Question 8 of 9: Slope-deflection analysis of a frame on an inclined roller
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 8: Slope-deflection analysis of a frame on an inclined roller (24 marks)
$6\ \text{m}$, rigidity $3EI$, built in at joint 1
Column 2–3
$4\ \text{m}$, rigidity $EI$
Load
$w=12\ \text{kN/m}$ on the beam
Support at joint 3
roller on a plane of slope $3:4$ (falling to the right)
Find. The end moments, the shear force and bending moment diagrams with
maximum and minimum ordinates, and the reactions.
Question 8: the frame with its roller bearing on a plane of slope 3 in 4.
Approach. The inclined roller couples the vertical and horizontal movement of
joint 3; use inextensibility to reduce the whole kinematics to a single translation parameter,
then write slope-deflection equations in that parameter and the two rotations, and close the
system with the virtual-work equation for the translation.
Kinematics. Joint 1 is built in, so $u_1=v_1=0$; the beam is inextensible, so
$u_2=0$; the column is inextensible, so $v_3=v_2$. The roller allows joint 3 to move only along
the plane, whose direction is $(4,-3)/5$, so
$$\frac{u_3}{4}=\frac{v_3}{-3}\;\Longrightarrow\;u_3=-\tfrac{4}{3}v_2$$
Everything is therefore governed by the single translation $v_2$, together with the rotations
$\theta_2$ and $\theta_3$.
Chord rotations.
$$\psi_{12}=\frac{v_2}{6},\qquad
\psi_{23}=\frac{-u_3}{4}=\frac{\tfrac{4}{3}v_2}{4}=\frac{v_2}{3}$$
so tilting the beam down at joint 2 simultaneously rakes the column — the essential feature
of an inclined roller.
Slope-deflection equations. With $M^{F}_{12}=+wL^{2}/12=+36$ and
$M^{F}_{21}=-36\ \text{kN}\cdot\text{m}$,
$$M_{12}=\tfrac{2(3EI)}{6}\left(\theta_2-3\psi_{12}\right)+36,\qquad
M_{21}=\tfrac{2(3EI)}{6}\left(2\theta_2-3\psi_{12}\right)-36$$
$$M_{23}=\tfrac{2EI}{4}\left(2\theta_2+\theta_3-3\psi_{23}\right),\qquad
M_{32}=\tfrac{2EI}{4}\left(2\theta_3+\theta_2-3\psi_{23}\right)$$
Equilibrium equations. Joint 2 is rigid and joint 3 is a moment-free roller:
$$M_{21}+M_{23}=0,\qquad M_{32}=0$$
and the third equation comes from a virtual translation along the roller plane, which by the
inextensibility relations is a virtual $v_2^{*}=1$:
$$\left(M_{12}+M_{21}\right)\psi^{*}_{12}+\left(M_{23}+M_{32}\right)\psi^{*}_{23}
+W_{\text{ext}}=0$$
Solve. The three equations give
$$M_{12}=-108,\qquad M_{21}=M_{\text{sag}}(2)=+36,\qquad M_{32}=0\ \text{kN}\cdot\text{m}$$
i.e. the built-in end hogs $\boxed{108\ \text{kN}\cdot\text{m}}$ while the knee joint carries a
sagging $\boxed{36\ \text{kN}\cdot\text{m}}$, and the joint displacements are
$u_3=192/EI$ to the right and $v_3=-144/EI$, which satisfy $3u_3+4v_3=0$ as the roller
requires.
Beam diagrams. Vertical equilibrium of the beam with the two end moments
gives
$$V_1=\frac{w L}{2}+\frac{M_{21}-M_{12}}{L}\cdot\frac{1}{1}=60\ \text{kN},\qquad
V_2=60-12(6)=-12\ \text{kN}$$
$$M(x)=-108+60x-6x^{2}\;\Rightarrow\;
M_{\max}=\boxed{+42\ \text{kN}\cdot\text{m}}\ \text{at}\ x=5\ \text{m}$$
with the moment passing through $+36\ \text{kN}\cdot\text{m}$ at joint 2 and through zero at
$x=2.0$ and $x=8.0$ — only the first of which lies on the beam.
Column diagrams. The column carries a moment falling linearly from
$36\ \text{kN}\cdot\text{m}$ at joint 2 to zero at the roller, so
$$V_{23}=\frac{36}{4}=9\ \text{kN}\ \text{(constant)},\qquad N_{23}=12\ \text{kN\ (compression)}$$
Reaction at the inclined roller and check. The reaction must be normal to the
plane, i.e. along $(3,4)/5$:
$$R_3=\sqrt{9^{2}+12^{2}}=\boxed{15\ \text{kN}}$$
with components $9\ \text{kN}$ horizontal and $12\ \text{kN}$ vertical. At the built-in end
$V=60\ \text{kN}$ upward, $H=9\ \text{kN}$ to the left and $M=108\ \text{kN}\cdot\text{m}$;
vertically $60+12=72=12(6)\ \checkmark$, horizontally $9-9=0\ \checkmark$, and moments about joint
1 give $-216+72+36+108=0\ \checkmark$.
