Structural compression members are critical elements in civil engineering structures, responsible for carrying axial compressive loads safely and efficiently. Their design and analysis must consider not only the direct compressive stresses but also the potential for buckling, which can lead to sudden failure. This chapter focuses on understanding the types of compression members commonly encountered in Kenyan construction projects and applying Euler's buckling theory to predict their load-carrying capacity under various conditions. Mastery of these concepts ensures safer, more economical structural designs in buildings, bridges, and other infrastructure.
Compression members in structures resist axial compressive forces and are prone to instability through buckling. In Kenya, typical structures such as multi-storey buildings, warehouses, and water tanks use columns and struts that must be properly classified and analyzed for buckling risk. This section covers the classification of columns based on slenderness and end conditions, followed by the application of Euler's buckling theory to determine critical buckling loads.
Columns vary widely depending on their geometry, end support conditions, and slenderness ratios, all of which influence their buckling behavior. Correct classification enables engineers to select appropriate design formulas and safety factors for compression members in civil engineering projects such as county government offices and hospitals.
The slenderness ratio is a key parameter that determines whether a column will fail by crushing or buckling. It is defined as the ratio of the effective length of the column to its radius of gyration.
The way a column is supported at its ends affects its effective length and buckling load. Common end conditions in Kenyan structures include:
Columns are classified depending on their slenderness and failure mode:
The material properties and cross-sectional shape influence column behavior:
A steel column in the headquarters of Safaricom PLC, Nairobi, has an effective length of 3.5 m and a radius of gyration of 25 mm.
Given:
- Effective length \(L_{eff} = 3.5\) m \(= 3500\) mm
- Radius of gyration \(r = 25\) mm
Formula:$$ \lambda = \frac{L_{eff}}{r} $$
Substitute:$$ \lambda = \frac{3500}{25} = 140 $$
Answer: 140 (Long column, prone to buckling)
A column in the Kenya Ports Authority warehouse is 4 m long with fixed-free end conditions.
Given:
- Actual length \(L = 4\) m
- End condition: fixed-free, coefficient = 2.0
Formula:$$ L_{eff} = L \times \text{coefficient} $$
Substitute:$$ L_{eff} = 4 \times 2.0 = 8\, \text{m} $$
Answer: 8 m
A concrete column at Moi Teaching and Referral Hospital has an effective length of 2.8 m and radius of gyration 40 mm.
Given:
- \(L_{eff} = 2.8\) m \(= 2800\) mm
- \(r = 40\) mm
Formula:$$ \lambda = \frac{2800}{40} = 70 $$
Classification:
- Short column: \(\lambda \leq 40\)
- Intermediate column: \(40 < \lambda \leq 100\)
- Long column: \(\lambda > 100\)
Answer: 70 (Intermediate column, may require empirical design)
Calculate the slenderness ratio of a steel column with effective length 3.6 m and radius of gyration 15 mm. (5 marks)
Determine the effective length of a column 4 m long with fixed-pinned end conditions. (5 marks)
Classify a column with slenderness ratio 60 and explain its likely failure mode. (5 marks)
Euler's buckling theory provides the fundamental basis for calculating the critical load at which a slender column will buckle elastically. It is essential for designing compression members in civil engineering to prevent sudden failure, especially in tall structures such as multi-storey office blocks and water towers.
Euler derived a formula to predict the maximum axial load a perfect, slender, pin-ended column can carry before buckling:
$$ P_{cr} = \frac{\pi^2 E I}{(L_{eff})^2} $$
where:
- \(P_{cr}\) = critical buckling load (N)
- \(E\) = modulus of elasticity of the material (Pa)
- \(I\) = least moment of inertia of the cross-section (m\(^4\))
- \(L_{eff}\) = effective length of the column (m)
Given: Steel column, \(E = 200 \times 10^9\) Pa, \(I = 8 \times 10^{-6}\) m\(^4\), \(L_{eff} = 3\) m.
$$P_{cr} = \frac{\pi^2 \times 200 \times 10^9 \times 8 \times 10^{-6}}{3^2}$$
$$P_{cr} = \frac{9.8696 \times 200 \times 10^9 \times 8 \times 10^{-6}}{9}$$
$$P_{cr} = \frac{1.5791 \times 10^6}{9} = 175,455.67 \text{ N}$$
Answer: 175.46 kN
Given: Concrete-filled steel column, \(E = 25 \times 10^9\) Pa, \(I = 5 \times 10^{-5}\) m\(^4\), \(L_{eff} = 4\) m.
$$P_{cr} = \frac{\pi^2 \times 25 \times 10^9 \times 5 \times 10^{-5}}{4^2}$$
$$P_{cr} = \frac{9.8696 \times 25 \times 10^9 \times 5 \times 10^{-5}}{16}$$
$$P_{cr} = \frac{12,337,000}{16} = 771,062.5 \text{ N}$$
Answer: 771.06 kN
Given: Timber column, \(E = 12 \times 10^9\) Pa, \(I = 2 \times 10^{-6}\) m\(^4\), \(L_{eff} = 2.5\) m.
$$P_{cr} = \frac{\pi^2 \times 12 \times 10^9 \times 2 \times 10^{-6}}{2.5^2}$$
$$P_{cr} = \frac{9.8696 \times 12 \times 10^9 \times 2 \times 10^{-6}}{6.25}$$
$$P_{cr} = \frac{236,870}{6.25} = 37,899.2 \text{ N}$$
Answer: 37.9 kN
Euler's formula assumes pinned-pinned ends; other end conditions require adjustment of effective length:
Given: Steel column, \(E = 200 \times 10^9\) Pa, \(I = 10 \times 10^{-6}\) m\(^4\), length \(L = 3.5\) m, fixed-fixed ends.
Calculate \(P_{cr}\).
$$L_{eff} = \frac{3.5}{2} = 1.75 \text{ m}$$
$$P_{cr} = \frac{\pi^2 \times 200 \times 10^9 \times 10 \times 10^{-6}}{1.75^2}$$
$$P_{cr} = \frac{9.8696 \times 200 \times 10^9 \times 10 \times 10^{-6}}{3.0625}$$
$$P_{cr} = \frac{19,739,200}{3.0625} = 6,446,835.52 \text{ N}$$
Answer: 6.45 MN
Given: Aluminum column, \(E = 70 \times 10^9\) Pa, \(I = 4 \times 10^{-6}\) m\(^4\), length \(L = 5\) m, fixed-free ends.
Calculate \(P_{cr}\).
$$L_{eff} = 2 \times 5 = 10 \text{ m}$$
$$P_{cr} = \frac{\pi^2 \times 70 \times 10^9 \times 4 \times 10^{-6}}{10^2}$$
$$P_{cr} = \frac{9.8696 \times 70 \times 10^9 \times 4 \times 10^{-6}}{100}$$
$$P_{cr} = \frac{2,765,488}{100} = 27,654.88 \text{ N}$$
Answer: 27.65 kN
Euler's theory applies to ideal slender columns under axial load without imperfections:
Given: A steel column with \(E = 210 \times 10^9\) Pa, \(I = 9 \times 10^{-6}\) m\(^4\), \(L_{eff} = 3\) m. Calculate Euler load and then adjust by a factor of 0.7 to account for imperfections.
$$P_{cr} = \frac{\pi^2 \times 210 \times 10^9 \times 9 \times 10^{-6}}{3^2}$$
$$P_{cr} = \frac{9.8696 \times 210 \times 10^9 \times 9 \times 10^{-6}}{9}$$
$$P_{cr} = \frac{18,650,000}{9} = 2,072,222.22 \text{ N}$$
Adjusted load:
$$P_{adj} = 0.7 \times 2,072,222.22 = 1,450,555.56 \text{ N}$$
Answer: 1.45 MN
Given: Column with \(E = 200 \times 10^9\) Pa, \(I = 5 \times 10^{-6}\) m\(^4\), \(L_{eff} = 2.5\) m. Calculate Euler load and apply a 15% reduction for initial imperfections and residual stresses.
$$P_{cr} = \frac{\pi^2 \times 200 \times 10^9 \times 5 \times 10^{-6}}{2.5^2}$$
$$P_{cr} = \frac{9.8696 \times 200 \times 10^9 \times 5 \times 10^{-6}}{6.25}$$
$$P_{cr} = \frac{9,869,600}{6.25} = 1,579,136 \text{ N}$$
Reduced load:
$$P_{red} = 0.85 \times 1,579,136 = 1,342,265.6 \text{ N}$$
Answer: 1.34 MN
Calculate the critical buckling load for a steel column 3 m long, pinned-pinned ends, \(E = 210 \times 10^9\) Pa, \(I = 7 \times 10^{-6}\) m\(^4\). (6 marks)
A concrete column has \(E = 30 \times 10^9\) Pa, \(I = 4 \times 10^{-5}\) m\(^4\), length 4 m, fixed-fixed ends. Find the Euler buckling load. (6 marks)
Determine the critical load for a timber column with fixed-free ends, \(L = 2\) m, \(E = 12 \times 10^9\) Pa, \(I = 1 \times 10^{-6}\) m\(^4\). (6 marks)
Calculate the adjusted buckling load for a steel column with Euler load 1 MN, considering a 20% reduction for imperfections. (4 marks)
This chapter focused on the analysis of structural compression members, beginning with the classification of columns based on their slenderness and behavior under load. Columns were categorized into short, intermediate, and long types, each exhibiting distinct failure modes and design considerations. Short columns primarily fail by material crushing, while intermediate columns experience a combination of crushing and buckling effects. Long columns are prone to buckling, which necessitates careful stability analysis. The chapter then introduced Euler's buckling theory, explaining its fundamental principles and how it predicts the critical load at which a long slender column will buckle. Application of Euler’s formula was demonstrated for practical structural design, highlighting the importance of effective length and boundary conditions. Understanding these concepts is essential for ensuring the safety and reliability of compression members in various civil engineering structures.
A steel column with pinned ends has a length of 3 m and a radius of gyration of 20 mm. Calculate the slenderness ratio of the column. (2 marks)
Determine the critical buckling load for a 4 m long concrete column with pinned ends, modulus of elasticity \(E = 25 \times 10^9\) Pa, and moment of inertia \(I = 8 \times 10^{-6} \, m^4\). (3 marks)
A short concrete column has a cross-sectional area of \(0.04 \, m^2\) and is subjected to an axial load of 400 kN. Calculate the compressive stress on the column. (2 marks)
Calculate the effective length of a column with one end fixed and the other end free. The actual length of the column is 2.5 m. (2 marks)
A steel column 5 m long with fixed-pinned end conditions has a radius of gyration of 15 mm. Calculate the slenderness ratio and determine whether it is short, intermediate, or long. (4 marks)
Using Euler’s formula, calculate the critical buckling load for a steel column with pinned ends, length 6 m, \(E = 200 \times 10^9\) Pa, and moment of inertia \(I = 1.2 \times 10^{-5} \, m^4\). (3 marks)
A reinforced concrete column has an effective length of 3.5 m and a radius of gyration of 30 mm. Calculate the slenderness ratio and classify the column type. (3 marks)
A steel column with length 4 m and fixed-fixed end conditions carries an axial load of 500 kN. The modulus of elasticity is \(210 \times 10^9\) Pa and moment of inertia is \(1.5 \times 10^{-5} \, m^4\). Calculate the critical buckling load using Euler’s formula. (4 marks)
A long column with length 7 m and radius of gyration 18 mm is subjected to an axial load of 350 kN. Calculate the slenderness ratio and discuss the risk of buckling. (3 marks)
A reinforced concrete column has a cross-sectional area of \(0.06 \, m^2\), length 3 m, modulus of elasticity \(30 \times 10^9\) Pa, and moment of inertia \(1.0 \times 10^{-5} \, m^4\). Calculate the Euler critical load and compare it with an axial load of 450 kN to determine if buckling will occur. (5 marks)
Formula for slenderness ratio:$$\lambda = \frac{L}{r}$$
Calculation:$$\lambda = \frac{3000}{20} = 150$$
Answer: The slenderness ratio is \(\boxed{150}\).
Euler’s critical buckling load:$$P_{cr} = \frac{\pi^2 E I}{L^2}$$
Calculate \(L^2\):$$L^2 = (4)^2 = 16\, m^2$$
Calculate \(P_{cr}\):$$P_{cr} = \frac{\pi^2 \times 25 \times 10^9 \times 8 \times 10^{-6}}{16}$$$$= \frac{9.8696 \times 25 \times 10^9 \times 8 \times 10^{-6}}{16}$$$$= \frac{9.8696 \times 25 \times 8 \times 10^{3}}{16}$$$$= \frac{9.8696 \times 200 \times 10^{3}}{16}$$$$= \frac{1,973,920}{16} = 123,370\, N = 123.37\, kN$$
Answer: The critical buckling load is \(\boxed{123.37\, kN}\).
Compressive stress:$$\sigma = \frac{P}{A}$$
Calculation:$$\sigma = \frac{400,000}{0.04} = 10,000,000\, Pa = 10\, MPa$$
Answer: The compressive stress is \(\boxed{10\, MPa}\).
Effective length factor \(K\) for fixed-free column is 2.0.
Effective length:$$L_{eff} = K \times L = 2.0 \times 2.5 = 5.0\, m$$
Answer: The effective length is \(\boxed{5.0\, m}\).
Calculate slenderness ratio:$$\lambda = \frac{K \times L}{r} = \frac{0.7 \times 5000}{15} = \frac{3500}{15} = 233.33$$
Classification:
- Short column \(\lambda \leq 40\)
- Intermediate column \(40 < \lambda \leq 100\)
- Long column \(\lambda > 100\)
Since 233.33 > 100, it is a long column.
Answer: Slenderness ratio is \(\boxed{233.33}\); column is classified as a long column.
Calculate effective length:$$L_{eff} = K \times L = 1.0 \times 6 = 6\, m$$
Calculate critical buckling load:$$P_{cr} = \frac{\pi^2 E I}{L_{eff}^2} = \frac{9.8696 \times 200 \times 10^9 \times 1.2 \times 10^{-5}}{6^2}$$$$= \frac{9.8696 \times 200 \times 10^9 \times 1.2 \times 10^{-5}}{36}$$
Calculate numerator:$$9.8696 \times 200 \times 1.2 = 9.8696 \times 240 = 2368.7$$$$2368.7 \times 10^{4} = 2.3687 \times 10^{7}$$ (since \(10^9 \times 10^{-5} = 10^4\))
Divide by denominator:$$P_{cr} = \frac{2.3687 \times 10^{7}}{36} = 657,975\, N = 657.98\, kN$$
Answer: The critical buckling load is \(\boxed{657.98\, kN}\).
Calculate slenderness ratio:$$\lambda = \frac{L_{eff}}{r} = \frac{3500}{30} = 116.67$$
Classification:
- Short column \(\lambda \leq 40\)
- Intermediate column \(40 < \lambda \leq 100\)
- Long column \(\lambda > 100\)
Since 116.67 > 100, it is a long column.
Answer: Slenderness ratio is \(\boxed{116.67}\); column is classified as a long column.
Calculate effective length:$$L_{eff} = K \times L = 0.5 \times 4 = 2\, m$$
Calculate critical buckling load:$$P_{cr} = \frac{\pi^2 E I}{L_{eff}^2} = \frac{9.8696 \times 210 \times 10^9 \times 1.5 \times 10^{-5}}{2^2}$$$$= \frac{9.8696 \times 210 \times 10^9 \times 1.5 \times 10^{-5}}{4}$$
Calculate numerator:$$9.8696 \times 210 \times 1.5 = 9.8696 \times 315 = 3108.9$$$$3108.9 \times 10^{4} = 3.1089 \times 10^{7}$$
Divide by denominator:$$P_{cr} = \frac{3.1089 \times 10^{7}}{4} = 7,772,250\, N = 7,772.25\, kN$$
Answer: The critical buckling load is \(\boxed{7,772.25\, kN}\).
Calculate slenderness ratio:$$\lambda = \frac{L}{r} = \frac{7000}{18} = 388.89$$
Since \(\lambda > 100\), it is a long column and at high risk of buckling under axial load.
Answer: Slenderness ratio is \(\boxed{388.89}\); the column is long and susceptible to buckling.
Calculate effective length:$$L_{eff} = K \times L = 1.0 \times 3 = 3\, m$$
Calculate Euler critical load:$$P_{cr} = \frac{\pi^2 E I}{L_{eff}^2} = \frac{9.8696 \times 30 \times 10^9 \times 1.0 \times 10^{-5}}{3^2}$$$$= \frac{9.8696 \times 30 \times 10^9 \times 1.0 \times 10^{-5}}{9}$$
Calculate numerator:$$9.8696 \times 30 = 296.09$$$$296.09 \times 10^{4} = 2.9609 \times 10^{6}$$
Divide by denominator:$$P_{cr} = \frac{2.9609 \times 10^{6}}{9} = 328,987\, N = 328.99\, kN$$
Compare \(P_{cr}\) with axial load \(P\):$$P = 450\, kN > P_{cr} = 328.99\, kN$$
Since \(P > P_{cr}\), buckling will occur.
Answer: Euler critical load is \(\boxed{328.99\, kN}\). Since applied load exceeds this, the column will buckle.
Given \(L_{eff} = 3\,m = 3000\,mm\), \(r = 15\,mm\)
\[
\lambda = \frac{L_{eff}}{r} = \frac{3000}{15} = 200
\]
Since \(\lambda = 200\), compare with limits (usually short < 50, intermediate 50-100, long >100), it is a long column.
Typical slenderness ratio limits:
Long columns: \(\lambda > 100\)
Given: \(L = 4\,m = 4\), \(E = 200 \times 10^9\,Pa\), \(I = 8 \times 10^{-6}\,m^4\), pinned-pinned so \(L_{eff}=L=4\,m\).
Euler’s critical load:
\[
P_{cr} = \frac{\pi^2 EI}{(L_{eff})^2} = \frac{\pi^2 \times 200 \times 10^9 \times 8 \times 10^{-6}}{4^2}
\]
Calculate denominator:
\[
4^2 = 16
\]
Calculate numerator:
\[
\pi^2 \times 200 \times 10^9 \times 8 \times 10^{-6} = 9.8696 \times 200 \times 10^9 \times 8 \times 10^{-6} = 9.8696 \times 200 \times 8 \times 10^{3} = 9.8696 \times 1600 \times 10^{3} = 15,791,360
\]
So,
\[
P_{cr} = \frac{15,791,360}{16} = 987,960\,N = \mathbf{987.96\,kN}
\]
Given \(r=40\,mm\), \(L_{eff} = 2.5\,m = 2500\,mm\)
\[
\lambda = \frac{2500}{40} = 62.5
\]
Since \(50 < 62.5 \leq 100\), it is an intermediate column.
Euler’s formula assumes buckling occurs before material yields. Short columns fail by crushing, not buckling, so Euler’s formula overestimates capacity and is not applicable.
For fixed-free column, effective length is:
\[
L_{eff} = 2 \times L = 2 \times 6 = 12\,m
\]
As slenderness ratio increases, the critical buckling load decreases, making the column more prone to buckling failure at lower loads.
Given both ends fixed:
\[
L = 3\,m, E = 210 \times 10^9\,Pa, I = 5 \times 10^{-6}\,m^4
\]
Effective length for fixed-fixed:
\[
L_{eff} = \frac{L}{2} = \frac{3}{2} = 1.5\,m
\]
Calculate critical load:
\[
P_{cr} = \frac{\pi^2 EI}{(L_{eff})^2} = \frac{9.8696 \times 210 \times 10^9 \times 5 \times 10^{-6}}{(1.5)^2}
\]
Calculate denominator:
\[
1.5^2 = 2.25
\]
Calculate numerator:
\[
9.8696 \times 210 \times 10^9 \times 5 \times 10^{-6} = 9.8696 \times 210 \times 5 \times 10^{3} = 9.8696 \times 1050 \times 10^{3} = 10,363,080
\]
Then,
\[
P_{cr} = \frac{10,363,080}{2.25} = 4,605,813\,N = \mathbf{4605.81\,kN}
\]
Given: \(A = 1500\,mm^2 = 1.5 \times 10^{-3}\,m^2\), \(L=2.5\,m\), \(r=20\,mm = 0.02\,m\), \(P=150\,kN = 150,000\,N\), \(E=210 \times 10^9\,Pa\)
Calculate slenderness ratio:
\[
\lambda = \frac{2.5}{0.02} = 125
\]
Calculate critical load:
\[
P_{cr} = \frac{\pi^2 EI}{L^2}
\]
Moment of inertia:
\[
I = A r^2 = 1.5 \times 10^{-3} \times (0.02)^2 = 1.5 \times 10^{-3} \times 4 \times 10^{-4} = 6 \times 10^{-7} m^4
\]
Calculate \(P_{cr}\):
\[
P_{cr} = \frac{9.8696 \times 210 \times 10^9 \times 6 \times 10^{-7}}{(2.5)^2} = \frac{9.8696 \times 210 \times 6 \times 10^{2}}{6.25} = \frac{9.8696 \times 1260 \times 10^{2}}{6.25} = \frac{12,433,296}{6.25} = 1,989,327\,N = 1989.3\,kN
\]
Since \(P = 150\,kN < P_{cr} = 1989.3\,kN\), column is safe under given load.
Given \(L=3\,m\), pinned-pinned so \(L_{eff}=3\,m\), \(I=12 \times 10^{-6}\,m^4\), \(E=25 \times 10^9\,Pa\).
Calculate critical load:
\[
P_{cr} = \frac{\pi^2 EI}{L^2} = \frac{9.8696 \times 25 \times 10^9 \times 12 \times 10^{-6}}{3^2} = \frac{9.8696 \times 25 \times 12 \times 10^{3}}{9} = \frac{9.8696 \times 300 \times 10^{3}}{9} = \frac{2,960,880}{9} = 328,987\,N = \mathbf{328.99\,kN}
\]
Question 11 (Compulsory - 20 marks)
A new wing of Kenyatta National Hospital requires steel columns to support a roof structure. One such column has length 4 m, is pinned at both ends, with cross-sectional moment of inertia \(I = 1.2 \times 10^{-5}\,m^4\), modulus of elasticity \(E = 200 \times 10^9\,Pa\) and radius of gyration \(r = 30\,mm\). The column carries an axial load of 500 kN.
a) Calculate the slenderness ratio and classify the column. (10 marks)
b) Determine the critical buckling load and state if the column is safe under the given load. (10 marks)
Question 12 (20 marks)
A steel column used in a retail mall in Mombasa has length 5 m, fixed at one end and pinned at the other. The moment of inertia is \(2 \times 10^{-5}\,m^4\), modulus of elasticity \(E = 210 \times 10^9\,Pa\), and radius of gyration \(r = 25\,mm\).
a) Calculate the effective length of the column for buckling analysis. (8 marks)
b) Find the critical buckling load. (12 marks)
Question 13 (20 marks)
A reinforced concrete column in a county government office in Nakuru is 3 m long. It has a cross-sectional area of \(0.09\,m^2\), radius of gyration 45 mm, and modulus of elasticity \(25 \times 10^9\,Pa\).
a) Determine the slenderness ratio and classify the column. (8 marks)
b) Calculate the critical buckling load for the column assuming pinned ends. (12 marks)
Question 14 (20 marks)
A steel column in a hotel in Eldoret is 3.5 m long with both ends fixed. The modulus of elasticity is \(210 \times 10^9\,Pa\), and the moment of inertia is \(7.5 \times 10^{-6} m^4\). Calculate:
a) The effective length of the column. (5 marks)
b) The critical buckling load. (15 marks)
Question 11
a) Given \(L=4\,m=4000\,mm\), \(r=30\,mm\)
\[
\lambda = \frac{4000}{30} = 133.33
\]
Since \(\lambda > 100\), column is long.
b) Critical load:
\[
P_{cr} = \frac{\pi^2 EI}{L^2} = \frac{9.8696 \times 200 \times 10^9 \times 1.2 \times 10^{-5}}{4^2} = \frac{9.8696 \times 200 \times 1.2 \times 10^{4}}{16} = \frac{9.8696 \times 2400 \times 10^{4}}{16} = \frac{23,687,040}{16} = 1,480,440\,N = 1480.44\,kN
\]
Since load = 500 kN < 1480.44 kN, column is safe.
Question 12
a) For fixed-pinned column, effective length is:
\[
L_{eff} = 0.7 \times L = 0.7 \times 5 = 3.5\,m
\]
b) Calculate critical load:
\[
P_{cr} = \frac{\pi^2 EI}{L_{eff}^2} = \frac{9.8696 \times 210 \times 10^9 \times 2 \times 10^{-5}}{(3.5)^2} = \frac{9.8696 \times 210 \times 2 \times 10^{4}}{12.25} = \frac{9.8696 \times 4200 \times 10^{4}}{12.25} = \frac{41,452,320}{12.25} = 3,384,890\,N = 3384.89\,kN
\]
Question 13
a) Given \(L=3\,m=3000\,mm\), \(r=45\,mm\)
\[
\lambda = \frac{3000}{45} = 66.67
\]
Classification: intermediate column.
b) Moment of inertia:
\[
I = A r^2 = 0.09 \times (0.045)^2 = 0.09 \times 0.002025 = 0.00018225\,m^4
\]
Critical load:
\[
P_{cr} = \frac{\pi^2 EI}{L^2} = \frac{9.8696 \times 25 \times 10^9 \times 0.00018225}{3^2} = \frac{9.8696 \times 25 \times 0.00018225 \times 10^{9}}{9} = \frac{44,970,750}{9} = 4,996,750\,N = 4996.75\,kN
\]
Question 14
a) Effective length for fixed-fixed ends:
\[
L_{eff} = \frac{L}{2} = \frac{3.5}{2} = 1.75\,m
\]
b) Critical load:
\[
P_{cr} = \frac{\pi^2 EI}{L_{eff}^2} = \frac{9.8696 \times 210 \times 10^9 \times 7.5 \times 10^{-6}}{(1.75)^2} = \frac{9.8696 \times 210 \times 7.5 \times 10^{3}}{3.0625} = \frac{15,527,880}{3.0625} = 5,069,100\,N = \mathbf{5069.1\,kN}
\]