Question 5 of 9: Slope-deflection analysis of a frame with a fabrication error

Question 5: Slope-deflection analysis of a frame with a fabrication error (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.

Find. The member end moments, and the shear and bending moment diagrams with their maximum and minimum ordinates.

Approach. Convert the fabrication error into a prescribed vertical displacement of joints 1 and 4, treat the beam as a continuous member with rotational springs supplied by the two inner columns, exploit the symmetry of both structure and imposed displacement, and solve two slope-deflection equations.

1. Turn the lack of fit into a support movement. Columns 1–5 and 4–8 are pinned at both ends, so they are two-force members carrying pure axial force and delivering no moment to the beam — hence $M_{12}=M_{43}=0$. They are also inextensible, so once forced into place their tops necessarily stand $0.048\text{ m}$ above the design level: $$v_1=v_4=+0.048\ \text{m},\qquad v_2=v_3=0 ,$$ the inner columns being inextensible and built in at the base. There is no sidesway, because the imposed displacement is symmetric and the two inner columns restrain horizontal movement symmetrically.
2. Compute the chord rotations. For a horizontal member $i\to j$ the chord rotation is $\psi_{ij}=(v_i-v_j)/L$, clockwise positive: $$\psi_{12}=\frac{0.048-0}{6}=+0.008,\qquad \psi_{23}=0,\qquad \psi_{34}=\frac{0-0.048}{6}=-0.008 .$$
3. Exploit symmetry. Both the structure and the imposed displacement are symmetric about the mid-point of span 2–3, so $\theta_3=-\theta_2$ and $\theta_4=-\theta_1$, and only two rotations remain.
4. Write the slope-deflection equations. With $k_1=2EI/6=1.6667\times10^{4}$, $k_2=2EI/4=2.500\times10^{4}$ and the inner columns contributing the rotational stiffness $4EI/6=3.3333\times10^{4}$, $$M_{12}=k_1\!\left(2\theta_1+\theta_2-3\psi_{12}\right)=0 ,$$ $$M_{21}=k_1\!\left(2\theta_2+\theta_1-3\psi_{12}\right),\qquad M_{23}=k_2\!\left(2\theta_2+\theta_3\right)=k_2\theta_2 ,$$ and joint 2 requires $M_{21}+M_{23}+\dfrac{4EI}{6}\theta_2=0$.
5. Solve the two equations. The first gives $\theta_1=\tfrac12\!\left(3\psi_{12}-\theta_2\right)=0.012-0.5\theta_2$. Substituting into the joint equation, $$1.5k_1\theta_2+\left(k_2+\tfrac{4EI}{6}\right)\theta_2 = 1.5k_1\frac{0.048}{6},$$ $$\boxed{\theta_2 = +0.00240\ \text{rad},\qquad \theta_1 = +0.01080\ \text{rad}}$$ (clockwise positive; by symmetry $\theta_3=-0.00240$ and $\theta_4=-0.01080$).
6. Back-substitute for the end moments. $$M_{21}=k_1\!\left(0.00480+0.01080-0.02400\right)=-140.0\ \text{kN}\cdot\text{m},\qquad M_{23}=k_2(0.00240)=+60.0\ \text{kN}\cdot\text{m},$$ $$M_{26}=\frac{2EI}{6}(2\theta_2)=+80.0\ \text{kN}\cdot\text{m},\qquad M_{62}=\frac{2EI}{6}\theta_2=+40.0\ \text{kN}\cdot\text{m} .$$ Joint 2 checks: $-140.0+60.0+80.0=0$. The mirror joint carries the same magnitudes with reversed signs, and $M_{12}=M_{43}=0$ as required by the pinned columns.
7. Convert to diagram ordinates. In the sagging-positive convention the bending moment at end $i$ of member $ij$ is $+M_{ij}$ and at end $j$ it is $-M_{ji}$, so the beam moment runs from zero at joint 1 to $+140.0\ \text{kN}\cdot\text{m}$ just left of joint 2, drops by the column moment of $80.0\ \text{kN}\cdot\text{m}$ to a constant $+60.0\ \text{kN}\cdot\text{m}$ across the middle span, and mirrors on the right. Each inner column runs from $-40.0\ \text{kN}\cdot\text{m}$ at its base to $+80.0\ \text{kN}\cdot\text{m}$ at its head.
8. Shears and axial forces. Span 1–2 carries no load, so its shear is constant: $$V_{12}=\frac{M_{12}+M_{21}}{-6}=+23.33\ \text{kN},$$ and span 2–3 carries zero shear (its two end moments are equal and opposite), which is why its moment is constant. Each inner column carries the horizontal shear $$H=\frac{M_{26}+M_{62}}{6}=\frac{80.0+40.0}{6}=20.0\ \text{kN},$$ the two acting in opposite directions and balanced through the axial force in the beam. The outer columns are therefore in compression at 23.33 kN (they jack the beam up) and the inner columns are in tension at 23.33 kN (they hold it down); the four axial forces sum to zero, as they must with no applied load.