07-Str-A1 · December 2015

Question 6 of 8: Three-hinged frame with a vertical load and a wind pressure

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examinations, December 2015, 07-Str-A1 — Elementary Structural Analysis. Three hours, closed book, approved Sharp or Casio calculator only. Six questions constitute a complete paper: answer ALL of Questions 1 to 4, then ONLY TWO of Questions 5, 6, 7 or 8. Marks are shown in the left margin (6 + 18 + 18 + 18 + 20 + 20 = 100). All eight questions are worked below, because the four alternatives exercise four different methods — moment distribution, three-hinged-frame statics, influence lines and virtual work — and the complete set is the more useful study resource.

Reference texts.

R. C. Hibbeler, Structural Analysis, 10th ed. — Ch. 2 (determinacy and stability), Ch. 3–5 (trusses, internal loadings, frames), Ch. 6 (influence lines), Ch. 8–9 (deflections and virtual work), Ch. 11–12 (slope-deflection and moment distribution).
A. Kassimali, Structural Analysis, 6th ed. — Ch. 3 (equilibrium and determinacy), Ch. 4 (plane trusses), Ch. 5 (beams and frames), Ch. 7 (virtual work), Ch. 8–9 (influence lines), Ch. 16–17 (slope-deflection and moment distribution).
K. Leet, C.-M. Uang and A. Gilbert, Fundamentals of Structural Analysis, 5th ed. — a parallel treatment of the same syllabus.
C. H. Norris, J. B. Wilbur and S. Utku, Elementary Structural Analysis, 4th ed. — the classical text this exam code is named after.

Check: the roller in Question 2(c). The roller under the inclined member of Question 2(c) is drawn on a bed ruled parallel to the member, so its reaction is taken normal to the member axis. A vertical-roller reading is also arithmetically self-consistent. The distinction changes only the reaction components and the axial force: the shear and bending-moment diagrams the question asks for are identical under either reading, and both sets of reactions are reported.

Question 6: Three-hinged frame with a vertical load and a wind pressure (20 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 three-hinged frame: two pinned supports at different levels, a horizontal top member carrying an internal hinge, and a horizontal pressure on the right-hand leg.

Given data for Question 6
Quantity	Value
Support A (left)	Pin at (0, 0)
Left knee B	(4.5, 6) m — leg AB is 7.5 m long on a 3 : 4 slope
Internal hinge	(8, 6) m, on the horizontal top member
Right knee C	(14.25, 6) m — top member BC is 9.75 m long
Support D (right)	Pin at (18, 1) m — leg CD is 6.25 m long, dropping 5 m over 3.75 m
Vertical load	80 kN downwards at the right knee C
Horizontal pressure	24 kN/m acting to the left over the 5 m vertical projection of leg CD, i.e. 120 kN at mid-height, $y = 3.5$ m

Find. The four reaction components, and the shear force and bending moment diagrams for each of the three members, with maximum and minimum ordinates.

Figure 9 — structure 6 with its reactions, and the shear and bending moment diagrams developed along the member axis A–B–C–D. Dashed lines mark the two knees; the moment passes through zero exactly at the hinge, which is the arithmetic check on the whole solution.

Approach. Two pins give four reaction components against three equations of equilibrium; the internal hinge supplies the fourth equation — the moment of everything on one side of it about the hinge vanishes. Solve the four together, then walk the members.

The hinge condition. No load acts on the part of the frame to the left of the hinge, so only the reaction at A appears. Taking moments about the hinge at (8, 6), $$ -8 A_y + 6 A_x = 0 \quad\Longrightarrow\quad A_y = 0.75\,A_x . $$ The ratio is simply the hinge coordinates: the reaction at A must act along the line joining A to the hinge.
Global equilibrium. The pressure resultant is $24 \times 5 = 120$ kN acting to the left at $y = 3.5$ m, so $$ \sum F_x = 0 : \quad A_x + D_x = 120, \qquad \sum F_y = 0 : \quad A_y + D_y = 80 , $$ and moments about A, with the 80 kN load at $x = 14.25$ m and the pressure at $y = 3.5$ m, give $$ 18 D_y - 1 D_x - 14.25(80) + 3.5(120) = 0 \quad\Longrightarrow\quad 18 D_y - D_x = 720 . $$
Solve the four equations. Substituting $D_x = 120 - A_x$ and $D_y = 80 - 0.75 A_x$, $$ 18\left(80 - 0.75 A_x\right) - \left(120 - A_x\right) = 720 \quad\Longrightarrow\quad 1320 - 12.5 A_x = 720, $$ $$ \boxed{A_x = 48\ \text{kN} \rightarrow, \quad A_y = 36\ \text{kN} \uparrow, \quad D_x = 72\ \text{kN} \rightarrow, \quad D_y = 44\ \text{kN} \uparrow} $$ Both horizontal reactions push to the right, which is what a pressure blowing to the left demands. A moment check about D closes to zero exactly.
Leg AB. No load acts along it, so the shear and axial force are constant. Resolving the reaction at A onto the normal $(-0.8,\,0.6)$ and the axis $(0.6,\,0.8)$ of the leg, $$ V_{AB} = 48(-0.8) + 36(0.6) = -16.8\ \text{kN}, \qquad N_{AB} = 48(0.6) + 36(0.8) = 57.6\ \text{kN compression}. $$ The bending moment grows linearly from zero at the pin to $$ M_B = -16.8 \times 7.5 = \boxed{-126\ \text{kN}\cdot\text{m}} $$ a hogging value that puts the outer (left-hand) face of the leg in tension.
Top member BC. Only the vertical reaction at A produces shear on this member, so $$ V_{BC} = A_y = 36\ \text{kN} \ \text{constant}, $$ and the sagging moment at a horizontal coordinate $x$ is $$ M(x) = 36x - 6(48) = 36x - 288 . $$ It is $-126$ kN·m at the knee B, exactly zero at $x = 8$ m — the hinge — and $$ M_C = 36(14.25) - 288 = \boxed{+225\ \text{kN}\cdot\text{m}} $$ at the right knee, a sagging value. The zero at the hinge is not an assumption: it falls out of numbers obtained without ever using it a second time, and it is the single best check available on this question.
Leg CD. Here the pressure acts, so both shear and moment vary. Taking the free body below a section at height $y$ and resolving onto the leg normal $(0.8,\,0.6)$, $$ V(y) = 0.8\left[72 - 24(y - 1)\right] + 0.6(44) = 84 - 19.2\,(y - 1), $$ so the shear runs from $+84$ kN at the pin D up to $-12$ kN at the knee C, crossing zero at $y = 5.375$ m — that is 0.781 m below C along the member. The bending moment about the same section is $$ M(y) = (18 - x)(44) - (1 - y)(72) - 12\,(y-1)^2, \qquad x = 14.25 + 0.75(6-y), $$ which is zero at the pin, $+225$ kN·m at the knee, and reaches its extreme value at the point of zero shear: $$ M_{\max} = \boxed{+229.69\ \text{kN}\cdot\text{m}} \quad\text{at } y = 5.375\ \text{m}. $$ The leg also carries an axial force running from 8 kN tension at D to 64 kN compression at C, the change being the along-axis component of the pressure.
Signs of the diagram segments. Leg AB hogs over its whole length. The top member hogs from B to the hinge at 8 m and sags from the hinge to C. Leg CD sags over its whole length. The extreme ordinates are therefore $+84$ kN and $-16.8$ kN on the shear diagram, and $+229.69$ kN·m and $-126$ kN·m on the bending moment diagram.

Question 6 — reactions and extreme ordinates
Item	Value
Reaction at A	$A_x = 48$ kN →, $A_y = 36$ kN ↑
Reaction at D	$D_x = 72$ kN →, $D_y = 44$ kN ↑
Leg AB (7.5 m)	$V = 16.8$ kN constant; $N = 57.6$ kN compression; $M$: 0 to −126 kN·m
Top member BC (9.75 m)	$V = 36$ kN constant; $M$: −126 → 0 at the hinge → +225 kN·m
Leg CD (6.25 m)	$V$: −12 kN at C to +84 kN at D; $M_{\max} = +229.69$ kN·m at 0.781 m below C
Overall extremes	Shear +84 / −16.8 kN; moment +229.69 / −126 kN·m

Topic. Statics of a three-hinged frame under combined vertical load and lateral pressure.

Key relations. Four unknowns, four equations: $\sum F_x = 0$, $\sum F_y = 0$, $\sum M = 0$ and the condition $\sum M_{\text{hinge}} = 0$ taken for one side only. For a pressure specified per metre of vertical projection, the resultant is $w h$ acting at mid-height of that projection. Shear and axial force on an inclined member are the components of the resultant of one side normal to and along the member axis.

Why this works. A hinge in the top member means the two halves of the frame can rotate relative to one another, so the moment there must vanish; that single extra equation converts an externally indeterminate frame (four reaction components) into a determinate one. Applying the condition to the side that carries no load, as here, reduces it to a statement about the line of action of one reaction — the reaction at A must point at the hinge — which is quicker and much less error-prone than writing the same equation for the loaded side. Developing the diagrams along the member axis rather than hanging them off the sloping members keeps the continuity of moment through each rigid knee visible as one unbroken curve.

Common pitfall. Taking the 24 kN/m along the 6.25 m sloping length of the leg instead of over its 5 m vertical projection: that gives 150 kN, and every reaction then comes out with no round numbers anywhere — the usual sign in this examination that a load has been read the wrong way. The second pitfall is assuming that the peak moment on leg CD sits at the knee; it sits at the point of zero shear, a little below.

Source. Hibbeler, Structural Analysis, 10th ed., §5.3–5.4 (frames and three-hinged arches); Kassimali, Structural Analysis, 6th ed., §5.6 and §5.8.

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