Question

Q1. [10 marks) Part of a piping system is subjected to static support loads as shown in Figure 1 below. The pipe is made of AISI 1030 (CD) steel and has an outer diameter of D and an inner diameter d. Determine the safety factor against static failure at location H using the distortional energy criterion (von-Mises criterion). Disregard any stress concentration effects. D=200 x (1 + P + P )/20) [mm] d=D/1.2 [mm] p1=5 p2=2 p3=1 p4=0 L = 500 x (1 + (P, +P/20) [mm] L = 1000 x (1 + (P, +P)/20) [mm] F = 20% (1 + (P, + P.)/20) [kN] F, -10% (1 + (P,+P)/20) [kN] p5=9 L- Figure 1.

Answer 1

20 20 given; D= 2008 (1+ Pit Ps) = 200x (1+14) = 340 mm. da Pe - 988.8 mm Ly=500x [ 1 + P2 + P3] - 500*(1 + 3) = 575 mm tx= 1000x [i+ Pit P4 ] = 1000 x [ 1 + 5 ] = 1250 mm F2 = 20x [it P₃ + P4] = dox [1+ 4.1 - KN Fy = 10xlit Pit P2 ] = Lox[i+ 7 = 13.5 KM AT point t o stresses due to fy = 13.5 ky; are :- rix Bending stress (06) - May = F x L x x D/2 32 fy x L x D A (04-04) - 104. d4) - 32X 13.5x103 x 1250 x 340 9,05 Mpa. xx (3404_288.34) 20 + stress is tensile in nature

= ty Direct shear stress = 1815 x 103 = 53989 stresses due to Fz are: - Torsional shear stress o I = I x = Fz X Ly xD x (04.04) a 16 fz XLYXD - 16 x 21x10²575 X 340 AX (04. d4 ) TX (3404_288.34) - 3.24 Mpq. direct shear a F² - 21x103 x (3402-285.32) = .823 Mpg No bending stress y 2o (on neutral axis). Net shear stress due to fz = 3.24 - 0823 Mpa = 2:417 MPa. Now, three stresses of = 8xca gios Mapa Izy =.53 MRa, Inz = 2.417 MP9 Area

Noval; ry=62- Distostian energy theory 82= 2 [ con-by)²+ loy-oz)² + 62-on) 2 +61 5wyt + Iyz? + Izu?) ] dy=0220 ; Iny = 0 : 5x [ 9.052 + 6x [1532+ 2.4172]] - 59.32 - or 7.7 M pq. for ALSI 1030 steel Syt = 440 Mapa. ts a Syt - 440 = 57014 + + 2.41 7.7