Question

6. For the cast iron link shown, the largest allowable stresses are 30 MPa in tension and 120 MPa in compression. Determine (a) the largest force P which can be applied, (b) the location of the neutral axis relative to point B. 90 mm A 20 mm P ID P' 40 mm 10 mm B B 10 mm 30 mm Section a-a

Answer 1

☺ Given that - allowable stress in tension St = 3ompa allowable stress in complession so=220 mpa K 90 mm P- 2 ol ki 5 = uo ja Agit AzY2) (AXA2) mm 38 mm >) A = (90) 20) = 1600 Y = (hoto) Az = (yo) (30) mm? 30mm 1200 mm² - = 50 Y2 = 40= 20mm y = 1800)(50) + (1200) (20) =38 mm 1800 + 1200 y 38mm nf forom forom B - Now transfer the load to the Centeroid with a load po & moment (P) (38-10) 0.028 P) Cams BL I M.

Comp. Co tensile OA 0.3766 ( T + YN R B Comp OB= Area moment of incabia of Section about Centroid Anis of the plane is I can It Iz (90) (20)3 + (90) Go) (50-387² 3+ (W) (30)(40)2+0) cu0) (3.6.20 Ich= 868X10 3 mmy - Anial Compressive stress -P-> KN Jaz Pa 103 (6.333 P) (mpa! ce A 3000 - Bending Storess at top my A I com (0.0 28XD) CIOC) (22) 868 X103 0.7096 P tensile stores N - Bending stsess at bottom. (06) in (0.028) (P) (1067 (38) «6& x co3 1.22540 Ls Compressive stress

- Resultant Stress at a JA = (06)A-oa = (0.7096) CAJ - 0.333(0) 0.3766(P) tensile stou 13 - Resultant storess at B OB= (Obbt oa com (1-225) CPJ + 0.333(P) 1.558 CP) -> Compress Casei when Coren) au = 30 mPa = GA >> la 3766) CP) = 30 P= 79.66 kN Check (Comp) SB= (1.558) (79.66) Cerca 124.11 mpa OB> Comp) anow PL 79166 Case o when o comp 120= JB = (10558) alow Pa 77.02 KN tensile) A = 6.3766) (99.02) = 29 mea TAL bensin) ahow

. allowable load on the beam is. p=77102 KN GA from the slope cavation 60-yor الاسم YN/A 60-YNIA YNIA 1.558 0.3766 YN/A 60-YNIA OB 248.22 - U. 137 (INIA) = Y(NIA) ܝܢ YNIA = 48.32 mm location of NA forom B