Shear force and bending moment diagrams

Why it matters

A gantry beam carrying a moving carriage, a conveyor frame, a robot's linear-axis profile or a sensor boom must be checked where the internal bending moment and shear force are largest. Shear force and bending moment diagrams show those values along the whole member at a glance. They feed directly into bending stress, shear stress and deflection calculations, so an error here propagates to every later result.

Key ideas

Internal forces. 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. They are found from the free body of either side of the cut, whichever is simpler.

Sign convention (used here).

• Shear force V is positive when the resultant of forces to the left of the section acts upward (equivalently, to the right acts downward).
• Bending moment M is positive (sagging) when it makes the beam concave upward, compressing the top fibres; negative moment is hogging.
• Reactions and loads to the left of the section: upward forces give positive shear and clockwise moments about the section give positive (sagging) moment. Any consistent convention works; mixing conventions between steps is the common source of error.

Load-shear-moment relations. For a distributed load w(x) (positive upward):

• dV/dx = w, so the slope of the SFD equals the load intensity; the change in shear between two points equals the net load between them.
• dM/dx = V, so the slope of the BMD equals the shear force; the change in moment equals the area of the SFD between two points. Consequences:
• No load on a segment: V constant, M linear.
• Uniformly distributed load: V linear, M parabolic (second degree).
• Linearly varying load: V parabolic, M cubic.
• Concentrated force: V jumps by the force; M has a kink (change of slope).
• Applied couple: M jumps by the couple; V unchanged.
• Maximum (or minimum) M occurs where V = 0 or changes sign, or at a point of concentrated load, support or end.

Point of contraflexure. A point where M changes sign (sagging to hogging). It appears in overhanging and fixed beams; it is where the beam's curvature reverses and is a good location for splices.

Standard results to remember.

• Simply supported, central point load P, span L: M_max = P·L/4 at mid-span, V = ±P/2.
• Simply supported, load P at a from A and b from B: M_max = P·a·b/L under the load.
• Simply supported, full UDL w: M_max = w·L²/8 at mid-span, V_max = w·L/2 at the supports.
• Cantilever, end load P: M = −P·L at the fixed end, V constant.
• Cantilever, full UDL w: M = −w·L²/2 at the fixed end, V_max = w·L.

Method. Find reactions; mark key points (supports, load points, start and end of distributed loads); find V just left and right of each; join using the shape rules; find M at key points (or by areas of the SFD); locate V = 0 points and compute M there.

Formulas

dV/dx = w(x), dM/dx = V(x)

• V: shear force (N); M: bending moment (N·m); w: distributed load intensity (N/m), positive upward; x: position (m).

M_B − M_A = ∫ V dx (area of the SFD from A to B)

• Valid between points without applied couples.

M_max = P·L/4, M_max = P·a·b/L, M_max = w·L²/8

• Simply supported beam of span L (m): central load P (N); load P at distances a, b from the supports; full uniform load w (N/m).

M_fixed = −P·L, M_fixed = −w·L²/2

• Cantilever with end load P or full UDL w.

Worked examples

Example 1 (standard): gantry beam with a carriage load and self-weight. Given: a simply supported beam AB, span 6 m, carries its own weight and fittings as a UDL of 2 kN/m over the full span and a carriage load of 10 kN at C, 2 m from A. Draw the SFD and BMD and find the maximum moment.

1. Reactions, moments about A: R_B × 6 = 2 × 6 × 3 + 10 × 2 = 56, so R_B = 9.333 kN. R_A = 12 + 10 − 9.333 = 12.667 kN.
2. Shear: just right of A, V = +12.667 kN. Just left of C: 12.667 − 2 × 2 = +8.667 kN. Just right of C: 8.667 − 10 = −1.333 kN. Just left of B: −1.333 − 2 × 4 = −9.333 kN (= −R_B, a check).
3. V changes sign at C, so M_max is at C.
4. M_C = R_A × 2 − 2 × 2²/2 = 25.333 − 4 = 21.33 kN·m. Check from the right: 9.333 × 4 − 2 × 4²/2 = 37.333 − 16 = 21.33 kN·m.
5. Shapes: SFD sloping straight lines (slope −2 kN/m) with a 10 kN drop at C; BMD made of two parabolic arcs, zero at A and B, peak at C. Answer: R_A = 12.67 kN, R_B = 9.33 kN, M_max = 21.3 kN·m (sagging) at C.

Example 2 (GATE level): overhanging beam with contraflexure. Given: beam ABC with supports at A and B (AB = 6 m) and an overhang BC = 2 m. A UDL of 10 kN/m acts on AB and a point load of 20 kN at the free end C. Find the reactions, the maximum sagging and hogging moments and the point of contraflexure.

1. Moments about A: R_B × 6 = 10 × 6 × 3 + 20 × 8 = 180 + 160 = 340, so R_B = 56.67 kN. R_A = 60 + 20 − 56.67 = 23.33 kN.
2. In AB, V(x) = 23.33 − 10x. V = 0 at x = 2.333 m.
3. Maximum sagging moment: M = 23.33 × 2.333 − 10 × 2.333²/2 = 27.22 kN·m.
4. Hogging moment at B (from the overhang): M_B = −20 × 2 = −40 kN·m. Check from the left: 23.33 × 6 − 10 × 6²/2 = 140 − 180 = −40 kN·m.
5. Contraflexure in AB: M(x) = 23.33x − 5x² = 0 gives x = 4.667 m from A.
6. Shear just left of B: 23.33 − 60 = −36.67 kN; just right of B: −36.67 + 56.67 = +20 kN, constant to C. Answer: R_A = 23.3 kN, R_B = 56.7 kN; M_max sagging = 27.2 kN·m at 2.33 m from A; M_max hogging = −40 kN·m at B; contraflexure 4.67 m from A. The hogging moment governs the design.

Common mistakes

• Using a distributed load's resultant to compute moments at sections inside the loaded length.
• Looking for M_max only at mid-span; it is where V = 0 or changes sign.
• Forgetting the jump in M at an applied couple, or treating a couple as a force.
• Changing sign convention between the left and right free bodies.
• Missing the hogging moment at an overhang support, which is often the largest.
• Saying the shear force "is zero" under a central point load; it jumps from +P/2 to −P/2 there.

For GATE ME

Expect reactions and SF or BM at a given section, location and value of maximum moment, points of contraflexure in overhanging beams, beams with applied couples, and identifying the correct SFD or BMD shape from a set of sketches. Practise the slope and area relations so you can draw diagrams without writing full equations.

Quick check

1. Simply supported 6 m span, central load 10 kN. What is M_max?
2. A cantilever 4 m long carries 5 kN/m over its full length. What is the fixing moment?
3. What shape is the BMD under a uniformly distributed load?
4. What happens to the BMD at a point where a couple is applied?
5. A simply supported 10 m beam has 30 kN at 3 m from the left support. What is the right reaction?

Answers: 1. 15 kN·m. 2. −40 kN·m (hogging). 3. Parabolic. 4. It jumps by the value of the couple. 5. 9 kN.