Solid Mechanics: Bending, Shear Centre, Torsion, Buckling and Combined Stresses

Everything in this chapter follows from one equation and the discipline of asking what it assumes. Simple bending theory says σ/y = M/I = E/R, and the middle term rearranged to σ = M/Z is the single most used line in the Civil paper. Its assumptions are worth naming because each one is an examinable failure: the material is elastic and the same in tension and compression, the section is symmetric about the plane of loading, the beam is initially straight, and plane sections remain plane. Break the second and the neutral axis leaves the centroid; break the last and you are in the territory of deep beams, where the formula quietly stops being true. Alongside bending goes shear, and the two are read differently — bending stress is largest at the extreme fibre and zero at the neutral axis — the flexural stress — while shear does exactly the reverse, so 1.5 V/A for a rectangle is a neutral-axis quantity and the extreme fibre carries no shear at all. Then the item Civil asks and Mechanical does not: the shear centre, the point a transverse load must pass through if the member is to bend without twisting — and for a channel it is not on the section. Uniform torsion gives τ = Tr/J and θ = TL/GJ, with the second worth as much as the first, because a shaft or a curved girder is usually governed by how much it rotates. Buckling is where a column stops being a strength problem: P_cr = π²EI/L_e² contains no strength term at all, so a stronger steel buys nothing and only E, the second moment of area and the effective length matter. And combined direct and bending stress, σ = P/A ± Pe/Z, delivers the middle-third rule — the eccentricity at which one face of a pier reaches exactly zero stress, beyond which it goes into tension.

Simple Bending Theory, and Why Shear Reads Backwards

The bending equation is three quantities set equal: σ/y = M/I = E/R. Take the first pair and the stress at a fibre distance y from the neutral axis is σ = My/I, largest at the extreme fibre where y = y_max, which is why the whole calculation is usually packed into the section modulus Z = I/y_max and read as σ = M/Z. Take the second pair instead and you have the curvature, 1/R = M/EI, which is the equation every deflection calculation integrates. The practical value of writing it as σ = M/Z is that it separates the two halves of a design decision: the loading fixes M and the section fixes Z, so choosing a beam is choosing a number. Return the 8 m beam of the previous chapter, which demanded M = 165 kN·m. Against an allowable stress of 165 MPa the requirement is Z = 165 × 10⁶/165 = 1.000 × 10⁶ mm³ exactly; drop the allowable to 150 MPa and the requirement rises to 1.100 × 10⁶. Nothing about the beam changed — only what you are willing to let the steel do.

Shear stress across the same section is τ = VQ/(Ib), in which Q is the first moment of the area beyond the fibre. Its shape is the exact opposite of the bending distribution, and that is the fact worth carrying rather than the formula. At the extreme fibre there is no area beyond it, so Q = 0 and the extreme fibre carries no shear at all; at the neutral axis Q is largest and so is τ. For a rectangle the distribution is parabolic and peaks at 1.5 V/A, so the familiar V/A is a mean and understates the worst point by half. Run the numbers on a 300 × 450 section carrying the beam’s V = 73 kN: the mean is 0.541 MPa and the neutral-axis maximum is 0.811 MPa. Both are small — which is the real lesson, because for a beam of ordinary proportions bending governs and shear does not, and shear only takes over when the span gets short relative to the depth or when the web gets thin, which is exactly why a plate girder is checked for web shear and a domestic floor joist is not.

Bending and shear on a 300 × 450 rectangle, under the previous chapter’s beam forces
QuantityWhere it is largestValue
Bending stress, M = 165 kN·mExtreme fibreM/Z = 16.30 MPa
Bending stressNeutral axiszero
Shear stress, V = 73 kNNeutral axis1.5 V/A = 0.811 MPa
Shear stressExtreme fibrezero
Mean shear for comparison—V/A = 0.541 MPa

The Shear Centre, and Why It Is Off the Section

Bending theory quietly assumes the load is applied so that the member bends and does not twist. For a section with two axes of symmetry — a rectangle, a circle, a standard I-beam — that happens automatically at the centroid, and the assumption is invisible. For a section with only one axis of symmetry, or none, it does not, and the point through which a transverse load must pass to produce bending without torsion has its own name: the shear centre. It is where the resultant of the shear flow acts, and it is a property of the shape alone — of the geometry, not of the load, the material or the magnitude.

🎯 A channel, and a point 37 mm out in mid-air
Take a thin channel with flanges 100 mm wide, 200 mm between flange centrelines and a uniform thickness of 10 mm, loaded parallel to the web. Follow the shear flow. It runs along the top flange from the tip towards the web, turns down the web, and runs back out along the bottom flange — so the two flange flows are a pair of equal, opposite horizontal forces separated by the depth: a couple. The web flow carries essentially the whole vertical shear and passes through the web, so it produces no twist about the web. Something must balance the flange couple, and the only way is for the applied load to act at an offset. Computing it: I_web = 6.667 × 10⁶ and the two flanges add 2.002 × 10⁷, giving I_x = 2.668 × 10⁷ mm⁴, and e = b²h²t/(4I_x) = (100²)(200²)(10)/(4 × 2.668 × 10⁷) = 37.48 mm. That is 0.375 of the flange width, measured from the web centreline and on the side away from the flanges — outside the metal entirely, in the air. So a channel loaded down its web, which is where a real load would naturally sit, is being loaded 37 mm off its shear centre and therefore twisting, no matter how carefully the bending calculation is done. This is not a curiosity: it is why channels are used in back-to-back pairs, why a purlin gets an anti-sag tie, and why an angle used as a lintel rotates as it deflects. Two consequences to keep. For any section built of thin parts that meet at a point — an angle, a tee, a cruciform — the shear centre is at that point, because every shear flow passes through it and no couple can form. And for a section with two axes of symmetry the shear centre coincides with the centroid, which is the only reason the assumption is safe to forget for an I-beam.

Transformation of Stress, and Uniform Torsion

A state of plane stress is three numbers — σ_x, σ_y and τ_xy — and they depend on the axes you chose, not on the material. Transformation answers what the same state reads on a plane rotated by θ, and Mohr’s circle is the whole answer drawn once: centre at (σ_x + σ_y)/2, radius √[((σ_x − σ_y)/2)² + τ_xy²]. The principal stresses are the two ends of the horizontal diameter, centre ± radius, and the maximum in-plane shear is the radius itself, on planes at 45° to the principal planes — which is the single most useful geometric fact in the chapter, because it explains why a ductile bar in tension necks on a 45° cone and a brittle one snaps square across.

🧠 One transformation, and the invariant that checks it
Take σ_x = 80 MPa, σ_y = 20 MPa, τ_xy = 40 MPa. Centre = (80 + 20)/2 = 50; radius = √(30² + 40²) = 50. So σ₁ = 100 MPa and σ₂ = 0 — exactly zero, which makes this a state of pure uniaxial tension seen from the wrong axes, and a good reminder that a state with three non-zero components can be one-dimensional in disguise. The in-plane maximum shear is the radius, 50 MPa. For the angles, tan 2θ_p = 2τ_xy/(σ_x − σ_y) = 80/60, so 2θ_p = 53.13° and θ_p = 26.57°, with the plane of maximum shear 45° away at 71.57°. Now the check that costs nothing: the sum of the normal stresses is invariant under rotation, so σ₁ + σ₂ must equal σ_x + σ_y, and 100 + 0 = 80 + 20 ✓. Run it every time — it catches a sign error in τ, which is the most common slip here, in one line. And note the third principal stress: in plane stress σ₃ = 0, so the absolute maximum shear is (σ₁ − σ₃)/2 = 50 MPa here, which happens to equal the in-plane value only because σ₂ is itself zero. When both in-plane principals are positive the absolute maximum shear exceeds the in-plane one, and a candidate who reports the in-plane figure has under-estimated the demand on a ductile material.

Uniform torsion of a circular shaft gives τ/r = T/J = Gθ/L, the same three-way equality as bending and read the same way. The stress form is τ_max = Tr/J, or 16T/πd³ for a solid circle; the stiffness form is θ = TL/GJ. Take d = 100 mm, T = 10 kN·m, G = 80 GPa over L = 2 m: J = πd⁴/32 = 9.82 × 10⁶ mm⁴, so τ_max = 50.93 MPa — both formulas agreeing, as they must — and θ = 0.0255 rad = 1.46°. For a civil structure the second number is usually the one that decides, because a curved bridge girder or an eccentrically loaded beam is limited by rotation and not by torsional stress. Note the one restriction that matters: this theory is for circular sections. A non-circular section warps out of plane when twisted, plane sections do not remain plane, and J is replaced by a torsion constant that is not the polar moment of inertia — which is why an open thin-walled section is dramatically worse in torsion than its polar moment would suggest.

Buckling, and Combined Direct and Bending Stress

A short column crushes and a long one buckles, and the two are governed by different equations with nothing in common. Euler’s load is P_cr = π²EI/L_e², and the striking feature is what is missing: no strength term appears anywhere in it. A column that buckles does so at a load fixed by stiffness and geometry, so specifying a stronger grade of steel buys exactly nothing, while making the section deeper or shortening the effective length buys a great deal. L_e is the effective length, and it is where the marks are: the same 6 m column with I = 9 × 10⁷ mm⁴ and E = 200 GPa carries 4935 kN pinned at both ends, 19 739 kN fixed at both — four times as much, because the load goes as 1/L_e² and L_e halved — and only 1234 kN as a fixed-free cantilever, a quarter of the pinned value. Fixing both ends of a column is therefore worth more than any change of material.

The same 6 m column, four end conditions (E = 200 GPa, I = 9 × 10⁷ mm⁴)
End conditionsL_eP_cr (kN)Relative
Both ends pinnedL4 9351.00
Both ends fixed0.5 L19 7394.00
One fixed, one pinned0.7 L10 1002.05
Fixed at base, free at top2 L1 2340.25
⚠️ The middle-third rule, and the eccentricity at which stress reaches exactly zero
When an axial load misses the centroid it produces both direct stress and bending, and the two superpose: σ = P/A ± Pe/Z. Take a 300 × 450 pier carrying 500 kN. The direct stress is 500 000/135 000 = 3.704 MPa, uniform. Put the load 50 mm off centre along the 450 mm dimension and the bending term is 500 000 × 50/(10.125 × 10⁶) = 2.469 MPa, so the two faces read 6.173 and 1.235 MPa — both still compression. Push the eccentricity out to e = 75 mm and the bending term becomes 3.704 MPa, exactly equal to the direct stress, so one face reads 7.407 and the other exactly zero. That eccentricity is d/6, and since it applies on both sides, the load may sit anywhere in the middle third of the section without producing tension anywhere. Go to e = 100 mm and the far face reads −1.235 MPa: tension. For steel that would be unremarkable. For masonry, plain concrete or a soil interface — none of which can be relied on in tension — it is the governing criterion, and it is the reason a gravity dam is checked for no tension at the heel, a masonry pier is proportioned rather than merely sized, and a footing is kept under a resultant inside the middle third. The trap in the arithmetic is Z: it must be the section modulus about the axis the eccentricity bends the member about, so an eccentricity along the 450 mm dimension uses bd²/6 with d = 450, and the same 100 mm offset along the 300 mm dimension is a far more serious matter.

Key takeaways

  • σ/y = M/I = E/R is one equation read three ways: σ = M/Z for stress, 1/R = M/EI for curvature. The ch01 beam’s 165 kN·m needs exactly 1.0 × 10⁶ mm³ of section modulus at 165 MPa.
  • Bending and shear read backwards to each other: bending is largest at the extreme fibre and zero at the neutral axis, shear the reverse. 1.5 V/A for a rectangle is a neutral-axis value, not an average.
  • The shear centre of a channel is off the section — 37.5 mm out for the one worked here — so loading a channel down its web twists it. Thin parts meeting at a point put the shear centre at that point; two axes of symmetry put it at the centroid.
  • Mohr’s circle in one line: centre (σ_x+σ_y)/2, radius √[((σ_x−σ_y)/2)² + τ²]. Always check the invariant σ₁ + σ₂ = σ_x + σ_y — it catches a sign error in τ immediately.
  • P_cr = π²EI/L_e² contains no strength term, so stronger steel does nothing for a buckling column. Fixing both ends quarters L_e² and quadruples the load; a fixed-free column carries only a quarter of the pinned value.
  • σ = P/A ± Pe/Z gives the middle-third rule: at e = d/6 one face reaches exactly zero and beyond it goes into tension. Decisive for masonry, plain concrete and soil, all of which cannot be trusted in tension.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. A beam section must resist a bending moment of 165 kN·m at an allowable bending stress of 165 MPa. The section modulus required, in units of 10³ mm³, is ______

    Numerical answer — type the value.

    Show answer

    Answer: 1000

    1000 × 10³ mm³, that is 1.0 × 10⁶ mm³. Straight from σ = M/Z, so Z = M/σ = (165 × 10⁶ N·mm)/(165 N/mm²) = 1.000 × 10⁶ mm³. The only trap is unit hygiene, and it is where most of the marks go: a moment in kN·m must become N·mm by multiplying by 10⁶, and MPa is already N/mm², so the answer arrives in mm³ with nothing else to do. Two observations worth more than the arithmetic. First, Z is a demand, not a property of any particular beam — it tells you which sections in a handbook are admissible and says nothing about which is economical, since a deep thin section and a squat thick one can share a Z. Second, tightening the allowable stress is expensive: at 150 MPa instead of 165 the same moment needs 1.100 × 10⁶ mm³, a 10 per cent larger section modulus for a 9 per cent lower stress. And note what Z does not cover — deflection depends on I, not Z, so a section chosen purely on strength can still fail a serviceability check, which is exactly why a design proceeds through both.

  2. In a rectangular beam section under transverse loading, the shear stress is

    1. maximum at the extreme fibres and zero at the neutral axis
    2. maximum at the neutral axis and zero at the extreme fibres, peaking at 1.5 V/A
    3. uniform across the depth and equal to V/A
    4. maximum at the neutral axis and equal to V/A there
    Show answer

    Answer: B — maximum at the neutral axis and zero at the extreme fibres, peaking at `1.5 V/A`

    Maximum at the neutral axis, zero at the extreme fibres, peaking at 1.5 V/A. The distribution comes from τ = VQ/(Ib), and Q is the first moment of the area beyond the fibre being examined. At the extreme fibre there is no area beyond it, so Q = 0 and the shear stress vanishes there; at the neutral axis Q is at its largest and so is τ. Integrating the parabola over the depth must return V, which forces the peak to 1.5 times the mean — so the answer is 1.5 V/A and not V/A. Option A is the bending distribution, which reads exactly the other way round, and stating the two together is how the pair is best remembered. Option C describes the assumption made in a rivet or a bolt in direct shear, where the section is small enough that a uniform average is the right idealisation; a beam web is not that case. Option D gets the location right and the magnitude wrong by a factor of 1.5, which is the answer that costs a mark most often. On the 300 × 450 section carrying V = 73 kN the mean is 0.541 MPa and the true maximum 0.811 MPa — and the fact that both are small is why bending, not shear, sizes an ordinary beam.

  3. For a channel section loaded parallel to its web, the shear centre lies

    1. at the centroid of the section
    2. on the web, midway between the flanges
    3. outside the section, on the side of the web away from the flanges
    4. at the junction of the web and either flange
    Show answer

    Answer: C — outside the section, on the side of the web away from the flanges

    Outside the section, on the far side of the web. Trace the shear flow: it runs inwards along one flange, down the web, and outwards along the other, so the two flange flows form a couple of equal and opposite horizontal forces separated by the depth, while the web carries nearly all the vertical shear. For the member to bend without twisting, the applied load must produce a moment that cancels that flange couple, and the only way is to act at an offset. For flanges 100 mm wide with 200 mm between flange centrelines and 10 mm thickness, I_x = 2.668 × 10⁷ mm⁴ and e = b²h²t/(4I_x) = 37.5 mm from the web centreline — about 0.375 of the flange width, and clear of any metal. Option A is the correct answer for a doubly symmetric section only, and the reason the concept can be ignored for an I-beam. Option B is where a real load actually sits, which is precisely the problem: a channel loaded on its web is loaded 37.5 mm off its shear centre and therefore twists. Option D is right for a different family of sections — an angle, a tee, a cruciform, anything built of thin parts meeting at a point, where every shear flow passes through that point and no couple can form. The practical consequence is visible on site: channels get used in back-to-back pairs, purlins get anti-sag ties, and an angle lintel rotates as it deflects.

  4. At a point, σ_x = 80 MPa, σ_y = 20 MPa and τ_xy = 40 MPa. The major principal stress, in MPa, is ______

    Numerical answer — type the value.

    Show answer

    Answer: 100

    100 MPa. Centre of Mohr’s circle = (80 + 20)/2 = 50; radius = √[((80 − 20)/2)² + 40²] = √(900 + 1600) = √2500 = 50. So σ₁ = 50 + 50 = 100 MPa and σ₂ = 50 − 50 = 0 — the minor principal stress is exactly zero, which means this apparently three-component state is uniaxial tension viewed from rotated axes. Check the invariant before writing anything down: σ₁ + σ₂ must equal σ_x + σ_y, and 100 + 0 = 100 ✓. That one line catches the commonest error here, a sign slip on τ, which would give a radius of the same magnitude but principal directions on the wrong side. The principal plane is at θ_p = ½ arctan[2(40)/(80 − 20)] = ½ arctan(1.333) = 26.57°, and the plane of maximum in-plane shear is 45° from it, at 71.57°, carrying τ = 50 MPa. One more thing the numbers make easy to see: since this is plane stress the third principal stress is σ₃ = 0, so the absolute maximum shear is (σ₁ − σ₃)/2 = 50 MPa, equal to the in-plane figure only because σ₂ happens to be zero. When both in-plane principals come out positive, the absolute maximum shear is larger than the in-plane one, and quoting the in-plane value then under-estimates the demand on a ductile material.

  5. Two identical columns differ only in their end conditions: one is pinned at both ends, the other fixed at both. The ratio of their Euler buckling loads (fixed to pinned) is

    1. 2
    2. 4
    3. 1, since the material and section are identical
    4. √2
    Show answer

    Answer: B — 4

    4. P_cr = π²EI/L_e², and fixing both ends halves the effective length from L to 0.5L. Since the load goes as 1/L_e², halving L_e quadruples P_cr. On the worked column — L = 6 m, E = 200 GPa, I = 9 × 10⁷ mm⁴ — that is 4935 kN pinned against 19 739 kN fixed, and the ratio is exactly 4.000. Option A is the answer you get by remembering that the load depends on effective length without remembering that it depends on the square of it. Option D is the same slip in the other direction. Option C is the trap worth understanding rather than merely avoiding, and it is subtler than it looks: the material and the section really are identical, and P_cr really does contain no strength term — but the end conditions change the buckled shape, and therefore the length over which the column is free to bow, which is what L_e measures. The same reasoning ranks the whole family: 0.5L fixed-fixed, 0.7L fixed-pinned, L pinned-pinned, 2L fixed-free. That last is worth its own note, because a cantilever column carries only a quarter of the pinned value — so a flagpole-like column with an unrestrained top is the worst arrangement available, and providing any lateral restraint at the top is the cheapest strengthening there is.

  6. A short masonry pier 300 mm × 450 mm carries a vertical load of 500 kN. The maximum eccentricity, measured along the 450 mm dimension, at which no tension develops anywhere in the section, in mm, is ______

    Numerical answer — type the value.

    Show answer

    Answer: 75

    75 mm, which is d/6. No tension requires the bending term to be no larger than the direct term: Pe/Z ≤ P/A, so e ≤ Z/A, and for a rectangle Z/A = (bd²/6)/(bd) = d/6 = 450/6 = 75 mm. Notice that P cancels — the limit is a property of the section alone, so the 500 kN in the question is deliberately irrelevant, and a candidate who tries to use it has misread what is being asked. The numbers make it concrete. Direct stress is 500 000/135 000 = 3.704 MPa. At e = 50 mm the bending term is 2.469 MPa and the two faces read 6.173 and 1.235 MPa, both compression. At e = 75 mm the bending term is 3.704 MPa, exactly equal to the direct stress, so one face reads 7.407 MPa and the other exactly zero. Past that, at e = 100 mm, the far face reads −1.235 MPa — tension. Since the same limit applies on both sides, the load may sit anywhere within the middle third of the section, which is where the rule gets its name. It is decisive precisely where tension cannot be relied on: masonry, plain concrete, and the interface between a footing and the soil, which is why a gravity dam is checked for no tension at the heel and a footing is proportioned to keep the resultant inside the middle third. One trap in the arithmetic: Z must be taken about the axis the eccentricity bends the pier about, so the same 75 mm along the 300 mm dimension would instead be 300/6 = 50 mm.

  7. A solid circular shaft of 100 mm diameter transmits a torque of 10 kN·m. The maximum shear stress, in MPa, is ______ (round to two decimals)

    Numerical answer — type the value.

    Show answer

    Answer: 50.93

    50.93 MPa. Either route works and getting the same answer twice is the check. Polar second moment J = πd⁴/32 = π(100)⁴/32 = 9.817 × 10⁶ mm⁴, so τ_max = Tr/J = (10 × 10⁶)(50)/(9.817 × 10⁶) = 50.93 MPa. Or straight to the closed form for a solid circle, τ_max = 16T/(πd³) = 16(10 × 10⁶)/(π × 10⁶) = 50.93 MPa ✓. Two things worth as much as the number. First, the stiffness side of the same equality: θ = TL/GJ, which at G = 80 GPa over a 2 m length is 0.0255 rad = 1.46°. For most civil applications — a curved bridge girder, an eccentrically loaded edge beam — that rotation is what governs the design, not the stress, so it is worth computing as a habit rather than only when asked. Second, the restriction: this theory holds for circular sections, where plane sections remain plane and radii stay straight. A non-circular section warps when twisted, so J is replaced by a torsion constant that is not the polar moment, and for an open thin-walled section — a channel, an I — that constant is very much smaller than the polar moment would suggest. Using πd⁴/32 reasoning on a rolled I-section overestimates its torsional capacity by an order of magnitude, which is why torsion in steel structures is designed out by closing the section rather than resisted.

  8. Which of the following statements about the Euler buckling load of a column are correct? (One or more)

    1. It is independent of the yield strength of the material
    2. It is inversely proportional to the square of the effective length
    3. For a given area, it is larger for a section with the material further from the centroid
    4. It governs the design of short, stocky columns
    Show answer

    Answer: A — It is independent of the yield strength of the material; B — It is inversely proportional to the square of the effective length; C — For a given area, it is larger for a section with the material further from the centroid

    A, B and C. Read P_cr = π²EI/L_e² and check each claim against it. A is right and is the most consequential fact in the chapter: no strength term appears in the expression at all, so a higher grade of steel raises the crushing capacity and does nothing whatever for the buckling load — money spent on strength is money wasted on a slender column. B is right, and the square is what makes end restraint so valuable: fixing both ends halves L_e and so quadruples P_cr, from 4935 kN to 19 739 kN on the column worked in this chapter. C is right and is how sections are actually chosen: I = Σ A y², so for a fixed area, pushing material away from the centroid raises I — which is why a hollow tube beats a solid rod of the same area, and why a built-up column is spread apart with lacing rather than bundled. D is false, and it is the inversion the question is testing. A short stocky column does not buckle; it crushes, and its capacity is σ_y A, a pure strength calculation. Euler governs only once the member is slender enough that the buckling load falls below the crushing load, and the slenderness at which the two coincide is the transition point. Real design curves join the two branches with an empirical fit — Rankine-Gordon, or the code’s buckling curves — precisely because neither pure theory describes an intermediate column, where initial crookedness and residual stresses put the true capacity below both.