Question 1 of 8: Classify each structure — unstable, statically determinate, or statically indeterminate (6 marks)

Question 1: Classify each structure — unstable, statically determinate, or statically indeterminate (6 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. Six independent structures. (a) a beam fixed at the left end, two interior roller supports, one internal hinge, and a roller at the right tip — carries UDL w. (b) a beam fixed at the left end, one internal hinge, then a roller support, then a short overhang to a free tip — carries UDL w. (c) a stepped rigid frame: a column from a pin support up to a UDL-loaded beam, stepping down through a second column (pin support) to a second UDL-loaded beam, stepping down again through a third column to a third pin support — three pin supports, all corners rigid. (d) a two-storey rigid portal: a top UDL beam and a lower UDL beam both framing into the same two outer legs, plus a third, central leg dropping from the lower beam to a third pin support — three pin supports, all joints rigid. (e) a triangulated truss, pin support at one end and roller at the other, loaded with three point loads P at the upper panel points. (f) a triangulated truss with three pin supports and an apex loaded with two P forces.

Find. For each structure: unstable / determinate / indeterminate, and the degree of indeterminacy where indeterminate.

Approach. For a beam or a beam-type frame, compare available equilibrium-plus-condition equations to the total reaction count; for a truss, compare the member-plus-reaction count to the classic $m+r=2n$ joint count.

1. (a) Fixed – roller – roller – hinge – roller. Reactions: fixed support contributes 3, each of the three rollers contributes 1, so $r = 3+3 = 6$. Equilibrium supplies 3 equations; the single internal hinge supplies 1 condition equation ($c=1$). Degree of static indeterminacy $\text{DSI}=r-3-c = 6-3-1=\boxed{2}$. Statically indeterminate to the 2nd degree.
2. (b) Fixed – hinge – roller – roller (cantilever + suspended span). $r = 3(\text{fixed})+1+1=5$. $c=1$ (one hinge). $\text{DSI}=5-3-1=\boxed{1}$. Statically indeterminate to the 1st degree.
3. (c) Stepped 3-leg rigid frame ("tree", no closed loop). Three pin supports give $r=2\times3=6$. Counting members/nodes for the general frame formula $\text{DSI}=3m+r-3n-c$: $m=5$ members (two columns split by the mid step, two beam segments, one more column), $n=6$ nodes, no internal hinges ($c=0$): $\text{DSI}=3(5)+6-3(6)-0=15+6-18=\boxed{3}$. Statically indeterminate to the 3rd degree — every extra pin beyond the one needed for a stable cantilevered tree adds 2 redundants, and there are 3 pins on a structure a single fixed support would stabilize.
4. (d) Two-storey rigid portal, 3 pins. This is a closed rectangular loop (top beam + two upper leg segments + lower beam) with a third leg hanging from the loop down to a third support. $m=8$, $n=8$, $r=6$ (three pins), $c=0$: $\text{DSI}=3(8)+6-3(8)-0=24+6-24=\boxed{6}$. Statically indeterminate to the 6th degree (one closed rigid loop contributes 3, and the 3 pin supports contribute 3 more beyond the 3 reactions a single support would need).
5. (e) Truss, pin + roller. Counting the drawn members: 4 bottom-chord panels, 2 upper-to-lower verticals/diagonals per side plus the two "X" diagonals into the bottom-centre joint, and the two long outer diagonals through the shoulder joints gives $m=11$ members and $n=7$ joints. $r=2(\text{pin})+1(\text{roller})=3$. $\text{DSI}=m+r-2n = 11+3-2(7)=14-14=\boxed{0}$. Statically determinate.
6. (f) Truss, 3 pins. Each of the three lower panel points is reached by exactly one diagonal or vertical from the fan at the apex plus one member of the lower chord (the crossing diagonals are not connected at the crossing, per the note): $m=8$, $n=7$. $r=2\times3=6$ (three pins). $\text{DSI}=m+r-2n=8+6-14=\boxed{0}$. Statically determinate.