Force Method · Problem 7 of 7All problems

Force Method ExampleGable frame force method

Problem. The gable frame is pinned at A and E. The columns are 4 m, the ridge is 2 m above the column tops, and the span is 8 m. A vertical load of 10 kN/m per member length acts on the rafters. EI is constant and axial deformation is neglected. Use the force method to find the support reactions, and draw the shear and moment diagrams.

Gable frame: pinned at A and E, 4 m columns, 2 m ridge height, 8 m span, 10 kN/m vertical load per member length on the rafters — Figure 1. The frame, its loads, and its dimensions.

Try it yourself before reading the solution. Show the results

$A_{x}$: 10.40 kN
$A_{y}$: 44.72 kN
$E_{x}$: −10.40 kN
$E_{y}$: 44.72 kN

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 horizontal reaction at A; the supports at A, E remain.

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

Primary structure: the released reactions are shown as X arrows

The unknown is the horizontal thrust. Making A a roller leaves a simply supported frame, and symmetry gives the vertical reactions right away.

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	0
B–C	0	-89.44
C–D	89.44	0
D–E	0	0

The load is per member length: $10 \cdot \sqrt{20} = 44.72$ kN on each rafter. The moment at C: 44.72 × 4 − 44.72 × 2 = 89.44.

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 horizontal 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	4.00
B–C	-4.00	6.00
C–D	-6.00	4.00
D–E	-4.00	0

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{269.25}{EI}

\delta_{10} = \dfrac{-2800}{EI}

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

269.25\,X_{1} = 2800

X_{1} = A_{x} = 10.40\ \text{kN}

Find $\delta_{11}$ by hand: columns 2 × 4³/3 = 42.67. Rafters 2 × $\sqrt{20}$ × (4² + 4 × 6 + 6²)/3 = 226.58. Use the sloped length, not the horizontal projection.

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} = 41.60\ \text{kN·m}

M_{BC} = -41.60\ \text{kN·m}

M_{CB} = -27.05\ \text{kN·m}

M_{CD} = 27.05\ \text{kN·m}

M_{DC} = 41.60\ \text{kN·m}

M_{DE} = -41.60\ \text{kN·m}

M_{ED} = 0\ \text{kN·m}

Equilibrium check: loads and support reactions together

Support	Rx (kN)	Ry (kN)	M (kN·m, ccw +)
A	10.40	44.72	0
E	-10.40	44.72	0

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

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

ΣFy89.44 (reactions) − 89.44 (loads) = 0.00 kN ✓

ΣM₀357.77 (reactions) − 357.77 (loads) = 0.00 kN·m ✓

✓ The reactions match the method's own reference, the axially rigid (EA ≈ ∞) matrix solution (largest difference 8.3 × 10⁻⁸ kN or kN·m).

Final moments by hand: at B, 10.40 × 4 = 41.6; at the ridge, 89.44 − 10.40 × 6 = 27.05.

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 gable frame: 41.60 at the column tops and 27.05 kN·m at the ridge — Figure 2. Bending moment (kN·m).

Shear diagram of the gable frame: 10.40 kN in the columns and 35.35 kN at the B end of the rafter — Figure 3. Shear force (kN).

Results

$A_{x}$: 10.40 kN
$A_{y}$: 44.72 kN
$E_{x}$: −10.40 kN
$E_{y}$: 44.72 kN

Matches the independent stiffness solution.

Common mistakes

When the load is given per member length, the total uses the sloped length: 10 × 2 × 4.47 = 89.44 kN. Using the horizontal projection, 10 × 8 = 80 kN, is wrong.
On a sloped member, the virtual work integral runs along the member axis, not over the horizontal length.
Both supports are held horizontally, so a horizontal reaction develops under vertical load too. The thrust here is 10.40 kN.

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