Shear and Moment · Problem 3 of 3All problems

Shear and Moment DiagramsThree-hinged frame example

Problem. The portal frame is pinned at A and E and has a hinge C at the middle of the beam. The columns are 5 m and the span is 8 m. A uniform load of 6 kN/m acts over the whole beam. Find the support reactions and hinge forces, and draw the shear and moment diagrams.

Three-hinged portal frame: pinned at A and E, hinge C at the middle of the beam, 5 m columns, 8 m span, 6 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

$A_{x}$: 9.60 kN
$A_{y}$: 24.00 kN
$E_{x}$: −9.60 kN
$E_{y}$: 24.00 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 (three-hinged frame: the hinge condition provides the fourth equation).

1Free-body diagrams and count

The hinges divide the structure into 2 segments: left segment (A–B–C) and right segment (C–D–E). 4 reactions + 2 hinge force components = 6 unknowns; 2 free bodies × 3 equations of equilibrium = 6 equations.

There are four reactions, and three overall equilibrium equations aren't enough. The hinge at C gives the fourth: M = 0 there.

2ΣM_A = 0 → E_y

Sum moments about A for the entire frame: this gives the vertical reaction at the other support.

\Sigma M_{A} = 0:\quad 8\,E_{y} - 24 \cdot 2 - 24 \cdot 6 = 0 \;\Rightarrow\; E_{y} = 24\ \text{kN}

3ΣF_y = 0 → A_y

ΣFy = 0 for the entire frame gives the other vertical reaction.

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

4ΣM_C = 0 → E_x

The moment at hinge C is zero, so summing moments about C for the right segment (C–D–E) gives the horizontal thrust by itself. Even a purely vertical load produces a horizontal thrust: that is the key idea of a three-hinged frame.

\Sigma M_{C} = 0:\quad 5\,E_{x} + 96 - 24 \cdot 2 = 0 \;\Rightarrow\; E_{x} = -9.60\ \text{kN}

Take the right part alone and write moments about C: V_E × 4 − 24 × 2 balances the thrust with its 5 m arm: 48 / 5 = 9.6.

5ΣF_x = 0 → A_x

ΣFx = 0 for the entire frame gives the remaining horizontal reaction.

\Sigma F_x = 0:\quad A_{x} - 9.60 = 0 \;\Rightarrow\; A_{x} = 9.60\ \text{kN}

6ΣF_x = 0 → C_x

From equilibrium of the right segment (C–D–E).

\Sigma F_x = 0:\quad - 9.60 - C_x = 0 \;\Rightarrow\; C_x = -9.60\ \text{kN}

7ΣF_y = 0 → C_y

From equilibrium of the right segment (C–D–E).

\Sigma F_y = 0:\quad 24 - C_y - 24 = 0 \;\Rightarrow\; C_y = 0\ \text{kN}

8Hinge forces (action and reaction)

A hinge transmits only force between two segments (no moment). The force acting on one segment acts on the other with the opposite sign (signs refer to the global axes: x positive to the right, y positive upward):

Hinge	Acts on	Fx (kN)	Fy (kN)
C	Left segment (A–B–C)	-9.60	0

9Free 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

Find the corner moment by hand: 9.6 × 5 = 48. At the hinge the beam moment is zero: 24 × 4 − 6 × 4 × 2 − 48 = 0.

10Check

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

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

ΣFy48.00 (reactions) − 48.00 (loads) = 0.00 kN ✓

ΣM₀192.00 (reactions) − 192.00 (loads) = 0.00 kN·m ✓

The equilibrium solution matches the matrix (direct stiffness) reactions (largest relative difference 5.9 × 10⁻¹⁶).

Bending moment diagram of the three-hinged frame: 48.00 kN·m at the corners and zero at the hinge — Figure 2. Bending moment (kN·m).

Shear diagram of the three-hinged frame: 9.60 in the columns and 24.00 kN at the corner end of the beam — Figure 3. Shear force (kN).

Results

$A_{x}$: 9.60 kN
$A_{y}$: 24.00 kN
$E_{x}$: −9.60 kN
$E_{y}$: 24.00 kN

Matches the independent stiffness solution.

Common mistakes

The M = 0 condition is written for the part on one side of the hinge, not for the whole structure.
A three-hinged system develops a horizontal reaction (thrust) under vertical load too.
A hinge carries no moment, so the diagram passes through zero there.

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