Trigonometric Functions and Constitutive Equations - Lab Report Example

Information: Semester: Due GSC336, Lab Report: Driving and Resisting Forces Lab Objective: To gain a deeper understanding of Driving and Resisting Forces, by building a simple model and applying the knowledge acquired in class. Lab Specific Goals: a. To learn and apply geomorphology’s unifying concept, driving and resisting forces. b. To review the various trigonometric functions and constitutive equations c. To learn how to build equations, simple models and graphical plots in Microsoft Excel. Lab Requirements: a. Computer, b. Microsoft Excel, c. Hand-held or online calculator, d. Printer. Procure 1. The Lab 1 Microsoft Excel worksheet was downloaded from the “Module 1” in Canvas. 2. The was opened in worksheet in Microsoft Excel. 3. The matrix elements B2, B3, B4, were filled with the corresponding density (ρ), gravitational acceleration (g) and thickness (z). The following constants were using in during the Lab: ρ = 1,000 kg m-3 g = 9.8 m s-2 z = 1 m 4. The angles values were then entered in column E2:E18, 0-1.57, in increments of tenths, e.g., 0, 0.1, 0.2,… In the last column, (E18), the value 1.57 was entered instead of 1.6. These angle values were expressed in radian units. 5. In columns F2:F18, the angles in radians were converted to degrees. A formula was developed in each cell of the Microsoft Excel to do the conversion. 6. Shear Stress was then calculated. In Column G, a constitutive equation that calculates the corresponding shear stress for each angle (degrees) listed in the worksheet was built. Learn 7. Normal Stress was also calculated. In Column H, a constitutive equation that calculates the corresponding normal stress for each angle (degrees) listed in the worksheet was developed. . 8. A scatterplot in was created in Excel. The y, or response axis, is stress (kPa). The x, or Slope angle (deg) and (2) Normal Stress vs. Slope angle (deg). The axes were properly labeled. 9. The matrix and scatter plot were saved as a pdf, and then printed out. Results and Discussion The Data Matrix obtained is as shown below: Slope angle (radians) Slope angle (Degrees) Shear stress (acts to impel material downslope) Normal stress (acts to hold material in place) F=M sinθ F=M cos θ 0 0 0 5092.958179 0 1000 0.1 5.729577951 127.111854 5067.514602 99.83342 995.00417 0.2 11.4591559 1011.814593 4991.438094 198.6693 980.06658 0.3 17.18873385 1505.072054 4865.488786 295.5202 955.33649 0.4 22.91831181 1983.291331 4690.925123 389.4183 921.06099 0.5 28.64788976 2441.694218 4469.491286 479.4255 877.58256 0.6 34.37746771 2875.700503 4203.39977 564.6425 825.33561 0.7 40.10704566 3280.973739 3895.309273 644.2177 764.84219 0.8 45.83662361 3653.46457 3548.298134 717.3561 696.70671 0.9 51.56620156 3989.451191 3165.833572 783.3269 621.60997 1 57.29577951 4285.576534 2751.737048 841.471 540.30231 1.1 63.02535746 4538.881814 2310.146077 891.2074 453.59612 1.2 68.75493542 4746.836086 1845.472889 932.0391 362.35775 1.3 74.48451337 4907.361541 1362.360347 963.5582 267.49883 1.4 80.21409132 5018.854262 865.6355506 985.4497 169.96714 1.5 85.94366927 5080.20025 360.2616098 997.495 70.737202 1.57 89.95437384 5092.956564 4.055658635 999.9997 0.7963267 Shear Stress, T=F/A Normal Stress, N=F/A F=M sinθ F=M cos θ whereby M=1000 whereby M=1000 Area (A)=PI() *(B4/2)^2 Area (A)=PI() *(B4/2)^2 T=M sinθ/PI() *(B4/2)^2 T=M sinθ/PI() *(B4/2)^2 Scatter Plot Helpful definitions: Matrix - a rectangular array of numbers, symbols, or expressions, arranged in rows and columns. The individual items in a matrix are called its elements or entries. Radian - standard unit of angular measure, used in many areas of mathematics. Constitutive equation - a relation between two physical quantities (especially kinetic quantities as related to kinematic quantities) that is specific to a material or substance, and approximates the response of that material to external stimuli, usually as applied fields or forces. Questions 1. Is shearing force a Driving or Resisting Force? Is normal force a Driving or Resisting force? Which is which? In your own words, justify your answers. (5 pts) 2. In your own words, explain Normal and Resisting Forces using a Boulder on a hillslope as an example. Hint: see you textbook, Chapter 1, for help. (5 pts) 3. Based on your scatter plot and knowledge acquired from your textbook, what is the critical angle, in degrees, that determines if the boulder would move, or not? Why did you choose this angle? (5 pts) 4. A local business informs you they have a 1 m-thick rock slab with a density of 1000 kg m-3 lying in their backyard that slopes at 25 degrees. Given normal weather conditions and unsaturated soils, will this rock slide and potentially destroy their business? Justify your answer in terms of angle, Driving and Resisting forces. (5 pts) Answer to the Questions 1. Is shearing force a Driving or Resisting Force? Is normal force a Driving or Resisting force? Which is which? In your own words, justify your answers. (5 pts) Shear force is a driving force because its effect is felt parallel to the slope. Normal force is a resisting force because it effects acts perpendicular to the slope (Lemke). 2. In your own words, explain Normal and Resisting Forces using a Boulder on a hillslope as an example. Hint: see you textbook, Chapter 1, for help. (5 pts) Normal forces or resting forces on a boulder cliff tends to restrict the movement of an object. The angle of the slope is a contributing factor to the movement of a boulder along a hill slope. Additionally, the stability of the slope also affects the movement of an object. In this case, factors such as friction and cohesion determine how fast an object moves (Lemke). 2. Based on your scatter plot and knowledge acquired from your textbook, what is the critical angle, in degrees, that determines if the boulder would move, or not? Why did you choose this angle? (5 pts) The critical angle for determining whether the boulder would move, or not is 45.840 degrees. At this point, the boulder would assume a stationery motion. The value is arrived by observing the point of intersection between the curves of shear and normal stress. 4. A local business informs you they have a 1 m-thick rock slab with a density of 1000 kg m-3 lying in their backyard that slopes at 25 degrees. Given normal weather conditions and unsaturated soils, will this rock slide and potentially destroy their business? Justify your answer in terms of angle, Driving and Resisting forces. (5 pts) At 25 degrees, the sheer force on the rock will be 984.807753 while the normal stress will be 173.6481777 (in the opposite direction). This means that the forces propelling the rock slab downwards are greater than the resisting forces. Assuming that the weather conditions remain at normal levels it is highly likely that the rock in question will inflict a considerable damage to the surrounding property. Conclusion The experiment is of great essence. This is because it gives the relationship between the weights of an object, the forces acting on it, the angle of inclination and the acceleration due to gravity. This concept can be, for instance, used by geomorphologists to establish the effects and extent of flooding on flood plains. Also the concept is applicable in monitoring the changes in river position and patterns, among other geomorphological applications. Work Cited Lemke, Karen A. “Slope Stability & Mass Wasting.” University of Wisconsin-Stevens Point, December 1, 2013. Web. January 26, 2015 < http://www4.uwsp.edu/geo/faculty/ lemke/geomorphology/lectures/10_mass_wasting.html> Read More