Ae1 19 20_class 2_09_10

cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
19/20 CLASS 2
cperez.eps@ceu.es
molina.eps@ceu.es
Hipódromo de la Zarzuela. Madrid 1931. Archiches y Domínguaz. Eduardo Torroja.
Assignment 1 year 18/19
STATICALLY DETERMINATE BEAMS AND FRAMES
DISPLACEMENTS

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
cperez.eps@ceu.es
molina.eps@ceu.es
3.- TEST 1 18/19
2.- TEST 1 17/18
1.- TEST 1 16/17
4.- ASSIGNMENT 1
Valle de los Caídos. Assignment 1 year 19/20

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
2017/2018 CLASS 2

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
cperez.eps@ceu.es
federico.prietomunoz@ceu.es
SIMILARITIES AND DIFFERENCES?

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
STATICALLY DETERMINATE? REACTION FORCES?

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
cperez.eps@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

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
cperez.eps@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

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
ARE BOTH STATICALLY DETERMINATE?

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
WILL BOTH MOVE HORIZONTALLY? TO THE LEFT OR TO THE RIGHT?

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
IN WHICH THE MAXIMUM BENDING MOMENT WILL BE HIGHER?

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

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
P* x uC =
𝑀⋅𝑀∗
𝐸𝐼
ⅆ𝑥
UNIT LOAD METHOD

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
REMEMBER!
WE ONLY CONSIDER MOVEMENTS PRODUCED BY BENDING
MOMENTS. WE NEGLECT SHEAR FORCE AND AXIAL FORCE
DEFORMATIONS. WE ASSUME BARS ARE INEXTENSIBLE

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
SIMPSON’S RULE

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 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

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 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

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 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

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
cperez.eps@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

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2SIMPLE BEAMS FORMULAS

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
WHERE IN THIS FRAME CAN WE APPLY EASILY THE SIMPLE BEAM FORMULAS?

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
𝑣(𝐸𝐹) =
−5𝑞 𝐿4
384𝐸𝐼
= -0,65 mm
v=-0,65 mm

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
cperez.eps@ceu.es

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
cperez.eps@ceu.es
19,53 mm
9,96 mm
v=-2,5 mm
u=9,96 mm

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
ARE ALL OF THEM STATICALLY DETERMINATE?

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
REMEMBER:
THE LOAD FOLLOWS THE SHORTEST WAY
TO REACH THE GROUND

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
WHICH BARS ARE GOING TO WORK IN EACH TRUSS?

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
AXIAL FORCES

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
DO THESE RESLTS MAKE SENSE?

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
CAN WE APPLY THE SUPERPOSITION PRINCIPLE?

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
TO CALCULATE THE HORIZONTAL DISPLACEMENT AT THE ROLLER WE MUST USE
AN AUXILIARY SYSTEM
CAN THIS SYSTEM BE THE SAME FOR EVERY TRUSS?

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
cperez.eps@ceu.es
BEAM 5
BEAM 6
LET’S COMPARE THESE BEAMS BEHAVIOUR

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
cperez.eps@ceu.es
BEAM 5
BEAM 6
BOTH BEAMS REACTION FORCES ARE EQUAL WHY?
WHAT WOULD HAPPEN IF A WAS FIXED SUPPORTED INSTEAD OF PINNED?

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
cperez.eps@ceu.es
BENDING MOMENTS DIAGRAM

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
cperez.eps@ceu.es
WHICH EFFECTS AFFECT THE VERTICAL DISPLACEMENT AT THE END OF THE CANTILEVER?
BEAM 5

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
THE EFFECT OF THE CANTILIVER IMPLIES A ROTATION 0 AT THE FIXED END

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
cperez.eps@ceu.es
TO CALCULATE THE ROTATION AT B … WHICH EFFECTS DO WE HAVE TO CONSIDER?
BEAM 5

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
THE EFFECT OF THE DISTRIBUTED LOAD PLUS THE EFFECT OF THE END MOMENT

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
THE VERTICAL DISPLACEMENT AT C IS THE ADDITION
OF THE ROTATION AT B EFFECT PLUS THE CANTILEVER LOAD EFFECT

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
BEAM 6
TO CALCULATE THE VERTICAL DISPLACEMENT AT C…
WHICH EFFECTS DO WE HAVE TO CONSIDER?

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 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

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
BEAM 6

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 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!

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
IS THIS A STATICALLY DETERMINATE SYSTEM?
WHAT HAPPENS IF WE REMOVE THE CABLE AF?

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
WHICH LOADS AFFECT THE CABLE AXIAL FORCE?

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
WHICH PART OF THE SYSTEM IS AFFECTED BY THE DISTRIBUTED LOAD q?

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
CABLE AXIAL FORCE?
Equilibrium equation:
MG = 0
- P x 6a + P x 4 a + N x 2a = 0
N = P
Gy = 3P

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
WHICH BARS ARE NOT GOING TO WORK?

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
BARS AXIAL FORCES? (JOINTS EQUILIBRIUM)

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
WHAT HAPPENS IN THIS KIND OF FOUNDATIONS?

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
2017/2018 CLASS 3
DO THESE RESLTS MAKE SENSE?

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
+
CAN WE APPLY THE SUPERPOSITION PRINCIPLE?

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
WHICH DISPLACEMENTS CAN WE CALCULATE MORE EASILY?
WHY ARE THESE TWO DIAGRAMS USEFUL
TO COMPUTE THE DISPLACEMENTS?

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
VERTICAL DISPLACEMENT AT F
ONLY THE CABLE AXIAL FORCE AFFECTS IT
vF = N x L / EA
vF = 3P x 6a / 65971,5
F goes up
vF varies from 1,81 mm to 3,8 mm

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2HORIZONTAL DISPLACEMENT AT G
ONLY THE COLUMN BG BENDING MOMENT AFFECTS IT
uG = 4,5qa x (6a + 4 x 4,5a + 3a)/(6x 52857) x 3a
uG varies between 11,49 and 55,37 mm

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2HORIZONTAL DISPLACEMENT AT D
ONLY THE COLUMN BG BENDING MOMENT AFFECTS IT
uD = 4,5qa2 x (3a + 4 x 1,5a + 0)/(6x 52857) x 3a
uD varies between 3,83 and 18,46 mm

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2VERTICAL DISPLACEMENT AT D
THE COLUMN BG AND THE BEAM CD BENDING MOMENTS AFFECTS IT
vD = 3qa2 + 4 x 3a + 3a)/(6x 52857) x 3a +
(4,5qa2 x 3a + 4 x 1,125qa2 x 1,5a + 0)/(6x11970) x 3a
vD varies between -16,1 and -77,7 mm

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2VERTICAL DISPLACEMENT AT J
THE TRUSS AND THE CABLE AXIAL FORCE AFFECT IT
vJ = S (N* x N x L / EA)for each bar + P x (-3) /65951
vJ varies between -12,21 and -25,67 mm

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2VERTICAL DISPLACEMENT AT I
THE TRUSS AND THE CABLE AXIAL FORCE AFFECT IT
vJ = S (N* x N x L / EA)for each bar + P x (+2) /65951
vJ varies between 4,81 and 10,11 mm

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
REMEMBER!
IN BEAMS A COLUMNS WE ONLY CONSIDER MOVEMENTS
PRODUCED BY BENDING MOMENTS. WE NEGLECT SHEAR
FORCE AND AXIAL FORCE DEFORMATIONS. WE ASSUME BARS
ARE INEXTENSIBLE
UNLESS CABLES OR TRUSSES ARE INVOLVED

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
IS IT UNSTABLE, STATICALLY DETERMINATE OR STATICALLY INDETERMINATE?

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
TWO DIFFERENT MODELS TO ANALYZE:
front elevation and side elevation

cperez.eps@ceu.es
molina.eps@ceu.es
Year 3 Term 1
19/20 CLASS 2
TRUSS ANALYSIS

Ae1 19 20_class 2_09_10

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Ae1 19 20_class 2_09_10