Shear force and bending moment diagrams

Why it matters

A wing is, to first order, a cantilever beam loaded by lift and by the weight of fuel, engines and structure. Its spar caps are sized from the bending moment and its webs from the shear force, so the shear force and bending moment diagrams are the first thing a structures engineer draws. They show where the critical sections are, how large the internal actions are, and where the moment changes sign.

Key ideas

Internal actions. Cut a beam at a section x. To keep either piece in equilibrium, the section must carry a shear force V (perpendicular to the axis) and a bending moment M. The SFD and BMD plot V and M along the beam.

Sign convention (the one used in most Indian textbooks).

• Shear force is positive when the resultant of forces to the left of the section acts upward (left-up, right-down).
• Bending moment is positive (sagging) when it makes the beam concave upward, i.e. compression at the top fibre. Negative moment is hogging. A cantilever with downward loads therefore has a negative (hogging) moment everywhere. Use one convention throughout a problem.

Differential relations (w = downward load intensity):

• dV/dx = −w: the slope of the SFD equals minus the load intensity.
• dM/dx = V: the slope of the BMD equals the shear force. Consequences: no load between points means constant V and straight-line M; a UDL gives a straight sloping V and a parabolic M; a linearly varying load gives a parabolic V and a cubic M. M is maximum or minimum where V = 0 or changes sign. A concentrated load makes V jump by the load; a concentrated couple makes M jump by the couple.

Integral relations. The change of shear between two sections equals minus the load between them; the change of moment equals the area of the SFD between them.

Boundary values. At a simple support or a free end, M = 0 (unless a couple is applied there). At a free end, V = 0 unless a point load acts there. At a fixed end, both V and M equal the reactions.

Point of contraflexure. A point where M changes sign (common in overhanging and fixed beams). It is where the curvature reverses and a natural place for splices.

Standard results worth memorising:

• Simply supported, central point load W: M_max = W·L/4 at midspan.
• Simply supported, full UDL w: M_max = w·L²/8 at midspan, V_max = w·L/2 at supports.
• Cantilever, end load W: M_max = −W·L at the fixed end.
• Cantilever, full UDL w: M_max = −w·L²/2 at the fixed end.

Formulas

dV/dx = −w V = shear force (N), w = distributed load intensity, downward positive (N/m), x = position (m).

dM/dx = V M = bending moment (N·m).

M_max = W·L / 4 Simply supported span L (m) with central point load W (N).

M_max = w·L² / 8, V_max = w·L / 2 Simply supported span with full UDL w (N/m).

M_max = W·a·b / L Simply supported span with point load W at distances a and b from the supports; occurs under the load.

M = −W·L, M = −w·L² / 2 Fixed-end moment of a cantilever with end load W or full UDL w (hogging).

M_max = w₀·L² / (9·√3) Simply supported span with a triangular load rising from 0 to w₀ (N/m); occurs at x = L/√3 from the zero end.

Worked examples

Example 1 (standard). A cantilever of length 3 m carries a UDL of 4 kN/m over its full length and a point load of 10 kN at the free end. Find the shear force and bending moment at the fixed end and sketch the diagrams.

1. Measure x from the free end. Shear (forces on the free-end side): V(x) = 10 + 4·x kN in magnitude.
2. At the fixed end, x = 3 m: V = 10 + 12 = 22 kN.
3. Moment: M(x) = −(10·x + 4·x²/2) kN·m (hogging).
4. At the fixed end: M = −(10 × 3 + 4 × 9/2) = −(30 + 18) = −48 kN·m.
5. Shape: V varies linearly from 10 kN to 22 kN; M is a parabola from 0 at the free end to −48 kN·m at the wall.

Answer: V = 22 kN and M = 48 kN·m (hogging) at the fixed end.

Example 2 (GATE level). Beam ABC is simply supported at A (x = 0) and B (x = 6 m) with an overhang BC of 2 m. A UDL of 10 kN/m acts over the whole 8 m. Find the reactions, the maximum sagging and hogging moments and the point of contraflexure.

1. Total load = 10 × 8 = 80 kN, acting at x = 4 m.
2. Moments about A: R_B × 6 = 80 × 4, so R_B = 53.33 kN; R_A = 80 − 53.33 = 26.67 kN.
3. In span AB: V(x) = 26.67 − 10·x. V = 0 at x = 2.667 m.
4. M(x) = 26.67·x − 5·x². At x = 2.667 m: M = 26.67 × 2.667 − 5 × 2.667² = 35.56 kN·m (maximum sagging).
5. At B, from the overhang side: M_B = −10 × 2²/2 = −20 kN·m (maximum hogging). Check from the left: 26.67 × 6 − 5 × 36 = −20 kN·m.
6. Contraflexure: 26.67·x − 5·x² = 0, so x = 26.67/5 = 5.333 m from A.

Answer: R_A = 26.67 kN, R_B = 53.33 kN; M_max sagging = 35.6 kN·m at 2.67 m; M_max hogging = 20 kN·m at B; contraflexure at 5.33 m from A.

Common mistakes

• Looking for M_max only under point loads; with distributed loads it is where V = 0.
• Mixing the left-side and right-side sign conventions in the same problem.
• Forgetting the jump in the BMD at an applied couple, or the jump in the SFD at a point load.
• Treating a UDL as a point load for the moment calculation but forgetting it acts at its centroid.
• Writing w·L²/2 instead of w·L²/8 for a simply supported beam, or vice versa for a cantilever.
• Missing the hogging moment over the support of an overhanging beam, which can govern the design.

For GATE AE

Expect quick numericals on reactions, maximum moment and its location, the point of contraflexure, and the shape of SFD and BMD for point loads, UDLs, triangular loads and applied couples. Conceptual questions test the relations dV/dx = −w and dM/dx = V. Practise reading a given SFD and recovering the loading, and use the standard results to check longer calculations.

Quick check

1. A simply supported beam of 4 m carries a central point load of 20 kN. What is M_max?
2. What is the slope of the BMD in a region where V = 5 kN?
3. What shape is the BMD under a UDL?
4. Where is M maximum on a simply supported beam with a full UDL?

Answers: 1. 20 kN·m. 2. 5 kN·m per m. 3. A parabola (second degree). 4. At midspan, where V = 0.