Statics practice questions

Answer: B) 2

• AIncorrect. That's a rigid body in 2-D.
• BCorrect. ΣFx = 0 and ΣFy = 0.
• CIncorrect. That's a rigid body in 3-D.
• DIncorrect. Both force components must sum to zero.

Answer: C) 10 kN

• AIncorrect. That halves one load.
• BIncorrect. It's shared.
• CCorrect. The total 20 kN load is shared equally.
• DIncorrect. Supports carry the load.

Answer: B) The negative of the load intensity

• AIncorrect. The slope of the moment diagram equals shear.
• BCorrect. Under no load, shear is constant.
• CIncorrect. That isn't the relationship.
• DIncorrect. That isn't the relationship.

Answer: C) 12 kN

Taking moments about the right support gives R_L = P·b/L = 20(6)/10 = 12 kN, and the right reaction is 8 kN.

• AIncorrect. Each support takes half the load only when the load is at midspan. Here the load is 4 m from the left support, not 5 m.
• BIncorrect. This is the right reaction: R_R = 20(4)/10 = 8 kN. The support farther from the load carries the smaller share.
• CCorrect. Sum moments about the right support: R_L(10) = 20(6), so R_L = 12 kN. The left support is closer to the load, so it carries more of it.
• DIncorrect. The two supports share the load. Vertical equilibrium requires R_L + R_R = 20 kN, so neither reaction alone can equal 20 kN.

Answer: A) m + r = 2j

Each joint of a planar truss gives two equilibrium equations. A stable planar truss is statically determinate when the unknowns (m + r) equal 2j; it is indeterminate when m + r > 2j.

• ACorrect. Each pin joint in a planar truss gives two equilibrium equations (ΣFx = 0, ΣFy = 0), so there are 2j equations. For determinacy these must equal the m member forces plus the r reactions.
• BIncorrect. This is the determinacy condition for a space (3-D) truss, where each joint gives three force-equilibrium equations.
• CIncorrect. This puts the reactions on the equation side. Support reactions are unknowns, just like member forces, so they belong with m: m + r = 2j.
• DIncorrect. This is the determinacy condition for a rigid-jointed plane frame. Each frame member carries three internal unknowns and each rigid joint gives three equations. Truss members carry only axial force.

Answer: C) 10 N·m

A couple's moment equals one force times the perpendicular distance between the two lines of action: 25 × 0.4 = 10 N·m. This is a free vector, so it has the same value about any point.

• AIncorrect. The two forces cancel, so the net force is zero, but their moments add up to a nonzero moment. A couple produces a pure moment.
• BIncorrect. This uses half the spacing (0.2 m), as if the moment were taken from a midpoint for only one force. The couple moment is one force times the full perpendicular distance between the lines of action.
• CCorrect. M = Fd = 25 N × 0.4 m = 10 N·m. The value is the same about every point in the plane.
• DIncorrect. This multiplies both forces by the full spacing (2Fd), which counts the moment twice.

Answer: A) 20 kN

• ACorrect. Each reaction is \(wL/2\).
• BIncorrect. The load is shared.
• CIncorrect. That isn't the calculation.
• DIncorrect. Multiply by length first.

Answer: A) 50.0 mm

The radius of gyration is r = √(I/A). Here I/A = 2,500 mm², so r = 50.0 mm.

• ACorrect. r = √(I/A) = √(12.5 × 10⁶ / 5,000) = √2,500 = 50.0 mm.
• BIncorrect. 2,500 is I/A, which has units of mm². The radius of gyration is the square root of that ratio.
• CIncorrect. This is √(2I/A) = √5,000. It doubles the moment of inertia, which the definition r = √(I/A) does not do.
• DIncorrect. This is √(I/2A) = √1,250. It halves the ratio, which the definition r = √(I/A) does not do.

Answer: A) 2.55 m

• ACorrect. \(4r/(3\pi)\).
• BIncorrect. That isn't the centroid.
• CIncorrect. That isn't the formula.
• DIncorrect. That isn't the calculation.

Answer: C) 50 N

• AIncorrect. That's the difference.
• BIncorrect. Forces add as vectors, not scalars.
• CCorrect. \(\sqrt{30^2 + 40^2} = 50\).
• DIncorrect. Use the vector sum.