STRUCTURAL GEOLOGY AND ROCK MECHANICS | QUIZ Quiz 1 // Mechanical Properties of Rocks Problem 1: Find the density of ethyl alcohol if 63.3 π occupies 80.0 ππΏ. Given: Required: π = 63.3 π π·πππ ππ‘π¦ (π) π = 80.0 ππ Note: 1 ππΏ = 1 ππ3 Solution: π= π π 63.3 π 80.0 ππ3 π π π = π. πππππ ≈ π. πππ π ππ πππ π= Problem 2: Specific gravity is the ratio between the density of a substance and density of water π (1 ππ3). Determine the volume of 200 π of carbon tetrachloride, for which specific gravity is 1.60. Given: Required: π = 200 π ππππ’ππ (π) ππππππππ πΊπππ£ππ‘π¦ = 1.60 π ππ π€ππ‘ππ = 1 π ππ3 Solution: ππππππππ πΊπππ£ππ‘π¦ = ππ π’ππ π‘ππππ ππ€ππ‘ππ π (π ) ππππππππ πΊπππ£ππ‘π¦ = ππ€ππ‘ππ 200 π ( π ) 1.60 = π 1 ππ3 π½πππππ = πππ πππ Problem 3: A solid cube made of a certain material has a mass of 1500 grams. The side length of the cube is 5 cm. What is the density of the material? Given: Required: π = 1500 π π·πππ ππ‘π¦ (π) π πππ = 5 ππ Note: ππππ’ππ = (π πππ)3 Solution: π= π= π π 1500 π (5 ππ)3 π = ππ π πππ π Problem 4: A fish tank is filled with 10,000 ππ3 of water. A metal object with a density of 7.5 ππ3 is placed into the tank, and the water level rises by 600 ππ3. What is the mass of the metal object? Given: Required: ππ‘πππ = 10000 ππ3 πππ π (π) π = 7.5 π ππ3 ππππ ππππππ = 600 ππ3 Note: An object immersed in a liquid displaces an amount of fluid equal to the object’s volume. Solution: π= 7.5 π π π π = 3 ππ 600 ππ3 π = ππππ π Problem 5: A geologist is studying two rock samples. Sample A has a total volume of 500 ππ3 and a pore space volume of 125 ππ3 . Sample B has a total volume of 400 ππ3, but the pore space is unknown. However, when both samples are combined, the average porosity of the combined samples is found to be 30%. What is the pore space volume of Sample B? Given: Sample A 500 ππ3 125 ππ3 Total Volume Pore Space Volume Combined Porosity (n) Sample B 400 ππ3 unknown 30% Solution: π= πππππ π ππππ (π΄ + π΅) ππ‘ππ‘ππ (π΄ + π΅) 0.30 = 125 ππ3 + π΅ 500 ππ3 + 400 ππ3 π© = πππ πππ Quiz 2 // Dynamic Properties of Rocks Problem 1: An iron rod 4.00 π long and 0.500 ππ2 in cross-section stretches 1.00 π when a mass of 225 ππ is hung from its lower end. Compute the Young’s modulus of the iron. Given: Cross-section Required: πΈ = πππ’ππ′ π ππππ’ππ’π πΏ = 4.00 π βπΏ = 1.00 π 4.00 π 0.500 ππ2 π΄ = 0.500 ππ2 = 5 × 10−5 π2 πΉ = 2207.25 π 1.00 π Solution: πΈ= 225 ππ πΈ= π πΉ = (225 ππ) (9.81 2 ) = 2207.25 π π πΉπΏ π΄βπΏ (2207.25 π)(4.00 π) (5 × 10−5 π2 )(1.00 π) π¬ = πππ, πππ, πππ π΅ π΅ ππ π. ππ × πππ π ππ π Problem 2: A load of 50 ππ is applied to the lower end of a steel rod 80 ππ long and 0.60 ππ in diameter. How much will the rod stretch? Young’s Modulus for steel is 190 πΊππ. Given: Cross-section 80 ππ 0.60 ππ Required: πΏ = 80 ππ = 0.80 π πΆβππππ ππ πππππ‘β βπΏ = πΈ = 190 πΊππ = 1.90 × 1011 π π2 π· = 0.60 ππ = 0.0060 π πΉ = 490.5 π βπΏ Solution: 50 ππ πΉ = (50 ππ) (9.81 π ) = 490.5 π π 2 π΄= ππ·2 4 π΄= π(0.0060 π)2 4 π΄ = 2.827 × 10−5 π2 πΈ= 1.90 × 1011 πΉπΏ π΄βπΏ (490.5 π)(0.80 π) π = (2.827 × 10−5 π2 )(βπΏ) π2 βπ³ = π. ππ × ππ−π π Problem 3: Two parallel and opposite forces, each 4000 π, are applied tangentially to the upper and lower faces of a cubical metal block 25 ππ on a side. Find the angle of shear and the displacement of the upper surface relative to the lower surface. The shear modulus of the metal is 80 πΊππ. βπ₯ Given: Required: πΏ = 25 ππ = 0.25 π πΆβππππ ππ πππππ‘β 4000 π βπ₯ = πΊ = 80 πΊππ = 8.0 × 1010 π π = π΄ππππ ππ π βπππ π2 πΉ = 4000 π 25 ππ π π π Solution: π πΊ= 8.0 × 1010 πΉπΏ π΄βπ₯ (4000 π)(0.25 π) π = (0.0625 π2 )(βπ₯) π2 βπ = π. π × ππ−π π 4000 π Apply Trigonometry to solve for π: tan π = πππππ ππ‘π βπ₯ 2.0 × 10−7 π = = ππππππππ‘ πΏ 0.25 π π = tan−1 ( 2.0 × 10−7 π ) 0.25 π π½ = π. πππ × ππ−π π«ππππππ Problem 4: Compute the volume change of a solid copper cube, 40 ππ on each edge, when subjected to a pressure of 20 πππ. The bulk modulus for copper is 125 πΊππ. Given: Required: πΏ = 40 ππ = 0.040 π βπ = πΆβππππ ππ πππππ‘β πΎ = 125 πΊππ = 1.25 × 1011 π = 20 πππ = 2.0 × 107 βπ π π2 π π2 π = πΏ3 = 6.4 × 10−5 π3 Solution: πΎ=− ( 40 ππ 1.25 × 1011 π βπ ) π π 2.0 × 107 2 π π = − βπ π2 ( ) 6.4 × 10−5 π3 βπ½ = −π. πππ × ππ−π ππ Problem 5: When a brass rod of diameter 6 ππ is subjected to a tension of 5 × 103 π, the diameter changes by 3.6 × 10−4 ππ. Calculate the longitudinal strain and Poisson’s ratio for brass given that Young’s Modulus for the brass is 9 × 1010 π/π². π΄π₯πππ ππ‘ππππ πΉππππ‘ ππππ€ 5 × 103 π 5 × 103 π πΏππ‘ππππ ππ‘ππππ 6 ππ π΅πππ: π³πππππππ ππππ = π¨ππππ 3.6 × 10−4 ππ Given: Required: π· = 6 ππ = 0.006 π ππ΄π₯πππ = π΄π₯πππ ππ‘ππππ π = 5 × 103 π π = ππππ π ππ′ π π ππ‘ππ βπ· = 3.6 × 10−4 ππ = 3.6 × 10−6 π πΈ = 9 × 1010 π/π² ππππ ππππ€ Solve for Lateral Strain (ππΏππ‘ππππ ): Solve for Axial Strain (ππ΄π₯πππ ): βπ· ππΏππ‘ππππ = π· πππ‘πππ π = ππΏππ‘ππππ = 3.6 × 10−6 π 0.006 π ππΏππ‘ππππ = 6.0 × 10−4 πΉ 5 × 103 π = π΄ 2.827 × 10−5 π2 π πππ‘πππ π = 1.77 × 108 2 π πΉπΏ βπΏ πΉ πΈ= → = = ππ΄π₯πππ π΄βπΏ πΏ π΄πΈ π 2 π ππ΄π₯πππ = ( ) π 10 9 × 10 π2 1.77 × 108 πΊπ¨ππππ = π. ππ × ππ−π π΄= ππ·2 π(0.006 π)2 = = 2.827 × 10−5 π2 4 4 Solution: π= πΏππ‘ππππ ππ‘ππππ ππΏππ‘ππππ = π΄π₯πππ ππ‘ππππ ππ΄π₯πππ π= 6.0 × 10−4 1.97 × 10−3 π = π. πππ
