Question 1 of 9: Minimum structural degrees of freedom for a slope-deflection analysis (8 marks)

Question 1: Minimum structural degrees of freedom for a slope-deflection analysis (8 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. Four plane frames, all members inextensible and rigidly jointed except where a support is drawn as a pin or a roller: (a) a two-storey frame on three fixed bases spaced $L$ and $2L$, with a horizontal load $P$ at each floor level on the left column; (b) a fixed-base column carrying an inclined member up to a beam that runs over a second fixed-base column and continues $L/2$ as a free overhang, with a downward $P$ at the apex and another at the overhang tip; (c) an upper beam on a roller carrying $w$, a column down to a second roller, and a lower beam carrying $w$ into a built-in end; (d) the same arrangement as (c) but with the upper beam built in at its left end, a pin at the foot of the column and a roller at the far end of the lower beam.

Find. For each frame, the minimum number $k$ of structural degrees of freedom (independent joint rotations plus independent joint translations) that a slope-deflection analysis must carry, and the arrows that identify them.

Approach. Count $k=n_{\theta}+n_{\Delta}$, taking $n_{\Delta}$ from the bar linkage that the inextensible members form and deleting every rotation at a joint whose moment is known to be zero.

1. State the counting rule. A slope-deflection analysis carries one unknown for every independent joint rotation and one for every independent joint translation, $$k=n_{\theta}+n_{\Delta}.$$ A rotation is not an unknown when the moment at that end is known in advance — the free end of an overhang, or a beam end sitting on a pin or roller support, where the modified stiffness $3EI/L$ absorbs it.
2. Get the sway count from the bar linkage. Replace every rigid joint by a pin and every inextensible member by a rigid bar. The number of independent translations is then the mobility of that linkage, $$n_{\Delta}=2n_{j}-n_{b}-n_{r},$$ where $n_{j}$ counts joints free to translate, $n_{b}$ the bars and $n_{r}$ the support components acting on those joints. (Bars running to a fixed or pinned base still count; the base itself is not a free joint.)
3. Frame (a). The free joints are the two roof joints and the three first-floor joints, so $n_{j}=5$; the bars are three lower column lengths, two upper column lengths, two first-floor beam lengths and the roof beam, so $n_{b}=8$; the three bases are fixed and carry no free joint, $n_{r}=0$. Hence $n_{\Delta}=2(5)-8-0=2$ — one sway per storey. Every one of the five joints is rigid and none has a known moment, so $n_{\theta}=5$ and $$\boxed{k_{(a)}=5+2=7}$$ The frame is not symmetric (the bays are $L$ and $2L$), so no symmetry reduction is available.
4. Frame (b). The $L/2$ length beyond the right column is a statically determinate cantilever: replace it by the force $P$ and the couple $P\,L/2$ that it delivers to the joint, and it disappears from the unknown list. That leaves the top of the left column, the apex and the top of the right column, so $n_{j}=3$ and $n_{b}=4$ (left column, inclined member, beam, right column), giving $n_{\Delta}=2(3)-4=2$. All three joints are rigid on fixed bases, so $$\boxed{k_{(b)}=3+2=5}$$
5. Frame (c). Free joints: the roller end of the upper beam, the corner, and the foot of the column where the lower beam starts, $n_{j}=3$; bars: upper beam, column, lower beam, $n_{b}=3$; restraints: each roller kills one vertical component, $n_{r}=2$. So $n_{\Delta}=6-3-2=1$ — the upper beam and the column head move horizontally together, while the lower beam is pinned in that direction by the built-in end. The roller end of the upper beam has $M=0$, so its rotation is absorbed by the $3EI/L$ modified stiffness and only two rotations survive: $$\boxed{k_{(c)}=2+1=3}$$
6. Frame (d). The same three free joints and three bars, but now the restraints are the pin at the foot of the column ($u=v=0$) and the roller at the far end of the lower beam ($v=0$), $n_{r}=3$; the built-in upper end also fixes the top of the upper beam. Hence $n_{\Delta}=6-3-3=0$: nothing can translate. The built-in end has $\theta=0$ and the roller end of the lower beam has $M=0$, so the unknowns are the two interior rotations, $$\boxed{k_{(d)}=2+0=2}$$
7. What to draw on the exam paper. On each frame put a curved counter-clockwise arrow at every joint counted in $n_{\theta}$ and a straight arrow at every level counted in $n_{\Delta}$: five curved plus two straight on (a), three curved plus two straight on (b), two curved plus one straight on (c), two curved and none on (d). The arrows are the answer — the number beside the frame simply totals them.