Force Method · Problem 6 of 7All problems

Force Method ExampleForce method portal frame, unequal stiffness

Problem. The portal frame is fixed at A and pinned at D. The columns are 4 m with I, and beam B–C is 8 m with 2I. A uniform load of 15 kN/m acts on the beam. Axial deformation is neglected. Use the force method to find the support reactions, and draw the shear and moment diagrams.

Portal frame: fixed at A, pinned at D, 4 m columns with I, 8 m beam with 2I, 15 kN/m uniform load on the beam — Figure 1. The frame, its loads, and its dimensions.

Try it yourself before reading the solution. Show the results

$A_{x}$: 13.64 kN
$A_{y}$: 58.64 kN
$M_{A}$: −10.91 kN·m
$D_{x}$: −13.64 kN
$D_{y}$: 61.36 kN

Solution

1Redundants and the primary structure

The frame is statically indeterminate to degree $2$ . Release 2 support reactions; 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 D; the vertical reaction at D; the support at A remains.

Redundants: $X_{1} = D_{x}$ $X_{2} = D_{y}$

EI differs between members, so the integrals are weighted by 1/EI (EI₀ as the reference): $EI_{A\text{–}B} = EI_0$ $EI_{B\text{–}C} = 2EI_0$ $EI_{C\text{–}D} = EI_0$

Primary structure: the released reactions are shown as X arrows

Removing D leaves a frame fixed at A with a free end at D. The two reactions at D are the two unknowns.

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	-480.00	480.00
B–C	-480.00	0
C–D	0	0

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 D)

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	-4.00
C–D	4.00	0

m_{2}

—

X_{2}=1

(the vertical reaction at D)

m₂ diagram: the unit redundant alone on the primary structure

Member	$m^{2}_{i}$ (kN·m)	$m^{2}_{j}$ (kN·m)
A–B	8.00	-8.00
B–C	8.00	0
C–D	0	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{106.67}{EI_0}

\delta_{12} = \dfrac{128}{EI_0}

\delta_{22} = \dfrac{341.33}{EI_0}

\delta_{10} = \dfrac{-6400}{EI_0}

\delta_{20} = \dfrac{-19{,}200.0}{EI_0}

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

106.67\,X_{1} + 128\,X_{2} = 6400

128\,X_{1} + 341.33\,X_{2} = 19{,}200.0

X_{1} = D_{x} = -13.64\ \text{kN}

X_{2} = D_{y} = 61.36\ \text{kN}

Find $\delta_{11}$ by hand: columns 4³/3 + 4³/3, beam 4² × 8 / 2. The beam's share is divided by 2 because it's 2I: 21.33 + 21.33 + 64 = 106.67.

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} = 10.91\ \text{kN·m}

M_{BA} = 43.64\ \text{kN·m}

M_{BC} = -43.64\ \text{kN·m}

M_{CB} = 54.55\ \text{kN·m}

M_{CD} = -54.55\ \text{kN·m}

M_{DC} = 0\ \text{kN·m}

Equilibrium check: loads and support reactions together

Support	Rx (kN)	Ry (kN)	M (kN·m, ccw +)
A	13.64	58.64	-10.91
D	-13.64	61.36	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 ✓

ΣM₀480.00 (reactions) − 480.00 (loads) = 0.00 kN·m ✓

✓ The reactions match the method's own reference, the axially rigid (EA ≈ ∞) matrix solution (largest difference 6.7 × 10⁻⁷ kN or kN·m).

$X_1$ is negative: the horizontal reaction at D acts opposite the assumed direction, pushing the frame inward. Horizontal equilibrium gives the same magnitude at A.

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 portal frame: 10.91 at A, 43.64 at B, 54.55 at C, and a maximum of 70.97 kN·m in the beam — Figure 2. Bending moment (kN·m).

Shear diagram of the portal frame: 58.64 at the B end of the beam, 61.36 at the C end, and 13.64 kN in the columns — Figure 3. Shear force (kN).

Results

$A_{x}$: 13.64 kN
$A_{y}$: 58.64 kN
$M_{A}$: −10.91 kN·m
$D_{x}$: −13.64 kN
$D_{y}$: 61.36 kN

Matches the independent stiffness solution.

Common mistakes

With unequal moments of inertia, every integral over the beam is divided by the I ratio. Without this the numbers look reasonable, but the result is wrong.
In $\delta_{12}$ and $\delta_{10}$ , reading the ordinate from the wrong diagram directly changes the result. An open-access textbook has a wrong answer for this reason.
Vertical equilibrium catches arithmetic slips: 58.64 + 61.36 = 15 × 8.

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