Question 8 of 9: Slope-deflection analysis of a frame with sidesway
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 07-Str-A4 Advanced Structural
Analysis, National Examinations, May 2013 — 3 hours, closed book (an approved
Sharp or Casio calculator is permitted). Nine questions are printed. The candidate must
answer Question 1 and Question 2, then two of Questions 3–5 and
two of Questions 6–9 — six questions constitute a complete paper
(8 + 12 + 2 × 16 + 2 × 24 =
100 marks). Marks are shown in the left margin of the paper.
All nine questions are solved below, because the set is intended as a study
resource rather than as a three-hour sitting.
Reference texts.
R. C. Hibbeler, Structural Analysis, 10th ed. — influence lines
(Ch. 6, 8), energy methods and Castigliano's theorems (Ch. 9), the force
(flexibility) method (Ch. 10), slope-deflection (Ch. 11) and moment distribution
(Ch. 12).
A. Kassimali, Structural Analysis, 6th ed. — Ch. 13 slope-deflection
including support settlement and sidesway; Ch. 8 influence lines for trusses.
A. Ghali, A. M. Neville and T. G. Brown, Structural Analysis: A Unified
Classical and Matrix Approach, 7th ed. — Ch. 3–5 force and displacement
methods, lack of fit and prescribed displacements.
J. C. McCormac and S. F. Csernak, Structural Analysis Using Classical and
Matrix Methods, 4th ed. — kinematic indeterminacy and the use of symmetry.
Sign conventions used throughout. Slope-deflection end moments
$M_{ij}$ are the moments the joint applies to the member, clockwise positive;
joint rotations $\theta$ are clockwise positive; the chord rotation
$\psi_{ij}=\Delta/L$ is positive when the member chord rotates clockwise. Bending moment
diagrams are plotted with sagging positive, so that the sagging moment at
end $i$ of member $ij$ is $+M_{ij}$ and at end $j$ it is $-M_{ji}$. All members are
inextensible unless stated otherwise, so axial deformation contributes nothing to the
compatibility equations.
Question 8: Slope-deflection analysis of a frame with sidesway (24 marks)
7 kN horizontal, applied to the right at joints 3 and 4 (the stub
tops)
Members
uniform $EI$, inextensible; sidesway is not prevented
Find. All member end moments, and the shear and bending moment diagrams
with the maximum and minimum ordinates for every member.
Q8 — the sway frame and its bending moment diagram. The two cantilever stubs deliver a 7 kN force and a 7 kN·m couple to each beam-level joint.
Approach. Replace each cantilever stub by the force and couple it
delivers to its joint, reduce the frame to a single-bay portal with three unknowns
($\theta_2$, $\theta_5$ and the sway $\Delta$), use anti-symmetry to collapse the two
rotations into one, and close the system with the storey-shear equation.
Statically reduce the cantilever stubs. Joints 3 and 4 are free
ends, so the stubs are determinate cantilevers. Each transmits to its base joint a horizontal
force of $7\text{ kN}$ and a couple
$$M_{\text{stub}} = 7\times1 = 7\ \text{kN}\cdot\text{m}\ \text{(clockwise)} .$$
The stub moment diagram is a straight line from zero at the tip to
$7\ \text{kN}\cdot\text{m}$ at the joint, with a constant shear of 7 kN.
Recognise the anti-symmetry. The reduced structure is a symmetric portal
carrying two equal rightward forces and two equal clockwise couples. Reflecting the frame
maps a rightward force to a leftward force and a clockwise couple to an anticlockwise couple,
so the loading is anti-symmetric and $\theta_2=\theta_5=\theta$. Both
columns then carry identical end moments and the beam bends in double curvature.
Write the slope-deflection equations. With $k=2EI/4$ for every member and
the column chord rotation $\psi=\Delta/4$,
$$M_{12}=k\!\left(\theta-3\psi\right),\qquad M_{21}=k\!\left(2\theta-3\psi\right),\qquad
M_{25}=k\!\left(2\theta+\theta\right)=3k\theta .$$
Joint equilibrium at 2. The stub couple is an applied moment on the
joint, so
$$M_{21}+M_{25}=7\ \text{kN}\cdot\text{m}\quad\Longrightarrow\quad
k\!\left(5\theta-3\psi\right)=7 .$$
Storey-shear equation. The total horizontal load carried above the base
is $2\times 7 = 14\text{ kN}$, and each column resists it through
$H = -\left(M_{12}+M_{21}\right)/4$, so
$$\left(M_{12}+M_{21}\right)+\left(M_{65}+M_{56}\right)=-14\times4=-56
\quad\Longrightarrow\quad k\!\left(3\theta-6\psi\right)=-28 .$$
Solve. Eliminating $\psi$ between the two equations gives
$7k\theta = 42$, hence $k\theta = 6$ and $k\psi = 23/3$. Back-substituting,
$$\boxed{M_{12}=M_{65}=-17.0\ \text{kN}\cdot\text{m},\quad
M_{21}=M_{56}=-11.0\ \text{kN}\cdot\text{m},\quad
M_{25}=M_{52}=+18.0\ \text{kN}\cdot\text{m}.}$$
The joint check is $-11.0+18.0 = 7.0\ \text{kN}\cdot\text{m}$, the stub couple, as required.
In closed form the answers are $M_{12}=-P(s+4h)/7$, $M_{21}=P(s-3h)/7$ and
$M_{25}=3P(2s+h)/7$ for a stub $s$, a column height $h$ and a load $P$, provided the beam
span equals the column height.
Member shears and axial forces. Each column shear is
$$H = -\frac{M_{12}+M_{21}}{4} = \frac{28}{4} = 7.0\ \text{kN},$$
so the two columns share the 14 kN storey shear equally, as anti-symmetry demands. The
beam shear follows from its two end moments:
$$V_{\text{beam}} = \frac{M_{25}+M_{52}}{4} = \frac{36}{4} = 9.0\ \text{kN},$$
which is also the axial force in each column — tension of 9.0 kN in the windward
column 1–2 and compression of 9.0 kN in the leeward column 6–5. Global
overturning confirms this: $14\times5 = 70\ \text{kN}\cdot\text{m}$ is resisted by
$2\times17.0$ of base fixity plus $9.0\times4$ of axial couple, i.e. $34+36=70$.
Diagram ordinates. Each column moment runs linearly from
$-17.0\ \text{kN}\cdot\text{m}$ at its base to $+11.0\ \text{kN}\cdot\text{m}$ at the beam,
with a point of contraflexure $2.43\text{ m}$ above the base. The beam moment runs linearly
from $+18.0$ at joint 2 to $-18.0\ \text{kN}\cdot\text{m}$ at joint 5, with
contraflexure exactly at midspan — the signature of the anti-symmetric response. Each
stub runs from zero at the free tip to $7.0\ \text{kN}\cdot\text{m}$ at its joint.
