Question 7 of 9: Frame with two columns and a loaded cantilever
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 2015 — three hours, closed book (an approved Sharp or Casio calculator is the only aid). Questions 1 and 2 are compulsory; the candidate then answers two of Questions 3, 4, 5, one of Questions 6, 7 and one of Questions 8, 9, so six questions totalling 100 marks constitute a complete paper. Marks are printed in the left margin. All nine questions are solved here, because the set is a study resource rather than an examination script.
Reference texts. R. C. Hibbeler, Structural Analysis (Pearson) — force/flexibility method, slope-deflection, moment distribution and influence lines; A. Kassimali, Structural Analysis (Cengage) — matrix stiffness formulation and support-settlement effects; J. C. McCormac, Structural Analysis: Using Classical and Matrix Methods (Wiley); CSA S16:19 Design of Steel Structures and CSA S6:19 Canadian Highway Bridge Design Code for the Canadian design setting in which these analyses are used.
Sign convention used throughout. Bending moments are reported as sagging positive and are plotted on the tension face of each member. Slope-deflection end moments follow the counter-clockwise-positive convention stated in Question 9: $$M_{ij}=\frac{2EI}{L}\left(2\theta_i+\theta_j-3\psi_{ij}\right)+M^{F}_{ij},\qquad \psi_{ij}=\frac{(\mathbf{D}_j-\mathbf{D}_i)\cdot \mathbf{e}_2}{L}$$ where $\mathbf{e}_2$ is the member axis rotated 90 degrees counter-clockwise, and the fixed-end moments of a downward uniform load are $M^{F}_{ij}=+wL^2/12$, $M^{F}_{ji}=-wL^2/12$. The sagging moment at the ends of a member is then $M(i)=-M_{ij}$ and $M(j)=+M_{ji}$.
Question 7: Frame with two columns and a loaded cantilever (24 marks)
A uniformly distributed load $w=8\ \text{kN/m}$ acts from joint 1 to joint 3.
Find. All end moments, the shear force and bending moment diagrams with their
maximum and minimum ordinates, and the reactions.
Question 7: the frame, with the hinged column base at joint 5 and the built-in base at joint 6.
Approach. Show that the frame cannot sway, reduce the cantilever to a known
couple applied at joint 3, and solve the two joint-rotation unknowns by slope-deflection with
modified stiffnesses for the two pin-ended members.
Prove there is no sway. The pin at joint 1 fixes the horizontal position of
that joint, and the beam is inextensible, so $u_2=u_3=u_4=0$. Each column is vertical and
inextensible, so $v_2=v_5=0$ and $v_3=v_6=0$. Every joint translation is therefore zero and only
the rotations $\theta_2$ and $\theta_3$ remain unknown (the rotations at the pin supports 1 and 5
are absorbed by modified stiffnesses).
Replace the cantilever. Member 3–4 is determinate: it delivers a
vertical force of $60\ \text{kN}$ and a known couple
$$M_{\text{cant}}=60(2)=120\ \text{kN}\cdot\text{m}$$ hogging at joint 3. It contributes no
stiffness, only this applied joint moment.
Fixed-end moments. For the two loaded spans, with a far end that is pinned in
the case of 1–2 the modified value applies:
$$M^{F}_{21}=\frac{wL^{2}}{8}=\frac{8(12)^{2}}{8}=144\ \text{kN}\cdot\text{m}\ \text{(propped
form)},\qquad
M^{F}_{23}=\frac{wL^{2}}{12}=96\ \text{kN}\cdot\text{m}$$
Stiffnesses. Using $k=EI/L$ with the relative values given,
$$k_{12}=\frac{3(2EI)}{12}=0.5EI\ \text{(far end pinned)},\qquad
k_{23}=\frac{4(2EI)}{12}=0.667EI$$
$$k_{25}=\frac{3EI}{6}=0.5EI\ \text{(far end hinged)},\qquad
k_{36}=\frac{4EI}{4}=1.0EI$$
with a carry-over of one half from joint 3 into the built-in base 6 and no carry-over into the two
pins.
Solve the two joint equations. Enforcing $\sum M=0$ at joints 2 and 3 and
solving simultaneously gives the beam moments (sagging positive)
$$M(2)_{\text{left}}=-127.5,\qquad M(2)_{\text{right}}=-111.0,\qquad
M(3)_{\text{beam}}=-99.0\ \text{kN}\cdot\text{m}$$
together with column head moments of $16.5\ \text{kN}\cdot\text{m}$ at joint 2 and
$21.0\ \text{kN}\cdot\text{m}$ at joint 3. Each joint balances, which is the arithmetic check:
$127.5-111.0=16.5$ and $120.0-99.0=21.0$.
Beam shears and span maxima. Span 1–2:
$$V_1=\frac{wL}{2}+\frac{M_{21}}{L}=48-\frac{127.5}{12}\cdot 1=37.375\ \text{kN},\qquad
V_{2^{-}}=-58.625\ \text{kN}$$
$$M_{\max}=\frac{V_1^{2}}{2w}=\frac{37.375^{2}}{16}=\boxed{+87.31\ \text{kN}\cdot\text{m}}
\ \text{at}\ x=4.672\ \text{m}$$
Span 2–3: $V_{2^{+}}=49.0$, $V_{3^{-}}=-47.0\ \text{kN}$ and
$$M(x)=-111+49x-4x^{2}\;\Rightarrow\;M_{\max}=\boxed{+39.06\ \text{kN}\cdot\text{m}}
\ \text{at}\ x=6.125\ \text{m}$$
Cantilever 3–4: constant shear $60\ \text{kN}$, moment from
$-120\ \text{kN}\cdot\text{m}$ at joint 3 to zero at the tip.
Columns. Column 2–5 carries a linear moment from
$16.5\ \text{kN}\cdot\text{m}$ at the head to zero at the hinge, so its shear is
$16.5/6=2.75\ \text{kN}$. Column 3–6 carries $21.0\ \text{kN}\cdot\text{m}$ at the head and
half of that, $10.5\ \text{kN}\cdot\text{m}$, of opposite sense at the built-in base, so its shear
is $(21.0+10.5)/4=7.875\ \text{kN}$ and it reverses curvature $2.667\ \text{m}$ below joint 3.
Reactions and checks.
$$R_1=37.375,\qquad R_5=107.625,\qquad R_6=107.0\ \text{kN (all upward)}$$
$$\sum V=37.375+107.625+107.0=252.0=8(24)+60\ \checkmark$$
Horizontally, $-5.125-2.75+7.875=0\ \checkmark$; the base moment at joint 6 is
$10.5\ \text{kN}\cdot\text{m}$.
Question 7: bending moment diagram, kN.m; note the step at each column joint.
Question 7: shear force diagram, kN.
The beam bending moment steps at each column joint, from $-127.5$ to $-111.0$ at joint
2 and from $-99.0$ to $-120.0$ at joint 3. Those steps are exactly the column head moments, 16.5
and 21.0 kN·m; they are joint equilibrium, not a drafting error.
Check: joint 5 is read from the drawing as a
hinged support — the paper labels the small circle there “typical
hinge”, and joint 6, by contrast, carries the conventional built-in hatching. If joint 5
were instead a rigid base the column head moment would rise and the beam moment at joint 2 would
change accordingly; the modified stiffness $3EI/L$ used above is the one consistent with the
hinge notation shown.
Question 7 — complete results
Quantity
Value
Moment at joint 2 (span 1–2 side)
$-127.5\ \text{kN}\cdot\text{m}$ — minimum ordinate in the frame
Moment at joint 2 (span 2–3 side)
$-111.0\ \text{kN}\cdot\text{m}$
Moment at joint 3 (span 2–3 side)
$-99.0\ \text{kN}\cdot\text{m}$
Moment at joint 3 (cantilever side)
$-120.0\ \text{kN}\cdot\text{m}$
Column head moments
$16.5\ \text{kN}\cdot\text{m}$ at joint 2; $21.0\ \text{kN}\cdot\text{m}$ at joint 3
