Simply Supported Beam — Concentrated Load(s)

Calculate reactions, shear forces, and bending moments for a simply supported beam under 1 or 2 concentrated loads

Inputs

W_1

Load 1

W_2

Load 2

l_1

Distance from A to load C

l_2

Distance from A to load D

l

Beam Span

Formula Interpretation

Support Reactions (Single Load)

\begin{aligned} R_A &= \dfrac{Wl_2}{l} \\[6pt] R_B &= \dfrac{Wl_1}{l} \end{aligned}

Taking moments about B gives R_A; then vertical equilibrium gives R_B. The load is shared inversely proportional to the distance from each support.

Shear Forces (Single Load)

\begin{aligned} F_1 &= -R_A \quad \text{(in AC)} \\[6pt] F_2 &= R_B \quad \text{(in CB)} \end{aligned}

The shear force is constant in each segment and jumps by the load magnitude W at the load point C. It changes sign at C, confirming C is the location of maximum moment.

Bending Moment (Single Load)

\begin{aligned} M_C &= R_A l_1 = \dfrac{Wl_1 l_2}{l} \\[6pt] M_{\max} &= M_C \end{aligned}

The bending moment is maximum at the load point C where the shear force changes sign. It is zero at both supports and increases linearly to the peak.

Reactions (Two Loads)

\begin{aligned} R_B &= \dfrac{W_1 l_1 + W_2 l_2}{l} \\[6pt] R_A &= W_1 + W_2 - R_B \end{aligned}

Moment equilibrium about A gives R_B directly. R_A is found from vertical equilibrium. The maximum moment occurs at the load point directly under whichever load creates the larger moment.

Knowledge Points

Shear Sign Rule for M_max

For a simply supported beam, the bending moment reaches its maximum where the shear force changes sign from positive to negative. This eliminates the need to evaluate the moment at every point — just find the sign-change location in the SFD.

Load Position and Moment

A single load W at the centre produces the largest possible Mmax = Wl/4. Moving the load toward either support reduces Mmax. For two equal loads placed symmetrically, the moment is uniform between them — the BMD has a flat top.

Simply Supported vs Cantilever

Unlike a cantilever where moment is zero at the free end and maximum at the fixed end, a simply supported beam has zero moments at both supports and a positive peak somewhere in between. This produces a concave BMD (hogging convention) or convex (sagging convention).

Worked Example

A simply supported beam of span l = 1000 mm carries W₁ = 5000 N at 300 mm from A and W₂ = 4000 N at 700 mm from A. Find the reactions, shear forces, and maximum bending moment.

Knowns

• Span: l = 1000 mm
• Load 1: W₁ = 5000 N at l₁ = 300 mm from A
• Load 2: W₂ = 4000 N at l₂ = 700 mm from A

Solution

Step 1 — Reaction $R_B$ (moments about A)

R_B = \dfrac{5000 \times 300 + 4000 \times 700}{1000} = \dfrac{1{,}500{,}000 + 2{,}800{,}000}{1000} = 4300 \text{ N}

Step 2 — Reaction $R_A$

R_A = 5000 + 4000 - 4300 = 4700 \text{ N}

Step 3 — Shear forces

F_1 = -R_A = -4700\,\text{N} \quad F_2 = -R_A + W_1 = 300\,\text{N} \quad F_3 = R_B = 4300\,\text{N}

Step 4 — Moment at C (x = 300 mm)

M_C = R_A \times l_1 = 4700 \times 300\,\text{N}{\cdot}\text{mm} = 1{,}410{,}000\,\text{N}{\cdot}\text{mm} = 1410\,\text{N}{\cdot}\text{m}

Step 5 — Moment at D (x = 700 mm)

M_D = R_B \times (l - l_2) = 4300 \times 300\,\text{N}{\cdot}\text{mm} = 1{,}290{,}000\,\text{N}{\cdot}\text{mm} = 1290\,\text{N}{\cdot}\text{m}

Step 6 — Maximum bending moment

M_{\max} = \max(M_C,\, M_D) = 1410\,\text{N}{\cdot}\text{m} \quad (\text{at load point C})

Result: $R_A$ = 4700 N, $R_B$ = 4300 N. Shear: −4700 / +300 / +4300 N. $M_{max}$ = 1410 N·m at C (300 mm from A).

Extended Knowledge

•Floor joists and bridge girders are designed as simply supported beams. Concentrated loads from columns or wheels create the characteristic triangular or step-shaped BMD seen in structural drawings.
•When a load moves across the span (like a vehicle), the reaction and moment at each section change continuously. The influence line for the maximum moment shows the load position that maximises M_max = at the ⅓ or midspan depending on load count.
•The reactions and moments for any number of loads can be found by superposition: calculate each load separately and add the results. This works because the beam behaves linearly (small deflections, elastic material).
•A simply supported beam under these loads has no point of contraflexure — the moment is zero only at the supports. Continuous beams and overhanging beams develop contraflexure points where moment changes sign.
•When the moment at M_max reaches the plastic moment M_p = Z_p × σ_y, a plastic hinge forms. One plastic hinge in a simply supported beam is enough to create a collapse mechanism — unlike fixed-end beams which require three hinges.