Shear force and bending moment diagrams

Why it matters

A beam is designed for the largest bending moment it carries (which sets its bending stress) and the largest shear force (which sets its shear stress and web or weld size). Shear force and bending moment diagrams (SFD and BMD) show where these maxima are, where the moment changes sign and where reinforcement or joints should go. They are also the starting point for deflection calculations.

Key ideas

Internal forces. Cut a beam at a section x. To keep each part in equilibrium the section must carry a shear force V (perpendicular to the axis) and a bending moment M.

Sign convention used here (common in Indian texts).

• Shear force at a section is positive when the resultant of forces to the left of the section acts upward (equivalently, to the right acts downward).
• Bending moment is positive (sagging) when it makes the beam concave upward; negative (hogging) when concave downward. For forces on the left part, clockwise moments about the section are sagging. Use one convention throughout and state it.

Procedure.

1. Find all support reactions from the FBD of the whole beam.
2. Mark key points: supports, point loads, applied couples, start and end of distributed loads.
3. Write V(x) and M(x) for each segment, or use the load–shear–moment relations to sketch.
4. Find M at every key point and where V = 0 (maximum or minimum M).
5. Find points of contraflexure, where M changes sign.

Load–shear–moment relations. With w = downward load intensity:

• dV/dx = −w: slope of the SFD equals minus the load intensity.
• dM/dx = V: slope of the BMD equals the shear force. So M has a maximum or minimum where V = 0 (or changes sign).
• The change in V between two points equals minus the area under the load diagram; the change in M equals the area under the SFD.

Shapes of diagrams.

Loading in the segment	SFD	BMD
No load	constant	straight line
Uniform load	straight sloping line	parabola (2nd degree)
Uniformly varying (triangular) load	parabola	cubic
Point load	sudden jump equal to the load	change in slope (kink)
Applied couple	no change	sudden jump equal to the couple

Standard results.

• Simply supported, central point load W, span L: M_max = WL/4 at centre; V = ±W/2.
• Simply supported, UDL w over full span: M_max = wL²/8 at centre; V_max = wL/2 at supports.
• Cantilever, point load W at free end: M_max = WL (hogging) at the fixed end; V = W throughout.
• Cantilever, UDL w over full length: M_max = wL²/2 at the fixed end; V_max = wL.
• Simply supported, point load W at a from A and b from B: M_max = Wab/L under the load.

At a free end or a pin/roller support with no applied couple, M = 0. At a fixed end, M equals the fixing moment.

Formulas

dV/dx = −w(x), dM/dx = V(x)

• V = shear force (N), M = bending moment (N·m), w = load per unit length (N/m), x = distance along beam (m).

M_max = W L / 4 (simply supported, central point load) M_max = W a b / L (simply supported, point load at distances a and b from the supports) M_max = w L² / 8 (simply supported, full UDL) M_max = W L (cantilever, end point load) M_max = w L² / 2 (cantilever, full UDL) M_max = w L² / (9√3) at x = L/√3 from the zero-load end (simply supported, triangular load rising from 0 to w)

Worked examples

Example 1 (standard). A simply supported beam AB of span 8 m carries a UDL of 10 kN/m over the left 4 m and a point load of 20 kN at 6 m from A. Draw the SFD and BMD and find the maximum bending moment.

1. Reactions: ΣM_A = 0: R_B × 8 = 40 × 2 + 20 × 6 = 200, so R_B = 25 kN; R_A = 60 − 25 = 35 kN.
2. Shear: V = 35 − 10x for 0 ≤ x ≤ 4, so V = 35 kN at A and −5 kN at x = 4 m. From 4 to 6 m, V = −5 kN. Just after the point load, V = −25 kN, constant to B (where R_B = 25 kN brings it to zero). ✓
3. V = 0 at x = 35/10 = 3.5 m.
4. Moments: M(3.5) = 35 × 3.5 − 10 × 3.5²/2 = 122.5 − 61.25 = 61.25 kN·m; M(4) = 35 × 4 − 40 × 2 = 60 kN·m; M(6) = 25 × 2 = 50 kN·m (from the right); M = 0 at A and B.
5. BMD: parabola from 0 to the peak at 3.5 m, then straight lines 60 → 50 → 0.

Answer: R_A = 35 kN, R_B = 25 kN; M_max = 61.25 kN·m at 3.5 m from A.

Example 2 (GATE level, overhang). Beam ABC is supported at A (pin) and B (roller), AB = 6 m, with an overhang BC = 2 m. It carries a UDL of 6 kN/m on AB and a point load of 12 kN at C. Find the reactions, the maximum sagging and hogging moments, and the point of contraflexure.

1. Reactions: R_B × 6 = 36 × 3 + 12 × 8 = 204, so R_B = 34 kN; R_A = 48 − 34 = 14 kN.
2. Between A and B: V = 14 − 6x, M = 14x − 3x².
3. V = 0 at x = 14/6 = 2.333 m: M_max,sag = 14 × 2.333 − 3 × 2.333² = 32.67 − 16.33 = 16.33 kN·m.
4. At B: M_B = 14 × 6 − 3 × 36 = −24 kN·m; check from the right: −12 × 2 = −24 kN·m. ✓ This is the maximum hogging moment.
5. Contraflexure: 14x − 3x² = 0, so x = 14/3 = 4.667 m from A.
6. SFD: 14 kN at A falling linearly to −22 kN just left of B, jumping up by 34 to +12 kN, constant to C.

Answer: R_A = 14 kN, R_B = 34 kN; maximum sagging 16.33 kN·m at 2.33 m; maximum hogging 24 kN·m at B; contraflexure at 4.67 m from A.

Common mistakes

• Placing the maximum moment under the largest load instead of where V = 0.
• Using the resultant of a UDL to compute moments inside the loaded length.
• Forgetting the jump in the BMD at an applied couple.
• Mixing sign conventions between the left and right parts of the cut.
• Missing the hogging moment over the support of an overhanging beam, which may govern the design.
• Assuming the shear force is zero at a simple support; it equals the reaction there.

For GATE PI

Expect maximum moment and its location for simply supported, cantilever and overhanging beams under point, uniform and triangular loads, shear at a given section, points of contraflexure, and identification of SFD or BMD shapes from load diagrams (or vice versa). Practise sketching with dV/dx = −w and dM/dx = V without writing full equations.

Quick check

1. Maximum moment for a simply supported beam of span 8 m with a UDL of 3 kN/m?
2. Where on a beam is the bending moment maximum?
3. What shape is the BMD in a segment carrying a uniform load?
4. Cantilever 5 m with 10 kN at the free end: moment at 3 m from the fixed end?

Answers: 1. 24 kN·m. 2. Where the shear force is zero or changes sign (or at a support for cantilevers and overhangs). 3. Parabola. 4. 20 kN·m hogging.