Three-hinged gable frame: 07-Str-A1, May 2015, Q8

Question 8: Three-hinged gable frame (22 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A gable frame pinned at node 1 and at node 5, with an internal hinge at the apex node 3. Taking node 1 as the origin, node 2 is 4 m above it, the apex node 3 is at $(12,\,9)$, node 4 at $(21,\,5.25)$ and the right-hand base node 5 at $(21,\,-3)$, that is 3 m below the level of node 1. Both rafters have the slope 5 in 12. A uniform load of 2.5 kN/m acts vertically over the horizontal projection of the left rafter, and a 30 kN horizontal force acts to the left on the right column, 3 m above node 5.

Given data — Question 8
Node	Coordinates (m)	Note
1	(0, 0)	pin
2	(0, 4)	left knee
3	(12, 9)	apex hinge
4	(21, 5.25)	right knee
5	(21, −3)	pin, 3 m below node 1
Loads: 2.5 kN/m on the projection 0–12 m (30 kN total); 30 kN horizontal, acting leftward, at (21, 0)

Find. The four reaction components and the shear force and bending moment diagrams for all four members, with the extreme ordinates.

Figure 12 — the three-hinged gable frame and its bending moment, developed along each half of the frame. Ordinates are plotted against distance measured along the member axis, so the continuity of moment through each rigid corner is visible as a single curve.

Approach. Four unknown reaction components need four equations: the three of overall equilibrium plus the condition that the bending moment at the apex hinge is zero, written as moments about the hinge for the right-hand portion alone.

Confirm determinacy. Two pins give $r = 4$; the apex hinge supplies one equation of condition, so $i = 4 - (3 + 1) = 0$ and statics alone suffices. The 2.5 kN/m acting over the 12 m projection totals 30 kN, applied through $x = 6$ m.
Moments about node 1 for the whole frame. The 30 kN horizontal force acts at the level of node 1, so it has no moment about that point; the right-hand reaction acts at $(21, -3)$. Hence $$ 21V_5 + 3H_5 = 30(6) = 180 . $$
Moments about the apex hinge for the right-hand portion. The right-hand portion carries only the 30 kN force and the reaction at node 5: $$ 9V_5 + 12H_5 = 30(9) = 270 . $$
Solve the pair. $$ \boxed{V_5 = 6.0\ \text{kN}\ (\uparrow), \qquad H_5 = 18.0\ \text{kN}\ (\rightarrow)} $$ and the two remaining equilibrium equations give $$ \boxed{V_1 = 24.0\ \text{kN}\ (\uparrow), \qquad H_1 = 12.0\ \text{kN}\ (\rightarrow)} $$ The horizontal check is $12 + 18 = 30$ kN, balancing the applied force, and the vertical check is $24 + 6 = 30$ kN, balancing the roof load.
Left column 1–2. It carries no transverse load, so the shear is the constant 12 kN and the moment grows linearly from zero at the pin to $$ M_2 = 12(4) = \boxed{48.0\ \text{kN}\!\cdot\!\text{m}} $$ with tension on the outer (left) face. The axial force is a constant 24 kN compression.
Left rafter 2–3. Taking the free body below and to the left of a section whose horizontal coordinate is $x$, and remembering that the distributed load acts on the projection, $$ M(x) = 48 - 19x + 1.25x^{2} $$ which is $+48$ kN·m at the knee, zero at the apex as the hinge demands, and stationary where $\mathrm{d}M/\mathrm{d}x = -19 + 2.5x = 0$, that is at $x = 7.6$ m, where $$ M = \boxed{-24.2\ \text{kN}\!\cdot\!\text{m}} $$ with tension on the inner (under) face. Resolving the net force $(12,\ 24-2.5x)$ perpendicular to the rafter, whose direction cosines are $12/13$ and $5/13$, the shear runs from $+17.54$ kN at the knee to $-10.15$ kN at the apex, vanishing at the same $x = 7.6$ m; the axial force runs from 20.31 kN to 8.77 kN compression.
Right column 5–4. Below the applied force the shear is the constant 18 kN and the moment grows to $$ M = 18(3) = \boxed{54.0\ \text{kN}\!\cdot\!\text{m}} $$ at the point of application, with tension on the inner (left) face. Above it the net horizontal force is $18 - 30 = -12$ kN, so the shear reverses to 12 kN and $M = 54 - 12y$ measured from that level, passing through zero 4.5 m above node 5 and reaching $$ M_4 = \boxed{-9.0\ \text{kN}\!\cdot\!\text{m}} $$ at the knee, now with tension on the outer (right) face. The axial force in the column is 6 kN compression throughout.
Right rafter 4–3. With no load on it the moment is linear, and $M = 12 - x$ in terms of the horizontal coordinate, so it runs from $-9.0$ kN·m at the knee to zero at the apex hinge, confirming the previous step from the other side. Its length is $\sqrt{9^2 + 3.75^2} = 9.75$ m, so the shear is the constant $9/9.75 = 0.92$ kN and the axial force is 13.39 kN compression.
Assemble the diagrams. The bending moment is zero at both pins and at the apex hinge, peaks at $+54$ kN·m where the 30 kN force is applied, reaches $+48$ kN·m at the left knee and dips to $-24.2$ kN·m on the left rafter. Those are the maximum and minimum ordinates of the whole frame.

Question 8 — reactions and extreme ordinates
Quantity	Value
Reaction at node 1	$H_1 = 12.0$ kN (to the right), $V_1 = 24.0$ kN upward
Reaction at node 5	$H_5 = 18.0$ kN (to the right), $V_5 = 6.0$ kN upward
Moment at the left knee (node 2)	48.0 kN·m, tension outside
Minimum moment on the left rafter	−24.2 kN·m at $x = 7.6$ m, tension inside
Moment at the apex hinge	zero
Maximum moment, right column	54.0 kN·m at the 30 kN level, tension inside
Moment at the right knee (node 4)	−9.0 kN·m, tension outside
Shear, left column / left rafter	12.0 kN constant / +17.54 to −10.15 kN
Shear, right column / right rafter	18.0 kN below, 12.0 kN above the load / 0.92 kN constant
Axial force, left column / left rafter	24.0 kN / 20.31 to 8.77 kN, compression
Axial force, right column / right rafter	6.0 kN / 13.39 kN, compression

Topic. Statics of a three-hinged gable frame with unequal column heights and bases at different levels.

Key relations. Three global equilibrium equations plus $\Sigma M_{\text{hinge}} = 0$ for one portion; for a member inclined at $(\cos\alpha,\ \sin\alpha)$ the shear and thrust are the components of the resultant of everything on one side, resolved perpendicular and parallel to the member; a load specified per metre of horizontal projection has resultant $w \times$ (projected length) acting through the centroid of the projection.

Why this works. The apex hinge converts a frame that would be indeterminate to the first degree into a determinate one by supplying the fourth equation, and that equation is best written for the smaller portion — here the right-hand side, which carries only two forces. Because the two bases are at different levels, the horizontal reactions have a moment about the reference point, so the equations do not uncouple in the familiar way and both must be solved simultaneously; the resulting reactions still come out in whole numbers, which is the sign that the geometry (5 in 12 rafters, 3 m base offset) was chosen deliberately. Plotting each half of the frame as a developed diagram, with the abscissa measured along the member axis, is what makes the continuity of moment at each rigid corner readable.

Common pitfall. Treating the 2.5 kN/m as acting per metre along the sloping rafter. It is drawn as a load block spanning the horizontal projection, so the resultant is $2.5 \times 12 = 30$ kN, not $2.5 \times 13 = 32.5$ kN, and the difference propagates into every reaction. The second common error is forgetting the 3 m level difference between the two bases, which contributes the $3H_5$ term to the moment equation about node 1 and, if dropped, gives $V_5 = 8.57$ kN and nothing checks afterwards.

Source. Hibbeler, Structural Analysis, 10th ed., §5.1–5.3 (statically determinate frames, three-hinged frames) and §4.5 (shear and moment diagrams for frames); Kassimali, Structural Analysis, 6th ed., §5.6 and §3.7.

Question 8 of 8: Three-hinged gable frame

Question 8: Three-hinged gable frame (22 marks)