Question

For the circle given below, find the stres values on points A and B. On a diagram, show the distrubition of stress on the section.

20 kN ф60 A B 80

Answer 1

If we transfer the All of IB load 20KN to the 20kn Action AB, Tensile load of 20kn and a moment of (20x80) (20kn x 80MM) RN-MM acts at section Ats tet yo yi Ri d Rn R Ro Given that d=60MM, R=80MM Load (P) = 20KN 200183 N = Bending moment (me) = (20x80) = 1600 KN-AM = 1600 x 103 at MM R= Rit it 80 - RiT GO 2 Ri = 50MM Ros Ritd 50+60 - Ro toMM

From desing data, In curred beams for circular section Neutral wris aurrature formula is there Rn= r² 2 (R-5R²-92 9160 60 = 30MM Rn= (30)? 2 ( 80 (80 80²-30² Rn 77.08 MM e= R-Rn = 80-77.08 e = 2.92 MM 77.08 - 50 Y;= Rn Ri Yia 27.08 MM 110 Yo= Ro - Rn 77.08 yo 32.92 MM 2 Area of the crossection (Al-Id 4 = I (60)2 A = 2827.43 MM

Р al Tensile stres at section AB C) А 20x10 3 2827.43 N/mm2 7.07356 N of For Curved beans of My A e R in Maximum bending stress at (obi) = my; M = yi inner fibre AC Ri = 1600x10 27.00 (2827:43)x2192)*(50) bi= 104.96 N/MM2 stress at Maximum bending outer fibre 6oo) - MY A e Ro = (600X10%)x(3292) (2827-43)*(2-42)*(110) bo) €57.99 Nimeme Resultant stress at inner fibre of tobi - 7.073 + 104.96 112.033 N/MM2 or - Tensile

Resultant stress at outer fibre (6)=of-obo 7:073-57.99 Los =-50.917 N/mm² Compressive should obi in innerfibre positive because becaur outer inner fibres tenda to tensile o bo should be negative in outer fibre fibres tends to compressive stress value at B (ori) = 112.033 alama Stress value at Aloo) - 50.917 relam? distribution for section AB i. stress 50.917N/mm2 compressive Tensile Neutral axis 112.033 N/mm2