Shear Force Diagram Calculator

The internal shear force $V$ at a section is the net transverse force carried across it. Cut the beam anywhere and sum everything to one side:

$V(x) = \sum F_{\text{left of the cut}}$

Before you can do that you need the support reactions, from the two statics equations:

$\sum F_y = 0, \qquad \sum M = 0$

Symbols and units:

• $V$ — internal shear force, newtons (N), usually kN in structures
• $R_A, R_B$ — support reactions, N
• $w$ — uniformly distributed load, newtons per metre (N/m)
• $x$ — distance from the left end, metres (m)

Shape rules that let you sketch $V(x)$ without integrating:

• an unloaded span gives a horizontal shear line
• a point load causes a vertical jump equal to that load
• a uniform load $w$ gives a straight line of slope $-w$

When it applies: statically determinate beams, where two equations are enough. Continuous or fixed-fixed beams are indeterminate and need compatibility conditions as well.

The assumption people forget: replace a distributed load by its resultant, $wL$ acting at its centroid, only when taking moments — never when drawing $V(x)$.

Once $V$ is known, the stress it produces on the cross-section follows. The quick estimate is the average shear stress:

$\tau_{\text{avg}} = \frac{V}{A}$

with $A$ the cross-sectional area in m² and $\tau$ in pascals. But shear stress is not uniform through the depth — it is zero at the top and bottom faces and peaks at the neutral axis. The exact distribution is

$\tau = \frac{VQ}{I\,t}$

where $Q$ is the first moment of the area above the level of interest (m³), $I$ is the second moment of area (m⁴) and $t$ is the width there (m). For a rectangular section this evaluates to a simple result:

$\tau_{\max} = \frac{3V}{2A} = 1.5\,\tau_{\text{avg}}$

and for a solid circular section $\tau_{\max} = 4V/(3A)$.

Design check: the largest $|V|$ almost always sits at a support, so that is where you check shear.

The assumption people forget: the $1.5$ factor is specific to rectangles. Using $\tau_{\text{avg}}$ alone underestimates the peak by $50\%$.

• Drawing the diagram before finding the reactions — every value on it depends on them.
• Missing the jump at a point load — the shear steps discontinuously by exactly the load.
• Getting the sign convention backwards — pick one (upward force on the left segment is positive shear) and hold it for the whole beam.
• Sloping the line the wrong way under a UDL — the slope is $-w$, so shear falls left to right.
• Using the resultant of a UDL when cutting — only the portion left of the cut, $wx$, acts.
• Assuming $\tau_{\max} = V/A$ — for a rectangle the peak is $1.5$ times that.
• Mixing mm and m in $\tau = V/A$ — a $50 \times 150$ mm section is $7.5 \times 10^{-3}$ m².