Problem 1A.3Four direct shear tests were performed in a 10 cm x 10cmsquare shear box for soil specimens with a similar densiityand the following data were obtained. Determine the angle oFinternal friction & and thesoil.cohesion component,c, OF theFvFn200272400324487632L0001500

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Problem 11.3 Four direct shear tests were per formed in a 10 cm x 10cm square shear box for soil specimens with a similar density and the following data were obtained. Determine the oangle oF internal friction s and the cohe sion component, c, OF the soil. Fv Fn 200 212 400 324 487 632 1000 1500

Problem 11.3 Four direct shear tests were per formed in a 10 cm x 10cm square shear box for soil specimens with a similar density and the following data were obtained. Determine the oangle oF internal friction s and the cohe sion component, c, OF the soil. Fv Fn 200 212 400 324 487 632 1000 1500

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Concept explainers

Fasteners and Design of Non Permanent Joints

Fasteners are described as mechanical components used to join or assemble two or more pieces of materials. The fasteners are used in various machines or structures based on their suitability to produce non-permanent, semi-permanent, and permanent joints.

Question

Problem 1A.3
Four direct shear tests were performed in a 10 cm x 10cm
square shear box for soil specimens with a similar densiity
and the following data were obtained. Determine the angle oF
internal friction & and the
soil.
cohesion component,c, OF the
Fv
Fn
200
272
400
324
487
632
L000
1500

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4) Calculations for the red dog food can rolling down the slope in the Rube Goldberg design are as follows (we will name it Step 1): Step 1 (calculations are given): Coefficient of friction → μ = 0.14 Mass of the object → m = 368 gm = 0.368kg Initial height of the object (red can on top of books) → h=8.89, cm = 0.0889 m Slope of the file folder → 0 = 14° Travelling Distance by the object = 11.5 inch = 0.292 m And length that the object will travel = h/sin 0 = 0.0889/ sin14° = 0.367 m So, the radius of the object → R=0.367 -0.292 = 0.075 m Initial Velocity of red can → u = 0 Velocity and Force Calculations for Step 1: -From total mechanical energy conservation: → Initial mechanical energy = final mechanical energy → mg - In case of pure rolling, the velocity of the center of mass: → V = Ro = 0.075 × 14.28 1.07 m/s. -Hence the change in force acting on the object for the travel: →F=mgsine = 0.368 × 9.81 × sin14° =0.89 N Step 2: The Selective Step (Step 2) in this design and for the…