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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

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

32 STRUCTURES II |

STRUCTURAL ANALYSIS IV

ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV

STRUCTURAL ANALYSIS IV

Axial

orces: Strength and Deformation

ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV

Strength

Axial Strength:
Pn = Fy x A

Bending Strength:

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

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

“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

Elastic Range

Rupture

Yielding

Plastic Range

Yield Strength

Min. tensile
Strength

Source: Krauss et al.: Grundlagen der Tragwerkslehre 1

St
re

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

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

E,A

N
F

Initial State

Deformed State

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 (L0,A) and the external forces (F)

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

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

L0
𝜀
𝜎
E
External Force

Internal Force

Stress in section

Section

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

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

𝐸 = 29,000 𝑘𝑠𝑖

F = 10 k

E,A

L
0

ΔL
𝜀
𝜎

ESteel = 29,000 ksi

0
.5

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
.

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

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

L
z
x
q

w(x)

Bending Deformation
L

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

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

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)

ARCH132 STRUCTURES II | STRUCTURAL ANALYSIS IV
STRUCTURAL ANALYSIS IV
Euler buckling cases
Initial state Deformed state
(axial elastic
deformation)

F2 = Fcrit

Buckling
(instability)

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

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

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

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

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

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