Force Method · Problem 5 of 7All problems

Force Method ExampleSupport settlement force method, two-span beam

Problem. The two-span beam has a pin at A and rollers at B and C. A uniform load of 10 kN/m acts over its full length, and support B settles 15 mm. EI = 30,000 kN·m². Use the force method to find the support reactions, and draw the shear and moment diagrams.

Continuous beam with two 6 m spans: pin at A, rollers at B and C, 10 kN/m uniform load, support B settling 15 mm — Figure 1. The beam, its loads, and its dimensions.

Try it yourself before reading the solution. Show the results

$A_{y}$: 28.75 kN
$B_{y}$: 62.50 kN
$C_{y}$: 28.75 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

Removing B leaves a 12 m simple beam. The settlement isn't an unknown; it's a known displacement.

2M₀ diagram (primary structure, actual loads)

M_{0,\,A\text{–}B}(\xi) = 60\xi - 5\xi^2

M_{0,\,B\text{–}C}(\xi) = 180 - 5\xi^2

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) = -0.50\xi

m_{1,\,B\text{–}C}(\xi) = -3 + 0.50\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{36}{EI}

\delta_{10} = \dfrac{-2700}{EI}

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

A support settles, so the total displacement in the released direction is not zero but equals the specified settlement of that support: $\bar\Delta_i = \Delta_i - \delta_{\text{rigid},i}$ (Δᵢ = settlement of the removed support, upward positive in this equation: the sign of downward settlement enters with a minus sign here; δ_rigid = rigid-body movement of the primary structure due to settlement of the supports that remain). The right-hand side is written multiplied by EI:

36\,X_{1} = 2250

X_{1} = B_{y} = 62.50\ \text{kN}

The settlement enters multiplied by EI: 30,000 × 0.015 = 450, so 2700 − 450 = 2250. With loads alone EI would cancel; here it doesn't.

5Support reactions (superposition) and check

A_{y} = 28.75\ \text{kN}

B_{y} = 62.50\ \text{kN}

C_{y} = 28.75\ \text{kN}

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

M_{B} = -7.50\ \text{kN·m}

Equilibrium check: loads and support reactions together

Support	Rx (kN)	Ry (kN)	M (kN·m, ccw +)
A	0	28.75	0
B	0	62.50	0
C	0	28.75	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 ✓

ΣM₀720.00 (reactions) − 720.00 (loads) = 0.00 kN·m ✓

✓ The reactions match the matrix (direct stiffness) solution (largest difference 4.1 × 10⁻⁹ kN or kN·m).

Without settlement, $V_B = 75$ and $M_B = -45$ . The 15 mm settlement unloads B by 12.5 kN, which moves to A and C.

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 beam with a settling support: −7.50 at B and a maximum of 41.33 kN·m in span — Figure 2. Bending moment (kN·m).

Shear diagram of the beam with a settling support: 28.75 at A and 31.25 kN just right of B — Figure 3. Shear force (kN).

Results

$A_{y}$: 28.75 kN
$B_{y}$: 62.50 kN
$C_{y}$: 28.75 kN

Matches the independent stiffness solution.

Common mistakes

With the wrong settlement sign, $V_B$ increases. When support B moves down, $V_B$ should drop: here from 75 kN to 62.5 kN.
EI cancels under load alone, but not with settlement. The settlement term is multiplied by EI, so its numerical value is needed.
Once the reactions are found, check vertical equilibrium: 28.75 + 62.5 + 28.75 = 10 × 12.

Same problem, another method

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