STATICALLY DETERMINATE STRUCTURES. DISPLACEMENTS.

1. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
cperez.eps@ceu.es
molina.eps@ceu.es
Hipódromo de la Zarzuela. Madrid 1931. Archiches y Domínguaz. Eduardo Torroja
STATICALLY DETERMINATE BEAMS AND FRAMES
DISPLACEMENTS

2. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
cperez.eps@ceu.es
molina.eps@ceu.es
3.- BEAM WITH CANTILEVERS
2.- TEST 0 16/17
1.- TEST 0 17/18
4.- ASSIGNMENT 1

3. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
cperez.eps@ceu.es
federico.prietomunoz@ceu.es

4. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
cperez.eps@ceu.es
federico.prietomunoz@ceu.es
BEAM 5
BEAM 6
LET’S COMPARE THESE BEAMS BEHAVIOUR

5. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
cperez.eps@ceu.es
federico.prietomunoz@ceu.es
BEAM 5
BEAM 6
BOTH BEAMS REACTION FORCES ARE EQUAL WHY?
WHAT WOULD HAPPEN IF A WAS FIXED SUPPORTED INSTEAD OF PINNED?

6. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
cperez.eps@ceu.es
federico.prietomunoz@ceu.es
BENDING MOMENTS DIAGRAM

7. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
cperez.eps@ceu.es
federico.prietomunoz@ceu.es
WHICH EFFECTS AFFECT THE VERTICAL DISPLACEMENT AT THE END OF THE CANTILEVER?
BEAM 5

8. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
THE EFFECT OF THE CANTILIVER IMPLIES A ROTATION 0 AT THE FIXED END

9. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
cperez.eps@ceu.es
federico.prietomunoz@ceu.es
TO CALCULATE THE ROTATION AT B … WHICH EFFECTS DO WE HAVE TO CONSIDER?
BEAM 5

10. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
THE EFFECT OF THE DISTRIBUTED LOAD PLUS THE EFFECT OF THE END MOMENT

11. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
THE VERTICAL DISPLACEMENT AT C IS THE ADDITION
OF THE ROTATION AT B EFFECT PLUS THE CANTILEVER LOAD EFFECT

12. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
BEAM 6
TO CALCULATE THE VERTICAL DISPLACEMENT AT C…
WHICH EFFECTS DO WE HAVE TO CONSIDER?

13. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
TO CALCULATE THE VERTICAL DISPLACEMENT AT C…
WHICH EFFECTS DO WE HAVE TO CONSIDER?
BEAM 6
WE MUST CONSIDER SAME AS BEAM (5)
PLUS THE EFFECT OF THE CABLE ELONGATION

14. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
BEAM 6

15. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
BEAM 6
WE WOULD GET SAME RESULT IF WE APPLY UNIT LOAD METHOD
WATCH OUT!!!!!
WE SHOULD CONSIDER
THE BENDING
MOMENTS OF THE
BEAM AND THE AXIAL
FORCE IN THE CABLE!

16. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
STRUCTURAL ANALYSIS I
2017/2018 CLASS 2
cperez.eps@ceu.es
federico.prietomunoz@ceu.es

17. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
ARE BOTH STATICALLY DETERMINATE?

18. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
WILL BOTH MOVE HORIZONTALLY? TO THE LEFT OR TO THE RIGHT?

19. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
IN WHICH THE MAXIMUM BENDING MOMENT WILL BE HIGHER?

20. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
IN WHICH THE MAXIMUM VERTICAL DISPLACEMENT WILL BE HIGHER?

21. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
STATICALLY DETERMINATE? REACTION FORCES?

22. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
STATICALLY DETERMINATE? REACTION FORCES?

23. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2

24. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2

25. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2

26. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2

27. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
P* x uC =
𝑀⋅𝑀∗
𝐸𝐼
ⅆ𝑥
UNIT LOAD METHOD

28. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
REMEMBER!
WE ONLY CONSIDER MOVEMENTS PRODUCED BY BENDING
MOMENTS. WE NEGLECT SHEAR FORCE AND AXIAL FORCE
DEFORMATIONS. WE ASSUME BARS ARE INEXTENSIBLE

29. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
SIMPSON’S RULE

30. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
P* x uC =
𝑀⋅𝑀∗
𝐸𝐼
ⅆ𝑥
UNIT LOAD METHOD
𝑀 ⋅ 𝑀∗
𝐸𝐼
ⅆ𝑥 =
SIMPSON’S RULE
=
𝑀1 ∙ 𝑀1
∗
+ 4𝑀2 ∙ 𝑀2
∗
+ 𝑀3 ∙ 𝑀3
∗
6𝐸𝐼
𝑥 𝐿

31. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
P* x uC =
𝑀⋅𝑀∗
𝐸𝐼
ⅆ𝑥
UNIT LOAD METHOD
𝑢𝐶 =
𝑀1 ∙ 𝑀1
∗
+ 4𝑀2 ∙ 𝑀2
∗
+ 𝑀3 ∙ 𝑀3
∗
6𝐸𝐼
𝑥 𝐿 =
4 𝑥6,125𝑞𝑎2 𝑥4𝑎 𝑥7𝑎
6𝐸𝐼
= 𝟏𝟏𝟒, 𝟑𝟑𝒒𝒂 𝟒
/𝑬𝑰

32. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
STRUCTURAL ANALYSIS I
2017/2018 CLASS 2

33. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
P* x uA =
𝑀⋅𝑀∗
𝐸𝐼
ⅆ𝑥 +
𝑀⋅𝑀∗
𝐸𝐼
ⅆ𝑥 +
𝑀⋅𝑀∗
𝐸𝐼
ⅆ𝑥
UNIT LOAD METHOD
𝑢𝐴 =
4 𝑥1,5𝑞𝑎2 𝑥1,71𝑎 + 3𝑞𝑎2 𝑥3,43𝑎 𝑥 17𝑎
6𝐸𝐼
+
3,43𝑞𝑎2
𝑥4𝑎 + 4𝑥6𝑞𝑎2
𝑥1,71𝑎 𝑥6𝑎
6𝐸𝐼
+
−4𝑥6,125𝑞𝑎2
𝑥2𝑎 𝑥7𝑎
6𝐸𝐼
= 𝟒, 𝟕𝟓𝒒𝒂 𝟒
/𝑬𝑰
INCLINED
SUPPORT
BEAM DE BEAM EF

34. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
cperez.eps@ceu.es
federico.prietomunoz@ceu.es
q=10 kN/m // a = 0,5 m // EI =30000 kNm2
2,38 mm-0,1 mm
u=1,19 mmu=1,19 mmu=1,19 mm
v=-0,34 mm

35. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2

36. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
WHERE IN THIS FRAME CAN WE APPLY EASILY THE SIMPLE BEAM FORMULAS?

37. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2

38. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
𝑣(𝐸𝐹) =
−5𝑞 𝐿4
384𝐸𝐼
= -0,65 mm
v=-0,65 mm

39. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
cperez.eps@ceu.es
federico.prietomunoz@ceu.es

40. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
cperez.eps@ceu.es
federico.prietomunoz@ceu.es
19,53 mm
9,96 mm
v=-2,5 mm
u=9,96 mm

41. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
cperez.eps@ceu.es
federico.prietomunoz@ceu.es
SIMILARITIES AND DIFFERENCES?

42. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
cperez.eps@ceu.es
federico.prietomunoz@ceu.es
STATICALLY DETERMINATE? REACTION FORCES?

43. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
cperez.eps@ceu.es
federico.prietomunoz@ceu.es
real bars elongation
Δ𝐿 =
𝑁𝐿
𝐸𝐴
It gives us information
about the contribution of
each member to the
displacement we want to
calculate
vC= 𝑁∗ 𝑥
𝑁𝐿
𝐸𝐴
=
−23,48𝑃𝑎
𝐸𝐴

44. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
cperez.eps@ceu.es
federico.prietomunoz@ceu.es
It gives us information
about the contribution of
each member to the
displacement we want to
calculate
vC = 𝑁∗ 𝑥
𝑁𝐿
𝐸𝐴
=
14,48𝑃𝑎
𝐸𝐴
real bars elongation
Δ𝐿 =
𝑁𝐿
𝐸𝐴

45. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
cperez.eps@ceu.es
federico.prietomunoz@ceu.es

46. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
cperez.eps@ceu.es
federico.prietomunoz@ceu.es
IS IT UNSTABLE, STATICALLY DETERMINATE OR STATICALLY INDETERMINATE?

47. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
cperez.eps@ceu.es
federico.prietomunoz@ceu.es
IS IT UNSTABLE, STATICALLY DETERMINATE OR STATICALLY INDETERMINATE?

48. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
cperez.eps@ceu.es
federico.prietomunoz@ceu.es
IS IT UNSTABLE, STATICALLY DETERMINATE OR STATICALLY INDETERMINATE?

49. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
cperez.eps@ceu.es
federico.prietomunoz@ceu.es

50. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
cperez.eps@ceu.es
federico.prietomunoz@ceu.es

51. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
cperez.eps@ceu.es
federico.prietomunoz@ceu.es

52. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
cperez.eps@ceu.es
federico.prietomunoz@ceu.es
𝑣 𝑞 = −
5 ⋅ 𝑞 ⋅ 𝐿4
384 ⋅ 𝐸𝐼
𝑣2𝑃 = −
2𝑃 ⋅ 𝐿3
48 ⋅ 𝐸𝐼
𝑣 𝑀𝑙𝑒𝑓𝑡 =
𝑀𝑙𝑒𝑓𝑡 ⋅ 𝐿2
16 ⋅ 𝐸𝐼
𝑣 𝑀𝑟𝑖𝑔ℎ𝑡 =
𝑀𝑟𝑖𝑔ℎ𝑡 ⋅ 𝐿2
16 ⋅ 𝐸𝐼
DEFLECTION IN THE MIDDLE OF THE SPAN

53. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
cperez.eps@ceu.es
federico.prietomunoz@ceu.es
𝑔𝐵𝑞 = −
𝑞 ⋅ 𝐿3
24 ⋅ 𝐸𝐼
𝑔𝐵2𝑃 = −
2𝑃 ⋅ 𝐿2
16 ⋅ 𝐸𝐼
𝑔𝐵 𝑀𝑙𝑒𝑓𝑡 =
𝑀𝑙𝑒𝑓𝑡 ⋅ 𝐿
3 ⋅ 𝐸𝐼
𝑔𝐵 𝑀𝑟𝑖𝑔ℎ𝑡 =
𝑀𝑟𝑖𝑔ℎ𝑡 ⋅ 𝐿
6 ⋅ 𝐸𝐼

54. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
cperez.eps@ceu.es
federico.prietomunoz@ceu.es
𝑣𝐿𝑒𝑓𝑡 𝑞 = −
𝑞 ⋅ 𝐿4
8 ⋅ 𝐸𝐼
𝑣𝐿𝑒𝑓𝑡 𝑃 = −
𝑃 ⋅ 𝐿3
3 ⋅ 𝐸𝐼
𝑣𝐿𝑒𝑓𝑡 𝑔𝐵 = 𝑔𝐵 ⋅ 𝐿𝑙𝑒𝑓𝑡

55. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
cperez.eps@ceu.es
federico.prietomunoz@ceu.es
𝑔𝐷 𝑞 = −
𝑞 ⋅ 𝐿3
24 ⋅ 𝐸𝐼
𝑔𝐷2𝑃 = −
2𝑃 ⋅ 𝐿2
16 ⋅ 𝐸𝐼
𝑔𝐷 𝑀𝑙𝑒𝑓𝑡 =
𝑀𝑙𝑒𝑓𝑡 ⋅ 𝐿
6 ⋅ 𝐸𝐼
𝑔𝐷 𝑀𝑟𝑖𝑔ℎ𝑡 =
𝑀𝑟𝑖𝑔ℎ𝑡 ⋅ 𝐿
3 ⋅ 𝐸𝐼
ROTATION AT D (ROLLER)

56. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
cperez.eps@ceu.es
federico.prietomunoz@ceu.es
𝑣𝑅𝑖𝑔ℎ𝑡 𝑞 = −
𝑞 ⋅ 𝐿4
8 ⋅ 𝐸𝐼
𝑣𝑅𝑖𝑔ℎ𝑡 𝑔𝐷 = 𝑔𝐵 ⋅ 𝐿𝑙𝑒𝑓𝑡
VERTICAL DISPLACEMENT AT THE END OF THE CANTILEVER

57. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 2
cperez.eps@ceu.es
federico.prietomunoz@ceu.es