24-Bld-A1 Elementary Structural Analysis · December 2017
Question 7 of 8: Virtual-work horizontal deflection of a hinged portal frame (22 marks)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams December 2017 — 07-BLD-A1 Elementary Structural Analysis, 3 hours, closed book. Answer ALL of Questions 1–5; answer ONLY ONE of Questions 6, 7 or 8 (this solution set, per pipeline convention, answers all three optional questions).
Given. Pin support at 1 (0,0); a hinge at 2 (0,10) connecting the left column to the top beam; the top beam runs 10+10=20 m to a rigid corner at 3 (20,10); a column drops to a pin support at 4 (20,0); a 20 kN point load at the beam's midspan.
Find. Horizontal deflection at joint 3.
Frame: pin at 1, hinge at 2, rigid corner at 3, pin at 4; 20 kN at beam midspan.
Check: the small circle drawn at joint 2 (and nowhere else on this frame's rigid corner at 3) is read as a genuine internal hinge, not just a joint marker — this is what the paper's own drafting convention uses elsewhere (e.g. Q1(a)'s "TYPICAL HINGE"). It is also the only reading that makes the frame statically determinate, matching a problem that asks for virtual work directly (no redundant to resolve first).
Approach. With a hinge at 2, column 1-2 is pinned at both ends (support at 1, hinge at 2) and so is a two-force axial member — it can only feed a vertical force into joint 2. That leaves the L-shaped bent 2-3-4 carried by this vertical "prop" at 2 plus the pin at 4: three reactions for three equilibrium equations, so the whole frame is determinate. Build the real M-diagram, then a virtual unit horizontal load at 3, and integrate $$Mm/EI$$ (flexural strain only, so the axial column 1-2 contributes nothing).
Real reactions. Taking moments about the pin at 4 (only vertical forces act, so lever arms are horizontal distances): $$\Sigma M_4=0:\ 20\,R_{2,vert}=20(10)\Rightarrow R_{2,vert}=10\text{ kN}$$ (column 1-2 in compression, propping joint 2 up), and $$\Sigma F_y=0:\ R_{4,y}=20-10=10\text{ kN},\qquad R_{4,x}=0$$ (no horizontal load anywhere).
Real moments along the beam (x from joint 2, hinge $$\Rightarrow M(0)=0$$): $$M(x)=10x\ (0\le x\le10),\qquad M(x)=10x-20(x-10)=200-10x\ (10\le x\le20)$$ giving $$M(10)=100\text{ kN}\cdot\text{m}$$ and $$M(20)=0$$ at joint 3. Because $$R_{4,x}=0$$, the column 3-4 carries zero moment.
Virtual system — unit horizontal load at 3: this is resisted entirely by the pin at 4 (the axial strut 1-2 cannot carry horizontal force), $$R_{4,x}=-1,\ R_{4,y}=0.5,\ R_{2,vert}=-0.5$$. A horizontal unit load is collinear with the (horizontal) beam, so it produces no bending in the beam directly — only the vertical prop reaction does: $$m(x)=-0.5x\ (0\le x\le20)$$ giving $$m(20)=-10$$, which continues linearly down column 3-4 to $$m=0$$ at the pin (4), consistent with $$R_{4,x}=-1$$ acting over the 10 m column height.
Integrate. Column 1-2: axial only, excluded. Column 3-4: real $$M\equiv0$$, contributes nothing regardless of $$m$$. Beam only: $$\int_0^{10}(10x)(-0.5x)\,dx+\int_{10}^{20}(200-10x)(-0.5x)\,dx = -1666.7-3333.3=-5000\text{ kN}^2\text{m}^3$$$$\Delta_3=\dfrac{-5000}{EI}=\dfrac{-5000}{2.0\times10^5}=-0.025\text{ m}=\boxed{25.0\text{ mm}}$$ (the negative sign shows joint 3 moves opposite to the assumed outward unit-load direction, i.e. toward joint 2).
