Question

100 KN 150 mm 450 mm ΑΙ B G С The three cylinder rods made of A-36 steel and have equal cross-sectional areas support a vertical load of 100kN as shown. The bar ABC is fully rigid. Determine: a) The force at A, B and C. b) The diameter of the rods consider safety factor of 1.5, rounded to the nearest even number. c) The displacement of point A, B, C. A 36 steel, Modulus of elasticity E = 200 GPa, Modulus of rigidity G = 75 GPa, Yield Strength Oy = 250 MPa, Ultimate Strength ou = 400 MPa 300 mm D E F 300 mm 300 mm

Answer 1

In this question we need to write one equation from deflection contrain of the rod as the bar ABC is rigid.

we will get total 3 equation to solve for forces coming on A, B and C.

see below complete solution.

solution: Griven, The Bar ABC is fully rigid since, the Bar ABC is rigid and Load is eccontric towards right. hence after deflection Let the system below. initial position becomes as 5 S S3 lon A B IL С. H part (a) 1 2 3 300mm 300mm from simillar triangle Ach and B CI. င် - ၆၊ S3-S2 600 300 V2 S3 = 252-5, 1-(i)

axial Now, the deflection in rod due to load is given as Phi = si= Deflection di AE pi L = Length AT Area of cross section E, Modulus o Elasticity, AT E, 1 and c at A B Let the forces coming and Pc Be PB PA Hince P P, P2 and P3, let be the forces PB=P2 coming on three rods. P= P3 Sza si= P, Li A,E, Sz= P₂₂ Ą & P₂ 23 Az Ez since, Lint=23 and A, = A₂ = A3 E = €₂=€3 => Si= PIL AE S₂ = P₂L AF 83= P₂L - 2 AE Pud these in i

We have. 2x P₂ P₂ - P₂ P3 41 100 KN Now Efya f co usomm nc Isomm B 100 KN 11 k IPit P₂ + P₂= 1 P₂ an taking EMATO 300mm 300 mm Pzx300 - 100x 450 + Pz X 600 =0 0 3P2 450 t6Pz = - (iv) => Pzt 2P3 - 150 using (ii) and (iv) Pz = 2xP2 - [100 - B - ] =) P2 = 2h - 100 + P2 +13 33.333 KN - P₂ = 100 3 P2 s 33.333 KN Puting in livl .P2 => P3 P₂ = 150 ISO-33.333 2

P₂ = 58.333 KN from Pit P₂ + P₂ = 100kn. KN a) Pi = 100 – 580333 – 33.333 KN. P,= 8.3333 hence force at A, B and c are. PA=P,= P,= 80333 KN P B = B = 33.333 KN 2 po P3 35 580 333 KN (b) Safety factor the Maximum force is coming onn rod c or 3 hence any design will be based on rod c. now Foss los given yielding as as failure criteria, taking by Рc. Ac 1.5

=> 3 250X10 58.333 da diameter ) 1- 5 17 15 d d=0.02110 m d= = 2lolo mm (neares res even number d= 22 mm (c) Risplacement of point A,B Ç PAL 8.333X10°x 300 SA-8,- = mm 3 AE X 200 X10 7 48 x 22² = > ISA=0.03288 mm 3 33.333 X 10 X 300 mm SBE PBL AE 7 x 222 x 200 x 103 SB = 0.13 15 mm 3 mm Sc = PCL 58.333x10ʻx 300 x 222 x 200 x103 AE Sc = 0.2301 mm