Force Method · Problem 2 of 7All problems

Force Method ExampleTwo-span continuous beam

Problem. The two-span beam has a pin at A and rollers at B and C. A uniform load of 12 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.

Continuous beam with 4 m and 6 m spans: pin at A, rollers at B and C, 12 kN/m uniform load over the full length — Figure 1. The beam, its load, and its dimensions.

Try it yourself before reading the solution. Show the results

$A_{y}$: 13.50 kN
$B_{y}$: 77.50 kN
$C_{y}$: 29.00 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 simply supported beam on A and C, with the intermediate supports removed.

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

Primary structure: the released reactions are shown as X arrows

One redundant is enough: remove B and what remains is a 10 m simply supported beam on A and C.

2M₀ diagram (primary structure, actual loads)

M_{0,\,A\text{–}B}(\xi) = 60\xi - 6\xi^2

M_{0,\,B\text{–}C}(\xi) = 144 + 12\xi - 6\xi^2

The two lines are one parabola: $60x - 6x^2$ . The second line is the same expression with $x = \xi + 4$ . The split comes from $m_1$ in the next step.

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) = -0.60\xi

m_{1,\,B\text{–}C}(\xi) = -2.40 + 0.40\xi

$m_1$ has a kink at B, so the integral is taken over two regions. The value at B: 4 × 6 / 10 = 2.4.

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{19.20}{EI}

\delta_{10} = \dfrac{-1488}{EI}

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

19.20\,X_{1} = 1488

X_{1} = B_{y} = 77.50\ \text{kN}

Get δ₁₁ by hand: two triangles of height 2.4, so 2.4² × 10 / 3 = 19.2.

5Support reactions (superposition) and check

A_{y} = 13.50\ \text{kN}

B_{y} = 77.50\ \text{kN}

C_{y} = 29\ \text{kN}

Support moments from the reactions, by cutting the beam (sagging positive):

M_{B} = -42\ \text{kN·m}

Equilibrium check: loads and support reactions together

Support	Rx (kN)	Ry (kN)	M (kN·m, ccw +)
A	0	13.50	0
B	0	77.50	0
C	0	29.00	0

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

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

ΣFy120.00 (reactions) − 120.00 (loads) = 0.00 kN ✓

ΣM₀600.00 (reactions) − 600.00 (loads) = 0.00 kN·m ✓

✓ The reactions match the matrix (direct stiffness) solution (largest difference 6.6 × 10⁻⁹ kN or kN·m).

Check with the three-moment equation: $20\,M_B = -12\,(4^3 + 6^3)/4$ , so $M_B = -42$ . Writing it from A gives the same: 13.5 × 4 − 12 × 4 × 2.

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 two-span beam: −42.00 at support B and a maximum of 35.04 kN·m in span — Figure 2. Bending moment (kN·m).

Shear diagram of the two-span beam: 13.50 at A and 43.00 kN just right of B — Figure 3. Shear force (kN).

Results

$A_{y}$: 13.50 kN
$B_{y}$: 77.50 kN
$C_{y}$: 29.00 kN

Matches the independent stiffness solution.

Common mistakes

$m_1$ can't be written as a single expression. It has a kink at B, so each span is a separate region.
In $\delta_{10}$ , reading the ordinate from the wrong diagram directly changes the result. An open-access textbook has a wrong answer for this reason.
Once the reactions are found, check vertical equilibrium: 13.5 + 77.5 + 29 = 12 × 10.

Same problem, another method

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