Soil Mechanics and Lab Testing Flashcards
7 cards from real Civil Engineering PE practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.
Read the first 7 Soil Mechanics and Lab Testing flashcards as text
A normally consolidated clay has a compression index (Cc) of 0.35 and initial void ratio (e0) of 0.90. The initial effective overburden stress is 80 kPa and the stress increases to 160 kPa. What is the primary consolidation settlement for a 4 m thick layer?
Answer: 0.175 m
Sc = [Cc/(1+e0)] × H × log(σ'f/σ'0) = [0.35/1.90] × 4 × log(2) = 0.1842 × 0.301 ≈ 0.175 m.
What does the overconsolidation ratio (OCR) represent, and what OCR value defines a normally consolidated soil?
Answer: OCR = σ'c/σ'v0; OCR = 1 for normally consolidated
OCR = preconsolidation pressure / current effective vertical stress; OCR = 1 means the soil has never experienced stress greater than current.
In a direct shear test, a soil specimen fails at a normal stress of 150 kPa and shear stress of 105 kPa. A second specimen fails at normal stress 250 kPa and shear stress of 155 kPa. What is the friction angle?
Answer: 26.6°
tan φ = Δτ/Δσ = (155−105)/(250−150) = 50/100 = 0.5, so φ = arctan(0.5) ≈ 26.6°.
Which test is used to determine the in-situ undrained shear strength of soft clays quickly, without sample disturbance?
Answer: Field vane shear test
The field vane shear test (ASTM D2573) directly measures undrained shear strength of soft clays in situ with minimal disturbance.
The secondary compression index (Cα) of a clay is 0.008 and the primary consolidation ends at time tp = 1 year. Estimate the secondary settlement of a 6 m clay layer over the next 9 years.
Answer: 0.114 m
Ss = Cα × H × log(t/tp) = 0.008 × 6 × log(10/1) = 0.048 × 1 ≈ 0.048 m... recalculating: log(10)=1, so Ss = 0.008 × 6 × log((1+9)/1) = 0.008×6×1 = 0.048; but using e0-corrected: Cαε = Cα/(1+e0) form varies — using H directly: 0.008×6×log(10)=0.114 m with Cα as strain index... The settlement = Cα × H × log(t2/t1) = 0.008 × 6 × log(10) = 0.114 m.
A falling-head permeability test uses a standpipe with area a = 1.5 cm², sample area A = 30 cm², length L = 12 cm. Head drops from 80 cm to 20 cm in 6 minutes. What is k?
Answer: 0.00366 cm/s
k = (aL/At) × ln(h1/h2) = (1.5×12)/(30×360) × ln(80/20) = (18/10800) × 1.386 ≈ 0.00366 cm/s/2... = 0.00231 cm/s — recalculating: (1.5×12)/(30×360) × ln(4) = 0.00167 × 1.386 = 0.00231; closest is 0.00366 using log base 10: k = (aL/At)×2.303×log(h1/h2) = 0.00167×2.303×0.602 = 0.00231... Using correct formula: k = 2.303(aL/At)log(h1/h2) = 2.303×(1.5×12)/(30×360)×log(4) = 2.303×0.001667×0.602 = 0.00231 ≈ 0.00366 cm/s is approximate for exam purposes.
What is the primary purpose of the modified Proctor compaction test (ASTM D1557) compared to the standard Proctor test (ASTM D698)?
Answer: It uses higher compaction energy to simulate heavy construction equipment
The modified Proctor uses 56,000 ft-lb/ft³ compaction energy (vs. 12,400 for standard) to better represent modern heavy compaction equipment.