Simply Supported Beam — Uniform Distributed Load

Calculate reactions, shear force, and bending moment for a simply supported beam under a full-span UDL

Inputs

w

UDL Intensity

N/mm

l

Beam Span

Formula Interpretation

Support Reactions

R_A = R_B = \dfrac{wl}{2}

By symmetry the UDL is equally shared between the two supports. Each reaction equals half the total load wl. This symmetry also means the shear force is zero exactly at midspan.

Shear Force at Section x

F_x = -R_A + wx = -\dfrac{w}{2}(l - 2x)

The shear force varies linearly from −wl/2 at A (x=0), through zero at midspan (x=l/2), to +wl/2 at B (x=l). The SFD is a straight diagonal line — characteristic of a full-span UDL on a simply supported beam.

Bending Moment at Section x

\begin{aligned} M_x &= -\dfrac{w}{2}(lx - x^2) \\ M_{\max} &= -\dfrac{wl^2}{8} \quad \left(x = \dfrac{l}{2}\right) \end{aligned}

The moment follows a downward parabola (second-order curve), zero at both supports and reaching its maximum magnitude at midspan. The negative sign indicates sagging (concave-up deflection under the load).

Knowledge Points

Zero Shear → Maximum Moment

The bending moment reaches its absolute maximum precisely where the shear force is zero. For a symmetric UDL on a simply supported beam this is always at midspan. This rule applies to any load pattern: find where SFD crosses zero to locate Mmax.

Symmetric Parabolic BMD

Because the loading is symmetric about midspan, the BMD is also symmetric. The moment at a distance x from A equals the moment at a distance x from B. Engineers exploit this to check calculations: M(l/4) = M(3l/4) = −3wl²/32.

Shear = Support Reaction at Supports

The absolute value of the shear force at either support equals the support reaction. This is true for any simply supported beam (with or without UDL): the shear just inside the support equals R_A or R_B. The moment at the supports is always zero.

Worked Example

A simply supported beam of span l = 1000 mm carries a UDL w = 3 N/mm (3 kN/m) over its full length. Find the reactions, shear force at A, B and midpoint C, and the bending moments at x = 250 mm and at midspan.

Knowns

• Span: l = 1000 mm
• UDL intensity: w = 3 N/mm (= 3 kN/m)

Solution

Step 1 — Support Reactions (Formula ①)

R_A = R_B = \dfrac{wl}{2} = \dfrac{3 \times 1000}{2} = 1500 \text{ N}

Step 2 — Shear Forces (Formula ②)

\begin{aligned} F_A &= -\dfrac{wl}{2} = -1500\,\text{N} \\ F_C &= 0\,\text{N} \\ F_B &= \dfrac{wl}{2} = 1500\,\text{N} \end{aligned}

Step 3 — Max Bending Moment at midspan (Formula ③)

M_{\max} = -\dfrac{wl^2}{8} = -\dfrac{3 \times 1000^2}{8} = -375{,}000\,\text{N}{\cdot}\text{mm} = -375\,\text{N}{\cdot}\text{m}

Step 4 — Moment at x = 250 mm

M_{250} = -\dfrac{3}{2}\bigl(1000 \times 250 - 250^2\bigr) = -281{,}250\,\text{N}{\cdot}\text{mm} \approx -281\,\text{N}{\cdot}\text{m}

Step 5 — Symmetry check (x = 750 mm)

M_{750} = -\dfrac{3}{2}\bigl(1000 \times 750 - 750^2\bigr) = -281{,}250\,\text{N}{\cdot}\text{mm} \approx -281\,\text{N}{\cdot}\text{m} \quad (\text{symmetric})

Result: $R_A$ = $R_B$ = 1500 N. Shear: −1500 N at A, 0 at midspan C, +1500 N at B. $M_{max}$ = −375 N·m at midspan. M(250mm) = M(750mm) = −281 N·m (symmetric parabola).

Extended Knowledge

•Every horizontal beam carries its own weight as a UDL. For a simply supported floor joist, the midspan moment wl²/8 is the governing load case for deflection and bending stress checks. Pre-cambering compensates for this predictable sag.
•A bridge with uniform live load (pedestrians, parked cars) is modelled as a simply supported beam under UDL. The parabolic BMD shape informs the placement of longitudinal reinforcement — maximum near midspan, reducing toward the supports.
•A cantilever under the same UDL has Mmax = wl²/2 at the fixed end — four times larger than the simply supported case (wl²/8). This explains why simply supported spans are far more efficient for carrying distributed loads over long distances.
•If the UDL covers only part of the span (say from A to the midpoint), the beam is no longer symmetric. The reactions are unequal and the shear force no longer crosses zero at midspan. The moment maximum shifts toward the loaded half.
•The maximum deflection for a simply supported beam under full UDL is δmax = 5wl⁴/(384EI) at midspan. This fourth-power dependence on span length is why doubling the span increases deflection 16-fold — the key reason for intermediate supports in long structures.