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Structural Mechanics Simulator

Shear & Moment Diagram Simulator — Simply Supported Beam Under Combined Loads

Real-time plotting of reactions, the shear force diagram and the bending moment diagram of a simply supported beam under combined point and distributed loads, with automatic detection of the maximum bending moment and its location.

Parameters
Span L
m
Point load P
kN
Position of P, a
m
Distributed load w
kN/m

The beam is simply supported at both ends. The point load P acts downward at position a, the distributed load w acts downward over the full span, and the external moment M_0 is fixed at 0 in this simulator (kept in the formula box for clarity). Position a is automatically clamped inside the span L.

Results
—
Left reaction R_A
—
Right reaction R_B
—
Maximum bending moment M_max
—
Location of M_max
Beam Model, SFD and BMD

Top = beam and loads (yellow arrow = point load, cyan arrows = distributed load, triangles = supports) / middle = SFD V(x) / bottom = BMD M(x), red marker = location of M_max

Theory & Key Formulas

For a simply supported beam of length L under a point load P at position a, a uniform distributed load w over the span, and an external moment M_0 at position b, the reactions follow from equilibrium and the internal forces are obtained section by section.

From the equilibrium of forces and moments, the right reaction R_B and the left reaction R_A:

$$R_B = \frac{P\,a + w\,L^2/2 + M_0}{L},\qquad R_A = P + w\,L - R_B$$

Shear force V(x), where H(·) is the Heaviside step function:

$$V(x) = R_A - w\,x - P\,H(x-a)$$

Bending moment M(x):

$$M(x) = R_A\,x - \frac{w\,x^2}{2} - P\,(x-a)\,H(x-a) + M_0\,H(x-b)$$

Extrema of M(x) occur where dM/dx = V(x) = 0, and that point is the cross-section with the largest bending stress in design.

What is the Shear & Moment Diagram Simulator

🙋
Professor, SFDs and BMDs always come up in structures class, but I never really got what they are good for.
🎓
Roughly speaking, the SFD is a map of "the force trying to slice the beam vertically", and the BMD is a map of "the force trying to bend it". Designers look at the peak of the BMD and decide, "the largest bending stress is right here, so I'll size the cross-section to survive it". Move P or w in the simulator and watch how the BMD hump shifts around.
🙋
What was the relationship between SFD and BMD again? They look similar but they're not the same.
🎓
The relationship is simple: dM/dx = V(x). The slope of the BMD is the value of the SFD. So where the SFD crosses zero, the BMD is at a peak or a trough. At the defaults (L=6, P=30, a=2.5, w=10), look at the SFD: it jumps from +22.5 to −7.5 at x=a=2.5. That sign change is exactly where the BMD reaches the peak M_max = 87.5 kN·m. Classic textbook result.
🙋
Oh I see! So if I set the point load to zero and leave only the distributed load, where does the peak go?
🎓
Try it. With P=0, w=10, L=6, you get R_A=R_B=30 kN by symmetry, the SFD drops linearly from +30 at the left to 0 at midspan and −30 at the right. The BMD peaks at x=3 with the textbook value wL²/8 = 10·36/8 = 45 kN·m. Always verify the simulator against a hand calculation in a clean case first, then layer on the combined loads. That's the safe way.
🙋
Got it. So if I move a so the point load is at midspan, the triangular BMD should peak in the middle, right?
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Exactly. With a=L/2, the point-load contribution alone gives PL/4 at midspan. With combined loads you see the triangular peak from the point load (at x=a) blended with the parabolic peak of the distributed load (at x=L/2). Watch the BMD summit slide toward the point-load side when you push a to the end and back toward the centre when a moves to midspan. That's the best way to build intuition about where the worst stress will appear.

Frequently Asked Questions

Downward applied loads are positive, and upward reactions are positive. The shear force V(x) is positive when the resultant on the left side of the section points up, and the bending moment M(x) is positive when the beam is bent concave-up (a "sagging" beam). When the BMD plot is on the positive side, the lower fibres of the beam are in tension and the upper fibres are in compression. Different textbooks and codes use slightly different conventions, so be careful when cross-checking results against other sources.
A concentrated load P is a force applied at a point, so the SFD changes by P instantaneously at that location. Because the BMD is the integral of the SFD, a step in the SFD makes the BMD continuous but with a sudden change of slope, i.e. a kink. Conversely, an externally applied moment M_0 causes the BMD to jump by M_0 while the SFD stays continuous. In real structures, "point loads" are an idealization: physical loads are always distributed over some small area.
Yes, it follows from a textbook-level calculation. From equilibrium, R_B = (30·2.5 + 10·6²/2)/6 = (75 + 180)/6 = 42.5 kN, and R_A = 30 + 60 − 42.5 = 47.5 kN. The shear is V(0+) = 47.5, V(2.5−) = 47.5 − 25 = 22.5, V(2.5+) = 22.5 − 30 = −7.5, V(6−) = −7.5 − 35 = −42.5, and finally V(6) = 0 once R_B is added. The maximum moment occurs at x = a = 2.5 where V changes sign, giving M = 47.5·2.5 − 10·2.5²/2 = 118.75 − 31.25 = 87.5 kN·m. The simulator should read R_A=47.5, R_B=42.5, M_max=87.5, x=2.50.
Once you have the maximum bending moment M_max, the bending stress is σ = M_max·c / I, where I is the second moment of area of the cross-section and c is the distance from the neutral axis to the outermost fibre. For a rectangular section of width b and height h, I = bh³/12 and c = h/2, so the section modulus is Z = I/c = bh²/6 and σ = M_max/Z. With the default M_max = 87.5 kN·m and, say, an H-section H-300×150 (Z ≈ 4.81×10⁻⁴ m³), σ ≈ 182 MPa, which exceeds the allowable 160 MPa for SS400 steel — so you would step up the section or change the member.

Real-World Applications

Sizing beam members: When designing a steel beam in a building frame or a wooden floor joist, you start by reading V_max and M_max from the SFD/BMD and then back out the minimum cross-section that satisfies the allowable stresses in bending and shear. From the steel-section tables you choose an H-section with a section modulus Z and second moment of area I that satisfy σ = M_max/Z ≤ σ_allow and τ = V_max/A_w ≤ τ_allow, where A_w is the web area.

Crane girders and bridge beams: For a crane main girder, the moving point load (the trolley plus the suspended load) is placed at many positions to build the envelope of M_max — the influence line. The design value is not the M_max for a single load position but the maximum over all possible positions. Sweeping the slider for a in this simulator is the first step of an influence-line study.

Sanity check for FEM analyses: Before and after a finite-element analysis of a complex beam-like structure, simplify part of it to a simple beam model and compare with the closed-form solution. If the FEM bending stress is off from the theoretical value by an order of magnitude, suspect mesh quality or boundary-condition errors. This tool serves as a quick analytical calculator for that sanity check.

Teaching and coursework: The integral relationship "shear → bending moment" and dM/dx = V are hard to feel from a paper textbook alone. Adding loads one by one (start with w only, then add P, then move a) reveals how the SFD and BMD change continuously and is an effective entry point for structural mechanics.

Common Misconceptions and Cautions

The most common misconception is to assume "the maximum bending moment is right under the load". That is only true for the textbook case of a simply supported beam with a single central point load. With combined loads, the location of the peak shifts: with an off-centre point load plus a full-span distributed load, M_max can fall at the point-load location, at midspan, or anywhere in between. This simulator searches V(x)=0 on a 401-point grid and additionally evaluates the point-load location to pin the peak down. Watching how M_max location moves as you slide a around is the surest way to internalize this.

The next most common mistake is to treat "the SFD and the BMD as two separate things to compute separately". They are a single pair tied together by dM/dx = V(x). The signed area under the SFD equals the difference of the BMD, and the sign-reversal points of the SFD are the extrema of the BMD. When solving by hand, you start from the reactions, draw the SFD, then integrate piecewise to draw the BMD. In the simulator, stack the SFD and BMD plots vertically and check that every zero crossing of the SFD aligns with a flat (horizontal-tangent) point on the BMD — that is your consistency check.

Finally, do not blur "simply supported at both ends" with "fixed at both ends". This simulator assumes a simply supported beam — a pin on the left and a roller on the right. A beam fixed at both ends develops fixing moments at the supports, and the maximum bending moment shifts to the supports. For the same load, the midspan moment in a fixed-fixed beam is about one third of the simply supported value, but a larger moment appears at the supports. Boundary conditions are the single most basic and important assumption in structural design, so judge carefully whether the real support is closer to a simple support or to a fixed connection.

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