Anti-Symmetric Analysis of a Portal Frame: 07-Str-A4, December 2016, Q8

Question 8: Anti-Symmetric Analysis of a Portal 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 symmetric single-bay portal with both bases built in, loaded by a pair of horizontal forces applied at mid-height of each column and both pointing the same way.

Given data
Quantity	Symbol	Value
Beam span (joints 2 to 3)	$b$	4.5 m
Column height (joints 1 to 2, 4 to 3)	$h$	6.0 m
Height of each applied load above the base	$h_L$	3.0 m
Applied horizontal load on each column	$P$	12 kN, both to the right
Flexural rigidity of every member	$EI$	the same, inextensible

Find. The member-end moments and the shear force and bending moment diagrams of every member, with maximum and minimum ordinates.

Question 8: symmetric portal under an anti-symmetric pair of horizontal loads. Both 12 kN forces act to the right at mid-height.

Approach. Establish that the loading is anti-symmetric, reduce the three unknowns $(\theta_2,\theta_3,\Delta)$ to two by $\theta_2=\theta_3$, write the joint equation and the storey-shear equation, and solve the $2\times2$ system.

Classify the loading. Reflecting the frame about its centre line maps the left column onto the right one and reverses the direction of a horizontal force. Both applied forces point right, so the reflected load set is the negative of the original: the loading is anti-symmetric. The response is therefore anti-symmetric as well — both joints rotate through the same angle $\theta_2=\theta_3=\theta$, both columns sway through the same $\Delta$, and the two columns carry identical internal forces. Three unknowns collapse to two.
Write the column and beam equations. Each column has a transverse point load at mid-height, so $\mathrm{FEM}_{12}=+Ph/8=+9.0$ and $\mathrm{FEM}_{21}=-9.0\ \text{kN}\cdot \text{m}$, and $\psi_{12}=-\Delta/h$. With $\theta_1=0$ at the built-in base, $$M_{12}=\frac{EI}{3}\left(\theta+\frac{\Delta}{2}\right)+9,\qquad M_{21}=\frac{EI}{3}\left(2\theta+\frac{\Delta}{2}\right)-9.$$ The beam has $\theta_3=\theta_2=\theta$ and no chord rotation (the columns are inextensible), so $M_{23}=M_{32}=6EI\theta/4.5=1.3333\,EI\theta$.
Joint equilibrium. $M_{21}+M_{23}=0$ gives $$2.0\,EI\theta+0.16667\,EI\Delta=9. \tag{i}$$
Storey-shear equation. The two base shears must equilibrate the total applied horizontal force of $24$ kN, and by anti-symmetry they are equal, so each base shear is $12$ kN. Taking moments on one column about its top joint, $$M_{12}+M_{21}+6H+36=0\quad\text{with}\quad H=-12 \;\Longrightarrow\;M_{12}+M_{21}=36,$$ which in terms of the unknowns is $$EI\theta+0.33333\,EI\Delta=36. \tag{ii}$$
Solve. Equations (i) and (ii) give $$\boxed{EI\theta=-6.0\ \text{kN}\cdot\text{m}^{2},\qquad EI\Delta=+126.0\ \text{kN}\cdot\text{m}^{3}},$$ so the frame sways $126/EI$ to the right while both joints rotate clockwise. Back substitution gives the member-end moments $$M_{12}=M_{43}=+28.0,\qquad M_{21}=M_{34}=+8.0,\qquad M_{23}=M_{32}=-8.0\ \text{kN}\cdot\text{m}.$$
Bending moment diagram. Each column runs from $\boxed{28.0\ \text{kN}\cdot\text{m}}$ at the base — the maximum ordinate anywhere in the frame — linearly to $-8.0$ at the load point $3.0$ m up, crossing zero at $2.33$ m, and then holds the constant $-8.0\ \text{kN}\cdot\text{m}$ over the whole upper $3.0$ m because the shear there is zero. The beam runs linearly from $+8.0$ at joint 2 to $-8.0$ at joint 3, passing through zero at midspan — the signature of an anti-symmetric response. Both columns carry identical diagrams, not mirror images.
Shear force diagram and equilibrium check. Each column carries $\boxed{12.0\ \text{kN}}$ of shear over its lower 3.0 m and exactly zero over its upper 3.0 m; the beam carries the constant $16.0/4.5=3.556$ kN, which is also the axial force in each column (tension on the left, compression on the right). The overturning check closes exactly: $$2(28.0)+3.556(4.5)=56.0+16.0=72.0=24\times3.0\ \text{kN}\cdot\text{m}.$$

Question 8: bending moment diagram (kN.m). Both columns carry identical diagrams, 28.0 at the base and a constant -8.0 above the load point; the beam passes through zero at midspan.

Question 8 — results
Quantity	Value
$EI\theta_2=EI\theta_3$	$-6.0\ \text{kN}\cdot\text{m}^{2}$
$EI\Delta$ (sway to the right)	$+126.0\ \text{kN}\cdot\text{m}^{3}$
Moment at each column base (maximum ordinate)	$28.0\ \text{kN}\cdot\text{m}$
Moment at the load point and along the upper column	$-8.0\ \text{kN}\cdot\text{m}$ (minimum ordinate)
Beam end moments	$\pm 8.0\ \text{kN}\cdot\text{m}$, zero at midspan
Column shear, lower 3.0 m / upper 3.0 m	$12.0$ kN / $0$
Beam shear (constant)	$3.556$ kN
Column axial force	$3.556$ kN (tension left, compression right)
Base reactions (each)	$12.0$ kN horizontal, $3.556$ kN vertical, $28.0\ \text{kN}\cdot\text{m}$
Height of zero moment in each column	$2.33$ m above the base

Topic. Anti-symmetric reduction of a symmetric sway frame analysed by slope-deflection.

Key relations. Anti-symmetry conditions $\theta_{\text{left}}=\theta_{\text{right}}$, equal sway $\Delta$ in both columns, zero vertical joint translation for inextensible columns; $\mathrm{FEM}=\pm Ph/8$ for a mid-height point load; joint equilibrium $\sum M_{ij}=0$; storey shear $\sum(M_{ij}+M_{ji})/h=\sum F$; overturning check $\sum M_{\text{base}}+N b=\sum F h_L$.

Why this works. Anti-symmetry is worth the effort of proving because it halves the problem and because it dictates the shape of the answer before any arithmetic is done. Under a reflection, a horizontal force reverses direction, so two like-directed horizontal loads form an anti-symmetric set; the response then has equal joint rotations rather than equal and opposite ones, and the two columns end up with identical diagrams. That is the fact most worth carrying away: a mirror-image pair of column diagrams would signal a symmetric response and therefore an error in the classification. The zero column shear above the load point follows from the same accounting — the base shear of $12$ kN is exactly cancelled by the $12$ kN applied at mid-height, so the upper half of each column carries a constant moment.

Common pitfall. The most common slip is to place the load at mid-height and then use $\mathrm{FEM}=\pm Ph/12$, which belongs to a distributed load; the point-load value is $\pm Ph/8$. The second is to omit the axial couple from the overturning check: the base moments alone give $56$ $\text{kN}\cdot\text{m}$ against an applied $72$, and the missing $16$ is the beam shear acting through the bay width. If the check does not close, the classification or the storey-shear equation is wrong.

Source. Hibbeler, Structural Analysis, 10th ed., §11.5 (slope-deflection for frames with sidesway) and §12.7–12.8 (symmetric and anti-symmetric moment distribution); Kassimali, 6th ed., §16.5; McCormac, Structural Analysis, 4th ed., Ch. 19.

Question 8 of 9: Anti-Symmetric Analysis of a Portal Frame

Question 8: Anti-Symmetric Analysis of a Portal Frame (22 marks)