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16-Civ-B1 Advanced Structural Analysis · December 2014

Question 4 of 9: Least work — moment and shear at joint (2) of a two-hinged portal (18 marks)

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

Notes on this paper

Paper format. National Examination, December 2014, 98-Civ-B1 Advanced Structural Analysis — 3 hours, closed book (approved Sharp or Casio calculator permitted). Nine questions over six pages: Questions 1 and 2 are compulsory (12 and 8 marks), then any two of Questions 3, 4, 5 (18 marks each) and any two of Questions 6, 7, 8, 9 (22 marks each). Six questions constitute a complete paper and total 100 marks; marks are printed in the left margin. All nine questions are solved here, because the set is a study resource rather than a three-hour sitting.

Reference texts.

Sign convention used throughout. End moments are counter-clockwise positive, the convention that matches the standard six-degree-of-freedom stiffness element. In it the slope-deflection equation is

$$M_{ij}=\frac{2EI}{L}\bigl(2\theta_i+\theta_j-3\psi_{ij}\bigr)+\mathrm{FEM}_{ij}$$

the chord rotation is $\psi_{ij}=\bigl[(\mathbf{d}_j-\mathbf{d}_i)\cdot\mathbf{e}_2\bigr]/L$ with $\mathbf{e}_2$ the member axis turned $+90^\circ$, the fixed-end moment of a downward UDL is $\mathrm{FEM}_{i}=+wL^{2}/12$ and $\mathrm{FEM}_{j}=-wL^{2}/12$, and the ordinary sagging bending moment is recovered as $M_{\text{sag}}(i)=-M_{ij}$ and $M_{\text{sag}}(j)=+M_{ji}$. Every translation degree of freedom is closed with the virtual-work statement

$$\sum_{\text{members}}\bigl(M_{ij}+M_{ji}\bigr)\,\psi^{*}_{ij}\;+\;\sum \mathbf{F}\cdot\mathbf{d}^{*}=0$$

in which $\mathbf{F}$ are the equivalent nodal loads and $\mathbf{d}^{*}$, $\psi^{*}$ belong to the unit sway pattern. Mixing this with the clockwise-positive convention of some textbooks produces clean-looking but wrong answers, which is the single most common failure on this subject.

All work below uses the drawing.

Question 4: Least work — moment and shear at joint (2) of a two-hinged portal (18 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 rectangular portal with pinned bases, carrying a uniform load on the beam.

Given data
QuantitySymbolValue
Column height, (1)–(2) and (4)–(3)$h$3 m
Beam span, (2)–(3)$L$6 m
Uniform load on the beam$w$12 kN/m downward
Supports at (1) and (4)—pinned (two reactions each)
Flexural rigidity$EI$the same for every member

Find. The bending moment at joint (2) and the shear force in the column immediately below joint (2).

12 kN/m 3 m 6 m (2) (3) (1) (4) H H
Question 4: pinned bases give four reactions for three equations, so the frame is one degree indeterminate and the horizontal thrust $H$ is the redundant.

Approach. Take the horizontal thrust as the redundant, write the internal moments in terms of it, and impose $\partial U/\partial H=0$ — the least-work condition that the thrust does no net work at a pinned base.

  1. Establish the degree of indeterminacy. Both bases are drawn as pins (a circle on a hatched triangle), so $r=4$ against three equations of statics: the frame is one degree indeterminate. Reading either base as a roller makes it determinate and least work would be unnecessary — the question's wording confirms the pins.
  2. Statics with the redundant carried symbolically. By symmetry the vertical reactions are $$V=\frac{wL}{2}=\frac{12(6)}{2}=36\ \text{kN}$$ and $H$ acts inward at each base. Measuring $y$ up a column and $x$ along the beam from joint (2), $$M_{\text{col}}(y)=Hy,\qquad M_{\text{beam}}(x)=Vx-\frac{wx^{2}}{2}-Hh$$
  3. Apply the least-work condition. With $\partial M_{\text{col}}/\partial H=y$ and $\partial M_{\text{beam}}/\partial H=-h$, $$\frac{\partial U}{\partial H}=\frac{2}{EI}\int_{0}^{h}(Hy)(y)\,\mathrm{d}y-\frac{h}{EI}\int_{0}^{L}\Bigl(Vx-\frac{wx^{2}}{2}-Hh\Bigr)\mathrm{d}x=0$$
  4. Evaluate the integrals. The first is $2Hh^{3}/3$; in the second, $\int_{0}^{L}(Vx-wx^{2}/2)\,\mathrm{d}x=wL^{3}/12$, so $$\frac{2Hh^{3}}{3}-h\left(\frac{wL^{3}}{12}-HhL\right)=0 \;\Longrightarrow\; H=\frac{wL^{3}}{8h^{2}+12hL}$$
  5. Substitute the numbers. $$H=\frac{12(6)^{3}}{8(3)^{2}+12(3)(6)}=\frac{2592}{72+216}=\frac{2592}{288}=\boxed{9.00\ \text{kN}}$$
  6. Bending moment at joint (2). The column is straight, unloaded along its length and pinned at the base, so its moment grows linearly to $$\boxed{M_2=Hh=9.00(3)=27.0\ \text{kN}\cdot\text{m}\ \text{(hogging at the beam end)}}$$
  7. Shear in the column below joint (2). The only transverse force on that column is the base thrust, so the shear is constant over its full height: $$\boxed{V_{\text{col}}=H=9.00\ \text{kN}}$$
  8. Check with the beam. At midspan, $$M_{\text{mid}}=V\frac{L}{2}-\frac{w}{2}\left(\frac{L}{2}\right)^{2}-Hh=108-54-27=+27.0\ \text{kN}\cdot\text{m}$$ so the frame divides the free-beam moment $wL^{2}/8=54\ \text{kN}\cdot\text{m}$ into equal hogging and sagging halves — a well-known result for this ratio of $h$ to $L$, and a strong confirmation that $H$ is right.

The thrust is what makes the portal efficient: it lifts 27 kN·m of the 54 kN·m free-beam moment off the midspan and puts it into the corners, where the column can share it.

Beam bending moment (sagging +) x from joint (2), m M (kN·m) -27.0 +27.0 -27.0
Question 4: beam bending moment. Equal hogging and sagging peaks of 27.0 kN·m are a consequence of $h/L=1/2$ with uniform $EI$.
Question 4 — results
QuantityValue
Vertical reaction at each base36.0 kN
Horizontal thrust (the redundant)9.00 kN inward at each base
Bending moment at joint (2)27.0 kN·m hogging
Shear in the column below joint (2)9.00 kN (constant over the height)
Beam midspan sagging moment27.0 kN·m
Free-beam check $wL^{2}/8$54.0 kN·m = 27.0 + 27.0