Question 6 of 9: Continuous beam with a settling support
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 6: Continuous beam with a settling support (24 marks)
1–2 : $3\ \text{m}$; 2–4 : $9\ \text{m}$ (joint 3 is an interior point at 6 m from joint 2)
Loading
$w=8\ \text{kN/m}$ over the full 12 m
Rigidity
$EI=5400\ \text{kN}\cdot\text{m}^{2}$
Support movement
joint 2 settles $\delta=0.01\ \text{m}$
Find. Support moments and reactions, the shear force and bending moment
diagrams with maximum and minimum ordinates, and the effect of the settlement.
Question 6: the continuous beam; supports at joints 1, 2 and 4 only.
Approach. The beam is one degree indeterminate; use the three-moment
(Clapeyron) equation with support-level terms, which handles the settlement directly, and confirm
the result by slope-deflection.
Identify the real structure. Supports exist only at joints 1, 2 and 4, so
this is a two-span continuous beam of $3\ \text{m}$ and $9\ \text{m}$ — joint 3 is simply a
point at which ordinates are to be reported. With three vertical reactions and two equations of
equilibrium for a beam, the structure is one degree indeterminate.
Three-moment equation. For spans $L_1$ and $L_2$ meeting at support B, with
the outer supports standing $h_L$ and $h_R$ above B,
$$M_A L_1+2M_B\left(L_1+L_2\right)+M_C L_2
=-\frac{w L_1^{3}}{4}-\frac{w L_2^{3}}{4}
+6EI\left(\frac{h_L}{L_1}+\frac{h_R}{L_2}\right)$$
The end supports are simple, so $M_A=M_C=0$, and settling B by $0.01\ \text{m}$ leaves both
neighbours $0.01\ \text{m}$ high.
Solve for the support moment.
$$2M_B(3+9)=-1512+144=-1368
\;\Longrightarrow\;
M_B=\boxed{-57\ \text{kN}\cdot\text{m}}$$
Without the settlement the same equation gives $-63\ \text{kN}\cdot\text{m}$, so the 10 mm
settlement relieves the hogging over support 2 by $6\ \text{kN}\cdot\text{m}$, about
9.5 per cent.
Reactions. Treating each span as simply supported and adding the support
moment,
$$R_1=\frac{wL_1}{2}+\frac{M_B}{L_1}=12-19=\boxed{-7\ \text{kN}}$$
$$R_4=\frac{wL_2}{2}+\frac{M_B}{L_2}=36-6.333=29.667\ \text{kN}$$
$$R_2=w(12)-R_1-R_4=96+7-29.667=\boxed{73.333\ \text{kN}}$$
The reaction at support 1 is negative: the short span is dragged upward by the long one
and the end must be held down with 7 kN.
Shear force diagram.
$$V(0)=-7,\quad V(3^{-})=-7-8(3)=-31,\quad V(3^{+})=-31+73.333=42.333,\quad
V(12)=-29.667\ \text{kN}$$
with straight lines between, and a jump of $73.333\ \text{kN}$ at support 2.
Bending moment diagram. In span 1–2, $M(x)=-7x-4x^{2}$, which is
hogging everywhere and reaches $-57\ \text{kN}\cdot\text{m}$ at support 2. In span 2–4,
measuring $s$ from support 2,
$$M(s)=-57+42.333\,s-4s^{2}$$
The shear vanishes at $s=42.333/8=5.292\ \text{m}$, giving
$$M_{\max}=-57+\frac{42.333^{2}}{16}=\boxed{+55.01\ \text{kN}\cdot\text{m}}$$
and at joint 3 ($s=6\ \text{m}$) the ordinate is $+53\ \text{kN}\cdot\text{m}$. The points of
contraflexure in the long span are at $s=1.581\ \text{m}$ and $s=9.0\ \text{m}$ (the latter being
the support itself is a coincidence of these numbers; the real second root lies beyond joint 4 and
is not on the beam).
Question 6: shear force (upper) and bending moment (lower) diagrams.
