Shear and Moment · Problem 2 of 3All problems

Shear and Moment DiagramsSloped beam shear and moment

Problem. The sloped beam is pinned at A and has a roller at B. Its horizontal projection is 6 m and its rise is 3 m. A vertical load of 10 kN/m per member length acts along it. Find the support reactions, and draw the shear and moment diagrams.

Sloped beam: pinned at A, roller at B, 6 m horizontal projection, 3 m rise, 10 kN/m vertical load per member length — Figure 1. The frame, its loads, and its dimensions.

Try it yourself before reading the solution. Show the results

$A_{y}$: 33.54 kN
$B_{y}$: 33.54 kN

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.

2ΣM_A = 0 → B_y

Sum moments about A; the reactions at A drop out of the equation.

\Sigma M_{A} = 0:\quad 6\,B_{y} - 67.08 \cdot 3 = 0 \;\Rightarrow\; B_{y} = 33.54\ \text{kN}

The total load uses the member length: $10 \cdot \sqrt{6^2 + 3^2} = 67.08$ kN. The resultant sits at mid-span horizontally, with a 3 m arm.

3ΣF_y = 0 → A_y

ΣFy = 0 gives the remaining vertical reaction.

\Sigma F_y = 0:\quad A_{y} + 33.54 - 67.08 = 0 \;\Rightarrow\; A_{y} = 33.54\ \text{kN}

4ΣF_x = 0 → A_x

ΣFx = 0 gives the horizontal reaction.

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

The roller holds vertically and there's no horizontal load, so H_A = 0.

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

The diagram shows 30 while the reaction is 33.54: shear is the component normal to the member, 33.54 × 6 / 6.71 = 30. The peak moment uses the equivalent load per horizontal length: 11.18 × 6² / 8 = 50.31.

6Check

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

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

ΣFy67.08 (reactions) − 67.08 (loads) = 0.00 kN ✓

ΣM₀201.25 (reactions) − 201.25 (loads) = 0.00 kN·m ✓

The equilibrium solution matches the matrix (direct stiffness) reactions (largest relative difference 2.1 × 10⁻¹⁶).

Bending moment diagram of the sloped beam: a maximum of 50.31 kN·m and zero at the ends — Figure 2. Bending moment (kN·m).

Shear diagram of the sloped beam: 30.00 kN at the ends and zero at the middle — Figure 3. Shear force (kN).

Results

$A_{y}$: 33.54 kN
$B_{y}$: 33.54 kN

Matches the independent stiffness solution.

Common mistakes

A load per member length shouldn't be treated as a load per horizontal length. The equivalent per horizontal length is 10 × 6.71 / 6 = 11.18 kN/m; using 10 kN/m gives a moment of 45 kN·m.
A support reaction isn't the shear force. The component of the vertical reaction normal to the member is the shear; the component along it is the axial force.
For a vertical force, the moment arm is the horizontal distance, not the sloped one.

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