1 × 3 – Rd × 5 = 0 Rd = 2.2 t ↑ ∑ Mp = 0 1 × 2 + 8 × 4 + 3 × 5 – Rb × 5 = 0 Rb = 6.93t ∑ Mq = 0 1 × 2 + 8 × 4 + 3 × 5 – Rc × 5 = 0 Rc = 6.93t 8t

18. Solution of Example 8 Method I 1- Part FD 1) ∑ MD = 0 8 × 2 = 4 YF YF = 4t 2) ∑Y = 0 YD = 8 – 4 YD = 4t 2- Part ECF 1) ∑ ME = 0 10×2.5+4×5= 4YC YC= 11.25t 2) ∑Y = 0 YE=10+4-11.25 YE=2.75t 3- Part ABE 1) ∑MA=0 12×3+2.75×6=5YB YB=10.5t 2) ∑Y = 0 YA= 2.75 + 12-10.5 YA= 4.25t

19. Solution of Example 8 “ Continued ” Method II 1)∑MF(right)= 0 8 × 2 – 4 YD = 0 YD = 4t 2) ∑ME(right)= 0 18×4.5 – 4×9 – 4YC= 0 YC=11.25t 3) ∑MA = 0 30×7.5 – 4×15 – 11.25× 10 - 5YB = 0 YB=10.5t 4) ∑Y = 0 30 – 4 – 11.25 – 10.5 – YA=0 YA=4.25t

20. = 0 = X d × 8 – 8 × 2 X d = 2 t  = 8 ×12 + 2.5 × 12 × 6 Xd - 2 × 8 - 10 ×Y d =0 Y d = 26 t ∑ Y = 0 = Y a + 26 – 2.5 × 12 – 8 Y a = 30 + 8 – 26 = 12 t. ∑ X = 0 = X a – 0.5 × 8 – 2 Xa = 4 + 2 = 6 t. ← Solution of Example 9

21. = 0 = Ma + 0.5 × 8 × 4 – 6 × 8 M a = 48 – 16 = 32 m.t (anticlockwise) Check : ∑Mc for part ac Xa Ya 32 + 0.5 × 8 × 4 + 2.5 × 12 × 4 – 6 × 8 – 12 × 10 = 32 + 16 + 120 – 48 – 120 = 0 Solution of Example 9 “Continued”

22. 1) ∑MF(right) = 0 RC cosα × 4.5 – 4 × 4.5 × 2.25 – 10 × 10.5 = 0 RC = 40.4 t 2) ∑MA = 0 YB×9 + XB × 2 + RC cosα × 18 + RC sinα × 8 –10×24 - 20×4.5 – 5 × 5 – 4 × 9 × 13.5 + 5 × 2 = 0 4.5 YB = 27.57 – XB (1) 3) ∑ME (right) = 0 4.5 YB - 6XB + RC cosα × 13.5 – 4 × 9 × 9 - 10 × 19.5 = 0 4.5 YB = 6XB + 82. 545 (2) Solution of Example 10

23. By solving equations (1) & (2) we get YB = 7.87 t. XB= - 7.85t 4) ∑Y = 0 YA = 36 + 10 + 20 + 5 – 32.33 – 7.87 YA = 30.8 t. 5) ∑ME(left) = 0 4.5 YA – 8 XA – 5 × 3 – 5 × 6.5 = 0 XA = 11.4 t. Solution of Example 10 “Continued”

25. Externally Statically Unstable Structures Geometrical Unstable Structures

26. Statically Determinate Structure is one in which t he reactions can be determined by application of the equations of equilibrium in addition to conditional equations(if any). Determinancy of Structures Statically Indeterminate Structure is one in which the number of unknown reactions is more than the number of available equations Stable-Once Statically Indeterminate Stable-Three Times Statically Indeterminate

27. Check the stability and determinacy for the following structures. Then, show how to modify it for stability and determinancy (a) (b) (c) (d) Solution of Example 11

28. Structure (a) : Is unstable because there are only two unknown reactions. The modification : Change either of the two roller supports to a hinge.

29. Structure (b) : Is unstable because the beam is rested on four link members connected at a point. The modification : Separate the link members Modification for structure (b)

30. Structure (C) : Is stable and three times statically indeterminate because there are six unknown support reactions. The modification : Add three intermediate hinges Modification for structure (C)

31. Structure (D) : Is stable and five times statically indeterminate because there are 8 unknown support reactions The modification : One of fixed supports is changed to a roller In addition to three intermediate hinges arranged as given in Struc. (c).