How to use this calculator

Pick whether support B is a cable or a roller. Enter the beam length, where B is measured from the pin A, and for a cable its angle above the horizontal and which way it runs. Put the beam’s own weight in the weight box (it acts at the middle), then add up to three downward point loads by their distance from A, and a uniform load if there is one. Leave unused loads at 0.

Formulas

Moments about the pinT sin θ · b = W L2 + Σ P_i a_i + w L²2

Horizontal forcesA_x = −T cos θ (cable running up to the right)

Vertical forcesA_y = W + Σ P_i + w L − T sin θ

For a roller at B, replace T sin θ with the roller force B_y; the roller has no horizontal part, so A_x = 0.

Worked example

A 6 m, 250 N beam with a 600 N load 4.5 m from the pin and a cable at 35° (the default)

Moments about A: 250 × 3 + 600 × 4.5 = 750 + 2700 = 3450 N·m.
T = 3450 / (6 × sin 35°) = 3450 / 3.441 = 1002.5 N.
A_x = −1002.5 × cos 35° = −821.2 N (the pin pushes left).
A_y = 250 + 600 − 1002.5 × sin 35° = 850 − 575.0 = 275.0 N.
Check, moments about B: A_y × 6 = 250 × 3 + 600 × 1.5 → 275 × 6 = 1650 = 750 + 900. ✓

Why sum moments about the pin first

Both pin forces A_x and A_y pass through A, so they have no moment about it. That leaves one unknown, the cable tension, in the moment equation. Once T is known, the two force equations each have one unknown left. Picking the point where the most unknowns meet is the standard shortcut for any rigid-body problem.

Assumptions and limits

The beam is rigid and horizontal, and all loads act straight down.
The cable is attached on the beam’s centerline, so its horizontal pull has no moment about A.
A cable can only pull and a roller can only push. If the answer comes out negative, the support cannot hold the beam that way.
For a beam with a pin and a roller at both ends and the internal shear and moment, use the shear and moment diagram calculator.

Frequently asked questions

Where does the beam’s weight go?

A uniform beam’s weight acts as one force at its center, half the length from either end. Mass in kg times 9.81 gives the weight in newtons.

Why is A_x negative?

The cable pulls the beam toward its anchor, so the pin has to push the other way to stop it sliding. A negative sign just means the force points in −x.

What if the cable is attached partway along the beam?

Enter that distance from A as b. The moment arm of the cable’s vertical pull becomes b instead of L.