Question

QUESTION 4. For the beam shown in the figure, the maximum allowable bending stress is 180 MPa and the maximum allowable shear stress is 120 MPa. Choose an appropriate Wide Flange Section (W) that can carry the distributed force and the point force safely. 10 kN 8 kN/m C B 4 m 2 m

Answer 1

Solution:- The given problem has been solved using the standard formulas and principles used in the subject of Mechanics of materials. Standard chart of wide flange section from standard textbooks has been followed to determine the section properties.

Solution- I s kolm 1จ 4 H A B С RA RR 2m H -um icting at I Taking 2 Let RA and RB be the upward reactions supports and B respectively E fyzo RAT RR= (8x6)+10 RATRB=58KN Ta moments about A. 10x6 t 8x6x6 RB X 4 4 R₂ = zou RB=51 Putting Re = 51 kN RA = 58-5) RA=7 KN Shear force cerluulentions! S.F. at just left of A= S.F at A = 7 KN just sight of S. F. ct just left of B=(7)+(-8x4) sif, at just right of B=(25)+(51) KN in equation ② gives O = -25 kn 26 KN

ulur S.F. aut just left of c= 26+(-8x2) = 10 KN Sif out just right of c= lot(-10)= 0kn from calculations, we came to know that Sf. changes its neatuse le. becomes positive to negative between A and B. so we will a new location D S.F. becounts zero. Let a be the distance D from A. S. f. at D=(7)+(-8 x 2) 7-8x x=1 금 01875 mt- 8 Bende ding Mourent calculations- B.M. at A= B.M. at D = D= [7X0.875)+ (-8X0.875% 0.875 of = 3.0625 KN-m B. M. at B=(7x4) + (-8x uxy - = 28+ (-64) - -36 Km B.M. at C=(746) + (-8X6X b) +(51x2) (42)+(-144)+(102) o kom

lokN 8kN/m А 01875 RA=7KNI um R9= SIEN ta LOADING DIAGRAM (26kN 7KN TO KN B SHEAR FORCE DIAGRAM 25KN 36 km 3,0625 kNam BENDING thoMENT DIAGRAM From the sheers force diengram we find that, maximum swear force is or Vmax=26kN from the bending moment dicigram we find that maximum bending moment is 36kN-m or Mmax = 36 kN-m 26 KN

Callow 07 M max = Sreqi = or Sree = or wide lange Maximum allowable beno stress bending 18 очра Gallow = 180x106 1801106 N/n? So, required sectibu modules the wide flange section will be is Srea 366103 E allow 180x106 2X104 m² 2x105 mm3 Sseq. 200 X los mm? So we need to select a section which should have Sxx above 200 x 103 mm² So, we select W 150x30 ullich has = 218 x 103 mm thickness web-tw=6.60mm= 6.6x103m depth=d= 157mm = 0.157m So, the maximum shear stress (max) developed in the section will ber Tmax Vmax 26x103 (6.6X153) (0.157) Tmax=250916181.14 N/m2 Tmax = Sxx Х of - twad 25.09 MPa

Allowable shear stress given is :- Tallow = 120 mpa Tmax < Tallow or 25.0g <120 Mpa Mpa So, selected Section is Diki Wis0X30 (mm) Aus