Question 1 of 9: Statical indeterminacy and structural degrees of freedom

Question 1: Statical indeterminacy and structural degrees of freedom (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 planar structures, all members prismatic and inextensible: (i) a three-span beam built in at the left-hand end, on rollers at each of the three remaining support points, with an internal hinge in the last span and a uniform load $w$ over the whole length; (ii) a symmetric two-bay gabled frame with the outer column bases built in, the centre column base pinned, rigid eaves, valley and apexes, and $w$ over the full horizontal projection; (iii) a single-bay portal of span $L$ and height $L$ with both bases built in, carrying a vertical point load $P$ on the beam a distance $L/4$ from the right-hand corner; and (iv) a symmetric two-bay single-storey frame, spans $L$ and $L$, height $L$, all three bases built in, carrying a horizontal point load $P$ at mid-height of the centre column.

Find. For each structure, the degree of statical indeterminacy $r$, and the minimum number $k$ of unknown joint rotations and independent joint translations that a slope-deflection analysis must carry.

Approach. Count $r$ from the reaction components less the available equations (three equilibrium equations plus one condition equation per internal release for beams, or $r=3m+r_{c}-3j-c$ for frames), then count $k$ as the unknown joint rotations plus the independent joint translations that survive inextensibility, symmetry or anti-symmetry, and the elimination of joints whose moment is known to be zero.

1. State the two counting rules. For a beam the degree of statical indeterminacy is$$r=n_{\text{reactions}}-\left(3+c\right)$$with $c$ the number of condition equations (one per internal hinge). For a rigid frame with $m$ members, $j$ joints (supports included) and $r_{c}$ reaction components,$$r=3m+r_{c}-3j-c .$$The kinematic count is different in kind: $k$ is the number of independent displacement unknowns, that is, the unknown joint rotations plus the independent joint translations permitted by axially rigid members. A member end whose moment is known to vanish (a pin support, or an end next to an internal hinge) is absorbed into a modified stiffness $3EI/L$ and contributes no unknown.
2. Structure (i) — the hinge makes the tail determinate. The supports supply $3+1+1+1=6$ reaction components and the single internal hinge adds one condition equation, so$$r=6-(3+1)=\boxed{2}.$$Kinematically there is no sway, because every support restrains vertical movement and the built-in end restrains horizontal movement. The internal hinge lies inside the last span, so the length from the hinge to the end roller is a simply supported span (zero moment at both ends) and the length from the third support to the hinge is a determinate overhang: both deliver known forces and a known moment to the third support without introducing unknowns. Only the rotations of the second and third supports remain, so $k=2$.
3. Structure (ii) — symmetry halves the work. With $m=7$ members, $j=8$ joints and $r_{c}=3+2+3=8$ reaction components and no internal release,$$r=3(7)+8-3(8)=\boxed{5}.$$The frame is symmetric about the centre column and the load is symmetric, so the response is symmetric. That kills the rotation of the valley joint ($\theta=-\theta\Rightarrow\theta=0$) and its horizontal movement, which in turn leaves the centre column carrying axial force only. On the left half, the eave and the apex rotations survive, and the four inextensible members reduce the six remaining translation components (three joints) to a single spreading parameter: the eave moves out by $u$, the apex moves out by $u/2$ and rises by $au/(2c)$. Hence $k=2+1=3$.
4. Structure (iii) — an off-centre load makes the portal sway. With $m=3$, $j=4$ and $r_{c}=6$,$$r=3(3)+6-3(4)=\boxed{3}.$$Both bases are built in, so the base rotations are zero; the two top-joint rotations are unknown, and the inextensible beam and columns leave exactly one storey translation. The load sits at $L/4$ from the right-hand corner, so it is neither symmetric nor anti-symmetric and the sway cannot be dismissed. Therefore $k=3$.
5. Structure (iv) — the load is purely anti-symmetric. With $m=5$, $j=6$ and $r_{c}=9$,$$r=3(5)+9-3(6)=\boxed{6}.$$Resolve the load into symmetric and anti-symmetric parts about the centre column: reflecting a rightward force on the axis of symmetry gives a leftward force at the same point, so the symmetric part is identically zero and the loading is purely anti-symmetric. The response is then anti-symmetric, which forces the two outer top-joint rotations to be equal. The centre top-joint rotation is not restricted by anti-symmetry (a direct-stiffness check gives a non-zero value), and the axially rigid beams leave one storey translation, so $k=1+1+1=3$.

The four counts are collected below. Note how loosely $r$ and $k$ are related: structure (iv) is twice as indeterminate as structure (iii) yet needs the same number of slope-deflection unknowns, which is exactly why the displacement method is attractive on highly redundant frames.