Question 4 of 9: Slope-deflection analysis of a frame with an imposed joint displacement
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 4: Slope-deflection analysis of a frame with an imposed joint displacement (18 marks)
$EI=18\,000\ \text{kN}\cdot\text{m}^{2}$, all members inextensible
Imposed displacement
joint 2 pulled down $\delta = 0.04\ \text{m}$; no applied loads
Find. All member end moments, the shear force and bending moment diagrams with
their maximum and minimum ordinates, and the support reactions.
Question 4: the frame, with joint 2 pulled down 0.04 m.
Approach. Establish that the only kinematic freedoms are the three joint
rotations, write the six slope-deflection equations with the chord rotations produced by the
imposed 0.04 m, and solve the three joint-equilibrium equations.
Degrees of freedom. The pin at joint 1 prevents horizontal movement and the
beam is inextensible, so $u_2=u_3=0$; the column is inextensible and built in at joint 4, so
$v_3=0$. There is no sway, and the unknowns are the three rotations
$\theta_1,\ \theta_2,\ \theta_3$.
Chord rotations from the imposed displacement. With
$\psi_{ij}=(\mathbf{D}_j-\mathbf{D}_i)\cdot\mathbf{e}_2/L$,
$$\psi_{12}=\frac{-0.04}{3}=-0.013333,\qquad
\psi_{23}=\frac{0-(-0.04)}{6}=+0.006667,\qquad \psi_{34}=0$$
Only the two beam members are distorted; the column merely follows joint 3.
Slope-deflection equations. With no span loads every fixed-end moment is
zero, so
$$M_{12}=\tfrac{2EI}{3}\left(2\theta_1+\theta_2-3\psi_{12}\right),\qquad
M_{21}=\tfrac{2EI}{3}\left(2\theta_2+\theta_1-3\psi_{12}\right)$$
$$M_{23}=\tfrac{2EI}{6}\left(2\theta_2+\theta_3-3\psi_{23}\right),\qquad
M_{32}=\tfrac{2EI}{6}\left(2\theta_3+\theta_2-3\psi_{23}\right)$$
$$M_{34}=\tfrac{2EI}{4}\left(2\theta_3\right),\qquad M_{43}=\tfrac{2EI}{4}\left(\theta_3\right)$$
Equilibrium equations. The pin carries no moment and joints 2 and 3 carry no
applied couple:
$$M_{12}=0,\qquad M_{21}+M_{23}=0,\qquad M_{32}+M_{34}=0$$
Solving the three equations,
$$\theta_1=-0.017500,\qquad \theta_2=-0.005000,\qquad
\theta_3=+0.005000\ \text{rad}$$
Back-substitute for the end moments.
$$M_{21}=+150,\quad M_{23}=-150,\quad M_{32}=-90,\quad M_{34}=+90,\quad
M_{43}=+45\ \text{kN}\cdot\text{m}$$
so the sagging moment at joint 2 is $\boxed{+150\ \text{kN}\cdot\text{m}}$, the beam hogs
$\boxed{90\ \text{kN}\cdot\text{m}}$ at joint 3, and the column base carries
$\boxed{45\ \text{kN}\cdot\text{m}}$.
Shears and reactions. Each member is unloaded, so its shear is constant:
$$V_{12}=\frac{150-0}{3}=50\ \text{kN},\qquad
V_{23}=\frac{-90-150}{6}=-40\ \text{kN},\qquad
V_{34}=\frac{90+45}{4}=33.75\ \text{kN}$$
Hence $R_1=50\ \text{kN}$ upward, $R_4=40\ \text{kN}$ upward with a horizontal thrust of
$33.75\ \text{kN}$ and a base moment of $45\ \text{kN}\cdot\text{m}$, and the roller at joint 2
must pull down with
$$R_2=50+40=\boxed{90\ \text{kN}\ \text{(downward)}}$$
Check: the support at joint 2 is read from the drawing as
a roller that is displaced 0.04 m downward, not as a settlement of a support that could
lift off. Because the analysis returns a 90 kN downward reaction there, the detail must be capable
of holding the beam down — a bearing that can only push would separate and the structure
would become the determinate propped beam 1–3.
