Beam Shear Force & Bending Moment

Calculate shear force and bending moment at any cross-section of a simply supported beam under a single concentrated load

Inputs

l

Beam Span

W

Concentrated Load

l_1

Load Position from A

x

Query Section Position x

Formula Interpretation

Reaction at A

R_A = \dfrac{W(l - l_1)}{l} = \dfrac{W \cdot l_2}{l}

Taking moments about B: $R_A = \dfrac{W(l - l_1)}{l} = \dfrac{W \cdot l_2}{l}$ eliminates R_B. R_A acts upward.

Reaction at B

R_B = \dfrac{W \cdot l_1}{l}

By vertical equilibrium: $R_B = \dfrac{W \cdot l_1}{l}$ . R_B acts upward.

Shear Force — AC

F(x) = +R_A \quad (0 \le x < l_1)

Only $R_A$ acts on the left part; resultant is upward, so F > 0.

Shear Force — CB

F(x) = R_A - W = -R_B \quad (l_1 < x \le l)

Left part carries $R_A$ (up) and $W$ (down); net is downward, so F < 0.

Bending Moment — AC

M(x) = R_A \cdot x \quad (0 \le x \le l_1)

Moment arm of $R_A$ is x; beam sags, so M > 0.

Bending Moment — CB

M(x) = R_A \cdot x - W(x - l_1) = R_B(l - x)

Equivalently $R_B(l - x)$ . Moment decreases linearly back to zero at B.

Maximum Bending Moment

M_{\max} = R_A \cdot l_1 = R_B \cdot l_2

Maximum sagging moment, occurring at the load point C.

Knowledge Points

Imaginary Cross-Section Method

To find the internal forces at a section, make an imaginary cut and apply equilibrium to either part. The shear force equals the algebraic sum of all transverse forces on one side; the bending moment equals the algebraic sum of all moments about the cut.

Sign Convention

Positive shear: the resultant of forces to the LEFT of the section acts UPWARD (left face up / right face down). Positive bending moment: the beam SAGS (concave upward, bottom fibre in tension). Both reverse when analysing from the right side.

SFD and BMD

The Shear Force Diagram (SFD) plots F(x) along the beam; it is piecewise constant for concentrated loads and jumps by W at each load point. The Bending Moment Diagram (BMD) plots M(x); it is piecewise linear and reaches its maximum (or minimum) where the shear force crosses zero.

Worked Example

A simply supported beam of span 1 m carries a concentrated load $W = 2000\,\text{N}$ at 400 mm from support A. Find the support reactions, and the shear force and bending moment at cross-sections X₁ (300 mm from A) and X₂ (700 mm from A).

Step 1 — Support Reactions

R_A = \dfrac{2000 \times (1000 - 400)}{1000} = \dfrac{2000 \times 600}{1000} = 1200 \text{ N}

R_B = 2000 - 1200 = 800 \text{ N}

Step 2 — Shear Forces

F_{X_1} = +R_A = +1200 \text{ N}

F_{X_2} = R_A - W = 1200 - 2000 = -800 \text{ N}

Step 3 — Bending Moments

M_{X_1} = R_A \cdot x_1 = 1200 \times 300 = 360{,}000\,\text{N}{\cdot}\text{mm} = 360\,\text{N}{\cdot}\text{m}

M_{X_2} = R_B(l - x_2) = 800 \times (1000 - 700) = 240{,}000\,\text{N}{\cdot}\text{mm} = 240\,\text{N}{\cdot}\text{m}

M_{\max} = R_A \cdot l_1 = 1200 \times 400 = 480{,}000\,\text{N}{\cdot}\text{mm} = 480\,\text{N}{\cdot}\text{m}

Result: $R_A$ = 1200 N, $R_B$ = 800 N; $F(X_1)$ = +1200 N, $F(X_2)$ = −800 N; $M(X_1)$ = 360 N·m (sagging), $M(X_2)$ = 240 N·m (sagging). Maximum moment = $480\,\text{N}{\cdot}\text{m}$ at C.

Extended Knowledge

•For beams with multiple concentrated loads, apply superposition: calculate the shear force and bending moment for each load separately, then add the results.
•When a uniformly distributed load (UDL) is present, the SFD becomes a straight sloped line and the BMD becomes a parabola. The maximum bending moment still occurs where F(x) = 0.
•In structural design, the maximum bending moment $M_{\max}$ governs the required section modulus: $\sigma_{\max} = M_{\max} / Z \le [\sigma]$ . Increasing the depth of the cross-section is the most effective way to reduce bending stress.