Force Method · Problem 4 of 7All problems

Force Method ExampleBeam with an overhang and a negative reaction

Problem. The beam is fixed at A, has rollers at B and C, and is free at D. A uniform load of 16 kN/m acts on span BC, and a 12 kN point load acts at D. EI is constant. Use the force method to find the support reactions and draw the shear and moment diagrams.

Beam fixed at A with rollers at B and C and a 2 m overhang: 16 kN/m uniform load on BC and a 12 kN point load at D — Figure 1. The beam, its loads, and its dimensions.

Try it yourself before reading the solution. Show the results

$A_{y}$: −11.08 kN
$M_{A}$: −18.46 kN·m
$B_{y}$: 61.23 kN
$C_{y}$: 57.85 kN

Solution

1Redundants and the primary structure

The beam is statically indeterminate to degree $2$ . Choose 2 redundants and remove them; what remains is the statically determinate primary structure.

Primary structure: a cantilever fixed at A, with the other supports removed.

Redundants: $X_{1} = B_{y}$ $X_{2} = C_{y}$

Primary structure: the released reactions are shown as X arrows

Remove B and C and what remains is a cantilever fixed at A. Working in from the free end, its moments are easy to write.

2M₀ diagram (primary structure, actual loads)

M_{0,\,A\text{–}B}(\xi) = -924 + 108\xi

M_{0,\,B\text{–}C}(\xi) = -384 + 108\xi - 8\xi^2

M_{0,\,C\text{–}D}(\xi) = -24 + 12\xi

Check the −924 kN·m at A by hand: 96 kN × 8 m + 12 kN × 13 m.

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) = 5 - \xi

m_{1,\,B\text{–}C}(\xi) = 0

m_{1,\,C\text{–}D}(\xi) = 0

m_{2}

X_{2}=1

(unit load at support C)

m_{2,\,A\text{–}B}(\xi) = 11 - \xi

m_{2,\,B\text{–}C}(\xi) = 6 - \xi

m_{2,\,C\text{–}D}(\xi) = 0

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{41.67}{EI}

\delta_{12} = \dfrac{116.67}{EI}

\delta_{22} = \dfrac{443.67}{EI}

\delta_{10} = \dfrac{-9300}{EI}

\delta_{20} = \dfrac{-32{,}808}{EI}

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

41.67\,X_{1} + 116.67\,X_{2} = 9300

116.67\,X_{1} + 443.67\,X_{2} = 32{,}808

X_{1} = B_{y} = 61.23\ \text{kN}

X_{2} = C_{y} = 57.85\ \text{kN}

Get $\delta_{11}$ by hand: $m_1$ is a triangle with base and height 5, so $\int m_1^2\,dx = 5 \cdot 5^2 / 3 = 41.67$ .

5Support reactions (superposition) and check

A_{y} = -11.08\ \text{kN},\quad M_{A} = -18.46\ \text{kN·m}

B_{y} = 61.23\ \text{kN}

C_{y} = 57.85\ \text{kN}

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

M_{B} = -36.92\ \text{kN·m}

M_{C} = -24\ \text{kN·m}

Equilibrium check: loads and support reactions together

Support	Ry (kN)	M (kN·m, ccw +)
A	-11.08	-18.46
B	61.23	0
C	57.85	0

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

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

ΣFy108.00 (reactions) − 108.00 (loads) = 0.00 kN ✓

ΣM₀924.00 (reactions) − 924.00 (loads) = 0.00 kN·m ✓

✓ The reactions match the matrix (direct stiffness) solution (largest difference 1.6 × 10⁻⁸ kN or kN·m).

$A_y$ came out negative, so support A pulls the beam down. That isn't an error: the load on BC lifts span AB about B.

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 an overhang: 18.46 at A, −36.92 at B, a maximum of 41.68 in span BC, and −24.00 kN·m at C — Figure 2. Bending moment (kN·m).

Shear diagram of the beam with an overhang: −11.08 in span AB, 50.15 just right of B, and 12.00 kN just right of C — Figure 3. Shear force (kN).

Results

$A_{y}$: −11.08 kN
$M_{A}$: −18.46 kN·m
$B_{y}$: 61.23 kN
$C_{y}$: 57.85 kN

Matches the independent stiffness solution.

Common mistakes

A negative $V_A$ is often taken for an arithmetic error. Under this loading, support A really does pull the beam down.
The 12 kN load at the tip of the overhang must be included. The −24 kN·m moment at C comes from it.
In $\delta_{10}$ and $\delta_{20}$ , reading the ordinate from the wrong diagram directly changes the result. An open-access textbook has a wrong answer for this reason.

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