Question 1 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. 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 1: Schematic shear force and bending moment diagrams (7 marks)
Given. Both structures have constant $EI$ and inextensible members.
Structure (a): encastré at A, rollers at B, C and D at spacing $L$, free tip E at $L/3$
beyond D; three patch loads of intensity $w$, each $L/3$ long, placed as shown.
Structure (b): a plane cantilever tree — column AB (length $L$, built in at A), arm BC
(length $L$) and the free hanging member CD (length $L$); a horizontal load $P$ acts at C.
Find. The shape of the shear force and bending moment diagrams for each
structure, with the ordinates that follow from statics identified.
Question 1(a): the encastré and roller-supported beam with three uniformly distributed patch loads.
Approach. Structure (b) is statically determinate, so both diagrams follow
from two free bodies; structure (a) is three degrees indeterminate, so the shapes are argued from
the loading pattern and the determinate overhang, and the exact ordinates are obtained from a
flexibility solution with the three roller reactions as redundants.
(a) The propped continuous beam
Count the redundants. The encastré end supplies three restraints and
each roller one, so $r=6$ against three equations of equilibrium:
$$\text{DSI}=r-3=6-3=\boxed{3}$$ The beam is three degrees statically indeterminate, and the
diagrams cannot be drawn from statics alone — except on the overhang.
Read the two determinate ordinates first. The overhang DE carries the whole
of the third patch, so working from the free tip E,
$$V_{D^{+}}=w\frac{L}{3}=0.3333\,wL,\qquad
M_D=-\frac{w}{2}\left(\frac{L}{3}\right)^{2}=-\frac{wL^{2}}{18}=-0.05556\,wL^{2}$$
Every diagram must pass through these two values, and $V$ and $M$ must both vanish at E.
Fix the shapes from the loading. On the unloaded lengths AB, the middle third
of BC and the whole of CD the shear is constant and the moment varies linearly; under each patch
the shear falls linearly at the rate $w$ and the moment is a parabola. At each roller the shear
jumps by the reaction there. The middle span BC carries a load pattern that is symmetric about its
own mid-point, so the sagging moment inside BC peaks close to mid-span.
Solve for the redundants. Taking the three roller reactions as redundants on
a primary cantilever fixed at A, the compatibility equations
$\sum_j f_{ij}R_j=-\Delta_{i0}$ (with $f_{ij}=\int m_i m_j\,dx/EI$ and
$\Delta_{i0}=\int M_0 m_i\,dx/EI$) give
$$R_B=0.4088\,wL,\qquad R_C=0.2635\,wL,\qquad R_D=0.3789\,wL$$
and hence, from vertical equilibrium of the whole beam,
$$R_A=wL-\left(R_B+R_C+R_D\right)=\boxed{-0.05128\,wL}$$
a downward reaction: the encastré end has to be held down.
Complete the moment ordinates. Working along the beam with these reactions,
$$M_A=+0.01709\,wL^{2},\quad M_B=-0.03419\,wL^{2},\quad M_C=-0.00997\,wL^{2},\quad
M_D=-0.05556\,wL^{2}$$ and the largest sagging moment, at $x=1.691L$ where the shear crosses
zero inside the second patch, is $M=+0.03780\,wL^{2}$.
Question 1(a): shear force (upper) and bending moment (lower) diagrams, ordinates in wL and wL squared.
Two features of the answer deserve comment. The built-in end carries a small sagging
moment rather than the hogging moment one instinctively draws, because span AB is unloaded and is
dragged upward by the loaded span next to it; and the largest hogging moment in the whole beam is
the determinate overhang value at D, not any interior support moment.
(b) The cantilever tree
Question 1(b): the cantilever tree, built in at A, with the horizontal load P at joint C.
Identify the unloaded branch. Member CD carries no load along its length and
its far end D is completely free, so all three internal actions vanish throughout CD:
$$N_{CD}=V_{CD}=M_{CD}=\boxed{0}$$
Pass the load along the arm. Cutting BC anywhere and taking the free body to
the right, the only external force is $P$, which acts along the axis of BC. The arm is
therefore in pure tension:
$$N_{BC}=P\ \text{(tension)},\qquad V_{BC}=0,\qquad M_{BC}=0$$
Bend the column. Transferring $P$ to the column, member AB carries a constant
transverse shear and a moment that grows linearly from the free joint B to the built-in end A:
$$V_{AB}=P\ \text{(constant)},\qquad M_B=0,\qquad M_A=\boxed{P\,L}$$
The column carries no axial force, since no vertical load acts anywhere on the structure.
Question 1(b): the bending moment diagram; the shear force is P in AB and zero elsewhere.
