Column Theory

Euler's critical (crippling) load is the smallest axial load at which an initially straight, slender column becomes unstable and buckles sideways.

Assumptions

Assumptions: the column is perfectly straight, of uniform section and homogeneous material, the load is perfectly axial, the material obeys Hooke's law, deflections are small, and the effect of shear and axial shortening is ignored.

Derivation

Let the column AB of length $L$ be hinged at both ends, with the load $P$ producing a small deflection $y$ at distance $x$ from A.

   P
   |  B
   |  /\
   |  \ y
   |  /
   A /
   P
 (A origin, x along column)

Bending moment at the section: $M = -Py$ (taking deflection that reduces curvature as negative). Using the differential equation of the elastic curve:

$$EI\frac{d^2y}{dx^2} = -Py \;\Rightarrow\; \frac{d^2y}{dx^2} + k^2 y = 0,\qquad k^2 = \frac{P}{EI}$$

The solution is

$$y = C_1\cos kx + C_2\sin kx$$

Boundary conditions:

• At $x = 0$, $y = 0 \Rightarrow C_1 = 0$.
• At $x = L$, $y = 0 \Rightarrow C_2\sin kL = 0$.

For a buckled shape $C_2 \ne 0$, so $\sin kL = 0$ and $kL = n\pi$. The smallest non-zero load is for $n = 1$:

$$k L = \pi \;\Rightarrow\; \frac{PL^2}{EI} = \pi^2$$ $$P_{cr} = \frac{\pi^2 EI}{L^2}$$

Here $I$ is the least moment of inertia, since the column buckles about the axis of least resistance. The shape is a half sine wave, $y = C_2\sin(\pi x/L)$, with maximum deflection at mid-span.

Euler's crippling load is the smallest axial load at which an initially straight, slender column buckles. For both ends fixed, the ends are held in position and direction.

Assumptions

Assumptions: the column is perfectly straight, of uniform section and homogeneous material, the load is perfectly axial, the material obeys Hooke's law, deflections are small, and the effect of shear and axial shortening is ignored.

Derivation

When the column buckles, the fixed ends develop an end moment $M_0$ (fixing moment) while the load $P$ stays axial. Take the origin at the lower fixed end A, with $y$ the lateral deflection at distance $x$.

   P
   |  B  (fixed, M0)
   |  /\
   | (  y
   |  \/
   A  fixed, M0
   P

The moment at the section is $M = M_0 - Py$, so

$$EI\frac{d^2y}{dx^2} = M_0 - Py \;\Rightarrow\; \frac{d^2y}{dx^2} + k^2y = \frac{M_0}{EI},\qquad k^2 = \frac{P}{EI}$$

The solution is

$$y = C_1\cos kx + C_2\sin kx + \frac{M_0}{P}$$

Boundary conditions:

• At $x=0$: $y = 0 \Rightarrow C_1 = -\dfrac{M_0}{P}$.
• At $x=0$: $\dfrac{dy}{dx} = 0 \Rightarrow C_2 k = 0 \Rightarrow C_2 = 0$.

So

$$y = \frac{M_0}{P}\,(1 - \cos kx)$$

• At $x = L$: $y = 0$ (the top end is at the same lateral position) gives $\cos kL = 1$.

The smallest non-zero solution is $kL = 2\pi$:

$$\frac{PL^2}{EI} = 4\pi^2$$ $$P_{cr} = \frac{4\pi^2EI}{L^2} = \frac{\pi^2EI}{(L/2)^2}$$

The effective length is $L_e = L/2$, so a fixed-fixed column carries 4 times the load of the same hinged column. The buckled shape has inflection points at $L/4$ and $3L/4$ from each end.

Euler's critical load for a column fixed at one end and hinged at the other is derived below.

Assumptions

Assumptions: the column is perfectly straight, of uniform section and homogeneous material, the load is perfectly axial, the material obeys Hooke's law, deflections are small, and the effect of shear and axial shortening is ignored.

Derivation

Let the column have its top end B hinged and its bottom end A fixed, length $L$. When it buckles, the fixed end develops a moment, and a horizontal reaction $H$ appears at the hinge to keep B in position. Take the origin at the hinged end B, with $x$ measured downward and $y$ the lateral deflection.

   P
   |  B  hinged (H)
   |  )
   |  ) y
   |  )
   A  fixed
   P

Moment at the section: $M = -Py + Hx$, so

$$EI\frac{d^2y}{dx^2} = -Py + Hx \;\Rightarrow\; \frac{d^2y}{dx^2} + k^2y = \frac{Hx}{EI},\qquad k^2 = \frac{P}{EI}$$

The solution is

$$y = C_1\cos kx + C_2\sin kx + \frac{H}{P}x$$

Boundary conditions:

• At $x=0$: $y = 0 \Rightarrow C_1 = 0$.
• At $x=L$ (fixed end): $y = 0 \Rightarrow C_2\sin kL + \dfrac{HL}{P} = 0$
• At $x=L$: $\dfrac{dy}{dx} = 0 \Rightarrow C_2k\cos kL + \dfrac{H}{P} = 0$

Eliminate $C_2$ and $H$ between the last two equations:

$$\tan kL = kL$$

This transcendental equation is solved by trial. The smallest non-zero root is

$$kL = 4.493 \;\Rightarrow\; \frac{PL^2}{EI} = 4.493^2 = 20.19$$ $$P_{cr} = \frac{20.19\,EI}{L^2} \approx \frac{2\pi^2EI}{L^2} = \frac{\pi^2EI}{(0.7L)^2}$$

The effective length is $L_e \approx 0.7L$. The inflection point lies at about $0.3L$ from the hinged end.

Euler's critical load for a column fixed at one end and hinged at the other is derived below.

Assumptions

Assumptions: the column is perfectly straight, of uniform section and homogeneous material, the load is perfectly axial, the material obeys Hooke's law, deflections are small, and the effect of shear and axial shortening is ignored.

Derivation

Let the column have its top end B hinged and its bottom end A fixed, length $L$. When it buckles, the fixed end develops a moment, and a horizontal reaction $H$ appears at the hinge to keep B in position. Take the origin at the hinged end B, with $x$ measured downward and $y$ the lateral deflection.

   P
   |  B  hinged (H)
   |  )
   |  ) y
   |  )
   A  fixed
   P

Moment at the section: $M = -Py + Hx$, so

$$EI\frac{d^2y}{dx^2} = -Py + Hx \;\Rightarrow\; \frac{d^2y}{dx^2} + k^2y = \frac{Hx}{EI},\qquad k^2 = \frac{P}{EI}$$

The solution is

$$y = C_1\cos kx + C_2\sin kx + \frac{H}{P}x$$

Boundary conditions:

• At $x=0$: $y = 0 \Rightarrow C_1 = 0$.
• At $x=L$ (fixed end): $y = 0 \Rightarrow C_2\sin kL + \dfrac{HL}{P} = 0$
• At $x=L$: $\dfrac{dy}{dx} = 0 \Rightarrow C_2k\cos kL + \dfrac{H}{P} = 0$

Eliminate $C_2$ and $H$ between the last two equations:

$$\tan kL = kL$$

This transcendental equation is solved by trial. The smallest non-zero root is

$$kL = 4.493 \;\Rightarrow\; \frac{PL^2}{EI} = 4.493^2 = 20.19$$ $$P_{cr} = \frac{20.19\,EI}{L^2} \approx \frac{2\pi^2EI}{L^2} = \frac{\pi^2EI}{(0.7L)^2}$$

The effective length is $L_e \approx 0.7L$. The inflection point lies at about $0.3L$ from the hinged end.

Limitations of Euler's formula

1. It applies only to long (slender) columns. In terms of stress, $\sigma_{cr} = \dfrac{\pi^2E}{\lambda^2}$ must not exceed the proportional limit, so $\lambda \ge \pi\sqrt{E/\sigma_p}$ (about 80 to 100 for mild steel). For short and intermediate columns it over-estimates the load.
2. It assumes a perfectly straight column, a perfectly axial load and homogeneous material. Real columns have initial crookedness and eccentricity, so they fail at a lower load.
3. It is valid only within the elastic range (Hooke's law), so $\sigma_{cr}$ cannot exceed the proportional limit.
4. It ignores the effect of direct compression, shear deformation and imperfect end conditions.
5. For very slender columns it is accurate, and for intermediate columns empirical formulas (Rankine, Johnson, Secant, straight-line) are used instead.

Given: round bar fixed at the bottom, free at the top, $L = 3$ m = 3000 mm. A horizontal force $H = 15$ kN at the top gives a horizontal deflection $\delta = 35$ mm.

Step 1: find $EI$ from the lateral load test

The bar acts as a cantilever under $H$:

$$\delta = \frac{HL^3}{3EI} \;\Rightarrow\; EI = \frac{HL^3}{3\delta} = \frac{15\times10^3\times3000^3}{3\times35} = 3.857\times10^{12}\ \text{N mm}^2$$

Step 2: buckling load (fixed-free)

Effective length $L_e = 2L = 6000$ mm.

$$P_{cr} = \frac{\pi^2EI}{L_e^2} = \frac{\pi^2\times3.857\times10^{12}}{(6000)^2} = 1.057\times10^6\ \text{N}$$

Answer: buckling load $P_{cr} = 1057$ kN (about 1.06 MN).

Given: the bar is clamped at the bottom and free at the top (cantilever column). The effective length is $L_e = 2$ m, so the actual length is $L = L_e/2 = 1$ m. A horizontal force $H = 300$ N at the top produces a deflection $\delta = 20$ mm.

Step 1: find $EI$ from the lateral load

Acting as a cantilever:

$$\delta = \frac{HL^3}{3EI} \;\Rightarrow\; EI = \frac{300\times(1000)^3}{3\times20} = 5\times10^{9}\ \text{N mm}^2$$

Step 2: buckling load

$$P_{cr} = \frac{\pi^2EI}{L_e^2} = \frac{\pi^2\times5\times10^9}{(2000)^2} = 12{,}337\ \text{N}$$

Answer: buckling load $\approx 12.34$ kN.

If the 2 m were taken as the actual length (so $L_e = 4$ m), $EI = 4\times10^{10}\ \text{N mm}^2$ and $P_{cr} = 24.67$ kN; the usual reading of "effective length" gives 12.34 kN.

Short column vs long column

Limiting slenderness ratio for Euler's formula

Euler's formula is valid while the critical stress does not exceed the limiting (proportional) stress $\sigma_{cr}$ given. The critical stress is

$$\sigma_{cr} = \frac{P_{cr}}{A} = \frac{\pi^2E}{\lambda^2}$$

Setting this equal to the limiting stress $330$ MPa:

$$\lambda = \pi\sqrt{\frac{E}{\sigma_{cr}}} = \pi\sqrt{\frac{210\times10^3}{330}} = \pi\times25.22 = 79.25$$

Answer: the limiting slenderness ratio is $\lambda \approx 79.3$. Euler's formula is valid for columns with $\lambda \ge 79.3$ (that is, the critical stress stays at or below 330 MPa); for smaller $\lambda$ it overestimates the load and Euler's formula must not be used.

Given: $L = 1.5$ m = 1500 mm, section $12\times5$ mm, hinged ends, $E = 2\times10^5\ \text{N/mm}^2$, yield stress $290\ \text{N/mm}^2$.

The bar buckles about the axis of least resistance (thickness 5 mm).

$$I_{min} = \frac{12\times5^3}{12} = 125\ \text{mm}^4,\qquad A = 60\ \text{mm}^2,\qquad Z = \frac{I}{y} = \frac{125}{2.5} = 50\ \text{mm}^3$$

Euler's buckling load

$$P_{cr} = \frac{\pi^2EI}{L^2} = \frac{\pi^2\times2\times10^5\times125}{1500^2} = 109.66\ \text{N}$$

Direct stress at this load:

$$\sigma_d = \frac{P_{cr}}{A} = \frac{109.66}{60} = 1.83\ \text{N/mm}^2$$

Central deflection at yield

After buckling, the load remains at $P_{cr}$ while the deflection $\delta$ grows. The maximum stress at mid-span is direct stress plus bending stress:

$$\sigma_{max} = \frac{P}{A} + \frac{P\,\delta}{Z}$$ $$290 = 1.83 + \frac{109.66\,\delta}{50}$$ $$\delta = \frac{(290 - 1.83)\times50}{109.66} = 131.4\ \text{mm}$$

Answer: Euler load = 109.7 N; maximum central deflection before yielding $\approx 131$ mm. This is an estimate, because the small-deflection theory is used.

Given: $L = 5$ m = 5000 mm, $d_i = 40$ mm, $t = 8$ mm, so $d_o = 40 + 2\times8 = 56$ mm. Both ends hinged ($L_e = L$), $E = 2.1\times10^5\ \text{N/mm}^2$, FOS = 3.

Section properties

$$I = \frac{\pi}{64}(56^4 - 40^4) = \frac{\pi}{64}(9834496 - 2560000) = 3.571\times10^5\ \text{mm}^4$$ $$A = \frac{\pi}{4}(56^2 - 40^2) = 1206.4\ \text{mm}^2$$

Radius of gyration $r = \sqrt{I/A} = 17.2$ mm, so $\lambda = 5000/17.2 = 290.6$, which is very slender, so Euler's formula applies.

Critical load

$$P_{cr} = \frac{\pi^2EI}{L^2} = \frac{\pi^2\times2.1\times10^5\times3.571\times10^5}{5000^2} = 29{,}604\ \text{N} = 29.60\ \text{kN}$$

Safe load

$$P_{safe} = \frac{P_{cr}}{FOS} = \frac{29.60}{3} = 9.87\ \text{kN}$$

Answer: critical load = 29.60 kN; safe load = 9.87 kN.

Euler's critical load for a column fixed at one end and free at the other is derived below.

Assumptions

Assumptions: the column is perfectly straight, of uniform section and homogeneous material, the load is perfectly axial, the material obeys Hooke's law, deflections are small, and the effect of shear and axial shortening is ignored.

Derivation

Let the column AB, length $L$, have its lower end A fixed and its upper end B free. Under the load $P$, the free end deflects by $\delta$. Take the origin at the fixed end A, with $y$ measured from the original axis.

      P
      v
   B  )  <- free end, deflection delta
      )
      ) y
      )
   A  fixed

At a section distance $x$ from A, the lever arm of $P$ is $(\delta - y)$, so $M = P(\delta - y)$:

$$EI\frac{d^2y}{dx^2} = P(\delta - y) \;\Rightarrow\; \frac{d^2y}{dx^2} + k^2y = k^2\delta,\qquad k^2 = \frac{P}{EI}$$

The solution is

$$y = C_1\cos kx + C_2\sin kx + \delta$$

Boundary conditions:

• At $x=0$: $y=0 \Rightarrow C_1 = -\delta$.
• At $x=0$: $\dfrac{dy}{dx} = 0 \Rightarrow C_2 = 0$.

So $y = \delta\,(1 - \cos kx)$.

• At $x=L$: $y = \delta \Rightarrow \delta = \delta(1 - \cos kL) \Rightarrow \cos kL = 0$

The smallest solution is $kL = \dfrac{\pi}{2}$:

$$\frac{PL^2}{EI} = \frac{\pi^2}{4}$$ $$P_{cr} = \frac{\pi^2EI}{4L^2} = \frac{\pi^2EI}{(2L)^2}$$

The effective length is $L_e = 2L$, so this column carries one quarter of the load of a hinged-hinged column of the same length.

Limitations of Euler's formula

1. It applies only to long (slender) columns. In terms of stress, $\sigma_{cr} = \dfrac{\pi^2E}{\lambda^2}$ must not exceed the proportional limit, so $\lambda \ge \pi\sqrt{E/\sigma_p}$ (about 80 to 100 for mild steel). For short and intermediate columns it over-estimates the load.
2. It assumes a perfectly straight column, a perfectly axial load and homogeneous material. Real columns have initial crookedness and eccentricity, so they fail at a lower load.
3. It is valid only within the elastic range (Hooke's law), so $\sigma_{cr}$ cannot exceed the proportional limit.
4. It ignores the effect of direct compression, shear deformation and imperfect end conditions.
5. For very slender columns it is accurate, and for intermediate columns empirical formulas (Rankine, Johnson, Secant, straight-line) are used instead.

Euler's critical (crippling) load is the smallest axial load at which an initially straight, slender column becomes unstable and buckles sideways.

Assumptions

Assumptions: the column is perfectly straight, of uniform section and homogeneous material, the load is perfectly axial, the material obeys Hooke's law, deflections are small, and the effect of shear and axial shortening is ignored.

Derivation

Let the column AB of length $L$ be hinged at both ends, with the load $P$ producing a small deflection $y$ at distance $x$ from A.

   P
   |  B
   |  /\
   |  \ y
   |  /
   A /
   P
 (A origin, x along column)

Bending moment at the section: $M = -Py$ (taking deflection that reduces curvature as negative). Using the differential equation of the elastic curve:

$$EI\frac{d^2y}{dx^2} = -Py \;\Rightarrow\; \frac{d^2y}{dx^2} + k^2 y = 0,\qquad k^2 = \frac{P}{EI}$$

The solution is

$$y = C_1\cos kx + C_2\sin kx$$

Boundary conditions:

• At $x = 0$, $y = 0 \Rightarrow C_1 = 0$.
• At $x = L$, $y = 0 \Rightarrow C_2\sin kL = 0$.

For a buckled shape $C_2 \ne 0$, so $\sin kL = 0$ and $kL = n\pi$. The smallest non-zero load is for $n = 1$:

$$k L = \pi \;\Rightarrow\; \frac{PL^2}{EI} = \pi^2$$ $$P_{cr} = \frac{\pi^2 EI}{L^2}$$

Here $I$ is the least moment of inertia, since the column buckles about the axis of least resistance. The shape is a half sine wave, $y = C_2\sin(\pi x/L)$, with maximum deflection at mid-span.

Limitations of Euler's formula

1. It applies only to long (slender) columns. In terms of stress, $\sigma_{cr} = \dfrac{\pi^2E}{\lambda^2}$ must not exceed the proportional limit, so $\lambda \ge \pi\sqrt{E/\sigma_p}$ (about 80 to 100 for mild steel). For short and intermediate columns it over-estimates the load.
2. It assumes a perfectly straight column, a perfectly axial load and homogeneous material. Real columns have initial crookedness and eccentricity, so they fail at a lower load.
3. It is valid only within the elastic range (Hooke's law), so $\sigma_{cr}$ cannot exceed the proportional limit.
4. It ignores the effect of direct compression, shear deformation and imperfect end conditions.
5. For very slender columns it is accurate, and for intermediate columns empirical formulas (Rankine, Johnson, Secant, straight-line) are used instead.

Euler's crippling load is the smallest axial load at which an initially straight, slender column buckles. For both ends fixed, the ends are held in position and direction.

Assumptions

Assumptions: the column is perfectly straight, of uniform section and homogeneous material, the load is perfectly axial, the material obeys Hooke's law, deflections are small, and the effect of shear and axial shortening is ignored.

Derivation

When the column buckles, the fixed ends develop an end moment $M_0$ (fixing moment) while the load $P$ stays axial. Take the origin at the lower fixed end A, with $y$ the lateral deflection at distance $x$.

   P
   |  B  (fixed, M0)
   |  /\
   | (  y
   |  \/
   A  fixed, M0
   P

The moment at the section is $M = M_0 - Py$, so

$$EI\frac{d^2y}{dx^2} = M_0 - Py \;\Rightarrow\; \frac{d^2y}{dx^2} + k^2y = \frac{M_0}{EI},\qquad k^2 = \frac{P}{EI}$$

The solution is

$$y = C_1\cos kx + C_2\sin kx + \frac{M_0}{P}$$

Boundary conditions:

• At $x=0$: $y = 0 \Rightarrow C_1 = -\dfrac{M_0}{P}$.
• At $x=0$: $\dfrac{dy}{dx} = 0 \Rightarrow C_2 k = 0 \Rightarrow C_2 = 0$.

So

$$y = \frac{M_0}{P}\,(1 - \cos kx)$$

• At $x = L$: $y = 0$ (the top end is at the same lateral position) gives $\cos kL = 1$.

The smallest non-zero solution is $kL = 2\pi$:

$$\frac{PL^2}{EI} = 4\pi^2$$ $$P_{cr} = \frac{4\pi^2EI}{L^2} = \frac{\pi^2EI}{(L/2)^2}$$

The effective length is $L_e = L/2$, so a fixed-fixed column carries 4 times the load of the same hinged column. The buckled shape has inflection points at $L/4$ and $3L/4$ from each end.

Limitations of Euler's formula

1. It applies only to long (slender) columns. In terms of stress, $\sigma_{cr} = \dfrac{\pi^2E}{\lambda^2}$ must not exceed the proportional limit, so $\lambda \ge \pi\sqrt{E/\sigma_p}$ (about 80 to 100 for mild steel). For short and intermediate columns it over-estimates the load.
2. It assumes a perfectly straight column, a perfectly axial load and homogeneous material. Real columns have initial crookedness and eccentricity, so they fail at a lower load.
3. It is valid only within the elastic range (Hooke's law), so $\sigma_{cr}$ cannot exceed the proportional limit.
4. It ignores the effect of direct compression, shear deformation and imperfect end conditions.
5. For very slender columns it is accurate, and for intermediate columns empirical formulas (Rankine, Johnson, Secant, straight-line) are used instead.

Buckling load

The buckling (crippling or critical) load is the axial compressive load at which a long, slender column just begins to buckle (deflect sideways). It is the highest load the column can carry in its straight, stable form.

Effective length

The effective length $L_e$ is the length of an equivalent column with both ends hinged that has the same buckling load. It equals the distance between the points of zero bending moment (inflection points) in the buckled shape:

For any end condition, $P_{cr} = \dfrac{\pi^2EI}{L_e^2}$.

Euler's formula for both ends hinged

Assumptions

Assumptions: the column is perfectly straight, of uniform section and homogeneous material, the load is perfectly axial, the material obeys Hooke's law, deflections are small, and the effect of shear and axial shortening is ignored.

Derivation

Let the column AB of length $L$ be hinged at both ends, with the load $P$ producing a small deflection $y$ at distance $x$ from A.

   P
   |  B
   |  /\
   |  \ y
   |  /
   A /
   P
 (A origin, x along column)

Bending moment at the section: $M = -Py$ (taking deflection that reduces curvature as negative). Using the differential equation of the elastic curve:

$$EI\frac{d^2y}{dx^2} = -Py \;\Rightarrow\; \frac{d^2y}{dx^2} + k^2 y = 0,\qquad k^2 = \frac{P}{EI}$$

The solution is

$$y = C_1\cos kx + C_2\sin kx$$

Boundary conditions:

• At $x = 0$, $y = 0 \Rightarrow C_1 = 0$.
• At $x = L$, $y = 0 \Rightarrow C_2\sin kL = 0$.

For a buckled shape $C_2 \ne 0$, so $\sin kL = 0$ and $kL = n\pi$. The smallest non-zero load is for $n = 1$:

$$k L = \pi \;\Rightarrow\; \frac{PL^2}{EI} = \pi^2$$ $$P_{cr} = \frac{\pi^2 EI}{L^2}$$

Here $I$ is the least moment of inertia, since the column buckles about the axis of least resistance. The shape is a half sine wave, $y = C_2\sin(\pi x/L)$, with maximum deflection at mid-span.

Given: simply supported beam, $L = 5$ m = 5000 mm, central load $W = 800$ N, maximum (central) deflection $\delta = 30$ mm.

Step 1: find $EI$

For a simply supported beam with a central point load:

$$\delta = \frac{WL^3}{48EI} \;\Rightarrow\; EI = \frac{WL^3}{48\delta} = \frac{800\times5000^3}{48\times30} = 6.944\times10^{10}\ \text{N mm}^2$$

Step 2: buckling load as a column

For one end fixed and the other hinged, the effective length is $L_e = 0.7L$ (equivalent to $P_{cr} = 2\pi^2EI/L^2$):

$$P_{cr} = \frac{2\pi^2EI}{L^2} = \frac{2\pi^2\times6.944\times10^{10}}{5000^2} = 54{,}831\ \text{N}$$

Using the standard textbook form $2\pi^2EI/L^2$ ($L_e \approx 0.7L$; the exact root of $\tan kL = kL$ gives about 56 kN):

Answer: buckling load $\approx 54.8$ kN.

Given: $L = 5$ m = 5000 mm, $d_i = 40$ mm, $t = 6$ mm so $d_o = 52$ mm. One end fixed, other hinged: $P_{cr} = \dfrac{2\pi^2EI}{L^2}$ ($L_e = 0.7L$). $E = 2\times10^5\ \text{N/mm}^2$, FOS = 3.

Section properties

$$I = \frac{\pi}{64}(52^4 - 40^4) = 2.332\times10^5\ \text{mm}^4$$ $$A = \frac{\pi}{4}(52^2 - 40^2) = 867.1\ \text{mm}^2$$

$r = \sqrt{I/A} = 16.4$ mm, and $L_e/r = 3500/16.4 = 213$, so the strut is slender and Euler's formula is valid.

Crippling load

$$P_{cr} = \frac{2\pi^2\times2\times10^5\times2.332\times10^5}{5000^2} = 36{,}832\ \text{N} = 36.83\ \text{kN}$$

Crippling stress

$$\sigma_{cr} = \frac{P_{cr}}{A} = \frac{36{,}832}{867.1} = 42.48\ \text{N/mm}^2$$

Safe load

$$P_{safe} = \frac{36.83}{3} = 12.28\ \text{kN}$$

Answer: crippling load = 36.83 kN; crippling stress = 42.48 N/mm²; safe load = 12.28 kN.