Objectives
In this lecture, the following will be learned:
Study Euler's theory of buckling.
Examine the effect of different end conditions on the buckling load.
Introduce the concept of "effective length."
Study struts with imperfections, including:
•Eccentricity
•Initial curvature
Calculate the critical buckling load for struts with imperfections.
Present the practical curves used in various design codes.
A strut is a member of structure which is in compression.
The term ‘strut’ is usually reserved for long slender members
that are liable to fail by buckling.
Buckling is a failure mode which happens suddenly.
Short, stocky columns fail by crushing.
Perfect (Ideal) struts
(1) straight without initial curvature.
(2) The load is concentric.
(3) Sections are isotropic.
(4) Free from residual stresses.
Pin-ended struts under axial load
Pin-ended struts under axial load
From Euler’s Theory:
𝑑2𝑦
𝐸𝐼 2 = 𝑀 = −𝑃𝑦 (1.1)
𝑑 𝑥
𝑑2𝑦 𝑃
∴ 2 + 𝑦=0 (1.2)
𝑑 𝑥 𝐸𝐼
𝑃
Put 𝐸𝐼 = 𝛼 2
𝑑2𝑦
+ 𝛼2𝑦 = 0 (1.2𝑎)
𝑑2𝑥
The solution is:
𝑦 = 𝐴 sin 𝛼𝑥 + 𝐵 cos 𝛼𝑥 (1.3)
At 𝑥 = 0 → 𝑦 = 0 → 𝐵 = 0
𝑦 = 𝐴 sin 𝛼𝑥 (1.3𝑎)
At 𝑥 = 𝑙 → 𝑦 = 0
𝐴 sin 𝛼𝑙 = 0 (1.4)
Either 𝐴 = 0 (not acceptable, No buckling!)
or sin 𝛼𝑙 = 0
𝝅 𝟐 𝝅𝟐
∴ 𝜶𝒍 = 𝝅 , 𝜶 = ,𝜶 = ,
𝒍 𝒍𝟐
𝑷 𝝅𝟐
= (1.5)
𝑬𝑰 𝒍𝟐
The value of P from (1.5) which causes the strut to buckle is
known as “Euler Crippling load”.
𝜋2 𝐸𝐼
𝑃𝑒 = (1.6)
𝑙2
where:
𝐼 is the minimum moment of inertia
from (3)
𝑦 = 𝐴 sin 𝛼𝑥
𝜋
𝑦 = 𝐴 sin 𝑥 (1.7)
𝑙
Direction fixed at both ends
𝑀𝑥 = 𝑀 − 𝑃𝑦 (2.1)
𝑑2 𝑦
∴ 𝐸𝐼 2 = −𝑃𝑦 + 𝑀 (2.2)
𝑑 𝑥
𝑑2𝑦 𝑃 𝑀
∴ 2 =− 𝑦+ (2.3)
𝑑 𝑥 𝐸𝐼 𝐸𝐼
𝑃
Put = 𝛼2
𝐸𝐼
𝑑2 𝑦 2
𝑀
2
+𝛼 𝑦 = (2.4)
𝑑 𝑥 𝐸𝐼
Complete solution of equation (2.4) is:
𝑀
𝑦 = 𝐴 sin 𝛼𝑥 + 𝐵 cos 𝛼𝑥 +
𝐸𝐼𝛼 2
𝑀
𝑦 = 𝐴 sin 𝛼𝑥 + 𝐵 cos 𝛼𝑥 + (2.5)
𝑃
𝑑𝑦
= 𝐴𝛼 cos 𝛼𝑥 − 𝐵𝛼 sin 𝛼𝑥 (2.6)
𝑑𝑥
𝑀
At 𝑥 = 0, 𝑦 = 0 → 𝐵 = −
𝑃
𝑑𝑦
At 𝑥 = 0, 𝑑𝑥 = 0 → 𝐴𝛼 = 0 → 𝐴 = 0 (𝛼 ≠ 0)
𝑀 𝑀
𝑦 = − cos 𝛼𝑥 +
𝑃 𝑃
𝑀
∴ 𝑦 = (1 − cos 𝛼) (2.7)
𝑃
At 𝑥 = 𝑙, 𝑦 = 0
𝑀
0 = (1 − cos 𝛼𝑙)
𝑃
2𝜋
∴ cos 𝛼𝑙 = 1 → 𝛼𝑙 = 2𝜋 → 𝛼 =
𝑙
𝑃 2𝜋
=
𝐸𝐼 𝑙
4𝜋 2
∴ 𝑃𝑒 = 2 𝐸𝐼 2.8
𝑙
𝜋2
𝑃𝑒 = 2 𝐸𝐼
𝑙
2
𝒍
This is equivalent to hinged strut with Length of 𝟐
.
𝑀𝑥 = 𝑀 − 𝑃𝑦 (𝑎)
𝑀𝑥 = 𝑃𝑎 − 𝑃𝑦 (𝑏)
𝑀𝑥 = 𝑃 𝑎 − 𝑦 (𝑐)
Mx M=P𝒂
𝒂 𝒚
𝒙
𝑷
𝑀𝑥 = 𝑃 𝑎 − 𝑦 (3.1)
𝑑2𝑦
∴ 𝐸𝐼 2 + 𝑃𝑦 = 𝑃𝑎 (3.2)
𝑑 𝑥
𝑑2𝑦 𝑃 𝑃
∴ 2 + 𝑦= 𝑎 (3.3)
𝑑 𝑥 𝐸𝐼 𝐸𝐼
𝑃
Put = 𝛼2
𝐸𝐼
𝑑2 𝑦 2 2
2
+ 𝛼 𝑦 = 𝛼 𝑎 (3.4)
𝑑 𝑥
The complete solution of equation (4) is:
𝑦 = 𝐴 sin 𝛼𝑥 + 𝐵 cos 𝛼𝑥 + 𝑎 (3.5)
𝑑𝑦
= 𝐴𝛼 cos 𝛼𝑥 − 𝐵𝛼 sin 𝛼𝑥 (3.6)
𝑑𝑥
At 𝑥 = 0, 𝑦 = 0 → 𝐵 = −𝑎
𝑑𝑦
At 𝑥 = 0, 𝑑𝑥 = 0 → 𝐴𝛼 = 0 → 𝐴 = 0
∴ 𝑦 = −𝑎 cos 𝛼𝑥 + 𝑎 = 𝑎 1 − cos 𝛼𝑥 (3.7)
𝑎𝑡 𝑥 = 𝑙, 𝑦 = 𝑎
𝑎 = 𝑎 1 − cos 𝛼𝑙
∴ cos 𝛼𝑙 = 0
𝜋 𝑃 𝜋
∴ 𝛼𝑙 = 𝑜𝑟 𝑙=
2 𝐸𝐼 2
𝑃 𝜋2
∴ = 2
𝐸𝐼 4𝑙
𝜋2 𝜋2
∴ 𝑃𝑒 = 𝐸𝐼 or 𝑃𝑒 = 𝐸𝐼 (3.8)
4𝑙 2 2𝑙 2
This is equivalent to hinged strut with length of 𝟐𝒍
𝑀𝑥 = −𝑃𝑦 − 𝑉𝑥 (4.1)
𝑑2𝑦
∴ 𝐸𝐼 2 + 𝑃𝑦 = −𝑉𝑥 (4.2)
𝑑 𝑥
𝑑2𝑦 𝑃 −𝑉
∴ 2 + 𝑦= 𝑥 (4.3)
𝑑 𝑥 𝐸𝐼 𝐸𝐼
𝑃
Put = 𝛼2
𝐸𝐼
𝑑2𝑦 2
−𝑉
2
+𝛼 𝑦 = 𝑥 (4.4)
𝑑 𝑥 𝐸𝐼
The complete solution is:
𝑉
𝑦 = 𝐴 sin 𝛼𝑥 + 𝐵 cos 𝛼𝑥 − 𝑥
𝐸𝐼
𝑉
At 𝑥 = 0, 𝑦 = 0 → 𝐵 = 0 → 𝑦 = 𝐴 sin 𝛼𝑥 − 𝑥
𝐸𝐼
𝑉
𝑎𝑡 𝑥 = 𝑙, 𝑦 = 0 → 𝐴 sin 𝛼𝑙 − 𝑙 = 0 (4.5)
𝐸𝐼
𝑑𝑦 𝑉
𝑎𝑡 𝑥 = 𝑙, = 0 → 𝐴𝛼 cos 𝛼𝑙 − =0 (4.6)
𝑑𝑥 𝐸𝐼
From (4.5) and (4.6)
tan 𝛼𝑙 = 𝛼𝑙 → 𝛼𝑙 = 4.493
2 2
𝑃 2
∴ 𝛼 𝑙 = 20.19 → 𝑙
𝐸𝐼
𝜋2
= 20.19 ≈
0.72
𝜋2
∴ 𝑃𝑒 = 2
𝐸𝐼 (4.7)
0.7𝑙
𝟐
Effective Length
𝝅
𝑷𝒆 = 𝟐
𝑬𝑰
𝑳𝒆
Slenderness Ratio
𝑳𝒆
𝝺= , 𝒓𝟐 = 𝑰/𝑨
𝒓
𝝅𝟐
𝑷𝒆 = 𝟐
𝑬𝑰
𝑳𝒆
𝑷𝒆 𝝅𝟐 𝑬𝑰 𝝅𝟐 𝑬 𝝅𝟐 𝑬
σ𝒆 = = = =
𝑨 𝑨 𝑳𝒆 𝟐 𝑳𝒆 /𝒓 𝟐 𝝺 𝟐
Slenderness Ratio
𝝅𝟐 𝑬 𝑪𝒐𝒏𝒔𝒕.
σ𝒆 = =
𝝺 𝟐 𝝺 𝟐
Struts with Imperfections
Buckling of struts with imperfections usually means
struts with initial curvature or/and eccentric loading
Struts with Eccentric Loads
Use the Euler theory.
Calculate the couple moment due to eccentricity (e).
No matter how small the load (P) and the eccentricity (e)
are, the couple (MA) will cause some bending of the column.
As the eccentric load increases, both the couple (MA) and
the axial force (P) increase, causing the column to bend
further.
Strut with an Eccentric Load
Note:
y measured from the line of action of the load
Strut with an Eccentric Load
From Euler Theory:
𝑑2𝑦
𝐸𝐼 2 = 𝑀 = −𝑃𝑦 (5.1)
𝑑𝑥
𝑑2𝑦 𝑃
∴ 2 + 𝑦=0 (5.2)
𝑑𝑥 𝐸𝐼
𝑃
Put = 𝛼2
𝐸𝐼
𝑑2𝑦 2𝑦 = 0
𝑑𝑥 2 + 𝛼 (5.2𝑎)
The solution is:
𝑦 = 𝐴 sin 𝛼𝑥 + 𝐵 cos 𝛼𝑥 (5.3)
At 𝑥 = 0 → 𝑦 = 𝑒 → 𝑩 = 𝒆
𝑙 𝑑𝑦
At 𝑥 = ,
2 𝑑𝑥
=0
𝑑𝑦 𝛼𝑙 𝛼𝑙
∴ = 0 = 𝐴𝛼 cos − 𝑒𝛼 sin
𝑑𝑥 2 2
𝜶𝒍
or 𝑨= 𝒆 𝒕𝒂𝒏 𝟐 (5.4)
𝛼𝑙
∴ 𝑦 = 𝑒 tan . sin 𝛼𝑥 + cos 𝛼𝑥 (5.5)
2
𝜶𝒍
Max. deflection at 𝒕𝒂𝒏 𝟐 =∞→𝒚=∞
2 2
𝛼𝑙 𝜋 𝜋 𝑃 𝜋
∴ = → 𝛼 2 = 2 𝑜𝑟 = 2
2 2 𝑙 𝐸𝐼 𝑙
𝝅𝟐 𝑬𝑰
∴ 𝑷𝒆 = 𝟐 (5.6)
𝒍
This the same crippling load (but not the same failure load !!!)
• Due to the additional moment, the strut will always fail by
compressive stress before the Euler load is reached (i.e. 𝑦
will not reach ∞)
𝑙 𝛼𝑙 𝛼𝑙 𝛼𝑙
• At 𝑥 = → 𝑦 = 𝑦,
ො 𝑦ො = 𝑒 tan . sin + cos
2 2 2 2
𝜶𝒍
ෝ = 𝒆 𝒔𝒆𝒄
∴𝒚 (5.7)
𝟐
𝜶𝒍
∴ 𝑴 = 𝑷ෝ
𝒚 = 𝑷𝒆 𝒔𝒆𝒄 (5.8)
𝟐
𝑷 𝑴𝒚
𝝈𝒎𝒂𝒙 = + (5.9)
𝑨 𝑰
Treating as a beam with initial curvature:
𝟏
𝑹𝒐 ≈ 𝟐 Τ 𝟐 ,
𝒅 𝒚 𝒅𝒙
1
𝑅𝑜 ≈ ,
𝑑 2 𝑦Τ𝑑𝑥 2
1 1
𝑀 = 𝐸𝐼( − ) (6.1)
𝑅 𝑅𝑜
𝑑2 𝑦 1 1
∴ 𝐸𝐼 = 𝑀 = 𝐸𝐼 − (6.2)
𝑑2 𝑥 𝑅 𝑅𝑜
𝑜𝑟
𝑑2 𝑦 𝑑 2 𝑦𝑜
𝐸𝐼 = −𝑃𝑦 + 𝐸𝐼 (6.3)
𝑑2 𝑥 𝑑2 𝑥
𝑑2 𝑦 𝜋2 𝜋𝑥
∴ + 𝛼2𝑦 = −𝐶 2 sin (6.4)
𝑑2 𝑥 𝑙 𝑙
The complete solution is:
𝜋2 𝜋𝑥
𝐶 2 sin
𝑙 𝑙
𝑦 = 𝐴 sin 𝛼𝑥 + 𝐵 cos 𝛼𝑥 + 𝜋2
(6.5)
2 −𝛼 2
𝑙
𝑙 𝑑𝑦
At 𝑥 = 0, 𝑦 = 0 → 𝑩 = 𝟎 & At 𝑥 = 2 , =0→𝑨=𝟎
𝑑𝑥
𝝅𝟐
𝑪. 𝟐 𝜋𝑥 𝑪.𝑷𝒆 𝜋𝑥
𝒍
∴𝑦= 𝝅𝟐 𝑷
sin →𝑦= sin (6.6)
− 𝑙 𝑷𝒆 −𝑷 𝑙
𝒍𝟐 𝑬𝑰
𝑷.𝑷𝒆
𝑀𝑚𝑎𝑥 = 𝑃𝑦𝑚𝑎𝑥 = 𝐶. (6.7)
𝑷𝒆 −𝑷
Limitations of Euler Theory
The challenge of buckling in struts with imperfections is not
simply about determining how long the column can stay
straight and stable as the load increases. Rather, it’s about
understanding how much bending the column can sustain
under this load without exceeding the allowable stress and
ensuring the maximum deflection remains within acceptable
limits.
Limitations of Euler Theory
• The stress to cause buckling from the Euler formula for
pined-ended strut is:
𝑷𝒆 𝝅𝟐 𝑬𝑰 𝝅𝟐 𝑬
𝝈𝒆 = = 𝟐
= 𝟐
𝑨 𝑨𝒍 𝒍ൗ
𝒌
• This gives the curve ABC.
𝒍
• For values of 𝒌
< 𝟏𝟐𝟎 the error in applying Euler theory is
too great to allow of its use.
• If 𝜎𝑒 > 𝜎𝑐 the strut fail by crushing along the line BD (short
column region)
• Allowing for imperfections of loading and strut, the
actual values at failure must lie within and below the line
ABD.
𝒍
• For values of < 𝟏𝟐𝟎 the error in applying Euler theory
𝒌
is too great to allow of its use.
• If 𝝈𝒆 > 𝝈𝒄 the strut fail by crushing along the line BD
(short column region).
• The practical strut formulae have been devised to cover
the intermediate zone between columns and struts and
to allow for imperfections.
Rankine-Gordon Formula
𝟏 𝟏 𝟏
= + (7.1)
𝝈 𝝈𝒄 𝝈𝒆
where 𝝈 is the actual stress to failure, 𝝈𝒄 is yield stress in
compression, 𝜎𝑒 is the Euler stress.
• Equation (1) will produce a curve which is tangential to 𝜎𝑐 as
𝑙 𝑙
→ 0, tangential to 𝝈𝒆 as → ∞.
𝑘 𝑘
• For intermediate values 𝝈 will be less than both 𝝈𝒆 , 𝝈𝒄 .
𝝈𝒄 .𝝈𝒆 𝝈𝒄
𝝈=
𝝈𝒄 +𝝈𝒆
= 𝟏+𝝈𝒄ൗ (7.2)
𝝈𝒆
For pin-ended strut:
𝝈𝒆 = 𝝅𝟐 𝑬𝒌𝟐 /𝒍𝟐
Hence,
𝝈𝒄
𝝈= 𝝈 (7.3)
𝟏+ 𝒄ൗ𝝈𝒆
𝝈𝒄
𝝈 = 𝝈𝒄 𝟐 (7.4)
𝟏+ ൗ𝝅𝟐 𝑬 𝒍ൗ𝒌
𝝈𝒄
Put =𝒂 constant .
𝝅𝟐 𝑬
𝝈𝒄
𝝈 = 𝟐
𝟏+𝒂 𝒍ൗ𝒌
• To make allowance for imperfections, the value of 𝒂
depending on the material and on the end conditions.
• The permissible load is given by:
𝝈𝒄 𝑨
𝑷 = 𝝈𝑨 = 𝟐
(7.5)
𝟏 + 𝒂 𝒍ൗ𝒌
• Typical values of 𝒂 :
Mild steel 1/7500 (pinned), 1/30,000 (fixed)
Cast iron 1/1600 (pinned), 1/6400 (fixed)
Timber 1/3000 (pinned), 1/12,000 (fixed).
• A factor of safety may be included in the Rankine formula by
reducing the value of 𝝈𝒄 , which becomes the working stress.
Questions???
Revise the followings:
• Struts (G H Ryder, p238-243)
• Example 1 (G H Ryder, p239-240)
• Example 2 (G H Ryder, p240)
• Examples 3 (G H Ryder, p241-242)
