07-Str-A1 · May 2018

Question 7 of 8: Frame with an Inclined Member by Slope Deflection

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 7: Frame with an Inclined Member by Slope Deflection (22 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Joint 1 is a fixed support 2.5 m below the beam level. Member 1–2 rises 2.5 m over a 6 m horizontal run, so its length is 6.5 m and its rigidity is $1.3EI$. From joint 2 a 3 m vertical member of rigidity $EI$ runs down to a pin at joint 5. The horizontal member 2–3 is 8 m of $1.4EI$ ending at a roller at joint 3, and it continues 3 m as an overhang of $1.4EI$ to a free end at joint 4. The uniformly distributed load of 10.8 kN/m is drawn over the entire horizontal projection, from directly above joint 1 to joint 4, and a 14.3 kN point load acts downward at joint 4. All members are inextensible.

Find. Shear-force and bending-moment diagrams for every member, with the maximum and minimum ordinates marked.

Approach. Show first that the frame cannot sway, so the only kinematic unknowns are the rotations at joints 2 and 3; reduce the overhang to a known moment at joint 3; then write four slope-deflection equations, solve the two joint-equilibrium equations, and build the diagrams member by member, plotting the inclined member developed along its own axis.

1. Rule out sidesway. Member 2–5 is vertical and inextensible, so joint 2 cannot move vertically; member 2–3 is horizontal and inextensible, so joints 2 and 3 share the same horizontal displacement; and member 1–2 is inextensible from a fixed base, so the displacement of joint 2 along its own axis is zero. With zero vertical movement, that last condition forces the horizontal movement to zero as well. Hence $\psi = 0$ throughout and only $\theta_2$ and $\theta_3$ are unknown.
2. Reduce the overhang. The 3 m cantilever beyond joint 3 carries $10.8 \times 3 = 32.4$ kN of distributed load and the 14.3 kN tip load, so it delivers to joint 3 a hogging moment of $$M_{34} = -\left[10.8(3)(1.5) + 14.3(3)\right] = -(48.6 + 42.9) = -91.5\ \text{kN}\cdot\text{m}$$ and a shear of $32.4 + 14.3 = 46.7$ kN. Being determinate it has no stiffness and takes no distribution factor.
3. Fixed-end moments. For the horizontal member, $$\mathrm{FEM}_{23} = \frac{wL^2}{12} = \frac{10.8 \times 64}{12} = 57.6\ \text{kN}\cdot\text{m}.$$ For the inclined member the load is uniform on the projection, so only its component perpendicular to the member bends it. Working through, $$q_{\perp} = w\left(\frac{L_h}{L}\right)^{2} = 10.8\left(\frac{6}{6.5}\right)^{2} = 9.2024\ \text{kN/m}, \qquad \mathrm{FEM}_{12} = \frac{q_{\perp}L^{2}}{12} = \frac{w L_h^{2}}{12} = \frac{10.8 \times 36}{12} = 32.4\ \text{kN}\cdot\text{m}.$$ The identity $q_{\perp}L^{2} = w L_h^{2}$ is worth remembering: a projection-uniform load on an inclined member gives the same fixed-end moment as the same intensity on a span equal to the projection.
4. Slope-deflection equations. With $2EI_m/L$ equal to $0.4EI$ for member 1–2 and $0.35EI$ for member 2–3, and the modified $3EI/L = 1.0EI$ for the pin-based column, $$M_{12} = 0.4EI\theta_2 - 32.4, \qquad M_{21} = 0.8EI\theta_2 + 32.4, \qquad M_{25} = 1.0EI\theta_2,$$ $$M_{23} = 0.7EI\theta_2 + 0.35EI\theta_3 - 57.6, \qquad M_{32} = 0.7EI\theta_3 + 0.35EI\theta_2 + 57.6.$$
5. Joint equilibrium and solution. Summing the member end moments at each joint, $$2.5EI\theta_2 + 0.35EI\theta_3 = 25.2, \qquad 0.35EI\theta_2 + 0.70EI\theta_3 = 33.9,$$ whence $$EI\theta_2 = 3.548, \qquad EI\theta_3 = 46.654.$$ Back-substituting, $$M_{12} = -30.98, \quad M_{21} = +35.24, \quad M_{25} = +3.55, \quad M_{23} = -38.79, \quad \boxed{M_{32} = +91.50\ \text{kN}\cdot\text{m}},$$ and $M_{32}$ balancing the overhang moment exactly is the arithmetic check on the whole solution.
6. Member 1–2, developed along its own axis. With beam-sign end moments of $-30.98$ and $-35.24$ kN·m and $q_{\perp} = 9.2024$ kN/m over 6.5 m, $$V_1 = \frac{-35.24 + 30.98 + 194.40}{6.5} = 29.25\ \text{kN}, \qquad V_2 = -30.56\ \text{kN},$$ with zero shear 3.179 m along the member and $$M_{\max} = -30.98 + \frac{29.25^{2}}{2(9.2024)} = \boxed{+15.51\ \text{kN}\cdot\text{m}}.$$
7. Member 2–3 and the roller reaction. With end moments $-38.79$ and $-91.50$ kN·m, $$V_2 = \frac{-91.50 + 38.79 + 345.6}{8} = 36.61\ \text{kN}, \qquad V_3 = -49.79\ \text{kN},$$ the zero-shear point is 3.390 m from joint 2 and $$M_{\max} = -38.79 + \frac{36.61^{2}}{2(10.8)} = \boxed{+23.27\ \text{kN}\cdot\text{m}}.$$ The jump in shear across joint 3 gives the roller reaction, $$R_3 = 46.70 - (-49.79) = \boxed{96.49\ \text{kN}\ (\uparrow)}.$$
8. Column 2–5. It carries no transverse load, a moment of 3.55 kN·m at joint 2 and zero at its pinned base, so its shear is the constant $3.55/3 = 1.18$ kN and its moment diagram is a single straight line. It is the lightest member in the frame precisely because the pin at its base gives it only the modified stiffness $3EI/L$.

A direct-stiffness model of the same frame — nodes at 1, 2, 3, 4 and 5, a fixed base, a pin, a vertical roller and the perpendicular component of the projection load applied to the inclined element — reproduces every end moment and the roller reaction to five figures. That independent check is what catches a wrong formula, as distinct from a wrong arithmetic step, on a question of this class.