Determinate and Indeterminate Structures Questions and Answers

Answer: B

For a space truss, which is stable and statically determinate internally, the number of members m and the number of joints j satisfy \(m = 3j - 6\). Each joint in space has three equations of equilibrium, which give 3j equations, and six of them are used by the overall rigid-body motion in space (3 translations and 3 rotations). For a plane truss the matching condition is \(m = 2j - 3\). The options with an inequality (j < 2m - 3 and m > 2j - 3) describe unstable or indeterminate plane trusses, and m = 2j - 1 is not a standard condition.

Answer: A

The degree of static indeterminacy is 6.

For a rigid-jointed plane frame, each member has three internal unknowns, and each joint gives three equilibrium equations.

\[ \begin{aligned} D_s &= 3m + r - 3j \\ &= 3(12) + 3 - 3(11) = 36 + 3 - 33 = 6 \end{aligned} \]

3 is the number of reactions, 2 and 8 come from using the truss formula \(m + r - 2j\) or slips in it. The rigid-frame formula is the one that applies here.

Answer: B

The beam is Determinate and unstable.

With rollers at both A and B, each support can only provide a reaction perpendicular to the beam, so there are just two unknown reactions, both vertical. The two vertical equilibrium conditions (forces and moments) find them, so in that sense the beam is determinate.

But the inclined load W has a horizontal component W cos θ along the beam, and no support can resist horizontal force. Horizontal equilibrium cannot be satisfied, so the beam would slide: it is unstable. A hinge at one end would make it a stable, determinate simply supported beam. It is not indeterminate, since it has fewer reactions than equations.

Answer: D

A propped cantilever is externally indeterminate to the first degree.

The fixed end supplies three reactions (horizontal, vertical and moment) and the prop adds one vertical reaction, four in all. A plane structure has only three equations of static equilibrium, so one reaction is redundant: 4 - 3 = 1.

A fixed beam (both ends fixed) would be third degree with six reactions, and second degree would correspond to five unknown reactions. Fourth degree has no basis for a single-span propped cantilever.

Answer: D

For a pin-jointed plane frame, static indeterminacy = (m + r) - 2j.

Each member of a pin-jointed truss carries one unknown, its axial force, and each support reaction is one unknown, giving m + r unknowns. At every joint the forces must satisfy two equilibrium equations (ΣF_x = 0 and ΣF_y = 0), giving 2j equations. The excess of unknowns over equations is (m + r) - 2j; zero means a statically determinate truss.

(m + r) - 3j is the form for space trusses, with three equations per joint. (3m + r) - 3j applies to rigid-jointed plane frames, where each member has three unknowns, and (6m + r) - 6j to rigid space frames.

Answer: D

The statically determinate structure is the three hinged arch.

A three hinged arch has four reaction components, two at each support. The three equations of equilibrium plus the condition that the bending moment is zero at the crown hinge give four equations, so all reactions can be found by statics alone.

A two hinged arch has the same four reactions but no crown hinge, so it is indeterminate to the first degree. A propped cantilever has one redundant reaction, and a continuous beam has one or more redundant supports, so all three need compatibility conditions as well.