Question 6 of 8: Moment Distribution Analysis of a Frame
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2018 — 07-Str-A1
Elementary Structural Analysis; 3 hours, CLOSED BOOK (an approved Casio or Sharp
calculator is permitted). Six questions constitute a complete paper: answer all of
Questions 1–5, then one only of Questions 6, 7 or 8. Marks are shown in the
left margin (6, 18, 18, 18, 18, then 22 for the optional question). Every question is
worked below, including all three alternatives, because the full set is the more useful
study resource.
Reference texts. R. C. Hibbeler, Structural Analysis (Ch. 2
determinacy and stability; Ch. 4 influence lines; Ch. 6–9 deflections and virtual work;
Ch. 12 moment distribution); A. Kassimali, Structural Analysis (Ch. 3, 6, 8, 13, 16);
K. Leet, C.-M. Uang and A. Gilbert, Fundamentals of Structural Analysis;
J. C. McCormac, Structural Analysis: Using Classical and Matrix Methods. For Canadian
practice the companion documents are the National Building Code of Canada (Part 4 load
combinations) and CSA S16 Design of Steel Structures; this paper is pure analysis, so no
design code is invoked in the answers.
Check — two figure readings that the printed figures leave open,
both stated where they are used. (i) In Question 2(c) the hatched plane under the
upper roller is drawn parallel to the inclined member, so the reaction is taken
normal to the member. That reading makes every result an integer
(R = 28 kN, shears 32 / −4 / −28 kN, axial forces
25 / 10 / 0 kN), whereas a vertical-plane reading gives nothing round; and because the two
readings differ only by a force acting along the member axis, the shear and moment
diagrams are identical either way. (ii) In Question 7 both wall supports are drawn as
triangles on hatching. The truss is analysable only if one of them releases a component;
taking the pin at U₂ and the horizontal-only roller at L₃ gives
δ = 29.0 mm, and the same value follows if the thin vertical
line between the two wall joints is a dimension witness line rather than a member. The
alternative (pin at L₃) would give 31.8 mm.
Question 6: Moment Distribution Analysis of a Frame (22 marks, optional)
48 kN downward at mid-span of 1-2 (4 m from joint 1)
Distributed load
12 kN/m downward over the whole 10 m of 2-3-4
Stiffness
same $EI$ for all members; members inextensible
Find. The member end moments by moment distribution, then the
shear and bending moment diagrams with the extreme ordinates labelled.
Question 6: roller at joint 1, pin support at joint 3, fixed base at joint 5, and a 2 m overhang beyond the pin.
Approach. Establish that no sidesway is possible, reduce the two
moment-free ends (the roller at 1 and the pin at 3) with modified stiffnesses $3EI/L$, transfer
the determinate overhang moment into joint 3, and then only joint 2 needs balancing — a
single distribution with no iteration.
Show that the frame cannot sway, which is what "inextensible" is telling
you. The beam 1-2-3-4 lies along one horizontal line and is axially rigid, and joint 3
is a pin support, so joints 1, 2 and 4 cannot move horizontally. The column 2-5 is also
inextensible with a fixed base, so joint 2 cannot move vertically; the roller at 1 prevents
vertical movement there. Every joint translation is therefore zero and the only unknowns are
the joint rotations — no sway correction pass is required.
Deal with the determinate overhang first. The 2 m length beyond the pin
carries $12 \times 2 = 24$ kN, so it delivers to joint 3 a known moment
$$M_{34} = -(12)(2)(1) = -24.0\ \text{kN}\cdot\text{m}$$
and a shear of 24.0 kN. Because a cantilever has no rotational stiffness, its distribution
factor is zero: it imposes a moment on joint 3 but never takes a share of the balancing moment.
Joint equilibrium at 3 then fixes $M_{32} = +24.0$ kN·m for good.
Fixed-end moments, modified for the two released ends. For member 1-2 with
the roller at 1 released, the propped-cantilever value applies:
$$M^F_{21} = \frac{3PL}{16} = \frac{3(48)(8)}{16} = +72.0\ \text{kN}\cdot\text{m}$$
For member 2-3 the encastre values are $\mp wL^2/12 = \mp 64.0$ kN·m; releasing end 3
from $+64.0$ to its true $+24.0$ carries half of the $-40.0$ change back to end 2:
$$M^F_{23} = -64.0 + \tfrac{1}{2}(24.0 - 64.0) = -84.0\ \text{kN}\cdot\text{m}$$
The column carries no load, so $M^F_{25} = M^F_{52} = 0$.
Distribution factors at joint 2. Using the modified stiffness $3EI/L$ for
the two members whose far ends are moment-free and $4EI/L$ for the column into the fixed base,
$$k_{21} = k_{23} = \frac{3EI}{8} = 0.375EI, \qquad k_{25} = \frac{4EI}{8} = 0.500EI,
\qquad \Sigma k = 1.25EI$$
$$DF_{21} = DF_{23} = 0.30, \qquad DF_{25} = 0.40$$
One balancing operation completes the distribution. The out-of-balance
moment at joint 2 is
$$M^F_{21} + M^F_{23} + M^F_{25} = 72.0 - 84.0 + 0 = -12.0\ \text{kN}\cdot\text{m}$$
so $+12.0$ kN·m is distributed. There is no carry-over to the released ends 1 and 3, and
half of the column's share goes to the fixed base:
$$M_{21} = 72.0 + 0.30(12.0) = \boxed{+75.6}, \qquad
M_{23} = -84.0 + 0.30(12.0) = \boxed{-80.4}$$
$$M_{25} = 0.40(12.0) = \boxed{+4.8}, \qquad M_{52} = \tfrac{1}{2}(4.8) = \boxed{+2.4}
\ \text{kN}\cdot\text{m}$$
with $M_{12} = 0$ at the roller. The joint check is exact:
$75.6 - 80.4 + 4.8 = 0$. (Running the classical iterative table with full $4EI/L$ stiffnesses
and repeated carry-overs converges on the same four numbers — 75.60, −80.40, 4.80,
2.40 — after four cycles, so the modified-stiffness shortcut costs nothing in accuracy.)
Convert to sagging moments and recover the shears. With the clockwise-positive
member convention, sagging $= +M_{ij}$ at a left-hand end and $-M_{ji}$ at a right-hand end, so
the beam moments are $0$ at joint 1, $-75.6$ kN·m at joint 2 in member 1-2,
$-80.4$ kN·m at joint 2 in member 2-3, and $-24.0$ kN·m at joint 3. The step of
4.8 kN·m across joint 2 is the column's moment. Member shears follow from statics:
$$V_1 = \frac{-75.6 + (48)(4)}{8} = 14.55\ \text{kN}, \qquad V_{2\text{(right)}}
= \frac{-24.0 + 80.4 + \tfrac{1}{2}(12)(8)^2}{8} = 55.05\ \text{kN}$$
giving $V = -33.45$ kN just left of joint 2, $-40.95$ kN just left of the pin, and 24.0 kN on
the overhang. The column shear is $(4.8 + 2.4)/8 = 0.90$ kN.
Reactions and peak sagging moments. Collecting the shear steps,
$$R_1 = 14.55\ \text{kN}, \qquad R_3 = 24.0 + 40.95 = 64.95\ \text{kN}, \qquad
R_5 = 55.05 + 33.45 = 88.50\ \text{kN}$$
which sum to 168.0 kN, exactly $48 + (12)(10)$. The horizontal reactions are the 0.90 kN column
shear at the base and its equal and opposite counterpart at the pin. The largest sagging moment
in 1-2 is under the point load, $14.55 \times 4 = 58.2$ kN·m, and in 2-3 it is at the
point of zero shear $x = 55.05/12 = 4.588$ m from joint 2:
$$M_{\max} = -80.4 + \frac{55.05^2}{2(12)} = \boxed{+45.87\ \text{kN}\cdot\text{m}}$$
$M_{21}$
$M_{23}$
$M_{25}$
$M_{52}$
$M_{32}$
$M_{34}$
Modified FEM
+72.0
−84.0
0
0
+24.0
−24.0
$DF$ at joint 2
0.30
0.30
0.40
—
—
—
Balance ($+12.0$)
+3.6
+3.6
+4.8
—
—
—
Carry-over
—
—
—
+2.4
—
—
Final (kN·m)
+75.6
−80.4
+4.8
+2.4
+24.0
−24.0
Question 6: developed shear and bending moment diagrams for the beam chain 1-2-3-4. The 4.8 kN.m step in moment at joint 2 is the column moment; the moment at the pin (joint 3) is the overhang value, -24.0 kN.m.
Question 6: the column carries a constant 0.90 kN shear and a moment running from +4.8 kN.m at joint 2 to -2.4 kN.m at the fixed base.
Quantity
Value
Reaction at the roller, joint 1
14.55 kN ↑
Reaction at the pin, joint 3
64.95 kN ↑ and 0.90 kN horizontal
Reaction at the fixed base, joint 5
88.50 kN ↑, 0.90 kN horizontal, 2.40 kN·m
Beam 1-2
V = +14.55 / −33.45 kN; M = +58.2 kN·m under the load, −75.6 kN·m at joint 2
Beam 2-3
V = +55.05 / −40.95 kN; M = −80.4 kN·m at joint 2, +45.87 kN·m at 4.59 m, −24.0 kN·m at the pin
Overhang 3-4
V = +24.0 to 0 kN; M = −24.0 kN·m at the pin to 0 at the tip
Column 2-5
V = 0.90 kN constant; M = +4.8 kN·m at joint 2 to −2.4 kN·m at the base; N = 88.50 kN compression
