A slender steel column buckles elastically. Which of the following changes would increase its critical load?

Replacing pinned ends with fixed ends, everything else unchanged

Redistributing the same cross-sectional area further from the centroid

Adding a lateral brace at mid-height

Replacing pinned ends with fixed ends, everything else unchanged

Heat-treating the steel to double its yield strength

Redistributing the same cross-sectional area further from the centroid

Adding a lateral brace at mid-height

A solid circular shaft is bored out so that the inner diameter is 0.6 times the outer, the outer diameter being unchanged. Its torsional strength and its weight, as fractions of the original, are respectively

0.8704 and 0.8704 — both scale with `J`

A beam of rectangular cross-section carries a transverse shear force V. The maximum transverse shear stress in the section is

`V/A`, uniformly across the depth

`1.5 V/A`, at the extreme fibres

`1.33 V/A`, at the neutral axis

Mechanics of Materials: Principal Stresses, Bending, Torsion and Buckling

This is the heaviest sub-item of Section 2 and the one whose formulae are most often remembered without their assumptions, which is exactly how marks get lost. Everything here is built from one idea: a load produces a stress, the stress produces a strain, and the constant between them is a property of the material and not of the part. For an isotropic material only two of the four elastic constants are independent, so E = 2G(1 + ν) = 3K(1 − 2ν), and any question giving you three of them is giving you a check rather than data. Principal stresses are the same state of stress read on rotated axes — Mohr’s circle is not an approximation but a picture of the exact algebra, with centre (σ_x + σ_y)/2 and radius √(((σ_x − σ_y)/2)² + τ_xy²). Bending gives σ = My/I, whose useful form is σ_max = M/Z with Z = I/y_max, and the trap lives in the plural: a section that is not symmetric about its neutral axis has two different section moduli, and the smaller one governs. Torsion is the same equation in polar coordinates, τ = Tr/J, and its most useful consequence is that material near the axis earns almost nothing — which is why hollow shafts win. Thin cylinders carry twice as much hoop as longitudinal stress, and the maximum shear stress in one is the question most often answered wrongly, because the third principal stress is zero and cannot be ignored. Columns fail by P_cr = π²EI/L_e², in which the material’s strength never appears at all — buckling is a stiffness failure, and the end conditions move the answer by a factor of sixteen.

Two Independent Constants, and a Circle That Is Not an Approximation

An isotropic elastic material has four constants in common use — E, G, K and ν — and exactly two degrees of freedom, so the other two are always derivable. The relations E = 2G(1 + ν), E = 3K(1 − 2ν) and E = 9KG/(3K + G) are three views of the same fact. Put E = 200 GPa and ν = 0.3 through them: G = 76.923 GPa, K = 166.667 GPa, and the third relation returns 200.000 GPa exactly, which is how you check your own arithmetic in an exam. The bounds on ν are worth reading off the same relations rather than memorising: K → ∞ as ν → 0.5, which is incompressibility, and K would go negative beyond it; ν = −1 would make G infinite. So −1 < ν < 0.5 for any real isotropic solid, and a question quoting ν = 0.6 is testing whether you noticed.

🎯 Mohr’s circle, worked once, with the 3D correction attached

Mohr’s circle is the whole of plane-stress transformation drawn once: plot the normal stress along one axis and the shear along the other, and every rotated axis system is a point on one circle. Take σ_x = 100 MPa, σ_y = −40 MPa, τ_xy = 50 MPa. The centre is the average, (100 − 40)/2 = 30 MPa, and the radius is √(70² + 50²) = 86.023 MPa. So the principal stresses are 116.023 and −56.023 MPa, and the principal plane is at θ_p = ½ arctan(2 × 50 / 140) = 17.769°. The in-plane maximum shear stress is the radius itself, 86.023 MPa, on planes 45° from the principal ones. Now the part that separates a 2-mark answer from a wrong one. This is plane stress, so the third principal stress is σ_3 = 0, and it is a genuine principal stress, not an absence of one. The absolute maximum shear stress in the material is (σ_max − σ_min)/2 taken over all three: here the three are 116.023, 0 and −56.023, so the extremes are 116.023 and −56.023 and the answer is 86.023 MPa — the in-plane value happens to be right, and it is right because σ₁ and σ₂ straddle zero. Change σ_y to +40 and the story changes completely: the principal stresses become 100 and 40 with σ_3 = 0 still there, the in-plane maximum shear is 30 MPa, and the absolute maximum is (100 − 0)/2 = 50 MPa. A candidate who reports the radius every time gets one of those two questions wrong.

Elastic constants at E = 200 GPa, and what each ν does
ν	`G = E/2(1+ν)`	`K = E/3(1−2ν)`	Reading
0	100 GPa	66.7 GPa	Cork-like: no lateral contraction
0.3	76.923 GPa	166.667 GPa	Steel — the default in every question
0.5	66.7 GPa	∞	Incompressible — rubber approaches this
> 0.5	positive	negative	Impossible for a real isotropic solid

Bending and Torsion: One Equation, Two Geometries, and the Plural in “Section Moduli”

The flexure formula σ/y = M/I = E/R assumes a lot and each assumption is examinable: plane sections stay plane, the material is linear-elastic, the beam bends about a principal axis of its cross-section, and the bending is pure. Under those, stress varies linearly from zero at the neutral axis — which passes through the centroid — to a maximum at the extreme fibre, and the design form is σ_max = M/Z with Z = I/y_max. Here is where a symmetric section flatters you: for a rectangle or a symmetric I-beam there is one Z and one answer. A T-section has two. Take a flange 90 mm × 30 mm sitting on a web 30 mm × 90 mm. Both areas are 2700 mm², so the centroid sits exactly halfway between their centroids, 75 mm from the bottom of a 120 mm section. Then I = 6.885 × 10⁶ mm⁴, and because y_top = 45 while y_bot = 75, Z_top = 153 000 mm³ and Z_bot = 91 800 mm³ — a ratio of exactly 5:3. The smaller section modulus governs, so on a sagging span the bottom fibre reaches its limit first: at 150 MPa, M = 13.77 kN·m, and the top fibre is then sitting at only 13.77 × 10⁶/153 000 = 90 MPa, well inside its allowance. Turn the beam over — or make the moment hogging — and the two swap, which is why a T-section is a deliberate choice and not just a shape.

The T-section worked end to end: flange 90×30 over web 30×90 mm
Quantity	Value	Where it comes from
Area	5400 mm²	2700 + 2700 — equal, which is why the arithmetic is clean
Centroid from bottom	75 mm	Midway between 45 and 105 because the areas are equal
`I` about the neutral axis	6.885 × 10⁶ mm⁴	2 632 500 + 4 252 500, each with its `A d²` term
`Z_top`	153 000 mm³	`I / 45`
`Z_bot`	91 800 mm³ — governs	`I / 75`; smaller `Z`, larger stress
Moment at 150 MPa	13.77 kN·m	`150 × 91 800`, using the smaller `Z`
Central load, 4 m simple span	13.77 kN	`M = WL/4`, so `W = 4M/L`
Top-fibre stress at that load	90 MPa	`M/Z_top` — 60 MPa of unused capacity

🎯 Deflection of beams is a second limit state, and it is not the one bending stress checks

A beam has to be strong enough and stiff enough, and those are two separate checks against two separate criteria: strength against an allowable stress, stiffness against an allowable deflection of beams usually quoted as a fraction of span. Integrating the flexure relation EI d²y/dx² = M twice gives the standard cases — a simply supported span with a central point load deflects WL³/48EI at midspan, a cantilever with an end load deflects WL³/3EI at the tip, and a simply supported span under a uniform load deflects 5wL⁴/384EI. Take the T-section beam already worked above at its full allowable moment: W = 13.77 kN at the centre of a 4 m span, E = 200 GPa, I = 6.885 × 10⁶ mm⁴. Then δ = 13 770 × 4000³/(48 × 200 000 × 6.885 × 10⁶) = 13.33 mm, which is exactly L/300 — and L/300 to L/360 is the usual serviceability limit, so this section arrives at both its strength limit and its stiffness limit at the same load. That coincidence is a designed one and it is worth noticing: a section that yields long before it sags is wasting stiffness, and one that sags long before it yields is wasting strength. Two comparisons that carry more than the formulae. First, the same load on a cantilever of the same length deflects sixteen times as far — WL³/3EI against WL³/48EI — which is why a cantilever is almost always governed by deflection and almost never by stress. Second, equal bending moment does not mean equal deflection: replace the point load with the uniform load that produces the identical midspan moment (w = 2W/L = 6.885 N/mm, since wL²/8 = WL/4) and the deflection rises to 16.67 mm, a quarter more. Moment is a local quantity and deflection is an integral of curvature over the whole span, so the load distribution matters to one and not the other.

🧠 Why hollow shafts win, in one line of arithmetic

Torsion of a circular shaft gives τ = Tr/J with J = π d⁴/32, so the stress is proportional to the distance from the axis and the material at the centre carries essentially nothing while weighing as much as the material at the surface. Quantify it. Bore out a shaft so that d_i/d_o = 0.6, keeping the outside diameter the same. The polar moment falls to 1 − 0.6⁴ = 0.8704 of the solid value, but the cross-sectional area — and so the weight — falls to 1 − 0.6² = 0.64. You have kept 87 per cent of the torque capacity for 64 per cent of the material, a 0.8704/0.64 = 1.36× improvement in strength per unit weight, and the shaft is stiffer per kilogram in the same ratio. This is the whole reason drive shafts, bicycle frames and aircraft spars are tubes. Two limits to keep honest about: the argument works for torsion and bending, where stress grows with distance from the neutral axis, and not for direct tension, where stress is uniform and a hollow bar is simply a weaker bar. And thin walls introduce a failure mode the formula does not see — local buckling — which is why the ratio is bored to 0.6 and not to 0.95.

Thin Cylinders, Columns and Thermal Stress: Three Places Intuition Misfires

A thin-walled cylinder under internal pressure carries a hoop stress σ_h = pd/2t and a longitudinal stress σ_l = pd/4t, exactly half of it, which is why pressure vessels split along their length and not around their girth. With p = 2 MPa, d = 1 m and t = 10 mm: σ_h = 100 MPa and σ_l = 50 MPa. Now the question that catches people. The in-plane maximum shear stress is (100 − 50)/2 = 25 MPa, but the radial principal stress at the outer surface is zero, and that is the third principal stress. The absolute maximum shear stress is therefore (100 − 0)/2 = 50 MPa — double the in-plane figure, and it is the one a Tresca yield check needs. On strains, the two are coupled by Poisson: ε_h = (σ_h − νσ_l)/E = (100 − 15)/200 000 = 4.25 × 10⁻⁴ and ε_l = (50 − 30)/200 000 = 1.00 × 10⁻⁴. The volumetric strain is 2ε_h + ε_l — two hoop strains, because the diameter change enters the volume twice — giving 9.50 × 10⁻⁴.

🎯 Buckling has no yield stress in it, and that is the whole point

Euler’s critical load is P_cr = π²EI/L_e². Read what is absent: the yield strength of the material never appears. A slender column does not fail because the stress reached a limit; it fails because a straight configuration stopped being the stable one. So heat-treating a long strut to double its yield strength buys nothing at all against buckling, while making it slightly fatter — I grows as the fourth power of a diameter — buys a great deal. With L = 3 m, E = 200 GPa and I = 4 × 10⁶ mm⁴, a pin-ended column goes at 877.3 kN. Change only the ends: fixing both halves the effective length and so quadruples the load to 3509 kN; fixed-pinned gives L_e = 0.7L and 2.04×; and a fixed-free cantilever column has L_e = 2L and takes only 219.3 kN, a quarter of the pinned case. Sixteen-fold between the best and worst end conditions on the same piece of steel, which is why a question that does not state the end conditions is incomplete rather than merely terse. One honesty note: Euler applies only while the column is slender enough that P_cr/A stays below the yield stress. Below that slenderness the column squashes rather than buckles, and the design curve becomes Rankine or a code formula — a real limit, not a footnote.

Euler buckling, L = 3 m, E = 200 GPa, I = 4 × 10⁶ mm⁴
End conditions	`L_e`	`P_cr`	Relative to pinned–pinned
Both ends fixed	0.5 L	3509.2 kN	4.00×
Fixed–pinned	0.7 L	1790.4 kN	2.04×
Both ends pinned	1.0 L	877.3 kN	1.00× — the reference
Fixed–free (flagpole)	2.0 L	219.3 kN	0.25×

⚠️ Thermal stress does not care how long the bar is

Thermal stress is produced not by heating but by heating something that is not allowed to move. Heat a bar that is free to expand and it develops strain αΔT and zero stress. Heat a bar that is fully restrained at both ends and it develops zero strain and a stress σ = αEΔT, compressive on heating. With α = 12 × 10⁻⁶ /°C, E = 200 GPa and ΔT = 50°C that is 12 × 10⁻⁶ × 200 000 × 50 = 120 MPa — a serious fraction of the yield stress of mild steel, produced by a temperature change a summer afternoon can supply. The length has cancelled, which is the counter-intuitive part and the reason expansion joints are needed on short bridges as much as long ones: a longer bar wants to expand further, but it is also more compliant, and the two effects cancel exactly. What does change the answer is partial restraint. If the supports yield by δ, the strain relieved is δ/L, so σ = E(αΔT − δ/L) — and now length is back, because a fixed gap is a smaller fraction of a longer bar. That is the version a 2-mark question asks, and the way to keep it straight is to treat thermal problems as compatibility problems: write down the total movement, set it to what the supports allow, and the stress follows.

Key takeaways

Only two elastic constants are independent. E = 2G(1+ν) = 3K(1−2ν) = 9KG/(3K+G); at E = 200 GPa and ν = 0.3, G = 76.923 and K = 166.667 GPa. −1 < ν < 0.5 follows from those relations, not from memory.
In plane stress the third principal stress is zero, not absent. For σ_x = 100, σ_y = −40, τ = 50 the radius 86.023 happens to be the absolute maximum shear; for σ_y = +40 the radius is 30 and the absolute maximum is 50.
A section unsymmetric about its neutral axis has two section moduli, and the smaller governs. The T-section here has Z_top = 153 000 and Z_bot = 91 800 mm³, so at 150 MPa the top fibre still has 60 MPa spare.
Hollow beats solid in torsion and bending, never in tension. Boring to d_i/d_o = 0.6 keeps 0.8704 of J for 0.64 of the weight — 1.36× the strength per kilogram.
A thin cylinder’s absolute maximum shear is σ_h/2, not (σ_h − σ_l)/2. At p = 2 MPa, d = 1 m, t = 10 mm: hoop 100, longitudinal 50, in-plane shear 25 but absolute shear 50 MPa.
Buckling is a stiffness failure — P_cr = π²EI/L_e² contains no yield stress. Same column, 877.3 kN pinned, 3509 kN fixed, 219.3 kN fixed-free: a factor of sixteen decided entirely by the ends.
Fully restrained thermal stress σ = αEΔT is independent of length — 120 MPa for steel at ΔT = 50°C. Partial restraint restores the length dependence as σ = E(αΔT − δ/L).

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.

For a material with E = 200 GPa and Poisson’s ratio 0.3, the shear modulus in GPa is ______ (round to two decimals)
Numerical answer — type the value.
Show answer
Answer: 76.92
76.92 GPa. From E = 2G(1 + ν), G = 200/(2 × 1.3) = 76.923 GPa. Because only two elastic constants are independent, the same two inputs also fix the bulk modulus: K = E/3(1 − 2ν) = 200/(3 × 0.4) = 166.667 GPa. Cross-check with the third relation, E = 9KG/(3K + G) = 9(166.667)(76.923)/(500 + 76.923) = 200.000 GPa ✓ — worth doing, because it costs one line and catches a slipped factor of 2. The relation is also the fastest route to the bounds on ν: K → ∞ as ν → 0.5 (incompressibility, which rubber approaches) and K would turn negative beyond that, so no real isotropic solid has ν > 0.5. The common error is using E = 3G(1 + ν) or 2G(1 + 2ν); the reliable way to remember which is which is that G is always the smallest of the three moduli for ordinary materials, so any answer larger than E has a factor upside down.
At a point in plane stress, σ_x = 100 MPa, σ_y = −40 MPa and τ_xy = 50 MPa. The major principal stress in MPa is ______ (round to two decimals)
Numerical answer — type the value.
Show answer
Answer: 116.02
116.02 MPa. Mohr’s circle has centre (100 − 40)/2 = 30 MPa and radius √(((100 + 40)/2)² + 50²) = √(70² + 50²) = √7400 = 86.023 MPa. So σ₁ = 30 + 86.023 = 116.023 and σ₂ = 30 − 86.023 = −56.023 MPa, and the principal plane is at θ_p = ½ arctan(100/140) = 17.769°. Two checks that cost nothing. The sum is invariant: σ₁ + σ₂ = 60 = σ_x + σ_y ✓, which catches almost every arithmetic slip. And σ₂ should be negative here, because a 50 MPa shear on a state that is already 140 MPa apart in normal stress is more than enough to pull the smaller principal stress below zero. The most common error is subtracting inside the radius as ((σ_x + σ_y)/2)²; the radius uses the half-difference, and the sign of σ_y matters — reading it as +40 gives a radius of 53.85 and a completely different state.
A thin cylinder of 1 m internal diameter and 10 mm wall thickness carries an internal pressure of 2 MPa. The absolute maximum shear stress in the wall is
1. 25 MPa
2. 50 MPa
3. 100 MPa
4. 75 MPa
Show answer
Answer: B — 50 MPa
50 MPa. The hoop stress is pd/2t = 2 × 1000/20 = 100 MPa and the longitudinal stress is half of it, 50 MPa. Now count the principal stresses properly: at the outer surface the radial stress is zero, so the state is (100, 50, 0) — three principal stresses, not two. The absolute maximum shear stress is (σ_max − σ_min)/2 over all three, (100 − 0)/2 = 50 MPa. Option A, 25 MPa, is the in-plane value (100 − 50)/2, and it is the answer you get by treating the problem as two-dimensional — which is exactly the trap, because "thin cylinder" invites you to forget the third direction even though nothing in the problem removed it. Option C is the hoop stress itself, confusing a normal stress with a shear stress. This matters beyond bookkeeping: a Tresca yield check compares 2τ_max = 100 MPa with the yield strength, so using 25 MPa would predict a factor of safety four times too large. Von Mises on the same state gives √(100² − 100 × 50 + 50²) = 86.6 MPa, which is the more accurate criterion and still nowhere near what the in-plane reading suggests.
A T-section has a flange 90 mm wide × 30 mm deep sitting on a web 30 mm wide × 90 mm deep. Bending about the horizontal neutral axis, the section modulus that governs the design, in mm³, is ______
Numerical answer — type the value.
Show answer
Answer: 91800
91 800 mm³. Both parts have area 2700 mm², so the centroid lands exactly midway between their individual centroids at 105 mm and 45 mm — that is 75 mm from the bottom of a 120 mm deep section, and the equal areas are why this comes out without a fraction. Then I = (90 × 30³/12 + 2700 × 30²) + (30 × 90³/12 + 2700 × 30²) = 2 632 500 + 4 252 500 = 6.885 × 10⁶ mm⁴. Two section moduli follow: Z_top = I/45 = 153 000 and Z_bot = I/75 = 91 800 mm³. The smaller one governs, because the same moment produces a larger stress on the fibre further from the neutral axis, and 75 > 45. Concretely: at a 150 MPa allowable, the section carries M = 150 × 91 800 = 13.77 kN·m, at which the top fibre is at only 13.77 × 10⁶/153 000 = 90 MPa — 60 MPa of capacity left unused, which is the honest cost of an unsymmetric section under sagging moment. Answering 153 000 is the trap and it overstates the capacity by 67 per cent. Note also that the answer flips if the moment is hogging: then the top fibre is in tension and Z_top is the one being tested, which is precisely why T-sections are put into cantilevers the other way up.
A slender steel column buckles elastically. Which of the following changes would increase its critical load?
1. Replacing pinned ends with fixed ends, everything else unchanged
2. Heat-treating the steel to double its yield strength
3. Redistributing the same cross-sectional area further from the centroid
4. Adding a lateral brace at mid-height
Show answer
Answer: A — Replacing pinned ends with fixed ends, everything else unchanged; C — Redistributing the same cross-sectional area further from the centroid; D — Adding a lateral brace at mid-height
A, C and D. Read P_cr = π²EI/L_e² and ask which symbol each option touches. A halves L_e from L to 0.5L, so P_cr quadruples — from 877.3 kN to 3509 kN for the column in this chapter. C raises I without changing the area, which is the entire logic of an I-beam or a tube: the second moment counts area times distance squared, so moving material outwards buys stiffness for free. D halves the unbraced length, quadrupling P_cr for the same reason as A. B is the one that does nothing, and it is the point of the question: the yield strength does not appear in Euler’s formula at all. Buckling is a loss of stability, not an overstress — the column deflects sideways while the material is still comfortably elastic, so a stronger steel buckles at exactly the same load. Two honest riders. First, doubling the yield strength does help the short column that squashes rather than buckles, so the answer depends on the word slender in the question. Second, E also barely varies between steels — all structural steels sit near 200 GPa regardless of grade — so there is no version of "better steel" that improves elastic buckling. Geometry is the only lever.
A solid circular shaft is bored out so that the inner diameter is 0.6 times the outer, the outer diameter being unchanged. Its torsional strength and its weight, as fractions of the original, are respectively
1. 0.8704 and 0.64
2. 0.64 and 0.8704
3. 0.36 and 0.1296
4. 0.8704 and 0.8704 — both scale with J
Show answer
Answer: A — 0.8704 and 0.64
0.8704 and 0.64. Torsional strength at a given allowable stress goes with J/r_o, and since r_o is unchanged it goes with J = π(d_o⁴ − d_i⁴)/32, giving a ratio of 1 − 0.6⁴ = 0.8704. Weight goes with cross-sectional area, 1 − 0.6² = 0.64. So the shaft keeps 87 per cent of its capacity on 64 per cent of the material — strength per unit weight improves by 0.8704/0.64 = 1.36×. That single line is why drive shafts, bicycle frames and aircraft spars are tubes: in torsion and bending, stress grows with distance from the axis, so the core is nearly idle payload. Option B swaps the two, which is the answer you get by pairing the fourth power with weight — and it is worth noticing that the swap is what makes it wrong, not the numbers. Option C computes what was removed rather than what remains. Option D forgets that weight follows the second power of diameter while stiffness and strength follow the fourth; if both scaled together there would be no advantage to a tube at all, and the whole practice would be pointless. One boundary: this argument does not transfer to direct tension, where stress is uniform across the section and removing the core simply removes capacity.
A steel bar is rigidly held between two immovable supports and heated through 50°C. With α = 12 × 10⁻⁶ /°C and E = 200 GPa, the magnitude of the stress induced, in MPa, is ______
Numerical answer — type the value.
Show answer
Answer: 120
120 MPa, compressive. Full restraint means the net strain is zero, so the thermal strain αΔT is exactly cancelled by an elastic strain −σ/E, giving σ = αEΔT = 12 × 10⁻⁶ × 200 000 × 50 = 120 MPa. Heating puts the bar in compression, because it wants to grow and the supports refuse. The striking feature is what is missing: the length of the bar never appears. A 200 mm bar and a 20 m bar develop the same 120 MPa under the same 50°C rise, because the longer bar wants to expand proportionally more but is also proportionally more compliant, and the two effects cancel exactly. That is why expansion joints are needed on short bridges as well as long ones. It also puts the number in context: 120 MPa is a substantial fraction of mild steel’s yield stress, produced by a temperature swing a single day can deliver. The version that restores length dependence is partial restraint: if the supports give way by δ, the relieved strain is δ/L and σ = E(αΔT − δ/L) — so a 1 mm gap does almost nothing for a 20 m bar and removes the stress entirely from a 200 mm one. Treat these as compatibility problems, not stress problems, and both cases fall out of the same line.
A beam of rectangular cross-section carries a transverse shear force V. The maximum transverse shear stress in the section is
1. V/A, uniformly across the depth
2. 1.5 V/A, at the neutral axis
3. 1.5 V/A, at the extreme fibres
4. 1.33 V/A, at the neutral axis
Show answer
Answer: B — `1.5 V/A`, at the neutral axis
1.5 V/A at the neutral axis. The shear formula τ = VQ/Ib puts the maximum where the first moment Q of the area beyond the cut is largest, and that is the neutral axis — the opposite end of the section from where bending stress peaks, which is the fact worth carrying away. For a rectangle the distribution is parabolic, zero at both extreme fibres (there is no area beyond them, so Q = 0) and 1.5 V/A at mid-depth. Option A is the average, and averaging a parabola underestimates its peak by exactly a third here. Option C puts the peak where bending stress peaks, confusing the two distributions — they are complementary, not aligned, and that complementarity is why a short deep beam can fail in shear at the neutral axis while its extreme fibres are nowhere near yield. Option D is the correct factor for a circular section, 4/3 V/A, which is a genuinely useful pair to remember together: 1.5 for a rectangle, 1.33 for a circle, and roughly V/A_web for a thin-web I-beam, where almost all the shear is carried by the web and the flanges contribute little. The practical reading: slender beams are governed by bending, stubby ones by shear, and which regime you are in is set by the span-to-depth ratio.