16-Civ-B1 Advanced Structural Analysis · December 2019
Question 2 of 9: Schematic shear force and bending moment diagrams
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examination, December 2019 —
16-Civ-B1 Advanced Structural Analysis, 3 hours, CLOSED BOOK (an
approved Casio or Sharp calculator only). Nine questions on seven pages:
Questions 1 and 2 are compulsory (8 and 12 marks); the candidate then answers
TWO of Questions 3, 4 or 5 (18 marks each) and TWO of Questions 6, 7, 8 or 9
(22 marks each), so six questions make a complete paper of 100 marks. Because
this set is a study resource, all nine questions are solved
here.
Reference texts. R. C. Hibbeler, Structural
Analysis, 10th ed. (Pearson) — Ch. 6 influence lines, Ch. 8–9
deflections and virtual work, Ch. 10 force (flexibility) method, Ch. 11
slope-deflection, Ch. 12 moment distribution, Ch. 15–16 matrix stiffness.
A. Kassimali, Structural Analysis, 6th ed. (Cengage) — Ch. 7
Castigliano and least work, Ch. 8 influence lines, Ch. 13 slope-deflection,
Ch. 17–18 matrix stiffness. A. Ghali, A. M. Neville and T. G. Brown,
Structural Analysis: A Unified Classical and Matrix Approach, 7th ed.
(CRC). Canadian context for the settlement question: NBC 2020 Part 4 and
CSA S6:19 treat differential support settlement as an imposed deformation to be
combined with the permanent loads, so the self-equilibrating moment set computed
in Question 6 is a real design action, not a curiosity.
Sign convention used throughout.
End moments are written counter-clockwise-positive on the member, which is the
convention that matches the standard six-degree-of-freedom element stiffness
matrix; the slope-deflection equation is then
$M_{ij} = \dfrac{2EI}{L}\left(2\theta_i + \theta_j - 3\psi_{ij}\right) +
\mathrm{FEM}_{ij}$ with $\psi$ the counter-clockwise chord rotation and, for a
downward UDL, $\mathrm{FEM}_{ij} = +wL^2/12$, $\mathrm{FEM}_{ji} = -wL^2/12$.
Ordinary sagging bending moments, which is what every diagram below
plots, are recovered as $M_{\text{sag}}(i) = -M_{ij}$ and
$M_{\text{sag}}(j) = +M_{ji}$. Mixing this with the clockwise-positive
convention of some textbooks produces clean-looking integers that are wrong.
Question 2: Schematic shear force and bending moment diagrams
(12 marks)
Given. Both structures have uniform $EI$ and inextensible
members. Structure (a) carries $w$ from the free tip to the roller and $P$
horizontally on the column; structure (b) carries $P$ vertically at the knee.
The exam prints no dimensions on either figure.
Find. The shape of the shear force and bending moment
diagrams for each structure, with the position of every zero, every jump and
every change of curvature identified.
Check: dimensions adopted. Because the
figures are dimensionless, the diagrams below are drawn to representative
dimensions taken from the drawn proportions — (a) overhang
$a = 2\ \text{m}$, pin-to-roller $L = 5\ \text{m}$, roller-to-corner
$c = 3\ \text{m}$, column $h = 2\ \text{m}$ with $P$ at mid-height,
$w = 12\ \text{kN/m}$, $P = 30\ \text{kN}$; (b) column $h = 3\ \text{m}$,
horizontal member $b = 5\ \text{m}$, inclined member 4 m run on 3 m fall
(true length 5 m), $P = 30\ \text{kN}$. Every shape feature and every
zero quoted below is independent of that choice, so nothing is lost;
only the numerical ordinates would move.
Approach. Establish the degree of redundancy first (both
frames are three times statically indeterminate, so no ordinate can be obtained
by statics alone), then reason the shape from the loading and the support
conditions, and finally pin the ordinates down with a direct-stiffness
solution.
Part (a) — determinacy and the features that are fixed by
inspection. The pin, the roller and the fixed base give
$R = 2 + 1 + 3 = 6$ against three equations, so $r = 3$. Three features
nevertheless follow from statics alone. The free tip carries no shear and no
moment, so both diagrams start at zero. Over the overhang the shear falls
linearly and the moment is the pure cantilever parabola, reaching
$$M_B = -\frac{wa^{2}}{2} = -\frac{12(2)^2}{2} = \boxed{-24.0\ \text{kN}\cdot\text{m}}$$
at the pin, hogging, whatever the rest of the structure does. And the segment
between the roller and the corner carries no load, so its shear is constant and
its moment varies linearly.
Part (a) — the ordinates. Solving the frame gives
reactions $R_B = 55.30\ \text{kN}$ and $R_C = 37.31\ \text{kN}$. The shear
therefore jumps from $-24.0$ to $+31.30\ \text{kN}$ at the pin, falls linearly
to $-28.70\ \text{kN}$ at the roller, jumps to $+8.61\ \text{kN}$ there and
holds that value to the corner. The moment consequently peaks in the main span
where the shear passes through zero, at
$x = 31.30/12 = 2.608\ \text{m}$ from the pin:
$$M_{\max} = -24.0 + 31.30(2.608) - \tfrac{1}{2}(12)(2.608)^2
= \boxed{+16.82\ \text{kN}\cdot\text{m}}$$
and it is $-17.50\ \text{kN}\cdot\text{m}$ over the roller and
$+8.33\ \text{kN}\cdot\text{m}$ at the corner.
Question 2(a): free tip, pin, roller, rigid corner and a column with a horizontal load at mid-height (adopted dimensions).
Question 2(a): shear force along the beam A to D. The shear jumps at each support and crosses zero 2.608 m past the pin.
Question 2(a): bending moment along the beam A to D. Zero at the free tip, hogging over both supports, sagging between them.
Part (a) — the column. The corner is rigid, so the
beam moment $+8.33\ \text{kN}\cdot\text{m}$ passes straight into the column
head. The column carries the horizontal load at mid-height, which makes its
shear step there from $-15.62\ \text{kN}$ above the load to
$+14.38\ \text{kN}$ below it (the two differ by exactly the applied
$30\ \text{kN}$), so the column moment diagram is two straight segments with a
kink at the load: $+8.33$ at the head, $-7.29$ at the load point and
$$M_{\text{base}} = \boxed{+7.08\ \text{kN}\cdot\text{m}}$$
The moment changes sign twice down the column — two points of
contraflexure — which is the characteristic signature of a member bent by
both a head moment and a mid-height force.
Part (b) — shape by inspection. Two built-in supports
give $R = 6$ and again $r = 3$. No member carries a distributed load, so
every shear is constant and every moment diagram is a straight
line: three straight segments meeting at the two rigid joints, where the moment
is continuous (the applied load at the knee is a force, not a couple).
Part (b) — the ordinates. Solving the frame,
$$M_{\text{base}} = \boxed{+27.12\ \text{kN}\cdot\text{m}},\qquad
M_{\text{corner}} = -21.93,\qquad M_{\text{knee}} = +19.01,\qquad
M_{\text{far support}} = -19.20\ \text{kN}\cdot\text{m}$$
with constant shears of $16.35\ \text{kN}$ in the column, $8.19\ \text{kN}$ in
the horizontal member and $7.64\ \text{kN}$ normal to the inclined member. The
horizontal reaction at each base is $16.35\ \text{kN}$ and the two vertical
reactions sum to the applied $30\ \text{kN}$, which is the arithmetic check
worth writing down.
Question 2(b): fixed base, rigid corner, horizontal member and an inclined member into a built-in support, loaded at the knee.
Question 2(b): bending moment plotted normal to each member. Every member is unloaded, so every diagram is a straight line.
What the marker is looking for. On (a): zero at the free
tip; a parabolic hogging arc to the pin; a shear jump at each support; a sagging
parabola peaking where the shear crosses zero; a linear moment from the roller
to the corner; continuity of moment round the rigid corner; a kink in the column
diagram at the horizontal load; and a non-zero base moment. On (b): three
straight moment segments, a sign change on the column and on the inclined
member, moment continuity at both rigid joints, and constant shear in every
member.
