Shear force and bending moment diagrams

In short

A shear force and bending moment diagram shows how the internal transverse force and internal bending couple vary along the length of a beam under a given set of loads and supports. They are the first step in every beam design: the maximum bending moment fixes the section modulus and the maximum shear fixes the web or stirrup requirement, so SFD and BMD feed directly into IS 800:2007 (steel) and IS 456:2000 (RCC) design checks. In exams this is the highest-scoring part of Strength of Materials - GATE ME/CE, ESE, SSC JE, state PSU and AMIE papers ask it almost every year, usually as "locate the point of contraflexure" or "find maximum BM". In interviews for design and structural roles (L&T, Tata Projects, Afcons, Godrej & Boyce, BHEL, ISRO/DRDO, Ashok Leyland chassis teams) a candidate is expected to sketch the diagrams for standard cases from memory.

Explained

Beams, supports and loads

• Beam: a member loaded transverse to its axis, carrying load mainly by bending. Classified by support: cantilever (one fixed end), simply supported (pin + roller), overhanging (simply supported with a projecting end), fixed, propped cantilever, continuous (more than two supports).
• Statically determinate beams (cantilever, simply supported, overhanging) have reactions found from ΣFx = 0, ΣFy = 0, ΣM = 0 alone. Indeterminate beams (fixed, propped, continuous) need compatibility - moment distribution, three-moment equation, or the coefficients of IS 456:2000 Table 12/13 for nearly equal continuous spans.
• Support reactions: roller - one vertical force; pin/hinge - vertical + horizontal; fixed - vertical, horizontal and a moment. An internal hinge carries shear but transmits zero bending moment - the extra equation that makes a hinged beam determinate.
• Load types: concentrated (point) load W in kN; uniformly distributed load (UDL) w in kN/m; uniformly varying load (UVL, triangular/trapezoidal); applied couple or moment M in kN·m; in practice characteristic loads come from IS 875 - Part 1 dead, Part 2 imposed, Part 3 wind, Part 4 snow, Part 5 combinations (seismic from IS 1893).
• A UDL over length a is replaced, for finding reactions only, by its resultant wa at the centroid of the load; never use the resultant when computing BM at a section that cuts the load.

Definitions and sign convention

• Shear force (V) at a section: the algebraic sum of all vertical forces on either side of that section. Unit kN.
• Bending moment (M) at a section: the algebraic sum of moments of all forces on either side, taken about that section. Unit kN·m.
• Standard sign convention (used in Indian textbooks and GATE): shear is positive when the left-hand portion tends to move up relative to the right (equivalently, upward forces to the left of the section are positive). Bending moment is positive (sagging) when it causes concavity upward - tension at the bottom fibre; negative (hogging) means tension at the top fibre.
• Consequence: a cantilever carrying downward load has hogging moment throughout (negative, maximum at the fixed end); a simply supported beam with downward load is sagging throughout.
• Free body method: cut the beam at distance x, keep one side, apply V and M to restore equilibrium. Write V(x) and M(x) for each span between discontinuities (a point load, a reaction, an applied couple, the start or end of a UDL) - these are the sections of the beam.

Relations between load, shear and bending moment

With x measured along the beam and downward distributed load w(x) taken positive:

• dV/dx = −w(x) - the slope of the SFD equals minus the load intensity.
• dM/dx = V(x) - the slope of the BMD equals the shear force.
• d²M/dx² = −w(x).
• Therefore: change in shear between two sections = −(area of load diagram) between them; change in BM = area of the SFD between them.
• Degree rule: if the load varies as xⁿ, the SFD is of degree n+1 and the BMD of degree n+2.
  • No load → SF constant (horizontal), BM linear (sloping).
  • UDL → SF linear, BM parabolic (2nd degree).
  • UVL → SF parabolic, BM cubic.
• Maximum/minimum BM occurs where V = 0 (or changes sign), because dM/dx = V. Also check BM at supports and at every point load and couple, since the true maximum may be at a discontinuity.
• Discontinuities: a point load W causes a vertical jump of W in the SFD (and only a kink in the BMD); an applied couple M causes a vertical jump of M in the BMD (and no change in the SFD).
• Point of contraflexure (inflexion): a section where BM = 0 and changes sign - the beam switches from sagging to hogging. It occurs only in overhanging, continuous or fixed beams and at a couple; it is where reinforcement is curtailed/lapped in RCC design.

Step-by-step procedure

1. Draw the free body: replace supports by reactions, show all loads with magnitudes and positions.
2. Find reactions: ΣM about one support = 0 gives the other reaction, then ΣFy = 0. Verify with ΣM about the second support.
3. Mark all section boundaries (point loads, reactions, couples, start/end of UDL).
4. For each section write V(x) and M(x) from the left portion (or right, whichever has fewer forces).
5. Compute V and M at every boundary - just left and just right where a jump occurs.
6. Plot the SFD to scale with the correct shape (straight/parabolic/cubic per the degree rule), closing back to zero at the free end or last reaction - the SFD must close, which is a built-in check.
7. Locate x where V = 0; substitute in M(x) for the maximum BM.
8. Plot the BMD; BM = 0 at a free end, at a simple support with no overhang, and at an internal hinge. Find points of contraflexure by solving M(x) = 0.
9. Mark maximum values with magnitude, sign and location; label units.

Standard cases to know by heart

• Cantilever, point load W at free end: SF = +W (hogging convention aside, magnitude W) constant; BM linear, 0 at free end to WL at fixed end (hogging).
• Cantilever, UDL w over whole span: SF linear 0 to wL; BM parabolic 0 to wL²/2.
• Cantilever, UVL (zero at free end, w at fixed end): SF parabolic 0 to wL/2; BM cubic 0 to wL²/6.
• Simply supported, point load W at centre: SF ±W/2; BM triangular, max WL/4 at centre.
• Simply supported, point load W at a and b from ends: reactions Wb/L and Wa/L; max BM Wab/L under the load.
• Simply supported, UDL w: SF +wL/2 to −wL/2 (zero at mid-span); BM parabolic, max wL²/8 at centre.
• Simply supported, triangular load, peak w at centre: max BM wL²/12.
• Simply supported, UVL zero at A to w at B: RA = wL/6, RB = wL/3; max BM wL²/(9√3) = 0.0642 wL² at x = L/√3 = 0.577L from A.
• Simply supported, couple M at mid-span: SF = M/L constant; BM jumps from −M/2 to +M/2 at the centre - contraflexure at the couple.
• Overhanging beam: hogging over the support region, sagging in the span; the overhang is often proportioned so hogging and sagging moments are equal, giving the lightest section.

Why it matters in design, and its limits

• Flexure check: σ = M·y/I; required section modulus Z = M/σ_permissible (working stress) or M_u ≤ M_d (limit state, IS 800:2007 and IS 456:2000).
• Shear check: average shear τ = V/A; for a rectangle τ_max = 1.5 V/A at the neutral axis; in RCC, nominal shear τ_v = V_u/(b·d) is compared with τ_c from IS 456 Table 19 and stirrups designed for the excess.
• Limitations: the elementary theory assumes a straight, prismatic, homogeneous, elastic beam with loads in a plane of symmetry and small deflections; it ignores local stress concentration at supports and under point loads, shear deformation in deep beams (span/depth less than about 2, treated separately in IS 456 Clause 29) and axial-flexure interaction.
• Diagrams are drawn for each load case; the design value comes from the worst load combination of IS 875 Part 5 / IS 456 Clause 36 (e.g. 1.5DL + 1.5LL at limit state).

Key points

• Shear force - algebraic sum of vertical forces on one side of a section, in kN.
• Bending moment - algebraic sum of moments about the section of all forces on one side, in kN·m.
• Sagging (+) - tension at bottom, concave up; hogging (−) - tension at top, typical of cantilevers and support regions.
• dV/dx = −w, dM/dx = V - slope of SFD is load intensity; slope of BMD is shear.
• Maximum BM where SF = 0 or changes sign - also check supports, point loads and couples.
• Change in BM = area of SFD between the two sections; change in SF = area of load diagram.
• Degree rule - point load: SF constant, BM linear; UDL: SF linear, BM parabolic; UVL: SF parabolic, BM cubic.
• Point load → jump in SFD; applied couple → jump in BMD, no jump in SFD.
• BM = 0 at a free end, at a simply supported end without overhang, and at an internal hinge.
• Point of contraflexure - BM changes sign; needed for bar curtailment in RCC and for splice location in steel.
• Cantilever maxima - WL (end load), wL²/2 (UDL): both at the fixed end.
• Simply supported maxima - WL/4 (central point load), wL²/8 (UDL): both at mid-span.
• Wab/L - max BM for an eccentric point load on a simple span.
• 0.0642 wL² at 0.577L - max BM for a triangular load zero at one end.
• The SFD must close to zero at the far end - a quick arithmetic check on reactions.
• Determinate vs indeterminate - only determinate beams can be solved by equilibrium alone; fixed/continuous beams need compatibility or IS 456 moment coefficients.
• For design - M gives bending stress and section modulus, V gives shear stress and stirrups.

Formulas and examples

Example 1 - Simply supported beam, eccentric point load. Span AB = 6 m, W = 20 kN at 2 m from A.

• ΣM_A = 0: R_B × 6 = 20 × 2 → R_B = 6.667 kN; R_A = 20 − 6.667 = 13.333 kN.
• SFD: +13.333 kN from A to just left of the load; drops by 20 kN to −6.667 kN; constant to B, where R_B closes it to zero.
• SF changes sign under the load → M_max there: M = 13.333 × 2 = 26.67 kN·m (check Wab/L = 20 × 2 × 4/6 = 26.67 ✓).
• BMD: two straight lines, 0 at A, peak 26.67 kN·m at 2 m, 0 at B. No contraflexure.

Example 2 - Simply supported beam with UDL. Span 8 m, w = 10 kN/m over the whole span.

• Total load = 80 kN → R_A = R_B = 40 kN.
• V(x) = 40 − 10x → zero at x = 4 m (mid-span); SFD is a straight line from +40 to −40 kN.
• M(x) = 40x − 10x²/2 = 40x − 5x²; M(4) = 160 − 80 = 80 kN·m (= wL²/8 = 10 × 64/8 ✓). BMD is a parabola, 0 at both ends.
• Design check: if σ_permissible = 165 N/mm², Z required = 80 × 10⁶ / 165 = 0.485 × 10⁶ mm³ - an ISMB 300 (Z ≈ 0.573 × 10⁶ mm³) suffices on bending.

Example 3 - Cantilever with UDL plus end point load. L = 3 m, w = 15 kN/m over the whole span, W = 20 kN at the free end.

• Max SF at the fixed end = 15 × 3 + 20 = 65 kN; SFD rises from 20 kN at the free end (step) linearly to 65 kN.
• Max BM at the fixed end = 15 × 3 × 1.5 + 20 × 3 = 67.5 + 60 = 127.5 kN·m, hogging.
• At 1.5 m from the free end: V = 20 + 15 × 1.5 = 42.5 kN; M = 20 × 1.5 + 15 × 1.5 × 0.75 = 30 + 16.875 = 46.875 kN·m hogging. BMD is parabola + line, zero at the free end.

Example 4 - Overhanging beam (contraflexure). Beam ABC, A at 0, support B at 8 m, free end C at 10 m; UDL 10 kN/m over the full 10 m.

• Total load 100 kN at 5 m from A. ΣM_A = 0: R_B × 8 = 100 × 5 = 500 → R_B = 62.5 kN; R_A = 37.5 kN.
• V(x) = 37.5 − 10x → zero at x = 3.75 m. M(x) = 37.5x − 5x²; M(3.75) = 140.625 − 70.3125 = 70.31 kN·m sagging (maximum positive).
• At B, from the right: M = −10 × 2 × 1 = −20 kN·m hogging (maximum negative).
• Point of contraflexure: 37.5x − 5x² = 0 → x = 0 or x = 7.5 m from A, i.e. 0.5 m left of support B.
• SFD: 37.5 → −42.5 kN just left of B, jumps up by 62.5 to +20 kN, then falls linearly to 0 at C ✓ (diagram closes).

Example 5 - Simply supported beam with an applied couple. Span 6 m, clockwise couple 30 kN·m applied at 2 m from A.

• ΣM_A = 0: R_B × 6 = 30 → R_B = +5 kN (up); ΣFy = 0 → R_A = −5 kN (down).
• SFD: constant −5 kN over the whole span (a couple does not change shear).
• BM just left of the couple = −5 × 2 = −10 kN·m; just right = +20 kN·m (jump of 30). Check from the right: 5 × 4 = 20 ✓. BM = 0 at both supports; contraflexure at the couple.

Where it is used

• Building and infrastructure design: RCC beams and slabs (IS 456:2000) and steel girders (IS 800:2007) at L&T, Tata Projects, Afcons, AECOM India, NHAI and Metro Rail consultants - BMD governs reinforcement and plate sizes, SFD governs stirrups and web stiffeners.
• Bridge engineering: influence lines and moving-load positions for maximum SF/BM under IRC vehicle loads; design of girders for RDSO/NHAI projects.
• Machine design: shafts in gearboxes, pumps and conveyors are analysed as beams - combined bending moment and torque give the equivalent moment for shaft diameter (Kirloskar, Thermax, Elecon, SKF applications).
• Automotive and rolling stock: truck chassis side-members and leaf springs (Ashok Leyland, Tata Motors, Volvo Eicher), rail coach underframes at ICF/RCF - treated as overhanging beams with the axle loads as point loads.
• Structural analysis software and roles: STAAD.Pro, ETABS, SAP2000, ANSYS and Ansys Mechanical all output SFD/BMD; graduate design engineers and FEA analysts must sanity-check software output against hand diagrams.
• Site and QA work: locating points of contraflexure for bar curtailment, lap and construction-joint positions; deciding scaffolding and formwork support spacing.

Interview questions

• Q: What is the relation between load, shear force and bending moment? A: dV/dx = −w and dM/dx = V, so the slope of the SFD is the load intensity and the slope of the BMD is the shear force.
• Q: Why does maximum bending moment occur where shear force is zero? A: Because dM/dx = V; the moment is stationary where its derivative, the shear force, vanishes or changes sign.
• Q: What happens to the SFD and BMD at a point load? A: The SFD shows a vertical jump equal to the load; the BMD has no jump, only a sudden change of slope (a kink).
• Q: What happens at an applied couple? A: The BMD jumps by the magnitude of the couple while the SFD is unchanged across that section.
• Q: What is a point of contraflexure and why does it matter? A: A section where the bending moment is zero and changes sign; in RCC it is where tension shifts from bottom to top, so bars are curtailed or lapped there.
• Q: For a simply supported beam with UDL, what are the maximum SF and BM? A: SF = wL/2 at the supports and BM = wL²/8 at mid-span.
• Q: Why is a cantilever's bending moment negative throughout under downward load? A: The load always causes tension in the top fibre and concavity downward, which is hogging by the standard convention.
• Q: If a UDL acts on a beam, can you replace it by its resultant to find the bending moment at a section inside the load? A: No - only for finding reactions. For a section inside the load, use the part of the UDL on that side acting at its own centroid.
• Q: A beam has an internal hinge. How does that help? A: The bending moment there is zero, which gives one extra equation and can make an otherwise indeterminate beam statically determinate.
• Q: Why is the BMD a parabola under UDL but cubic under triangular load? A: The load varies as xⁿ, so shear is of degree n+1 and moment of degree n+2 - n = 0 gives a parabola, n = 1 gives a cubic.
• Q: In an overhanging beam, how do you get the most economical section? A: Choose the overhang length so that the maximum hogging moment at the support equals the maximum sagging moment in the span, minimising the design moment.
• Q: How do SFD and BMD connect to the actual design check? A: M_max fixes the required section modulus through σ = M/Z, and V_max fixes shear capacity - stirrups in RCC per IS 456 Table 19, web checks in steel per IS 800.

Common mistakes

• Using the resultant of a UDL when writing M(x) for a section that cuts the UDL, instead of integrating only the part on that side.
• Getting reactions wrong by not verifying with a second moment equation, so the SFD does not close to zero at the far end.
• Forgetting that the maximum BM may be at a point load, a couple or a support rather than where the computed V = 0.
• Mixing sign conventions mid-problem, or drawing a cantilever BMD as sagging.
• Drawing a UDL region as a straight-line BMD, or a no-load region as a curve - ignoring the degree rule.
• Confusing a point of contraflexure (BM = 0, sign change) with the point of maximum BM (SF = 0).
• Treating fixed or continuous beams as determinate and solving them by equilibrium alone; and mixing units (N·mm with kN·m, kN/m with kN).

Related topics

Strength of materials: stress and strain · Bending, torsion and columns · Deflection of beams · Principal stresses and theories of failure · Strain energy and impact loading · Structural analysis of indeterminate beams and frames