Mechanics of Materials practice questions

Answer: B) Higher than the nominal stress

• AIncorrect. Stress concentrates at the hole.
• BCorrect. A stress concentration.
• CIncorrect. It peaks there.
• DIncorrect. Stress concentrates at the hole.

Answer: A) 0.30

Poisson's ratio is the negative of the lateral strain divided by the axial strain: ν = −(−0.0003)/0.0010 = 0.30.

• ACorrect. ν = −ε_lateral/ε_axial = −(−0.0003)/0.0010 = 0.30.
• BIncorrect. This inverts the ratio (axial over lateral). Poisson's ratio is lateral strain over axial strain, and for isotropic materials it cannot exceed 0.5.
• CIncorrect. This doubles the lateral strain to account for two transverse directions. Poisson's ratio uses the strain in a single lateral direction, so ν = 0.0003/0.0010 = 0.30.
• DIncorrect. This is roughly the volumetric strain, ε_axial + 2ε_lateral = 0.0010 − 0.0006. It is not Poisson's ratio.

Answer: A) 0.041 rad

• ACorrect. \(\theta = TL/(JG)\).
• BIncorrect. That's off by a factor of 10.
• CIncorrect. That's half the value.
• DIncorrect. That's double the value.

Answer: B) 2.25 MPa

• AIncorrect. That's the average; use 1.5V/A.
• BCorrect. \(\tau_{max} = 1.5V/A = 1.5(30{,}000)/20{,}000\).
• CIncorrect. Recheck the factor.
• DIncorrect. Recheck the calculation.

Answer: A) 12,000 mm²

In a transformed-section analysis, steel is converted to equivalent concrete by multiplying its area by the modular ratio n = Es/Ec. Here n = 8, so the transformed area is 8 × 1,500 = 12,000 mm².

• ACorrect. n = Es/Ec = 200/25 = 8, and the transformed area is n·As = 8 × 1,500 = 12,000 mm².
• BIncorrect. (n−1)·As adjusts for concrete displaced by steel that lies in uncracked concrete, such as compression steel. In a cracked section the concrete around the tension steel is ignored, so the factor is n.
• CIncorrect. This divides As by n. Steel is stiffer than concrete, so its equivalent concrete area must be larger than As, not smaller.
• DIncorrect. This leaves the steel area untransformed. That ignores the difference in elastic moduli, which is the reason for using a transformed section.

Answer: C) \(q = T/(2A_m)\)

• AIncorrect. That's for beam shear.
• BIncorrect. That gives shear stress in solid shafts.
• CCorrect. It's constant around the tube.
• DIncorrect. That's axial stress.

Answer: D) 2.0

• AIncorrect. That's for fixed-pinned.
• BIncorrect. That's for pins at both ends.
• CIncorrect. That's for both ends fixed.
• DCorrect. The column buckles like half of a pin-ended column.

Answer: B) 200 MPa

• AIncorrect. Recheck the units.
• BCorrect. \(\tau = V/A = 20{,}000/100\) N/mm².
• CIncorrect. Recheck the units.
• DIncorrect. That would be double shear.

Answer: C) 40.7 MPa

• AIncorrect. Recheck the formula.
• BIncorrect. Use 16T/(πd³).
• CCorrect. \(\tau = 16T/(\pi d^3) = 16(10^6)/(\pi \cdot 125{,}000)\).
• DIncorrect. Recheck the formula.

Answer: A) 1.5 mm

For a prismatic bar under axial load, δ = PL/(AE). With consistent units (N, mm, MPa), δ = 100,000 × 3,000 / (1,000 × 200,000) = 1.5 mm.

• ACorrect. δ = PL/(AE) = (100,000 N × 3,000 mm)/(1,000 mm² × 200,000 N/mm²) = 1.5 mm.
• BIncorrect. This is off by a factor of 10, typically from a unit slip such as taking E as 2,000,000 MPa or L as 300 mm.
• CIncorrect. This is too large by a factor of 10, typically from using E = 20,000 MPa (misreading GPa) or A = 100 mm².
• DIncorrect. This uses half the load (P/2), as in the self-weight elongation formula. A concentrated end load is carried along the full length, so the full P is used.