Force Method · Problem 1 of 7All problems

Force Method ExamplePropped cantilever force method

Problem. The beam is fixed at A and has a roller at B. A uniform load of 8 kN/m acts over its full length. EI is constant. Use the force method to find the support reactions, and draw the shear and moment diagrams.

6 m beam: fixed at A, roller at B, 8 kN/m uniform load over the full length — Figure 1. The beam, its loads, and its dimensions.

Try it yourself before reading the solution. Show the results

$A_{y}$: 30.00 kN
$M_{A}$: 36.00 kN·m
$B_{y}$: 18.00 kN

Solution

1Redundants and the primary structure

The beam is statically indeterminate to degree $1$ . Choose 1 redundant and remove it; what remains is the statically determinate primary structure.

Primary structure: a cantilever fixed at A, with the other supports removed.

Redundants: $X_{1} = B_{y}$

Primary structure: the released reactions are shown as X arrows

Removing B leaves a cantilever fixed at A, and its moment is easy to write from the free end. Taking the moment at A as the redundant would leave a simple beam, with the same result.

2M₀ diagram (primary structure, actual loads)

M_{0,\,A\text{–}B}(\xi) = -144 + 48\xi - 4\xi^2

Check −144 by hand: 8 × 6 × 3.

3Unit-load (virtual) diagrams

A unit value of each redundant is applied to the primary structure separately; the resulting $m_i$ moment diagrams are:

m_{1}

X_{1}=1

(unit load at support B)

m_{1,\,A\text{–}B}(\xi) = 6 - \xi

4Flexibility coefficients and compatibility

By virtual work, $\delta_{ij}=\int\frac{m_i m_j}{EI}dx$ , $\delta_{i0}=\int\frac{M_0 m_i}{EI}dx$ (with constant EI, the 1/EI cancels): EI = 30,000 kN·m²

\delta_{11} = \dfrac{72}{EI}

\delta_{10} = \dfrac{-1296}{EI}

Compatibility $\sum_j \delta_{ij}X_j + \delta_{i0}=0$ :

72\,X_{1} = 1296

X_{1} = B_{y} = 18\ \text{kN}

Find both by hand: $\delta_{11} = 6^3/3 = 72$ , $\delta_{10} = -8 \cdot 6^4/8 = -1296$ .

5Support reactions (superposition) and check

A_{y} = 30\ \text{kN},\quad M_{A} = 36\ \text{kN·m}

B_{y} = 18\ \text{kN}

Equilibrium check: loads and support reactions together

Support	Rx (kN)	Ry (kN)	M (kN·m, ccw +)
A	0	30.00	36.00
B	0	18.00	0

Equilibrium check (ΣFx, ΣFy, and ΣM about the origin):

ΣFx0.00 (reactions) + 0.00 (loads) = 0.00 kN ✓

ΣFy48.00 (reactions) − 48.00 (loads) = 0.00 kN ✓

ΣM₀144.00 (reactions) − 144.00 (loads) = 0.00 kN·m ✓

✓ The reactions match the matrix (direct stiffness) solution (largest difference 1.4 × 10⁻⁸ kN or kN·m).

Check $M_A$ by hand: 8 × 6 × 3 − 18 × 6 = 36.

6Member equilibrium check (free-body diagrams)

Each member is cut free: starting from the reactions, equilibrium gives its end moments and shears ( $\sum M = 0,\ \sum F = 0$ ). The arcs and arrows are the end moments and forces acting on the member (kN, kN·m), and each box is in equilibrium on its own.

Bending moment diagram of the propped cantilever: −36.00 at A and a maximum of 20.25 kN·m in span — Figure 2. Bending moment (kN·m).

Shear diagram of the propped cantilever: 30.00 at A and −18.00 kN at B — Figure 3. Shear force (kN).

Results

$A_{y}$: 30.00 kN
$M_{A}$: 36.00 kN·m
$B_{y}$: 18.00 kN

Matches the independent stiffness solution.

Common mistakes

The primary structure must stay stable. If A is changed to a pin and B is also removed, nothing is left to carry the load.
The largest span moment isn't necessarily at midspan. It occurs where the shear is zero: here x = 30 / 8 = 3.75 m and M = 20.25 kN·m.
Once the reactions are found, check vertical equilibrium: 30 + 18 = 8 × 6.

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