Force Method · Problem 3 of 7All problems

Force Method ExampleForce method frame example: L-shaped frame

Problem. The frame has a roller at A and is fixed at C. Column A–B is 4 m and beam B–C is 6 m. A lateral uniform load of 6 kN/m acts on the column, from left to right. EI is constant and axial deformation is neglected. Use the force method to find the support reactions, and draw the shear and moment diagrams.

L-shaped frame: roller at A, fixed at C, 6 kN/m lateral uniform load on the 4 m column, 6 m beam — Figure 1. The frame, its loads, and its dimensions.

Try it yourself before reading the solution. Show the results

$A_{y}$: 12.00 kN
$C_{x}$: −24.00 kN
$C_{y}$: −12.00 kN
$M_{C}$: 24.00 kN·m

Solution

1Redundants and the primary structure

The frame is statically indeterminate to degree $1$ . Release 1 support reaction; what remains is the statically determinate primary structure. Axial deformation is neglected ( $EA \approx \infty$ , the usual hand-method assumption).

Primary (determinate) structure: remove the vertical reaction at A; the support at C remains.

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

Primary structure: the released reactions are shown as X arrows

The redundant is the vertical reaction at A. Removing it leaves a cantilever fixed at C.

2M₀: primary structure, actual loads

The primary structure is determinate, so M₀ is found from statics (end moments clockwise + on the member, as in the other methods). In the sketch the diagram is plotted perpendicular to each member; end and peak values are in kN·m.

M₀ diagram: actual loads on the primary structure

Member	$M^{0}_{i}$ (kN·m)	$M^{0}_{j}$ (kN·m)
A–B	0	48.00
B–C	-48.00	48.00

Check the 48 at B by hand: 6 × 4 × 2. The lateral load's moment doesn't stay in the column; it passes from B into the beam unchanged.

3Unit-load cases

A unit value of each redundant is applied to the primary structure separately ( $m_i$ end moments):

m_{1}

—

X_{1}=1

(the vertical reaction at A)

m₁ diagram: the unit redundant alone on the primary structure

Member	$m^{1}_{i}$ (kN·m)	$m^{1}_{j}$ (kN·m)
A–B	0	0
B–C	0	-6.00

4Flexibility coefficients and compatibility

Virtual work is summed along the member axes: $\delta_{ij}=\sum_{el}\int \frac{m_i m_j}{EI}\,ds$ , $\delta_{i0}=\sum_{el}\int \frac{M_0 m_i}{EI}\,ds$ :

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

\delta_{10} = \dfrac{-864}{EI}

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

72\,X_{1} = 864

X_{1} = A_{y} = 12\ \text{kN}

$m_1$ is zero in the column, so the column's $M_0$ doesn't enter the integral. Only the beam counts: $\delta_{11} = 6^3/3 = 72$ , $\delta_{10} = -48 \cdot 6^2/2 = -864$ .

5Result: M diagram, reactions, and check

Superposition: $M = M_0 + \sum_i X_i\,m_i$ . The diagram is now the response of the actual (indeterminate) frame.

M_{AB} = 0\ \text{kN·m}

M_{BA} = 48\ \text{kN·m}

M_{BC} = -48\ \text{kN·m}

M_{CB} = -24\ \text{kN·m}

Equilibrium check: loads and support reactions together

Support	Rx (kN)	Ry (kN)	M (kN·m, ccw +)
A	0	12.00	0
C	-24.00	-12.00	24.00

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

ΣFx-24.00 (reactions) + 24.00 (loads) = 0.00 kN ✓

ΣFy0.00 (reactions) + 0.00 (loads) = 0.00 kN ✓

ΣM₀48.00 (reactions) − 48.00 (loads) = 0.00 kN·m ✓

✓ The reactions match the method's own reference, the axially rigid (EA ≈ ∞) matrix solution (largest difference 4.5 × 10⁻⁷ kN or kN·m).

The vertical reaction at C comes out negative, so C pulls the beam down. With no vertical load, $V_A$ and $V_C$ balance each other.

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 L-shaped frame: 48.00 at corner B and 24.00 kN·m at C — Figure 2. Bending moment (kN·m).

Shear diagram of the L-shaped frame: 0.00 at the base of the column and 12.00 kN in beam B–C — Figure 3. Shear force (kN).

Results

$A_{y}$: 12.00 kN
$C_{x}$: −24.00 kN
$C_{y}$: −12.00 kN
$M_{C}$: 24.00 kN·m

Matches the independent stiffness solution.

Common mistakes

The moment from the lateral load on the column passes through corner B at the same magnitude and must be carried into the beam.
The vertical unit load acts along the column axis, so it causes no moment there. $m_1$ is zero in the column.
Check horizontal equilibrium as well. The horizontal reaction at C carries the whole column load: 6 × 4 = 24 kN.

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