Question

Figure 2: The bar has a diameter of 40 mm. Determine the state of stress at point B and show the results on a differential volume element located at this point Determine the shear stress (T.xy) B? 16.2 MPa O 0.509 MPa -0.679 (MPa

Answer 1

200 mm 1: 200mm IB : 1200 SOON N Answer: 0 .. FBD for section at B - - - , 1200 N 1200 (Vy) >2.800 sino 4 SOONa :-800 COS O. (0 frees 53.13° ) Internal Efre = o. if2 -1200 50. Na = 1200 N Efyzo, fy=0; Ef2 = 0; Vz+800 EO . Vza-800N Emge 30;, 3 mg = mx + 800 ( 2 )x si CH3 =-boitem 712.45 Nam m2 = 800 CO S65 3:13) . ... "zoo am 25 .X60 2:

Here cross section is circular, (40 mm zd) te td ? = A xooi 2 = 1.2566X10-3m2 t = x (0.02) + Ty = I2 = Etre 140 J = Ir4 1,25664107 = 2 x 6.02) .2.5.132 X 107 my +0.02m To avj @B), = {'a'- cum X 0:02) = 4X0:02 x 10.02)? :::31 5. 33x10-6m 3; CQBy) = 0.1 . . 1 Here

Normal stress (of . : B.tm O=OB - . Niger A .may . IX. Jy = 1200 140.04) -69.60)*(-0.02) - [email protected]) ii.... - (245 .+.- 12.45)X (0) *** * SỞ N Ô +)”. 9547.76.86 N/m2! = 0.95mpą stress shear Zv=VQ . It (zaizlig -- (20), 2(7F. : :.= -800X COS(53-13) *[5133 x1000) ..: (40 X10-9m4x) *(0.004) F:-15.99003.805 :- 1.59mpa and (Zry) a =(2v) = v) ay n ompa] It B co Answer = impal .. . . . :