Calculus of Variations: Euler–Bernoulli Beam Problems and Worked Solutions

CV08 showed that a functional containing a second derivative generally produces a fourth-order Euler–Lagrange equation and a boundary term containing both the endpoint variation η and its derivative η′. The Euler–Bernoulli beam is one of the clearest physical realizations of this structure. In the small-deflection theory of a slender elastic beam, curvature is approximated by w′′(x) and the bending strain energy is quadratic in that curvature [3, 2]. The resulting potential-energy functional provides a direct bridge from higher-order calculus of variations to structural mechanics.

Throughout this companion set, w(x) denotes transverse deflection, E Young’s modulus, I the second moment of area, and

(1)

is the flexural rigidity. When D is constant, it will simply be written EI.

For a beam under a distributed load q(x), with the sign convention that positive q acts in the positive w direction, the total potential energy is

(2)

The small-slope Euler–Bernoulli model neglects transverse shear deformation and rotary inertia; those assumptions should be kept distinct from the variational method itself [3, 4].

1 Exercises

Exercise 1: derive the Euler–Bernoulli equation from total potential energy

Consider

(3)

1. Compute the first variation δΠ[w; η].
2. Integrate by parts twice and identify the interior equation.
3. Identify the two natural boundary quantities conjugate to η and η′.
4. Specialize the differential equation to constant EI.

Exercise 2: uniformly loaded cantilever

A beam of length L and constant flexural rigidity EI is clamped at x = 0 and free at x = L. It carries a constant distributed load q > 0.

1. State the essential conditions at the clamped end.
2. Derive the natural conditions at the free end.
3. Solve the beam equation for w(x).
4. Find the free-end slope w′(L) and deflection w(L).

Exercise 3: uniformly loaded simply supported beam

A uniform beam has simple supports at x = 0 and x = L and carries the same constant load q.

1. Explain why w = 0 is essential at each support while the zero-moment condition is natural.
2. Solve for w(x).
3. Find the maximum deflection and its location.

Exercise 4: clamped–clamped beam under uniform load

A uniform beam is clamped at both ends and carries constant q.

1. State the four essential endpoint conditions.
2. Solve for w(x).
3. Compare the midspan deflection with the simply supported result from Exercise 3.

Exercise 5: cantilever with a concentrated tip force

A uniform cantilever has no distributed load. A force P > 0 acts at the free tip in the positive w direction. Use

(4)

1. Derive the differential equation and free-end natural conditions directly from δΠ = 0.
2. Solve for w(x).
3. Find the tip slope and tip deflection.
4. Explain how the endpoint force appears through a boundary condition rather than through the interior differential equation.

Exercise 6: variable flexural rigidity with a tip moment

A cantilever has

(5)

and a terminal moment M applied at x = L. There is no distributed load. Use

(6)

1. Derive the free-end conditions.
2. Show that the bending moment D(x)w′′(x) is constant.
3. For α≠0, determine w′(x) and w(x) when w(0) = w′(0) = 0.
4. Check the limit α → 0.

Exercise 7: endpoint springs and generalized natural conditions

A beam has no distributed load near its right end, but the endpoint is attached to a translational spring of stiffness k_t and a rotational spring of stiffness k_r. An external tip force P also acts there. The endpoint contribution to the potential is

(7)

For constant EI, derive the two natural conditions at x = L when both w(L) and w′(L) are free.

Exercise 8: beam under axial tension

Consider the functional

(8)

1. Derive the Euler–Lagrange equation.
2. Derive the two natural boundary conditions at a completely free end.
3. Explain physically why the quantity conjugate to endpoint displacement is no longer simply −EIw′′′.

Exercise 9: one-parameter Rayleigh–Ritz estimate for a tip-loaded cantilever

Return to the cantilever of Exercise 5. Instead of solving the exact fourth-order boundary-value problem, restrict the admissible set to

(9)

This trial family satisfies the clamped essential conditions but does not impose the free-end natural conditions in advance.

1. Substitute w_a into the exact potential energy and find Π(a).
2. Minimize with respect to a.
3. Compute the Ritz estimate of the tip deflection.
4. Compare with the exact result from Exercise 5.
5. Explain why essential conditions must be built into a Ritz trial function but natural conditions need not be.

2 Worked solutions

Solution 1: derive the Euler–Bernoulli equation from total potential energy

The integrand is

(10)

Therefore

(11)

The first variation is

(12)

Integrate the second term by parts once:

(13)

Integrate the remaining integral by parts again:

(14)

Hence

δΠ	= 0L	(15)
	+ ∫ 0L η dx.	(16)

For arbitrary interior variations, the Fundamental Lemma gives

(17)

The two boundary quantities are therefore

(18)

as the quantity conjugate to the slope variation η′, and

(19)

as the quantity conjugate to the displacement variation η. The symbols M_b and Q_b are deliberately called moment-like and shear-like because detailed structural sign conventions differ. The variational statement itself fixes their signs once the positive directions of w, q, endpoint force, and endpoint moment have been chosen.

For constant EI,

(20)

This is the static Euler–Bernoulli beam equation in the present sign convention [3, 2].

Solution 2: uniformly loaded cantilever

At the clamp,

(21)

These are essential conditions, so

(22)

At the free end there is no applied endpoint moment or force. Both η(L) and η′(L) are arbitrary, so the boundary term requires

(23)

Thus

(24)

The differential equation is

(25)

Integrate four times:

w′′′	= x + C1,	(26)
w′′	= x2 + C 1x + C2,	(27)
w′	= x3 + C1x2 + C 2x + C3,	(28)
w	= x4 + C1x3 + C2x2 + C 3x + C4.	(29)

The free-end conditions give

(30)

while the clamped conditions give

(31)

Therefore

(32)

Differentiate:

(33)

At x = L,

(34)

and

(35)

Solution 3: uniformly loaded simply supported beam

A simple support prevents transverse displacement but does not prescribe the slope. Hence

(36)

are essential conditions, giving

(37)

The slope variations η′(0) and η′(L) remain free. Therefore the coefficients of those variations must vanish:

(38)

Thus the zero bending moment at a simple support emerges as a natural boundary condition.

Solving

(39)

with

(40)

gives

(41)

By symmetry, the maximum occurs at

(42)

Substitution gives

(43)

Solution 4: clamped–clamped beam under uniform load

Both displacement and slope are prescribed at both ends:

(44)

All four endpoint variations therefore vanish. There are no natural conditions to derive because no endpoint kinematic variable is free.

Solving EIw′′′′ = q gives

(45)

At midspan,

(46)

The simply supported maximum from Exercise 3 was

(47)

Thus the clamped–clamped midspan deflection is only one fifth as large:

(48)

The stronger kinematic restrictions make the beam substantially stiffer under the same loading.

Solution 5: cantilever with a concentrated tip force

The potential is

(49)

Its variation is

(50)

After two integrations by parts,

δΠ	= 0L	(51)
	+ ∫ 0LEIw′′′′η dx − Pη(L).	(52)

The clamped end removes the x = 0 terms. Since there is no distributed load,

(53)

At the free end, η(L) and η′(L) are independent. Their coefficients give

(54)

and

(55)

Equivalently,

(56)

Solving with w(0) = w′(0) = 0 gives

(57)

Therefore

(58)

and

(59)

The key variational point is that the tip force is represented by the endpoint potential −Pw(L). It therefore modifies the boundary equation but not the load-free interior equation EIw′′′′ = 0.

Solution 6: variable flexural rigidity with a tip moment

Now

(60)

The first variation is

δΠ	= 0L	(61)
	+ ∫ 0L(Dw′′)′′η dx − Mη′(L).	(62)

At the free end,

(63)

and

(64)

The interior equation is

(65)

Hence Dw′′ is linear in x. The zero-shear condition says its derivative vanishes at L, so the linear function is actually constant. The moment condition then gives

(66)

throughout the beam.

Therefore

(67)

For α≠0, integrate once and use w′(0) = 0:

(68)

Integrate again and use w(0) = 0:

(69)

As α → 0,

(70)

so

(71)

which is the familiar constant-curvature result.

Solution 7: endpoint springs and generalized natural conditions

The beam contribution at x = L is

(72)

The endpoint potential contributes

(73)

Collect the coefficients of the independent endpoint variations.

For η′(L):

(74)

Thus

(75)

For η(L):

(76)

Hence

(77)

These are mixed, or Robin-type, natural boundary conditions. The endpoint springs interpolate continuously between free and strongly restrained endpoint behavior.

Solution 8: beam under axial tension

The integrand is

(78)

Therefore

(79)

The second-order Euler–Lagrange equation is

(80)

For constant T and EI,

(81)

Thus

(82)

The boundary term is

(83)

At a completely free end with no endpoint loads,

(84)

and

(85)

The displacement-conjugate boundary quantity now contains both the bending contribution −EIw′′′ and the transverse component associated with axial Tension, Tw′. This is exactly what the variational boundary term predicts.

Solution 9: one-parameter Rayleigh–Ritz estimate for a tip-loaded cantilever

Choose

(86)

Then

(87)

and

(88)

Substitute into

(89)

Π(a)	= ∫ 0LEI(2a)2dx − PaL2	(90)
	= 2EILa2 − PL2a.	(91)

Stationarity in the one-dimensional trial space requires

(92)

Hence

(93)

The Ritz approximation is therefore

(94)

with tip deflection

(95)

The exact result from Exercise 5 is

(96)

Thus

(97)

The one-parameter approximation underestimates the compliance by 25%.

This last point is fundamental. Essential boundary conditions define the admissible function space, so every trial function must satisfy them. Natural boundary conditions arise from stationarity of the functional itself. A Ritz approximation should therefore be allowed to discover them only approximately through energy minimization rather than impose them as artificial kinematic restrictions [2].

3 What the beam examples teach about higher-order variational problems

The beam equations make several abstract points from CV08 concrete.

1. A functional depending on w′′ naturally produces a fourth-order differential equation.
2. The endpoint variables w and w′ are independent kinematic channels.
3. Their conjugate boundary quantities arise directly from repeated integration by parts.
4. Loads can enter either through the interior functional, as q(x) does, or through endpoint potentials, as P and M do.
5. A boundary condition is not “essential” or “natural” because of its physical name; the distinction is variational. Essential data are imposed on the admissible class. Natural data emerge from free endpoint variations.
6. Variable stiffness is handled without changing the variational principle: one simply retains derivatives of D(x)w′′ rather than replacing them prematurely by EIw′′′′.
7. Rayleigh–Ritz methods are finite-dimensional restrictions of the same energy principle.

4 Common mistakes

• Replacing (Dw′′)′′ by Dw′′′′ when D varies with x. That simplification is valid only for constant flexural rigidity.
• Imposing zero slope at a simple support. A simple support fixes displacement but ordinarily leaves rotation free.
• Imposing w′′ = w′′′ = 0 at every endpoint. Those are free-end natural conditions for a uniform beam with no endpoint loads, not universal beam conditions.
• Forgetting endpoint work. A tip force or moment changes the natural boundary condition even when the interior loading is zero.
• Using the wrong sign convention for shear or moment without declaring it. The variational derivation is internally consistent, but engineering sign conventions differ across texts.
• Building natural conditions into every Ritz trial function. Trial functions must satisfy essential conditions; natural conditions generally follow from stationarity.
• Assuming Euler–Bernoulli theory is exact for every beam. Short, thick beams or cases with important transverse shear require more refined theories such as Timoshenko beam theory.

Summary

For the Euler–Bernoulli potential

(98)

stationarity gives

(99)

For constant rigidity,

(100)

The associated boundary term is

(101)

This compact expression contains the essential/natural boundary-condition logic for clamped, simply supported, free, elastically restrained, force-loaded, and moment-loaded endpoints. It is one of the most useful physical examples of the higher-order calculus of variations.

References