Mechanics of Materials II: Shear Force and Bending Moment, Bending and Shear Stresses, Shear Centre, Deflection, Torsion, Columns and Castigliano’s Theorems
1. Shear force and bending moment diagrams
For a beam with distributed load w(x) downward, equilibrium of a slice gives dV/dx = −w and dM/dx = V: the slope of the shear-force diagram is the (negative) load intensity and the slope of the bending-moment diagram is the shear force. So the moment is largest where V crosses zero, a point load makes V jump by its size and leaves M with a change of slope, and an applied couple makes M jump. The area under the V diagram between two sections is the change in M. A point of contraflexure is where M changes sign, and there the beam’s curvature reverses.
| Beam and load | Maximum moment | Where |
|---|---|---|
| Simply supported, central point load P | PL/4 | midspan |
| Simply supported, uniform load w per unit length | wL²/8 | midspan |
| Cantilever, end load P | PL | fixed end |
| Cantilever, uniform load w | wL²/2 | fixed end |
A simply supported beam of 6 m span with 10 kN/m has reactions of 30 kN each, V falling linearly from +30 to −30 kN, and M_max = 10 × 6²/8 = 45 kN·m at midspan. Draw the diagrams from the left, taking a section at x and writing V and M from the forces on one side; the shear diagram is a straight line for a uniform load and the moment diagram a parabola.
2. Bending and shear stresses, and the shear centre
Plane sections stay plane in pure bending, so the strain varies linearly through the depth and the neutral axis passes through the centroid. The flexure formula is M/I = σ/y = E/R: σ = My/I, largest at the extreme fibre, with the section modulus Z = I/y_max. I = bh³/12 for a rectangle about its centroidal axis and πd⁴/64 for a circle. For a 100 mm wide, 200 mm deep rectangle, I = 6.667 × 10⁷ mm⁴, and a moment of 20 kN·m gives σ_max = 20 × 10⁶ × 100/(6.667 × 10⁷) = 30 MPa.
The transverse shear stress at a level y₁ is τ = V·Aȳ/(I b), where Aȳ is the first moment about the neutral axis of the area beyond the level and b the width there. It is zero at the top and bottom fibres and greatest at the neutral axis: τ_max = 1.5 V/A for a rectangle (parabolic), 4V/(3A) for a solid circle, and for an I-section the web carries most of the shear, close to V/(web area). For V = 30 kN on the 100 × 200 mm section, τ_max = 1.5 × 30 000/20 000 = 2.25 MPa. Bending stress is greatest where shear is zero and the reverse.
The shear centre is the point in a cross-section through which a transverse load must act for the beam to bend without twisting. It lies on any axis of symmetry, and at the centroid for a doubly symmetric section such as an I-section or a rectangle. For an unsymmetric section such as a channel it does not coincide with the centroid: the shear flow in the flanges and web has a resultant that acts at a distance outside the web, on the side away from the flanges’ tips, and a load through the centroid would twist the channel. For an angle or a tee it lies at the junction of the legs.
3. Deflection of beams
The elastic curve obeys EI d²y/dx² = M(x) for small slopes. Integrate twice, fitting the constants to the supports (zero deflection at a pin, zero slope and deflection at a built-in end); Macaulay’s method writes one expression across point loads, and moment-area, conjugate-beam and superposition give the same results. The results worth memorising: cantilever, end load P: δ = PL³/(3EI), θ = PL²/(2EI); cantilever, uniform load: δ = wL⁴/(8EI); simply supported, central load: δ = PL³/(48EI); simply supported, uniform load: 5wL⁴/(384EI).
For a 2 m cantilever with a 1 kN tip load, E = 200 GPa and I = 4 × 10⁻⁶ m⁴, EI = 8 × 10⁵ N·m² and δ = 1000 × 2³/(3 × 8 × 10⁵) = 3.33 × 10⁻³ m = 3.33 mm. Because the deflection scales with L³ (end load) or L⁴ (uniform load), doubling a cantilever’s length multiplies the tip deflection by 8 (or 16); because it falls as 1/I and I as h³, doubling the depth of a rectangular section cuts it eightfold. Superposition adds the deflections of separate loads on a linear beam.
4. Torsion of circular shafts
For a circular shaft in torsion, plane sections remain plane and radii remain straight, so the shear strain grows linearly with radius and T/J = τ/r = Gθ/L. τ = Tr/J is zero on the axis and greatest at the surface, the polar moment is J = πd⁴/32 for a solid shaft and π(D⁴ − d⁴)/32 for a hollow one, and the angle of twist is θ = TL/(GJ), the ratio GJ/L being the torsional stiffness. For a solid shaft of 50 mm diameter carrying 1 kN·m, τ_max = 16T/(πd³) = 16 × 1000/(π × 0.05³) = 40.7 × 10⁶ Pa = 40.7 MPa.
The power transmitted is P = Tω = 2πNT/60 with N in rpm: 200 N·m at 1500 rpm is 2π × 1500 × 200/60 = 31.4 kW. The same torque at half the speed needs twice the shaft strength for the same power. A hollow shaft is more efficient per unit mass than a solid one, because the material near the axis is barely stressed. Shafts in series carry the same torque and add their twists; shafts in parallel share the same twist and add their stiffnesses, so the torque divides in proportion to GJ/L.
5. Euler’s theory of columns and Castigliano’s theorems
A slender column loaded axially buckles at the Euler critical load P_cr = π²EI/L_e², where the effective length L_e depends on the ends: L for pinned–pinned, 0.5L for fixed–fixed, 0.7L for fixed–pinned and 2L for fixed–free. The buckling load is inversely proportional to L_e², so a fixed–fixed column carries (2L/0.5L)² = 16 times the load of a fixed–free one. With E = 200 GPa, I = 2 × 10⁻⁶ m⁴ and L = 4 m pinned at both ends, P_cr = π² × 200 × 10⁹ × 2 × 10⁻⁶/16 = 246.7 kN. Buckling happens about the axis of the least I, and the slenderness ratio is L_e/r with r = √(I/A); Euler’s formula holds for large slenderness ratios and overestimates the strength of short columns, which fail by yielding first.
Castigliano’s first theorem: the deflection of a point in the direction of a load P is the partial derivative of the strain energy with respect to that load, δ = ∂U/∂P, with U = ∫M²/(2EI)dx + ∫T²/(2GJ)dx + ∫F²/(2AE)dx (bending, torsion, axial; shear usually negligible). For a cantilever with a tip load, M = −Px, U = ∫₀ᴸ P²x²/(2EI)dx = P²L³/(6EI), and δ = ∂U/∂P = PL³/(3EI), the known result. Where no load acts at the point wanted, add a dummy load Q, differentiate and set Q = 0. Castigliano’s second theorem (least work): for a statically indeterminate structure the redundant reaction R makes the strain energy a minimum, ∂U/∂R = 0 when the corresponding displacement is zero. For a cantilever of length L fixed at A with a prop at B under a uniform load w, this gives R_B = 3wL/8, so 24 kN for w = 16 kN/m and L = 4 m.
Key takeaways
- dV/dx = −w, dM/dx = V; M_max is at V = 0; simply supported with uniform load: wL²/8; central load: PL/4.
- σ = My/I; τ_max = 1.5V/A (rectangle), 4V/3A (circle); the shear centre of a channel lies outside its web.
- EI y″ = M; cantilever PL³/3EI, simply supported central PL³/48EI, uniform load 5wL⁴/384EI.
- Torsion: T/J = τ/r = Gθ/L; P = 2πNT/60; hollow shafts are more efficient per unit mass.
- Euler: P_cr = π²EI/L_e² with L_e = L, 0.5L, 0.7L, 2L; Castigliano: δ = ∂U/∂P, and ∂U/∂R = 0 for a redundant R.
Practice questions (16)
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 beam carrying a distributed load w(x), the relation between the shear force V and the bending moment M is:
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Answer: A — dM/dx = V
The slope of the bending-moment diagram is the shear force, so the moment is extreme where V = 0, while dV/dx = −w gives the load. The other options mix the two relations.A simply supported beam of span 6 m carries a uniformly distributed load of 10 kN/m over its full length. The maximum bending moment is ____ kN·m.
Numerical answer — type the value.
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Answer: 45
M_max = wL²/8 = 10 × 36/8 = 45 kN·m at midspan, where V = 0. wL²/2 = 180 is the cantilever’s value, and PL/4 has no meaning for a distributed load.A point of contraflexure in a beam is a point where:
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Answer: A — the bending moment changes sign
At a point of contraflexure M passes through zero and the curvature reverses from sagging to hogging. Zero shear marks a maximum or minimum of M, not necessarily a zero of it.A beam of rectangular cross-section 100 mm wide and 200 mm deep carries a bending moment of 20 kN·m. The maximum bending stress is ____ MPa.
Numerical answer — type the value.
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Answer: 30
I = bh³/12 = 100 × 200³/12 = 6.667 × 10⁷ mm⁴ and y_max = 100 mm, so σ = My/I = 20 × 10⁶ × 100/(6.667 × 10⁷) = 30 MPa. Swapping b and h, so that the beam is 200 mm wide and 100 mm deep, gives 60 MPa.The rectangular beam of 100 mm width and 200 mm depth carries a transverse shear force of 30 kN. The maximum shear stress in the section is ____ MPa.
Numerical answer — type the value.
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Answer: 2.25
τ_max = 1.5V/A = 1.5 × 30 000/(100 × 200) = 2.25 MPa, at the neutral axis. The average V/A = 1.5 MPa understates it by a third.The shear centre of a channel section lies:
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Answer: A — outside the section, on the side of the web away from the flanges
The shear flow in the two flanges forms a couple about the web, so the resultant shear force must act at an offset from the web, outside the section. A load through the centroid would twist the channel. Only a doubly symmetric section has its shear centre at the centroid.A cantilever of length 2 m carries a point load of 1 kN at its free end. Given E = 200 GPa and I = 4 × 10⁻⁶ m⁴, the tip deflection is ____ mm.
Numerical answer — type the value.
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Answer: 3.33
δ = PL³/(3EI) = 1000 × 8/(3 × 200 × 10⁹ × 4 × 10⁻⁶) = 8000/(2.4 × 10⁶) = 3.33 × 10⁻³ m = 3.33 mm. PL³/(48EI) belongs to a simply supported beam and would give 0.21 mm.The length of a cantilever carrying a tip load is doubled with the same section and material. The tip deflection becomes:
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Answer: A — 8 times
δ = PL³/(3EI) varies as L³, so 2³ = 8. Sixteen times would be the result for a uniformly distributed load, which varies as L⁴ (with the same w).A solid circular shaft of diameter 50 mm transmits a torque of 1 kN·m. The maximum shear stress at its surface is ____ MPa.
Numerical answer — type the value.
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Answer: 40.7
τ = 16T/(πd³) = 16 × 10⁶ N·mm/(π × 50³ mm³) = 16 × 10⁶/(3.927 × 10⁵) = 40.7 MPa. Using 32T/(πd³) (the bending-type constant) would double it.A shaft transmits a torque of 200 N·m at 1500 rpm. The power transmitted, correct to one decimal place, is ____ kW.
Numerical answer — type the value.
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Answer: 31.4
P = 2πNT/60 = 2π × 1500 × 200/60 = 31 416 W = 31.4 kW. Using N directly as ω (1500 rad/s) would give 300 kW, mixing rpm with rad/s.A slender steel column pinned at both ends has length 4 m, E = 200 GPa and least second moment of area I = 2 × 10⁻⁶ m⁴. The Euler buckling load, to the nearest kilonewton, is ____ kN.
Numerical answer — type the value.
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Answer: 247
P_cr = π²EI/L² = 9.8696 × 200 × 10⁹ × 2 × 10⁻⁶/16 = 9.8696 × 4 × 10⁵/16 = 246.7 × 10³ N, so 247 kN. Taking L_e = 2L (a fixed–free end condition) would give a quarter of it, 61.7 kN.The ratio of the Euler buckling load of a column fixed at both ends to that of an otherwise identical column fixed at one end and free at the other is ____.
Numerical answer — type the value.
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Answer: 16
P_cr ∝ 1/L_e². The fixed–fixed effective length is 0.5L and the fixed–free one is 2L, so the ratio is (2L/0.5L)² = 4² = 16. Comparing the effective lengths themselves, without the square, gives 4.A cantilever of length 4 m fixed at A carries a uniformly distributed load of 16 kN/m and is propped at its free end B. Using Castigliano’s second theorem (zero deflection at B), the prop reaction is ____ kN.
Numerical answer — type the value.
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Answer: 24
With the prop reaction R, M(x) = Rx − wx²/2 measured from B, and ∂U/∂R = ∫₀ᴸ M(∂M/∂R)/EI dx = 0 gives RL³/3 = wL⁴/8, so R = 3wL/8 = 3 × 16 × 4/8 = 24 kN. Equating the prop to the whole load wL/2 = 32 kN ignores the fixed end’s share.Which of the following statements about Euler’s theory of columns are true?
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Answer: A — The critical load is proportional to EI and inversely proportional to the square of the effective length; B — The effective length of a fixed–free column is twice its actual length; C — Euler’s formula overestimates the load capacity of short columns
(A) P_cr = π²EI/L_e². (B) L_e = 2L for fixed–free. (C) A short column fails by yielding at a load below the Euler value, and the formula gives a stress above the yield. (D) False: elastic buckling depends on stiffness EI, not strength.Which of the following statements about the torsion of circular shafts are true?
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Answer: A — The shear stress varies linearly with radius, being zero on the axis; B — For the same mass and length a hollow shaft can carry more torque than a solid one; C — T/J = τ/r = Gθ/L
(A) γ = rθ/L. (B) Material moved outwards raises J for the same area. (C) The torsion equation. (D) False: θ = TL/(GJ) grows in proportion to L.According to Castigliano’s first theorem, the deflection of the point of application of a load P, in the direction of P, is:
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Answer: A — the partial derivative of the total strain energy with respect to P
δ = ∂U/∂P with U expressed in terms of the loads. The derivative with respect to the deflection gives the load, which is the converse (Castigliano’s theorem in its displacement form). For a linear system U = ½Pδ, so U/P is only half of δ.