Shear and Moment · Problem 1 of 3All problems

Shear and Moment DiagramsFrame shear and moment diagrams: L-shaped frame

Problem. The L-shaped frame is fixed at A and free at C. Column A–B is 4 m and beam B–C is 5 m. A lateral uniform load of 6 kN/m acts on the column from left to right, and a 12 kN vertical load acts at C. Find the support reactions, and draw the shear and moment diagrams.

L-shaped frame: fixed at A, free at C, 6 kN/m lateral uniform load on the 4 m column, 12 kN vertical load at the tip of the 5 m beam — Figure 1. The frame, its loads, and its dimensions.

Try it yourself before reading the solution. Show the results

$A_{x}$: −24.00 kN
$A_{y}$: 12.00 kN
$M_{A}$: 108.00 kN·m

Solution

For a statically determinate structure the equations of equilibrium are enough on their own; the steps below follow the order of a hand solution.

1Structure and unknowns

The structure is a single rigid body, so the equations of equilibrium are enough. 3 reactions = 3 unknowns; 1 free body × 3 equations of equilibrium = 3 equations.

A fixed support gives three reactions, and three equations are enough. You could also start from the free end; here the reactions come first.

2ΣF_x = 0 → A_x

ΣFx = 0 gives the horizontal reaction directly.

\Sigma F_x = 0:\quad A_{x} + 24 = 0 \;\Rightarrow\; A_{x} = -24\ \text{kN}

3ΣF_y = 0 → A_y

ΣFy = 0 gives the vertical reaction.

\Sigma F_y = 0:\quad A_{y} - 12 = 0 \;\Rightarrow\; A_{y} = 12\ \text{kN}

4ΣM_A = 0 → M_A

Summing moments about the support leaves only the moment reaction at the fixed support.

\Sigma M_{A} = 0:\quad M_{A} - 24 \cdot 2 - 12 \cdot 5 = 0 \;\Rightarrow\; M_{A} = 108\ \text{kN·m}

The lateral load's resultant is 24 kN, 2 m above A. The 12 kN has a 5 m arm: 48 + 60 = 108.

5Free bodies and moment sketch

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.

Loads and support reactions together

Walk to the corner from the free end: at the B end of the beam, M = 12 × 5 = 60. There's no applied moment at the corner, so the column top has 60 too.

6Check

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

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

ΣFy12.00 (reactions) − 12.00 (loads) = 0.00 kN ✓

ΣM₀108.00 (reactions) − 108.00 (loads) = 0.00 kN·m ✓

The equilibrium solution matches the matrix (direct stiffness) reactions (largest relative difference 2.5 × 10⁻¹⁴).

Bending moment diagram of the L-shaped frame: 108.00 at A and 60.00 kN·m at corner B — Figure 2. Bending moment (kN·m).

Shear diagram of the L-shaped frame: 24.00 at the base of the column and 12.00 kN in the beam — Figure 3. Shear force (kN).

Results

$A_{x}$: −24.00 kN
$A_{y}$: 12.00 kN
$M_{A}$: 108.00 kN·m

Matches the independent stiffness solution.

Common mistakes

The bending moment passes through the corner unchanged. The beam's shear becomes the axial force in the column.
The resultant of the column load is 24 kN and acts at the column's midheight, 2 m from A.
The moment diagram is a parabola in the loaded column and a straight line in the unloaded beam.

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