Compiled by: Samuel Wachira
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. For mild steel, the boundary between intermediate and long columns is the limiting slenderness ratio at which the Euler buckling stress equals the yield stress:
λ_lim = π × √(E ÷ f_y)
λ_lim = 3.1416 × √(200,000 ÷ 250)
λ_lim = 3.1416 × 28.28
λ_lim ≈ 89
| Column type | Typical slenderness (mild steel) | Mode of failure | Method of analysis |
|---|---|---|---|
| Short | Below about 30 | Crushing of the material | Crushing load = stress × area |
| Intermediate | About 30 to 89 | Combined crushing and buckling (inelastic buckling) | Empirical formulas such as Rankine-Gordon |
| Long | Above about 89 | Elastic buckling | Euler's formula |
The exact limits vary with the material and design code, so the values above are guides for mild steel only.
The material properties and cross-sectional shape influence column behavior:
A short column is a compression member whose length is small compared with its cross-sectional dimensions, so that it fails by crushing of the material before any buckling occurs. Its strength depends only on the cross-sectional area and the crushing (compressive) strength of the material, and it does not need a buckling check. Typical short columns include stocky concrete pedestals, column stubs, piers and short timber posts. Example: the 900 mm high concrete pedestals supporting a steel water tank at a primary school in Machakos were designed as short columns, since their large cross-section made buckling impossible.
P_crush = σ_c × A
A concrete pedestal 300 mm × 300 mm and 0.9 m high has a crushing strength of 20 N/mm². Check that it is short and find its crushing load.
r = h ÷ √12
r = 0.3 ÷ 3.464
r = 0.0866 m
λ = L_eff ÷ r
λ = 0.9 ÷ 0.0866
λ = 10.4, so the pedestal is a short column.
P_crush = σ_c × A
P_crush = 20 × (300 × 300)
P_crush = 20 × 90,000
P_crush = 1,800,000 N = 1.8 MN
An intermediate column has a slenderness between that of short and long columns, so it fails by a combination of crushing and buckling, often called inelastic buckling, in which parts of the section yield before the column bows sideways. Neither the simple crushing formula nor Euler's formula gives safe results, so empirical formulas such as the Rankine-Gordon formula or the design curves in codes are used. Most practical building columns, including steel stanchions and reinforced concrete columns of normal storey height, fall into this range. Example: the steel columns of a two-storey warehouse for a farmers' cooperative in Trans Nzoia had slenderness ratios of about 60 and were designed using the code buckling curves for intermediate columns.
Rankine-Gordon formula: P = σ_c × A ÷ (1 + a × λ²)
where σ_c is the crushing stress and a is the Rankine constant (for mild steel, σ_c ≈ 320 N/mm² and a ≈ 1 ÷ 7,500).
A mild steel column has a cross-sectional area of 2,000 mm² and a slenderness ratio of 60. Take σ_c = 320 N/mm² and a = 1 ÷ 7,500.
P = σ_c × A ÷ (1 + a × λ²)
P = 320 × 2,000 ÷ (1 + 60² ÷ 7,500)
P = 640,000 ÷ (1 + 3,600 ÷ 7,500)
P = 640,000 ÷ 1.48
P = 432,432 N ≈ 432 kN
For comparison, Euler's formula would give σ_E = π² E ÷ λ² = 9.8696 × 200,000 ÷ 3,600 = 548 N/mm², so P_E = 548 × 2,000 = 1,096 kN, which greatly overestimates the true strength. This shows why Euler's formula must not be used for intermediate columns.
A long column is a compression member whose length is large compared with its cross-section, so it fails by elastic buckling, bending sideways suddenly at a load far below the crushing load while the material is still elastic. Its strength depends on the stiffness of the material (E), the least moment of inertia of the section and the effective length, but not on the material's strength, and it is calculated using Euler's formula. Long columns include slender steel struts, tall timber posts, bracing members and slender tower legs. Example: the slender bracing struts of a telecommunication mast at a hotel in Nanyuki were checked as long columns, because their buckling load, not crushing, controlled their design.
A mild steel strut has an area of 2,000 mm² and a slenderness ratio of 150. Take E = 200,000 N/mm² and f_y = 250 N/mm².
σ_E = π² E ÷ λ²
σ_E = 9.8696 × 200,000 ÷ 150²
σ_E = 1,973,920 ÷ 22,500
σ_E = 87.7 N/mm²
P_E = σ_E × A
P_E = 87.7 × 2,000
P_E = 175,400 N ≈ 175 kN
The crushing load would be 250 × 2,000 = 500,000 N = 500 kN, so the strut fails by buckling at about one third of its crushing load.
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 = π² E I ÷ (L_eff)²
where:
Given: Steel column, E = 200 × 10⁹ Pa, I = 8 × 10⁻⁶ m⁴, L_eff = 3 m.
P_cr = π² E I ÷ (L_eff)²
P_cr = π² × 200 × 10⁹ × 8 × 10⁻⁶ ÷ 3²
P_cr = 9.8696 × 1.6 × 10⁶ ÷ 9
P_cr = 15,791,360 ÷ 9
P_cr = 1,754,596 N
Answer: 1.75 MN
Given: Concrete-filled steel column, E = 25 × 10⁹ Pa, I = 5 × 10⁻⁵ m⁴, L_eff = 4 m.
P_cr = π² E I ÷ (L_eff)²
P_cr = 9.8696 × 25 × 10⁹ × 5 × 10⁻⁵ ÷ 4²
P_cr = 12,337,000 ÷ 16
P_cr = 771,062.5 N
Answer: 771.06 kN
Given: Timber column, E = 12 × 10⁹ Pa, I = 2 × 10⁻⁶ m⁴, L_eff = 2.5 m.
P_cr = π² E I ÷ (L_eff)²
P_cr = 9.8696 × 12 × 10⁹ × 2 × 10⁻⁶ ÷ 2.5²
P_cr = 236,870 ÷ 6.25
P_cr = 37,899.2 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 × 10⁹ Pa, I = 10 × 10⁻⁶ m⁴, length L = 3.5 m, fixed-fixed ends.
Calculate P_cr.
L_eff = L ÷ 2
L_eff = 3.5 ÷ 2 = 1.75 m
P_cr = π² E I ÷ (L_eff)²
P_cr = 9.8696 × 200 × 10⁹ × 10 × 10⁻⁶ ÷ 1.75²
P_cr = 19,739,200 ÷ 3.0625
P_cr = 6,445,453 N
Answer: 6.45 MN
Given: Aluminum column, E = 70 × 10⁹ Pa, I = 4 × 10⁻⁶ m⁴, length L = 5 m, fixed-free ends.
Calculate P_cr.
L_eff = 2L
L_eff = 2 × 5 = 10 m
P_cr = π² E I ÷ (L_eff)²
P_cr = 9.8696 × 70 × 10⁹ × 4 × 10⁻⁶ ÷ 10²
P_cr = 2,763,488 ÷ 100
P_cr = 27,634.9 N
Answer: 27.63 kN
Euler's theory applies to ideal slender columns under axial load without imperfections:
Given: A steel column with E = 210 × 10⁹ Pa, I = 9 × 10⁻⁶ m⁴, L_eff = 3 m. Calculate Euler load and then adjust by a factor of 0.7 to account for imperfections.
P_cr = π² E I ÷ (L_eff)²
P_cr = 9.8696 × 210 × 10⁹ × 9 × 10⁻⁶ ÷ 3²
P_cr = 18,653,544 ÷ 9
P_cr = 2,072,616 N
Adjusted load:
P_adj = 0.7 × P_cr
P_adj = 0.7 × 2,072,616
P_adj = 1,450,831 N
Answer: 1.45 MN
Given: Column with E = 200 × 10⁹ Pa, I = 5 × 10⁻⁶ m⁴, L_eff = 2.5 m. Calculate Euler load and apply a 15% reduction for initial imperfections and residual stresses.
P_cr = π² E I ÷ (L_eff)²
P_cr = 9.8696 × 200 × 10⁹ × 5 × 10⁻⁶ ÷ 2.5²
P_cr = 9,869,600 ÷ 6.25
P_cr = 1,579,136 N
Reduced load:
P_red = 0.85 × P_cr
P_red = 0.85 × 1,579,136
P_red = 1,342,265.6 N
Answer: 1.34 MN
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:
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:
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:
Formula:$$ \lambda = \frac{2800}{40} = 70 $$
Classification:
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)
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:
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:
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:
\[
P_{cr} = \frac{\pi^2 EI}{L_{eff}^2} = \frac{9.8696 \times 210 \times 10^9 \times 7.5 \times 10^{-6}}{(1.75)^2} = \frac{9.8696 \times 210 \times 7.5 \times 10^{3}}{3.0625} = \frac{15,527,880}{3.0625} = 5,069,100\,N = \mathbf{5069.1\,kN}
\]