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2 Figure 1 shows the stress state of an element. Determine: a) The average and maximum shearing stress magnitude. b) Draw and

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Given Data: Ex = by = -80 MPa 60 mpa Zay = 40 mpa Principal stress 10/12 = (**74) + (50 -32) +zep 4= (-30460) + (-50250) + 40Average stoess, (cave): 3+52 70.622-90-622 have = have = 10 mpa - ( zman) Manimum shear stress, zmeine = 67-72 70-622-(-90.6254 - १.5११ १ and normal stress along perpendicular +7 - 6-51 tor (28) 2 -80+ 60 न - 2 -80-60 - 05(283-13 2 - -29.5११ ११ 4

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