axial force, deformation,buckling
The
Cooper Union Name of Student:
Irwin S. Chanin School of Architecture ______________________________________
Arch 132 Structures II Spring Semester 2021
Thorsten Helbig, Associate Professor
Florian Meier, Instructor
Class Meets: Monday: 2:00PM-04:50PM (Zoom)
Assignment 2
1 of 6
Assignment 2: Axial Force, Deformation, Buckling
1) Determine the axial force N and the deformation ΔL of the steel rod shown below:
Ashley Fu
The Cooper Union Name of Student:
Irwin S. Chanin School of Architecture ______________________________________
Arch 132 Structures II Spring Semester 2021
Thorsten Helbig, Associate Professor
Florian Meier, Instructor
Class Meets: Monday: 2:00PM-04:50PM (Zoom)
Assignment 2
2 of 6
2) A) Determine the axial force N and the deformation ΔL of the W-profile shown below.
Determine the axial strength Pn of the section and compare with the load P. Do not consider
the possibility of buckling yet.
Ashley Fu
The Cooper Union Name of Student:
Irwin S. Chanin School of Architecture ______________________________________
Arch 132 Structures II Spring Semester 2021
Thorsten Helbig, Associate Professor
Florian Meier, Instructor
Class Meets: Monday: 2:00PM-04:50PM (Zoom)
Assignment 2
3 of 6
B) Now determine the critical Euler-buckling load Pcrit of the system and analyze whether a
buckling problem exists or not (buckling problem exists if Pcrit is lower than Pn). Be careful:
Consider buckling in both directions (around the section’s y-y and z-z axis)
Ashley Fu
The Cooper Union Name of Student:
Irwin S. Chanin School of Architecture ______________________________________
Arch 132 Structures II Spring Semester 2021
Thorsten Helbig, Associate Professor
Florian Meier, Instructor
Class Meets: Monday: 2:00PM-04:50PM (Zoom)
Assignment 2
4 of 6
C) How would you modify the system to prevent the buckling problem or at least improve the
Euler buckling strength (choose one)? Calculate the critical buckling load for the new system
(length, section and load remains the same, only support conditions change).
□
□
□
Ashley Fu
Ashley Fu
The Cooper Union Name of Student:
Irwin S. Chanin School of Architecture ______________________________________
Arch 132 Structures II Spring Semester 2021
Thorsten Helbig, Associate Professor
Florian Meier, Instructor
Class Meets: Monday: 2:00PM-04:50PM (Zoom)
Assignment 2
5 of 6
D) What other parameter could be changed in order to prevent buckling and how? Give 2
suggestions:
3) A) Determine the critical buckling load Pcrit of the system below and analyze whether a
buckling problem exists or not (buckling problem exists if Pcrit is lower than Pn).
The moment of Inertia for a round section can be calculated with:
𝐼𝑦 = 𝐼𝑧 =
𝜋
4
× 𝑅4 where R is the outer radius of the circle
Ashley Fu
The Cooper Union Name of Student:
Irwin S. Chanin School of Architecture ______________________________________
Arch 132 Structures II Spring Semester 2021
Thorsten Helbig, Associate Professor
Florian Meier, Instructor
Class Meets: Monday: 2:00PM-04:50PM (Zoom)
Assignment 2
6 of 6
B) The system is modified to a circular hollow profile. The profile is produced with the same
resources/tonnage of steel (same area A). Determine the new critical buckling load Pcrit of the
system below and analyze whether a buckling problem still exists or not.
Ashley Fu
ARCH
1
32 STRUCTURES II |
STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
Axial
F
orces: Strength and Deformation
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
F
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
Strength
Axial Strength:
Pn = Fy x A
Bending Strength:
M
n = Fy x S
(with S = I/zmax)
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
Strength
Axial Strength:
Pn = Fy x A
Bending Strength:
Mn = Fy x S
(with S = I/zmax
)
Stiffne
ss
Axial Stiffness:
E x A
Bending Stiffness:
E x I
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
Robert Hooke (1635 – 1703),
British physicist
Image Source: Wikipedia
ut tensio, sic vi
s
“as the extension, so the force“ or
“the extension is proportional to the force”
Hooke‘s Law
ca. 1676
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
Steel: Elastic and Plastic
Deformation
Stra
in
Elastic Range
Rupture
Yielding
Plastic Range
Yield Strength
Min. tensile
Strength
Source: Krauss et al.: Grundlagen der Tragwerkslehre 1
St
re
ss
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
09/24/20189
Simplified Bilinear Stress Strain Curve
Yield Strength
Simplified Diagram
Source: Krauss et al.: Grundlagen der Tragwerkslehre
1
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
Elastic Stress Strain Curve
Source: Krauss et al.: Grundlagen der Tragwerkslehre 1
Young’s Modulus E for steel: 210,000 Mpa = 29,000 ksi
(Comparison: E-Modul Wood E = ca. 8,000-16,000 Mpa)
𝐸 =
𝜎
𝜀
𝜀
=
ൗ𝐹 𝐴
ൗ∆𝐿
𝐿0
𝑐𝑜𝑛𝑠𝑡.
𝐸 is the Young’s modulus (modulus of elasticity)
𝐹 is the force exerted on an object under tension
𝐴 is the actual cross-sectional area, which equals the area of the
cross-section perpendicular to the applied force
Δ𝐿 is the amount by which the length of the object changes (Δ𝐿 is
positive if the material is stretched , and negative when
the material is compressed)
𝐿0 is the original length of the object
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
F
E,A
N
F
Initial State
Deformed State
ΔL
P
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
F
E,A
The deformation of an element depends on the material properties
(E), the system properties (L0,A) and the external forces (F)
L0
External Force
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
F
E,A
The deformation of an element depends on the material properties
(E), the system properties (L0,A) and the external forces (F)
The Young´s Modulus E describes the relation between strain and
stress
𝐸 =
𝜎
𝜀
with 𝜎 =
𝐹
𝐴
and 𝜀 =
∆𝐿
𝐿0
L0
𝜀
𝜎
E
External Force
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
F
E,A
The deformation of an element depends on the material properties
(E), the system properties (L,A) and the external forces (P)
The Young´s Modulus E describes the relation between strain and
stress
𝐸 =
𝜎
𝜀
with 𝜎 =
𝐹
𝐴
and 𝜀 =
∆𝐿
𝐿0
P
F
σ
L0
𝜀
𝜎
E
External Force
Internal Force
Stress in section
Section
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
F
E,A
The deformation of an element depends on the material properties
(E), the system properties (L,A) and the external forces (P)
The Young´s Modulus E describes the relation between strain and
stress
𝐸 =
𝜎
𝜀
with 𝜎 =
𝐹
𝐴
and 𝜀 =
∆𝐿
𝐿0
Knowing all required properties, the deformation can be calculated as
follows:
𝐸 =
𝜎
𝜀
=
ൗ𝐹 𝐴
ൗ∆𝐿 𝐿0
=
𝐹 × 𝐿0
𝐴 × ∆𝐿
P
F
σ
L0
𝜀
𝜎
E
External Force
Internal Force
Stress in section
Section
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
F
E,A
The deformation of an element depends on the material properties
(E), the system properties (L,A) and the external forces (P)
The Young´s Modulus E describes the relation between strain and
stress
𝐸 =
𝜎
𝜀
with 𝜎 =
𝐹
𝐴
and 𝜀 =
∆𝐿
𝐿0
Knowing all required properties, the deformation can be calculated as
follows:
𝐸 =
𝜎
𝜀
=
ൗ𝐹 𝐴
ൗ∆𝐿 𝐿0
=
𝐹 × 𝐿0
𝐴 × ∆𝐿
follows: ∆𝑳=
𝑭×𝑳𝟎
𝑬×𝑨
for the example to the left
P
F
σ
L0
𝜀
𝜎
E
F
External Force
Internal Force
Deformation
Stress in sectionSection
Δ
L
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
F
E,A
The deformation of an element depends on the material properties
(E), the system properties (L,A) and the external forces (P)
The Young´s Modulus E describes the relation between strain and
stress
𝐸 =
𝜎
𝜀
with 𝜎 =
𝐹
𝐴
and 𝜀 =
∆𝐿
𝐿0
Knowing all required properties, the deformation can be calculated as
follows:
𝐸 =
𝜎
𝜀
=
ൗ𝐹 𝐴
ൗ∆𝐿 𝐿0
=
𝐹 × 𝐿0
𝐴 × ∆𝐿
follows: ∆𝑳=
𝑭×𝑳𝟎
𝑬×𝑨
for the example to the left
General formula: ∆𝐿 = 0
𝐿
𝜀 𝑥 𝑑𝑥 = 0
𝐿 𝑃 𝑥 𝑑
𝑥
𝐸𝐴
P
F
σ
L0
𝜀
𝜎
E
External Force
Internal Force
Stress in sectionSection
F
Deformation
ΔL
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
EXAMPLE
Calculating the deformation of a load hanging on a steel rod
F= 10 kip = 10,000 lbf
𝐴 = 0.5𝑖𝑛 × 0.5𝑖𝑛 = 0.25 𝑖𝑛
2
𝐸 = 29,000 𝑘𝑠𝑖
F = 10 k
ip
z
x
E,A
L
0
=
3
.2
ft
ΔL
𝜀
𝜎
ESteel = 29,000 ksi
0
.5
in
0.5 in
Section
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
EXAMPLE
Calculating the deformation of a load hanging on a steel rod
F=10,000 lbf
𝐴 = 0.5𝑖𝑛 × 0.5𝑖𝑛 = 0.25 𝑖𝑛2
𝐸 = 29,000 𝑘𝑠𝑖
∆𝑳=
𝑭 × 𝑳𝟎
𝑬 × 𝑨
=
10𝑘𝑖𝑝 × 38.4 𝑖𝑛
29,000𝑘𝑠𝑖 × 0.25𝑖𝑛2
=
384𝑘𝑖𝑝 − 𝑖𝑛
7,250𝑘𝑖𝑝
= 0.05𝑖𝑛
z
x
0
.5
in
0.5 in
P
=5
0
k
ip
𝜀
𝜎
ESteel = 29,000 ksi
Section
Internal Force
F = 10 kip
E,A
L
0
= 3
.2
ft
= 3
8
.
4
in
ΔL
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
EXAMPLE
Calculating the deformation of a load hanging on a steel rod
F=10,000 lbf
𝐴 = 0.5𝑖𝑛 × 0.5𝑖𝑛 = 0.25 𝑖𝑛2
𝐸 = 29,000 𝑘𝑠𝑖
∆𝑳=
𝑭 × 𝑳𝟎
𝑬 × 𝑨
=
10𝑘𝑖𝑝 × 38.4 𝑖𝑛
29,000𝑘𝑠𝑖 × 0.25𝑖𝑛2
=
384𝑘𝑖𝑝 − 𝑖𝑛
7,250𝑘𝑖𝑝
= 0.05𝑖𝑛
z
x
0
.5
in
0.5 in
P
=5
0
k
ip
𝜀
𝜎
ESteel = 29,000 ksi
Section
Internal Force
F = 10 kip
E,A
L
0
= 3
.2
ft
ΔL
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
EXAMPLE
Calculating the deformation of a load hanging on a steel rod
F=10,000 lbf
𝐴 = 0.5𝑖𝑛 × 0.5𝑖𝑛 = 0.25 𝑖𝑛2
𝐸 = 29,000 𝑘𝑠𝑖
∆𝑳=
𝑭 × 𝑳𝟎
𝑬 × 𝑨
=
10𝑘𝑖𝑝 × 38.4 𝑖𝑛
29,000𝑘𝑠𝑖 × 0.25𝑖𝑛2
=
384𝑘𝑖𝑝 − 𝑖𝑛
7,250𝑘𝑖𝑝
= 0.05𝑖𝑛
Additional question:
Is the steel section still OK? Assume 50 ksi yield strength
z
x
0
.5
in
0.5 in
P
=5
0
k
ip
𝜀
𝜎
ESteel = 29,000 ksi
Section
Internal Force
F = 10 kip
E,A
L
0
= 3
.2
ft
ΔL
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
EXAMPLE
Calculating the deformation of a load hanging on a steel rod
F=10,000 lbf
𝐴 = 0.5𝑖𝑛 × 0.5𝑖𝑛 = 0.25 𝑖𝑛2
𝐸 = 29,000 𝑘𝑠𝑖
∆𝑳=
𝑭 × 𝑳𝟎
𝑬 × 𝑨
=
10𝑘𝑖𝑝 × 38.4 𝑖𝑛
29,000𝑘𝑠𝑖 × 0.25𝑖𝑛2
=
384𝑘𝑖𝑝 − 𝑖𝑛
7,250𝑘𝑖𝑝
= 0.05𝑖𝑛
Additional question:
Is the steel section still OK? Assume 50 ksi yield strength
𝑃 = 10𝑘𝑖𝑝
𝑃𝑛 = 50𝑘𝑠𝑖 × 𝐴 = 12.5 𝑘𝑖𝑝
𝑃/𝑃𝑛 =
10
12.5
= 0.8 < 1.0 𝑂𝐾
z
x
0
.5
in
0.5 in
P
=5
0
k
ip
𝜀
𝜎
ESteel = 29,000 ksi
Section
Internal Force
F = 10 kip
E,A
L
0
= 3
.2
ft
ΔL
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
Elastic Deformation: Bending
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
Bending Deformation
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
Bending Deformation
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
The deformation of a beam is related to the material properties E, the system
parameters I,A,L and the external loads q.
For EI=const
𝑑2𝑤 𝑥
𝑑𝑥2
=
𝑀(𝑥)
𝐸𝐼
q
L
z
x
q
w(x)
M
Bending Deformation
L
fm
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
The deformation of a beam is related to the material properties E, the system
parameters I,A,L and the external loads q.
For EI=const
𝑑2𝑤 𝑥
𝑑𝑥2
=
𝑀(𝑥)
𝐸𝐼
For a simple beam the following formula can be used to calculate the
deformation:
𝑤 𝑥 =
𝑞𝑙4
24𝐸𝐼
× (
𝑥
𝑙
− 2
𝑥
𝑙
3
+
𝑥
𝑙
4
)
𝒇𝒎 =
𝟓𝒒𝒍𝟒
𝟑𝟖𝟒𝑬𝑰
Deformation at mid-span
Bending Deformation
q
L
z
x
q
w(x)
M
L
fm
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
The deformation of a beam is related to the material properties E, the system
parameters I,A,L and the external loads P.
For EI=const
𝑑2𝑤 𝑥
𝑑𝑥2
=
𝑀(𝑥)
𝐸𝐼
For a simple beam the following formula can be used to calculate the
deformation:
𝑤 𝑥 =
𝐹𝑙3
48𝐸𝐼
× (3
𝑥
𝑙
− 4
𝑥
𝑙
3
)
𝑓𝑚 =
𝐹𝐿3
48𝐸𝐼
Deformation at mid-span
Bending Deformation
L
z
x
w(x)
M
fm
F
F
F
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
Precambering
q
L
z
x
q
M
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
Precambering
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
Stability: Euler
Buckling
Theory
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
Strength
Axial Strength:
Pn = Fy x A
Stiffness
Axial Stiffness:
E x A
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
Strength
Axial Strength:
Pn = Fy x A
Stiffness
Axial Stiffness:
E x A
…true for tensile forces!
For slender columns under compression, the
critical buckling load Pcrit may be less than the
actual axial strength Pn
NEW: Stability
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
Euler buckling cases
Initial state
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
Euler buckling cases
Initial state Deformed state
(axial elastic
deformation)
F1
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
Euler buckling cases
Initial state Deformed state
(axial elastic
deformation)
F2 = Fcrit
Buckling
(instability)
F1
ARCH132 STRUCTURES II | BASICS OF STRUCTURAL ANALYSIS IV
BASICS OF STRUCTURAL ANALYSIS IV
F1 < Fcrit
ARCH132 STRUCTURES II | BASICS OF STRUCTURAL ANALYSIS IV
BASICS OF STRUCTURAL ANALYSIS IV
F2 = FcritF1 < Fcrit
ARCH132 STRUCTURES II | BASICS OF STRUCTURAL ANALYSIS IV
BASICS OF STRUCTURAL ANALYSIS IV
F2 = Fcrit F3 < FcritF1 < Fcrit
ARCH132 STRUCTURES II | BASICS OF STRUCTURAL ANALYSIS IV
BASICS OF STRUCTURAL ANALYSIS IV
F1
F2 < Fcrit
ARCH132 STRUCTURES II | BASICS OF STRUCTURAL ANALYSIS IV
BASICS OF STRUCTURAL ANALYSIS IV
F2 = Fcrit
F1 F2
F2 < Fcrit
ARCH132 STRUCTURES II | BASICS OF STRUCTURAL ANALYSIS IV
BASICS OF STRUCTURAL ANALYSIS IV
F2 < Fcrit
F2 = Fcrit
F3 < Fcrit
F1 F2 F3
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
O N E O C E A N P A V I L I O N – E X P O 2 0 1 2 Y E O S U K O R E A S O M A
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
O N E O C E A N P A V I L I O N – E X P O 2 0 1 2 Y E O S U K O R E A S O M A
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
O N E O C E A N P A V I L I O N – E X P O 2 0 1 2 Y E O S U K O R E A S O M A
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
O N E O C E A N P A V I L I O N – E X P O 2 0 1 2 Y E O S U K O R E A S O M A
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
O N E O C E A N P A V I L I O N – E X P O 2 0 1 2 Y E O S U K O R E A S O M A
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
O N E O C E A N P A V I L I O N – E X P O 2 0 1 2 Y E O S U K O R E A S O M A
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
O N E O C E A N P A V I L I O N – E X P O 2 0 1 2 Y E O S U K O R E A S O M A
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
O N E O C E A N P A V I L I O N – E X P O 2 0 1 2 Y E O S U K O R E A S O M A
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
O N E O C E A N P A V I L I O N – E X P O 2 0 1 2 Y E O S U K O R E A S O M A
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
O N E O C E A N P A V I L I O N – E X P O 2 0 1 2 Y E O S U K O R E A S O M A
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
O N E O C E A N P A V I L I O N – E X P O 2 0 1 2 Y E O S U K O R E A S O M A
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
R E S E A R C H P A V I L I O N 2 0 1 0 I T K E S T U T T G A R T , P R O F . K N I P P E R S
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
R E S E A R C H P A V I L I O N 2 0 1 0 I T K E S T U T T G A R T , P R O F . K N I P P E R S
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
R E S E A R C H P A V I L I O N 2 0 1 0 I T K E S T U T T G A R T , P R O F . K N I P P E R S
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
Buckling
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
Leonhard Euler (1707 –1783),
Swiss Mathematician
Euler‘s Critical Load
𝑃𝑐𝑟 =
𝜋2𝐸𝐼
𝐾𝐿 2
𝑃𝑐𝑟 Euler’s critical load
E Modulus of Elasticit
y
I Minimum Area Moment of Inertia of
the Column Cross Section
L Length of Column
K Column Effective Length Factor
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
F
Euler buckling cases
𝑃𝑐𝑟 =
𝜋2𝐸𝐼
𝐾𝐿 2
𝑃𝑐𝑟 Euler’s critical load
E Modulus of Elasticity
I Minimum Area Moment of Inertia of
the Column Cross Section
L Length of Column
K Column Effective Length Factor
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
F
F F F
Euler buckling cases
Case 1 Case 2 Case 3
Case 4
L
𝑃𝑐𝑟 =
𝜋2𝐸𝐼
𝐾𝐿 2
𝑃𝑐𝑟 Euler’s critical load
E Modulus of Elasticity
I Minimum Area Moment of Inertia of
the Column Cross Section
L Length of Column
K Column Effective Length Factor
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
F F F
Euler buckling cases
L s
s
s
F
s
Case 1
s = 2 L
K = 2
Case 2
s = L
K = 1
Case 3
s = 0.7 L
K = 0.7
Case 4
s = 0.5 L
K = 0.5
𝑃𝑐𝑟 =
𝜋2𝐸𝐼
𝐾𝐿 2
𝑃𝑐𝑟 Euler’s critical load
E Modulus of Elasticity
I Minimum Area Moment of Inertia of
the Column Cross Section
L Length of Column
K Column Effective Length Factor
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
Internal Forces
Euler Curve
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
Quest Forum, Rosenheim
Behnisch + Knippers Helbig
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
Quest Forum, Rosenheim
Behnisch + Knippers Helbig
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
y
z
3 in
3
in
0.25 in
0
.7
5
in
0
.7
5
in
x
z
EXAMPLE 1
Compute Euler‘s critical buckling force for the
system to the left. Can the column resist the
force F?
L=40 ft
F=5kip
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
y
z
3 in
3
in
0.25 in
0
.7
5
in
0
.7
5
in
1
x
z
L=40 ft
F=5kip
EXAMPLE 1
Compute Euler‘s critical buckling force for the
system to the left. Can the column resist the
force F?
𝐼1 =
𝑤 × ℎ3
12
=
3𝑖𝑛 × 4.5𝑖𝑛 3
12
= 22.8 𝑖𝑛4
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
y
z
3 in
3
in
0.25 in
0
.7
5
in
0
.7
5
in
2 2
1
x
z
L=40 ft
F=5kip
EXAMPLE 1
Compute Euler‘s critical buckling force for the
system to the left. Can the column resist the
force F?
𝐼1 =
𝑤 × ℎ3
12
=
3𝑖𝑛 × 4.5𝑖𝑛 3
12
= 22.8 𝑖𝑛4
𝐼2 =
1.375𝑖𝑛 × 3𝑖𝑛 3
12
= 3 𝑖𝑛4
𝐼𝑦 = 𝐼1 − 2 × 𝐼2 = 16.8 𝑖𝑛
4
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
y
z
3 in
3
in
0.25 in
0
.7
5
in
0
.7
5
in
x
z
L=40 ft
F=5kip
EXAMPLE 1
Compute Euler‘s critical buckling force for the
system to the left. Can the column resist the
force F?
𝐼𝑦 = 16.8 𝑖𝑛
4
𝐼1 =
𝑤 × ℎ3
12
=
0.75𝑖𝑛 × 3𝑖𝑛 3
12
= 1.69 𝑖𝑛4
𝐼2 =
3 × 0.25𝑖𝑛 3
12
= 0.004 𝑖𝑛4
𝐼𝑧 = 𝐼2 + 2 × 𝐼1 = 3.38 𝑖𝑛
4
2
1
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
EXAMPLE 1
𝐼𝑦 = 16.8 𝑖𝑛
4
𝐼𝑧 = 3.38 𝑖𝑛
4 (governing!)
𝐸 = 29,000 𝑘𝑠𝑖 (Steel)
KL = 40 ft
𝑃𝑐𝑟 =
𝜋2𝐸𝐼
𝐾𝐿 2
=
3.142×29,000𝑘𝑠𝑖×3.38𝑖𝑛4
(1×480𝑖𝑛)2
=
=
966,437𝑘𝑖𝑝 − 𝑖𝑛2
230,400 𝑖𝑛2
= 4.2 𝑘𝑖𝑝
𝑃/𝑃𝑐𝑟 =5kip/4.2kip=1.19 > 1.0 NOT OK
x
z
L=40 ft
F=5 kip
s
Case 2
s = L
K = 1
F=5 kip
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
y
x
L=40 ft
F = 5kip
s
Case 3
s = 0.7 L
K = 0.7
F = 5kip
EXAMPLE 2
Revise the calculation for the system to the left
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
y
x
L=40 ft
F = 5kip
s
Case 3
s = 0.7 L
K = 0.7
F = 5kip
EXAMPLE 2
Revise the calculation for the system to the left
𝐼𝑦 = 16.8 𝑖𝑛
4
𝐼𝑧 = 3.38 𝑖𝑛
4 (governing!)
𝐸 = 29,000 𝑘𝑠𝑖 (Steel)
𝐾𝐿 = 0.7 × 40 𝑓𝑡 = 28𝑓𝑡
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
y
x
L=40 ft
F = 5kip
s
Case 3
s = 0.7 L
K = 0.7
F = 5kip
EXAMPLE 2
Revise the calculation for the system to the left
𝐼𝑦 = 16.8 𝑖𝑛
4
𝐼𝑧 = 3.38 𝑖𝑛
4 (governing!)
𝐸 = 29,000 𝑘𝑠𝑖 (Steel)
𝐾𝐿 = 0.7 × 40 𝑓𝑡 = 28𝑓𝑡
𝑃𝑐𝑟 =
𝜋2𝐸𝐼
𝐾𝐿 2
=
3.142×29,000𝑘𝑠𝑖×3.38𝑖𝑛4
(336)2
=
=
966,437𝑘𝑖𝑝 − 𝑖𝑛2
112,896 𝑖𝑛2
= 8.56 𝑘𝑖𝑝
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
y
x
L=40 ft
F = 5kip
s
Case 3
s = 0.7 L
K = 0.7
F = 5kip
EXAMPLE 2
Revise the calculation for the system to the left
𝐼𝑦 = 16.8 𝑖𝑛
4
𝐼𝑧 = 3.38 𝑖𝑛
4 (governing!)
𝐸 = 29,000 𝑘𝑠𝑖 (Steel)
𝐾𝐿 = 0.7 × 40 𝑓𝑡 = 28𝑓𝑡
𝑃𝑐𝑟 =
𝜋2𝐸𝐼
𝐾𝐿 2
=
3.142×29,000𝑘𝑠𝑖×3.38𝑖𝑛4
(336)2
=
=
966,437𝑘𝑖𝑝 − 𝑖𝑛2
112,896 𝑖𝑛2
= 8.56 𝑘𝑖𝑝
𝑃/𝑃𝑐𝑟 =5kip/8.56kip=0.58 < 1.0 OK
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV