Shear and moment diagrams
One cut gives one number, and the number you want is the worst one. Rather than cutting a hundred times, let the position stay a symbol — and then stop solving altogether, because the three diagrams are each other's slopes and areas.
Cutting a beam at one station and solving for what crosses the cut is a complete method, and it answers a question nobody asked. The beam does not fail at the station you chose; it fails at whichever station is worst, and until you have looked everywhere you do not know where that is. The fix begins as a small change of habit — leave the position as a letter — and ends somewhere unexpected, with the whole calculation replaced by reading areas off a graph.
Functions of position
The procedure for internal forces never used the fact that the cut was at any particular place. Reactions first, cut, keep a side, three equations. If the cut is at \(x\) instead of at 5 m, every step survives unchanged and the answers come out with an \(x\) in them:
Plotting those against position gives the axial force, shear and bending moment diagrams. Conventionally they are stacked beneath a sketch of the loaded beam, on the same horizontal scale, so that a feature of the loading sits directly above its consequence. That alignment is not decoration — half of what follows is read off it by eye.
For most beams the axial diagram is uninteresting, because the loads are transverse and \(N\) is zero everywhere. It matters for columns, for inclined members, and whenever a load has a component along the member, and it is worth a glance to confirm it is zero rather than assuming. The two that carry the design are \(V\) and \(M\).
What the diagrams are for is finding the maxima and where they occur. The bending moment usually governs the size of the section and the shear usually governs its detailing near the supports, so a beam is checked at \(M_{\max}\) and at \(V_{\max}\), and those two are rarely at the same place. A diagram that is qualitatively right but numerically vague is still useful; a single number with no idea where it came from is not.
Take a beam that will run through this whole page: 8 m, simply supported, with 10 kN/m spread over the left 4 m and a 20 kN load hung at 6 m. Moments about the left support give \(8B = 40(2) + 20(6) = 200\), so \(B = 25\) kN and \(A = 35\) kN. Cut at 2 m and you get \(V = 15\) kN, \(M = 50\) kN·m; cut at 7 m and you get \(V = -25\) kN, \(M = 25\) kN·m. Neither is the maximum, and neither cut tells you so.
Cut at two metres and you get one pair of numbers.
Section by section
One expression will not cover the whole beam, and the reason is worth being precise about. \(V(x)\) and \(M(x)\) are sums over everything to one side of the cut. Slide the cut past a point load and that load joins the sum; slide it past the end of a distributed load and the amount of the distribution you are carrying stops growing. The list of terms changes, so the formula changes.
A new region therefore begins at every discontinuity in the loading: at each point load, at each applied couple, at each support reaction, and at each end of a distributed load — including where a distribution changes its rate. Between those points the loading is smooth and one pair of expressions holds throughout.
The worked beam has three regions. For \(0 \le x \le 4\) the cut carries the reaction and the part of the distributed load to its left, which is \(10x\) acting at \(x/2\):
For \(4 \le x \le 6\) the whole 40 kN is behind the cut, permanently at 2 m, so \(V = -5\) and \(M = 60 - 5(x-4)\). For \(6 \le x \le 8\) the 20 kN load joins as well, giving \(V = -25\) and \(M = 50 - 25(x-6)\), which reaches zero at the right-hand support exactly as it must.
Two errors account for most lost marks here. The first is carrying the full distributed load into a region where the cut is still inside it: only the portion to the left of the cut is on your free body, and it acts at the centroid of that portion, not of the whole. The second is trying to stretch one expression across a discontinuity, which produces a smooth curve where the real diagram has a corner or a jump, and quietly misses the maximum.
Done honestly, this is arithmetic with no insight in it, and for three regions it takes a page. There is a much better way, and it comes from asking what happens between two cuts that are almost in the same place.
Cut under the distributed load and carry its part along.
The graphical relations
Cut twice, a distance \(dx\) apart, and consider the slice between the cuts. On its left face are \(V\) and \(M\); on its right face, having moved a little along the beam, are \(V + dV\) and \(M + dM\). On top is whatever load the slice carries, which is \(w\,dx\) if the intensity there is \(w\). The slice is a body like any other, so it balances.
Vertically, taking the positive shear convention as it stands:
And taking moments about the right-hand face, the load on the slice contributes \(w\,dx \cdot dx/2\), which is second-order and vanishes in the limit, leaving:
These two lines are the heart of the topic. They say the three diagrams form a chain in which each is the slope chart of the one before it: the load diagram gives the slope of the shear diagram at every point, and the shear diagram gives the slope of the moment diagram. Nothing here is specific to any beam, and neither relation required solving anything.
The minus sign in the first is a consequence of the sign convention, not of physics, and it is the one thing worth committing to memory rather than rederiving: downward load, falling shear. Note also that both relations fail exactly where the loading is discontinuous, since the diagrams have no slope at a jump. That is not a defect; it is why the regions of the previous section exist, and the relations apply within each region rather than across the breaks.
One immediate consequence deserves its own line. Wherever \(V = 0\), the moment diagram is momentarily flat, so the maximum moment in any region is at a point where the shear passes through zero or at one of the region’s ends. Finding \(M_{\max}\) stops being a search and becomes a matter of looking at where the shear diagram crosses the axis.
Take a slice of width dx and put everything on it.
Areas under the curve
Derivatives run the wrong way for drawing. Integrate both relations between two stations and they turn into something you can use directly:
In words, and this is how they are actually used: the change in shear between two points is minus the area under the load diagram between them, and the change in moment is the area under the shear diagram. Both are areas of shapes you have already drawn — rectangles and triangles, mostly — so the diagrams can be built left to right by adding areas, with no expressions written down at all.
On the worked beam: start at the left support, where the reaction lifts the shear straight to \(+35\). Across the first 4 m the load area is \(10 \times 4 = 40\), so the shear falls by 40 to \(-5\), linearly, crossing zero at \(x = 3.5\). There is no load between 4 and 6, so the shear holds at \(-5\). The 20 kN load drops it to \(-25\), where it stays until the 25 kN reaction closes it back to zero at the far end.
Now integrate that. The moment starts at zero at a simple support. The shear area from 0 to 3.5 is a triangle, \(\tfrac12 (3.5)(35) = 61.25\), so \(M(3.5) = 61.25\) kN·m — the maximum, because that is where the shear crossed zero. From 3.5 to 4 the area is \(-1.25\), bringing \(M(4)\) to 60. From 4 to 6 it is \(-5 \times 2 = -10\), giving 50. From 6 to 8 it is \(-25 \times 2 = -50\), giving zero.
That last figure is the check, and it is free. On a beam with no applied couple at its ends, the moment must return to zero at a simple support or a free end, so the positive and negative shear areas must cancel exactly. If they do not, either a reaction is wrong or an area has been misread, and you know this before doing any of the work that depends on it. Reactions found from equilibrium and diagrams closed by areas are two independent routes to the same answer.
Start from the reaction and work rightwards.
Drawing them by inspection
Everything above collapses into a handful of habits, and with them the diagrams for an ordinary beam can be drawn in about a minute, with the algebra kept in reserve for the few numbers that actually need it.
A concentrated force produces a jump in \(V\) equal to the force and in the same sense, and no jump in \(M\) — the moment diagram merely acquires a corner there, because its slope has changed suddenly. A concentrated couple does the mirror image: it steps \(M\) by its own magnitude and leaves \(V\) completely unaffected. A reaction is a concentrated force like any other, which is why the shear diagram always begins and ends with a vertical jump at the supports.
Where the load is smooth, count degrees. Each integration raises the order of the curve by one, so under no load \(V\) is constant and \(M\) is straight; under a uniform load \(V\) is straight and \(M\) is parabolic; under a triangular load \(V\) is parabolic and \(M\) is cubic. Knowing the shape before you have a single value is most of the drawing done.
The curvature follows too, and this is the part people most often get backwards. Since \(dV/dx = -w\), a downward load makes the shear diagram slope down, and the steeper the load the steeper the slope. Since \(dM/dx = V\), the moment diagram rises while the shear is positive and falls while it is negative, so under a downward distributed load the moment curve is always concave downward. A sketched parabola bulging the wrong way is the commonest visual error on the page.
Then locate the peaks. Find where the shear crosses zero — usually one short linear interpolation on the shear diagram — and evaluate \(M\) there by accumulated area. Check the ends. If the beam is simply supported and the moment does not close at zero, something upstream is wrong. If it is a cantilever, the free end must show \(V = 0\) and \(M = 0\), and the built-in end must show the full reaction and the full fixing moment.
Once you can draw them, checking them is worth doing on something that will not agree with you out of politeness. The beam calculator takes any arrangement of supports and loads and draws all four diagrams at once — shear, moment, and the deflected shape — so a hand sketch can be held against it, peak by peak.
The habit these diagrams build is worth more than the diagrams. They are the first place in statics where the answer is a function rather than a number, and where the shape of that function — where it peaks, which way it curves, how it responds to moving a load — is the thing being designed for. Everything that follows in mechanics of materials starts from \(M(x)\) and asks what stress it causes.