Statics practice questions

Answer: A) 20 kN at 2 m

• ACorrect. Area of the load diagram at its centroid.
• BIncorrect. Multiply by the length.
• CIncorrect. It acts at the center.
• DIncorrect. Recheck the total.

Answer: B) \(\mu = \tan\theta\)

• AIncorrect. Use the tangent.
• BCorrect. At impending slip on its own slope.
• CIncorrect. Use the tangent.
• DIncorrect. Use the tangent.

Answer: D) 5 kN·m

A pure couple produces no force reactions, but moment equilibrium requires the fixed support to supply an equal and opposite moment of 5 kN·m.

• AIncorrect. Moment equilibrium (ΣM = 0) requires the support to resist the entire applied couple, not half of it.
• BIncorrect. Although there are no applied forces (so the force reactions are zero), the applied couple must still be balanced by a moment reaction.
• CIncorrect. A couple is a free vector; the reaction needs only to equal it, so doubling has no basis in equilibrium.
• DCorrect. ΣM = 0 gives a support moment equal in magnitude and opposite in sense to the 5 kN·m applied couple, regardless of beam length.

Answer: B) 0.0018 m⁴

• AIncorrect. That's \(bh^3/12\).
• BCorrect. \(bh^3/3\).
• CIncorrect. Divide by 3.
• DIncorrect. That isn't the calculation.

Answer: B) 9 kN·m

For a cantilever under a uniform load, the resultant wL acts at L/2. The fixed-end moment is therefore wL²/2 = 9 kN·m.

• AIncorrect. 6 kN is the total resultant load (wL), which is a force. To get the moment, multiply it by its lever arm of L/2.
• BCorrect. The resultant wL = 6 kN acts at L/2 = 1.5 m from the support, so M = wL²/2 = 2(3)²/2 = 9 kN·m.
• CIncorrect. This uses wL², which places the resultant at the free end. It actually acts at mid-span (L/2).
• DIncorrect. This uses wL²/4, the wrong lever arm. The resultant acts at L/2, not L/4.

Answer: A) 50 mm

The radius of gyration is k = √(I/A). Here I/A = 2,500 mm², so k = 50 mm.

• ACorrect. k = √(I/A) = √(4.5 × 10⁶ / 1,800) = √2,500 = 50 mm.
• BIncorrect. 2,500 is I/A, in mm². You still have to take the square root to get a length.
• CIncorrect. This is √I. It leaves out the division by the area.
• DIncorrect. This is √(A/I), the ratio flipped upside down. The radius of gyration is √(I/A).

Answer: A) 50 N·m

• ACorrect. The couple equals F × d.
• BIncorrect. Multiply, don't divide.
• CIncorrect. Multiply by the distance.
• DIncorrect. A couple is needed.

Answer: B) 707 N

• AIncorrect. The cables are inclined.
• BCorrect. Each vertical component carries 500 N.
• CIncorrect. The load is shared.
• DIncorrect. That doubles the correct value.

Answer: D) 20 kN·m

• AIncorrect. That isn't the calculation.
• BIncorrect. That isn't the case here.
• CIncorrect. That isn't the calculation.
• DCorrect. Each reaction is 10 kN, and M = 10 × 2.

Answer: A) 82.7 mm

Find the composite centroid with the area-weighted average ȳ = ΣAy/ΣA. The larger flange area pulls the centroid above the simple average of the two centroids, to about 82.7 mm.

• ACorrect. Flange area = 2400 mm², web area = 2000 mm². ȳ = ΣAy/ΣA = (2400×110 + 2000×50)/4400 = 364,000/4400 ≈ 82.7 mm.
• BIncorrect. This is the simple average of 110 and 50. It ignores the fact that the flange has more area than the web, so the centroid sits closer to the flange.
• CIncorrect. This comes from swapping the areas, using 2000 mm² with the flange and 2400 mm² with the web. That pulls the result toward the web.
• DIncorrect. This is half the 120 mm total height. That would only be true for a section symmetric about mid-height, which a T-section is not.