FLEXURAL BUCKLING

1. For the compression members shown in Figure 1 (a) (b) and (c) determine
the elastic buckling load Pcr (i) using the neutral equilibrium approach and
(ii) the energy approach. For the column in Figure 1 (b) μL = tan μL when
μL = 4.493 radians

1a) i) neutral equilibrium method

+𝜇 =0

Assume 𝑤 = 𝐶𝑐𝑜𝑠(𝜇𝑥) + 𝐵𝑠𝑖𝑛(𝜇𝑥) + 𝑅𝑥 + 𝑆

= 𝑤 = 𝐵𝜇𝑐𝑜𝑠(𝜇𝑥) − 𝐶𝜇𝑠𝑖𝑛(𝜇𝑥) + 𝑅

= 𝑤 = −𝐵𝜇 𝑠𝑖𝑛(𝜇𝑥) − 𝐶𝜇 𝑐𝑜𝑠(𝜇𝑥)

At x=0, w=0 → C+S=0 → C=-S

w’=0 → Bμ+R=0 → R=-Bμ
At x=L, w=0 → 𝐶𝑐𝑜𝑠(𝜇𝐿) + 𝐵𝑠𝑖𝑛(𝜇𝐿) + 𝑅𝐿 + 𝑆 = 0 … (A)
w’=0 → 𝐵𝜇𝑐𝑜𝑠(𝜇𝐿) − 𝐶𝜇𝑠𝑖𝑛(𝜇𝐿) + 𝑅 = 0 … (B)
substituting S=-C, R=-Bμ in (A) & (B)
𝐶𝑐𝑜𝑠(𝜇𝐿) + 𝐵𝑠𝑖𝑛(𝜇𝐿) − Bμ𝐿 − C = 0 → 𝐶(𝑐𝑜𝑠(𝜇𝐿) − 1) + 𝐵(𝑠𝑖𝑛(𝜇𝐿) − μ𝐿) = 0 …(C)
𝐵𝜇𝑐𝑜𝑠(𝜇𝐿) − 𝐶𝜇𝑠𝑖𝑛(𝜇𝐿) − Bμ = 0 → −𝐶𝜇𝑠𝑖𝑛(𝜇𝐿) + 𝐵𝜇(𝑐𝑜𝑠(𝜇𝐿) − 1) …(D)
C=B=0 (trivial solution) or determinant of coeff=0
i.e. 𝜇(𝑐𝑜𝑠(𝜇𝐿) − 1) + 𝜇𝑠𝑖𝑛(𝜇𝐿)(𝑠𝑖𝑛(𝜇𝐿) − 1) = 0
μ=0 → P=0 – can be ignored!
Or 𝑐𝑜𝑠 (𝜇𝐿) − 2𝑐𝑜𝑠(𝜇𝐿) + 1 + 𝑠𝑖𝑛 (𝜇𝐿) − 𝜇𝐿𝑠𝑖𝑛(𝜇𝐿) = 0
2 1 − 𝑐𝑜𝑠(𝜇𝐿) − 𝜇𝐿𝑠𝑖𝑛(𝜇𝐿) = 0
4𝑠𝑖𝑛 − 2𝜇𝐿𝑠𝑖𝑛 𝑐𝑜𝑠 =0
4𝑠𝑖𝑛 sin − 𝑐𝑜𝑠 =0
4𝑠𝑖𝑛 =0 → =𝜋 or 𝑡𝑎𝑛 = → = 4.493
So lowest root is given by =𝜋
4𝜋 𝐸𝐼
𝑃 = 𝜇 𝐸𝐼 =
𝐿
ii) energy method
∫
∫ 𝑃 𝑑𝑥 = ∫ 𝐸𝐼 𝑑𝑥 i.e. 𝑃=
∫

Let 𝑤(𝑥) = 𝑤 1 − 𝑐𝑜𝑠 𝑥 , =𝑤 𝑠𝑖𝑛 𝑥

At x=0, 𝑐𝑜𝑠 𝑥 =1 & 𝑠𝑖𝑛 𝑥 =0 → w=0 & w’=0
At x=L 𝑐𝑜𝑠 𝐿 =1 & 𝑠𝑖𝑛 𝐿 =0 → w=0 & w’=0
Now =𝑤 =𝑤 𝑐𝑜𝑠 𝑥
2𝜋 2𝜋
𝐸𝐼 ∫ 𝑤 𝑐𝑜𝑠 𝑥 𝑑𝑥
𝑃= 𝐿 𝐿
2𝜋 2𝜋
∫ 𝑤 𝐿 𝑠𝑖𝑛 𝐿 𝑥 𝑑𝑥

𝑤 𝐸𝐼 ∫ 𝑐𝑜𝑠 𝑥 𝑑𝑥 = 𝑤 𝐸𝐼 ∫ 𝑑𝑥 = 𝑤 𝐸𝐼 ∫ 1 + cos 𝑥 𝑑𝑥
4𝜋
2𝜋 1 𝑠𝑖𝑛 𝐿 𝑥 2𝜋 𝐿
=𝑤 𝐸𝐼 𝑥+ =𝑤 𝐸𝐼
𝐿 2 4𝜋 𝐿 2
𝐿
Similarly
𝑤 ∫ 𝑠𝑖𝑛 𝑥 𝑑𝑥 = 𝑤

So 𝑃 = = 𝐸𝐼 = 𝐸𝐼
 1b) i) Neutral equilibrium method

𝑑 𝑤 𝑑 𝑤
+𝜇 =0
𝑑𝑥 𝑑𝑥
Assume 𝑤 = 𝐶𝑐𝑜𝑠(𝜇𝑥) + 𝐵𝑠𝑖𝑛(𝜇𝑥) + 𝑅𝑥 + 𝑆

𝑑𝑤
= 𝑤 = 𝐵𝜇𝑐𝑜𝑠(𝜇𝑥) − 𝐶𝜇𝑠𝑖𝑛(𝜇𝑥) + 𝑅
𝑑𝑥
𝑑 𝑤
= 𝑤 = −𝐵𝜇 𝑠𝑖𝑛(𝜇𝑥) − 𝐶𝜇 𝑐𝑜𝑠(𝜇𝑥)
𝑑𝑥
At x=0, w=0 → C+S=0 → C=-S
w’=0 → Bμ+R=0 → R=-Bμ
At x=L, w=0 → 𝐶𝑐𝑜𝑠(𝜇𝐿) + 𝐵𝑠𝑖𝑛(𝜇𝐿) + 𝑅𝐿 + 𝑆 = 0 … (A)
w’=0 → −𝐵𝜇 𝑠𝑖𝑛(𝜇𝐿) − 𝐶𝜇 𝑐𝑜𝑠(𝜇𝐿) = 0
−𝐵𝑠𝑖𝑛(𝜇𝐿) − 𝐶𝑐𝑜𝑠(𝜇𝐿) = 0 … (B)
(A)+(B) → RL+S=0 → S=-RL
C=-S=RL=-BμL
Substitute C =-BμL in (B)
−𝐵𝑠𝑖𝑛(𝜇𝐿) + BμL𝑐𝑜𝑠(𝜇𝐿) = 0
μL𝑐𝑜𝑠(𝜇𝐿) − 𝑠𝑖𝑛(𝜇𝐿) = 0
→ μL − tan(μL) = 0 → tan(μL) = μL → μL = 4.493
4.493 20.19
𝑃 = 𝜇 𝐸𝐼 = 𝐸𝐼 = 𝐸𝐼
𝐿 𝐿
ii) Energy method
assume an appropriate polynomial deflection function
𝑤(𝑥) = 𝑎 + 𝑏𝑥 + 𝑐𝑥 + 𝑑𝑥 + 𝑒𝑥
𝑤 ( ) = 𝑏 + 2𝑐𝑥 + 3𝑑𝑥 + 4𝑒𝑥
𝑤′′(𝑥) = 2𝑐 + 6𝑑𝑥 + 12𝑒𝑥
At x=0 w=0 → a=0 At x=0 w’=0 → b=0
At x=L w=0 → 𝐿 (𝑐 + 𝑑𝐿 + 𝑒𝐿 ) = 0
w’’=0 → 2(𝑐 + 3𝑑𝐿 + 6𝑒𝐿 ) = 0
hence, 2𝐿𝑑 + 5𝐿 𝑒 = 0 → 𝑑=− , 𝑐 = 𝐿 𝑒 (from w’’)
𝑤(𝑥) = 𝑒 𝐿 𝑥 − 𝐿𝑥 + 𝑥
𝑤 (𝑥) = 𝑒 3𝐿 𝑥 − 𝐿𝑥 + 4𝑥
𝑤′′(𝑥) = 𝑒(3𝐿 − 15𝐿𝑥 + 12𝑥 )
From Eq 2.44
∫ ( )
∫ 𝑃 𝑑𝑥 = ∫ 𝐸𝐼 𝑑𝑥 → 𝑃=
∫ ( ( ))
after integration
𝐸𝐼
𝑃 = 21
𝐿

1c) neutral equilibrium method +𝜇 =0

+𝜇 =0 (shear force balance)

Assume 𝑤 = 𝐶𝑐𝑜𝑠(𝜇𝑥) + 𝐵𝑠𝑖𝑛(𝜇𝑥) + 𝑅𝑥 + 𝑆

= 𝑤 = 𝐵𝜇𝑐𝑜𝑠(𝜇𝑥) − 𝐶𝜇𝑠𝑖𝑛(𝜇𝑥) + 𝑅
𝑑 𝑤
= 𝑤 = −𝐵𝜇 𝑠𝑖𝑛(𝜇𝑥) − 𝐶𝜇 𝑐𝑜𝑠(𝜇𝑥)
𝑑𝑥
𝑑 𝑤
= 𝑤 = −𝐵𝜇 𝑐𝑜𝑠(𝜇𝑥) + 𝐶𝜇 𝑠𝑖𝑛(𝜇𝑥)
𝑑𝑥
At x=0 → w=0 → C+S=0 → C=-S … (1)
w’=0 → 𝐵𝜇 + 𝑅 = 0 → 𝑅 = −𝐵𝜇 … (2)
At x=L → w’’=0 → −𝐵𝜇 𝑠𝑖𝑛𝜇𝐿 − 𝐶𝜇 𝑐𝑜𝑠𝜇𝐿 = 0 … (3)
𝑤 +𝜇 𝑤 = 0 → −𝐵𝜇 𝑐𝑜𝑠𝜇𝐿 + 𝐶𝜇 𝑠𝑖𝑛𝜇𝐿 + 𝜇 (𝐵𝜇𝑐𝑜𝑠𝜇𝐿 − 𝐶𝜇𝑠𝑖𝑛𝜇𝐿 + 𝑅) = 0 … (4)
From (4) → R=0 → B=0, S=-C, −𝐶𝜇 𝑐𝑜𝑠𝜇𝐿 = 0
𝜇≠0 → 𝑐𝑜𝑠𝜇𝐿 = 0
𝜇𝐿 =
𝑃 = 𝜇 𝐸𝐼 =
2. For the continuous columns shown in Figure 2 (a) and (b) determine the buckling load P cr using the energy approach. Assume the
columns to be pin ended and the beams to be simply supported. Assess the accuracy of your solution in relation to known exact
solutions.

2a) ∫ 𝑃 𝑑𝑥 = ∫ 𝐸𝐼 𝑑𝑥 (2.44)
Assuming AB and BC pin ended
In AB let 𝑤(𝑥) = 𝑤 𝑠𝑖𝑛 (0 ≤ 𝑥 ≤ 𝐿)
In BC let 𝑤(𝑥) = 𝑤 𝑠𝑖𝑛 (0 ≤ 𝑥 ≤ 𝐿)
At B , 𝑤 = 𝑤 × 0 = 𝑤 × 0 = 0
[ ] [ ]
𝑤 = =𝑤 𝑐𝑜𝑠 =𝑤 𝑐𝑜𝑠 → 𝑤 = −𝑤

Using Eq (2.44) for AB and BC together

𝜋 𝜋𝑥 𝜋 𝜋𝑥 𝜋 𝜋𝑥 𝜋 𝜋𝑥
2𝑃𝑤 𝑐𝑜𝑠 𝑑𝑥 + 𝑃𝑤 𝑐𝑜𝑠 𝑑𝑥 = 𝐸𝐼𝑤 sin 𝑑𝑥 + 𝐸𝐼𝑤 sin 𝑑𝑥
𝐿 𝐿 𝐿 𝐿 𝐿 𝐿 𝐿 𝐿

Knowing 𝑤 =𝑤 and ∫ 𝑐𝑜𝑠 𝑑𝑥 = ∫ sin 𝑑𝑥

𝜋 𝐿 𝜋 𝐿
(2𝑃 + 𝑃)𝑤 = (𝐸𝐼 + 𝐸𝐼)𝑤
𝐿 2 𝐿 2
2 𝜋 𝐸𝐼
𝑃 =
3 𝐿
In AB, load = 2𝑃 = = (∝ )
where ∝= = 0.866

In BC, load = 𝑃 = = (∝ )
where ∝= = 1.225
Buckling iniated in AB, ∝ between pinned -fixed ()>&) and pinned- pinned.
In BC, ∝ in between pinned-pinned (1) and Fixed-free(2)
2b) compression in BC is P, in AB is 2P
Let 𝑤(𝑥) = 𝑤 𝑠𝑖𝑛 (0 ≤ 𝑥 ≤ 2𝐿) in AB
𝑤(𝑥) = 𝑤 𝑠𝑖𝑛 (0 ≤ 𝑥 ≤ 𝐿) in BC
At B, 𝑤 = 𝑤 (0) = 𝑤 (0)
𝑤 = 𝑤 𝑐𝑜𝑠 | = 𝑤 𝑐𝑜𝑠 |
→ 𝑤 = −2𝑤 for continuity

Using Eq. (2.44) for AB and BC together

𝜋 𝜋𝑥 𝜋 𝜋𝑥 𝜋 𝜋𝑥 𝜋 𝜋𝑥
2𝑃𝑤 𝑐𝑜𝑠 𝑑𝑥 + 𝑃𝑤 𝑐𝑜𝑠 𝑑𝑥 = 2𝐸𝐼 𝑤 𝑠𝑖𝑛 𝑑𝑥 + 𝐸𝐼𝑤 𝑠𝑖𝑛 𝑑𝑥
2𝐿 2𝐿 𝐿 𝐿 2𝐿 2𝐿 𝐿 𝐿
𝜋 𝜋 𝐿 𝜋 𝜋 𝐿
(2𝑃)(4𝑤 ) (𝐿) + (𝑃)(𝑤 ) = (2𝐸𝐼)(4𝑤 ) (𝐿) + (𝐸𝐼)(𝑤 )
4𝐿 𝐿 2 16𝐿 𝐿 2
𝑃 + = + → 𝑃 =
( )
In AB, load = 2𝑃 = = (∝ )
where ∝= = 0.791
( )
In BC, load = 𝑃 = = (∝ )
where ∝= = 1.581
Buckling is initiated in AB, ∝ between pin-fixed(0.7) and pin-pin(1)
In BC, ∝ is between pinned-pinned(1) and fixed-free (2)
3. Assuming L = 4 m, E = 210 000 N/mm2, σo = 350 N/mm2, A = 130 cm2 and I = 6000 cm4, determine the ultimate load of the
continuous columns shown in Figures 2 (a) and (b).
3a) From question 2(a), buckling is initiated in AB, at load = 10.363 × 10 𝑁
Squash load =σoA=4.55×106 N
𝐿 =𝐿 = 3.464𝑚, 𝑟= = 67.94𝑚𝑚
𝜂 = 0.003 = 0.1530
Let ultimate load in AB be Pu
then (𝑃 − 𝑃 )(𝑃 − 𝑃 ) = 𝜂𝑃 𝑃
𝑃 − (𝑃 + 𝑃 )𝑃 + 𝑃 𝑃 = 𝜂𝑃 𝑃
𝑃 − (𝑃 + (1 + 𝜂)𝑃 )𝑃 + 𝑃 𝑃 = 0
𝑃 − (16.499 × 10 )𝑃 + 47.152 × 10 = 0
𝑃 = × 10 16.499 ± √16.499 − 4 × 47.152 = 3.678 × 10 𝑁 (lowest root)
But load in AB is 2P; loading in beam has 𝑃 = 1.839 × 10 𝑁(dived by 2 cause 2P)

3b) buckling initiated in AB at load = 6.218 × 10 𝑁

Squash load =σoA=4.55×106 N
𝐿 =𝐿 = 6.325𝑚, 𝑟= = 96.08𝑚𝑚
𝜂 = 0.003 = 0.1975
Let ultimate load in AB be Pu
then (𝑃 − 𝑃 )(𝑃 − 𝑃 ) = 𝜂𝑃 𝑃
𝑃 − (𝑃 + 𝑃 )𝑃 + 𝑃 𝑃 = 𝜂𝑃 𝑃
𝑃 − (𝑃 + (1 + 𝜂)𝑃 )𝑃 + 𝑃 𝑃 = 0
𝑃 − (11.996 × 10 )𝑃 + 28.292 × 10 = 0
𝑃 = × 10 11.996 ± √11.996 − 4 × 28.292 = 3.226 × 10 𝑁 (lowest root)
But loading in AB is 2P, so loading on beam has 𝑃 = 1.613 × 10 𝑁 (dived by 2 cause 2P)
TORSIONAL AND LATERAL BUCKLING

1. A simply supported steel column of length L = 8 m, σo = 275 N/mm2, G = 80 000 N/mm2 , E = 210 000 N/mm2 has the cross section
dimensions shown in Figure 1. Determine the squash load Ps and buckling loads Pcr for flexural buckling about the y and z axes and
pure torsional buckling in the y-z plane.

𝐴 = 2𝑏 𝑡 + 𝑑 𝑡 = 2(150 × 10) + (150 × 10)4.5 × 10 𝑚𝑚

1 𝑑 1 1 150 1
𝐼 = 2𝑏 𝑡 𝑡 + + 𝑑 𝑡 = 2(150 × 10) 10 + + (150 × 10) = 19.7125 × 10 𝑚𝑚
12 2 12 12 4 12
1 1 1 1
𝐼 =2 𝑏 𝑡 + 𝑑 𝑡 =2 150 × 10 + × 150 × 10 = 5.6375 × 10 𝑚𝑚
12 12 12 12
𝐼 = 𝐼 + 𝐼 = (19.7125 + 5.6375) × 10 = 25.35 × 10 𝑚𝑚
1
𝐽 = ∑𝑏𝑡 = 3 × × 150 × 10 = 150 × 10 𝑚𝑚
3
𝑏 𝑑 150 150
𝐶 =𝑡 = 10 = 31.641 × 10 𝑚𝑚
12 2 12 2
Squash load 𝑃 = 𝜎 𝐴 = 275 × (4.5 × 10 )𝑁 = 1.2375 × 10 𝑁
Using Eq (3.27)
𝜋 𝐸𝐼 𝜋 (210 × 10 ) × (19.7125 × 10 )
𝑃 = = = 0.6384 × 10 𝑁
𝐿 (64 × 10 )
𝜋 𝐸𝐼 𝜋 (210 × 10 ) × (5.6275 × 10 )
𝑃 = = = 0.1826 × 10 𝑁
𝐿 (64 × 10 )
𝐴 𝜋
𝑃 = 𝐸𝐶 + 𝐺𝐽
𝐼 𝐿
4.5 × 10 𝜋
= (210 × 10 ) × (31.641 × 10 ) × + (80 × 10 )(150 × 10 ) = 2.312 × 10 𝑁
25.35 × 10 (64 × 10 )

2. The continuous beam shown in Figure 2 is simply supported at A and C, laterally and torsionally restrained at A, B and C, and
subjected to a uniform moment M about its major principal axis. Determine the elastic lateral buckling moment of the beam M cr
using the energy approach.

Using the method of section 3.3.3 in the notes

In AB, let 𝛿𝑣 = 𝐶 𝑠𝑖𝑛 , 𝛿𝑤 = 0, 𝛿𝜃 = 𝐶 𝑠𝑖𝑛 , (0 ≤ 𝑥 ≤ 𝐿)
In BC, let 𝛿𝑣 = 𝐶 𝑠𝑖𝑛 , 𝛿𝑤 = 0, 𝛿𝜃 = 𝐶 𝑠𝑖𝑛 , (0 ≤ 𝑥 ≤ 2𝐿)
At B 𝛿𝑣 = 𝐶 (0) = 𝐶 (0) and 𝛿𝜃 = 𝐶 (0) = 𝐶 (0)
𝛿𝑣′ = 𝐶 𝑐𝑜𝑠(𝜋) = 𝐶 𝑐𝑜𝑠(0) so 𝐶 = −2𝐶 for continuity
𝛿𝜃′ = 𝐶 𝑐𝑜𝑠(𝜋) = 𝐶 𝑐𝑜𝑠(0) so 𝐶 = −2𝐶 for continuity
Using Eq (3.28) for AB and BC together:
𝜋 𝜋𝑥 𝜋𝑥 𝜋 𝜋𝑥 𝜋𝑥 𝐺𝐽 𝜋 𝜋𝑥
𝑀(−𝐶 ) 𝑠𝑖𝑛 × (𝐶 )𝑠𝑖𝑛 𝑑𝑥 + 𝑀(2𝐶 ) 𝑠𝑖𝑛 × (−2𝐶 )𝑠𝑖𝑛 𝑑𝑥 + (𝐶 ) 𝑐𝑜𝑠 𝑑𝑥
𝐿 𝐿 𝐿 2𝐿 2𝐿 2𝐿 2 𝐿 𝐿
𝐺𝐽 𝜋 𝜋𝑥 𝐸𝐼 𝜋 𝜋𝑥 𝐸𝐼 𝜋 𝜋𝑥
+ (−2𝐶 ) 𝑐𝑜𝑠 𝑑𝑥 + (−𝐶 ) 𝑠𝑖𝑛 𝑑𝑥 + (2𝐶 ) 𝑠𝑖𝑛 𝑑𝑥
2 2𝐿 2𝐿 2 𝐿 𝐿 2 2𝐿 2𝐿
𝐸𝐶 𝜋 𝜋𝑥 𝐸𝐶 𝜋 𝜋𝑥
+ (−𝐶 ) 𝑠𝑖𝑛 𝑑𝑥 + (2𝐶 ) 𝑠𝑖𝑛 𝑑𝑥 = 0
2 𝐿 𝐿 2 2𝐿 2𝐿
− −
Now ∫ 𝑠𝑖𝑛 𝑑𝑥 = ∫ 1+𝑐𝑜𝑠 𝑑𝑥 = and ∫ 𝑠𝑖𝑛 𝑑𝑥 = ∫ 1+𝑐𝑜𝑠 𝑑𝑥 = 𝐿
𝑐𝑜𝑠 𝑐𝑜𝑠
→ −𝑀𝐶 𝐶 + 𝐺𝐽𝐶 + 𝐸𝐼 𝐶 + 𝐸𝐶 𝐶 =0

→ −4𝑀𝐶 𝐶 + 2𝐺𝐽𝐶 + 𝐸𝐼 𝐶 + 𝐸𝐶 𝐶 =0
𝐸𝐼 −2𝑀 𝐶
→ [𝐶 𝐶] =0
𝐶
−2𝑀 𝐸𝐶 + 2𝐺𝐽
For a non-trivial solution, the determinant of the matrix must be zero
 → 𝐸𝐼 𝐸𝐶 + 2𝐺𝐽 − 4𝑀 = 0

→ 𝑀 = 𝐸𝐼 𝐸𝐶 + 2𝐺𝐽

3. If the beam in Figure 2 has L = 6 m, σo = 275 N/mm2, G = 80 000 N/mm2, E = 210 000 N/mm2, bf = 150 mm, tf = 10 mm, dw = 440 mm
and tw = 8 mm determine the ultimate moment resistance Mu.
1 1 1 1
𝐼 =2 𝑏 𝑡 + 𝑑 𝑡 =2 150 × 10 + × 440 × 8 = 5.644 × 10 𝑚𝑚
12 12 12 12
1 1 1 1
𝐽 = ∑𝑏𝑡 = 2 𝑏 𝑡 + 𝑑 𝑡 = 2 × × 150 × 10 + × 440 × 8 = 175.1 × 10 𝑚𝑚
3 3 3 3
𝑏 𝑑 150 440
𝐶 =𝑡 = 10 = 272.25 × 10 𝑚𝑚
12 2 12 2
Using the result of question (2)

𝑀 = 𝐸𝐼 𝐸𝐶 + 2𝐺𝐽

= (210 × 10 )(5.644 × 10 ) (210 × 10 )(272.25 × 10 ) + 2(80 × 10 )(175.1 × 10 ) = 59.58 × 10 𝑁𝑚𝑚

× ×

𝑀 =𝜎 𝑡 𝑏 𝑑 +𝑡 = 275 10 × 150 × 440 + 8 × = 287.98 × 10 𝑁𝑚𝑚

Using Eq (3.35) with 𝜆 = 0.6
𝑀 − 𝑀 + (1 + 𝜆)𝑀 𝑀 + 𝑀 𝑀 = 0

i.e. 𝑀 − (287.98 × 10 + (1 + 0.6)59.58 × 10 )𝑀 + 287.98 × 10 × 59.58 × 10 = 0

→ 𝑀 = × 10 383.308 ± √383.308 − 4 × 17157.848 = 51.7510 𝑁𝑚𝑚 ( taking smaller root)

MOMENT RESISTING FRAMES

1. Analyse the structures shown in Figure 1 (a) to (c) using linear elastic matrix analysis and plot their bending moment diagrams. All members
have uniform E, I and A

Ignoring axial effect, each member has element stiffness matrix.

⎡ 12 6 −12 6 ⎤
𝑘 𝑘 ⎢ 6 4 −6 2 ⎥
⎢ ⎥
𝑘 𝑘 ⎢−12 −6 12 −6 ⎥
⎢ ⎥
⎣ 6 2 −6 4 ⎦

Global stiffness in

𝑘 ( ) 𝑘 ( ) 0
𝑘 ( ) 𝑘 ( ) +𝑘 ( ) 𝑘 ( )
0 𝑘 ( ) 𝑘 ( )

Deleting row and column corresponding to the centroid freedom w 1, w3

⎡ 4 −6 2 0 ⎤
𝜃 𝑀 0
⎢−6 24 0 6 ⎥
⎢ ⎥ 𝑤 𝑃 𝑉
= =
⎢ 2 0 8 2 ⎥ 𝜃 𝑀 0
⎢ ⎥ 𝜃 𝑀 0
⎣ 0 6 2 4 ⎦

From symmetry, θ2=0 and θ3=- θ1; so, equation reduce to:
𝐸𝐼 𝐸𝐼
4 −6 𝜃
𝑎 𝑎 0
=
𝐸𝐼 𝐸𝐼 𝑤 𝑉
−12 24
𝑎 𝑎

From first Eq, 𝜃 = 𝑤

Sub in second Eq, (24 − 18) 𝑤 =𝑉

→𝑤 = , 𝜃 = ,𝜃 = − , 𝜃 =0

End force in member (1)

 ⎡ 12 6 −12 6 ⎤ 0 ⎡− ⎤
𝑃
⎢ 6 4 −6 2 ⎥⎡ ⎤ ⎢ 0 ⎥
𝑀 ⎥⎢ ⎥ = ⎢
=⎢ ⎢ ⎥ ⎥
𝑃 ⎢−12 −6 12 −6 ⎥⎢ ⎥ ⎢ ⎥
𝑀 ⎢ ⎥ ⎣ ⎦ ⎢− ⎥
0 ⎣ ⎦
⎣ 6 2 −6 4 ⎦

Member (2) end force found by symmetry.

(Arrow show the direction)

Bending moment diagram

b)

(i)first find fixed end forces corresponding to the displacement vector (udl).

(ii) Now reverse these fixed end forces and use matrix analysis

The support combine W 1, W2, W3 and θ3; so only remaining freedom are θ1, θ2

4 2 𝜃 𝑀
Member (1) = ; member (2) 4 [𝜃 ] = [𝑀 ]
2 4 𝜃 𝑀

4 2 𝜃 𝑀 +
Combing: = =
2 8 𝜃 𝑀
−

From 1st Eq. 𝜃 = − 𝜃

Sub in 2nd Eq. 8 − 2 × 𝜃 =− −2 =

→𝜃 =− , so 𝜃 = + =

End force in member (1) add fixed end force from (i) member (2)

⎡6 6 ⎤ ⎡ 𝑝𝑎 ⎤
⎡− 𝑝𝑎⎤ ⎡6 ⎤ ⎡− 𝑝𝑎 ⎤
𝑃 𝑃
⎢ 4 2 ⎥ ⎢ 𝑝𝑎 ⎥ ⎢ 0 ⎥ ⎢ 4 ⎥ ⎢− 𝑝𝑎 ⎥
𝑀 𝑀
=⎢ ⎥ =⎢ ⎥→⎢ ⎥ =⎢ ⎥ − =⎢ ⎥
𝑃 ⎢−6 −6 ⎥ − ⎢− 𝑝𝑎 ⎥ ⎢− 𝑝𝑎⎥ 𝑃 ⎢−6 ⎥ ⎢ 𝑝𝑎 ⎥
𝑀 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
𝑝𝑎 ⎦
𝑀 ⎢ ⎥ ⎢ ⎥
⎣ 2 4 ⎦ ⎣− 𝑝𝑎 ⎦ ⎣ ⎣ 2 ⎦ ⎣− 𝑝𝑎 ⎦

end force, arrow show true direction

 Shear Force Diagram

Bending Moment Diagram

c)

(i) fixed end force due to displacement vector (udl)

(ii) reverse fixed end forces and use matrix analysis.

Assuming member are inexterile, only unectrained freedom is θ 2

Member (1) 4 [𝜃 ] = [𝑀 ] = − → 𝜃 =−

⎡6 ⎤ ⎡ 𝑝𝑎 ⎤ ⎡ − 𝑝𝑎 ⎤
𝑃
⎢2 ⎥ ⎢ 𝑝𝑎 ⎥ ⎢− 𝑝𝑎 ⎥
𝑀
End force in member (1); =⎢ ⎥ − =⎢ ⎥ → add end force from (i) ⎢ ⎥
𝑃 ⎢−6 ⎥ ⎢ − 𝑝𝑎 ⎥ ⎢ − 𝑝𝑎 ⎥
𝑀 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣4 ⎦ ⎣− 𝑝𝑎 ⎦ ⎣ 𝑝𝑎 ⎦

⎡6 ⎤ ⎡ − 𝑝𝑎 ⎤
𝑃
⎢ 4 ⎥ ⎢− 𝑝𝑎 ⎥
𝑀
Member (2) ; =⎢ ⎥ − =⎢ ⎥
𝑃 ⎢−6 ⎥ ⎢ 𝑝𝑎 ⎥
𝑀 ⎢ ⎥ ⎢ ⎥
⎣ 2 ⎦ ⎣− 𝑝𝑎 ⎦

End force, showing arrow in true direction (axial force not shown)

Shear Force Diagram (clockwise +) Bending Moment Diagram (sagging +)

2. Determine the elastic buckling loads of the structures shown in Figure 2 (a) to (d) using matrix analysis. All members have uniform E, I and A.

a) Degree of freedom are 𝑊 , 𝜃

12 −6 −
[𝑘] = , 𝑡𝑘 = −𝑃
−6 4 −

𝐸𝐼 12 𝐸𝐼 1
12 −𝑃 −6 +𝑃
𝑘 + 𝑡𝑘 = 𝑎 10𝑎 𝑎 10 = 0
𝐸𝐼 1 𝐸𝐼 4𝑎
−6 + 𝑃 4 −𝑃
𝑎 10 𝑎 30

→ 48 − + (𝑃) + (𝑃 ) − 36 + − (𝑃 ) = 0

→ 𝑃 − (𝑃) + (12) =0

→ 𝑃= 52 ± √52 − 4 × 15 × 12 = (52 ± 44.54)

Taking lower root 𝑃 = 2.486

(Compare with exact result 𝑃 = = 2.467 ; discrepancy due to use of just one element)

b) add a central node Degree of freedom are 𝑊 , 𝜃

96 −24 −
member (1) [𝑘] = , 𝑡𝑘 = −𝑃
−24 8 −

96 24
member (2) [𝑘] = , 𝑡𝑘 = −𝑃
24 8

192 − 𝑃 0
combining, 𝑘 + 𝑡𝑘 =
0 16 − 𝑃

set 𝑘 + 𝑡𝑘 =0 → 192 − 𝑃 = 0 or 16 − 𝑃=0

→ 𝑃 = 40 or 𝑃 = 120

Taking lower root 𝑃 = 40

(This given a symmetric mode with 𝑊 ≠ 0, 𝜃 = 0. The other root given an antisymmetric mode with 𝑊 = 0, 𝜃 ≠ 0)

[compare with the exact result 𝑃 = 4 = 39.47 ]

c) find end force for beam

total load=2P

load/unit length=

reverse fixed end forces and apply matrix analysis

assuming member are inexterivile; cannot sway; degrees of freedom are 𝜃 , 𝜃 (with 𝜃 = −𝜃 )

member (1) [𝑘] = 4 , 𝑡𝑘 = −𝑃 4 member (3) the same

4 2
member (2) [𝑘] = , 𝑡𝑘 = 0 leaver no axial load
2 4

6 −4
combining 𝑘 + 𝑡𝑘 =
6 −4

now setting 𝑘 + 𝑡𝑘 =0

36 − (𝑃𝑎) + (𝑃𝑎) − =0

→ 𝑃= ± −4× × 35 = ±

Taking the lower root 𝑃 = 37.5

d) assume member are inextille

free to sway; degree of freedom are 𝑊 , 𝜃 , 𝑊 , 𝜃 ; with 𝜃 = 𝜃

column have compression P

beam have compression P

𝑝 12 −6 𝑊 −
member (1) 𝑀 = , 𝑡𝑘 = −𝑃
−6 4 𝜃 −

𝑀 4 2 𝜃 −
member (2) = , 𝑡𝑘 = −𝑃
𝑀 2 4 𝜃 −

combining and using 𝜃 = 𝜃

𝐸𝐼 12 𝐸𝐼 1
12 − 𝑃 −6 + 𝑃
𝑘 + 𝑡𝑘 = 𝑎 10𝑎 𝑎 10
𝐸𝐼 1 𝐸𝐼 𝑃𝑎
−6 + 𝑃 7 − 16
𝑎 10 𝑎 30
now setting 𝑘 + 𝑡𝑘 =0

𝐸𝐼 84 192 𝐸𝐼 192 𝐸𝐼 12 𝐸𝐼 1
84 − + (𝑃) + (𝑃 ) − 36 + (𝑃) − (𝑃 ) = 0
𝑎 10 30 𝑎 300 𝑎 10 𝑎 100

→ 𝑃 − (𝑃) + 48 =0

→ 𝑃 = 10 × 136 ± √136 − 4 × 63 × 48 = 10 × [136 ± 80]

Taking smaller root 𝑃 = 10 × =

3. For the structures shown in Figure 2 (c) and (d) determine the rigid plastic collapse load and Merchant Rankine collapse load assuming all
members have uniform plastic moment resistance Mp.

Beam collapse

Plate moment resistance=4𝑀 𝛿𝜃

Average displacement of beam = 𝛿𝑣 = 𝑎𝛿𝜃

→ Work done by applied load=2𝑃 𝑎𝛿𝜃 = 𝑃 𝛿𝜃

Equating gives 𝑃 = 4

Merchant Rankine = + where 𝑃 = 37.5

→ 𝑃 =
.

Problem (d): beam collapse gives 𝑃 = 4 (as for problem (c))

Sway collapse

Plate moment resistant=4𝑀 𝛿𝜃; working by applied load= 2𝑃𝛿𝑤 = 2𝑃 𝛿𝜃

Equating gives 𝑃 = 2

Combine collapse.

Plate resistance=6𝑀 𝛿𝜃

Work done by applied load= 2𝑃𝛿𝑤 + 2𝑃 𝛿𝑤 = 3𝑃 𝛿𝜃

Equating gives 𝑃 = 2

Lowest plate collapse load 𝑃 = 2 ;𝑃 =

Merchant Rankine = +

→ 𝑃 =