For the block loaded triaxially as described in Prob. 2, find the uniformly distributed load that must be added in the x direction to produce no deformation in the y direction. 2 - A rectangular steel block issubjected to a triaxial loading of threeuniformly distributed forces. If v = 0.30and E = 200 GPa, determine the singleuniformly distributed load in the xdirection that would produce the samedeformation in the z direction as theoriginal loading. Fx = 10kN, Fy = 15 kN,F.F2 = 20 kN, L = 600 mm, W = 400 mm, &2- axisD = 800 mm.X-axis * Solution : • Dimensions of the block:D = 800 mmL= 600 mm ;• Force acting in x, y and z direction:Fx = 10KN ; Fy = 15 kN , Fz = 20 kN• Poisson's Ratio :• Modulus of Elasticity : E = 200 GPa = 200 X 103 N/mm2W= 400 mmV= 0. 30Stress in the x, y and z-direction :10 x 10 N(800mm) x (400 mm)Fx0.03125 Mpa (tension )+ 6x%3D%3DDXW15x 10 N(400 mm) x (600 mm)• 5y =Fy0.06250 MPa (Tension )%3D%3DwXL- 20 X 10 N(800mm) x (600 mm)FzMPa =-0.04167 MPa24DXL( Compressin )• Stzain in the z- direction:Ez%3D[( + %) xA - 3] ->(200 x 1o N/mm2)Ezx240. 30 x (0.031 25 + 0.06 250)Ez =-3.489583333 x 10 7force is tequired in the x-directionEz is negati ve, thus tensileto produce the same deformation in the ydirection as the originalforces.• For equivalent single forre in the x-direction :K = - EzEx.: V x €x = - Ez= - Ez6x =- Ez XE-(-3.489583333 x 167)× C200x103)0.306x =0. 2326388889 MPu

Answer: The single uniformly distributed load in the x-direction is 70 kN (Compressive).

Answered: For the block loaded triaxially as…

For the block loaded triaxially as described in Prob. 2, find the uniformly distributed load that must be added in the x direction to produce no deformation in the y direction.

Mechanics of Materials (MindTap Course List)

9th Edition

ISBN:9781337093347

Author:Barry J. Goodno, James M. Gere

Publisher:Barry J. Goodno, James M. Gere

Chapter3: Torsion

Section: Chapter Questions

Problem 3.4.8P: Two sections of steel drill pipe, joined by bolted flange plates at Ä are being tested to assess the...

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