20 ft -200 ft- Exercise Figure 8.1 8.1). The concrete has a unit weight of 2.4 Mg/m³ (150 lb/ft3). The height of the dam is 67.1 m (220 ft), the crest is 6.1 m (20 ft) wide, and the base is 61 m (200 ft) wide. Note: 1 ft = 0.3048 m. a. What is the average pressure exerted on the rock. foundation in MPa (lb/ft2)? What is the maximum pressure? Explain. b. Refer to the different sedimentary rocks listed in Table 8.1. Using average compressive strength and shear strength values for these sedimentary rocks compare them to the average pressure calculated in part (a). Are any so low to be of concern? Explain. c. Assume that the pressure from the weight of the dam is dissipated within 30.3 m (100 ft) below the rock surface and the full pressure acts over this distance. How much settlement will occur in the center of the dam because of the gravity force for the different sedimentary rocks listed in Table 8.1 (use average values)? 7. Excavation for a gravity dam foundation revealed a unit of poorly cemented sandstone 30.3 m (100 ft) wide (plan view) within a massive dolomite sequence. The dolomite has an E value of 5.5 x 104 MPa (8 x 106 psi) and the sandstone has an E value of 6.9 x 103 MPa (1 x 106 psi). a. What would be the concern if both rocks were loaded equally by the gravity dam? What would be the effect on the dam? b. How could this problem be alleviated? Consider both replacement and strengthening possibilities. 8. A 333.3-m (1100-ft) long, horseshoe-shaped tunnel is to be driven through massive granite with a unit weight of 2.72 Mg/m³ (170 lb/ft³). The tunnel is 9.1 m (30 ft) high and 9.1 m (30 ft) wide in cross section (Exercise Figure 8.2). The maximum rock cover above the tunnel in the middle of the mountain is 366.5 m (1200 ft). Note: 1 ft = 0.3048 m. Engineering Properties of Rocks H 15 ft -30 ft Exercise Figure 8.2 a. What is the total overburden stress at the midway point in the tunnel in MPa (psi and psf)? b. Because of arching effects that transfer load around the tunnel opening, the roof or crown of the tunnel does not commonly support the full overburden stress, but instead is typically much less. If the roof stress was found to be 0.12 MPa (2500 psf), how many meters (feet) of rock above the tunnel does this represent? How do you suppose this load would be supported in the tunnel? c. If the tunnel excavation was completed in 80 work- ing days with two shifts per day, how many lineal meters (feet) of tunnel were excavated per day? How many meters (feet) per shift? What was the average volume of rock removed per day, per shift? (Hint: Calculate the area of the cross section.) 179 d. The tunnel project was bid at $1,878,000. What would be the cost per lineal foot of the tunnel? Right abutment 9. An arch dam transfers much of the water load onto the abutments (the rock mass on the sides of the dam) (see Exercise Figure 8.3 for a plan view). The E value for the concrete is 4.14 x 104 MPa (6 x 106 psi). a. What would occur if the abutments were shale with an E of 2.07 x 104 (3 x 106 psi)? b. What if the abutments were diabase with an E of 9.66 x 104 MPa (14 x 106 psi)? Which of the two rocks would be preferred, the shale or the diabase? Explain why. Exercise Figure 8.3 Upstream Downstream Left abutment
20 ft -200 ft- Exercise Figure 8.1 8.1). The concrete has a unit weight of 2.4 Mg/m³ (150 lb/ft3). The height of the dam is 67.1 m (220 ft), the crest is 6.1 m (20 ft) wide, and the base is 61 m (200 ft) wide. Note: 1 ft = 0.3048 m. a. What is the average pressure exerted on the rock. foundation in MPa (lb/ft2)? What is the maximum pressure? Explain. b. Refer to the different sedimentary rocks listed in Table 8.1. Using average compressive strength and shear strength values for these sedimentary rocks compare them to the average pressure calculated in part (a). Are any so low to be of concern? Explain. c. Assume that the pressure from the weight of the dam is dissipated within 30.3 m (100 ft) below the rock surface and the full pressure acts over this distance. How much settlement will occur in the center of the dam because of the gravity force for the different sedimentary rocks listed in Table 8.1 (use average values)? 7. Excavation for a gravity dam foundation revealed a unit of poorly cemented sandstone 30.3 m (100 ft) wide (plan view) within a massive dolomite sequence. The dolomite has an E value of 5.5 x 104 MPa (8 x 106 psi) and the sandstone has an E value of 6.9 x 103 MPa (1 x 106 psi). a. What would be the concern if both rocks were loaded equally by the gravity dam? What would be the effect on the dam? b. How could this problem be alleviated? Consider both replacement and strengthening possibilities. 8. A 333.3-m (1100-ft) long, horseshoe-shaped tunnel is to be driven through massive granite with a unit weight of 2.72 Mg/m³ (170 lb/ft³). The tunnel is 9.1 m (30 ft) high and 9.1 m (30 ft) wide in cross section (Exercise Figure 8.2). The maximum rock cover above the tunnel in the middle of the mountain is 366.5 m (1200 ft). Note: 1 ft = 0.3048 m. Engineering Properties of Rocks H 15 ft -30 ft Exercise Figure 8.2 a. What is the total overburden stress at the midway point in the tunnel in MPa (psi and psf)? b. Because of arching effects that transfer load around the tunnel opening, the roof or crown of the tunnel does not commonly support the full overburden stress, but instead is typically much less. If the roof stress was found to be 0.12 MPa (2500 psf), how many meters (feet) of rock above the tunnel does this represent? How do you suppose this load would be supported in the tunnel? c. If the tunnel excavation was completed in 80 work- ing days with two shifts per day, how many lineal meters (feet) of tunnel were excavated per day? How many meters (feet) per shift? What was the average volume of rock removed per day, per shift? (Hint: Calculate the area of the cross section.) 179 d. The tunnel project was bid at $1,878,000. What would be the cost per lineal foot of the tunnel? Right abutment 9. An arch dam transfers much of the water load onto the abutments (the rock mass on the sides of the dam) (see Exercise Figure 8.3 for a plan view). The E value for the concrete is 4.14 x 104 MPa (6 x 106 psi). a. What would occur if the abutments were shale with an E of 2.07 x 104 (3 x 106 psi)? b. What if the abutments were diabase with an E of 9.66 x 104 MPa (14 x 106 psi)? Which of the two rocks would be preferred, the shale or the diabase? Explain why. Exercise Figure 8.3 Upstream Downstream Left abutment
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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