Question 1 of 9: Structural degrees of freedom for a slope-deflection analysis

Question 1: 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 (a)–(d), all with inextensible members of uniform $EI$, are to be analysed by the slope-deflection method. Frame (a) is a symmetric two-bay, two-storey frame on three fixed bases carrying a horizontal load $P$ at each floor; frame (b) is a fixed-base column, an inclined member and a horizontal member built into a wall, carrying a uniformly distributed load $w$; frame (c) is a three-column frame in which two inclined members and the middle column form a closed triangle, loaded by $P$ at the triangle apex and by $P$ at the top of the right-hand column; frame (d) is a stepped beam-and-column assembly on a pin, a roller and a roller, carrying $w$ on each horizontal member.

Find. The minimum number of structural (kinematic) degrees of freedom — independent joint rotations plus independent joint translations — for each of the four structures.

Approach. Count one rotation for every rigid joint whose moment is not known in advance, count the independent joint translations that survive the inextensibility constraints of the members, then delete the rotations that symmetry, anti-symmetry or a known-zero-moment end makes dependent or redundant.

1. State the counting rule. The number of structural degrees of freedom is $$n_{\text{dof}} = n_{\theta} + n_{\Delta},$$ where $n_{\theta}$ counts the independent rotations of the rigid joints and $n_{\Delta}$ counts the independent joint translations. A member end at a simple support (a pin or a roller carrying only one member) has a known zero moment, so the modified slope-deflection equation $M_{nf}=\dfrac{3EI}{L}\left(\theta_n-\psi\right)$ eliminates that rotation and it is not counted. Every inextensible member imposes one constraint on the translations of its two ends, which is what makes $n_{\Delta}$ far smaller than the raw count of free joint displacements.
2. Frame (a): use anti-symmetry. The frame is geometrically symmetric about the centre column. Because the floor beams are inextensible, a horizontal load applied at a floor may be redistributed arbitrarily among the joints of that floor without changing the response — only the total storey shear enters the equilibrium equations, and a horizontal joint force contributes nothing to joint moment equilibrium. Replacing each $P$ by $P/2$ at the left column and $P/2$ at the right column produces a loading that reverses sign under reflection, so the response is anti-symmetric: mirror-image joints rotate by equal amounts in the same sense, and the centre column (which lies on the axis) carries zero axial force but is free to bend. Hence $\theta_{A1}=\theta_{C1}$ and $\theta_{A2}=\theta_{C2}$, leaving four independent rotations $\theta_{A1},\ \theta_{A2},\ \theta_{B1},\ \theta_{B2}$, plus one sway per storey: $$\boxed{n_{\text{dof}}^{(a)} = 4 + 2 = 6.}$$ Without the symmetry argument the count would have been $6+2=8$.
3. Frame (b): chase the inextensibility chain. Both ends of the structure are built in, so no rotation is eliminated by a zero-moment end and the two kinks give $\theta_1$ and $\theta_2$. For the translations, let joint 1 (the top of the vertical column) move by $(a,\,0)$ — the vertical column being inextensible fixes its vertical displacement — and let joint 2 (the kink at the horizontal member) move by $(0,\,b)$, because the horizontal member runs into the wall. The inclined member ties the two together through $$\left(\mathbf{u}_2-\mathbf{u}_1\right)\cdot\mathbf{e}_{12}=0 \quad\Longrightarrow\quad -a\,c_x + b\,c_y = 0 ,$$ so $b$ follows from $a$ and only one translation is independent: $$\boxed{n_{\text{dof}}^{(b)} = 2 + 1 = 3.}$$
4. Frame (c): the closed triangle kills the sway. Four rigid joints rotate: the triangle apex $A$, the top of the middle column $B$, the point $C$ where the lower inclined member meets the middle column, and the top of the right-hand column $D$; none of them is a known-zero-moment end. Members $AB$, $AC$ and $CB$ form a closed inextensible triangle, so $A$, $B$ and $C$ move as a rigid body — two translations and one rotation. The left column pins $A$ vertically ($A_y=0$) and the lower part of the middle column pins $C$ vertically ($C_y=0$); since $A$ and $C$ have different abscissae these two conditions force both the rigid-body rotation and the rigid-body vertical translation to vanish, leaving a pure horizontal translation. The right-hand column then fixes $D_y=0$ and the inextensible top beam fixes $D_x=B_x$, so $D$ carries no independent translation: $$\boxed{n_{\text{dof}}^{(c)} = 4 + 1 = 5.}$$ A rank count on the seven inextensibility constraints acting on the eight translational freedoms of $A,B,C,D$ confirms a single independent translation.
5. Frame (d): delete the zero-moment ends, keep the sway. The pin at the left end of the upper beam and the roller at the right end of the lower beam are single member ends carrying zero moment, so their rotations are condensed out by the modified stiffness. Only the two rigid corners remain, giving $\theta_F$ and $\theta_G$. For the translations: the upper beam ties $F$ horizontally to the pin, the vertical member ties $F$ vertically to $G$, and the roller under $G$ fixes $G_y=0$, so $F$ is completely immobilised — but nothing restrains $G$ horizontally, because the lower beam simply carries $H$ along with it and the roller at $H$ offers no horizontal reaction. The vertical member therefore sways: $$\boxed{n_{\text{dof}}^{(d)} = 2 + 1 = 3.}$$