Question

The bar has a diameter of 40 mm. If it is subjected to a force of 800N as shown figure, determine the stress components that act at point A and B. 150 mm 200 mm 30 800N

Answer 1

solution: - steps: shift all loads at the given cross-section and write Normal, Shear and bending moment:- fy= Boosin30 = woon Fz= 80060S30 = 696N z N 200mm fox Now UZ Fy ; U = Fz Cinternal shear force) = 696N Mb y x NT :) (N=fy=400N] (internal normal force.) iiij Mo=FqX0:2 = 696 XD.2 M6= 139-2)-(internat bending moment) Now we can show them on cross-section: - id=40mm N Mb step:- Now calculate state of of stress at point A con neutral anis) j Normal stresses: Tay on to BA A is on N.A, hence AG TBA ТА Neatral anis. B TA= N A

7 400 TIX(0.04.13 > [ony = 0.318 MPa = ry Tx= =82=0 iij shear stress:- .: point a ês on neutral anis, hence transverse shear stress will be maximum. TA = 4 xt = x 696 Txo-su- ITA= 1. = 0.738 MPa TA=Tyz=0 0.738 and Exy - Txz=0 step3:- calculate state of stresses at point Bi- i Normal stress: - TB = TN + BB B is at extreme lower fiber, hence - 32 Mb JBIB will be compressive π ď as per the direction of my - Jo= 0-318X106_ 32x 139.2 TX0-043 => DB = 0.3 -318-22-1543X106 TB=-21.836 MPa Negative sign shows, it js compressive => DB = N A 3 1

foy= -21.836 I Tx=O2=0 Shear stress:- : B is at lower fiber (Extreme), hence the transverse shear stress e tero Tyz = Txy = Txzzo Any
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