Question

The state of stress on a certain elementos presented by Morse 90 M 60 50. 4.7 None 15. The maximum sharing stress and corresponding orientation angle (0.00 a. 80.9, 50.9° b. 90, 10.4 c. 30, 26.6 16. The principal stresses (...) are: a. 90, 60 b. 95.5, -65.8 c. 150, -30 d. 60, 90 17. The corresponding orientation angle (0,} for a. 21.8 cw b. 10.9° cw e. 10.9 cew d. 21.8 cew e. None 18. The normal stress condition (cx', oy') after rotating the element by 15 counterclockwise (CCW) are respectively in MPa): a. +34.95, 64.95 b. 90, 60, 15 c. -34.95, 64.95 d. -45, -15 e. 60,90 19. The shear stress condition {Txy) after rotating the element by 15° counterclockwise (CCW) are respectively (in MPa): a. 63.48 b. 15 c. -34.95 d.-15 e. 30 mm A composite beam is constructed by bonding brass and aluminum

Answer 1

90 MPa → 30 MPa 60 MPa 9o MPa ry = x = -60 MPa Tay = 30 MPa (16 Principal stresses Emax/min Rtg + sfgaft Pay? = -4** J-1)+(37 2 15 † 80.78 option B is correct Lomax = 95.78 MPa omin=-65.78 MPa shear stress Maximum omax - Emin (15 2 Timex 80.78 MPa

Orientation of maximum shearing stress tan 20 max of-ry ziny -Go-90 2.30 Olma = 34.1° correct None option is 0 = 34.1° as

KT Tan 28p = złny on-ry 121-13 Op = -21.8° CW Op = 21.8°CW option A is correct 0 = +15° Isº CCW (18) 0 = + Gal: Entry 2 Ex-by cos 20 + Try sinzo 2 łny = = 30Mla Tu = - 60 MPa бу? ?• MP4 Go' = 15 - 64.95 + 15 n's 34.95 MPa 5' = 57,3 - 07-02 Cos 28 on-by (os 20 - Try sin 20 64.95 MPal og'a option C is correct (-34.95, 64.95) 2

8 = + Is® 19 Tuy - ) sin 20 - Тъз оз10 — 34S + 25:58 łny - 63.48 MPа