Tools · Structural analysis
Mohr’s moment-area theorems
Step through the two theorems, then drag points A and B along the beam. The tangents, the shaded M/EI area and its centroid move with them, and every number is worked out below.
Step 1 of 5
The idea: curvature comes from bending moment
A beam bends into an elastic curve. Its curvature at any section is d²y/dx² = M/EI. So if we know the M/EI diagram, we know how fast the slope is changing everywhere. Mohr turned this into two simple rules about areas.
Beam
Points A and B
You can also drag the A and B markers on the drawing.
Theorem 1: change of slope
- Area of the M/EI diagram from A to B∫ M/EI dx from x = 0 to 6 m
Area = −9.9875 × 10⁻³ rad
- So the tangents turn by that muchθB − θA = Area
θA = 0 rad, θB = −0.01 rad
θB − θA = −0.01 rad ✓ equals the area (−0.573°)
Theorem 2: tangential deviation
- Centroid of the area, measured from Bx̄B = moment of area about Barea
x̄B = −0.046592−9.9875 × 10⁻³ = 4.665 m
- Deviation of B from the tangent at At(B/A) = Area × x̄B
t(B/A) = −9.9875 × 10⁻³ × 4.665 = −46.5917 mm
Check on the curve: yB − yA − θA(xB − xA) = −46.6667 − (0) − (0) × 6 m = −46.6667 mm ✓
The two deviations are generally different, because the same area has a different lever arm about A and about B.
Applying it: deflection at B
- Theorem used
At the fixed end the tangent to the elastic curve stays horizontal, so the deflection at any point equals the tangential deviation from that tangent.
Mohr’s 2nd theorem: deviation = (moment of the M/EI diagram between the two points, taken about the point) .
- Area of the BM diagram between x = 0 and 6 m
A = ∫ M dx = −159.7334 kN·m²
Its centroid is at x̄ = 1.3356 m, i.e. 4.6644 m from the point
- Deflection
EI·y = A × distance = −159.7334 × 4.6644 = −745.0669 kN·m³
y = −745.066916000 = −46.567 mm (downward)
Actual deflection at B from the elastic curve: -46.667 mm. For any loading and support, see the full SFD and BMD solver.