Question

1.) Where is the reference point located when taking the moment? How did they get the sum of the moments equal to the solution given? ------I'm confused on how they got Fcos30(.025m/2) why did they divide the 0.025 by 2? where is the reference point?

2.) Why did they make the combined stress equal to the yield stress 276MPa?

3.) why is the maximum deflection negative for -PL^3/3EI

MECHANICS OF MATERIALS Columns 3 m А 25 mm 30° What force, F, will cause plastic deformation? (A) 1900 N (B) 3770 N (C) 3900 N (D) 7530 N 75 mm diameter aluminum: E = 69 GPa Syield = 276 MPa Sultimate = 434.7 MPa PPIPPI2PASS.COM © 2019 Kaplan, Inc. 94 A KAPLAN COMPANY MECHANICS OF MATERIALS Columns Solution Calculate the force that will cause plastic deformation. For an eccentrically loaded beam-column, The moment is max + M = Mwall = Fcos 30° (0.025 m) + F sin 30°(3 m) Faxial MC А 0.866 F TT(0.0375 m) 276 x 106 Pa= (1.511 m (F)(0.0375 m) (H) (0.0375 m)* = (1.511 m) F=7525 N (7530 N) Answer is D. The axial load is Faxin = F cos 30° = 0.866F To prevent plastic deformation, Omax < 276 MPa PP PPI2PASS.COM © 2019 Kaplan, Inc. 95 ANAMAN