axial force, deformation,buckling
Alex289
S2021-04-02SeminarAnalysisIV.pdf
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
Axial Forces: 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: Mn = 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)
Stiffness
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 vis "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
Strain
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
ΔLP
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 sectionSection
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 sectionSection
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
ΔL
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 𝑑𝑥
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 𝑖𝑛2
𝐸 = 29,000 𝑘𝑠𝑖
F = 10 kip
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 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
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
