Short Answer:

For a simply supported beam of span carrying a uniformly distributed load (UDL) of intensity (force per unit length), the maximum bending moment occurs at the midspan and equals

Reactions at the supports are , shear is zero at midspan, and the bending moment varies parabolically from supports to midspan.

In plain words: a UDL spreads the load evenly along the beam; the supports share the load equally, and the largest bending effect is in the middle. That peak moment is the number engineers use to size the beam so it does not bend too much or fail.

Maximum Bending Moment for Simply Supported Beam with UDL

Detailed Explanation :

Problem setup and assumptions
Consider a straight, prismatic beam of length simply supported at its two ends. A uniformly distributed load of intensity acts along the entire span (units: N/m or kN/m). We assume linear-elastic material behavior, small deflections, and that loads are vertical — standard beam theory conditions.

Support reactions
Take equilibrium of the whole beam. Total load = . For symmetry (UDL and symmetric supports) each support carries half the load:

These reactions are needed to compute internal shear and moment.

Shear force (V) and bending moment (M) functions
Cut the beam at a distance from the left support ( ). Consider the left segment. Sum of vertical forces gives shear :

Integrate shear (or take moments) to get bending moment measured at the cut:

This is a quadratic (parabolic) function of . It is positive for most of the span (sagging moment if using sign convention).

Location and value of maximum moment
To find the maximum, set shear because . Solve:

So the maximum bending moment occurs at midspan . Substitute into :

Interpretation of the formula

grows with the square of the span: doubling span quadruples the moment for the same .
For a given span, doubling the load intensity doubles the moment.
Compare with a point load at midspan on a simply supported beam where . A UDL of intensity over length exerts the same total load , and its peak moment equals , which is half of — showing distributed loading produces a smaller peak moment than concentrating the same total load at midspan.

Shear and moment diagrams (conceptual)

Shear force diagram (SFD): a straight line dropping from at left support to at right support, crossing zero at midspan.
Bending moment diagram (BMD): a parabola peaking at midspan with value , zero at supports.

Design implications

The maximum bending moment is the key input for checking bending strength: bending stress at outer fiber is or (where is section modulus).
To limit deflection, flexural rigidity and span must be chosen so midspan deflection under UDL is acceptable. For a simply supported beam with UDL, midspan deflection .
Shear checks are also necessary: maximum shear at supports is , and web or section must resist it.

Worked numeric example
If and :

Support reactions . Shear at left support is , zero at midspan.

Special cases and notes

If UDL is over only part of span, the maximum moment location may shift; you must derive piecewise expressions and check extrema.
For continuous beams (more than two supports) or fixed ends, moments change and you must use indeterminate analysis — the simple applies only to a single simply supported span under full-length UDL.
Units: ensure and units give consistent moment units (e.g., N/m × m² = N·m).

Why engineers care
The compact formula is quick, reliable, and widely used in preliminary sizing. It gives a conservative estimate of bending demand for final checks, together with shear and deflection checks.

Conclusion

For a simply supported beam carrying a uniformly distributed load over its full span, the maximum bending moment occurs at midspan and equals . Knowing this value is essential for calculating bending stresses, selecting an appropriate cross-section (via section modulus ), checking deflection limits, and ensuring the beam safely carries the applied load.