Beams: SFD and BMD, Bending and Shear Stresses, Combined Stresses, Deflection and Indeterminate Beams
2. Axial and bending stresses: the flexure formula
Under pure bending of a straight, homogeneous, linearly elastic beam with a plane of symmetry, plane sections stay plane and the strain varies linearly with the distance y from the neutral axis, which passes through the centroid. This gives σ = My/I and M/I = σ/y = E/R, with I the second moment of area about the neutral axis and R the radius of curvature. The extreme-fibre stress is σmax = M/Z with section modulus Z = I/ymax. Sign: for sagging M the stress is compressive above the neutral axis and tensile below.
| Section | I | Z |
|---|---|---|
| Rectangle b wide, h deep | bh³/12 | bh²/6 |
| Solid circle, diameter d | πd⁴/64 | πd³/32 |
| Hollow circle, D and d | π(D⁴ − d⁴)/64 | π(D⁴ − d⁴)/(32D) |
Why depth matters most. For a rectangle Z ∝ bh², so doubling the depth at the same width and moment cuts the extreme stress to one-fourth, while doubling the width halves it. Material far from the neutral axis does the work, which is why I sections and hollow tubes are efficient. Worked example: a hollow circular section with D = 100 mm and d = 80 mm carries M = 5 kN·m. I = π(100⁴ − 80⁴)/64 = 2.898 × 10⁶ mm⁴, so σmax = 5 × 10⁶ × 50/2.898 × 10⁶ = 86.3 MPa.
3. Shear stress in beams
The transverse shear stress at a level y₁ above the neutral axis is τ = VQ/(Ib), where Q is the first moment about the neutral axis of the area lying beyond that level, b is the width at that level and I is the second moment of the whole section. It is zero at the free top and bottom surfaces. For a rectangle the distribution is a parabola, τ = (3V/2bh)[1 − (2y/h)²], with maximum 1.5 V/A at the neutral axis, i.e. 1.5 times the average. For a solid circle the maximum at the neutral axis is 4V/3A.
I section. Take flanges 100 mm × 20 mm, web 10 mm thick, total depth 200 mm (web depth 160 mm), V = 100 kN. I = 100 × 200³/12 − 90 × 160³/12 = 35.95 × 10⁶ mm⁴. At the neutral axis Q = 100 × 20 × 90 + 10 × 80 × 40 = 212 000 mm³, so τmax = 100 × 10³ × 212 000/(35.95 × 10⁶ × 10) = 59.0 MPa. At the web side of the web–flange junction Q = 180 000 mm³ and τ = 50.1 MPa; at the flange side b = 100 mm so the same Q gives only 5.0 MPa, a tenfold jump because b changes.
4. Combined stresses: axial with bending, bending with torsion
When stresses of the same kind act on the same point, superpose them. Eccentric axial load P at eccentricity e from the centroid is equivalent to P plus a moment Pe, so σ = P/A ± Pe y/I. For a rectangle h deep with the load on the depth axis, σ = P/A ± 6Pe/(bh²) and the whole section stays in compression only if e ≤ h/6 (the middle third). Worked: b = 100, h = 150 mm, P = 90 kN compressive at e = 40 mm gives P/A = 6.0 MPa and 6Pe/bh² = 9.6 MPa, so the stress is 15.6 MPa compressive at one edge and 3.6 MPa tensile at the other (e = 40 mm exceeds h/6 = 25 mm).
Bending with torsion in a solid circular shaft of diameter d gives a surface point with σ = M/Z, Z = πd³/32, and τ = T/(2Z), since the polar section modulus is twice Z. Example: d = 40 mm, M = 400 N·m, T = 300 N·m: Z = 6283 mm³, σ = 63.7 MPa, τ = 23.9 MPa. Principal stresses σ1,2 = σ/2 ± √[(σ/2)² + τ²] = 31.8 ± 39.8, so σ1 = 71.6 MPa, σ2 = −8.0 MPa and τmax = 39.8 MPa. Von Mises: √(σ² + 3τ²) = 75.9 MPa. A shaft therefore sees an equivalent bending moment ½[M + √(M² + T²)] by the maximum principal stress theory.
5. Deflection by double integration and standard results
For small slopes the curvature is 1/R = d²y/dx² and EI d²y/dx² = M(x) when y is positive upward and M is positive sagging. Integrate twice and fix the constants from the boundary conditions: y = 0 at a pin or roller, y = 0 and dy/dx = 0 at a fixed end, and continuity of y and dy/dx where the moment expression changes. The sign check is physical: a sagging beam has a negative y (downward) under this convention, and standard results are quoted as magnitudes.
| Beam and load | Maximum deflection | End slope |
|---|---|---|
| Cantilever, tip load P | PL³/3EI | PL²/2EI |
| Cantilever, uniform w | wL⁴/8EI | wL³/6EI |
| Cantilever, tip couple M₀ | M₀L²/2EI | M₀L/EI |
| Simply supported, central P | PL³/48EI | PL²/16EI |
| Simply supported, uniform w | 5wL⁴/384EI | wL³/24EI |
Worked example. A simply supported steel beam (E = 200 GPa) of span 4 m with I = 4 × 10⁷ mm⁴ carries w = 16 kN/m (16 N/mm). δ = 5 × 16 × 4000⁴/(384 × 200 000 × 4 × 10⁷) = 6.67 mm, which is L/600, well within the usual L/360 serviceability limit. By scaling, deflection grows as L⁴ for a uniform load and L³ for a point load, and falls linearly with I, so doubling the depth of a rectangle (I × 8) cuts it to one-eighth. Superposition lets several loads be added.
6. Beams with one redundant: compatibility method
A beam with one more reaction than statics can supply is indeterminate to the first degree. Method: choose one reaction (or internal moment) as the redundant, remove it to leave a determinate released beam, and require that the deflection or slope the released beam suffers at that point under the real loads, plus the effect of the redundant acting alone, equals the true boundary value (zero at a rigid prop). The standard results above supply both deflections.
Propped cantilever, uniform w over L (fixed at A, roller prop at B). Remove the prop: the tip deflection is wL⁴/8EI downward and the prop force R gives RL³/3EI upward. Setting them equal gives R = 3wL/8. Then the fixed-end reaction is 5wL/8 and the fixed-end moment is wL²/8 (hogging). The maximum sagging moment, at 3L/8 from the prop where V = 0, is 9wL²/128. For w = 12 kN/m and L = 4 m these are R = 18 kN, 30 kN, 24 kN·m and 13.5 kN·m. For a central point load P the prop reaction is 5P/16 and the fixed-end moment is 3PL/16.
Two equal spans L under uniform w: take the middle reaction as the redundant. Remove it: the beam becomes simply supported over the full length 2L and deflects δ = 5w(2L)⁴/384EI = 5wL⁴/24EI at the middle, and the central force gives R(2L)³/48EI = RL³/6EI. Equating, RB = 5wL/4, the end reactions are 3wL/8 each, and the support moment is wL²/8 hogging. The fixed-pinned beam is the same problem as the propped cantilever.
Key takeaways
- Use dV/dx = −w and dM/dx = V with sagging positive; maximum moment sits where V = 0 or changes sign, and a couple jumps the BMD only.
- Flexure formula σ = My/I with Z = I/ymax; rectangle Z = bh²/6, solid circle πd³/32, and depth enters squared.
- Shear stress τ = VQ/(Ib): zero at the free surfaces, maximum at the neutral axis, 1.5 × average for a rectangle and 4/3 × average for a solid circle; the web of an I section carries nearly all of V.
- Combined loads: superpose P/A ± Pey/I (middle third rule e ≤ h/6) and for bending with torsion find principal stresses from σ = M/Z and τ = T/2Z.
- Deflection: EI y″ = M; memorise PL³/3EI, wL⁴/8EI, PL³/48EI, 5wL⁴/384EI and the scaling with L and I.
- One redundant: remove it, equate the released-beam deflection to the redundant's deflection (propped cantilever R = 3wL/8; two equal spans middle R = 5wL/4).
Practice questions (18)
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.
A beam carries a distributed load w(x), positive downward, and the bending moment M is taken positive when sagging, with shear force V positive when the left part is pushed up. Which pair of relations holds?
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Answer: B — dV/dx = −w and dM/dx = V
Taking the upward forces on the left part as positive V, a downward load w dx on a slice reduces V, so dV/dx = −w. Moments about the right face of the slice give dM = V dx for the sagging-positive M. The pair with dM/dx = −V results from the opposite moment convention, and dV/dx = w would apply if w were taken upward.A simply supported beam carries a clockwise couple C applied at its mid-span and no other load. Just on either side of the mid-span section the shear force and bending moment behave as follows:
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Answer: C — the shear force is constant on the whole span and the bending moment jumps upward by C at the couple
A couple on a simply supported beam produces equal and opposite reactions C/L, so the shear is constant (no change at the couple since V is a force balance), while M varies linearly on each side of the couple and jumps by C at the couple, upward for a clockwise couple. Setting V to zero everywhere is wrong because the reactions C/L are nonzero, and a jump of C in the shear force confuses a couple with a point load.A simply supported beam of span 5 m carries a single 30 kN vertical point load at 2 m from the left support and its self-weight is neglected. The rectangular section is 100 mm wide and 200 mm deep, bending about its horizontal centroidal axis. The maximum bending stress in the beam, in MPa, is ____ (give one decimal place).
Numerical answer — type the value.
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Answer: 54.0
RA = 30 × 3/5 = 18 kN, so the maximum moment under the load is 18 × 2 = 36 kN·m. Z = bh²/6 = 100 × 200²/6 = 666 667 mm³, so σmax = 36 × 10⁶/666 667 = 54.0 MPa. Using the wrong reaction (taking 12 kN, the right-hand one) gives 36.0 MPa.A cantilever of rectangular section carries a given tip load. If the depth of the section is doubled while the width, the length and the load are unchanged, the maximum bending stress at the fixed end becomes:
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Answer: C — one-fourth of its original value
M = PL is unchanged, and σmax = 6M/(bh²) varies inversely as h², so doubling h gives one-fourth. One-eighth is the factor by which the second moment I, and hence the tip deflection, change; M being unchanged does not mean the stress is unchanged.For a beam of rectangular cross-section carrying a transverse shear force V, the shear stress across the depth is:
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Answer: B — parabolic, zero at the top and bottom surfaces and 1.5 V/A at the neutral axis
τ = VQ/(Ib) with Q a parabolic function of y gives zero at the free surfaces and the maximum 1.5 V/A at the neutral axis. The linear distribution with zero at the neutral axis describes the bending stress, which is the tempting confusion.A symmetric I section has flanges 120 mm wide and 16 mm thick and a web 8 mm thick, the overall depth being 200 mm (so the web is 168 mm deep between the flanges). It carries a transverse shear force of 80 kN. Using τ = VQ/(Ib), the maximum shear stress in the section, in MPa, is ____ (give one decimal place).
Numerical answer — type the value.
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Answer: 57.3
I = 120 × 200³/12 − 112 × 168³/12 = 80.0 × 10⁶ − 44.25 × 10⁶ = 35.74 × 10⁶ mm⁴. At the neutral axis Q = 120 × 16 × 92 + 8 × 84 × 42 = 176 640 + 28 224 = 204 864 mm³, b = 8 mm. τmax = 80 × 10³ × 204 864/(35.74 × 10⁶ × 8) = 57.3 MPa. Taking b = 120 mm by mistake gives 3.8 MPa.Which of the following statements about the flexure formula σ = My/I for a straight, homogeneous, linearly elastic beam in pure bending are correct?
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Answer: A — P: The neutral axis passes through the centroid of the section.; C — R: Plane cross-sections remain plane and the strain varies linearly with y.
Zero net axial force with a linear strain field puts the neutral axis at the centroid (P), and plane sections remaining plane is the basic assumption (R). The stress is zero, not maximum, at the neutral axis and varies linearly, so Q and S are wrong.A short column of rectangular section 120 mm deep (in the plane of bending) and 80 mm wide carries a compressive load applied on the depth axis at eccentricity e from the centroid. For no tensile stress anywhere in the section, the largest permissible e is:
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Answer: D — 20 mm
The stress on the far edge is P/A − 6Pe/(bh²) = (P/A)(1 − 6e/h), which is non-negative only when e ≤ h/6 = 120/6 = 20 mm, the middle-third rule. The value 13.3 mm would come from using the width 80 mm in h/6, the wrong dimension, and 40 mm is h/3.A solid circular shaft of diameter 50 mm transmits a torque of 0.9 kN·m while it is also under a bending moment of 1.2 kN·m at the section considered. Treat the surface point of maximum bending tension, with σ = M/Z, Z = πd³/32 and τ = T/(2Z). The maximum in-plane shear stress at that point, in MPa, is ____ (give one decimal place).
Numerical answer — type the value.
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Answer: 61.1
Z = π × 50³/32 = 12 272 mm³, σ = 1.2 × 10⁶/12 272 = 97.8 MPa and τ = 0.9 × 10⁶/(2 × 12 272) = 36.7 MPa. τmax = √[(σ/2)² + τ²] = √(48.9² + 36.7²) = 61.1 MPa. Quoting σ/2 = 48.9 MPa alone, i.e. ignoring the torsion, is the usual slip.A short rectangular column, 120 mm wide and 200 mm deep, carries a 160 kN compressive load acting on the depth axis at an eccentricity of 50 mm from the centroid. The maximum compressive stress in the section, in MPa, is ____ (give one decimal place).
Numerical answer — type the value.
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Answer: 16.7
P/A = 160 000/(120 × 200) = 6.667 MPa and the bending part is 6Pe/(bh²) = 6 × 160 000 × 50/(120 × 200²) = 10.0 MPa, so the maximum compression is 6.667 + 10.0 = 16.7 MPa (the far edge has 3.3 MPa tension since e = 50 mm exceeds h/6 = 33.3 mm).A cantilever of length L, uniform EI, carries a total load W = wL. Compare the tip deflection when W is spread uniformly over the length with the tip deflection when the same W acts as a point load at the free end. The ratio (uniform : point) is:
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Answer: A — 3 : 8
Uniform load gives wL⁴/8EI = WL³/8EI; the tip load gives WL³/3EI. The ratio is (1/8)/(1/3) = 3/8. The ratio 8:3 inverts it, and 1:2 and 5:16 are not ratios that arise from these two cases.A simply supported steel beam of span 4 m (E = 200 GPa, I = 4 × 10⁷ mm⁴) carries a uniformly distributed load of 16 kN/m over the whole span, neglecting self-weight. The maximum deflection at mid-span, in mm, is ____ (give two decimal places).
Numerical answer — type the value.
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Answer: 6.67
δ = 5wL⁴/(384EI) with w = 16 N/mm, L = 4000 mm, EI = 200 000 × 4 × 10⁷ = 8 × 10¹² N·mm²: δ = 5 × 16 × 2.56 × 10¹⁴/(384 × 8 × 10¹²) = 2.048 × 10¹⁶/3.072 × 10¹⁵ = 6.67 mm. Using the point-load coefficient 1/48 on wL would give a different number.A beam of uniform EI and length 2L rests on three simple supports at its two ends and its mid-point and carries a uniformly distributed load w over the whole length. The reaction at the middle support is:
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Answer: D — 5wL/4
Removing the middle support and equating 5w(2L)⁴/384EI to R(2L)³/48EI gives R = 5wL/4. The value wL is what two independent simply supported spans would give, ignoring continuity, while 3wL/8 is the reaction at each end support.A propped cantilever of span L, uniform EI, is fixed at one end and held by a rigid roller at the other, and carries a single point load P at mid-span. The reaction at the roller (the prop) is:
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Answer: D — 5P/16
Releasing the prop, the tip deflection from the mid-span load is 5PL³/48EI downward and the prop force R gives RL³/3EI upward, so R = 5P/16. The value 11P/16 is the fixed-end reaction (P − 5P/16), and P/2 is the answer for a simply supported span.A propped cantilever of span 4 m is fixed at the left end and supported by a rigid roller at the right end, with uniform EI. It carries a uniformly distributed load of 12 kN/m over the entire span. The magnitude of the maximum sagging bending moment in the span, in kN·m, is ____ (give one decimal place).
Numerical answer — type the value.
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Answer: 13.5
Prop reaction R = 3wL/8 = 3 × 12 × 4/8 = 18 kN. From the prop, V = 18 − 12x is zero at x = 1.5 m (= 3L/8), where M = 18 × 1.5 − 12 × 1.5²/2 = 27 − 13.5 = 13.5 kN·m (= 9wL²/128). The fixed-end moment 24 kN·m is hogging and larger, but it is not the sagging maximum.For a symmetric I-section beam carrying a transverse shear force, which of the following statements are correct?
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Answer: A — P: The web carries by far the larger share of the shear force.; B — Q: The shear stress is greatest at the neutral axis.
Q grows toward the neutral axis and the web is thin, so the web carries about 90 per cent of V and τ peaks at the neutral axis. The web stress varies parabolically though mildly, so R is not exact, and the stress is zero on the free top surface, so S is wrong.Which of the following statements about a propped cantilever of uniform EI under a given vertical load are correct?
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Answer: A — P: It is indeterminate to the first degree for the vertical reactions and the support moment.; C — R: Removing the prop leaves a determinate cantilever, and the compatibility condition is zero deflection at the prop.
There are three reactions in the plane problem (two forces and a moment), exceeding the two independent vertical and moment equations, so the degree is one (P) and the compatibility condition at the rigid prop is zero deflection (R). For a rigid prop and a single uniform-EI beam the reactions depend only on loading and geometry, so S is wrong, and Q is false precisely because the beam is indeterminate.A cantilever of length 3 m carries a uniformly distributed load of 5 kN/m over its whole length and a 6 kN point load at the free end. Taking sagging as positive, the bending moment at the fixed end is:
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Answer: A — −40.5 kN·m, hogging
Moment of the loads about the fixed end is 6 × 3 + 5 × 3 × 1.5 = 18 + 22.5 = 40.5 kN·m, and a downward load on a cantilever causes hogging (tension on top), so M = −40.5 kN·m. The values 22.5 and 18 each omit one of the two loads, and the positive sign ignores the direction of curvature.