Knowing that sigma 220 MPa and tauall 100 MPa select the m
Solution
solution:
1) as we konw that point B and C are support and they are subjected to equal bending moment,hence design has to be withstand only bending stresses and shear stresses are negligible
2)from static equillibrium taking moment about B and C we get
Rby=629.14 kN
Rcy=-629.14 kN
3)where bending moment inside the bar is at point B and point C,as we have to design for maximum but both of them are equal in strength
Mb=Mc=385*10^3*1.3*1000=500.5*10^6 Nmm
4)hence maximum bending stresses in bar at maximu distance is given by
Sd=Mb*y/I
I=b*d^3/12
y=maximum distance from neutral line
hence
3) for first choice of W360*216
Sb1=500.5*10^6*108/360*216^3/12=178.78 N/mm2
4) for W690*125
Sb2=500.5*10^6*62.5/690*125^3/12=278.53 N/mm2
5)for W760*147
Sb3=500.5*10^6*73.5/760*147^3/12=182.85 N/mm2
6)as Sb1<Sb3<Sb2
hence Sb1 eans first choice W360*216 is most economic and Sb1<200,hence design will besafe
7)maximum shear force is 385 KN,shear force in first choice is
t1=385*10^3/360*670=1.596 N/mm2<100 MPa
8)as design W360*216 is safe and most economic hence it is siutable for given loading
