Question 5 of 9: Uniform temperature rise in a three-column frame (16 marks)

Question 5: Uniform temperature rise in a three-column frame (16 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. A symmetric two-bay portal on three 6 m columns, unloaded, with every member lengthening by the same amount.

Find. Shear and bending moment diagrams with the maximum and minimum ordinate on every member.

Approach. The elongations prescribe every joint translation once symmetry fixes the centre joint, so the only unknown left is one joint rotation.

1. Use symmetry to lock the centre joint. Geometry, supports and loading are all symmetric about the centre line through joints 3 and 6. A symmetric response has $u=0$ and $\theta=0$ on the axis of symmetry, so $$u_{3}=0,\qquad \theta_{3}=0,\qquad \theta_{4}=-\theta_{2}.$$
2. Turn the elongations into joint displacements. Each column grows 0.006 m along its own axis, which is vertical, so every top joint rises: $v_{2}=v_{3}=v_{4}=+0.006$ m. Each beam grows 0.006 m horizontally, and with $u_{3}=0$ that spreads the frame symmetrically: $$u_{2}=-0.006\ \text{m},\qquad u_{4}=+0.006\ \text{m}.$$ Nothing else is free — the whole translation field is prescribed and no sway equation is needed.
3. Compute the chord rotations. A chord rotation is the transverse relative displacement divided by the length. Both beams have $v_{2}=v_{3}=v_{4}$, so $\psi_{23}=\psi_{34}=0$; the centre column has $u_{3}=u_{6}=0$, so $\psi_{63}=0$; only the outer columns rotate, $$\psi_{12}=\frac{u_{2}-u_{1}}{6}=-0.001\ \text{rad},\qquad \psi_{54}=+0.001\ \text{rad}$$ (clockwise positive). This is the whole of the thermal loading.
4. Write the slope-deflection equations. Joints 1 and 5 are pins, so use the modified form on the outer columns and the standard form on the beams: $$M_{21}=\frac{3EI}{6}\bigl(\theta_{2}-\psi_{12}\bigr),\qquad M_{23}=\frac{2EI}{6}\bigl(2\theta_{2}+\theta_{3}\bigr)=\frac{2EI}{3}\theta_{2}.$$ The centre column has $\theta_{3}=\theta_{6}=0$ and $\psi_{63}=0$, so $M_{36}=M_{63}=0$ before anything is solved — the centre column is completely unstressed.
5. Solve the one joint equation. Moment equilibrium at joint 2 gives $$\frac{3EI}{6}\bigl(\theta_{2}+0.001\bigr)+\frac{2EI}{3}\theta_{2}=0\;\Rightarrow\;\theta_{2}\left(\frac{3}{6}+\frac{4}{6}\right)=-\frac{3(0.006)}{6^{2}},$$ that is $\tfrac{7}{6}\theta_{2}=-5.000\times10^{-4}$, so $$\boxed{\theta_{2}=-4.2857\times10^{-4}\ \text{rad}}$$ (clockwise positive; joint 4 rotates by the same amount the other way). Joint 3 is automatically in balance because $M_{32}$ and $M_{34}$ are equal and opposite by symmetry.
6. Back-substitute for the end moments. With $EI=2.1\times10^{5}$ kN·m², $$M_{21}=\frac{3(2.1\times10^{5})}{6}\bigl(-4.2857\times10^{-4}+1.000\times10^{-3}\bigr)=\boxed{60.0\ \text{kN}\cdot\text{m}}$$ and the beam end moments follow as $M_{23}=-60.0$, $M_{32}=-30.0$ kN·m (magnitudes 60.0 and 30.0), with joint 2 in balance: $60.0-60.0=0$.
7. Shears and reactions. Each outer column carries $$V=\frac{60.0}{6}=10.0\ \text{kN},$$ directed inwards at the base, so the pins push back with 10.0 kN horizontally (equal and opposite at joints 1 and 5, sum zero — the horizontal check). Each beam carries $$V=\frac{60.0+30.0}{6}=15.0\ \text{kN},$$ so the outer columns are in 15.0 kN compression and the centre column in 30.0 kN tension — a hold-down of 30.0 kN at joint 6, balanced by 15.0 kN of uplift resistance at each pin.
8. Diagram shapes. No member carries a span load, so every shear diagram is a constant and every bending moment diagram is a straight line. The outer columns run from zero at the pin to 60.0 kN·m at the top; each beam runs from 60.0 kN·m hogging at its outer joint to 30.0 kN·m at the centre joint, with a point of contraflexure two-thirds of the way along; the centre column is a flat line at zero. Maximum ordinate 60.0 kN·m, minimum ordinate 0.