Question

J OILIJ 2020. ap 2 (part 1) Spr 2020.pdf Chap 2 (part 1) Spr 2020.pdf (63.9 KB) A+ Alternative formats 0 - The rod with a cross sectional area (Ao) at 25 mm² is under an axial load of P = 10 Newton. Find out the values of normal and shear stress at cross section at 0 = 0º, 15°, 30°, 45°, 60°, and 80°. Label stresses in colors and in proportion for each case. HP ENVY x360

Answer 1

Answera- page 0 Given Area (cross sectional) = 25 mm load (P) = 10N - 10 мро - 23 mpa = 0.umpa we know from complex stresses to and to = (9740g) + (03-) cos20 + T sin 20 To = ( y) sin20 - T 00820 Here Ty = 0; Z=0; x = 0.4 O To 60-440 ) +(04-0) coso + O sinzco) 2 ou to =o.umpa To = (0.4-0) sin2(o) - 0 cosa (O) = ompa

page ② à 0=15°; - 0.4+0) S2(15) tosin 265) = = 0-240:2 () +0 0.2(+B) € 0. 3732 MPa To = 0.4-0 ) sin 2015) - o cas2015) = 0.2 ( ) - 0 = 0.1 MPa no=30°; Do = (0-470) + (014-0) Cos2(30) + osin2(30) = 0.2 +0.2 COS2(30) to = 0.2+0 2 ( ² ) = 0.2 (1+²) = 0.3 MPa to = (0.4-0) sinz (30) - Ocos 2(30) = 0.2 sin2 (30) - 0 = 0.1732 Mpa

page ③ a o=45°; To 0:4 +0) 4 (0-40 ) Cos 2 (45) + asio2cus) - 0.2 +0.2 COS2(45) to = 0.2 to -0.2 Mpa to = (0.4-0 ) Sinz(45) - OCOS2 (45) = (0.2) sin 2 (45) - o = 0.2(1) = 0.2 MPa 70=60°; ° 0 – 0.440 )+ (04-0) Cos2(60 )+ o sinz (60) -0.2 +0.2 C052 (60) to = 0.2+0.2 (-1) = 0.2(1-2) = 0.1 MPa To = (0.4-0) sin 2 (60) - 0 cos2 (60) = = 0.2 sin 2 (60) - O 0.2 (3) = 0.1732 MPa

page 9 * 0=80°) To = (0.4 to ) +(0.4-0 ) Cos2 (80) + osin 2(80) = 0.2 + 0.2 Cos2 (80) to = 0·01206 mpa To = 0.4-0) sîn2(80) – o cos2 (80) = 0.2 sin 2(80) -0 = 0.0684 Mpa. To (Green) to (Red) 0.4 0. 3732 0.3 0.1732 0.2 0.2 0.1 O. 1732 0.01206 0.0684