Shear-force and bending-moment diagrams

Reactions first. Before any internal force can be found, treat the whole beam as one free body and solve $\sum F_y = 0$ and $\sum M = 0$ for the support reactions. Every later step is checked against these numbers.

Pick a sign convention and stick to it. The standard choice: $V$ is positive if the net transverse force to the left of a section acts upward; $M$ is positive if it produces sagging — concave up, tension in the bottom fibre. Getting this backwards flips every diagram you draw.

The two governing relations. Between concentrated loads and couples, shear and moment vary smoothly according to

where: $w(x)$ is the distributed load intensity (force per length, positive downward), $V(x)$ is the shear force at position $x$, and $M(x)$ is the bending moment at position $x$. The first equation says shear falls at a rate equal to the local load; the second says moment rises at a rate equal to the local shear, so $M$ is literally the area swept under the $V$ diagram. At a concentrated force $P$, $V$ jumps by $\mp P$; at a concentrated couple, $M$ jumps by its magnitude.

Reading the shapes. No load between two points means $V$ is constant and $M$ is a straight line. Uniform load means $V$ is a straight line (slope $-w$) and $M$ is a parabola. A triangular load makes $V$ parabolic and $M$ cubic. Recognizing these shapes on sight is faster than re-deriving them each time.

Critical points. The bending moment is a maximum (in magnitude) wherever $V = 0$, because that is where the slope of the $M$ curve vanishes — this is where a beam is most likely to fail in bending and is the point every design check starts from.

A simply supported beam $AB$, span $L = 8\ \text{m}$, carries a single concentrated load $P = 20\ \text{kN}$ at midspan ($4\ \text{m}$ from $A$). Draw the SFD and BMD and state the peak shear and moment.

Step 1. Reactions from overall equilibrium: by symmetry, $R_A = R_B = P/2 = 10\ \text{kN}$.

Step 2. Shear diagram: from $A$ to just before the load, $V = +R_A = 10\ \text{kN}$ (constant, no distributed load). At $x = 4\ \text{m}$, $V$ jumps down by $P = 20\ \text{kN}$, giving $V = 10 - 20 = -10\ \text{kN}$, which stays constant to $B$ — consistent with $R_B = 10\ \text{kN}$.

Step 3. Moment diagram: $M(0) = 0$, then $M$ rises linearly with slope $V = 10\ \text{kN}$ (using $dM/dx = V$) up to $x = 4\ \text{m}$:

From $x = 4$ to $x = 8$, $M$ falls linearly with slope $V = -10\ \text{kN}$, returning to $M(8) = 0$ as required at the pin support $B$.

Step 4. Peak values: $V_{max} = \pm 10\ \text{kN}$ everywhere except the point load, and $M_{max} = 40\ \text{kN}\,\cdot\,\text{m}$ at midspan, exactly where $V$ changes sign. Answer: $V_{max} = \pm 10\ \text{kN}$, $M_{max} = 40\ \text{kN}\,\cdot\,\text{m}$ at midspan.

Competitive stretch

A cantilever $AB$, length $L = 4\ \text{m}$, fixed at $A$, carries a uniformly distributed load $w = 5\ \text{kN/m}$ over its whole length. Taking $x$ from the free end $B$, the shear at a section a distance $x$ from the free end equals the load carried outboard of it: $V(x) = -wx$, linear, reaching $V(4) = -20\ \text{kN}$ at the fixed end. Integrating with $dM/dx = V$ gives $M(x) = -wx^2/2$, a downward-opening parabola reaching $M(4) = -(5)(16)/2 = -40\ \text{kN}\,\cdot\,\text{m}$ at the fixed support, matching the reactive moment $M_A = -(wL)(L/2) = -40\ \text{kN}\,\cdot\,\text{m}$ found from overall equilibrium of the fixed end. The distributed-load SFD is a straight line and its BMD is genuinely curved — a common place students wrongly sketch a straight moment diagram instead.

• Sketching the BMD as straight lines under a distributed load, when $dM/dx = V$ forces it to be curved — parabolic under uniform load, cubic under triangular load.
• Forgetting that the moment must return to zero at any free end or simple/pin support; a nonzero value there flags an arithmetic or reaction error.
• Missing the reactive moment at a fixed support under distributed or offset load — a cantilever needs both a force reaction and a moment reaction.
• Assuming the maximum moment sits at midspan or at a support by habit, instead of locating it properly at $V = 0$.

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